text:
# Let $L: \mathbb{R}^2\rightarrow \mathbb{R}^2$ be a linear operator such that $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, then find $L((7,5)^T)$.

Let $$L: \mathbb{R}^2\rightarrow \mathbb{R}^2$$ be a linear operator. If $$L((1,2)^T)=(-2,3)^T$$ and $$L((1,-1)^T)=(5,2)^T$$, find the value of $$L((7,5)^T).$$

Is there a way to solve these kinds of problems? I only know if $$L(\alpha v_1+\beta v_2)=\alpha L(v_1)+\beta L(v_2)$$, then the vector space is said to be a linear transformation.

Let your transformation matrix $$L=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$. Then $$L((1,2)^T)=(-2,3)^T$$ and $$L((1,-1)^T)=(5,2)^T$$ give \begin{align}&a+2b=-2 &c+2d=3\\&a-b=5 &c-b=2.\end{align} Solving we get $$L=\begin{bmatrix}\dfrac{8}{3}&\dfrac{-7}{3}\\\dfrac{7}{3}&\dfrac{1}{3}\end{bmatrix}$$. $$\:$$So $$L((7,5)^T)=\begin{bmatrix}7\\18\end{bmatrix}$$.
• @ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $m\times n$ – Yadati Kiran Nov 18 '18 at 10:37
Right, so you have to find $$\alpha$$ and $$\beta$$ with $$\alpha(1,2)+\beta(1,-1)=(7,5)$$


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# In the complex number system, why can't 1+1 = 0 ?

1. Feb 14, 2014

### pondzo

in the following i will demonstrate a 'proof' that 1+1=0

1+1=√1 +1
=√(-1)(-1) +1
= (√(-1))(√(-1)) +1
= (i)(i) +1
= i2 +1
= -1 + 1
= 0

I know I'm not the first to come up with this 'proof', and i have been told that the problem lies with splitting the radical between the 2nd and 3rd lines of working. But in the complex number system there is no problem with going from, say, √(-4) = √(-1)(4) = 2i so why is there a problem with the above working, isn't it the same line of thought?

thanks, Michael.

2. Feb 14, 2014

### Staff: Mentor

3. Feb 14, 2014

### phinds

Doesn't simple common sense tell you this is nonsense?

4. Feb 14, 2014

### pondzo

yes, which is why i asked the question. it just seemed mathematically sound to me so i wanted to see why we disregard it. But it seems i have a lot to catch up on with complex exponentiation!

5. Feb 14, 2014

### phinds

Ah ... yeah, I see from rereading your question that you aren't quite saying that you believe it, just that you are puzzled by it.

6. Feb 18, 2014

### crownedbishop

The problem is that for complex numbers, √(a+bi) has two complex roots. With real numbers, one is positive and the other is negative. We can assign the positive one to equal the positive square root. However, with complex numbers it isn't so simple. A complex root is neither negative nor positive, sense it doesn't make any sense to say a complex number is positive. So, it doesn't make sense to have a "positive" square root. This is where the issue arises.

7. Feb 18, 2014

### Seydlitz

Here's the explanation:

The fun thing is you can go 'crazy' with the rule if you know it.

say:

$i^{-i} = e^{\frac{\pi}{2}}$

8. Feb 18, 2014

### Curious3141

Umm, that actually is valid. One of the values of $i^{-i}$ is $e^{\frac{\pi}{2}}$.

9. Feb 19, 2014

### pondzo

okay, why are we allowed to go from ((4)*(-1))0.5 to (4)0.5*(-1)0.5 (which is equal to 2i) but we aren't allowed to go from ((-1)(-1))0.5 to (-1)0.5*(-1)0.5?

10. Feb 19, 2014

### pondzo

and on another note, in my most recent post, isn't the first example a negative number and the second a positive (ie (4)(-1)= -ve and (-1)(-1)= +ve)

11. Feb 19, 2014

### Seydlitz

Yes it is valid, I don't think I imply otherwise.

12. Feb 19, 2014

### Curious3141

Oh, OK. Sorry that I misunderstood your post.

13. Feb 19, 2014

### lendav_rott

It's not possible. If 1+1 = 0, then +1 would have to be equal to -1, but it can't be equal to -1 and +1 at the same time. Same way you could try and prove for any "A" that A+A = 0

14. Feb 19, 2014

### gopher_p

Alright, here's the rub. This isn't actually a "valid" operation. It just looks like what's going on. The notation convention used by every mathematician and math educator that I know is that $\sqrt{a}$ denotes the non-negative solution of $x^2=a$ if $a$ is a non-negative real number and $\sqrt{a}$ denotes the root $i\sqrt{-a}$ of $x^2=a$ when $a$ is a negative real number (so we've already agreed on what $\sqrt{-a}$ means), where $i$ denotes either (a) some made-up thing with the property that $i^2=-1$ or (b) the principle complex root (we singled one of them out and said "you're it") of $x^2+1=0$ depending on where you are in your math studies. There are no rules of exponents or multiplicative identities of roots or anything of that sort going on. It's all just definitions of notation; these are the symbols that we use to talk about these abstract mathematical concepts.

In other words, $\sqrt{-4}=i\sqrt{-(-4)}=i\sqrt{4}=2i$ because somebody a long time ago decided that's what those symbols mean, and everybody else went along with it. It only looks like we're using $\sqrt{-4}=\sqrt{-1\cdot 4}=\sqrt{-1}\cdot\sqrt{4}=i\cdot2$ because either (a) you were clever enough to notice that might be the case based on your experience using rules that you learned in precalc or (b) you had a teacher (or tutor or helper on the interwebs) that told you a little white lie to convince you that the answer they gave was correct without realizing (or caring) that it might later cause serious strife for you and other teachers (or tutors or helpers on the interwebs) or (c) you had a teacher (or tutor or helper on the interwebs) who was incompetent tell you this is what was going on without understanding that it really wasn't.

15. Feb 19, 2014

### pondzo

if the same exponential laws applied to negative numbers as they do positive.

a+a=√(a^2 )+a
=√((-a)(-a) )+a
=√(-a) √(-a)+a
=√((-1)(a) ) √((-1)(a) )+a
=(√(-1))(√a)(√(-1))(√a)+a
=(i)(√a)(i)(√a)+a
=i^2 a+a
=-a+a
=0

16. Feb 19, 2014

### pondzo

Isn't that pretty much equivalent to what your doing when you say √(-4) = i√(-(-4)) because your just taking √(i)2 out of the radical, which is the same as √(-1). so really your doing the exact same thing as me?

17. Feb 19, 2014

### Staff: Mentor

No, he is not doing the same thing. He is saying that we use the sequence of symbols "$\sqrt{a}$" as a convenient shorthand for "the non-negative solution of $x^2=a$ if $a$ is a non-negative real number; and the root $i\sqrt{-a}$ of $x^2=a$ when $a$ is a negative real number, where $i$ denotes some made-up thing with the property that $i^2=-1$"

That's a definition, and we can use that definition to show that if $a$ and $b$ are both non-negative reals, then $\sqrt{ab} = \sqrt{a}\sqrt{b}$. However, we can also use this definition to show that, in general, this equality does not hold for negative numbers.

18. Feb 19, 2014

### gopher_p

No it's not. Read again what I wrote regarding the definition of the meaning of the collection of symbols $\sqrt{a}$ when $a$ is a negative real number:

If $a$ is a negative real number, the $\sqrt{a}=i\sqrt{-a}$.

There is nothing in between the left and right hand side that explains why it is true. There is not proof of this identity. It is not a matter of mathematics, but a matter of mathematical notation. It is simply true by definition of the notation being used. I'm not "taking √(i)2 out of the radical". The $i$ essentially materializes out of thin air, just like the $\times$ does in the definition of the exponential notation $x^2=x\times x$.

19. Feb 19, 2014

### pondzo

okay i think i understand what you said, sort of.

Are you saying that 'the square root' symbol is used to refer to only the principle (positive) root: ie √(a2) = |a| where a can be +ve or -ve but the radical implies the absolute (positive) value of it?

what i was getting most confused about was that we can go from $√(-4) = √(-1)(4) = √(-1)√(4)$ but we cant go from $√16 = √(-4)(-4) = √(-4)√(-4)$ I realise now that its because the property $√ab = √a√b$ holds only if a and b are positive or one of them is negative. It does not hold if BOTH a and b are negative as this leads to inconsistencies.

But what if we weren't to classify these as inconsistencies and hence didn't ban the property √ab = √a√b for a negative a and negative b. could this mean that there would then be solutions to the system of equations (for example): (sorry this may seem off track, but this is why i started considering my original question on this thread)
a - 2b + 3c = -2
-a + b - 2c = 3
2a -b + 3c = 1
putting into a co efficient matrix:
$$\begin{pmatrix} 1 & -2 & 3 & -2\\ -1 & 1 & -2 & 3\\ 2 & -1 & 3 & 1 \end{pmatrix}$$
putting into row echelon form:
$$\begin{pmatrix} 1 & -2 & 3 & -2\\ 0 & -1 & 1 & 1\\ 0 & 0 & 0 & 8 \end{pmatrix}$$
the last row says 0=8 which is not possible and leads this system to have no solutions.
what if we said that 0=8 is an identity that is necessary for there to be a solution to this system of equations and provable by (im now assuming that $√ab = √a√b$ for negative a and negative b is not treated as an inconsistency):
0 = -4 + 4
0 = 4i2 + 4
0 = (2i)(2i) + 4
0 = √(4)(-1)√(4)(-1) + 4
0 = √(-4)(-4) + 4
0 = 4 + 4
0 = 8
similarly it can be shown the 0=-8 which yields -8=0=8
now, going back to the matrix in row echelon form, treating 'c' as a free variable the solution space is ${(-4-c,c-1,c)}$
lets say c = 2, then a=-6 and b=1 and plugging these values into the three equations we get:
a - 2b +3c = -2 <-------- this is true for these values
-a + b - 2c = 3 <-------- this is true for these values
2a - b + 3c = 1
2(-6) - 1 + 3(2) = 1
-7 = 1 <----- since we showed that -8=0=8 this is true also (under the assumptions that i stated)

so what im getting at is; if √ab = √a√b for negative a and b was not treated as an inconsistency and was allowed then could this mean that certain systems of equations that as it stands in modern mathematics have 'No solution' could actually have a solution?

20. Feb 19, 2014

### pondzo

ehhhh scratch all that i just realised that if -8=0=8 then by the same logic every single number is equal to every other number, which is definitely an inconsistency!!

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# Does a generalized difference of powers formula exist?

The identity $$\left(\frac{n+1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2=n$$ can be used to represent any number as difference of two squares. (Note that this formula gives integer values when $$n$$ is odd.) I have been looking for a similar formula for cubes, quartics, etc., but have been unsuccessful in finding them, which leads me to believe that this problem is related to Fermat's Last Theorem, and can be disproven with it.

I have been using the similar structure and attempting to find one by trial and error. Here's some example attempts: (that don't work) $$\left(\frac{n+1}{3}\right)^3-\left(\frac{n-1}{3}\right)^3=n.$$ $$\left(\frac{n!+1/3}{n!}\right)^3-\left(\frac{n!-1/3}{n!}\right)^3=n.$$ After much effort, it seems formulas for higher powers do not exist, so how might one prove it?

• It does not exist for $n\ge3$ by FLT: a cube of an integer cannot be a difference of two non-zero cubes. Apr 7, 2022 at 17:48
• @markvs How is that relevant? Apr 7, 2022 at 18:21
• Any difference of cubes formula would involve the cubic terms cancelling, but then how would you force the quadratic terms to cancel? What kind of expressions are you even looking for / functions are you allowing to be in the parentheses? I mean, obviously $(\sqrt[3]{n^3+1})^3-1^3=n$ works, for example. No linear functions of $n$ in parentheses can work (exercise!). Apr 7, 2022 at 18:24
• @markvs True, I guess that rules out formulas which produce only integers in the parentheses. (I meant relevant to the problem, not to me personally.) Apr 7, 2022 at 18:28
• Following up, here's the best I've been able to do. It's not quite the difference of cubes, but it has the fun of requiring only small obfuscation to make everything on the left a 3. $$\left( \frac{3}{3-\frac33}\right) \left( \left(\frac{x+3}{3}\right)^3 - \left(\frac{x-3}{3}\right)^3 \right) -3 = x^2$$ Yes, we could replace the denominators with 18^(1/3), but that's obviously outside the spirit of what we're trying to do. Apr 8, 2022 at 0:21

$$\large\bf{Proposition}$$

If $$P(x)$$ and $$Q(x)$$ are any polynomials with coefficients in $$\mathbb{C}$$ which are not both constant, then the degree of $$R(x) = P(x)^n - Q(x)^n$$ will be at least $$n-1$$ if it is non-zero.

$$\large\bf{Corollary}$$

The above proposition implies that there are no polynomials with $$P(x)^n - Q(x)^n = x$$, since $$P(x)$$ and $$Q(x)$$ cannot both be constant and the degree $$1$$ of the RHS is not at least $$n-1 \ge 2$$.

$$\large\bf{Proof}$$

Assume that the degree of $$R(x)$$ is at most $$n-2$$. Start by noting that, over the complex numbers, there is af factorization

$$R(x) = P(x)^n - Q(x)^n = \prod_{k=1}^{n} (P(x) - \zeta^k Q(x))$$

where $$\zeta = e^{2 \pi i/n}$$. Since the degree of this product must equal the degree of $$R(x)$$, which lies in $$[0,\ldots,n-2]$$ by assumption, it follows that all the factors are non-zero and at least two of the factors (say for $$k = i$$ and $$k = j$$) must be constant. But this means that there are constants $$c_i$$ and $$c_j$$ such that:

$$P(x) - \zeta^i Q(x) = c_i,$$ $$P(x) - \zeta^j Q(x) = c_j,$$

But since $$\zeta^i \ne \zeta^j$$ we can solve for $$P(x)$$ and $$Q(x)$$ and see that they are both constant, a contradition.

For variations and generalizations of this argument, see Mason–Stothers theorem.

• Maybe I am missing something, but this argument doesn't seem to work. If we use $P(x)=Q(x)$ with $\deg(P(x))>2$ then the argument implies that $\deg(R(x))>1$ but clearly $R(x)=0$. Edit: thinking about it it seems like it works if you are more careful with zeros, i.e. if $P(x)^n$ is not $Q(x)^n$ the argument is fine. Apr 9, 2022 at 0:12
• @ClydeKertzer If there are no solutions with complex polynomials, there are certainly no solutions in polynomials with rational coefficients either. Apr 9, 2022 at 16:34
• How do we know that If $P(x)$ and $Q(x)$ are any polynomials with coefficients in $\mathbb{C}$ which are not both constant, then the degree of $$R(x)=P(x)^n−Q(x)^n$$ is at least $n-1$? Apr 11, 2022 at 15:30
• @ClydeKertzer Did you actually try reading the answer? The entire point of my answer is exactly to prove this claim. The "first line" is the statement, the "second line" is the remark that this answers your question, and then the rest of the answer is giving a proof of the first line. Apr 15, 2022 at 15:25
• @MasonJar: It may be that Clyde Kertzer didn't immediately follow the argument well enough to see that, and was still working their way through it. It may be helpful (though not strictly speaking necessary) to give markers like "Proposition" and "Proof" to further guide them, but that's just one point of view. Apr 16, 2022 at 17:02

It's not clear what restrictions you put on the expressions, but if you for example consider polynomials over rationals, then it's not hard to show that is indeed impossible. The argument below uses Fermat's last theorem, but you can completely avoid it, see Mason's answer.

Assume for a contradiction that $$k\geq 3$$ and we have polynomials $$f,g\in \mathbb{Q}[x]$$ such that $$f(n)^k-g(n)^k=n$$ for all integers $$n$$ (then in fact it must hold for all complex numbers $$x$$). Choose any integer $$a\neq 0$$ and put $$n=a^k$$. Let $$f(n)=\frac{p}{q}$$, $$g(n)=\frac{r}{s}$$, $$q,s\neq 0$$ be the rational representations, we have $$\Bigl(\frac{p}{q}\Bigr)^k-\Bigl(\frac{r}{s}\Bigr)^k=a^k.$$ Multiplying by $$(qs)^k$$ we get $$(ps)^k-(rq)^k=(aqs)^k.$$ The terms are all integers and by Fermat's last theorem this is only possible if one them is $$0$$. Since $$a,q,s \neq 0$$, this means either $$p=0$$ or $$r=0$$, in other words $$f(n)=0$$ or $$g(n)=0$$. Since $$a$$ was chosen arbitrary, this means one of $$f,g$$ attains $$0$$ at infinitely many points, hence it must be the zero polynomial. Say $$g(x)=0$$, then we have $$f(x)^k=x$$, impossible by simple polynomial degrees comparison on both sides. Similarly if $$f(x)=0$$. Hence no such polynomials $$f,g$$ exist. $$\tag*{\square}$$

Very similarly we can also resolve case where $$f,g$$ are polynomials in $$n!$$ (to cover one of your attempts). Indeed, let $$f(n!)^k-g(n!)^k=n$$ and the same argument as above implies one of $$f,g$$ is zero infinitely often (since $$n!$$ is increasing for $$n>1$$, $$f$$ or $$g$$ is actually zero on infinitely many distinct (important!) values), hence again we have the zero polynomial. Then $$f(n!)^k=n$$ or $$-g(n!)^k=n$$ for all $$n$$, which leads to contradiction (for example in the former $$n=2$$ implies $$\sqrt[k]{2}$$ is rational, impossible). You can think of generalizing the factorial away, but you need to make sure the function you replace it with takes infinitely many distinct values, otherwise the argument does not go through, I will leave that to you...

Without any such restrictions on $$f,g$$, you can construct counterexamples by letting $$f(n)$$ be arbitrary and then just put $$g(n)=\sqrt[k]{f(n)^k-n}$$, but that is not very interesting...

• This is just strictly worse than the other answer, using Fermat's Last Theorem instead of unique factorization to prove a weaker result. Apr 15, 2022 at 22:57
• @ThePhoenix I agree, I've made that clear in the text now
– Sil
Apr 16, 2022 at 5:56


text:
• Volume 91, Issue 1

July 2018

• General editorial on publication ethics

• Cloaks for suppression or enhancement of scattering of diffuse photon density waves

Enhancement of wave-like characteristics of heavily damped diffuse photon density waves in a random medium by amplification can induce strongly localised resonances. These resonances can be used to either suppress or enhance scattering from an inhomogeneity in the random medium by cloaking the inhomogeneous region by a shell of random medium with the correct levels of absorption or amplification. A spherical core–shell structure consisting of a shell of a random amplifying medium is shown to enhance or suppress specific resonant modes. A shell with an absorbing random medium is also shown to suppress scattering which can also be used for cloaking the core region.

• Influence of structural and doping parameter variations on Si and $\rm{Si_{1−x} Ge_{x}}$ double gate tunnel FETs: An analysis for RF performance enhancement

This paper deals with the effect of structural and doping parameter variations on RF parameters for Si and $\rm{Si_{1−x}Ge_{x}}$ -based double gate (DG) tunnel FETs (TFETs). For the first time, asymmetric gate oxide is introduced in the gate-drain overlap and compared with that of DG TFETs. The DC parameter subthreshold swing (SS) and RF parameter metrics, unity gain cut-off frequency ($f_{t}$) and maximum oscillation frequency ($f_{max}$) are extracted by varying structural parameters, gate length ($L_{g}$), gate oxide thickness ($t_{ox}$), channel thickness ($t_{ch}$), doping parameters, channel doping ($N_{ch}$), drain doping ($N_{d}$) and source doping ($N_{s}$) in and around their nominal value. For a channel thickness of 15 nm, a very less SS of 8 mV/dec is achieved in $\rm{Si_{1−x}Ge_{x}}$ -based DG TFETs with gate-drain overlap. Variations of gate oxide thickness offer better RF performance enhancement for Si-based asymmetric gate oxide devices. This could be achieved because of the higher tunnelling rate of electrons occurring at the source side of asymmetric gate oxide devices.

• Tunable microwave generation based on frequency quadrupling

To generate linearly chirped microwave signals with large frequency tunable range, a photonic approach is proposed. A dual-output dual-parallel Mach–Zehnder modulator followed by a polarisation beam combiner and an optical filter are utilised to generate orthogonally polarised $\pm$ second-order optical sidebands. A polarisation modulator is employed to achieve phase modulation of the two wavelengths. The balanced detection is applied to suppress the distortion and background noise. The central frequency of the generated signal is four times that of the local oscillator frequency. Simulation results show that a linear pulse is produced with time-bandwidth as well as a compression ratio for the pulse of 11 and 9.3 respectively. Moreover, a peak-to-sidelobe ratio of 7.4 dB is generated. The system has both good reconfigurability and tunability, and its frequency can be continuously adjusted from about 10 GHz to as much as 50 GHz in principle.

• Strange non-chaotic attractors in a state controlled-cellular neural network-based quasiperiodically forced MLC circuit

In this paper, we report the dynamical transitions to strange non-chaotic attractors in a quasiperiodically forced state controlled-cellular neural network (SC-CNN) based MLC circuit via two different mechanisms, namely the Heagy–Hammel route and the gradual fractalisation route. These transitions were observed through numerical simulations and hardware experiments and confirmed using statistical tools, such as maximal Lyapunov exponent spectrum and its variance and singular continuous spectral analysis. We find that there is a remarkable agreement of the results from both numerical simulations as well as from hardware experiments.

• All possible tripartitions of $^{236}\rm{U}$ isotope in collinear configuration

Using the recently proposed unified ternary fission model (UTFM), the tripartition of $^{236}\rm{U}$ isotope was studied for all possible fragmentations, in which the interacting potential barrier is taken as the sum of the Coulomb and proximity potentials with fragments in collinear configuration. The highest yield is obtained for the fragmentation $^{48}\rm{Ca}+^{58}\rm{Ti}+^{130}\rm{Sn}$ and next highest yield is found for $^{58}\rm{Cr}+^{46}\rm{Ar}+^{132}\rm{Sn}$,which stress the importance of doubly magic or near doubly magic nuclei in the tripartition of $^{236}\rm{U}$ isotope. The formation of $^{68}\rm{Ni}$ and $^{70}\rm{Ni}$ as the edge fragments linking the doubly magic nucleus $^{132}\rm{Sn}$ by the isotope of Si is in good agreement with experimental and theoretical studies, in the collinear cluster tripartition of $^{236}\rm{U}$ isotope which reveals the reliability of our model (UTFM) in ternary fission.

• Deterministic assisted cloning of an unknown single-particle four-dimensional quantum state

In this paper, we present a scheme which can produce a perfect copy of an unknown single-particle four-dimensional quantum state with assistance from a state preparer. Two stages were included in this scheme. The first stage requires the usual teleportation, after Alice’s (the state sender) generalised Bell state measurement. Bob (the state receiver) can get the original state with unit probability. In the second stage, after having received Victor’s (the state preparer) classical message, and using the rest resource of the teleportation process, the perfect copy of an original unknown state can be produced in Alice’s place. To realise the scheme, several novel sets of measuring basis were introduced. It must be pointed out that, in the present scheme, the total success probability for assisted cloning of a perfect copy of the unknown state can reach 1.

• The structure of ion-acoustic waves in a low-frequency three-component electron–ion space plasma with two-electron populations

Low-frequency ion-acoustic waves are analysed on the ion time-scale, in a three-component electron–ion space plasma. The solitary waves propagate in the positive $x$ direction relative to an ambient magnetic field $\overrightarrow{B}_{0}$ which forms static background for a configuration consisting of cool fluid ions and both warm and hot Boltzmann distributed electrons with temperatures $T_{ic}$, $T_{ew}$ and $T_{eh}$, respectively. We derive linear dispersion relation for the waves by introducing first-order density, pressure and velocity perturbations into the ion fluid equations. Additionally, the variation in the nonlinear structure of the waves are investigated by carrying out a full parametric analysis utilising our numerical code. Our results reveal that ion-acoustic waves exhibit well-defined nonlinear spikes at speeds of $M \geq 2.25$ and an electric field amplitude of $E_{0} = 0.85$. It is also shown that low wave speeds ($M \leq 2$), higher densities of the hot electrons, antiparallel drifting of the cool fluid ions, and increased ion temperatures all lead to significant dispersive effects. The ion-acoustic plasma waves featured in this paper have forms that are consistent with those classified as the type-A and type-B broadband electrostatic noise (BEN) observed in the data obtained from earlier satellite missions.

• Geometry of magnetic rotational (MR) band-crossing phenomenon in MR bands

A semiclassical (SC) approach is proposed to calculate the $B(M1)$ transition rates in the band-crossing region of two magnetic rotational (MR) bands. In the present work, a geometry is suggested for the shear blades to govern its behaviour during the band-crossing. In the crossing region, gradual alignment of two nucleons is responsible for the crossing behaviour and it must give a quantised resultant angular momentum. As an example, it is successfully implemented for the MR bands in the mass $A = 110$ and $A = 200$ regions. A good agreement of the present semiclassical calculations with the experimental values is presented and furthermore, it is seen that the present proposal is also helpful to see the core contribution in the MR phenomenon.

• Exact solutions to (2 + 1)-dimensional Chaffee–Infante equation

In this paper, the canonical-like transformation method and trial equation method are applied to (2+1)- dimensional Chaffee–Infante equation, and some exact solutions are obtained. In particular, a new solution in terms of elliptic functions is given.

• Applicability of strange nonchaoticWien-bridge oscillators for secure communication

In this paper, experimental realisation of synchronisation of strange nonchaotic attractor inunidirectionally coupled Wien-bridge oscillator is presented. Preliminary work of chaotic and strange nonchaotic dynamics with diode as the nonlinear element is published in J. Nonlinear Dyn. 2015, 1 (2015). In this work, two Wien-bridge oscillators are coupled at the aperiodic dynamics for application in secure communication. The dynamics of the coupled system is explored by experimental and numerical studies. Employing the simple communication masking method with aperiodic dynamics offers some advantages rather than using chaotic attractors, i.e., the range of control parameter value is smaller than the range for which the system exhibits chaotic behaviour. The performance of the system is evaluated and it is found that experimental result agrees with numerical results.

• Carpet oscillator: A new megastable nonlinear oscillator with infinite islands of self-excited and hidden attractors

In this paper, we design a new two-dimensional nonlinear oscillator with infinite number of coexisting limit cycles distributed in a plane. One-third of these limit cycles are self-excited attractors while two-third of them are hidden attractors. Modifying this new system to its forced version, we obtain a new nonlinear system with infinite number of coexisting torus attractors and limit cycle attractors.

• Coexistence of attractors in autonomous Van der Pol–Duffing jerk oscillator: Analysis, chaos control and synchronisation in its fractional-order form

In this paper, a Van der Pol–Duffing (VdPD) jerk oscillator is designed. The proposed VdPD jerk oscillator is built by converting the autonomous two-dimensional VdPD oscillator to a jerk oscillator. Dynamical behaviours of the proposed VdPD jerk oscillator are investigated analytically, numerically and analogically. The numerical results indicate that the proposed VdPD jerk oscillator displays chaotic oscillations, symmetrical bifurcations and coexisting attractors. The physical existence of the chaotic behaviour found in the proposed VdPD jerk oscillator is verified by using Orcad-PSpice software. A good qualitative agreement is shown between thenumerical simulations and the PSpice results. Moreover, the fractional-order form of the proposed VdPD jerk oscillator is studied using stability theorem of fractional-order systems and numerical simulations. It is found that chaos, periodic oscillations and coexistence of attractors exist in the fractional-order form of the proposed jerk oscillator with order less than three. The effect of fractional-order derivative on controlling chaos is illustrated. It is shown that chaos control is achieved in fractional-order form of the proposed VdPD jerk oscillator only for the values of linear controller used. Finally, the problem of drive–response synchronisation of the fractional-order form of the chaotic proposed VdPD jerk oscillators is considered using active control technique.

• Numerical simulation of unsteady 3D magneto-Sisko fluid flow with nonlinear thermal radiation and homogeneous–heterogeneous chemical reactions

The chemically reactive unsteady 3D Sisko fluid flow over a bidirectional stretched surface in the presence of nonlinear thermal radiation and magnetic field is considered. The numerical scrutiny of the transformed nonlinear ODEs is performed with the help of a numerical technique known as bvp4c. The mechanism of heat and mass transfer by the applications of chemical reaction and radiation is presented using temperature and concentration graphs. A remarkable enhancing results are estimated in the presence of nonlinear thermal radiation parameter $R_{d}$ and temperature ratio parameter $\theta_{w}$ for the temperature field. Another significant decreasing outcomes are found while plotting the concentration profile with variation of homogeneous and heterogeneous parameters ($k_{1}, k_{2}$). Computational results of the local skin friction and local Nusselt number are tabulated under the influence of physical parameters which governs the flow. Decrease in skin friction with varying values of $\lambda$ is observed. On the other hand, the impact of $R_{d}$ and $\theta_{w}$ on the tabular values of local Nusselt number is found in increasing order. All the graphical results and tabular values are illustrated while testing both cases of power-law fluids including pseudoplastic (0 & lt; n & lt; 1) and dilatant (n & gt; 1) behaviours. A comparison of the bvp4c and shooting technique with RK-45 Fehlberg is presented which shows excellent agreement. Additionally, the present results are compared with the results in the existing literature and both are found to be in very good correlation.

• Aspects of improved heat conduction relation and chemical processes in 3D Carreau fluid flow

This article communicates the numerical consideration of 3D Carreau liquid flow under the impact of chemical responses over a stretched surface. Moreover, the heat transfer exploration is carried out with a view to improve the heat flux relation. This phenomenon is established upon the theory of Cattaneo–Christov heat flux relation that contributes by the thermal relaxation. On exploitation of an appropriate transformation a system of nonlinear ODEs is attained and then elucidated numerically by means of bvp4c scheme. The descriptions of temperature and concentration fields equivalent to the frequent somatic parameters are graphically scrutinised.Our analysis carries that the concentration of the Carreau liquid displays similar tendency and decline as theheterogeneous–homogeneous reaction parameters ($k_{2}, k_{1}$) augment. Furthermore, it is notable that for shear thinning ($n$ & lt; 1) liquid, the influence of local Weissenberg numbers ($We_{1}, We_{2}$) are absolutely conflicting compared with the instance of shear thickening ($n$ & gt; 1) liquid. Additionally, validation of numerical results is done via benchmarking with previously stated limiting cases with two different schemes namely, homotopy analysis method (HAM) and bvp4c scheme. These comparisons initiate a superb correspondence with these outcomes.

• Effect of tensile and compressive strain on the magnetic ordering of the $\rm{(LaMnO_{3})_{1}/(SrTiO_{3})_{1}}$ superlattice

We study the effects of tensile and compressive strain on the $\rm{(LaMnO_{3})_{1}/(SrTiO_{3})_{1}}$ superlattice from density functional theory using Quantum-Espresso open source code. In the unstrained superlattice, electron interactions in out-of-plane Mn–O–Ti chains are dominated by superexchange interactions, giving rise to ferromagnetic and half-metallic conducting characters. We found that the most stable magnetic configuration is G-type antiferromagnetic configuration for strong compressive strain and for strong tensile strain it isA-type antiferromagnetic configuration. The results are in accordance with the experimental observations which show that the superlattices can be grown on different substrates, and due to the difference in lattice parameters of the substrate and the main layer, there are also changes in the amount of strain applied to the superlattice.

• # Pramana – Journal of Physics

Current Issue
Volume 93 | Issue 6
December 2019

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
# If LIBOR rates declined by 0.1% each period from the January 1, 20X4 reset rate of 7%, what would be...

Assessing the effectiveness of a fixed for variable interest rate swap. Tesker International has loaned $20,000,000 to a foreign company that Tesker has a 20% equity interest in. Interest, paid June 30 and December 31, and principal on the loan is to be paid in U.S. dollars. The loan matures on December 31, 20X5, and has a fixed interest rate of 8%. Tesker anticipates that variable interest rates will be increasing in the near term and is therefore considering an interest rate swap. Tesker’s bank has agreed to swap a variable rate, based on LIBOR, plus 1% in exchange for receipt of a fixed rate through the maturity date of the foreign loan. The swap will cover a notional amount of$10,000,000. Reset dates will be January 1 and July 1, with net settlement occurring at the same time. Assumed rates and values are as follows:

 Reset Dates LIBOR Rates for Next Period Assumed Value of Swap January 1, 20X4 70% July 1, 20X4 72 \$27,698 January 1, 20X5 74 37,614 July 1, 20X5 73 14,402

1. Tesker’s management is uncertain as to whether or not they should engage in a swap. In order to assist them in their decision, complete the following schedule for all interest periods based on the assumed rates and values.

 Value of Value of Unrealized Unrealized Fixed Variable Interest Interest Note Gain (Loss) Gain (Loss) Interest Interest Period Swap Receivable on Derivative on Note Cash Income Income

2. Summarize the effect on Tesker’s income with and without the interest rate swap.

3. If LIBOR rates declined by 0.1% each period from the January 1, 20X4 reset rate of 7%, what would be the overall effect on earnings of this scenario?

## Plagiarism Checker

Submit your documents and get free Plagiarism report

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text:
# Does Earth get in Hubble's way?

Hubble takes very long exposures for its images so that it can collect enough light to see very distant objects. These exposures can be several hours, or even several days, long. I often read that the Hubble Ultra Deep Field image was taken with a 1,000,000 second exposure.

It takes Hubble about 97 minutes to orbit Earth once, so while it's orbiting Earth how do they manage to keep it focused on a single point? When Hubble goes behind Earth on its orbit, wouldn't Earth get in the way of the image?


text:
# Egbert Rijke: A higher encode decode method

Datum objave: 7. 4. 2021
Seminar za temelje matematike in teoretično računalništvo
četrtek
8
april
Ura:
17.30 - 19.00
Lokacija:
Spletni seminar

Note, that we meet at a non-standard time at 17:30. The talk will be given as a part of the Homotopy Type Theory Electronic Seminar.

Abstract: The Postnikov tower of a pointed type X induces a fiber sequence $K(G,n+1) -> |X|_{n+1} -> |X|_n$ for every n, where G is the (n+1)-st homotopy group of X. This fiber sequence suggests a general approach to computations of higher homotopy groups of types, by encoding a family of n-connected (n+1)-truncated types over the type $|X|_n$. While I do not yet have any finished computations of homotopy groups of types via this approach, it does suggest a fruitful line of research with interesting intermediate results.


text:
# Normal ordering macro in LyX

I want to define a macro for the normal ordering of operators as done in this question.

The LaTeX code produced by LyX for this macro is:

\global\long\def\norder#1{\vcentcolon#1\vcentcolon}

However, when I invoke it:

$\norder{\hat{a}\hat{a}^{\dagger}}$

it gives an error:

...ven by $\norder{\hat{a}\hat{a}^{\dagger}}$ The control sequence at the end of the top line of your error message was never \def'ed.

I don't understand why it gives the error. I have more macros in the document and they work perfectly. amsmath and amssimb are loaded through LyX's "Math Options" (and I have checked that I can use commands such as \mathbb inside macros). Looks like the problem is the \vcentcolon command but I don't see why or how to fix it. Thank you!

PS. This is what I have before the begin{document} in the LaTeX source generated by LyX.

\documentclass[english]{article}
\usepackage{mathpazo}
\renewcommand{\familydefault}{\rmdefault}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=3cm,lmargin=3cm,rmargin=2cm}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{esint}
\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\usepackage{braket}
\makeatother
\usepackage{babel}

The command \ventcolon has not been defined. Load the package mathtools; you can load it instead of amsmath. Moreover, wrap the stuff between the colons into a \mathrel command to improve the spacing.

\usepackage{mathtools}
\newcommand\norder[1]{\vcentcolon\mathrel{#1}\vcentcolon}

\documentclass{article}
\usepackage{mathtools}
\newcommand\norder[1]{\vcentcolon\mathrel{#1}\vcentcolon}
\begin{document}
$A \norder{\hat{a}\hat{a}^{\dagger}} B$
\end{document}

• Adding mathtools solved the issue. Thanks! – VictorSeven May 1 '17 at 15:28


text: A ball is kicked at a speed of 15.0 m/s when t=0. Theinitial velocity vector of the ball makes an angle of 60 degreeswith the horizontal direction. Calculate the instants when thevelocity vector of the ball makes an angle of 45 degrees withthe horizontal direction. I keep plugging the numbers into but I cannot come up with the answerin the back of the book. What am I missing? If someone could helpme I would appreciate it. Thank you   P.S. In thedenominator, the equation Im using is only g not 2 g.


text:
# Lower Bounds on the Query Complexity of Non-uniform and Adaptive Reductions Showing Hardness Amplification

Research paper by Sergei Artemenko, Ronen Shaltiel

Indexed on: 26 Jan '13Published on: 26 Jan '13Published in: Computational Complexity

#### Abstract

Hardness amplification results show that for every Boolean function f, there exists a Boolean function Amp(f) such that if every size s circuit computes f correctly on at most a 1 − δ fraction of inputs, then every size s′ circuit computes Amp(f) correctly on at most a $${1/2+\epsilon}$$ fraction of inputs. All hardness amplification results in the literature suffer from “size loss” meaning that $${s' \leq \epsilon \cdot s}$$. We show that proofs using “non-uniform reductions” must suffer from such size loss.A reduction is an oracle circuit $${R^{(\cdot)}}$$ which given oracle access to any function D that computes Amp(f) correctly on a $${1/2+\epsilon}$$ fraction of inputs, computes f correctly on a 1 − δ fraction of inputs. A non-uniform reduction is allowed to also receive a short advice string that may depend on both f and D. The well-known connection between hardness amplification and list-decodable error-correcting codes implies that reductions showing hardness amplification cannot be uniform for $${\epsilon < 1/4}$$. We show that every non-uniform reduction must make at least $${\Omega(1/\epsilon)}$$ queries to its oracle, which implies size loss. Our result is the first lower bound that applies to non-uniform reductions that are adaptive, whereas previous bounds by Shaltiel & Viola (SICOMP 2010) applied only to non-adaptive reductions. We also prove similar bounds for a stronger notion of “function-specific” reductions in which the reduction is only required to work for a specific function f.


text:
# RS Aggarwal solutions for Class 6 Mathematics chapter 4 - Integers [Latest edition]

#### Chapters

Chapter 1: Number System

Chapter 2: Factors and Multiples

Chapter 3: Whole Numbers

Chapter 4: Integers

Chapter 5: Fractions

Chapter 6: Simplification

Chapter 7: Decimals

Chapter 8: Algebraic Expressions

Chapter 9: Linear Equation in One Variable

Chapter 10: Ratio, Proportion and Unitary Method

Chapter 11: Line Segment, Ray and Line

Chapter 12: Parallel Lines

Chapter 13: Angles and Their Measurement

Chapter 14: Constructions (Using Ruler and a Pair of Compasses)

Chapter 15: Polygons

Chapter 16: Triangles

Chapter 18: Circles

Chapter 19: Three-Dimensional Shapes

Chapter 20: Two-Dimensional Reflection Symmetry (Linear Symmetry)

Chapter 21: Concept of Perimeter and Area

Chapter 22: Data Handling

Chapter 23: Pictograph

Chapter 24: Bar Graph

## Chapter 4: Integers

Exercise 4AExercise 4BExercise 4CExercise 4DExercise 4EExercise 4FTest Paper 4
Exercise 4A [Pages 63 - 64]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4A [Pages 63 - 64]

Exercise 4A | Q 1.01 | Page 63

Write the opposite of the following: An increase of 8

Exercise 4A | Q 1.02 | Page 63

Write the opposite of the following: A loss of Rs 7

Exercise 4A | Q 1.03 | Page 63

Write the opposite of the following: Gaining a weight of 5 kg

Exercise 4A | Q 1.04 | Page 63

Write the opposite of the following: 10 km above sea level

Exercise 4A | Q 1.05 | Page 63

Write the opposite of the following: 5°C below the freezing point

Exercise 4A | Q 1.06 | Page 63

Write the opposite of the following: A deposit of Rs 100

Exercise 4A | Q 1.07 | Page 63

Write the opposite of the following: Earning Rs 500

Exercise 4A | Q 1.08 | Page 63

Write the opposite of the following: Going 6m to the east

Exercise 4A | Q 1.09 | Page 63

Write the opposite of the following: 24

Exercise 4A | Q 1.1 | Page 63

Write the opposite of the following: -34

Exercise 4A | Q 2.1 | Page 63

Indicate the following using '+' or '−' sign: A gain of Rs 600

• +

• -

Exercise 4A | Q 2.2 | Page 63

Indicate the following using '+' or '−' sign: A loss of Rs 800

• +

• -

Exercise 4A | Q 2.3 | Page 63

Indicate the following using '+' or '−' sign: 7°C below the freezing point

• +

• -

Exercise 4A | Q 2.4 | Page 63

Indicate the following using '+' or '−' sign: Decrease of 9

• +

• -

Exercise 4A | Q 2.5 | Page 63

Indicate the following using '+' or '−' sign: 2 km above sea level

• +

• -

Exercise 4A | Q 2.6 | Page 63

Indicate the following using '+' or '−' sign: 3 km below sea level

• +

• -

Exercise 4A | Q 2.7 | Page 63

Indicate the following using '+' or '−' sign: A deposit of Rs 200

• +

• -

Exercise 4A | Q 2.8 | Page 63

Indicate the following using '+' or '−' sign: A withdrawal of Rs 300

• +

• -

Exercise 4A | Q 3.1 | Page 64

Mark the following integers on a number line: -5

Exercise 4A | Q 3.2 | Page 64

Mark the following integers on a number line: -2

Exercise 4A | Q 3.3 | Page 64

Mark the following integers on a number line: 0

Exercise 4A | Q 3.4 | Page 64

Mark the following integers on a number line: 7

Exercise 4A | Q 3.5 | Page 64

Mark the following integers on a number line: -13

Exercise 4A | Q 4.1 | Page 64

Which number is larger in the following pair?

• 0

• -2

Exercise 4A | Q 4.2 | Page 64

Which number is larger in the following pair?

• -3

• -5

Exercise 4A | Q 4.3 | Page 64

Which number is larger in the following pair?

• -5

• 2

Exercise 4A | Q 4.4 | Page 64

Which number is larger in the following pair?

• -16

• 8

Exercise 4A | Q 4.5 | Page 64

Which number is larger in the following pair?

• -365

• -913

Exercise 4A | Q 4.6 | Page 64

Which number is larger in the following pair?

• -888

• 8

Exercise 4A | Q 5.1 | Page 64

Which number is smaller in the following pair?

• 6

• -7

Exercise 4A | Q 5.2 | Page 64

Which number is smaller in the following pair?

• 0

• -1

Exercise 4A | Q 5.3 | Page 64

Which number is smaller in the following pair?

• -13

• -27

Exercise 4A | Q 5.4 | Page 64

Which number is smaller in the following pair?

• -26

• 17

Exercise 4A | Q 5.5 | Page 64

Which number is smaller in the following pair?

• -317

• -603

Exercise 4A | Q 5.6 | Page 64

Which number is smaller in the following pair?

• -777

• 7

Exercise 4A | Q 6.1 | Page 64

Write all integers between: 0 and 6

Exercise 4A | Q 6.2 | Page 64

Write all integers between: -5 and 0

Exercise 4A | Q 6.3 | Page 64

Write all integers between: -3 and 3

Exercise 4A | Q 6.4 | Page 64

Write all integers between: -7 and -5

Exercise 4A | Q 7.1 | Page 64

Fill in the blanks by appropriate symbol

0 ...... 7

• >

• <

Exercise 4A | Q 7.2 | Page 64

Fill in the blanks by appropriate symbol

0 ...... −3

• >

• <

Exercise 4A | Q 7.3 | Page 64

Fill in the blanks by appropriate symbol

−5 ...... −2

• >

• <

Exercise 4A | Q 7.4 | Page 64

Fill in the blanks by appropriate symbol

-15 ...... 13

• >

• <

Exercise 4A | Q 7.5 | Page 64

Fill in the blanks by appropriate symbol

−231 ...... −132

• >

• <

Exercise 4A | Q 7.6 | Page 64

Fill in the blanks by appropriate symbol

-6 ...... 6

• >

• <

Exercise 4A | Q 8.1 | Page 64

Write the following integers in the increasing order: 5, −7, −2, 0, 8

Exercise 4A | Q 8.2 | Page 64

Write the following integers in the increasing order: −23, 12, 0, −6, −100, −1

Exercise 4A | Q 8.3 | Page 64

Write the following integers in the increasing order: −17, 15, −363, −501, 165

Exercise 4A | Q 8.4 | Page 64

Write the following integers in the increasing order: 21, −106, −16, 16, 0, −2, −81

Exercise 4A | Q 9.1 | Page 64

Write the following integers in the decreasing order: 0, 7, −3, −9, −132, 36

Exercise 4A | Q 9.2 | Page 64

Write the following integers in the decreasing order: 51, −53, −8, 0, −2

Exercise 4A | Q 9.3 | Page 64

Write the following integers in the decreasing order: −71, −81, 36, 0, −5

Exercise 4A | Q 9.4 | Page 64

Write the following integers in the decreasing order: −365, −515, 102, 413, −7

Exercise 4A | Q 10.1 | Page 64

Using the number line, write the integer which is: 4 more than 6

Exercise 4A | Q 10.2 | Page 64

Using the number line, write the integer which is: 5 more than −6

Exercise 4A | Q 10.3 | Page 64

Using the number line, write the integer which is: 6 less than 2

Exercise 4A | Q 10.4 | Page 64

Using the number line, write the integer which is: 2 less than −3

Exercise 4A | Q 11.1 | Page 64

For the following statement, write (T) for true and (F) for false:

The smallest integer is zero.

• True

• False

Exercise 4A | Q 11.2 | Page 64

For the following statement, write (T) for true and (F) for false:

Zero is not an integer.

• True

• False

Exercise 4A | Q 11.3 | Page 64

For the following statement, write (T) for true and (F) for false:

The opposite of zero is zero.

• True

• False

Exercise 4A | Q 11.4 | Page 64

For the following statement, write (T) for true and (F) for false:

−10 is greater than −6.

• True

• False

Exercise 4A | Q 11.5 | Page 64

For the following statement, write (T) for true and (F) for false:

The absolute value of an integer is always greater than the integer.

• True

• False

Exercise 4A | Q 11.6 | Page 64

For the following statement, write (T) for true and (F) for false:

0 is larger than every negative integer.

• True

• False

Exercise 4A | Q 11.7 | Page 64

For the following statement, write (T) for true and (F) for false:

Every negative integer is less than every natural number.

• True

• False

Exercise 4A | Q 11.8 | Page 64

For the following statement, write (T) for true and (F) for false:

The successor of −187 is −188.

• True

• False

Exercise 4A | Q 11.9 | Page 64

For the following statement, write (T) for true and (F) for false:

The predecessor of −215 is −214.

• True

• False

Exercise 4A | Q 12.1 | Page 64

Find the value of: |−9|

Exercise 4A | Q 12.2 | Page 64

Find the value of: |−36|

Exercise 4A | Q 12.3 | Page 64

Find the value of: |0|

Exercise 4A | Q 12.4 | Page 64

Find the value of: |15|

Exercise 4A | Q 12.5 | Page 64

Find the value of: −|−3|

Exercise 4A | Q 12.6 | Page 64

Find the value of: 7 + |−3|

Exercise 4A | Q 12.7 | Page 64

Find the value of: |7−4|

Exercise 4A | Q 12.8 | Page 64

Find the value of: 8 −|−7|

Exercise 4A | Q 13.1 | Page 64

Write five negative integers greater than −7.

Exercise 4A | Q 13.2 | Page 64

Write five negative integers less than −20.

Exercise 4B [Pages 68 - 69]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4B [Pages 68 - 69]

Exercise 4B | Q 1.1 | Page 68

Use the number line and add the following integers: 9 + (−6)

Exercise 4B | Q 1.2 | Page 68

Use the number line and add the following integers: (−3) + 7

Exercise 4B | Q 1.3 | Page 68

Use the number line and add the following integers: 8 + (−8)

Exercise 4B | Q 1.4 | Page 68

Use the number line and add the following integers: (−1) + (−3)

Exercise 4B | Q 1.5 | Page 68

Use the number line and add the following integers: (−4) + (−7)

Exercise 4B | Q 1.6 | Page 68

Use the number line and add the following integers: (−2) + (−8)

Exercise 4B | Q 1.7 | Page 68

Use the number line and add the following integers: 3 + (−2) + (−4)

Exercise 4B | Q 1.8 | Page 68

Use the number line and add the following integers: (−1) + (−2) + (−3)

Exercise 4B | Q 1.9 | Page 68

Use the number line and add the following integers: 5 + (−2) + (−6)

Exercise 4B | Q 2.1 | Page 68

Fill in the blanks:

(−3) + (−9) = .......

Exercise 4B | Q 2.2 | Page 68

Fill in the blanks:

(−7) + (−8) = .......

Exercise 4B | Q 2.3 | Page 68

Fill in the blanks:

(−9) + 16 = .......

Exercise 4B | Q 2.4 | Page 68

Fill in the blanks:

(−13) + 25 = .......

Exercise 4B | Q 2.5 | Page 68

Fill in the blanks:

8 + (−17) = .......

Exercise 4B | Q 2.6 | Page 68

Fill in the blanks:

2 + (−12) = .......

Exercise 4B | Q 3.1 | Page 68

Exercise 4B | Q 3.2 | Page 68

Exercise 4B | Q 3.3 | Page 68

Exercise 4B | Q 3.4 | Page 68

Exercise 4B | Q 4.1 | Page 68

Exercise 4B | Q 4.2 | Page 68

Exercise 4B | Q 4.3 | Page 68

Exercise 4B | Q 4.4 | Page 68

Exercise 4B | Q 5.01 | Page 68

Find the sum of: 137 and −354

Exercise 4B | Q 5.02 | Page 68

Find the sum of: 1001 and −13

Exercise 4B | Q 5.03 | Page 68

Find the sum of: −3057 and 199

Exercise 4B | Q 5.04 | Page 68

Find the sum of: −36 and 1027

Exercise 4B | Q 5.05 | Page 68

Find the sum of: −389 and −1032

Exercise 4B | Q 5.06 | Page 68

Find the sum of: −36 and 100

Exercise 4B | Q 5.07 | Page 68

Find the sum of: 3002 and −888

Exercise 4B | Q 5.08 | Page 68

Find the sum of: −18, + 25 and −37

Exercise 4B | Q 5.09 | Page 68

Find the sum of: −312, 37 and 192

Exercise 4B | Q 5.1 | Page 68

Find the sum of: −51, −203, 36 and −28

Exercise 4B | Q 6.1 | Page 68

Find the additive inverse of: -57

Exercise 4B | Q 6.2 | Page 68

Find the additive inverse of: 183

Exercise 4B | Q 6.3 | Page 68

Find the additive inverse of: 0

Exercise 4B | Q 6.4 | Page 68

Find the additive inverse of: −1001

Exercise 4B | Q 6.5 | Page 68

Find the additive inverse of: 2054

Exercise 4B | Q 7.1 | Page 68

Write the successor of the following: 201

Exercise 4B | Q 7.2 | Page 68

Write the successor of the following: 70

Exercise 4B | Q 7.3 | Page 68

Write the successor of the following: −5

Exercise 4B | Q 7.4 | Page 68

Write the successor of the following: −99

Exercise 4B | Q 7.5 | Page 68

Write the successor of the following: −500

Exercise 4B | Q 8.1 | Page 68

Write the predecessor of the following: 120

Exercise 4B | Q 8.2 | Page 68

Write the predecessor of the following: 79

Exercise 4B | Q 8.3 | Page 68

Write the predecessor of the following: −8

Exercise 4B | Q 8.4 | Page 68

Write the predecessor of the following: −141

Exercise 4B | Q 8.5 | Page 68

Write the predecessor of the following: −300

Exercise 4B | Q 9.1 | Page 69

Simplify: (−7) + (−9) + 12 + (−16)

Exercise 4B | Q 9.2 | Page 69

Simplify: 37 + (−23) + (−65) + 9 + (−12)

Exercise 4B | Q 9.3 | Page 69

Simplify: (−145) + 79 + (−265) + (−41) + 2

Exercise 4B | Q 9.4 | Page 69

Simplify: 1056 + (−798) + (−38) + 44 + (−1)

Exercise 4B | Q 10 | Page 69

A car travelled 60 km to the north of Patna and then 90 km to the south from there. How far from Patna was the car finally?

Exercise 4B | Q 11 | Page 69

A man bought some pencils for Rs 30 and some pens for Rs 90. The next day, he again bought some pencils for Rs 25. Then, he sold all the pencils for Rs 20 and the pens for Rs 70. What was his net gain or loss?

Exercise 4B | Q 12.1 | Page 69

For the following statement write (T) for true and (F) for false:

The sum of two negative integers is always a negative integer.

• True

• False

Exercise 4B | Q 12.2 | Page 69

For the following statement write (T) for true and (F) for false:

The sum of a negative integer and a positive integer is always a negative integer.

• True

• False

Exercise 4B | Q 12.3 | Page 69

For the following statement write (T) for true and (F) for false:

The sum of an integer and its negative is zero.

• True

• False

Exercise 4B | Q 12.4 | Page 69

For the following statement write (T) for true and (F) for false:

The sum of three different integers can never be zero.

• True

• False

Exercise 4B | Q 12.5 | Page 69

For the following statement write (T) for true and (F) for false:

| −5| < |−3|

• True

• False

Exercise 4B | Q 12.6 | Page 69

For the following statement write (T) for true and (F) for false:

|8 − 5| = |8| + |−5|

• True

• False

Exercise 4B | Q 13.1 | Page 69

Find an integer a such that: a + 6 = 0

Exercise 4B | Q 13.2 | Page 69

Find an integer a such that: 5 + a = 0

Exercise 4B | Q 13.3 | Page 69

Find an integer a such that: a + (−4) = 0

Exercise 4B | Q 13.4 | Page 69

Find an integer a such that: −8 + a = 0

Exercise 4C [Pages 70 - 71]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4C [Pages 70 - 71]

Exercise 4C | Q 1.01 | Page 70

Subtract: 18 from −34

Exercise 4C | Q 1.02 | Page 70

Subtract: −15 from 25

Exercise 4C | Q 1.03 | Page 70

Subtract: −28 from −43

Exercise 4C | Q 1.04 | Page 70

Subtract: 68 from −37

Exercise 4C | Q 1.05 | Page 70

Subtract: 219 from 0

Exercise 4C | Q 1.06 | Page 70

Subtract: −92 from 0

Exercise 4C | Q 1.07 | Page 70

Subtract: −135 from −250

Exercise 4C | Q 1.08 | Page 70

Subtract: −2768 from −287

Exercise 4C | Q 1.09 | Page 70

Subtract: 6240 from −271

Exercise 4C | Q 1.1 | Page 70

Subtract: −3012 from 6250

Exercise 4C | Q 2 | Page 70

Subtract the sum of −1050 and 813 from −23.

Exercise 4C | Q 3 | Page 70

Subtract the sum of −250 and 138 from the sum of 136 and −272.

Exercise 4C | Q 4 | Page 70

From the sum of 33 and −47, subtract −84.

Exercise 4C | Q 5 | Page 70

Add −36 to the difference of −8 and −68.

Exercise 4C | Q 6 | Page 70

Simplify: [37 − (−8)] + [11 − (−30)]

Exercise 4C | Q 6.2 | Page 70

Simplify: [−13 − (−17)] + [−22 − (−40)]

Exercise 4C | Q 7 | Page 70

Find 34 − (−72) and (−72) − 34. Are they equal?

Exercise 4C | Q 8 | Page 70

The sum of two integers is −13. If one of the numbers is 170, find the other.

Exercise 4C | Q 9 | Page 70

The sum of two integers is 65. If one of the integers is −47, find the other.

Exercise 4C | Q 10.1 | Page 70

Is the following statement true or false?

The sum of two integers is always an integer.

• True

• False

Exercise 4C | Q 10.2 | Page 70

Is the following statement true or false?

The difference of two integers is always an integer.

• True

• False

Exercise 4C | Q 10.3 | Page 70

Is the following statement true or false?

−14 > −8 − (−7)

• True

• False

Exercise 4C | Q 10.4 | Page 70

Is the following statement true or false?

−5 − 2 > −8

• True

• False

Exercise 4C | Q 10.5 | Page 70

Is the following statement true or false?

(−7) − 3 = (−3) − (−7)

• True

• False

Exercise 4C | Q 11 | Page 71

The point A is on a mountain which is 5700 metres above sea level and the point B is in a mine which is 39600 metres below sea level. Find the vertical distance between A and B.

Exercise 4C | Q 12 | Page 71

On a day in Srinagar, the temperature at 6 p.m. was 1°C but at midnight that day, it dropped to −4°C. By how many degrees Celsius did the temperature fall?

Exercise 4D [Pages 72 - 73]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4D [Pages 72 - 73]

Exercise 4D | Q 1.01 | Page 72

Multiply: 15 by 9

Exercise 4D | Q 1.02 | Page 72

Multiply: 18 by −7

Exercise 4D | Q 1.03 | Page 72

Multiply: 29 by −11

Exercise 4D | Q 1.04 | Page 72

Multiply: –18 by 13

Exercise 4D | Q 1.05 | Page 72

Multiply: –56 by 16

Exercise 4D | Q 1.06 | Page 72

Multiply: 32 by –21

Exercise 4D | Q 1.07 | Page 72

Multiply: –57 by 0

Exercise 4D | Q 1.08 | Page 72

Multiply: 0 by –31

Exercise 4D | Q 1.09 | Page 72

Multiply: –12 by –9

Exercise 4D | Q 1.1 | Page 72

Multiply:  (–746) by (–8)

Exercise 4D | Q 1.11 | Page 72

Multiply: 118 by −7

Exercise 4D | Q 1.12 | Page 72

Multiply: −238 by −143

Exercise 4D | Q 2.1 | Page 73

Find the product: (−2) × 3 × (−4)

Exercise 4D | Q 2.2 | Page 73

Find the product: 2 × (−5) × (−6)

Exercise 4D | Q 2.3 | Page 73

Find the product: (−8) × 3 × 5

Exercise 4D | Q 2.4 | Page 73

Find the product: 8 × 7 × (−10)

Exercise 4D | Q 2.5 | Page 73

Find the product: (−3) × (−7) × (−6)

Exercise 4D | Q 2.6 | Page 73

Find the product: (−8) × (−3) × (−9)

Exercise 4D | Q 3.1 | Page 73

Use convenient groupings and find the value of: 18 × (−27) × 30

Exercise 4D | Q 3.2 | Page 73

Use convenient groupings and find the value of: (−8) × (−63) × 9

Exercise 4D | Q 3.3 | Page 73

Use convenient groupings and find the value of: (−17) × (−23) × 41

Exercise 4D | Q 3.4 | Page 73

Use convenient groupings and find the value of:  (−51) × (−47) × (−19)

Exercise 4D | Q 4.1 | Page 73

Verify the following: 18 × [9 + (−7)] = 18 × 9 + 18 × (−7)

Exercise 4D | Q 4.2 | Page 73

Verify the following: (−13) × [(−6) + (−19)] = (−13) × (−6) + (−13) × (−19)

Exercise 4D | Q 5 | Page 73

Complete the following multiplication table:

 x -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
Exercise 4D | Q 6.1 | Page 73

Which of the following statements are true and which are false?

The product of a positive integer and a negative integer is negative.

• True

• False

Exercise 4D | Q 6.2 | Page 73

Which of the following statements are true and which are false?

The product of two negative integers is a negative integer.

• True

• False

Exercise 4D | Q 6.3 | Page 73

Which of the following statements are true and which are false?

The product of three negative integers is a negative integer.

• True

• False

Exercise 4D | Q 6.4 | Page 73

Which of the following statements are true and which are false?

Every integer when multiplied with −1 gives its multiplicative inverse.

• True

• False

Exercise 4D | Q 7.1 | Page 73

Simplify: (−9) × 6 + (−9) × 4

Exercise 4D | Q 7.2 | Page 73

Simplify: 8 × (–12) + 7 × (–12)

Exercise 4D | Q 7.3 | Page 73

Simplify: 30 × (–22) + 30 × (14)

Exercise 4D | Q 7.4 | Page 73

Simplify: (–15) × (–14) + (–15) × (–6)

Exercise 4D | Q 7.5 | Page 73

Simplify: 43 × (–33) + 43 × (–17)

Exercise 4D | Q 7.6 | Page 73

Simplify: (–36) × (72) + (–36) × 28

Exercise 4D | Q 7.7 | Page 73

Simplify: (–27) × (–16) + (–27) × (–14)

Exercise 4E [Page 75]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4E [Page 75]

Exercise 4E | Q 1.01 | Page 75

Divide: 85 by −17

Exercise 4E | Q 1.02 | Page 75

Divide: −72 by 18

Exercise 4E | Q 1.03 | Page 75

Divide: −80 by 16

Exercise 4E | Q 1.04 | Page 75

Divide: –121 by 11

Exercise 4E | Q 1.05 | Page 75

Divide: 108 by –12

Exercise 4E | Q 1.06 | Page 75

Divide: –161 by 23

Exercise 4E | Q 1.07 | Page 75

Divide: –76 by –19

Exercise 4E | Q 1.08 | Page 75

Divide: –147 by –21

Exercise 4E | Q 1.09 | Page 75

Divide: –639 by –71

Exercise 4E | Q 1.1 | Page 75

Divide: –15625 by –125

Exercise 4E | Q 1.11 | Page 75

Divide: 2067 by –1

Exercise 4E | Q 1.12 | Page 75

Divide: 1765 by –1765

Exercise 4E | Q 1.13 | Page 75

Divide: 0 by –278

Exercise 4E | Q 1.14 | Page 75

Divide: 3000 by –100

Exercise 4E | Q 2.1 | Page 75

Fill in the blanks:

80 ÷ (......) = −5

Exercise 4E | Q 2.2 | Page 75

Fill in the blanks:

(−84) + (......) = −7

Exercise 4E | Q 2.3 | Page 75

Fill in the blanks:

(......) ÷ (−5) = 25

Exercise 4E | Q 2.4 | Page 75

Fill in the blanks:

(......) ÷ 372 = 0

Exercise 4E | Q 2.5 | Page 75

Fill in the blanks:

(......) ÷ 1 = −186

Exercise 4E | Q 2.6 | Page 75

Fill in the blanks:

(......) ÷ 17 = −2

Exercise 4E | Q 2.7 | Page 75

Fill in the blanks:

(......) ÷ 165 = −1

Exercise 4E | Q 2.8 | Page 75

Fill in the blanks:

(......) ÷ (−1) = 73

Exercise 4E | Q 2.9 | Page 75

Fill in the blanks:

1 ÷ (......) = −1

Exercise 4E | Q 3.1 | Page 75

Write (T) for true and (F) for false for the following statement:

0 ÷ (−6) = 0

• True

• False

Exercise 4E | Q 3.2 | Page 75

Write (T) for true and (F) for false for the following statement:

(−8) ÷ 0 = 0

• True

• False

Exercise 4E | Q 3.3 | Page 75

Write (T) for true and (F) for false for the following statement:

15 ÷ (−1) = −15

• True

• False

Exercise 4E | Q 3.4 | Page 75

Write (T) for true and (F) for false for the following statement:

(−16) ÷ (−4) = −4

• True

• False

Exercise 4E | Q 3.5 | Page 75

Write (T) for true and (F) for false for the following statement:

(−7) ÷ (−1) = 7

• True

• False

Exercise 4E | Q 3.6 | Page 75

Write (T) for true and (F) for false for the following statement:

(−18) ÷ 9 = −2

• True

• False

Exercise 4E | Q 3.7 | Page 75

Write (T) for true and (F) for false for the following statement:

20 ÷ (−5) = −4

• True

• False

Exercise 4E | Q 3.8 | Page 75

Write (T) for true and (F) for false for the following statement:

(−10) ÷ 1 = −10

• True

• False

Exercise 4E | Q 3.9 | Page 75

Write (T) for true and (F) for false for the following statement:

(−1) ÷ (−1) = −1

• True

• False

Exercise 4F [Pages 75 - 76]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Exercise 4F [Pages 75 - 76]

Exercise 4F | Q 1.1 | Page 75

Which of the following is a true statement?

−4 > −3

• True

• False

Exercise 4F | Q 1.2 | Page 75

Which of the following is a true statement?

−4 < −3

• True

• False

Exercise 4F | Q 1.3 | Page 75

Which of the following is a true statement?

−4 and −3 are non-comparable

• True

• False

Exercise 4F | Q 2 | Page 75

2 less than −3 is

• -1

• 1

• -5

• 5

Exercise 4F | Q 3 | Page 76

4 more than −5 is

• 9

• -9

• -1

• 1

Exercise 4F | Q 4 | Page 76

2 less than −7 is

• -9

• -5

• 5

• none of these

Exercise 4F | Q 5 | Page 76

7 + |−3| = ?

• 4

• 10

• -10

• none of these

Exercise 4F | Q 6 | Page 76

(−42) + (−35) = ?

• -7

• 7

• -77

• none of these

Exercise 4F | Q 7 | Page 76

(−37) + 6 = ?

• -43

• -31

• 31

• none of these

Exercise 4F | Q 8 | Page 76

49 + (−27) = ?

• -73

• 73

• 22

• none of these

Exercise 4F | Q 9 | Page 76

The successor of −18 is

• -19

• 17

• -17

• 19

Exercise 4F | Q 10 | Page 76

The predecessor of −16 is

• -15

• -17

• 15

• 17

Exercise 4F | Q 11 | Page 76

The additive inverse of −5 is

• 5

• 0

• -4

• 6

Exercise 4F | Q 12 | Page 76

−12 − (−5) = ?

• -17

• -7

• 7

• none of these

Exercise 4F | Q 13 | Page 76

−5 − (−8) = ?

• 3

• 13

• -3

• none of these

Exercise 4F | Q 14 | Page 76

The sum of two integers is −25. If one of them is 30 then the other is

• 55

• 5

• -55

• none of these

Exercise 4F | Q 15 | Page 76

The sum of two integers is 20. If one of them is −5 then the other is

• 25

• -25

• 15

• none of these

Exercise 4F | Q 16 | Page 76

The sum of two integers is −13. If one of them is 8 then the other is

• -5

• -21

• 21

• none of these

Exercise 4F | Q 17 | Page 76

On subtracting −8 from 0, we get

• -8

• 8

• 0

• none of these

Exercise 4F | Q 18 | Page 76

8 + (−8) = ?

• 16

• -16

• 0

• none of these

Exercise 4F | Q 19 | Page 76

(−6) + 4 − (−3) = ?

• -5

• -1

• 1

• none of these

Exercise 4F | Q 20 | Page 76

6 − (−4) = ?

• 2

• -10

• 10

• none of these

Exercise 4F | Q 21 | Page 76

(−7) + (−9) + 12 + (−16) = ?

• -20

• 20

• -12

• none of these

Exercise 4F | Q 22 | Page 76

On subtracting 8 from −4, we get

• 4

• 12

• -12

• none of these

Exercise 4F | Q 23 | Page 76

On subtracting −9 from −6, we get

• -15

• -3

• 3

• none of these

Exercise 4F | Q 24 | Page 76

On subtracting −5 from 10, we get

• 5

• -15

• 15

• none of these

Exercise 4F | Q 25 | Page 76

(−6) × 9 = ?

• 54

• -54

• 0

• none of these

Exercise 4F | Q 26 | Page 76

(−9) × 6 + (−9) × 4 = ?

• -90

• 90

• -18

• 18

Exercise 4F | Q 27 | Page 76

36 ÷ (−9) = ?

• 4

• -4

• 0

• none of these

Test Paper 4 [Pages 78 - 79]

### RS Aggarwal solutions for Class 6 Mathematics Chapter 4 Integers Test Paper 4 [Pages 78 - 79]

Test Paper 4 | Q 1.1 | Page 78

What are integers?

Test Paper 4 | Q 1.2 | Page 78

Write all integers from −5 to 5.

Test Paper 4 | Q 2.1 | Page 78

In the pair given below, find the larger integer.

• 0

• -3

Test Paper 4 | Q 2.2 | Page 78

In the pair given below, find the larger integer.

• -4

• -6

Test Paper 4 | Q 2.3 | Page 78

In the pair given below, find the larger integer.

• -99

• 9

Test Paper 4 | Q 2.4 | Page 78

In the pair given below, find the larger integer.

• -385

• -615

Test Paper 4 | Q 3 | Page 78

Write the following integers in increasing order:

−18, 16, 0, −5, 8, −36, −1, 1

Test Paper 4 | Q 4.1 | Page 78

Find the value of: 9 − |−6|

Test Paper 4 | Q 4.2 | Page 78

Find the value of: 6 + |−4|

Test Paper 4 | Q 4.3 | Page 78

Find the value of: −8 − |−3|

Test Paper 4 | Q 5.1 | Page 78

Write four integers less than −6

Test Paper 4 | Q 5.2 | Page 78

Write four integers greater than −6.

Test Paper 4 | Q 6.1 | Page 78

Evaluate: 8 + (−16)

Test Paper 4 | Q 6.2 | Page 78

Evaluate: (−5) + (−6)

Test Paper 4 | Q 6.3 | Page 78

Evaluate: (−6) × (−8)

Test Paper 4 | Q 6.4 | Page 78

Evaluate: (−36) ÷ 6

Test Paper 4 | Q 6.5 | Page 78

Evaluate: 30 − (−50)

Test Paper 4 | Q 6.6 | Page 78

Evaluate: (–40) ÷ (–10)

Test Paper 4 | Q 6.7 | Page 78

Evaluate: 8 × (–5)

Test Paper 4 | Q 6.8 | Page 78

Evaluate: (–30) – 15

Test Paper 4 | Q 7 | Page 78

The sum of two integers is −12. If one of them is 34, find the other.

Test Paper 4 | Q 8.1 | Page 78

Simplify: (−24) × (68) + (−24) × 32

Test Paper 4 | Q 8.2 | Page 78

Simplify: (−9) × 18 − (−9) × 8

Test Paper 4 | Q 8.3 | Page 78

Simplify: (−147) ÷ (−21)

Test Paper 4 | Q 8.4 | Page 78

Simplify: 16 ÷ (−1)

Test Paper 4 | Q 9 | Page 78

The successor of −89 is

• -90

• -88

• 90

• 88

Test Paper 4 | Q 10 | Page 78

The predecessor of −99 is

• -98

• -100

• 98

• 100

Test Paper 4 | Q 11 | Page 78

• (-1)/23

• (1)/23

• 23

• -23

Test Paper 4 | Q 12 | Page 78

If (-13 + 6) ____ -25 - (-9) then the correct symbol in the place holder is

• <

• >

• =

• none of these

Test Paper 4 | Q 13 | Page 78

? + (−8) = 12

• -4

• -20

• 20

• 4

Test Paper 4 | Q 14 | Page 78

The integer which is 5 more that (−7) is

• -12

• 12

• -2

• 2

Test Paper 4 | Q 15 | Page 78

What should be added to 16 to get (−31)?

• 15

• -15

• 47

• -47

Test Paper 4 | Q 16 | Page 78

When 34 is subtracted from −36, we get

• 2

• -2

• 70

• -70

Test Paper 4 | Q 17.1 | Page 78

Fill in the blanks.

−23 − (?) = 15.

Test Paper 4 | Q 17.2 | Page 78

Fill in the blanks.

The largest negative integer is ...... .

Test Paper 4 | Q 17.3 | Page 78

Fill in the blanks.

The smallest positive integer is ...... .

Test Paper 4 | Q 17.4 | Page 78

Fill in the blanks.

(−8) + (−6) − (−3) = ...... .

Test Paper 4 | Q 17.5 | Page 78

Fill in the blanks.

The predecessor of −200 is ...... .

Test Paper 4 | Q 18.1 | Page 79

Write 'T' for true and 'F' for false for the following:

0 is neither positive nor negative.

• True

• False

Test Paper 4 | Q 18.2 | Page 79

Write 'T' for true and 'F' for false for the following:

−(−36) − 1 = − 37.

• True

• False

Test Paper 4 | Q 18.3 | Page 79

Write 'T' for true and 'F' for false for the following:

On the number line −10 lies to the right of −6.

• True

• False

Test Paper 4 | Q 18.4 | Page 79

Write 'T' for true and 'F' for false for the following:

0 is an integer.

• True

• False

Test Paper 4 | Q 18.5 | Page 79

Write 'T' for true and 'F' for false for the following:

−|−15| = −15.

• True

• False

Test Paper 4 | Q 18.6 | Page 79

Write 'T' for true and 'F' for false for the following:

|−40| + 40 = 0.

• True

• False

## Chapter 4: Integers

Exercise 4AExercise 4BExercise 4CExercise 4DExercise 4EExercise 4FTest Paper 4

## RS Aggarwal solutions for Class 6 Mathematics chapter 4 - Integers

RS Aggarwal solutions for Class 6 Mathematics chapter 4 (Integers) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 6 Mathematics solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RS Aggarwal textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 6 Mathematics chapter 4 Integers are Concept of Integers, Addition of Integers, Concept for Ordering of Integers, Representation of Integers on the Number Line, Concept for Natural Numbers, Concept for Whole Numbers, Concept of Negative Numbers, Additive Inverse, Subtraction of Integers, Addition of Integers on Number line.

Using RS Aggarwal Class 6 solutions Integers exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RS Aggarwal Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 6 prefer RS Aggarwal Textbook Solutions to score more in exam.

Get the free view of chapter 4 Integers Class 6 extra questions for Class 6 Mathematics and can use Shaalaa.com to keep it handy for your exam preparation


text:
Series in mathematics are fascinating. Some series converge to a value, and other series diverge. According to a story about the mathematician Gauss, his teacher gave the class the task to calculate the sum $1 + 2 + 3 + … + 100$. The teacher prepared himself to relax while the pupils struggled with the task. Gauss found a mathod to calculate such a sum instantly, and gave the teacher no rest.

Well, let’s examing that sum like

S1 = 1
S2 = 1 + 2
S3 = 1 + 2 + 3
...

or better:

S100 = 100 + 99 + 98 + ... + 1
S99 =  99 + 98 + 97 + ... + 1
...
S1 = 1

or even better:

S100 = 100 + S99
S99 =  99 + S98
...
S1 = 1

or in general:

S(n) = n + S(n-1)
S(1) = 1

The last approach may serve as a model for the recursive function implementation in (e.g) Python. Here is the code:

def sum(val):
if val == 1:
return 1 # Yes, this is the last row in our approach above
else:
return val + sum(val - 1)

print(sum(100))

The output: $5\,050$, the number Gauss calculated in minutes.

I think recursion is like magic, we add (or any other operation) something not yet calculated. The result is obtained when the program reaches the start condition (or end condition?).

Well, there are plenty of series waiting to be calculated. Another example is the alternating harmonic series: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+…$$ Believe it or not, this series equals the irrational number $\ln 2$, that is, the natural logaritm of 2. It’s a bit harder to implement it as a recursion due to the alternation och signs between the terms. I will handle it with the modulo operator, % in Python:

import math as m # For reference value
def ln2(val):
if val == 1:
return 1
else:
return 1/val + (-1**(val % 2)) * ln2(val - 1) # ** means powered to

for n in [11, 31, 51, 71, 91, 191, 291, 391]:
print(f"S{n} = {ln2(n):.8f}")

print(f"\nReference: ln(2) ≈ {m.log(2):.8f}")

Here is the sum the first n terms with some different values of n:

S11  = 0.73654401
S31  = 0.70901620
S51  = 0.70285500
S71  = 0.70013985
S91  = 0.69861150
S191 = 0.69575813
S291 = 0.69486244
S391 = 0.69442432

Reference: ln(2) ≈ 0.69314718

I have choosen some odd values for the sequence number. In this implemantation the program goes through the sequence from the higher to the lower ordinals. In order to get the + and - signs in right places we have to do so as a quick and dirty fix 😅.

We see that the series converges really slow, 391 terms just give us 2 correct decimals. It seems that the default maximum recursion depth is 995, and this number of terms gives us 0.69364944. A little bit closer. We may increase the maximum recursion depth with sys.setrecursionlimit(), but I think there are good reasons not to do.

As I said, I’m fascinated about recursion and recursive functions. Nevertheless, in some cases there are strong reasons to use recursion in programming for simplifying tasks as well as there are situations there recursion should be avoided.


text:
# What is the noise whitening process?

I came across this process when working on OFDM in wireless communication. It's mentioned if we have two copies of a signal resulted from SIMO system (Single Rx Multiple Tx), we can process the noise whitening, after the FFT, in order to improve the quality of the signal.

What does mean the noise whitening, and what's its process?

Here is an example, in the paper.

After taking the DFT in OFDM, we supposed to get frequency response, so how w[k] is obtained?

• en.wikipedia.org/wiki/Whitening_transformation – Marcus Müller Aug 2 '18 at 9:04
• "it's mentioned": Who mentions that, in what context? Cite, and give context. – Marcus Müller Aug 2 '18 at 9:05
• @MarcusMüller .. 1- Why do you usually vote -1 for all my posts,? that really hurts. .. 2- I've read for example here: sndgw.snd.elec.keio.ac.jp/~sanada/TRANSACTION/… in sections 2.1 and similarly in other papers also. but this one which I'm interested in understanding it. – New_student Aug 2 '18 at 9:26
• I don't usually vote -1 on your posts. This one was at least apparently underresearched and didn't give context. That's a severe question quality problem, so I downvoted; I'm sorry if it hurts you, but you might just want to post well-defined questions presenting your own research to avoid that hurt. Please go ahead and edit your question to include that reference. – Marcus Müller Aug 2 '18 at 9:34
• Having read the section and especially the paragraphs after "As already stated, v(t) in Eq. (2) is the filtered noise.", I wonder what your specific question or unclarity is? – Marcus Müller Aug 2 '18 at 9:46

Consider a WSS (wide sense stationary) discrete-time random process (RP) with i.i.d. samples $X[n,s]$ whose auto-correlation sequence (ACF function) is $\phi_{XX}[m]$. We call a random process as white-noise if its ACF is the following:

$$\phi_{XX}[m] = \begin{cases} {\sigma_X^2 ~~~~~, \text{ for } m=0 \\ 0 ~~~~~~~~, \text{ otherwise} }\end{cases}$$

This also means that the RP is uncorrelated. If a RP does not have this property then it's a colored noise. And it means its samples are correlated.

When a RP is colored, it may be possible to find an LTI filter which will try to remove the correlation between the samples of the noise and hence make it whitened. Such a filter is called either as a decorrelator or equivalently as whitening filter.

It's best described in the context of AR-p (Auto-Regressive) random processes generation through all pole filtering, where the whitening filter will be the inverse of that all-pole AR filter which generates the colored noise from an input white noise.

• Hello .. OK, so let me explain the question, when taking the DFT in OFDM demodulator, we got two copies of signal, and it's described that noise whitening is done. how can we perform that noise whitening? is it by that whitening filter or decorrelator ? – New_student Aug 2 '18 at 11:19
• Since almost all communication engineering optimizations are based on the assumption of additive white noise on the received signals, when the noise is not white, their performance degrades from the theoretical maximum. A suggestion to improve the situation is to perform some processing on the received signal to make it whiter. And that's what the noise whitening block tries to achieve... In your case, the algorithm uses fractional oversampling for OFDM receiver which colors the received noise and therefore it's tried to be whitened to improve the reception performance. – Fat32 Aug 2 '18 at 12:34
• Thank you so much .. So, what's the process of noise whitening? do you have any link which explains that OR matlab code example? – New_student Aug 3 '18 at 13:50
• @Eng.Badr the simplest example is the inverse system of an AR process generator... Literature s abundant on this topic. – Fat32 Aug 3 '18 at 14:00
• so, whitening the noise should improve the performance of signal, is that right? and the other thing, noise whitening can include any algorithm, there is not a specific algorithm or process which is called noise whitening .. is it right? – New_student Aug 3 '18 at 15:10


text:
# Proof $\sum\limits_{n \le k/2} \frac 1 n < \log k$ to show Pólya's inequality

In Introduction to Analytic Number Theory, theorem 8.21, Apostol's proves Pólya's inequality. The last step of the proof requires showing:

$$\sum\limits_{n \le k/2} \frac 1 n < log \; k$$

my proof of the theorem shows that we need $$k > 1$$. Is this a standard inequality? What is an easy way to prove it? I have to code it into a theorem prover!

Here is the proof of the book:

• You can bound this sum by something like $\int_1^{k/2+1}\frac{1}{x} dx=\log(k/2+1)$ – Julian Mejia May 28 at 17:37
• If $k = 1$ then both left and right parts are $0$, so inequality doesn't hold. Even if $k = 2$, right part is less than $1$, and left part is $1$. From calculator, for $k = 8$ right part is $2.079$ and left part is $2.083$, and for $k=10$ it's already $2.30$ and $2.28$. Are you sure it isn't enough that it holds for large enough $k$? (then left part is $\log k + \gamma - 1 + O(\frac{1}{k})$, and as $\gamma < 1$, this inequality holds - however, as difference is small, we need sophisticated enough tools) – mihaild May 28 at 17:47
• @JulianMejia it will be $\log(k / 2) + 1$ - first addend is $1$, $i$-th is at most $\log(i + 1) - \log i$, and this bound isn't enough. For example, if $k = 2$, this sum is $1$ and your bound is $\log(2) < 1$. – mihaild May 28 at 17:49
• @mihaild , yes you are right. I can just get bound $\log k+1-\log 2$. – Julian Mejia May 28 at 17:52
• @mihaild i posted the complete proof, it doesn't say anything about large k – Javier May 28 at 17:58

Let's write $$H_m = \sum\limits_{n=1}^m \frac{1}{n}$$. Then we need to show:

$$(1) \; k = 2m + 1 \implies H_m < \log(2m + 1)$$ and $$(2) \; k = 2m + 2 \implies H_m < \log(2m + 2)$$

It's enough to prove $$H_m < \log(2m + 1)$$.

We have $$\log(2m + 1) = \log(2 + \frac{1}{m}) + \log m = \log 2 + \log m + \log(1 + \frac{1}{2m})$$.

From, for example, Approximating Euler's constant by Hirschhorn (we actually need a weaker result than theirs, but it's the only reference I found) we have $$H_m < \log m + \gamma + \frac{1}{2m}$$. So we need $$\gamma + \frac{1}{2m} \leqslant \log2 + \log(1 + \frac{1}{2m})$$ or $$\log 2 - \gamma + \log(1 + \frac{1}{2m}) - \frac{1}{2m} \geqslant 0$$.

Derivative of left side is $$\frac{1}{2m^2}\cdot\left(\frac{1}{-1 + 1 / 2m} + 1\right) > 0$$, so if the inequality holds for some $$m$$, it holds for all greater $$m$$. From wolfram, it holds for $$m = 2$$. For $$m = 1$$, we can manually check $$H_1 = 1$$ and $$\log(3) > \log(e) > 1$$.

(I would like to see some proof involving less calculations, but as I commented we at least need to somehow get $$\gamma < 1$$)

Let me combine all the ideas I find to prove that $$\forall k \ge 1. H_m < \log(2*m+1)$$. As noted by @mihaild this is sufficient for the proof of the theorem. We can proceed by induction.

For $$k = 1$$ , we have $$1 < \log(3) = 1.09\ldots$$ which holds.

For $$k > 1$$, we assume $$H_m < \log(2*m+1)$$ and would like to show $$H_{m+1} < \log(2*(m+1)+1)$$.

We have $$H_{m+1} = H_m + \frac 1 {m+1} < \log(2m+1) + ?$$. To fill this placeholder we note that $$\log(2m+1) + \log(x) = \log(2*(m+1)+1)$$ when $$x = \frac{2m+3}{2m+1}$$ and in that case we would just need to prove that $$\frac 1 {m+1} < \log(1+\frac 2 {2m+1})$$. But this is true since the function $$f(x) = (x+1) \log(1+ \frac 2 {2x+1}) - 1 > 0$$ in $$]\frac{-1} 2,+\infty[$$. To see this is for the cases we are interested is useful to remember that $$e$$ is the unique number for which $$\Big(1+\frac 1 {x}\Big)^{x} < e < \Big(1+ \frac 1 {x}\Big)^{x+1}$$ for all $$x > 0$$ (see wikipedia).

Application on Polya's inequality

In fact, what is needed to conclude the proof of Polya's inequality is:

For $$k \ge 2$$ we have

1. If $$k$$ is odd then $$H_{\frac k 2} < \log(k)$$.

2. If $$k$$ is even then $$H_{\frac k 2 - 1} + \frac 1 k < \log(k)$$

For 1. the proof follows easily from the above.

For 2. one notes $$h(\frac k 2 - 1) < \log(2*(\frac k 2 - 1)+1) = \log(k-1)$$ by the above. Then $$\log(k-1) + \frac 1 k < \log(k)$$ since the function $$x*\log(1+\frac 1 {x-1}) - 1$$ is again positive in $$[1,+\infty[$$. The same inequality as before is used to establish this.

• $\frac 1 {m+1}< \log(1+\frac 2 {2m+1})$ comes directly from $e<(1+\frac{1}{x})^{x+\frac{1}{2}}$ . – user90369 Jun 4 at 8:44

The corrected inequation with $$~k\geq 3~$$: $$\sum\limits_{n\leq\frac{k-1}{2}}\frac{1}{n} < \ln k$$

It’s enough to proof $$~\displaystyle \sum\limits_{v=1}^m\frac{1}{v} < \ln(2m+1)~$$ for $$~m\geq 1~$$.

We can use $$~\displaystyle e<\left(1+\frac{1}{x}\right)^{x+\frac{1}{2}}~$$ for $$~x>0~$$ . A short proof for that is my note here .

Let $$~\displaystyle x:=v-\frac{1}{2}~$$ with $$~v\in\mathbb{N}~$$ .

It’s $$~\displaystyle e<\left(1+\frac{1}{v-\frac{1}{2}}\right)^v~$$ and it follows:

$$\displaystyle \frac{1}{v} < \ln\left(1+\frac{2}{2v-1}\right) = \ln\left(1+\frac{1}{2v-1}\right) + \ln\left(1+\frac{1}{2v}\right)$$

We get the proof:

$$\sum\limits_{v=1}^m\frac{1}{v} < \sum\limits_{v=1}^m\ln\left(1+\frac{2}{2v-1}\right) = \sum\limits_{v=1}^{2m}\ln\left(1+\frac{1}{v}\right) = \ln(2m+1)$$


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When do these methods for PDEs apply?

1. Nov 4, 2009

logarithmic

I'm doing a course in PDEs, where the lecturer hasn't really explain when all these methods for solving PDEs are suitable.

How do you decide which method to use to solve PDEs? Can someone explain for which class of PDEs do the following methods work:
- similarity solutions
- separation of variables
- method of characteristics
- Laplace transform
- Fourier transform
- Green's function.

2. Nov 4, 2009

HallsofIvy

Staff Emeritus
When they work!

For a better answer you need to look into the separation of partial differential equations as "parabolic", "elliptic", or "hyperbolic" equations. (And with variable coefficients you can have equations that are "hyperbolic" for some values of the variables and "parabolic" or "elliptic" for others.)

3. Nov 4, 2009

jasonRF

Halls of Ivy is of course right, and you should not hesitate to ask your professor for his/her take on it. If they are teaching that class they likely have some real insight into the topic (at least you would hope so).

Here is a stab at a couple of these from someone on the practical side (I am not a mathematician who knows deep theory of pdes. I actually just know very elementary stuff) ...

Separation of variables: As a bare minimum, the pde has to be separable, of course, and the boundaries need to be along constant coordinate surfaces. Thus, if you have a 2-D Laplace equation to solve in the region bounded on the inside by a circle and the outside by a square, this technique will not work! EDIT: Also, since you will almost always need a sum of eigenfunctions to satisfy the boundary conditions, the pde needs to be linear.

Characteristics: I have personally only found these useful when dealing with hyperbolic equations (ie wave equations). This can be a very nice approach for some nonlinear wave problems as well.

Fourier transforms: This can somewhat depend upon what you count as a Fourier transform (complex argument? real argument only? distributions? bounded domain? etc.), but usually this means infinite or semi-infinite in some dimension. Limitation is that this approach will only find solutions that have a Fourier transform! Likewise for Laplace transforms (bilateral? unilateral? etc.) EDIT: forgot to mention that if the pde isn't linear, then the transformed equation is in general not so nice. If the coefficients are not constant in the variable you are transforming, the technique can still be useful but you end up with derivatives after the transform (this is usually only helpful if the resulting pde is of lower order than the original).

Green's functions: The basic idea is that to solve a nonhomogenous pde with what can be a complicated source term, you can instead find the "simple" solution to a relatively simple source (delta function), approximate the complicated source term as a sum of these simple sources, and then sum up the simple solutions just like you did the sources. This of course requires linearity. And the sum is really an integral. Sometimes you actually just care about a delta function source in order to understand some fundamental source (like a simple dipole antenna located above the ground, for example), and you have no intention of using the Green's function in an integral to solve for a more complicated source.

Green's functions can also be useful for turning a differential equation into an integral equation. This is often done in order to apply numerical methods for solving integral equations, such as the method of moments that is often used in electromagnetics.

If your prof has any good insight please post it. Thanks!

Good luck,

Jason

Last edited: Nov 5, 2009


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Common and phylogenetically widespread coding for peptides by bacterial small RNAs
• +3
• Robin C Friedman,
• Benno Schwikowski,
• Olivia Doppelt-Azeroual,
• Awaiting Activation,
• Nico,
• Awaiting Activation
Robin C Friedman
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Benno Schwikowski
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Olivia Doppelt-Azeroual
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## Abstract

Background:

While eukaryotic noncoding RNAs have recently received intense scrutiny, it is becoming clear that bacterial transcription is at least as pervasive. Bacterial small RNAs and antisense RNAs (sRNAs) are often assumed to be noncoding, due to their lack of long open reading frames (ORFs). However, there are numerous examples of sRNAs encoding for small proteins, whether or not they also have a regulatory role at the RNA level.

Results:

Here, we apply flexible machine learning techniques based on sequence features and comparative genomics to quantify the prevalence of sRNA ORFs under natural selection to maintain protein-coding function in 14 phylogenetically diverse bacteria. A majority of annotated sRNAs have at least one ORF between 10 and 50 amino acids long, and we conservatively predict that $$188\pm 25.5$$ unannotated sRNA ORFs are under selection to maintain coding, an average of 13 per species considered here. This implies that overall at least $$7.5\pm 0.3\%$$ of sRNAs have a coding ORF, and in some species at least 20% do. $$84\pm 9.8$$ of these novel coding ORFs have some antisense overlap to annotated ORFs. As experimental validation, many of our predictions are translated in ribosome profiling data and are identified via mass spectrometry shotgun proteomics. B. subtilis sRNAs with coding ORFs are enriched for high expression in biofilms and confluent growth, and S. pneumoniae sRNAs with coding ORFs are involved in virulence. sRNA coding ORFs are enriched for transmembrane domains and many are novel components of type I toxin/antitoxin systems.

Conclusions:

We predict over a dozen new protein-coding genes per bacterial species, but crucially also quantified the uncertainty in this estimate. Our predictions for sRNA coding ORFs, along with novel type I toxins and tools for sorting and visualizing genomic context, are freely available in a user-friendly format at http://disco-bac.web.pasteur.fr. We expect these easily-accessible predictions to be a valuable tool for the study not only of bacterial sRNAs and type I toxin-antitoxin systems, but also of bacterial genetics and genomics.


text:
arctan integral

Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4)

according to the book the answer is sqrt(3) * arctan(2u/sqrt(3))
but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function.

The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra?

1
$$\frac{3/2}{u^2+3/4}$$

2
$$3/2 * \frac{1}{(4u^2+3)/4}$$

3
$$3/2*4 * \frac{1}{4u^2+3}$$

4
$$6* \frac{1}{3(4/3u^2+1)}$$
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Mentor
 Quote by jesuslovesu Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4) according to the book the answer is sqrt(3) * arctan(2u/sqrt(3)) but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function. The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra? 1 $$\frac{3/2}{u^2+3/4}$$ 2 $$3/2 * \frac{1}{(4u^2+3)/4}$$ 3 $$3/2*4 * \frac{1}{4u^2+3}$$ 4 $$6* \frac{1}{3(4/3u^2+1)}$$

Careful. You know that

$$\int{ \frac{1}{v^2 + 1} dv = arctan(v) + c.$$

So, for your question, what is $v$?
 something was wrong in step 4. becuse $$\int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c$$ here p > 0. thus,the correct answer is $$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$

Recognitions:
Homework Help

arctan integral

 Quote by jesuslovesu Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4) according to the book the answer is sqrt(3) * arctan(2u/sqrt(3)) but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function. The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra? 1 $$\frac{3/2}{u^2+3/4}$$ 2 $$3/2 * \frac{1}{(4u^2+3)/4}$$ 3 $$3/2*4 * \frac{1}{4u^2+3}$$ 4 $$6* \frac{1}{3(4/3u^2+1)}$$
Fine still step 4. :)
You know that:
$$\int \frac{du}{u ^ 2 + 1} = \arctan (u) + C$$, right?
And now, you've gotten to:
$$\int \frac{6 dx}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 dx}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$
Now, to find the anti-derivative of this expression, you can use the substitution: $$v = \frac{2}{\sqrt{3}} u$$.
Can you go from here? :)
 OK,let`s see it $$\int \frac{6 du}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$ and now we use the substitution: $$p = \frac{2}{\sqrt{3}} u$$ then we get this equation: $$u =\frac{\sqrt{3}}{2} p$$ so: $$\int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}=\int \frac{2}{p^2+1}d(\frac{\sqrt{3}}{2} p) = \sqrt{3}actan(p) + c$$ finally we get the answer: $$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$
 Recognitions: Homework Help Yup, it's correct, congratulations. :)

 Similar discussions for: arctan integral Thread Forum Replies General Math 23 Calculus & Beyond Homework 7 Calculus & Beyond Homework 4 Introductory Physics Homework 4 Calculus 7


text:
# SuanShu, a Java numerical and statistical library

com.numericalmethod.suanshu.stats.factor.implicitmodelpca

## Class ImplicitModelPCA

• java.lang.Object
• com.numericalmethod.suanshu.stats.factor.implicitmodelpca.ImplicitModelPCA

• public class ImplicitModelPCA
extends Object
Given a (de-meaned) time series of vectored observations, we decompose them into a reduced dimension of linear sum of implicit factors. The factors are orthogonal. Specifically, we have $$R = \bar{R} + {B}'F + E$$ R is the time series of vectored observation. Its size is N x T. R_bar is the average of each subject's value over time, one entry per subject. It size is N. We copy the vector T times by columns to form a matrix for computation. F is the time series of factor values. Its size is K x T, where K is the number of implicit factors. B' is the matrix of factor loadings. Its size is N x K. Each row is the factor loadings for each subject. E is the residual matrix. In general, we have: The bigger T is, the more accurate B is. Assuming B is correct, the bigger N is, the more accurate F is; the smaller E (the noise) is, the more accurate F is. Therefore, we first need T big enough to accurately estimate B then need N big enough to accurately F (and E).
"Li, H, Ke, Y., "Using Explicit and Implicit Factors to Explain Stock Returns, 2017."
• ### Nested Class Summary

Nested Classes
Modifier and Type Class and Description
static class  ImplicitModelPCA.Result
the regression results
• ### Constructor Summary

Constructors
Constructor and Description
ImplicitModelPCA(Matrix R)
Constructs an implicit-model that will have one and only one implicit factors.
ImplicitModelPCA(Matrix R, double varExplained)
Constructs an implicit-model that will have the number of implicit factors such that the variance explained is bigger than a threshold
ImplicitModelPCA(Matrix R, int K)
Constructs an implicit-model that will have K implicit factors.
• ### Method Summary

All Methods
Modifier and Type Method and Description
ImplicitModelPCA.Result run()
Runs the regression.
• ### Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Constructor Detail

• #### ImplicitModelPCA

public ImplicitModelPCA(Matrix R,
int K)
Constructs an implicit-model that will have K implicit factors.
Parameters:
R - the time series of observations
K - the number of factors
• #### ImplicitModelPCA

public ImplicitModelPCA(Matrix R)
Constructs an implicit-model that will have one and only one implicit factors.
Parameters:
R - the time series of observations
• #### ImplicitModelPCA

public ImplicitModelPCA(Matrix R,
double varExplained)
Constructs an implicit-model that will have the number of implicit factors such that the variance explained is bigger than a threshold
Parameters:
R - the time series of observations
varExplained - the percentage of variance explained
• ### Method Detail

• #### run

public ImplicitModelPCA.Result run()
Runs the regression.
Returns:
the regression results


text:
# How do you graph y=-x+2 by plotting points?

Jul 12, 2018

Graph:

$y = - x + 2$

Choose values for $x$ and solve for $y$.

$x = - 2$, $y = 4$

$x = - 1$, $y = 3$

$x = 0$, $y = 2$

$x = 1$, $y = 1$

$x = 2$, $y = 0$

Plot the points and draw a straight line through the points.

graph{y=-x+2 [-11.02, 11.48, -3.51, 7.74]}


text:
# [SPM Physics Trial Penang 2019 P1Q27 Solution] How to answer this question?

2 Likes

### SPM Physics Trial 2019 Penang P1 Q27

1. Diagram 21 shows a refraction of light ray \mathrm{R} when hitting point \mathrm{O} after entering glass prism. M is an angle from refracted light ray to normal line.
Rajah 21 menunjukkan pembiasan alur cahaya R apabila menghentam titik O selepas memasuki prisma kaca. M ialah sudut dari sinar biasan ke garis normal.

Diagram 21
Rajah 21
What happen to angle M when the incident angle, \theta is increased ? Apakah yang berlaku kepada sudut M apabila sudut tuju, \theta ditingkatkan ?
A. Increases.
Bertambah.
B. Unchanged.
Tidak berubah.
C. Decreases.
Berkurang.

### SPM Physics Trial 2019 Penang P1 Q27 Solution

Hi Farhaini,

First we need to identify the relevant Physics concept related to this question. Notice the keywords like “refracted light ray”, “prism”, “normal line”, we know that it is related to the concept of total internal reflection.

In diagram below, when we shine a light ray at a small incident angle (red line), the light ray will undergo normal refraction. As we gradually increases the incident angle (purple line), the light ray will be refracted at same line at the wall of the prism. This is known as the critical angle. If we increases the light ray incident angle to be larger than critical angle (green line), then total internal reflection will occur.


text:
A Python package with tools for the PRMS hydrologic model.

## Project description

PRMS-Python provides a Python interface to PRMS data files and manages PRMS simulations. This module aims to improve the efficiency of PRMS simulations while providing “pythonic” tools to do scenario-based PRMS simulations. By “scenario-based” we mean testing model hypotheses associated with model inputs, outputs, and model structure. For example, parameter sensitivity analysis, where each “scenario” is an iterative perturbation of one or many parameters. Another example “scenario-based” modeling exercise would be climate scenario modeling: what will happen to modeled outputs if the input meteorological data were to change?

## Project details

Uploaded source
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