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# Reduction Formulae Question : In= ∫x(cos^n(x))

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1. Apr 12, 2017

### k1902

1. The problem statement, all variables and given/known data
Let In = ∫x(cos^n(x)) with limits between x=π/2, x=0 for n≥0
i) Show that nIn=(n-1)In-2 -n^-1 for n≥2
ii) Find the exact value of I3

2. Relevant equations
∫u'v = uv-∫uv' is what I use for these questions

3. The attempt at a solution
Rewritten as ∫ xcos^n-1(x) cosx
u'=cosx v= xcos^n-1x
u= sinx v'= cos^n-1 -x(n-1)sinxcos^n-2x
But I can't seem to write it in the form it asks for.

2. Apr 12, 2017

### LCKurtz

Your questions are indecipherable. Strange notation and no parentheses.

3. Apr 12, 2017

### Staff: Mentor

Part of the problem is that if you type i in brackets, the browser things you mean the change the font type to italics. Besides that, it's hard to tell exactly what the problem is you're trying to solve.

Here's what I think you meant.
Let $I_n = \int_0^{\pi/2} x \cos^n(x) dx$, with $n \ge 0$
i) Show that $n I_n = (n - 1)I_{n-2} - n^{n - 1}$, for $n \ge 2$.
ii) Find the exact value of $I_3$.
Is this anywhere close to what you're asking?

PS - I used LaTeX to format what I wrote. We have a tutorial here: https://www.physicsforums.com/help/latexhelp/. This is under the INFO menu, under Help/How-to.

4. Apr 13, 2017

### k1902

Sorry, I'm new here and have no idea how to format the text.
But yes that's what I was trying to write , except part i) is i) Show that $n I_n = (n - 1)I_{n-2} - n^{- 1}$, for $n \ge 2$.
Thanks for the help with formatting :)

5. Apr 13, 2017

### k1902

Apologies for that, here's the cleaned up, comprehensible version of the question (thanks to Mark44):
Let $I_n = \int_0^{\pi/2} x \cos^n(x) dx$, with $n \ge 0$
i) Show that $n I_n = (n - 1)I_{n-2} - n^{- 1}$, for $n \ge 2$.
ii) Find the exact value of $I_3$.

6. Apr 13, 2017

### Staff: Mentor

I would try integration by parts twice, starting with $u = \cos^n(x), dv = xdx$. After the second integration by parts, you should have an equation that you can solve algebraically for $I_n$ in terms of $I_{n - 2}$ and other terms.

format: plain


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# Balancing Redox Reactions: Half-Reaction Method

Credit: User:Tuandoanvan/Wikimedia Commons
Source: http://commons.wikimedia.org/wiki/File:AscendingThunderDolphin.jpg

What makes the paint color orange?

The picture above shows one of the two Thunder Dolphin amusement ride trains. This train has an orange stripe while its companion train has a yellow stripe. Pigments of these colors are often made with a dichromate salt (usually sodium or potassium dichromate). These brightly colored compounds serve as strong oxidizing agents in chemical reactions.

### Balancing Redox Equations: Half-Reaction Method

Another method for balancing redox reactions uses half-reactions. Recall that a half-reaction is either the oxidation or reduction that occurs, treated separately. The half-reaction method works better than the oxidation-number method when the substances in the reaction are in aqueous solution. The aqueous solution is typically either acidic or basic, so hydrogen ions or hydroxide ions are present.

In general, the half-reactions are first balanced by atoms separately. Electrons are included in the half-reactions. These are then balanced so that the number of electrons lost is equal to the number of electrons gained. Finally, the two half-reactions are added back together. The example is the oxidation of Fe 2+ ions to Fe 3+ ions by dichromate (Cr 2 O 7 2− ) in acidic solution. The dichromate ions are reduced to Cr 3+ ions.

Credit: Ben Mills (Wikimedia: Benjah-bmm27)
Source: http://commons.wikimedia.org/wiki/File:Potassium-dichromate-sample.jpg

Potassium dichromate (K 2 Cr 2 O 7 ) [Figure2]

Step 1: Write the unbalanced ionic equation .

$\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow \text{Fe}^{3+}(aq)+\text{Cr}^{3+}(aq)$

Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved by the balancing method.

Step 2: Write separate half-reactions for the oxidation and the reduction processes . Determine the oxidation numbers first, if necessary.

$& \text{Oxidation:} \ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) \\& \text{Reduction:} \ \overset{\underset{\mathrm{+6}}{}}{\text{Cr}_2} \ \text{O}^{2-}_7(aq) \rightarrow \text{Cr}^{3+}(aq)$

Step 3: Balance the atoms in the half-reactions other than the hydrogen and oxygen . In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms.

$\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)$

Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation . For the reduction half-reaction above, seven H 2 O molecules will be added to the product side.

$\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance . In this example, fourteen H + ions will be added to the reactant side.

$14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Step 5: Balance the charges by adding electrons to each half-reaction . For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a 2+ total charge on both sides.

$\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq)+e^-$

There is a total charge of 12+ on the reactant side of the reduction half-reaction (14 – 2). The product side has a total charge of 6+ due to the two chromium ions (2 × 3). To balance the charge, six electrons need to be added to the reactant side.

$6e^-+14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Now equalize the electrons by multiplying everything in one or both equations by a coefficient . In this example, the oxidation half-reaction will be multiplied by six.

$6\text{Fe}^{2+}(aq) \rightarrow 6\text{Fe}^{3+}(aq)+6e^-$

Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. If necessary, cancel out H 2 O or H + that appear on both sides.

$6\text{Fe}^{2+}(aq) & \rightarrow 6\text{Fe}^{3+}(aq)+ \cancel{6e^-} \\\cancel{6e^-}+14\text{H}^+(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) & \rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l) \\\hline14\text{H}^+(aq)+6\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) & \rightarrow 6\text{Fe}^{3+}+2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Step 7: Check the balancing. In the above equation, there are 14 H, 6 Fe, 2 Cr, and 7 O on both sides. The net charge is 24+ on both sides. The equation is balanced.

#### Summary

• The half-reaction method of balancing redox equations is described.

#### Practice

Do all half-reaction equation balancing problems at the link below:

#### Review

1. Why do you need the unbalanced ionic equation?
2. What does balancing each half-reaction separately tell you?
3. How do you balance hydrogens in an acidic medium?


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(n-w)(w-OT-i) _ , ) . v

2.2n-i M -r- â– â€¢ j v<w

Every ordinary harmonic of degree n is expressible as a linear func- tion of the system of 2Â» + i zonal, tesseral and sectorial harmonics of degree Â«; thus the general form of the surface harmonic is

Â«  OoPâ€ž (m) + 2 (on cos m<Â£ +6 m sin m\$) Pâ€ž 00 â€¢ (5)

m = i In the present notation we have

P.U)+22iâ„¢-Â£Â£-j-^P lt (Ai)cosÂ»Â»(*-a) I

if we put = 0, we thus have

n

(cos 0+i sin cos <*>)" = Pâ€ž(cos 0) +22 t '"' / iâ€ž\| Pâ€ž(cos 0) cos Â»Â»#,

1 â–

m

from this we obtain expressions for Pâ€ž(cos 0), P n (cos 0) as definite integrals

Pâ€ž(cos 0) =- y (cos + i sin cos <t>) n d<t> j

iP., (cos 0) =- i" (cos + i sin cos #)"cos nt(t>d<t>. \

by Differentiation, â€” The

(6)

1 W+*n

4. Derivation of Spherical Harmonics linear character of Laplace's equation shows that.lrom any solution, Others may be derived by differentiation with respect to the variables x, y, z; or, more generally, if

f(-^^-)

J \dx dy dz]

denote any rational integral operator,

J \dx' W dz) v

is a solution of the equation, if V satisfies it. This principle has been applied by Thomson and Tait to the derivation of the system of any integral degree, by operating upon i/r, which satisfies Laplace's equation. The operations may be conveniently carried out by means of the following differentiation theorem. (See papers by Hobson, in the Messenger of Mathematics, xxiii. 115, and Proc. Lond. Math. Soc. vol. xxiv.)

L I J -._! _2n+l

' n \dx' dy' dz) r '

+ :

2"nl r" r 4 V 4

-!â– -

-...(/.<-

2 .\.2n â€” l 2n â€” 3 which is a particular case of the more general theorem

f (A, A, A) F(r) = \ 2Â»-^Â£ +HT. 2 f ~'F 7 * + }n \dx dy dz) t ( r >-} 2 d(r*)" + TT Jp)^ 1 + â– â–

y,z)

(7)

V-

d"-'F

ry+--. \fn(x,y,z)

(7 1 ),

of the singular point. Let a singular point of degree zero, and strength e , be on an axis hi, at a distance ao from the origin, and also suppose that the origin is a singular point of strength â€” e ; let eo be indefinitely increased, and 00 indefinitely diminished, but so that the product Â«o<>o is finite and equal to eo ! the origin is then said to be a singular point of the first degree, of strength e u the axis being hi. Such a singular point is frequently called a doublet. In a similar manner, by placing two singular points of degree, unity and strength, e u â€”ei, at a distance 01 along an axis h 2 , and at the origin respectively, when Â«i is indefinitely increased, and 01 diminished so that dai is finite and=e 2 , we obtain a singular point of degree 2, strength e% at the origin, the axes being hi, hi. Proceeding in this manner we arrive at the conception of a singular point of any degree n, of strength eâ€ž at the origin, the singular point having any n given axes hi, }h,. . .hâ€ž. If e n -i 4>n-i (x, y, 2) is the potential due to a singular point at the origin, of degree n â€” 1, and strength eâ€ž_i, with axes hi, h*,â€”h*-i, the potential of a singular point of degree n, the new axis of which is h n , is the limit of

s\ d(r 2 )Â»"

where /â€ž(*, y, z) is a rational integral homogeneous function of degree n. The harmonic of positive degree n corresponding to that of degree â€” n â€” I in the expression (7) is

} 2.2Â»-I T 24.2(l-1.2Â«-3 ) J X ' â€¢" '

It can be verified that even when n is unrestricted, this expression satisfies Laplace's equation, the sole restriction being that of the convergence of the series.

5. Maxwell's Theory of Poles.â€” Before proceeding to obtain by means of (7), the expressions for the zonal, tesseral and sectorial harmonics, it is convenient to introduce the conception, due to Maxwell (see Electricity and Magnetism, vol. i. ch. ix.), of the poles of a spherical harmonic. Suppose a sphere of any radius drawn with its centre at the origin ; any line whose direction-cosines are /, m, n drawn from the origin, is called an axis, and the point where this axis cuts the sphere is called the pole of the axis. Differ- ent axes will be denoted by suffixes attached to the direction-cosines ; the cosine (l,x-\-m,y+niz)jr of the angle between the radius vector r to a point (x, y, z) and the axis (h, m,, ni), will be denoted by Xi; the cosine of the angle between two axes is Uy+m,m y -\-nMy, which will be denoted by n,y. The operation

when

this limit is

a

dx dy dz

I'-zz+m^+nr

eâ€ž_i <Â£n-i (* â€” JÂ»a, yâ€” m n a, zâ€”nâ€ža) â€” eâ€ž_! 4>â€ž_i (x, y, z) ; Lo = o, Le n _i = o>, Laen_i = e n ;

â€ž /; 5><Â£Â»-l_LÂ«, d0â€ž_i 1 â€ž 1

d<t>Â«-i 1 â€ž d<t>*-i \ -. ,

-dy- +nn ^r)' 0T _e "a

-<t>n-l-

dx dy ' "" dz J'"' - n dhJ

Since <j>e = i/r, we see that the potential V, due to a singular point at the origin of strength e,â€ž and axes hi, hi, . . .h n is given by

^=^ n ^hidZ..dh j (8)

6. Expression for a Harmonic with given Poles. â€” The result of performing the operations in (8) is that VÂ» is of the form

where Yâ€ž is a surface harmonic of degree n, and will appear as a function of the angles which r makes with the n axes, and of the angles these axes make with one another. The poles of the n axes are defined to be the poles of the surface harmonics, and are also frequently spoken of as the poles of the solid harmonics Yâ€žr", Ynr - " -1 . Any spherical harmonic is completely specified by means of its poles.

In order to express Yâ€ž in terms of the positions of its poles, we apply the theorem (7) to the evaluation of Vâ€ž in (8). On putting r=n f n {x, y, z) â€”Xl(l,x+mry+n r z), we have t =1 (2n!) I /. rV , r 4 V 4

performed upon any function of x, y, z, is spoken of as differentiation with respect to the axis (L, m>, Â«i), and is denoted by d/dh,. The potential function Vo = ft>/r is defined to be the potential due to a singular point of degree zero at the origin ; eo is called the strength

V _ (2"0 1 /. fV I " 2"Â»!n! r n \ 2.2n~i

2 .\.2n â€” 1 .in

=i-)

X

n (l,x +m r y+n,z).

1 By ZGi'X" -25 ) we shall denote the sum of the products of 5 of the quantities p, and n â€” 2s of the quantities X; in any term each suffix is to occur once, and once only, every possible order being taken. We find

Mlx+my+nz) =2(X")r\ A 2 n(te +my+nz) =22( 1 u 1 X' , - 2 )r"- 2 , and generally

A 2m n(lx+my+nz) =2 m m ! 2(^" , X"- 2 "')r n - :!m ;

thus we obtain the following expression for Y n , the surface har- monic which has given poles hi, hi, . . ,hâ€ž;

d" Â±

r

Yn = r n+lL-Lt

n\ dhidhi. . .dhâ€ž

(-1)"

, (2Â«â€” 2m)! 2"~'"n\ (nâ€”m)

,2(XÂ»

V") \

(9)

m =

where S denotes a summation with respect to m from m = o to m = \n, or \(n â€” 1), according as n is even or odd. This is Maxwell's general expression (loc. cit.) for a surface harmonic with given poles.

If the poles on a sphere of radius r are denoted by A, B, C. . ., we obtain from (9) the following expressions for the harmonics of the first four degrees: â€”

~ Yi = cos PA, Y s = J(3 cos PA cos PB-cos AB),

YÂ« = 4(J5 cos PA cos PB cos PC -cos PA cos BC-cos PB cos CA

--cos PC cos AB), Y 4 = K35 cos PA cos PB cos PC cos PD -52 cos PA cos PB cos CD

+ 2 cos AB cos CD), 7. Poles of Zonal, Tesseral and Sectorial Harmonics. â€” Let the n axes of the harmonic coincide with the axis of z, we have then by (8) the harmonic

( â€” i)"rÂ» +1 dÂ» I. n ! dz' r '


text:
# Variance of a summation of Gaussians

## Homework Statement

I am trying to follow a step in the text book but I don't understand.

$var\left(\frac{1}{N}\sum_{n=0}^{N-1}w[n]\right)\\ =\frac{1}{N^2}\sum_{n=0}^{N-1}var(w[n])$
where $w[n]$ is a Gaussian random variable with mean = 0 and variance = 1

## Homework Equations

$Var(X) = \operatorname{E}\left[X^2 \right] - (\operatorname{E}[X])^2.$

## The Attempt at a Solution

The mean is 0 because a summation of Gaussian is Gaussian.
But squaring the whole expression doesn't seem right as there seems to be a trick used to go from line 1 to 2.

Last edited:

jbunniii
Homework Helper
Gold Member
Two facts are being used here:

1. If X is a random variable and c is a constant, then $\text{var}(cX) = c^2\text{var}(X)$.
2. If X and Y are uncorrelated random variables, then $\text{var}(X + Y) = \text{var}(X) + \text{var}(Y)$. From this, it's an easy induction to handle the sum of N uncorrelated random variables.

Both of these facts are straightforward to prove and should be found in any probability book.

Thanks a lot. Haven't touched random variables for a while and the summation threw me off.

The proofs for those facts are indeed very straightforward.


text:
# Generating/spanning modules and submodules ... ...

Gold Member

## Main Question or Discussion Point

In Chapter 1 of his book: "Modules and Rings", John Dauns (on page 7) considers a subset $T$ of an R-module $M$ and defines the R-submodule generated by $T$ ... for which he uses the notation $\langle T \rangle$ ... ... as follows:

Now, note that Dauns (in Section 1-2.5) also defines $\sum M_i = \langle \cup M_i \rangle$ ... and so it follows (I think) that if the family of submodules, $\{ M_i \}_I$ spans or generates $M$ ... then we have

$\{ M_i \}_I$ generates/spans $M \Longrightarrow M = \sum M_i = \langle \cup M_i \rangle$ ... ... ... ... ... (1)

Note that on page 8, under the heading Observations, Dauns states:

" ... ... if $1 \in R, \langle T \rangle = \sum \{ tR \mid t \in T \}$ ... ... ... ... ... (2)

Now, we have that

(1) (2) $\Longrightarrow M = \sum M_i = \langle \cup M_i \rangle = \sum \{ tR \mid t \in \cup M_i \}$ ... ... ... ... ... (3)

But ... how do we reconcile Dauns' definitions with Bland's Definition 4.1.2 which states

" ... ... An R-module M is said to be generated by a set $\{ M_\alpha \}_\Delta$ of R-modules if there is an epimorphism $\bigoplus_\Delta M_\alpha \to M$. ... ... "

The complete Definition 4.1.2 by Bland reads as follows:

Can someone please explain how to reconcile Dauns' and Bland's definitions ...

Just a note ... I feel that Dauns definition has more the "feel" of something being generated ...

To give readers of the above post the context including the notation of Dauns approach I am providing the text of Sections 1-2.4 to 1-2.8 ... as follows ...

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fresh_42
Mentor
The only essential difference between the two definitions is, that Bland uses a sum of any modules $M_\alpha$, which are external to $M$, and Dauns starts with elements, which are already in $M$. So we should first make them comparable. Let's take Bland's notation and write
$$\bigoplus_{\alpha \in \Delta}M_\alpha = \{\,(m_\alpha)_{\alpha\in \Delta}\,|\,m_\alpha \in M_\alpha\,\}\text{ and } \iota_\alpha\, : \,M_\alpha \longrightarrow \bigoplus_{\alpha \in \Delta}M_\alpha\, , \,\iota(m_\alpha)=(0,\ldots,m_\alpha,\ldots,0,\ldots)$$
We get Dauns' definition now by setting $T:=\cup \iota_\alpha(M_\alpha)\subseteq M$. And in the other direction, if we already have $M=\sum_\alpha M_\alpha$ then the projection is clear.

Math Amateur
Gold Member
The only essential difference between the two definitions is, that Bland uses a sum of any modules $M_\alpha$, which are external to $M$, and Dauns starts with elements, which are already in $M$. So we should first make them comparable. Let's take Bland's notation and write
$$\bigoplus_{\alpha \in \Delta}M_\alpha = \{\,(m_\alpha)_{\alpha\in \Delta}\,|\,m_\alpha \in M_\alpha\,\}\text{ and } \iota_\alpha\, : \,M_\alpha \longrightarrow \bigoplus_{\alpha \in \Delta}M_\alpha\, , \,\iota(m_\alpha)=(0,\ldots,m_\alpha,\ldots,0,\ldots)$$
We get Dauns' definition now by setting $T:=\cup \iota_\alpha(M_\alpha)\subseteq M$. And in the other direction, if we already have $M=\sum_\alpha M_\alpha$ then the projection is clear.

Thanks fresh_42 ...

Still reflecting on your post ...

Peter

The definitions of Bland and Dauns are the same, compare definition 1.4.3. of Bland with definition 1-2.4 of Dauns. Use proposition 1.4.4. of Bland and 1-2.5 of Dauns to see that the definitions are the same. So let’s concentrate on Bland.
You could first read post #2 of
https://mathhelpboards.com/linear-abstract-algebra-14/modules-generated-sets-submodules-bland-problem-1-problem-set-4-1-a-24243.html

(Recall: If $(M_\alpha)_\Delta$ is a family of R-modules such that $M_\alpha = M$ for each $\alpha \in \Delta$ then $\Pi_\Delta M_\alpha$ and $\bigoplus_\Delta M_\alpha$ will be denoted by $M^\Delta$ and $M^{(\Delta)}$, respectively).

Generated vs. spanned.
(def.1.4.3.p29) Let $S \subset M$, $M$ is generated by (the elements of) $S$ if each element $x \in M$ can be expressed as a finite sum $x = \Sigma x_\alpha a_\alpha$ with $x_\alpha \in S$ and $a_\alpha \in R$.

(p.104) Let $\mathscr{S} = \{N_\alpha\}_\Delta$ be a set of submodules of M such that $M = \Sigma_\Delta N_\alpha = \{\text{ finite sums } \Sigma x_\alpha a_\alpha \text{ with } x_\alpha \in N_\alpha \text{ and } a_\alpha \in R \}$ then $\mathscr{S}$ is said to span $M$. $\mathscr{S}$ is called the spanning set.

Thus $M$ is generated by elements of $M$, and $M$ is spanned by submodules of $M$.

(p.28, p.51) Clear is that if $M$ is generated by $\{x_1, \cdots, x_n\}$ then $M$ is spanned by the submodules $\{x_1R, \cdots, x_nR \}$, and conversely:

$M = \langle x_1, \cdots, x_n \rangle$ $\Longleftrightarrow$ $M = \Sigma x_i R$.

You can prove that yourself. This is also valid for infinite sets of generators and spanning sets.

For instance $\mathbb{R}^3$ is spanned by $x\mathbb{R}$, $y\mathbb{R}$, and $z\mathbb{R}$, the x-axis, y-axis, and z-axis, respectively.
If $x \in M$ then $x\mathbb{R}$ is a submodule of $M$, and it is a very special submodule of $M$. You can wonder if $M$ can be spanned by more general submodules of $M$ ?. Can $M$ be generated by modules that are outside of $M$, i.e., that are not submodules of $M$ ?. For instance, can the real plane $\mathbb{R}^2$ be spanned by a sphere or a torus? To answer these questions, Bland introduced a “new” definition of generating a module.

Bland Definition 4.1.2.
(1) An $R$-module $M$ is said to be generated by a set $(M_\alpha)_\Delta$ of R-modules (or $(M_\alpha)_\Delta$ generates $M$) if there is an epimorphism $\bigoplus_\Delta M_\alpha \to M$.

(2) An $R$-module $M$ is said to generate an $R$-module $N$ if there is an epimorphism $M^{(\Delta)} \to N$ for some set $\Delta$.

Annoying is that Bland (and all other authors) now speak of generating a module with (other) modules instead of spanning a module with modules. We have to live with this confusing mix of terms.

Let us now go back to free modules and recall this important theorem:
Bland Proposition 2.2.6.p.54
Every $R$-module $M$ is the homomorphic image of a free $R$-module. Furthermore, if $M$ is finitely generated, then the free module can be chosen to be finitely generated.
Recall that every module $M$ has at least one set of generators, namely $M$ itself.
What does this theorem say? It says (you can check it in your textbook):
For each module $M$ there exists an index set $\Delta$ $\text{ AND }$ an epimorpism $F:R^{(\Delta)} \to M$.

That’s it. The set $\Delta$ may be finite or infinite.

Now compare this with part (2) of definition.4.1.2. above.

I think this reconciles this new definition with the former definitions of Bland.

Of course part (1) of definition 4.1.2 is a generalization of part (2).

Math Amateur
fresh_42
Mentor
Thus $M$ is generated by elements of $M$, and $M$ is spanned by submodules of $M$.
This is a bit too artificial for my taste. There is not really a clear definition for either of them, except that generate extends to non-linear structures as groups and spanned usually refers to linear structures as modules and vector spaces. They are often used equivalently, so what looks like a definition above is by no means general and might even be inconsistent within Bland's book. Whether a module is generated by something or spanned by something doesn't matter.

So the terms might only apply to: (see next post)

Last edited:
Math Amateur
It is not my definition, there are definitions of Bland.

In def.1.4.3 Bland defines generated by elements
On p.104 Bland defines spanned by submodules
In def.4.1.2 Bland defines generated by modules
Mathamateur is struggling with these notions and asked for “reconciliation”

Last edited:
Math Amateur


text:
Katok's conjecture on entropy and periodic orbits for generic $C^1$ diffeomorphisms

Let $M$ be a compact finite-dimensional manifold and $f\colon M\to M$ be a diffeomrphism. By $P_n(f)$ we denote the number of periodic points of $f$ with period $n$, that is, the number of fixed points of $f^n$. Katok Lyapunov exponents, entropy and periodic orbits for diffeomorphisms, Publications Mathématiques de l'IHÉS 51, no. 1 (1980), pp. 137-173 showed that the topological entropy of a $\mathcal{C}^{1+\varepsilon}$ diffeomorphism of a compact surface obeys the following inequality: $$\limsup_{n\to\infty} \frac{\log P_n(f)}{n}\ge h(f).$$ He also conjectured (see paragraph 5 page 141 of the article cited above) that this inequality holds in any dimension generically in the $\mathcal{C}^1$-topology. In other words: There is a dense $G_\delta$ set $\mathcal{G} \subset \text{Diff}^1 (M)$ such that for any $f\in\mathcal{G}$ one has $$\limsup_{n\to\infty} \frac{\log P_n(f)}{n}\ge h(f).$$ Is this conjecture still open? I would be grateful for any references.

The main point is that $C^1$-far from homoclinic tangencies the result must be true due to the existence of a maximal entropy measure (see here, here, here and references therein) where the splitting has one dimensional center (and so Katok type results can be carried out, see here). There is a subtelty here which is that the central exponent may be a priori equal to zero, in which case one would like to use the genericity hypothesis to show that the estimate holds (I think the concept of principal simbolic extension in the above references should be helpful for this). Here there might be some work to be done.


text:
# Monthly Archives: September 2016

## Owen Jones meets Ha-Joon Chang | The economic argument against neoliberalism

The Guardian has the interview clip Owen Jones talks to Ha-Joon Chang: ‘austerity is based on lies’ – video interview.

The full interview is on YouTube Owen Jones meets Ha-Joon Chang | The economic argument against neoliberalism. That full addition is embedded below.

Ha-Joon Chang is a professor of economics at Cambridge University and a best-selling international author. I asked him to provide an argument against much of today’s prevailing economic thought. Isn’t austerity necessary? Do we have capitalism for the rich and socialism for the poor? Is taxation theft? Does welfare encourage laziness and fecklessness? What’s wrong with inequality? And can we defeat neoliberlism?

The more ways you hear this explained the more likely you are to get it. The accent may get in the way for some of you to understand as well as you might otherwise. Ha-Joon Chang also has a web site. He has also written very understandable books.

## ‘Quantum teleportation’ breakthrough by DARPA-funded physicists

Ironically it is RT that has published this article ‘Quantum teleportation’ breakthrough by DARPA-funded physicists

This is the best explanation I have seen. The explanation that accompanies the cartoon is necessary to complete the understanding. It is not until the Alice tells Bob what the state of his photon will be that any real information is passed at a distance. So the secret that Alice tells the Bob is the state of the Bob’s photon. The fact that Bob can verify that what Alice said is true, is the accomplishment in this experiment. Of course, since there were only two choices, the Alice had a 50% chance of being right even if she had not known the correct answer. It would take a significant string of right answers before the level of trust could be established with enough confidence.

By the way, I think some of the details in the cartoon explanation are a little off, but the gist of it is correct.

## Jill Stein Event Calendar for Central Massachusetts

Here is a link to the calendar.

## The Charade That Deters the Use of “Direct Spending” to Fund Federal Operations

Naked Capitalism has the article The Charade That Deters the Use of “Direct Spending” to Fund Federal Operations.

In the article the author imagines a speech that Bernie Sanders could have given.

I, Bernie Sanders, believe the primary purpose of the state, the strategic mission of the federal government, is to empower the basic units of our social fabric: our households, communities, and local economies. Big businesses and corporations can take care of themselves, and have much to offer and contribute—but it is only by nurturing the health and vigor of the basic units of our society that we can, in fact, create a more perfect union. And this is not a task that big business, pursuing corporate profits, is disposed or interested in undertaking.

In another part of the article, the author provides an interesting take on historically remembered era of great inflation.  The crudest definition of inflation is too many dollars (or whatever currency applies) chasing too few goods.

… historical analysis demonstrates that the root causes of hyper-inflation—the infamous Weimar Republic or Zimbabwe, for example—have not been caused by “money printing” but, instead, by the collapse of the production capabilities creating things for the money to buy.

In other words too few goods to supply the demand of too many dollars.  In fact it was Republicans, I believe,  who coined the term of supply side economics as a way of stopping the inflation that this country saw in the 1970s.  In other words fix the shortage of goods instead of controlling the excess of  dollars.  Of course, that is not actually how they stopped inflation.  They did it by bringing on the worst economic decline of consumer demand, up until then,  since the great depression.  That did in fact control the “excess” money.

## Lessons Learned From the 2016 Presidential Campaign

Now that I am working to promote the Jill Stein/Ajamu Baraka campaign for the Presidency, I think I see some lessons to be learned.

I started on the Bernie Sanders campaign before May, 2015. I know that date, because that is when I started to go to meetings of the “We Want Bernie – Worcester” group.

The Jill Stein campaign, with less than two months to go until the election, is at about the same place as the Sanders’ campaign was 15 months ago. How did this happen?

My conjecture is that there was not such a tough primary contest in the Green Party to cause Jill Stein to gear up back then the way Bernie Sanders did. Moreover, Bernie was getting so much attention, that he sucked all the air out of whatever efforts the Green Party and Jill Stein could have mustered.

Of course, had Bernie not run, there wouldn’t have been 13 million votes cast for him and his platform. So, without Bernie, there would not be any where near as much interest in the platform as there is now.

Here is where we can see some strategic flaws in what Bernie Sanders did. In order to get credibility with Democratic Party voters, he made a promise that he would support the nominee of the party in November no matter who that turned out to be. Without that promise, he undoubtedly felt that he could not be trusted to be an actual Democrat after his years of being an independent who caucused with the Democrats.

Was there another strategy he could have chosen, and how do we know it would have been successful had he chosen it? Donald Trump showed that there was a better strategy, although you could not have known that in advance. Donald Trump made threats of going the third party route if the Republican party did not treat him well. There were efforts from high ranking party members to thwart him, but not the rampant vote rigging, voter suppression, and other shenanigans deployed against Bernie Sanders by the Democratic Party. We now know that Donald Trump actually became the Republican nominee whereas Bernie Sanders failed to get the Democratic nomination. Maybe the Democratic Party “leaders” would have shown some restraint with respect to Bernie Sanders if there were a threat of a third party run hanging over them.

OK, so we see that there probably was a better strategy for Bernie Sanders to have chosen at the outset. However, given that he made the choices that he did, we need to look at the ramifications.

Since Bernie Sanders did not create an escape hatch for himself, and being a man of his word, he has no other choice but to strongly support the Hillary Clinton candidacy. His initial promise prevents him from even hinting that he is not enthusiastic or suggesting his former backers should be switching their support to Jill Stein.

Had Bernie been able to switch his support to a Jill Stein ticket, or perhaps even joined the ticket, or perhaps even headed that ticket (all of which were opportunities that were presented to him), the effort to build a Progressive movement would be light-years ahead of where it is today.

Jill Stein has just announced that she thinks that upping her ground game is very important. She has just hired three people for the entire country to “explode” the size of her ground game less than two months before the election. We former Bernie supporters can see how far behind the curve that this effort is. We can see what the campaign should have been doing for the last 15 months which we were doing when we were on the other team. We can see why this effort didn’t happen.

Some of us were warned by Green Party supporters of past years how hopeless the effort would be when we switched over. I felt there was a certain amount of irony in that these people were deserting the party just at the moment when it had its greatest chance of success.

The lesson is not that they were right, and we were wrong. The lesson to be learned is what we have to continue to do up to November and beyond. We need to infuse the Green Party with the experience of what it takes to attract huge numbers of voters. We need to build the size of the party as much as we can for this election, so that we have some momentum going forward to future elections. We have to try to make sure that this wasn’t the greatest opportunity the party ever had that was muffed because of certain decisions that seemed right under the circumstances.

## Bill Moyers: We, the Plutocrats vs. We, the People: Saving the Soul of Democracy

Naked Capitalism has reposted Bill Moyers: We, the Plutocrats vs. We, the People: Saving the Soul of Democracy.

Taking a cue from what Moyers wrote, the ideas are best summed up by quoting the preamble to our constitution.

We the People of the United States, in Order to form a more perfect Union, establish Justice, insure domestic Tranquility, provide for the common defense, promote the general Welfare, and secure the Blessings of Liberty to ourselves and our Posterity, do ordain and establish this Constitution for the United States of America.

There seems to be very little of that idealism in what passes for our political system of today. Bill Moyers explores where it went and why we need it back.

## Sen. Bob Graham: FBI Covered Up Role of Bandar and Saudis in 9/11 Attacks

The Real News Network has a recent interview Sen. Bob Graham: FBI Covered Up Role of Bandar and Saudis in 9/11 Attacks.

If there was such a cover up as described above, and there is continuing effort to get this information ouf of the FBI, CIA, and other agencies, then it must be that under Obama’s direction this cover up continues. If Obama vetoes the legislation that has passed Congress to let Americans sue Saudi Arabia over 9/11, then he must know reasons for this that the Congress does not know. In other words, continuing obstruction of information being given to the American public. If Obama knows, you can bet Hillary Clinton knows. And we are expected to vote for a continuation of the cover-up.

Think Progress has the article Obama Expected To Veto 9/11 Bill Because It Sets A Dangerous Precedent.

Obama said in April he would veto the bill. The White House says that the bill’s enactment could put American officials overseas in danger.

By opening up the prospect of victims suing governments (or states), the United States could be opening itself up to law suits from individuals who feel that the country has committed crimes in their nation — like victims of drones in Pakistan, or civilians killed by the U.S.-backed Saudi coalition in Yemen.

I think the above video hints at much more likely reasons why Obama wants to veto that bill.

The article goes on further to state:

The bill passed on Friday despite the release of the notorious 28 pages — a missing section of the congressional 9/11 report. The 28 pages that were released in July did not provide the evidence that some suspected would tie the financing of the 9/11 attacks to the Saudi royal family.

From the video we know what utter hogwash it is for even Think Progress to publish claims tof such failure to make links.

Vote for Jill Stein to get the to bottom of the cover up.

## Goodby to Bernie Sanders Activists

The deletion of this conversation thread by the group moderator is what made me decide to leave the Bernie Sanders Activists group on Facebook.

## Beautiful Math

Just got an update to the Jetpack software in WordPress for use on this web site.

One of the features is called Beautiful Math. Here is an example of what it can do.

$latex i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$

I am sure this will come in handy whenever I want to discuss the politics of physics. 🙂


text:
## anonymous 4 years ago what is the limit of F (x) = √(〖(y〗^2-9))/√(2y^2+7y+3) as x approaches -3?

1. anonymous

its hard to put this one the radical is only 1 in the question but i cant put it 1 so i decide to put 2

2. anonymous

$\sqrt({y^{2}-9})/(2y^2+7y+3)$

3. anonymous

the sqrt is for all

ahaaa...wait one second

5. anonymous

a. 0 b -sqrt 6/5 c. sqrt 6/5 d. sqrt 5/6

6. anonymous

c


text:
# Comments the way to do ?

This topic is 4857 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi So I am mostly coded for myself, and I didn't do anything special with comments, I written them as I understood the code, and in the parts, where I thought it was necessary. But what is the right way for Software Companies to do ? Where the code should be, where not, and what should be written in what way ? I am looking especially advice from people who are working for commercial companies, but anyone else is also welcomed :) Thanks.

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I am programming in Java (commercially) and C++ (more or less for myself).

I comment my code extensively. All classes have a "class comment", each method has a comment (partly not only about _what_ they do but also their rationales), and each field has one. Additionally, comments are inserted into the method bodies. Often algorithms are such complex that I (as the author) would not understand them 2 monthes after writing, if especially tricky parts would not be commented.

I use JavaDoc for commenting Java source, of couse, and doxygen (under linux) for C++. So I use also more or less standardized formatting for comments (e.g. @param in JavaDoc or \pre in doxygen).

I use assertions in the way suggested by Meyer's Eiffel language: "Design by Contract."

In medium to large scale projects also unit tests are essential. I write them also for my private libraries. They not only give one a safety of "all is correct" but also early points out some bugs and inefficiencies. And reworks are much less painful to verify.

Well, partitioning the code into appropriate compilation units is another question. I don't assume that there is a golden way to go. Ok, in Java one has not such freedom as in C++. I make "packages" in Java (of course) but also in C++ (ok, at an amount of 500+ classes one would die on searching otherwise). Besides that I _normally_ have one compilation unit per class.

In a consequence all this is a question of how many effort one wants to spend now and in the future. I've often had to read code written by other people, and often other people have to read my code. I know how painful that may but need not to be ...

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These are mostly not my tips but I have found them useful:

If you use java, the best thing to do is make sure that the javadocs are complete for your program. Since your sig indicates you use C# then I would probably do the MS-version of a javadoc (since I assume MS would make at least a half-clone of javadoc)

When you use C++ make sure that each function is commented with a description of what the inputs and outputs are and what task is performed. Also make sure that each 3 to 5 lines of "non-trivial" code have a comment attached describing any non-obvious parts of those lines.

And be careful not to write pointless or stupid comments, like:

...x++; //increment xi = sin(M_PI/2) //take the sine of pi over 2...

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Thank you all.

Ok how about writing about something complicated that is going on in the function ?

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Quote:
 Original post by etothexAnd be careful not to write pointless or stupid comments, like:...x++; //increment xi = sin(M_PI/2) //take the sine of pi over 2...

Just to expand from this as a base point...

<blatentopinion>

Comments should not be used to say what a given line of code is doing, this should be obvious from the code itself (self documenting code). Comments should address things like why, and to what end. If a section of code is nonobvious in purpose I prefer to factor that out into an appropriately named function (self documenting code) or using appropriately named temporaries, rather than commenting - this also leads to more reuse.

Basically, things like:

      ... box.right - box.left ...   => ... width( box ) ...or => scalar_type box_width = box.right - box.left;      ... box_width ...

An alternative good use for comments is things like listing side-effects - e.g. std::vector::* functions when a resize triggers a realloc will invalidate existing iterators.

</blatentopinion>

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My general rule is, if there is code that could not be quickly read and understood by someone familiar(but not an expert in) C++ and what your program does, then it needs a comment. Basically, any time something that is not immediately clear or uses some non-obvious or non-standard techniques.

Consider a function computeAngleBetweenVectors(Vector v1, Vector v2). In the body of the function you could take the dot product and figure the angle that way. The code, IMHO, should be commented something to the tune of "use the dot product to calculate the angle" or even "calc. the dot product". Not much more is needed, because it will help someone reading your code interpret it, without giving them any more information that wasn't easily gained from the code itself.

BTW, using smart, descriptive variable and function names _really_ helps here - because that provides a lot of info in the first place.

Another rule of thumb is to assume that the readers of your code know a lot of basic, common knowledge - but have forgotten all of it. Following the computeAngle example, you should assume that the reader is familar with using the dot product but would not recognize the code for such on sight. The comment should clearly explain the high-level view on the code - the reader should see the code, read the comment, and see what you are doing, and if necessary know the terms to search under if they are unfamiliar with the dot product.

Obviously, the dot product example is a bit contrived because it is so simple. But the same principle applies to more complex algorithms that may not be as intuitively readable in the code.

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Here is an example. Not very complicated stuff, but with a little trick. However, more or less my method bodies look this way as soon as enough lines come together.

boolBoundingSphere::consider(const Point3_t& point) {	// classifying the point (initially _sqRadius is -1, so that the first candidate is ever classified as an outlier)	const Point3_t difference(_center,point);	scalar_t distance = difference.squaredNorm2(); // in fact the squared distance	const bool isOutlier = distance > _squaredRadius;	// if the point is an outlier the volumes geometry must be adapted ...	if(isOutlier) {		distance = XMath<scalar_t>::sqrt(distance);		// faking on the first point ...		if(_radius<0) {			// due to this little trick the newRadius will become 0, and the center will be translated by 1 multiple of			// the difference vector from Point3_t::ZERO on, so yielding in the point as center of a sphere w/ radius 0			_radius = -distance;		}		// computing the new radius to include the sphere so far as well as the outlier		const scalar_t newRadius = (_radius+distance)*0.5f;		// shifting the sphere "half-way" towards the outlier		if(distance) {			_center.add(difference,(newRadius-_radius)/distance);		}		// scaling the sphere		_radius = newRadius;		_squaredRadius = newRadius*newRadius;	}	// done	return isOutlier;}

IMHO the comments should not paraphrase the code but (a) may explain the flow of control, and (b) should make clear what effects code has.

Quote:
 Original post by etothexBTW, using smart, descriptive variable and function names _really_ helps here - because that provides a lot of info in the first place.

I absolutely agree (see above ;-)

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If you have a moderately long function that you can't really break into pieces, it is good to summarize it by writing small "headings" in front of code blocks, even if they are otherwise easy to understand. A control function for some resource loader might look like this:
/* Check that arguments are valid */
...

(open file, small read loop, close file, error handling)

/* Parse contents */
(call parse function, handle errors)

/* Run through SomeFilter if type is .xyz */
(call(s) to SomeFilter, error handling)

/* Cache to SomePlace if caching is turned on */
(call(s) to cache function(s), error handling)

/* Replace existing data structure atomically */
(some compare-and-swap code)

/* Handle common errors */
(end of a long try-catch block for instance)

This way a reader can scan the code much more quickly because there is no need to read any of the programming language while looking for something.

edit: use plenty of whitespace too.

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I know this doesn't help, but it seems like no one at my company believes in commenting, bc when I am at my company and I am trying to find bugs in code that someone else wrote I typically end up spending too much time trying to figure out what they're trying to acheive and how they did it.

that's my rant.
~guyaton

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Haegarr, which test framework are you using for C++? I'm having a hard time finding something easy to use, e.g. most frameworks require you to use a (test_)main() function which interferes with my WinMain() etc.

PS: Selm, hehe... only half an hour away!

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### Physics - Electronic Devices

#### Question - 1

The correct relation for the cohesive energy is

• A cohersive energy = free energy of atoms - crystal energy
• B cohersive energy = free energy of atoms + crystal energy
• C cohersive energy = $\frac { free\quad energy\quad of\quad atoms }{ crystal\quad energy }$
• D cohersive energy = free energy of atoms x crystal energy

#### Question - 2

In an unbaised p-n junction, holes diffuse from the p-region to n-region because

• A free electrons in the n-region attract them
• B they move across the junction by the potential difference
• C hole concentration in p-region is more as compared to n-region
• D all the above

#### Question - 3

A spectrum of silicon is to be made p-type semiconductor. For this one atom of indium, on an average, is doped in 5 x ${ 10 }^{ 7 }$silicon atoms. If the number density of silicon is 5 x ${ 10 }^{ 28 }$atoms/${ m }^{ 2 }$, then the number of acceptor atoms per ${ cm }^{ 3 }$will be

• A 2.5 x${ 10 }^{ 30 }$
• B 1.0 x ${ 10 }^{ 13 }$
• C 1.0 x ${ 10 }^{ 15 }$
• D 2.5 x ${ 10 }^{ 36 }$

#### Question - 4

In the ration of the concentration of electrons and of holes in a semiconductor is 7/5 and the ration of currents is 7/4, then what isthe ration oftheir drift velocities?

• A 4/7
• B 5/8
• C 4/5
• D 5/4

#### Question - 5

The conductivity of a semiconductor increases with increase in temperature because

• A number density of free current carriers increase
• B relaxation time increases
• C both number density of carriers and relaxation time increase
• D number density of current carriers increases, relaxation time decreases but effect of decerese in relaxation time is much less than increase in number density

#### Question - 6

The ratio of electron and hole currents in a semiconductor is $\frac { 5 }{ 4 }$ and the ration of drift velocities of electrons and holes is $\frac { 7 }{ 4 }$, then the ration of concentrations of electrons and holes will be

• A $\frac { 25 }{ 49 }$
• B $\frac { 49 }{ 25 }$
• C $\frac { 7 }{ 5 }$
• D $\frac { 5 }{ 7 }$

#### Question - 7

The resistivity of an n-type extrinsic semiconductor is 0.25 $\Omega$-m. If the electron mobility is 8.25 ${ m }^{ 2 }$/v-s, then the concentration of donor atoms will be (in ${ m }^{ -3 }$)

• A 3.0 x $10^{ 16 }$
• B 3.0 x $10^{ 17 }$
• C 3.0 x $10^{ 18 }$
• D 3.0 x $10^{ 19 }$

#### Question - 8

Assume that the number of hole- electron pairs in am intrinsic semiconductor is proportional to $e^{ -\triangle E/2kT }$. Here , $\triangle E$ = energy. gap and k = 8.62 x ${ 10 }^{ -5 }$eV/K. The ration of electron-hole pairs at 300 k and 400 k, is

• A ${ e }^{ -5.31 }$
• B ${ e }^{ +5 }$
• C e
• D ${ e }^{ +3 }$

#### Question - 9

THe contribution in the total current flowing through a semiconductor due to electrons and holes are $\frac { 3 }{ 4 }$ and $\frac { 1 }{ 4 }$. If the drit velocity of the electron is $\frac { 5 }{ 2 }$ times that of holes at this temperature, then the ratio of concentration of electrons and holes is

• A 6 : 5
• B 5 : 6
• C 3 : 2
• D 2 : 3

#### Question - 10

The number of densities of electrons and holes in pure silicon at ${ 27 }^{ 0 }$C are equal and its value is 1.5 x ${ 10 }^{ 16 }$${ m }^{ -3 }$. On doping with indium, the hole density increase to 4.5 x ${ 10 }^{ 27 }$${ m }^{ -3 }$. The electron density in doped silicon will be

• A 50 x ${ 10 }^{ 9 }$${ m }^{ -3 }$
• B 5 x ${ 10 }^{ 9 }$${ m }^{ -3 }$
• C ${ 10 }^{ 8 }$${ m }^{ -3 }$
• D ${ 10 }^{ 7 }$${ m }^{ -3 }$


text:
Easiest groups/lineages of machines to port software between?
05-10-2015, 09:17 PM
Post: #1
 Dave Britten Senior Member Posts: 1,928 Joined: Dec 2013
Easiest groups/lineages of machines to port software between?
What programmable calculators have the advantage of being able to run programs written for previous or lesser-equipped contemporaries with little or no software changes required?

For instance, you could forward port something from the 41C to a 41CX or 41CV, and probably a 42S if external hardware or the clock isn't needed, and I think 11C to 15C would be a relatively easy conversion as well.

What about the 67? Does that translate forward to anything easily? (Excepting the hardware assistance from the card reader for the 41C.)

Then there's the 32S family, with a rather different approach to labels and register naming, so there's usually a good amount of overhaul to bring something over, but going from 32S to 32Sii to 33S to 35S is relatively painless.

And in TI land, you can go from the 58C to the 59 with no changes, and also to the 66 if the program doesn't use any ROM modules, and you can adapt any printer usage (watch out for the CE behavior difference though). But the 66 doesn't have enough memory to handle large 59 programs, nor does it have a card reader.

What other porting family trees are there? Mostly I want to have an idea what I've got the best shot of running something on if I find an interesting program on the museum DVDs or elsewhere.
05-10-2015, 09:58 PM
Post: #2
 Paul Dale Senior Member Posts: 1,726 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
The 34C would go to the 15C in a straightforward manner. What about the 25 to the 33E or 33C?

The 34S was designed to just about run 42S (and hence 41C) programs as is. It isn't quite there but it is close. We changed the bitwise operations (for the better I think) which broke easy compatibility with the 16C. It should also run 34C, 11C programs as is and 15C programs that avoid some matrix operations.

- Pauli
05-10-2015, 10:13 PM (This post was last modified: 05-10-2015 10:15 PM by Csaba Tizedes.)
Post: #3
 Csaba Tizedes Senior Member Posts: 492 Joined: May 2014
RE: Easiest groups/lineages of machines to port software between?
(05-10-2015 09:17 PM)Dave Britten Wrote:  going from 32Sii to 35S is relatively painless.

I have a hydrostatics example where the program code totally same for the two calculator and the 32Sii can to SOLVE that program and after few seconds gives a result but 35s can't.
05-10-2015, 10:29 PM
Post: #4
 Dave Britten Senior Member Posts: 1,928 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-10-2015 10:13 PM)Csaba Tizedes Wrote:
(05-10-2015 09:17 PM)Dave Britten Wrote:  going from 32Sii to 35S is relatively painless.

I have a hydrostatics example where the program code totally same for the two calculator and the 32Sii can to SOLVE that program and after few seconds gives a result but 35s can't.

Now that's interesting, though I guess not totally surprising, given all the other, let's say, "quirks" of the 35S.

And I forgot that the 35S changed the indirection register setup, so there's a potential incompatibility.
05-13-2015, 01:00 PM
Post: #5
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-10-2015 10:13 PM)Csaba Tizedes Wrote:  I have a hydrostatics example where the program code totally same for the two calculator and the 32Sii can to SOLVE that program and after few seconds gives a result but 35s can't.

Then obviously the 35s program contains an error. There are some significant differences between these two calculators, so 32s code may not work on the 35s and 35s code may not work on the 32s. For instance the 35s supports direct line addressing while the 32s doesn't.

If you post the 35s program here I'm sure the error will be found.

(05-10-2015 10:29 PM)Dave Britten Wrote:  Now that's interesting, though I guess not totally surprising, given all the other, let's say, "quirks" of the 35S.

And I forgot that the 35S changed the indirection register setup, so there's a potential incompatibility.

The mentioned problem most probably is not due to a "quirk" in the 35s. Indeed there are significant differences that make both less than 100% compatible. Using indirection is just one example: the 32s uses positive indices to address the variables A–Z. Which is ok due to its limited memory. The 35s uses the same indices for its huge extended memory (800+ variables), while the variables A–Z are addressed by negative indices.

Dieter
05-13-2015, 01:47 PM
Post: #6
 TASP Senior Member Posts: 401 Joined: Mar 2015
RE: Easiest groups/lineages of machines to port software between?
For purposes of this thread would the 41/71 translator module for the 71 be considered cheating?

I have the module, but have never used it.

2speed HP41CX,int2XMEM+ZEN, HPIL+DEVEL, HPIL+X/IO, I/R, 82143, 82163, 82162 -25,35,45,55,65,67,70,80
05-13-2015, 01:55 PM
Post: #7
 Dave Britten Senior Member Posts: 1,928 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 01:00 PM)Dieter Wrote:  The mentioned problem most probably is not due to a "quirk" in the 35s. Indeed there are significant differences that make both less than 100% compatible. Using indirection is just one example: the 32s uses positive indices to address the variables A–Z. Which is ok due to its limited memory. The 35s uses the same indices for its huge extended memory (800+ variables), while the variables A–Z are addressed by negative indices.

I wouldn't be so sure without seeing the program. I was reading the 35S bug list yesterday, and it mentioned the silver going into an infinite loop with certain programs containing looping. This could be one of those cases. Would have to see the code to tell, of course.

(05-13-2015 01:47 PM)TASP Wrote:  For purposes of this thread would the 41/71 translator module for the 71 be considered cheating?

I have the module, but have never used it.

Yes, but still worth mentioning. Impressive that it could convert RPN keystroke programs to BASIC.
05-13-2015, 07:01 PM
Post: #8
 Jake Schwartz Senior Member Posts: 314 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 01:55 PM)Dave Britten Wrote:  Yes, but still worth mentioning. Impressive that it could convert RPN keystroke programs to BASIC.

I think you may have it backwards - HP41 programs were able to be run on the 71B but conversion to BASIC was not involved.

Jake
05-13-2015, 09:37 PM
Post: #9
 Dave Britten Senior Member Posts: 1,928 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 07:01 PM)Jake Schwartz Wrote:
(05-13-2015 01:55 PM)Dave Britten Wrote:  Yes, but still worth mentioning. Impressive that it could convert RPN keystroke programs to BASIC.

I think you may have it backwards - HP41 programs were able to be run on the 71B but conversion to BASIC was not involved.

Jake

Oh, so it's not like the 41C card reader that would convert 67 programs on the fly? Was it a bytecode interpreter, or would it convert them to actual 71 machine code that could be run independently?
05-14-2015, 05:00 AM (This post was last modified: 05-14-2015 05:01 AM by HrastProgrammer.)
Post: #10
 HrastProgrammer Member Posts: 151 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 09:37 PM)Dave Britten Wrote:  Was it a bytecode interpreter, or would it convert them to actual 71 machine code that could be run independently?

It's FORTH, actually. There is a separate HP-41 environment where you can execute most HP-41 functions/instructions + you also have FORTH here. It is not an emulator at CPU level - no synthetic programming etc., just the regular HP-41 code.

http://www.hrastprogrammer.com/hrastwerk/
http://hrastprogrammer.bandcamp.com/
05-29-2015, 09:14 AM
Post: #11
 Csaba Tizedes Senior Member Posts: 492 Joined: May 2014
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 01:00 PM)Dieter Wrote:  If you post the 35s program here I'm sure the error will be found.

Sorry for the long time delaying, now I have a little time to post this short article about that hydrostatics example - only for Hungarian, but I think it's not a big issue to translate the text to English.

This is an example of a centrifugal process of air. What is the required rotation speed (omega in rad/s) if we want to produce pressure at the (A) point 200'000 Pa. The temperature is constant.

In the paper you can found an analitycal solution, an Euler method and a Predictor-Corrector method also.

For SOLVING program I you need to set variable H, which is a distance step and according to your distance step selection the number of divisions in program line I06.

If you have any question, I'll try to answer

Try to use these programs on 32SII and on 35s. The 32SII will produce the right result and 35s can not.

Csaba
05-30-2015, 06:22 AM
Post: #12
 Csaba Tizedes Senior Member Posts: 492 Joined: May 2014
RE: Easiest groups/lineages of machines to port software between?
(05-13-2015 01:00 PM)Dieter Wrote:  Then obviously the 35s program contains an error. If you post the 35s program here I'm sure the error will be found.

So, Dieter, what is your opinion? (Don't afraid from Hungarian sentences, try to imagine you're in Los Alamos during WWII.)

As you can see, no direct GTO, no indirect addressing, only a simple Euler method with SOLVE.
I think it's a real bug in the SOLVE routine, therefore the real question is that what is the solution if I want to use my calc for these type problem solving? With this bug this calculator is useless when I try to solve more complex examples than "A+B=C".

Csaba
05-30-2015, 02:40 PM
Post: #13
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-10-2015 10:13 PM)Csaba Tizedes Wrote:  I have a hydrostatics example where the program code totally same for the two calculator and the 32Sii can to SOLVE that program and after few seconds gives a result but 35s can't.

(05-29-2015 09:14 AM)Csaba Tizedes Wrote:  Try to use these programs on 32SII and on 35s. The 32SII will produce the right result and 35s can not.

This can reproduced with a simpler program:
Code:
I001 LBL I I002 INPUT F I003 INPUT O I003 1.001 I004 STO I I005 0 G001 LBL G G002 RCL+ 0 G003 ISG I G004 GTO G001 G004 RCL- F G005 RTN

You can test the program with:

XEQ I ENTER
F? 140 R/S
0? 100 R/S
-40

To solve for O you can run:

FN= I
100 STO O
200 SOLVE 0
F? 140 R/S

After this the calculator just displays:

SOLVING

The solver returns immediately with the correct result when lines G003-G004 are removed:

O=140

Cheers
Thomas
05-30-2015, 03:29 PM (This post was last modified: 05-30-2015 03:30 PM by Thomas Klemm.)
Post: #14
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
Addendum: Easiest groups/lineages of machines to port software between?
This program is a little shorter:
Code:
F001 LBL F F002 1 F003 STO K F004 RCL A F005 RCL- X F006 DSE K F007 GTO F005 F008 RTN

To test the program you can use:

140 STO A
100 STO X
XEG F ENTER

This should return:

40

Use the solver to solve $$a-k\cdot x=0$$ for x:

FN= F
SOLVE X

The calculator hangs and just displays:

SOLVING

Cheers
Thomas
05-30-2015, 03:40 PM
Post: #15
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
From item 20 of the HP-35s bug list:
Quote:Solve infinite loops for some program based functions. Assign FN=D and SOLVE the following for X. The solve doesn't finish. This seems to be related to using a loop in the function being solved.

Code:
D001 LBL D D002 10 D003 STO N D004 DSE N D005 GTO D004 D006 RCL X D007 COS D008 RTN

A workaround for this is to include an always false conditional after any INPUT statements with a RTN instruction and a STOP. For example:

Code:
B001 LBL B B002 INPUT T B003 INPUT X B004 CF 1 B005 FS? 1 B006 RTN B007 FS? 1 B008 XEQ C001 B009 0.2xX-T B010 RTN C001 LBL C C002 STOP C003 RTN

Not sure how this workaround should be applied to the example above.

Cheers
Thomas
05-31-2015, 02:30 PM
Post: #16
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Easiest groups/lineages of machines to port software between?
(05-30-2015 06:22 AM)Csaba Tizedes Wrote:  So, Dieter, what is your opinion?

Your problem is caused by a bug in the 35s solver. It simply does not solve programs with loops. #-\

In this way the 35s solver is different from all other HP solvers I know of. I cannot imagine this behaviour was intended, so I would call it a bug

(05-30-2015 06:22 AM)Csaba Tizedes Wrote:  I think it's a real bug in the SOLVE routine, therefore the real question is that what is the solution if I want to use my calc for these type problem solving? With this bug this calculator is useless when I try to solve more complex examples than "A+B=C".

The solver itself is quite powerful. In equation mode it can even solve simple equations symbolically (if the desired variable appears only once). But indeed it has this limitation: no loops, no DSE, no ISG.

Dieter
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text:
(Analysis by Dhruv Rohatgi and Brian Dean )

Intuitively, a clockwise fence will "tend" to turn right, and a counterclockwise fence will tend to turn left. More concretely, for every two adjacent fence segments, one can compute the angle turned at that corner: either no turn (if the two segments are in the same direction), or a turn by $90^{\circ}$ counterclockwise (e.g. if the first segment is E and the second is N), or a turn by $-90^{\circ}$ counterclockwise (e.g. if the first segment is E and the second is S). Doing a few examples by hand will reveal that the sum of these angles, over all corners of the fence, is precisely $360^{\circ}$ if the fence is counterclockwise, and $-360^{\circ}$ if clockwise (other sums can be achieved if the fence is allowed to intersect itself, but we don't have to worry about this complication in this problem).

This fact can be proven a variety of ways, such as by inducting on the area of the region enclosed by the fence. It gives a linear-time algorithm, shown below.

Dhruv's code:

#include <iostream>
#include <string>
#include <cassert>
using namespace std;

int angle_from_direction(char a)
{
if(a == 'E') return 0;
if(a == 'N') return 90;
if(a == 'W') return 180;
if(a == 'S') return 270;
}

int angle_change(char a,char b)
{
int theta1 = angle_from_direction(a);
int theta2 = angle_from_direction(b);
if(theta2 == (theta1 + 90)%360) return 90;
else if(theta2 == theta1) return 0;
else if(theta2 == (theta1 + 270)%360) return -90;
else assert(false);	//fence should not backtrack on itself
}

void test(string s)
{
int total_change = 0;
for(int i=0;i<s.size();i++)
total_change += angle_change(s[i],s[(i+1)%s.size()]);
if(total_change == 360) cout << "CCW\n";
else cout << "CW\n";
}

int main()
{
int N;
string s;
cin >> N;
for(int i=0;i<N;i++)
{
cin >> s;
test(s);
}
}

There are several other ways to effectively approach this problem. For example, one can look at the direction of any "boundary" segment. That is, look any horizontally-oriented segment that is as far north as possible --- if it is directed east, the entire path must be clockwise, and if directed west, the entire path must be counterclockwise. In fact, we can draw any horizontal or vertical line through the scene and deduce from the direction of the segments along this cross section the overall orientation of the fence. For example, if we draw a horizontal line through the scene, it will be cut by some vertical segments pointing north and some pointing south (in fact, these will alternate in direction along the cross section); if the westmost segment points north, the entire path has a clockwise orientation, and vice versa.

The shoelace formula for computing areas of polygons also leads to a solution for this problem, since it delivers a "signed" area, being positive or negative depending on the orientation of the polygon in question.


text:
# Why CAPS LOCK in AMS mode in LyX?

I am trying to use the AMS article in LyX. I want my output to look like the following (note margins are the same for the entire paragraph and not all letters are capitalized):

Here is my source code:

\begin{doublespace}

\section{Introduction. AbCdEfGEEEEEEEEEEEEEEE EE EE EE}\end{doublespace}

However, here is what I have:

and here is the PDF:

The first 6 letters I typed were AbCdEf, yet they are all capitalized in my document. How can I edit this so that not all letters are capitalized (for each section and subsection)? I also don't want all the letters in the word "Introduction" to be capitalized, but LyX does this once I click "Section."

This is the default for the AMS article document class - sections are set in small caps. These defaults are set in what is called a layout (and stored in a .layout file). Layouts are available in the in the Resources/layouts subfolder of your LyX folder, if you want to take a gander. Specifically, the AMS article layout is available in amsart.layout.

If you wish to change parts of a layout, you can do so by means of a local layout. Follow Document > Settings... > Local Layout and add the following:

Format 49
Style Section
Align        Center
Font
Series     Medium
Shape      Default
Size       Normal
EndFont
TocLevel 1
End

Style Section*
Align        Center
Font
Series     Medium
Shape      Default
Size       Large
EndFont
End

The above updates the Section and Section* styles to use a Default shape (formerly SmallCaps). Note that this will only change the display within LyX, and not the PDF output generated via pdfLaTeX. For this you also need to change the way the \section and \section* titles are handled under the AMS class. Therefore, additionally add the following to your Document > Settings... > LaTeX preamble:

\makeatletter
\renewcommand{\section}{%
\@startsection{section}
{1}
\z@
{.7\linespacing \@plus \linespacing}
{.5\linespacing}
{\normalfont\centering}% \scshape
}
\makeatother

The above preamble addition will change the font for both \section and \section* to be void of \scshape on the LaTeX side.

LyX input:

LaTeX output:

• I edited the preamble. When I enter your code into the local layout, it says "layout is invalid." – The Substitute Nov 8 '15 at 6:47
• @TheSubstitute: What version of LyX are you using? Perhaps try without the line "Format 49". – Werner Nov 8 '15 at 6:49
• That worked! I am using 2.0. Now let me reread your answer and see what happens. Thanks in advance! – The Substitute Nov 8 '15 at 6:50
• @TheSubstitute: LyX is currently on version 2.1.4. I'd suggest an upgrade, just because. – Werner Nov 8 '15 at 6:54


text:
# Probability of Equally Likely Outcome

## Interactive practice questions

List all possible outcomes when a coin is flipped.

A

red

B

tails

C

black

D

A

red

B

tails

C

black

D
Easy
Less than a minute
A card is selected from a standard deck of $52$52 cards (no Jokers included). How many colour outcomes are there in the sample space?
$\left\{1,3,3,6,6,6,8,8,8,8,10,10,10,10,10\right\}${1,3,3,6,6,6,8,8,8,8,10,10,10,10,10}