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# Voltage Drop

Voltage drop is calculated in accordance with CENELEC technical report CLC/TR 50480 "Determination of cross-sectional area of conductors and selection of protective devices", dated February 2011.

The voltage drop (as a percentage) is given by:

$ΔU= b( Rcosφ+Xsinφ ) I b U o ⋅100$

where
ΔU         = cable voltage drop, %
Uo             = nominal  line to neutral voltage, V
R           = cable resistance, Ω
X           = cable reactance, Ω
Ib           = cable design current, A
b            = circuit factor (=2 for d.c. and single-phase, =1 for three-phase)

Note: within myCableEngineering, we use complex arithmetic and the above is evaluated as:

$ΔU= b( R+jX ) I b U o ⋅100$

Note: R and X are per line conductor.  For example, the resistance of a single-phase two core circuit (live and neutral) would be 2R (assuming the live and neutral circuits are of equal resistance).  For parallel cables, R and X are the per line resultant values. For example, nph cables in parallel, with the resistance of one line conductor is Rc0ph then the R = Rc0ph/nph.

The above equation), has the relative voltage drop expressed as a percentage of Uo.  Multiplying by Uo will give the actual voltage drop for d.c. and single-phase circuits and for three-phase circuits, this needs to be multiplied by √3.

Adjusting the CENELEC equation to take into account referring the voltage drop to line-line voltage for three-phase systems, setting R, X in ohms, and using complex forms gives:

ΔU, V/m ΔU, % per m
d.c.  systems
a.c.  systems, single-phase
a.c systems, three-phase

-  the resistance of a single conductor, Ω/m
- the reaction of a single conductor,  Ω/m
- the impedance of single conductor , Ω/m
- cable design current, A
- the nominal line to neutral/earth voltage (for single-phase a.c. or d.c.), V
- the nominal line to line voltage (for three-phase systems), V

### BS 7671  Voltage Drop Tables

BS7671 "Requirements for Electrical Installations", the IET Wiring Regulations, Appendix 4 voltage drop tables, values are given in mV/A/m (or effectively mΩ).  However, these tables are related to the line to neutral voltage for single-phase circuits and line to line voltage for three-phase circuits.
To convert the three-phase table values so that they relate to the input resistance (or reactance) required by CENELEC 50480 it is necessary to divide the BS 7671 three-phase values by √3 (the square root of three).  No adjustment is required for single phase values.  Given values of voltage drop for three-phase balanced systems are related to the line voltage.
Note: the values in the tables are given at the maximum conductor operating temperature.  myCableEngineering uses circuit theory (see Impedance) to calculate impedance and voltage drops are calculated using CENELEC 50480.  Our testing shows that the difference in resistance calculation by circuit theory and that given in BS 7671 is small.


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## anonymous one year ago check my work ??

1. anonymous

nah

2. anonymous

$\frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0$ Let $y(x)=e^{rx}$$\implies r^2e^{rx}+2re^{rx}+e^{rx}=0$$e^{rx}(r^2+2r+1)=0$ On solving the polynomial, $r=-1$ therefore, $y(x)=e^{-x}$

3. anonymous

if $y(x)=e^{rx}$ then $\frac{dy}{dx}=re^{rx}$ and $\frac{d^2y}{dx^2}=e^{rx}+r^2e^{rx}$ so we get: $r^2e^{rx}+2re^{rx}+2e^{rx}=0$ the polynomial you will get is slightly different: $r^2+2r+2=0$

4. anonymous

why is that ??

5. anonymous

oh yeah ur right!!! let me try again

6. anonymous

noo wait the polynomial is correct

7. anonymous

I think your $\frac{d^2y}{dx^2}$ is wrong

8. IrishBoy123

careful you have a **repeated root** !!!

9. anonymous

Oh sorry I forgot the constant $y(x)=Ce^{-x}$

10. anonymous

for the second derivative we need to use the chain rule, or derivative of a product.

11. IrishBoy123
12. anonymous

yeah but r is constant so it's derivative would be 0 and u will only get 1 term

13. anonymous

ooppss!! haha! you are right! r is constant, my bad there!! sorry... take a look at the link @IrishBoy123 posted

14. anonymous

Yeh im checking it

15. anonymous

There is 1 more solution ??

16. IrishBoy123

the general solution for repeated roots is $$y = (Ax + B ) e^{ax}$$

17. anonymous

ok I read the article now so...the solution should be I think $y(x)=C_{1}e^{-x}+C_{2}xe^{-x}$

18. IrishBoy123

i don't know how you were shown how to to these, but with a differential operator D = d/dx, you take (D-a)(D-a)y = 0 and let u = (D-a)y, so (D-a)u = 0, giving $$u = A e^{ax}$$ from there $$( D-a)y = A e^{ax},\ \frac{dy}{dx} - ay =A e^{ax}$$, solve with integrating factor

19. IrishBoy123

yes, Ax + B appears in general solution for repeated roots

20. anonymous

I only knew one method u take the polynomial and with the roots u do $y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}$

21. IrishBoy123

ok, so now also remember that, with repeated roots, that rule changes :p

22. anonymous

kk

format: plain


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# Use discriminant for linear and quad. function.

1. May 21, 2016

### Evangeline101

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Back again :) Is this correct?

Thanks.

#### Attached Files:

File size:
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Views:
72
2. May 21, 2016

### Mastermind01

Again you are correct.

3. May 21, 2016

### Staff: Mentor

Yes. Only that I have $(-1.09 ; 1.73)$ and $(2.76 ; 13.27)$. But the path is correct.
It's always better to have exact numbers until the end and calculate the roots in a final step.

4. May 21, 2016

### Ray Vickson

In thread after thread you ask "is this correct?...". That is a bad habit: as a matter of standard operating procedure you should check your answers for yourself (as you would need to do in an exam setting). Then, if you find something does not "work", you will know you have made an error somewhere, and asking for help then makes sense.

5. May 22, 2016

### Mastermind01

I agree , I have already given the OP a way to check her work in a related similar thread.

6. May 22, 2016

### Evangeline101

I understand, but can you suggest a less time consuming method than graphing, if I am doing an exam I won't have time to graph anything to check my answer. For this specific question, is there a way I can put the points back into the equation to verify, well maybe in this case I would have to graph since they are points of intersection.

Also I should be more clear when asking my question ( I will make sure to do that in future questions), in this case I would like to know if my process for part b is correct, does it make sense to use the linear equation to solve for y-values, and if so, is it because the linear equation is set equal to y?

7. May 22, 2016

### Ray Vickson

To save time, don't graph anything; just check if your (x,y) values satisfy the requirement that they lie on both the line and the curve. That is easy: just plug in the numbers and calculate.

I have not checked your calculations for part (b), but I see that you have used the correct formulas. Again, you can check the solutions for yourself, by plugging in the numerical values and doing the calculations to see if everything works as it should.

Again: this is all just standard practice that you should get in the habit of doing.

8. May 22, 2016

### Staff: Mentor

Sure. You can substitute the points you found into each equation. If the point is on the graph, the equation should be a true statement.
In your work above, you found one solution to be $x = \frac{5 + \sqrt{133}}{6}$. The corresponding y value is $y = \frac{15 + \sqrt{133}}{2}$ If you substitute this x value into the right sides of the equations $y = 3x^2 - 2x - 4$ and $y = 3x + 5$ you should get the y value I show above. Note that the values I show are the exact values. If you approximate the x values using a calculator, the y values you get will be only as precise as the rounding you do.


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Select Board & Class

# Board Paper of Class 10 2014 Maths - Solutions

(1) Check the question paper for fairness of printing. If there is any lack of fairness, inform the Hall Supervisor immediately.(2) Use Blue or Black ink to write and underline and pencil to draw diagrams.Note : This question paper contains four sections.
• Question 1
If {(7, 11), (5, a)} represents a constant function, then the value of 'a' is:

(a) 7

(b) 11

(c) 5

(d) 9 VIEW SOLUTION
• Question 2
If a, b, c are in A.P. then $\frac{a-b}{b-c}$ is equal to :

(a) $\frac{a}{b}$

(b) $\frac{b}{c}$

(c) $\frac{a}{c}$

(d) 1 VIEW SOLUTION
• Question 3
If the nth term of an A.P. is tn = 3 – 5n, then the sum of the first n terms is:

(a) $\frac{n}{2}\left[1-5n\right]$

(b) n(1 – 5n)

(c) $\frac{n}{2}\left(1+5n\right)$

(d) $\frac{n}{2}\left(1+n\right)$
VIEW SOLUTION
• Question 4
If ${x}^{2}+5kx+16=0$ has no real roots, then :

(a) $k>\frac{8}{5}$

(b) $k>-\frac{8}{5}$

(c) $-\frac{8}{5}

(d) $0 VIEW SOLUTION
• Question 5
The system of equations x – 4y = 8, 3x – 12y = 24

(a) has infinitely many solutions

(b) has no solution

(c) has a unique solution

(d) may or may not have a solution VIEW SOLUTION
• Question 6
If A is of order 3 × 4 and B is of order 4 × 3 then the order of BA is :

(a) 3 × 3

(b) 4 × 4

(c) 4 × 3

(d) not defined VIEW SOLUTION
• Question 7
The x and y intercepts of the line 2x – 3y + 6 = 0, respectively are :

(a) 2, 3

(b) 3, 2

(c) –3, 2

(d) 3, –2 VIEW SOLUTION
• Question 8
Area of the triangle formed by the points (0, 0), (2, 0) and (0, 2) is

(a) 1 sq. unit

(b) 2 sq. units

(c) 4 sq. units

(d) 8 sq. units VIEW SOLUTION
• Question 9
In the figure, if ∠PAB = 120° then ∠BPT =

(a) 120°

(b) 30°

(c) 40°

(d) 60° VIEW SOLUTION
• Question 10
The perimeter of two similar triangles ΔABC and ΔDEF are 36 cm and 24 cm respectively. If DE = 10 cm then AB is :

(a) 12 cm

(b) 20 cm

(c) 15 cm

(d) 18 cm VIEW SOLUTION
• Question 11
sin (90° – θ) cos θ + cos (90° – θ) sin θ =

(a) 1

(b) 0

(c) 2

(d) –1 VIEW SOLUTION
• Question 12
9 tan2 θ – 9 sec2θ =

(a) 1

(b) 0

(c) 9

(d) –9 VIEW SOLUTION
• Question 13
Curved surface area of solid sphere is 24 cm2. If the sphere is divided into two hemispheres, then the total surface area of one of the hemispheres is:

(a) 12 cm2

(b) 8 cm2

(c) 16 cm2

(d) 18 cm2 VIEW SOLUTION
• Question 14
For any collection of n items, $\mathrm{\Sigma }\left(x-\overline{)x}\right)$ =

(a) Σx

(b) $\overline{)x}$

(c) $n\overline{)x}$

(d) 0 VIEW SOLUTION
• Question 15
Probability of sure event is :

(a) 1

(b) 0

(c) 100

(d) 0.1 VIEW SOLUTION
• Question 16
If A = {4, 6, 7, 8, 9}, B = {2, 4, 6} and C = {1, 2, 3, 4, 5, 6} then find A⋃(B ⋂ C) VIEW SOLUTION
• Question 17
Let X = {1, 2, 3, 4}. Examine whether the relation given below is a function from X to X or not. Explain.
f = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)} VIEW SOLUTION
• Question 18
Find the 17th term if the A.P. : 4, 9, 14 .......... VIEW SOLUTION
• Question 20
Form the quadratic equation whose roots are . VIEW SOLUTION
• Question 21
If then verify AI = IA = A, where I is the unit matrix of order 2. VIEW SOLUTION
• Question 22
The coordinates of the midpoint of the line segment joining the points (2a + 2,3) and (4, 2b + 1) are (2a, 2b). Find the values of a and b. VIEW SOLUTION
• Question 23
AB and CD are two chords of a circle which intersect each other internally at P. If CP = 4 cm, AP = 8 cm, PB = 2 cm, then find PD. VIEW SOLUTION
• Question 25
Find the angular elevation (angle of elevation from the ground level) of the sun when the length of the shadow of a 30 m long pole is $10\sqrt{3}$ m. VIEW SOLUTION
• Question 26
A solid right circular cylinder has radius of 14 cm and height of 8 cm. Find its total surface area. VIEW SOLUTION
• Question 27
How many litres of water will a hemispherical tank hold whose diameter is 4.2 m? VIEW SOLUTION
• Question 28
Find the standard deviation of the first 10 natural number. VIEW SOLUTION
• Question 29
Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice. VIEW SOLUTION
• Question 30
(a) Construct a 2 × 3 matrix A = [aij] whose elements are given by aij = |2i – 3j|

OR

(b) Find  the equations of the straight lines parallel to the coordinate axes and passing through the point (3, –4). VIEW SOLUTION
• Question 31
Use Venn diagrams to verify De Morgan's law for set difference A\(B ⋂ C) = (A\B) ⋃ (A\C). VIEW SOLUTION
• Question 32
A function f : [1, 6) → R is defined as follows:

Find the value of
(a) f(5)
(b) f(3)
(c) f(1)
(d) f(2) – f(4)
(e) 2f(5) – 3f(1) VIEW SOLUTION
• Question 33
Find the total area of 14 squares whose sides are 11 cm, 12 cm, 13 cm, .............., 24 cm. VIEW SOLUTION
• Question 34
The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and returns downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream. VIEW SOLUTION
• Question 35
Simplify : $\frac{{a}^{2}-16}{{a}^{3}-8}×\frac{2{a}^{2}-3a-2}{2{a}^{2}+9a+4}÷\frac{3{a}^{2}-11a-4}{{a}^{2}+2a+4}$ VIEW SOLUTION
• Question 37
Find the area of the quadrilateral formed by the points (–4, –2), (–3, –5), (3, –2) and (2, 3). VIEW SOLUTION
• Question 38
Find the equation of the perpendicular bisector of the straight line segment joining the points (3, 4) and (–1, 2). VIEW SOLUTION
• Question 39
State and prove Basic Proportionality theorem. VIEW SOLUTION
• Question 40
A person in an helicopter flying at a height of 500 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. $\left(\sqrt{3}=1.732\right)$ VIEW SOLUTION
• Question 41
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4 π cm. If the depth is 14 cm, then find its volume. VIEW SOLUTION
• Question 42
Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere. VIEW SOLUTION
• Question 43
Calculate the standard deviation of the following data:
 x 3 8 13 18 23 f 7 10 15 10 8
VIEW SOLUTION
• Question 44
Two unbiased dice are rolled once. Find the probability of getting :
(a) a sum 8
(b) a doublet
(c) a sum greater than 8 VIEW SOLUTION
• Question 45
(a) If S1, S2 and S3 are the sum of first n, 2n and 3n terms of a geometric series respectively, then prove that S1(S3 – S2) = (S2 – S1)2.

OR

(b) If α and β are the roots of the equation 3x2 – 4x + 1 = 0 form a quadratic equation whose roots are . VIEW SOLUTION
• Question 46
(a) Draw a circle of radius 3 cm. From an external point 7 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
OR

(b) Construct a cyclic quadrilateral ABCD with AB = 7 cm, ∠A = 80°, AD = 4.5 cm and BC = 5 cm. VIEW SOLUTION
• Question 47
(a) Draw the graph of y = 2x2 and hence solve 2x2 + x – 6 = 0.

OR

(b) The cost of the milk per litre is Rs 15. Draw the graph for the relation between the quantity and cost. Hence find:
(i) the proportionality constant
(ii) the cost of 3 litres of milk.
VIEW SOLUTION
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Syllabus


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# chainer.functions.dilated_convolution_2d¶

chainer.functions.dilated_convolution_2d(x, W, b=None, stride=1, pad=0, dilate=1, cover_all=False)[source]

Two-dimensional dilated convolution function.

This is an implementation of two-dimensional dilated convolution in ConvNets. It takes three variables: the input image x, the filter weight W, and the bias vector b.

Notation: here is a notation for dimensionalities.

• $$n$$ is the batch size.
• $$c_I$$ and $$c_O$$ are the number of the input and output, respectively.
• $$h$$ and $$w$$ are the height and width of the input image, respectively.
• $$k_H$$ and $$k_W$$ are the height and width of the filters, respectively.
Parameters: x (Variable) – Input variable of shape $$(n, c_I, h, w)$$. W (Variable) – Weight variable of shape $$(c_O, c_I, k_H, k_W)$$. b (Variable) – Bias variable of length $$c_O$$ (optional). stride (int or pair of ints) – Stride of filter applications. stride=s and stride=(s, s) are equivalent. pad (int or pair of ints) – Spatial padding width for input arrays. pad=p and pad=(p, p) are equivalent. dilate (int or pair of ints) – Dilation factor of filter applications. dilate=d and dilate=(d, d) are equivalent. cover_all (bool) – If True, all spatial locations are convoluted into some output pixels. It may make the output size larger. Output variable. Variable

The two-dimensional dilated convolution function is defined as follows. Then the DilatedConvolution2D function computes correlations between filters and patches of size $$(k_H, k_W)$$ in x. Patches here are extracted at intervals of the dilation factor. Note that correlation here is equivalent to the inner product between expanded vectors. Patches are extracted at intervals of the dilation factor and at positions shifted by multiples of stride from the first position -pad for each spatial axis. The right-most (or bottom-most) patches do not run over the padded spatial size.

Let $$(s_Y, s_X)$$ be the stride of filter application, $$(p_H, p_W)$$ the spatial padding size, and $$(d_Y, d_X)$$ the dilation factor of filter application. Then, the output size $$(h_O, w_O)$$ is determined by the following equations:

$\begin{split}h_O &= (h + 2p_H - k_H - (k_H - 1) * (d_Y - 1)) / s_Y + 1,\\ w_O &= (w + 2p_W - k_W - (k_W - 1) * (d_X - 1)) / s_X + 1.\end{split}$

If the bias vector is given, then it is added to all spatial locations of the output of convolution.


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# Condition for Commensurability of Roots of Quadratic Equation

## Theorem

$(1): \quad a x - x^2 = \dfrac {b^2} 4$

Then $x$ and $a - x$ are commensurable if and only if $\sqrt{a^2 - b^2}$ and $a$ are commensurable.

In the words of Euclid:

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.

And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length.

### Lemma

In the words of Euclid:

If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.

## Proof

We have that:

 $\ds x \paren {a - x} + \paren {\frac a 2 - x}^2$ $=$ $\ds a x - x^2 + \frac {a^2} 4 - 2 \frac a 2 x + x^2$ $\ds$ $=$ $\ds \frac {a^2} 4$ simplifying $\ds \leadsto \ \$ $\ds 4 x \paren {a - x} + 4 \paren {\frac a 2 - x}^2$ $=$ $\ds a^2$ $\ds \leadsto \ \$ $\ds b^2 + \paren {a - 2 x}^2$ $=$ $\ds a^2$ from $(1)$ $\ds \leadsto \ \$ $\ds a^2 - b^2$ $=$ $\ds \paren {a - 2 x}^2$ $\ds \leadsto \ \$ $\ds \sqrt {a^2 - b^2}$ $=$ $\ds a - 2 x$

Let $a \frown b$ denote that $a$ is commensurable with $b$.

### Necessary Condition

Let $\paren {a - x} \frown x$.

$a \frown x$
$x \frown 2 x$

So:

 $\ds a$ $\frown$ $\ds 2 x$ Commensurability is Transitive Relation $\ds$ $\frown$ $\ds \paren {a - 2 x}$ Commensurability of Sum of Commensurable Magnitudes $\ds$ $\frown$ $\ds \sqrt {a^2 - b^2}$

$\Box$

### Sufficient Condition

Let $a \frown \sqrt {a^2 - b^2}$.

Then:

 $\ds a$ $\frown$ $\ds \sqrt {a^2 - b^2}$ $\ds$ $\frown$ $\ds \paren {a - 2 x}$ $\ds$ $\frown$ $\ds 2 x$ Commensurability of Sum of Commensurable Magnitudes
$x \frown 2 x$
$a \frown x$
$\paren {a - x} \frown x$

$\blacksquare$

## Historical Note

This proof is Proposition $17$ of Book $\text{X}$ of Euclid's The Elements.


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# Page 2 Quantitative Reasoning Study Guide for the ISEE

#### Types of Numbers

Numbers can be classified as real or imaginary. Real numbers include whole, rational (numbers that can be written as fractions or ratios) or irrational (such as pi and most roots) numbers.

Familiarity with various types of numbers will facilitate answering questions such as this example:

“Which of the following choices correctly sorts these numbers from the least to the greatest value?”

$-\frac{3}{5}$, $0.5$, $-\frac{6}{9}$, $\pi$, $\sqrt{5}$, $3^{-1}$, $2 \;x \;10^{-4}$

#### Ratio and Proportion

Ratios are used to compare values or to represent a scale or proportion. When comparing two or more values, part-to-part ratios or part-to-whole ratios can be used. Most ratio questions can be solved with logic and very minimal computation, such as in this example:

The ratio of boys to girls in a class of 50 students is 2:3. The students need to create 10 groups where each group must have an equal number of boys as there are girls. How many girls will not be included in the groupings?

The ratio of boys to girls to total is 2 : 3 : 5
From this ratio and the given total, we know that there are 20 boys : 30 girls : 50 total
To create 10 groups with equal number of boys and girls 20 boys : 20 girls : 40 total
Obviously, there will be a surplus of 10 girls who can’t be grouped as specified.

When we say that a ratio is in proportion (or is proportional) to another, we mean that the two ratios are equal. The corresponding sides of similar triangles, for example, are proportional. Knowing the dimensions of a triangle will, therefore, help solve the dimensions of a triangle similar to it. This is the same principle used for scaling on maps.

#### Fractions

Part-to-whole ratios can also be written as fractions. In the previous example, the ratio of the boys for every 5 students is “2 : 5”. This can also be stated as “$\frac{2}{5}$ of the class of 50 students is made up of boys”. The numerator 2 represents the part, while the denominator 5 represents the whole.

The fraction $\frac{2}{5}$ is a proper fraction because the numerator is less than the denominator (and the value of the fraction is less than 1).

Learn, as well, about improper, mixed, and equivalent fractions.

It is also important to know how to determine the greatest common factor (GCF); this is used when simplifying fractions.

Learn how to find the least common multiple (LCM); this is useful when comparing fractions and when performing addition and subtraction of fractions.

#### Decimals

The concept of decimals is an extension of the subject of fractions. The fraction $\frac{1}{10}$ is the same as the decimal 0.10. Make sure you can convert decimals to fractions and percentage with ease. Without a calculator, some questions that include decimals can be more easily solved using fractions. Multiplying $\frac{1}{4}$ by $\frac{3}{4}$ is far simpler and quicker than multiplying 0.25 by 0.75 manually.

Your math skills must also include the ability to add, subtract, multiply, divide, and raise decimals to powers.

#### Percents

Percent means per hundred, so 1% means “1 per hundred” or “$\frac{1}{100}$” or “0.01”. It could also refer to a scale or ratio of “1:100”. You must develop the skill of translating from percent to ratios, fractions and decimals, depending on the context in which they are given in the question.

#### Complex Numbers

Numbers that are not real are called imaginary numbers, and are indicated by the notation i. The imaginary unit is equal to the square root of -1, or:

When imaginary numbers combine with real numbers, they are called complex numbers. Try this question:

Which of the following choices gives the correct value of $3i^{26}$?

Before you start with the tedious repeated multiplication (26 times!), let’s get acquainted with this unique property of i: When i is multiplied by itself (or raised to an exponent), it follows this repetitive 4-point cycle:

Then $i^5 = i$, which starts another cycle which ends with $i^8 = 1$.

The seemingly difficult sample question above, then, becomes an easy one with this solution:

Take note that when we raise i to a multiple of 4, the value will always be equal to 1.

#### Vectors and Matrices

Vectors are values with magnitude and direction. Vectors are written in a row or column. This is a vector:

(3, 4)

And here is another:

(7, 3)

The two vectors can be added to get the resultant vector below:

(10, 7)

Without the visuals, though, we ought to know now that we add vectors by adding the corresponding elements in the row or column:

(3, 4) + (7, 3) = (10, 7)

Matrices, on the other hand, are an array of values in rows and columns. There can be one or more rows, and one or more columns. So a vector is also a matrix. The determinant is a number computed from a square matrix, such as a 2x2 or a 3x3 matrix.

Vectors, matrices, and determinants are useful in solving systems of equations. A system of linear equations, for example, can be solved using matrices, instead of the elimination or substitution methods.

These concepts also find application in calculus, physics and astronomy. At this level, you are expected to know how to perform the basic operations using vectors and matrices, and to calculate the determinant.

Review further about topics such as adding and subtracting matrices, multiplying a matrix by a scalar, multiplying a matrix by another matrix, dividing (or multiplying by an inverse of a matrix), transposing a matrix, and calculating the determinant of a matrix.

#### Number Theory

The concepts mentioned involving natural numbers and integers are part of the study of numbers, called the Number Theory. Explore the properties of odd and even numbers, primes and composites, and prime factorization. You will find these number properties useful when reasoning out mathematically. Other types of numbers, albeit rarely seen in this test, are perfect, triangular and Fibonacci numbers.


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# zbMATH — the first resource for mathematics

The invariance of Poincaré’s generating function for canonical transformations. (English) Zbl 0235.70008

##### MSC:
 70H15 Canonical and symplectic transformations for problems in Hamiltonian and Lagrangian mechanics 70F15 Celestial mechanics
Full Text:
##### References:
 [1] Abraham, R., Marsden, J.: Foundations of mechanics. New York: W. A. Benjamin 1967. [2] Meyer, K. R.: Generic bifurcation of periodic points. Trans. Amer. Math. Soc.149, 95–107 (1970). · Zbl 0198.42902 [3] Poincaré, H.: Les méthodes nouvelles de la mécanique céleste, vol. 3. Paris: Gauthier-Villars 1892. [4] Tougeron, J.-Cl.: Idéaux de fonctions differentiables I. Ann. Inst. Fourier, Grenoble18, 177–240 (1968). · Zbl 0188.45102 [5] Wall, C. T. C.: Lectures onC stability and classification. In: Proceedings of Liverpool Singularities-Symposium. I, pp. 178–206. Berlin-Heidelberg-New York: Springer 1971. [6] Weinstein, A.: Perturbation of periodic manifolds of hamiltonian systems. Bull. Amer. Math. Soc.77, 814–818 (1971). · Zbl 0218.58006 [7] Weinstein, A.: Periodic orbits of hamiltonian systems via critical point theory (to appear). [8] Weinstein, A.: Symplectic structures on Banach manifolds. Bull. Amer. Math. Soc.75, 1040–1041 (1969). · Zbl 0179.50104 [9] Weinstein, A.: Symplectic manifolds and their lagrangian submanifolds. Advances in Math.3, 329–346 (1971). · Zbl 0213.48203
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
## Decode SQLite database blobs: how to start?

2

Updated: see bottom of this question.

Not sure if I am at the right channel here but I have good hopes someone here might be able to help me.

I am trying to process and analyse data from a Footscan system that is exported from the Footscan 9 Gait Essentials software to a .rsdb database. This database is a standard SQLite database with a different file extension.

I have no problems accessing the data via Python but the most interesting data (the raw data from the sensor) is stored as a 'blob' (see sqlite documentation for more details).

Exporting this data from SQLitePro to .csv shows that the data in the blob looks like this (although much longer):

<9c4e79cb cbd09de5 c380fc78 a74e819f ... ab2172c7 d9cf311a 357aebac>

My question is: what would be my first steps to try to discover how this data is encoded? And does anyone maybe recognize this type of encoding?

Extra hints: The data represents all forces of a force plate with 64x64 force sensors, sampled at 300Hz, so it's most probably a multidimensional array.

Update

I am making some progress. As commented below, the values are most probably hexadecimal values, but I'm not sure if the values are base 16: most values seem to be >10^9...

Furthermore: I found a table in the database called 'Contact' which linkes to entries in the blob table which contain the most hexadecimal values (>40000). My guess is that these are the entries which are recognized as actual steps. I extracted the data from one of these steps here (Contact id 2; blob id 24). This blob contains 50792 hexadecimal values.

This is the entry for this step in 'Contacts':

id: {35d9a73e-608d-4ba3-9823-de0540f11c93}
timestamp: 2
deleted: NULL
FrameWidth: 64
Frames: 24
FrameOffset: 12204
FrameCount: 214
FrameHeight: 64
FrameStoreWithContacts_Contacs: {5d10546e-5ef3-4469-b898-d7b9bd58458f}
OriginalId: {32ae0f73-8022-4151-92e7-510c0e86d09d}
orphan: false

Full export of the blob table can be downloaded here. The full .rsdb can be found here (you might have to rename to .sqlite). The tables 'Foot' and 'Region' in the full database might contain useful data too, but I'm not yet sure how...

2Do you mean the encoding beyond the hexadecimal system of encoding bytes, or have you never seen hexadecimal before? – Spacedman – 2016-02-13T16:15:36.193

I have seen hexadecimal before but I am not sure if I understand your question completely. Can you elaborate? – Aart Goossens – 2016-02-14T11:15:13.780

1Convert the data to binary. If the data's not private, post it and I can show you what I mean. The first few bytes you provided don't suggest a file type, but emacs thinks it might be Japanese/JIS text. – None – 2016-02-14T15:49:07.197

1Its not clear from your Q whether you understand that "9c4e79cb" means anything to you (ie you understand its hexadecimal). So there's (at least) two "encodings" going on - first the representation of the data as hex, and then however the force plate measurements are arranged into that data. It could just be 64x64xN bytes. There's not really enough context in the Q to figure it out... – Spacedman – 2016-02-14T23:22:52.130

1I think its a bit tricky ... blobs can be of any structure inside. It helps that you expect a certain array of data (64x64) sampled at 300Hz, though how long is the data stored for? If you know how many seconds it is stored, and you know the blob size, you can then estimate the individual point resolution. – Marcus D – 2016-02-15T13:22:40.260

Thanks for all you comments so far! I understand that some information is lacking: I will post more data and some more details I found about the data tomorrow. – Aart Goossens – 2016-02-15T20:44:50.553

I updated my question with extra information, actual data and new insights. Any help is appreciated. I will keep you updated on my progress. I might also post my progress as a separate answer to not pollute the question too much. – Aart Goossens – 2016-02-16T20:44:49.300

The blobs are assorted lengths which aren't multiples of 64. The blob data looks random, with no obvious pattern or repeat or structure. That makes me think the blob data is compressed or encoded in some unknown manner, which makes sense since there's likely to be a lot of redundancy in the original measurements. Its possibly a raw chunk of gzipped data, but any of a number of compression algorithms might have been used. Have you asked the vendor? – Spacedman – 2016-02-17T12:59:05.813

Hi @Spacedman thanks for your reply. I emailed the vendor but didn't get a response yet. I am still hoping that they just 'cropped' the data to the sensor area where the foot actually was. In the table 'Foot' there are columns AreaWidth, AreaHeight and FrameCount but these values multiplied give >4 times the expected number of data points. – Aart Goossens – 2016-02-17T16:57:12.357


text:
Question

What are strong acids and weak acids?a.) An acid which dissociates completely or almost completely in water is called a strong acid whereas an acid which only partially does so is referred to as a weak acid.b.) An acid which dissociates completely is called a strong acid whereas an acid which dissociates almost completely in water is referred to as a weak acid.c.) An acid which does not dissociate completely in water is referred to as a strong acid whereas an acid which dissociates almost completely in water is called a weak acid.d.) An acid that dissociates only partially in water is referred to as a strong acid whereas an acid that does not completely dissociate in water is called a weak acid.

Hint: Examples of strong acids include: HCl, ${{H}_{2}}S{{O}_{4}}$ etc.
Examples of weak acids include: HCOOH, $C{{H}_{3}}COOH$ etc.
Think of what distinct chemical property differentiates these two classes of acids.

Let us discuss both these terms: strong acids and weak acids; one at a time.
A strong acid is one which is virtually 100% ionised in solution and splits into aqueous versions of its constituent cation and anion.
Some common strong acids include sulphuric acid, hydrochloric acid and nitric acid.

Here is the equation for the ionisation of Hydrochloric Acid (HCl) written in a simplified form:
$HC{{l}_{(aq)}}\to {{H}^{+}}_{(aq)}+C{{l}^{-}}_{(aq)}$

This shows the hydrogen chloride dissolved in the water splitting to give hydrogen ions in solution and chloride ions in solution.
Now let us discuss what a weak acid really is.
By contrast, a weak acid is one which doesn't ionise fully when it is dissolved in water.
Ethanoic acid is a typical weak acid. It reacts with water to produce hydronium ions and ethanoate ions, but the back reaction is more successful than the forward one. The ions react very easily to reform the acid and the water.
$C{{H}_{3}}COOH+{{H}_{2}}O \longrightarrow C{{H}_{3}}CO{{O}^{-}}+H_3{{O}^{+}}$
At any one time, only about 1% of the ethanoic acid molecules have converted into ions. The rest remain as simple ethanoic acid molecules.
Therefore, from the above analysis and explanation we can conclude that the answer to this question is a).

NOTE: Please be as clear as possible with the concept of ionisation in water of different compounds as the knowledge of this concept is essential for you to be able to solve this question easily.


text:
How to fix \useshorthands \defineshorthand usage on bookmark entries?

\documentclass[10pt,a5paper]{abntex2}
\usepackage[brazil]{babel}
\usepackage{hyphenat}

\useshorthands{~}\defineshorthand{~=}{\hyp{}}

\begin{document}

\chapter{First~=Chapter}

\end{document}

Instead of ~= to become - (hyphen) on my bookmarks entries, it become = (a space and an equal sign):

hyperref doesn't let you do that, apparently. A bit hackish, but try this (this assumes that whatever ~ in a section title will become a valid pdf string):

\documentclass[10pt,a5paper]{abntex2}
\usepackage[brazil]{babel}
\usepackage{hyphenat}

\makeatletter
\def\HyPsd@LetUnexpandableSpace#1{%
\if\noexpand~\noexpand#1%
\else
\expandafter\@firstofone
\fi{\expandafter\futurelet\expandafter#1\expandafter\@gobble\space\relax}}
\makeatother

\useshorthands{~}
\defineshorthand{~=}{\hyp{}}

\begin{document}

\chapter{First~=Chapter}

\end{document}


text:
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 Izv. RAN. Ser. Mat.: Year: Volume: Issue: Page: Find

 Izv. RAN. Ser. Mat., 2001, Volume 65, Issue 4, Pages 49–66 (Mi izv347)

A'Campo–Gusein-Zade diagrams as partially ordered sets

G. G. Ilyuta

Independent University of Moscow

Abstract: The real analogues of many results about complex monodromies of singularities can be formulated and proved in terms of partial orderings on A'Campo–Gusein-Zade diagrams, the real versions of Coxeter–Dynkin diagrams of singularities. In this paper it is proved that the only diagrams among the A'Campo–Gusein-Zade diagrams of singularities that determine partially ordered sets of finite type (in the sense of representations of a quiver) are the diagrams of simple singularities. To encode the real decompositions of a singularity the analogue of Vasilev invariants turn out to be surjections of a partially ordered set onto a chain. Formulae are proved for Arnold $(\operatorname{mod}2)$-invariants of plane curves in terms of the corresponding A'Campo–Gusein-Zade diagrams. We define, in the context of higher Bruhat orders, higher partially ordered sets and we describe their connection with the higher $M$-Morsifications $A_n$. We also consider certain previously known results about real singularities from the point of view of partially ordered sets.

DOI: https://doi.org/10.4213/im347

Full text: PDF file (1902 kB)
References: PDF file   HTML file

English version:
Izvestiya: Mathematics, 2001, 65:4, 687–704

Bibliographic databases:

MSC: 06A99

Citation: G. G. Ilyuta, “A'Campo–Gusein-Zade diagrams as partially ordered sets”, Izv. RAN. Ser. Mat., 65:4 (2001), 49–66; Izv. Math., 65:4 (2001), 687–704

Citation in format AMSBIB
\Bibitem{Ily01} \by G.~G.~Ilyuta \paper A'Campo--Gusein-Zade diagrams as partially ordered sets \jour Izv. RAN. Ser. Mat. \yr 2001 \vol 65 \issue 4 \pages 49--66 \mathnet{http://mi.mathnet.ru/izv347} \crossref{https://doi.org/10.4213/im347} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1857710} \zmath{https://zbmath.org/?q=an:1026.58031} \elib{https://elibrary.ru/item.asp?id=13672138} \transl \jour Izv. Math. \yr 2001 \vol 65 \issue 4 \pages 687--704 \crossref{https://doi.org/10.1070/IM2001v065n04ABEH000347} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-33746837383}

• http://mi.mathnet.ru/eng/izv347
• https://doi.org/10.4213/im347
• http://mi.mathnet.ru/eng/izv/v65/i4/p49

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This publication is cited in the following articles:
1. G. G. Ilyuta, “Interpolation by symmetric functions and alternating higher Bruhat orders”, Izv. Math., 67:5 (2003), 849–880
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It just depends on what you want to study, my understanding is that it doesn't make a difference. I had applied to MAcc but after the interview we changed it to MBT and it wasn't an issue at all. I wanted to concentrate in tax so if that is the case you should apply for MBT. The best way to increase your chances is to apply early. I had a 2.8 GPA with a 700 GMAT and was accepted to the USC MBT w/ some \$ and I had applied back in December.
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text:
# How do I load a model in SARAH ?

After the installation, the package is loaded in Mathematica via

<<"SARAH.m"

if it was put into the applications directory of Mathematica or by using the entire path:

<<"[$PATH]/SARAH.m" A model is initialized by Start["Modelname"]; Here, Modelname is the name of the corresponding model file, e.g. for the minimal supersymmetric standard model the command would read Start["MSSM"]; or for the next-to-minimal supersymmetric standard model in CKM basis Start["NMSSM","CKM"]; is used. In both cases, after about 1 minute the following message should appear All Done.$MODEL is ready!

# How do I get a spectrum generator based on SPheno for a new model?

To obtain the SPheno output run in Mathematica after SARAH is loaded and the model is initialized the command

MakeSPheno[];

When executing MakeSPheno, SARAH calculates first all information which it needs, i.e. it is not necessary that the user has done the calculation of vertices or RGEs before. When SARAH is done, the source code for SPheno is stored in $SPATH/SARAH/Output/$MODEL/EWSB/SPheno/. The compilation of this codes is done as follows: one has to enter the directory of the SPheno installation, a new sub-directory has to be created and the code must copied into this directory:

$cd$PATH/SPHENO
$mkdir$MODEL
$cp$PATH/SARAH/Output/$MODEL/EWSB/SPheno/*$MODEL/

Afterwards, the code is compiled via

$make Model=$MODEL

and a new executable SPheno$MODEL is available. An input file, by default called LesHouches.in.$MODEL, is needed to run SPheno$MODEL. SARAH writes a template for that file which has been copied to the $MODEL subdirectory of SPheno together with the Fortrancode. You can move it to the root directory of SPheno

$cp \$MODEL/LesHouches.in.\$MODEL .  By doing this we can work now from the SPheno main directory and we don’t have to give the file as argument when running SPheno. Thus, SPheno can be called via $ ./bin/SPheno\$MODEL Alternatively, one can keep the Les Houches file in the $MODEL and work with it via

$./bin/SPheno\$MODEL \$MODEL/LesHouches.in.\$MODEL

Of course, before you can run SPheno you must first fill the template written by SARAH with numbers.

# What are the features of a SPheno version generated like this?

A spectrum generator based on SARAH /SPheno provides the same precision as dedicataed MSSM codes do for a huge variety of models. The NMSSM version provides even a more precise Higgs mass calculation than other public NMSM codes do. The main features are

• Full two-loop running of all parameters and all masses at one-loop
• Complete one-loop thresholds at $M_Z$ from BSM to gauge and Yukawa couplings
• Two-loop corrections to Higgs masses
• calculation of flavour and precision observables at full one-loop
• calculation of decay widths and branching ratios
• interface to HiggsBounds and HiggsSignals
• estimate of electroweak Fine-Tuning

# How can I check points with HiggsBounds and HiggsSignals ?

In the same directory in which the SPheno spectrum file is located, also all other input files for HiggsBounds and HiggsSignals are saved by SPheno. The (relative) path to this directory has to be given as last argument to HiggsBounds when executing it. Thus, working from the directory $PATH, HiggsBounds is started via: $ ./HiggsBounds/HiggsBounds LandH effC #neutral #charged SPHENO/

From other directories, one can use absolute paths:

PATH/HiggsBounds/HiggsBounds LandH effC #neutral #charged $PATH/SPHENO/ $\#neutral$ and $\#charged$ are the number of (physical) neutral and charged Higgs bosons in the model. HiggsSignals is the complement to HiggsBounds and checks how good a point reproduces the Higgs mass and rate measurements. The syntax is very similar to HiggsBounds : $ ./HiggsSignals/HiggsSignals latestresults peak 2 effC #neutral #charged SPHENO/

# How do I implement a model in MadGraph and link it to SPheno

A new model in MadGraph can be implemente via the UFO format which is also supported by other tools like Herwig++ or Sherpa. The command to generate the UFO files is

MakeUFO[]

All files written by SARAH have to be copied to a new sub-directory in MadGraph ’s model directory:

cd $PATH/MADGRAPH/ mkdir models/$MODEL
cp $PATH/SARAH/Output/$MODEL/EWSB/UFO/* models/$MODEL The model can be loaded in MadGraph via import model$MODEL -modelname

The option -modelname is used to keep the names of the particles as given in the model files. This prevents conflicts with internal MadGraph conventions.

# How do I implement a model in WHIZARD/O’Mega and link it to SPheno

The model files for WHIZARD/O’Mega are obtained by

MakeWHIZARD[]

After the interface has completed, the generated files are stored in the directory

$PATH/SARAH/Output/$MODEL/EWSB/WHIZARD_Omega/

In order to use the model with WHIZARD and O’Mega, the generated code must be compiled and installed. In most cases this is done by

$cd$PATH/SARAH/Output/$MODEL/EWSB/WHIZARD_Omega$ ./configure --prefix=$PATH/WHIZARD/ WO_CONFIG=$PATH/WHIZARD/bin/
$make$ make install

If WHIZARD has not been installed globally in the home directory of the current user, WHIZARD won’t be able to find the binaries. Thus, the WO_CONFIG environment variable was used to point explicitly to the binaries. By default, the configure script would install the compiled model into .whizard in the home directory of the user. If the user wants to have several WHIZARD installations or install WHIZARD locally, it might be better to provide a model just for one installation. For these cases the installation path has been defined via the --prefix option of the configure script. More information on the available options is shown with the command

./configure  --help

To link WHIZARD and SPheno, all SPheno modules created by SARAH write the information about the parameters and masses into an additional file. This file is written in the WHIZARD specific format and can be directly read by WHIZARD. One just has to make sure that the corresponding flag is turned on the Les Houches input for SPheno to get this output:

[style=file,numbers=none,title=\hspace{12cm}LesHouches.in.\$MODEL] Block SPhenoInput # SPheno specific input ... 75 1 # Write WHIZARD files  The parameter file can then be included in the Sindarin input file for WHIZARD via model =$model_sarah
...
include("$PATH/SPHENO/WHIZARD.par.$MODEL")

# How can I calculate the relic density with MicrOmegas

The calculation of the cross section and all necessary decay widths are done by CalcHep which comes together with MicrOmegas. Thus, a new model in MicrOmegas is implemented by providing the corresponding CalcHep model files. That means, one can use the SARAH output for CalcHep to work with MicrOmegas :

MakeCHep[];

When SARAH is finished with MakeCHep, the CalcHep model files are located in the directory

$PATH/SARAH/Output/$MODEL/EWSB/CHep/

To implement the model in MicrOmegas, a new project has to be created and the files have to be copied in the working directory of this project:

$cd$PATH/MICROMEGAS
$./newProject MODEL$ cd $MODEL$ cp $PATH/SARAH/Output/$MODEL/EWSB/CHep/*  work/models

Steering files written by SARAH were copied together with all model files into the working directory of the current project. You can move them to the main project directory and compile it

$mv work/models/CalcOmega_with_DDetection.cpp .$ make main=CalcOmega_with_DDetection.cpp

A new binary CalcOmega_with_DDetection is now available. The only missing piece are the input parameters.
Providing the numerical parameters is pretty easy because MicrOmegas /CalcHep can read the SPheno spectrum file. However, the user must make sure that no complex rotation matrices show up in the spectrum file: in the case of Majorana matrices and no CP violation, there are two equivalent outputs: (i) all Majorana masses are positive, but some entries of the corresponding rotation matrices are complex; (ii) all mixing matrices are real, but some masses are negative. CalcHep can just handle the second case with real matrices. Hence, one has to use the flag

[style=file,numbers=none,title=\hspace{12cm}LesHouches.in.BLSSM]
Block SPhenoInput   # SPheno specific input
...
50 0               # Majorana phases: use only positive masses

to get the spectrum according to that convention. Afterwards, the spectrum file just has to be moved to the same directory as CalcOmega_with_DDetection. Copy it there and start the calculation:

$cp$PATH/SPHENO/SPheno.spc.$MODEL .$ ./CalcOmega_with_DDetection

# How do I implement new model in Vevacious ?

MakeVevacious[];


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# Is x+y equal to 3.6 ?

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Is x+y equal to 3.6 ?  [#permalink]

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10 Aug 2013, 12:02
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Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

it's a refresher. Will upload solution soon

_________________

Asif vai.....

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Re: Is x+y equal to 3.6 ?  [#permalink]

### Show Tags

10 Aug 2013, 12:45
Asifpirlo wrote:
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

it's a refresher. Will upload solution soon

(1) $$x > 1.7 or x = 1.7$$
INSUFFICIENT as we dont know y

(2) $$y > 1.9$$
dont know x HENCE insufficient.

combining
least value of $$x=1.7$$least value of $$y$$ will be $$>$$ than $$1.9$$
therefore $$x+y> 3.6$$
hence sufficient

hence C
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Re: Is x+y equal to 3.6 ?  [#permalink]

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11 Aug 2013, 12:28
blueseas wrote:
Asifpirlo wrote:
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

it's a refresher. Will upload solution soon

(1) $$x > 1.7 or x = 1.7$$
INSUFFICIENT as we dont know y

(2) $$y > 1.9$$
dont know x HENCE insufficient.

combining
least value of $$x=1.7$$least value of $$y$$ will be $$>$$ than $$1.9$$
therefore $$x+y> 3.6$$
hence sufficient

hence C

cool and simple work by blueseas again....................
great
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Re: Is x+y equal to 3.6 ?  [#permalink]

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27 Jan 2015, 12:28
If y exceedes 1.9.....then our x should be 1.7 or less for 36....not possible with confirmation ---- C
Intern
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Re: Is x+y equal to 3.6 ?  [#permalink]

### Show Tags

09 Oct 2015, 22:22
blueseas wrote:
Asifpirlo wrote:
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

it's a refresher. Will upload solution soon

(1) $$x > 1.7 or x = 1.7$$
INSUFFICIENT as we dont know y

(2) $$y > 1.9$$
dont know x HENCE insufficient.

combining
least value of $$x=1.7$$least value of $$y$$ will be $$>$$ than $$1.9$$
therefore $$x+y> 3.6$$
hence sufficient

hence C

Correct me if I am wrong but

if x>1.7 and
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

If x can be greater than 1.7 then x can be 2.0
if y is greater than 1.9 then y can be 2.0 too

x+y can equal 4
The question is asking if x+y is equal to 3.6, in this case it is not, it is greater than 3.6
Math Expert
Joined: 02 Sep 2009
Posts: 49320
Re: Is x+y equal to 3.6 ?  [#permalink]

### Show Tags

11 Oct 2015, 05:32
hesham08 wrote:
blueseas wrote:
Asifpirlo wrote:
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

it's a refresher. Will upload solution soon

(1) $$x > 1.7 or x = 1.7$$
INSUFFICIENT as we dont know y

(2) $$y > 1.9$$
dont know x HENCE insufficient.

combining
least value of $$x=1.7$$least value of $$y$$ will be $$>$$ than $$1.9$$
therefore $$x+y> 3.6$$
hence sufficient

hence C

Correct me if I am wrong but

if x>1.7 and
Is x+y equal to 3.6 ?

(1) x > 1.7 or x = 1.7
(2) y > 1.9

If x can be greater than 1.7 then x can be 2.0
if y is greater than 1.9 then y can be 2.0 too

x+y can equal 4
The question is asking if x+y is equal to 3.6, in this case it is not, it is greater than 3.6

It's not clear what you are trying to say here. When we combine the statements we can say that x + y is NOT equal to 3.6, so the answer to the question is a definite NO, thus sufficient and answer is C.
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Re: Is x+y equal to 3.6 ?  [#permalink]

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11 Oct 2015, 05:47
Statement 1) No info on Y...Insufficient
Statement 2) No info on X...Insufficient

Both the statement shows x+y>=3.6. Ans: C
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Is x + y equal to 3.6 ?  [#permalink]

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10 Mar 2017, 01:48
Is x + y equal to 3.6 ?

(1) x ≥ 1.7

(2) y > 1.9
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Re: Is x + y equal to 3.6 ?  [#permalink]

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10 Mar 2017, 03:34
Top Contributor
STAT1: x ≥ 1.7 Not sufficient as we do know the value of y to find x+y

STAT2: y > 1.9 Not sufficient as we do know the value of x to find x+y

Both together, x ≥ 1.7 and y > 1.9 => Min x is 1.7 and min y is greater than 1.9 so x+y > 1.7 + 1.9 => x + y > 3.6
So, x+y is NOT equal to 3.6. So, SUFFICIENT
Answer will be C
Hope it helps!

vikasp99 wrote:
Is x + y equal to 3.6 ?

(1) x ≥ 1.7

(2) y > 1.9

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Re: Is x+y equal to 3.6 ?  [#permalink]

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10 Mar 2017, 03:34
vikasp99 wrote:
Is x + y equal to 3.6 ?

(1) x ≥ 1.7

(2) y > 1.9

Please search before posting.
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Re: Is x+y equal to 3.6 ?  [#permalink]

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10 Mar 2017, 09:58
(1) x≥ 1.7.
No information about y. Hence, Insufficient.

(2) y>1.9
No information about x. Hence, Insufficient.

(1) & (2) together,
x≥ 1.7 & y>1.9
so, x + y > 1.7+ 1.9
=>x + y > 3.6. Sufficient.

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Re: Is x+y equal to 3.6 ? &nbs [#permalink] 10 Mar 2017, 09:58
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# Theoretical vs Experimental flow rate of fluid coming out of a water bottle with a hole in it

Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).

According to Bernoulli's law the velocity $v$ of the water flowing out is equal to $\sqrt{2gh}$

Therefore for the setup above, $v=\sqrt{2*9.81\ m/s^2*0.078\ m} = 1.24\ m/s$

Using this, the flow rate can be calculated as $Q\ =\ Av\ =\ π(0.002\ m)^2*1.24\ m/s = 0.000016\ m^3/s = 16\ mL/s$

This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?)

I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.

Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.)

Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?

• should be v=√(2*9.81*0.078). 7.8 cm is 0.078 m not 0.78 m – pentane Feb 17 '18 at 3:18
• Fixed and edited accordingly, narrows down the difference in values a lot. – cjeccjec Feb 17 '18 at 4:23
• The velocity based on the Torricelli law is that of water that reached atmospheric pressure, which is not true for the water near the orifice, but may be true for water further downstream, farther from the orifice. To calculate flow rate, one could multiply this Torricelli velocity by cross-section area of the jet $A'$ there, but it is not easy to determine this $A'$. In practice I would expect $A'$ to be smaller than cross-section of the orifice. Search for vena contracta and coefficient of contraction/reduction. – Ján Lalinský Feb 17 '18 at 23:53

You have two issues at hand. The first is that Bernoulli's law gives the instantaneous flow rate: as water leaves the bottle the height of the water column above the hole also changes. So you compared the flow rate at $t=0$ to the average flow rate over a 5 second interval. The more accurate comparison would be to calculate the volume of water lost after 5 seconds and compare that to your measured 40mL loss. To do that you need to solve the differential equation:

$$\frac{dV}{dt} = A_1 \sqrt{\frac{2g(V_0-V)}{A_2}}$$

Where $V$ is the amount of volume the water bottle has lost, $V_0$ is the original amount of volume above the hole, $A_1$ is the area of the hole, and $A_2$ is the cross-sectional area of the bottle. Separating the differentials and integrating both sides: $$\int_0^V (V_0-V)^{-1/2} dV = \int_0^t A_1 \sqrt{\frac{2g}{A_2}}t dt$$ $$\sqrt{V_0} - \sqrt{V_0-V} = \left( A_1 \sqrt{\frac{g}{2A_2}}\right) t$$ I had to estimate $A_2$ from your information provided and I assume it is close to 32 cm^2. In that case, $V_0$ = 251.6mL, t = 5s, and solving for $V$ = 72mL, with an average rate of 14 mL/s. Still not much of an improvement in correctly predicting 8 mL/s, which brings me to my second point:

Think of Bernoulli's equation as the best case scenario, analogous to free-fall without air resistance. You get further from this idealization the more:

I think item #3 is the largest factor from you realizing the best case scenario. You might try injecting a few drops of food coloring into the water and see if you see turbulence around the hole. Fixing #3 is all about the geometry of your container, so it could be massaged into steady flow by avoiding sharp edges near the fluid flow, etc.

Poiseuille's law applies to the pressure drop for a fluid traveling down a long straight pipe (like fluid flowing in a medical catheter) . I don't believe it will apply to any of your setup here.

• I understand the right hand side of the differential equation, however I don't understand the left hand side of it. How did you end up with 72mL? – cjeccjec Feb 17 '18 at 13:38
• @cjeccjec I edited my response to give the steps I used to solve the DE. $dV/dt$ is the instantaneous flow rate in units of $m^3 / s$. I used separation of variables to solve the differential equation. – cms Feb 17 '18 at 15:59

Given $v\ =\ \sqrt{2gh}\$ and that the change is volume over time $dV/dt$ is given by the surface area of the leaking hole $a$ times the speed of water through the hole $v$ (negative sign to show that it is decreasing): $$\frac{dV}{dt}\ =\ -av\ =\ -a \sqrt{2gh}$$ since the volume $V$ at any given height $h$ assuming the cross section of the tank is a constant $A$ is given by $V\ =\ Ah$, if we take the derivative of both sides with respect to time we arrive at: $$\frac{dV}{dt}\ =\ A \frac{dh}{dt}\$$ Combining this with the previous equation gives: $$-a \sqrt{2gh} = A \frac{dh}{dt}$$ $$\frac{dh}{\sqrt{h}} = -\frac{a \sqrt{2g}}{A} dt$$ Integrating both sides: $$\int_{h_0}^{h_f}\frac{dh}{\sqrt{h}} = -\frac{a \sqrt{2g}}{A}\int_{0}^{t} dt$$

$$(2h^{1/2}) \Big|_{h_0}^{h_f}\ =\ -\frac{a \sqrt{2g}}{A}t$$ To make it easier to look at, define $k\ =\ \frac{a \sqrt{2g}}{A}$ then: $$(2h^{1/2}) \Big|_{h_0}^{h_f}\ =\ -kt$$ Expanding and rearranging to isolate $h_f$: $$h_f\ =\ \left(-\frac{k}{2}t+\sqrt{h_0} \right)^2$$

We know that the change in volume after time t is: $\Delta V\ =\ A\Delta h\ =\ A\left[ h_f\ -\ h_0 \right]\ =\ A\left[\left(-\frac{k}{2}t+\sqrt{h_0} \right)^2 - h_0 \right]$

where
$A$ = $0.003226\ m^2$
$k$ = $0.0172\ m^{1/2}s^{-1}$
$t$ = $5\ s$
$h_0$ = $0.078\ m$

Plugging everything in results in $\Delta V = -71.5\ mL$


text:
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# Systems of equations with graphing: chores

Videotranscript
Use graphing to solve the following problem. Abby and Ben did household chores last weekend. Together they earned $50, and Abby earned$10 more than Ben. How much did they each earn? So let's define some variables here. Let's let A equal Abby's earnings. And let's let B equal Ben's earnings. Then they tell us how these earnings relate. They first tell us that together they earned $50. So that statement can be converted mathematically into-- well, together, that means the sum of the two earnings. So A plus B needs to be equal to$50. Abby's plus Ben's earnings is $50. And then they tell us Abby earned$10 more than Ben. So we could translate that into Abby's earnings is equal to Ben's earnings plus 10. Abby earned $10 more than Ben. So we have a system of two equations and actually with two unknowns. And then they say, how much did each earn? So to do that, and they want us to solve this graphically. There's multiple ways to solve it, but we'll do what they ask us to do. Let me draw some axes over here. And I'll be in the first quadrant since we're dealing with earnings, so neither of their earnings can be negative. And let me just define the vertical axis as Abby's axis or the Abby's earnings axis. And let me define the horizontal axis as Ben's axis or Ben's earnings axis. And let me just graph each of these equations. And to do that, I'm going to take this first equation, and I'm going to put it in the equivalent of slope-intercept form. It might look a little unfamiliar to you, but it really is slope-intercept form. Let me rewrite it first. So we have A plus B is equal to 50. We can subtract B from both sides. So let's subtract B from both sides. And then we get A is equal to negative B plus 50. So if you think about it this way, when B is equal to 0, A is going to be 50. So we know our A intercept, we could call it. We normally would call that a y-intercept, but now this is the A axis. So this right here, let me call this 10, 20, 30, 40, and 50. So if Ben made$0, then Abby would have to make $50 based on that first constraint. So we know that that's a point on the line right over there. And we also know that the slope is negative 1, that B is the independent variable, the way I've written it over here, and this coefficient is negative 1. Or another way to think about it is if A is 0, then B is going to be 50. If Abby made no money, then Ben would have to make$50. And that falls purely out of this equation right here. If Abby made nothing, then Ben would have to make $50, so 10, 20, 30, 40, 50. So those are those two situations and every point in between will satisfy this first constraint. So let me connect the dots. So it would look something like that. That's due to this first constraint, due to the fact that together they earned$50. Now, let's think about the second one. Abby earned $10 more than Ben. So that's this equation right here. And it's really already in our slope-intercept form. If Ben made$0, then Abby would make $10. So that's our A intercept. So it's right over there. We could keep doing that. Our slope is going to be 1 here. If Ben makes$10, then Abby's going to make $20. If Ben makes$20, Abby's going to make $30. We could keep going, but I think this gives us the general direction. It already hints at a point of intersection. So just eyeballing it, so we've graphed to the two constraints, together they earned$50. That's the magenta constraint right over here. Abby earned $10 more than Ben. That's this green constraint right over here. And it looks like we have a point of intersection. And it looks like we have a point of intersection at Ben earning$20. Let me label this as 10, 20, 30, 40, and 50. So this is Ben earning $20 and Abby earning-- this is 10, 20, and 30. And Abby earning$30. So just eyeballing it off of this, it looks like A is 30 and B is 20. And let's go verify, make sure that these levels of earnings for Abby and Ben actually satisfy both constraints. So the first constraint is that Abby plus Ben have to make $50. Well,$30 plus $20 is$50. So it meets our first constraint. The second constraint is that Abby earned $10 more than Ben, that Abby is equal to Ben plus 10. Well, once again, over here, Abby is making$10 more than Ben. So it meets our second constraint. And we only have two of them. So it meets both of them. So that's our solution. Abby earned $30. Ben earned$20.


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# Axons: Physics and chemistry at work in the body

## Problem

In a nerve cell, ion channels open and close based on their environment and other factors. When a channel is open, it behaves as an Ohmic resistor (V=IR). When it is closed, an insulator. The currents through the ion channels are determined by the electrical potential and ion concentration differences between the cytoplasm inside the cell and the surrounding interstitial fluid. If only sodium and potassium ions were present, the “reversal potential,” V, start subscript, m, end subscript, would be predicted by the Goldman-Hodgkin-Katz equation, (Equation 1).
V, start subscript, m, end subscript, equals, 59, start text, space, m, V, end text, dot, l, o, g, left parenthesis, start fraction, P, start subscript, N, a, start superscript, plus, end superscript, end subscript, open bracket, N, a, start superscript, plus, end superscript, close bracket, start subscript, o, u, t, end subscript, plus, P, start subscript, K, start superscript, plus, end superscript, end subscript, open bracket, K, start superscript, plus, end superscript, close bracket, start subscript, o, u, t, end subscript, divided by, P, start subscript, N, a, plus, end subscript, open bracket, N, a, start superscript, plus, end superscript, close bracket, start subscript, i, n, end subscript, plus, P, start subscript, K, start superscript, plus, end superscript, end subscript, open bracket, K, start superscript, plus, end superscript, close bracket, start subscript, i, n, end subscript, end fraction, right parenthesis
Equation 1. Reversal potential
The reverse potential is the potential that, if it were applied to the cell, would result in potassium and sodium crossing the membrane at rates that yield no net current. Each permeability, P, start subscript, X, end subscript, is proportional to the number ion X channels open at that node.
Table 1 shows the typical concentrations of sodium and potassium ions inside and outside of the cell.
Table 1. Typical ion concentrations, [mM]
inside celloutside cell
[Nastart superscript, start text, plus, end text, end superscript]15150
[Kstart superscript, start text, plus, end text, end superscript]1505
As the ions enter and exit the cytoplasm, they change the electric field within the axon, creating a new electrical potential. The cytoplasm conducts a current down its length in response to that electrical potential, much like a copper wire.
If we treat the interstitial fluid and cytoplasm as two separate wires and the channels as resistors in an equivalent circuit diagram, would adding more of these transmembrane “resistors” increase or decrease the total equivalent resistance at the node?


text:
## anonymous one year ago Which complex number has a distance of √17 from the origin on the complex plane? A. 2 + 15i B.17 + i C. 20 - 3i D. 4 - i

• This Question is Open
1. anonymous

r=$\sqrt{a ^{2+}}b ^{2}$ Where a is the real part and b is imaginary (doesn't matter for this one actually)

2. anonymous

wait what @muscrat123

3. anonymous

@horsegirl27

4. anonymous

? @muscrat123


text:
# AMAT using Local miss rate, global miss rate, local hit rate, global hit rate

Consider the following scenario as shown in image:

I have summarized the above Memory layout in terms of Miss rates and hit rates:

.---.----------------.------------.------------.------------.--------.
|   |                |     0      |     1      |     2      |   3    |
:---+----------------+------------+------------+------------+--------:
|   |                | LMR        | GMR        | LHR        | GHR    |
:---+----------------+------------+------------+------------+--------:
| 0 | L1 cache       | 10/100     | 10/100     | 90/100     | 90/100 |
:---+----------------+------------+------------+------------+--------:
| 1 | L2 cache       | 3/10       | 3/100      | 7/10       | 7/100  |
:---+----------------+------------+------------+------------+--------:
| 2 | Main Memory    | 0/3        | 0/100      | 3/3        | 3/100  |
:---+----------------+------------+------------+------------+--------:
| 3 | Sum of columns | Not useful | Not useful | Not useful | 1      |
'---'----------------'------------'------------'------------'--------'

Here LMR - local miss rate, GMR - global miss rate, LHR - local hit rate, GHR - global hit rate. From the above table I deduced the following insights:

1. Sum of the Global Hit rates (GHR) of all caches and Main Memory is 1.
2. The formula for AMAT(Average Memory Access Time) can be interpreted as follows:

AMAT = t1 + m1t2 + m1m2t3

(or)

AMAT = t1 + M1t2 + M2t3

. where m1,m2 are local miss rates of Level 1 and Level 2 caches respectively. M1,M2 are global miss rates of Level 1 and Level 2 caches respectively. t1, t2, t3 are access times for level 1 cache, level 2 cache and Main Memory. For the given example,

AMAT = t1 + 10/100t2 + 3/100t3

1. From point 2 it can be said that

M1 = m1; M2 = m1*m2; and so on

1. Local Hit rate of Main Memory (or final level cache) is always 1. Here it is assumed that required word is surely found in Main Memory ( or final level cache).
• Local Hit rate of level 1 cache = Global Hit rate of level 1 cache.

• Local Miss rate of level 1 cache = Global Miss rate of level 1 cache.

Now my question is "Are these points correct?". I did not find any information relevant to the above points. Also are there any more insights I can get from the table? Like relation between GHR and lmr, lhr.


text:
# LRFTubes example: side resonator insertion loss¶

This example shows the computation of the insertion loss by adding a side resonator to a pipe. The situation is as follows:

The impedance at the rightmost side is equal to $z_0 / S$, such that no reflections come back from that side. This way, the insertion loss can be computed by the ratio of the acoustic pressure on the right side by the incoming wave on the left side. As we would expect, the insertion loss peaks at the quarter wave resonance frequency of the side pipe.

In [1]:
# Import necessary stuff
%pylab inline
#%pylab
from lrftubes import Air, System, PrsDuct

Populating the interactive namespace from numpy and matplotlib

In [2]:
# Define the geometry
r = 0.15
r1 = r # Radius of duct 1
r2 = r # Radius of duct 2
r3 = 0.4*r # Smaller radius for the side resonator
S1 = np.pi*r1**2
S2 = np.pi*r2**2
S3 = np.pi*r3**2

L1 = 1.
L2 = 1.
L3 = 0.6

d1 = PrsDuct(L=L1,S=S1,cs='circ')
d2 = PrsDuct(L=L2,S=S2,cs='circ')
d3 = PrsDuct(L=L3,S=S3,cs='circ')

mat = Air()

# Create the system, add the segments and boundary conditions below
sys = System(mat)

sys.addSeg(0,1,d1) # Duct 1 gets nodes 0 and 1. Node 1 is the branch node of the side resonator
sys.addSeg(1,2,d2) # Duct 2 connects to node 1 and has free node 2
sys.addSeg(3,1,d3) # Duct 3 is connected with its right side to the branch

z0 = mat.z0 # Characteristic impedance (lossless), rayls
sys.addBc('Z',2,z0/S2) # Set a characteristic impedance on the acoustic volume flow at node 3

freq = np.linspace(10,1000,500) # Create frequency array

sol = sys.solve(freq) # Solve the system

In [3]:
# Compute the incoming wave at the left side
figure(figsize=(12,6))
A = 0.5*(sol[d1]['pL'] + z0*sol[d1]['UL']/S1)
abspR_ov_A = np.abs(sol[d2]['pR']/A)
IL = -20*np.log10(abspR_ov_A)
semilogx(freq, IL)
title('Insertion loss of the side resonator as a function of the frequency',size=18)
xlabel('Frequency [Hz]')
_=ylabel('IL [dB]')

In [4]:
# Comparison with
print("First resonance frequency insertion loss according to the model:", end='')
quarter_wave_freq_model = freq[np.argmax(-abspR_ov_A)]
print(' %0.2f Hz' %quarter_wave_freq_model)
print("First resonance frequency insertion loss according to first principles:", end='')
quarter_wave_freq_fp = mat.c0/L3/4
print(' %0.2f Hz' %quarter_wave_freq_fp)

First resonance frequency insertion loss according to the model: 142.93 Hz
First resonance frequency insertion loss according to first principles: 143.33 Hz


text:
## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE

 Options: Use Forum View Use Monospaced Font Show Text Part by Default Show All Mail Headers Message: [<< First] [< Prev] [Next >] [Last >>] Topic: [<< First] [< Prev] [Next >] [Last >>] Author: [<< First] [< Prev] [Next >] [Last >>]

 Subject: Re: ideal future document processing From: "Michel Lavaud Orleans (France)" <[log in to unmask]> Reply To: Mailing list for the LaTeX3 project <[log in to unmask]> Date: Wed, 25 Jun 1997 18:08:20 +0000 Content-Type: text/plain Parts/Attachments: text/plain (41 lines)
>  > The structure of the document in its first stages is indicated
>  > with punctuation marks, and typesetting indications such as
>  > \smallskip, \hspace{2cm}, \bf or whatever extend the set of
>  > punctuation marks available, and provide a convenient visual way
>  > to refine the structure. This mimicks exactly what one does when
>  > writing an article by hand.
> um, speak for yourself. I have tried to train myself *not* to write
> articles simply as a mimic of how i used to write by hand. on the
> contrary, when i write by hand these days i put in generic markup'

You have cut this citation from its context. I did not
say that I never use structured commands and use only \hspace{2cm}.
I said that in the early stages of _creating_ a scientific article,
it is totally unproductive to try to separate structure from
formatting, because the structure is largely undefined: when one does
research work, one does not know the final results _by definition_ so
one knows still much less the final structure of the document.

If the history of a scientific article would be done from conception
to publishing, I would say it starts as a pure Plain TeX document, as
there are mostly mathematical formulas plus some unstructured
comments explaining  what the formulas are about, separated by
commas, points and blank lines. That's all.

Then, during the refinement process, as
mathematical / physical / whatever new results are found and
accumulate,  the document evolves toward a more and more structured
document, either with LaTeX or with Plain TeX/AMSTeX and personal
macros (for many mathematicians).

Ultimately, when the document reaches _you_the_publisher_ in its
definitive form, then it may be transformed into SGML or anything
else you like (PostScript or PDF?). But trying to write a scientific
article from the scratch directly into SGML would be absurd, in my
opinion. A secretary, maybe, but not a researcher. I hope that only
those who never used \noindent or \medskip in an article to tune the
aspect of the output, instead of modifying the class of their
document (or the DTD for SGML addicts) will throw me some electronic
stones at the head :-)
`


text:
# Given a symmetric matrix $A$, find $P$ such that $P^T A P$ is a diagonal matrix

Given $$A =\begin{pmatrix} 0 & 3 & 0 \\ 3 & 0 & 4 \\ 0 & 4 & 0\end{pmatrix}$$ find a matrix $P$ such that $P^T A P$ orthogonally diagonalizes $A$. Verify that $P^TAP$ is diagonal.

I'm not sure how to approach this problem. Any help is appreciated. Thanks!

• Presumably you should find eigenvalues and eigenvectors? – Ted Shifrin Jul 21 '16 at 17:36
• – Moo Jul 21 '16 at 17:36
• @TedShifrin: Do you have any relations to Lenoid Shifrin? :) – Hosein Rahnama Jul 21 '16 at 17:46
• @H.R. Not that I know of, although we could be distant cousins or something. Same is true of some non-mathematicians of whom I'm aware with the same patronym. – Ted Shifrin Jul 21 '16 at 17:50

You want to find the eigenvalues and eigenvectors of $A$. Since $A$ is a real symmetric matrix, the eigenvectors should be orthogonal if the eigenvalues are distinct (which it is in this case): if they were not distinct you might have to use the Gram-Schmidt procedure to make the eigenvectors orthogonal. Divide each by its length, so they are orthonormal. Then form the matrix $P$ whose columns are those orthonormal eigenvectors.
This should be easy even without calculating any eigenvalues or eigenvectors explicitly. Note that $A=5(uv^T+vu^T)$, where $u=(0,1,0)^T$ and $v=\frac15(3,0,4)^T$ are already two orthonormal vectors. So, if you take $w=\frac15(4,0,-3)^T$, then $Q=(u,v,w)$ is real orthogonal and $$A=5Q\pmatrix{0&1\\ 1&0\\ &&0}Q^T.$$ Thus the problem boils down to the orthogonal diagonalisation of $\pmatrix{0&1\\ 1&0}$. Yet this $2\times2$ matrix geometrically represents the reflection about the line $x=y$. So, it is immediate that $$A=5Q\pmatrix{\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\ \frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\\ &&1}\pmatrix{1\\ &-1\\ &&0}\pmatrix{\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\ \frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\\ &&1}Q^T.$$ Absorb the factor $5$ into the diagonal matrix in the middle, you are done.


text:
# Ruggedizing a home-made High-Current SSPC

After establishing that we're not going to use our load bank in PWM mode (see here), I'm now trying to draft up a circuit that would be controlled by an µC (Arduino or else) and a multiplexer (this).

I'm also trying to protect the input from reverse voltage and overvoltage / making it able to cater with 1,8-24V digital input.

OK so here's my design thinking

1. 1N4148 for reverse voltage protection
2. 1,8V Zener diode for overvoltage protection (able to take 24V as a digital signal)
3. Resistor for current protection (500R ==> 10mA@5V)
4. This one I'm really not sure about : a 0,1µF in order for the whole contraption not to turn-off between two "cycles" of the multiplexer. I know that'd definitely be a no-go in a fast-switching application but that's not the case there... I'm also guessing that the MUX chip will not change an output state between two update cycles but...

So anyway, I hope I'm not completely off on this and wait for your feedback.

So far what I take away :

• My // diode is useless
• I could use a P-channel transistor instead of a diode for reverse protection (avoids the diode Vf). Say something like this

But it also brings one question to my mind : would such an input protection be compatible with some high-speed input? Wouldn't the rise time of this p-channel transistor also add to the whole chain?

• I cannot use a resistor to limit the current since my input voltage has a wide range

Where I'm still lost :

• What should I use to limit current ? :-)
• Is a zener diode used this way (through it's breakdown voltage) will really work at 10-40mA? If I refer to the datasheet Zener diode datasheet the Vz is way above the nominal 1,7V when above it's nominal 50µA. It's way above 3V at only 10mA
• Would maybe just using a LDO such as this one solve all of that?

Thanks for the help!

Hi everyone,

OK so I'm still not confident in my zener overvoltage protection...

After some reading, I found that I have to make sure the zener is always going to be conducting some current or it might not be very effective at all... So I'm going to add another resistor in // with the zener to ensure this minimum current. Not just completely sure that's the right way to do it.

• Last I checked, Zeners work just fine in a forward direction. You could easily cut the 1N4148. – Dave Dec 28 '17 at 23:29
• Hi @Dave. You're right, I completely missed out on this... Thanks for the comment! – VoltsAndNuts Dec 29 '17 at 7:54
• Interest only: Zeners tend to have a noticeably higher forward voltage than standard silicon diodes. From memory 1V fwd drop is not unexpected. With some devices the higher voltage may matter (but almost certainly not here). – Russell McMahon Dec 29 '17 at 9:34
• Very minor detail: if you make the schematic horizontally more compact (shorter traces and shorter space between components, U1/Q1 for example), the image will probably look bigger and clearer. (Or you can use the embedded editor) – Wesley Lee Dec 29 '17 at 23:07

## Input

1. Input from 3.3V or 5V uC output.
2. input reverse voltage protection.
3. 1,8~24V digital input ( if possible)

## Output

1. External 12V supply to series low-side switch , galvanic isolated from input.
2. Output switch RdsOn ~0.0012ohm

## IR opto LED characteristics

All LEDs can be simplified by $V_f = V_{th}+I_f*ESR$ From the (typ) graph, $V_{th}=1.36V~ @T_c=25'C ~~ ESR = 2.86Ω~$(= 0.1V/0.035A)
However there is a large tolerance on this as indicated by VOM1271 datasheet table LED forward voltage $V_f= 1.2V_{min}, ~1.4V_{typ}, ~1.6V_{max} @I_f = 10 mA ~~$

• It is the wide tolerance on ESR that affects Vf vs If as well as Temperature shift in threshold voltage Vth, that makes using a fixed series R not suitable for a wide input voltage range when a limited current range is desired.
• An input 3.3V to 5V is possible but one must analyze the ESR of the logic voltage and include this in the voltage drop using Vol/Iol.
• for reverse voltage protection rather than shunt diodes, you use a series Shottky diode or seriously prevent reverse voltage by design.

## Recommendation:

Review example specs for Automotive Smart low Side switches, there are lots of these with better performance.
Then revise your design specs as req'd and start over. eg.

https://www.digikey.com/en/product-highlight/i/infineon/high-and-low-side-switches

• Hi Tony. Thanks for the answer. A couple of precisions though : I'm trying to make the input compatible to the widest voltage possibles, from 1,8V to 24V. But as I see it with ESR tolerance problem, that's gonna be tough... Also, the 10mA was pretty much out of the blue, I'm seeing that it can take up to 40mA and in that case, I'll up the current as long as I don't risk frying it, it should also speed-up turn-on... Regarding the Output, I'm controlling a 36V / 35A load which is too high for these automotive low-side switches... Too bad because they're pretty great ! – VoltsAndNuts Dec 29 '17 at 7:48
• Regarding reverse voltage protection, could you explain in what way is a series schottky superior to a // shunt diode? I guess in the shunt diode, you risk frying the supply, right? But in series, you have to swallow the diode Vf... :/ I was also looking at this way of doing with a small P-channel transistor, what's your take? – VoltsAndNuts Dec 29 '17 at 7:50
• Correct. better with low RdsOn losses – Tony Stewart Sunnyskyguy EE75 Dec 29 '17 at 15:26


text:
# Stacking the dual tetrahedron in an ordinary tetrahedron. Is it possible?

I'm thinking about the following problem.

## Introduction

First let me introduce the problem with a 2D example. The area of the triangle constructed by connecting the midpoints of a triangle is 1/4 of the the area of the full triangle.

Very straightforward and a picture is a nice clarification. I like how the center triangle can be found four times in the full triangle.

## Onto 3 dimensions

Consider a tetrahedron, then the volume of the dual tetrahedron is 1/27 of the full tetrahedron.

This seems a lot less obvious then the 2D example. I printed the model using my 3D printer for better visualizing, but still it is not obvious.

I would like to stack the dual tetrahedron 27 times to visualize this quantity, but thus far I'm unable to do so. My first approach to divide each side in three and then connect corresponding points doesn't seem successful as there are pyramids with a square base to be found.

I guess I should start like in the picture below, but it's unclear how to proceed. Which makes me think it's not possible to stack the smaller tetrahedron in a similar way.

Is it possible to stack the tetrahedron? And if so, could someone provide me with some pointers on how to proceed.

No, tetrahedrons cannot fill space. They can stack with regular octahedrons to fill space. Your figures justify the statement that the volume of the dual is $1/27$ the volume of the original because you find the centroid of the faces of the original tetrahedron. As the centroid is $2/3$ of the way along the bisector, the altitude of the dual is $1/3$ the altitude of the original. As the volume scales with the cube of the size, the volume is $1/3^3=1/27$ of the original. But you can't take $27$ of the duals and stack them together to make a copy of the original.

• hm, that's a bit disappointing, thanks for the info though. Any ideas on how to visualize the number 27? My best visualization yet seems to be the 11 tetrahedra and the 4 regular octahedrons. – dietervdf Jun 27 '16 at 20:41
• @dietervdf: just what I gave-the side length is $1/3$ and you cube that for volumes. 3D packing is much more constrained than 2D. In 2D, every polygon can be cut into a finite number of pieces and reassembled to make any polygon of equal area. That is not true in 3D. The Dehn invariant makes an issue – Ross Millikan Jun 27 '16 at 20:42
• Right and sorry, deleted my comment just before your above comment. Please consider deleting yours too. – Narasimham Jun 27 '16 at 20:44
• I know it's pretty easy to prove, but I find it difficult to grasp visualizing the tetrahedron. Looking at the object it's hard to imagine that the volume is 1/27 of the full tetrahedron. Math can be mind-boggling ;). – dietervdf Jun 27 '16 at 20:44

If the vertices of the tetrahedron are $A,B,C,D$, the vertices of the dual tetrahedron are $$V_D=\frac{A+B+C}{3},\quad V_C=\frac{A+B+D}{3},\quad V_B=\frac{A+C+D}{3},\quad V_A=\frac{B+C+D}{3}$$ and the centroid $G$ is the same: $$G = \frac{A+B+C+D}{4}=\frac{V_A+V_B+V_C+V_D}{4}.$$ It follows that a dilation with centre at $G$ and ratio $\lambda=-3$ brings $V_X$ to $X$ for any $X\in\{A,B,C,D\}$, so the ratio between the volume of $ABCD$ and the volume of $V_A V_B V_C V_D$
is $|\lambda|^3 = \color{red}{27}$.

As already pointed by Ross Millikan, a tetrahedron is not a space-filling solid, since neither the angle between faces nor the solid angle in a vertex are rational multiples of a radian/steradian.

You may also use a pretty elegant argument based on Dehn's invariant, solving Hilbert's third problem.

• I like the "angle" reasoning, thanks. I didn't know of the link with Hilbert's third problem, cool! – dietervdf Jun 27 '16 at 20:47


text:
# Nilpotent group with rationally powered abelianization need not be rationally powered

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) does not always satisfy a particular subgroup property (i.e., quotient-powering-faithful subgroup)
View subgroup property satisfactions for subgroup-defining functions $|$ View subgroup property dissatisfactions for subgroup-defining functions

## Statement

It is possible to have a nilpotent group $G$ such that the abelianization of $G$ is a rationally powered group (which in this case means it is a vector space over $\mathbb{Q}$) but such that $G$ is not rationally powered over any prime. In fact, we can select $G$ to be such that it has $p$-torsion for every prime $p$.

This also shows that the derived subgroup of a nilpotent group need not be a quotient-powering-faithful subgroup.

## Proof

Further information: quotient of UT(3,Q) by a central Z

Let $G$ be the quotient group of $UT(3,\mathbb{Q})$ (the unitriangular matrix group of degree three over the field of rational numbers) by a subgroup $\mathbb{Z}$ inside its center (which is a copy of $\mathbb{Q}$). Explicitly, we can think of $G$ as matrices of the form:

$\{ \begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix} \mid a_{12},a_{23} \in \mathbb{Q}, \overline{a_{13}} \in \mathbb{Q}/\mathbb{Z} \}$

with the matrix multiplication defined as:

$\begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix}\begin{pmatrix} 1 & b_{12} & \overline{b_{13}} \\ 0 & 1 & b_{23} \\ 0 & 0 & 1 \\\end{pmatrix} = \begin{pmatrix} 1 & a_{12} + b_{12} & \overline{a_{12}b_{23}} + \overline{a_{13}} + \overline{b_{13}} \\ 0 & 1 & a_{23} + b_{23} \\ 0 & 0 & 1 \\\end{pmatrix}$

where $\overline{a_{12}b_{23}}$ is understood to be the image of $a_{12}b_{23}$ under the quotient map $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$.

The center of $G$ coincides with its derived subgroup, and is:

$\left \{ \begin{pmatrix} 1 & 0 & \overline{a_{13}} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{pmatrix} \mid \overline{a_{13}} \in \mathbb{Q}/\mathbb{Z} \right \}$

The inner automorphism group and the abelianization are therefore both isomorphic to $\mathbb{Q} \times \mathbb{Q}$, which is rationally powered. However, the group as a whole has $p$-torsion for all primes $p$.


text:
# Enthalpy and stoichiometry

1. ### Agent M27

171
1. The problem statement, all variables and given/known data
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.5 Kj. If the change in enthalpy is 5074.2 Kj, how much work is done during the combustion? Find work in Kj.

2. Relevant equations

$$\Delta$$E=$$\Delta$$H + (-P$$\Delta$$V)

w=-P$$\Delta$$V

3. The attempt at a solution

$$\Delta$$E=5084.5Kj
$$\Delta$$H=5074.2Kj

5084.5-5074.2=-1atm($$\Delta$$V)
-10.3=$$\Delta$$V

w=-(1)(-10.3)=10.3

10.3L*atm x 101.3J=1043.39J

$$\frac{1043.39}{1000}$$= 1.04339 Kj

This is an online homework set, so when I input the answer it usually will give me a hint if I am close, but I am not getting anything. Can anyone spot my error? Thanks in advance.

Joe

2. ### nessiejo236

1
I don't know if you have already figured this out, but I thought I'd let you know I found your problem. The way you solved it, your change in volume ended up in kJ/atm when it should be in liters. I just did a track and took 10300J and divided it by 101.3 J to end up with about 101.68 L.

10.3kJ/atm x 1000J/1kJ x 1 Latm/101.3J= 101.68 L.

Hopefully that helped if you haven't already figured it out!


text:
### Just Touching

Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles?

 A hexagon is inscribed inside a circle. The sides of the hexagon are alternately $a$ and $b$ units in length. What is the radius of the circle?


text:
## Publications

Export 5 results:
Sort by: Author Title Type [ Year ]
2017
Dimitrova, I., Vítor H. Fernandes, and J. Koppitz. "A note on generators of the endomorphism semigroup of an infinite countable chain." Journal of Algebra and its Applications (DOI: 10.1142/S0219498817500311). 16 (2017): 1750031 (9 pages). AbstractWebsite

In this note, we consider the semigroup $O(X)$ of all order endomorphisms of an infinite chain $X$ and the subset $J$ of $O(X)$ of all transformations $\alpha$ such that $|Im(\alpha)|=|X|$. For an infinite countable chain $X$, we give a necessary and sufficient condition on $X$ for $O(X) = < J >$ to hold. We also present a sufficient condition on $X$ for $O(X) = < J >$ to hold, for an arbitrary infinite chain $X$.

2016
Fernandes, Vítor H., and Teresa M. Quinteiro. "A note on bilateral semidirect product decompositions of some monoids of order-preserving partial permutations." Bull. Korean Math. Soc.. 53.2 (2016): 495-506. AbstractWebsite

In this note we consider the monoid $PODI_n$ of all monotone partial permutations on $\{1,\ldots,n\}$ and its submonoids $DP_n$, $POI_n$ and $ODP_n$ of all partial isometries, of all order-preserving partial permutations and of all order-preserving partial isometries, respectively. We prove that both the monoids $POI_n$ and $ODP_n$ are quotients of bilateral semidirect products of two of their remarkable submonoids, namely of extensive and of co-extensive transformations. Moreover, we show that $PODI_n$ is a quotient of a semidirect product of $POI_n$ and the group $\mathcal{C}_2$ of order two and, analogously, $DP_n$ is a quotient of a semidirect product of $ODP_n$ and $\mathcal{C}_2$.

2008
Fernandes, Vítor H. "Normally ordered semigroups." Glasg. Math. J.. 50 (2008): 325-333.Website
2002
Fernandesh, V. U. "A new class of divisors of semigroups of isotone mappings of finite chains." Izv. Vyssh. Uchebn. Zaved. Mat. (2002): 51-59.
1998
Fernandes, Vitor H. "Normally ordered inverse semigroups." Semigroup Forum. 56 (1998): 418-433.Website