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# Calculus, how to find the value of delta in Epsilon-delta limit proof

Problem: Show that $\lim_{x\to 4}x^3=64$.

My steps:

$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
$\sqrt[3]{64-\epsilon}<x<\sqrt[3]{64+\epsilon}$
$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$
So $\delta<\sqrt[3]{64+\epsilon}-4$
Let $\epsilon>0$. Pick $\delta$ so that $\delta<\sqrt[3]{64+\epsilon}-4$.
Suppose $0<|x-4|<\delta$. Then $|x-4|<\sqrt[3]{64+\epsilon}-4$.
Thus, $|x|<\sqrt[3]{64+\epsilon}$ and cube to get $|x^3|<64+\epsilon$.
Then $|x^3-64|<\epsilon$ which completes the proof.

But this is what the answer said:

Let $\epsilon>0$. Pick $\delta$ so that $\delta<1$ and $\delta<\frac{\epsilon}{61}$.
Suppose $0<|x−4|<\delta$. Then $4−\delta<x<4+\delta$.
Cube to get $(4−\delta)^3 <x^3 <(4+\delta)^3$.
Expanding the right-side inequality, we get
$x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64$
$=64+\epsilon$.
The other inequality is similar.

So my questions are:
1. How can I get $\delta<\frac{\epsilon}{61}$?
2. Why $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64=64+\epsilon$?
3. When we need to pick $\delta=min(1,\frac{\epsilon}{61}$) in this case?
4. Does my proof has problems ?

I am just a newbie to calculus... Thanks in advance.

• I think your proof is equivalent, but it just doesn't follow the typical order of $\epsilon$-$\delta$ -proof. In $\epsilon$-$\delta$ proof one discovers what $\delta$ one needs, by "looking in the future" in the proof. So you do things starting from the "Suppose $0 < |x-4| < \delta$" and then decide on the $\delta+12\delta+48\delta+64$ part, what $\delta$ you need in order to make the conclusion that you wish, which is $x^3 < 64 + \epsilon$. – mavavilj Mar 22 '18 at 14:40
• Notice what the line just before $64+\epsilon$ reads if you plug in $\epsilon/61$. Thus this becomes the constraint that you want to use for $\delta$. – mavavilj Mar 22 '18 at 14:46

Well, one is not supposed to use inverse functions like cube roots in expression for $\delta$ otherwise this leads to a circular argument.

The typical approach is to analyze your goal namely $$|x^3-64|<\epsilon$$ which has to be achieved using a specific kind of means namely $0<|x-4|<\delta$. So one first tries to simplify our goal by rewriting it as $$|x-4||x^2+4x+16|<\epsilon$$ The first factor in LHS can be controlled using our means but the second factor needs more work. One just needs to bound this factor suitably (again using the same means). We note that if $0<|x-4|<1$ then $3<x<5, x\neq 4$ and hence $|x^2+4x+16|<61$. Further we can also start with the bound $0<|x-4|<2$ and get $|x^2+4x+16|<76$. The important thing to observe here is that this factor $|x^2+4x+16|$ is bounded. Let's choose the simpler bound $61$ which is possible if $0<|x-4|<1$.

Now we have the obvious inequality $$|x^3-64|=|x-4||x^2+4x+16|<61|x-4|$$ provided that $0<|x-4|<1$. If the RHS of the above inequality is less than $\epsilon$ then the LHS will also be less than $\epsilon$ and hence our goal is achieved if we have $61|x-4|<\epsilon$ or $|x-4|<\epsilon/61$. Combining the above we see that if $0<|x-4|<\min(1,\epsilon /61)$ then $|x^3-64|<\epsilon$. And hence $\delta$ can be taken as $\min(1,\epsilon /61)$. In case you have read the last paragraph carefully you can immediately see that the $\delta$ can also be taken as $\min(2,\epsilon /76)$.

Such problems do not have a unique answer and most importantly they are not supposed to be solved via algebraic manipulation of inequalities.

• Thank you for the explanation ! – ethanlevy97 Mar 24 '18 at 9:04
1. for any given (strictly) positive $\epsilon$ just pick a positive number that is smaller than $\epsilon/64$ such a number exist, because this quotient is (strictly) positive.
1. Because of 3 ( if $\delta < 1$ then $\delta^3 <\delta^2 < \delta$
2. I suppose that should be "why" : for 2 to be true
3. I'll leave that one to another

Note that in you proof is ok from here

$$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$$

you should set

$$\delta=min\{4-\sqrt[3]{64-\epsilon},\sqrt[3]{64+\epsilon}-4\}=\sqrt[3]{64+\epsilon}-4$$

and what you find is the optimal value for $\delta$.

Note that by binomial expansion

$$\sqrt[3]{64+\epsilon}-4=4\sqrt[3]{1+\epsilon/64}-4\approx\frac{\epsilon}{48}$$

then for $\epsilon$ small also $\frac{\epsilon}{61}<\frac{\epsilon}{48}$ fulfills the condition.

The other limit is obtained by

$$|x^3-64|<\epsilon\iff |x-4||x^2+4x+16|<\epsilon\iff |x-4|<\frac{\epsilon}{ |x^2+4x+16|}=\delta$$

thus for $0<|x-4|<1 \implies3< x<5 \quad x\ne 4$ the minimum value for $\delta$ is attained for $x=5$ that is

$$\delta_{min}=\frac{\epsilon}{ |5^2+4\cdot5+16|}=\frac{\epsilon}{61}$$

then the condition is satisfy for $\delta=\min\left(1,\frac{\epsilon}{61}\right)$.

• Using cube roots involves the fact that cube roots exist. And existence of cube roots is a direct consequence of continuity of $x^3$ in which case the limit in question is already available so this is circular. In general one should not use inverse functions to find $\delta$ in terms of $\epsilon$. – Paramanand Singh Mar 22 '18 at 15:18
• @ParamanandSingh so the first approach is not correct? I will look to your answer, Thanks! – gimusi Mar 22 '18 at 15:21
• Also see math.stackexchange.com/a/2104039/72031 (especially last paragraph) and math.stackexchange.com/a/2487862/72031 – Paramanand Singh Mar 22 '18 at 15:25
• @ParamanandSingh Thanks, I've tried to make the second part more clear, I'll take a look. Your help is always valuable. – gimusi Mar 22 '18 at 15:45

In regards to finding a value for $\delta$ that satisfies the proof, we only really care about values of $x$ that are close to $4$, so setting $\delta < 1$ and $\delta < \frac{\epsilon}{61}$ is like saying the distance between $x$ and $4$ must be no more than $1$ or $\frac{\epsilon}{61}$.

By setting these restrictions on $\delta$, we can then make our lives easier in terms of solving this proof because the values acceptable for $x$ also become restricted, i.e if $\delta \lt 1 \rightarrow\lvert x - 4 \rvert \lt 1$ then its safe to make the claim that $3 \lt x \lt 5; x \not=4$ (because the difference between $x$ and $4$ cannot be greater than 1).

So, to answer question 1, "How can I get $\delta \lt \frac{\epsilon}{61}$ ?", it's the result we get when we restrict $\delta \lt 1$.

\begin{align} \lvert ~f(x) - 64~\rvert &\lt \epsilon\\ \lvert ~x^3 - 64~\rvert &\lt \epsilon && \text{factor x^3 - 64} \\ \lvert ~(x - 4)(x^2 + 4x + 16)~| &\lt \epsilon && \text{isolate (x - 4)}\\ \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert} \end{align}

Knowing that when $\delta \lt 1$, then $3 \lt x \lt 5; x \not=4$, so our goal is to apply this inequality to $x^2 + 4x + 16$ and see what we get.

\begin{align} 3 \lt x &\lt 5 && \text{let's find the min and max of x^2}\\ 9 \lt x^2 &\lt 25 \\\\ 3 \lt x &\lt 5 && \text{let's find the min and max of 4x}\\ 12 \lt 4x &\lt 20 \\\\ 9 \lt x^2 &\lt 25 && \text{let's bring this all together now}\\ 9 + 12 \lt x^2 + 4x &\lt 25 + 20 \\ 21 \lt x^2 + 4x &\lt 45 && \text{all that's left to add is 16}\\ 21 + 16 \lt x^2 + 4x + 16 &\lt 45 + 16\\\\ 37 \lt x^2 + 4x + 16 &\lt 61 \end{align}

What we just showed is that when $\delta \lt 1$, then $37 \lt x^2 + 4x + 16 \lt 61$ which in turn means that, \begin{align} \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert} \lt \frac{\epsilon}{61} \\\\ \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{61} && \text{\square} \end{align}

For your second question: "Why $x^3<\delta^3+12\cdot \delta^2 + 48\cdot \delta + 64 < \delta + 12\delta + 48\delta + 64 = 64 + \epsilon$ ?", if think about the value for $\delta$, it's always less than $1$ so when we when multiply two or more values that are less than $1$, we get a smaller number. This explains why, $$\delta^3+12\delta^2 + 48\delta + 64 < \delta + 12\delta + 48\delta + 64$$

In regards to how $\delta + 12\delta + 48\delta + 64 = 64 + \epsilon$, we just simplify the right side of the equation to get \begin{align} \delta + 12\delta + 48\delta + 64 &= 61\delta + 64 && \text{sub. in \frac{\epsilon}{61} for \delta}\\ 61 \cdot \frac{\epsilon}{61} + 64 &= \epsilon + 64 && \text{\square} \end{align}

Unfortunately, for your third question, I don't really know what your asking. However, what I can say about knowing when to set $\delta$ to either $1$ or $\frac{\epsilon}{61}$ is that it depends on the value for $\epsilon$; if $\epsilon \lt 61$ then set $\delta = \frac{\epsilon}{61}$, otherwise set $\delta = 1$.

Lastly, in terms of your proof having problems, these questions are not easy to begin with, so making a misstep in the beginning of your proof will very easily lead you astray. The good thing is that you tried to get the $\epsilon$ inequality to resemble the $\delta$ inequality which is the right thing to do, the problem is how you chose to do it.

Trying to isolate for $x$ and then manipulating the inequality to resemble the $\delta$ inequality isn't really an effective approach because things can get complicated very quickly (especially in this case). I would try and factor first and see where that takes me. That being said, when it comes to $\epsilon-\delta$ proofs, there is no silver bullet approach to finding the solution. You will need to practice them and with experience, you will start to have an idea of what to do.

As a side note, here's a reference to a tutorial on Limits I wrote a couple months ago for Mathbook.

• Thank you! Your answer is really clear and helped me a lot =) – ethanlevy97 Mar 24 '18 at 9:05
• You very welcome. I'm glad I could be of help. – JetJet13 Mar 24 '18 at 17:40


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# Combination: Proof

• Feb 24th 2009, 01:56 AM
Khorne
Combination: Proof
Hey all!

I have a question, which I've been pondering over for a few hours, and just can't seem to grasp it.

It goes:

(m+n)C(3) = (m)C(3) + (m)C(2)*(n)C(1) + (m)C(1)*(n)C(2) + (n)C(3)

I've tried simplify the LHS down to (m+n)(m+n-1)(m+n-2)/3! But this didn't help, and I've tried simplify, and canceling down the RHS too, but I always end up with a huge expansion (18 terms (Worried)).

Could anyone lend a helping hand? Or some insight into the problem?

Thanks!
• Feb 24th 2009, 06:59 AM
red_dog
We have $(1+x)^{m+n}=(1+x)^m(1+x)^n$

The coefficient of $x^3$ in the LHS is $C_{m+n}^3$

The coefficient of $x^3$ in the RHS is $C_m^3+C_m^2C_n^1+C_m^1C_n^2+C_n^3$

But the coefficients are the same, so we have the equality.
• Feb 24th 2009, 11:58 AM
Khorne
Hello, and thank you.

However, could you just clarify the first line? I don't comprehend how you structure? or deduced it.
• Feb 24th 2009, 12:16 PM
Plato
Quote:

Originally Posted by Khorne
could you just clarify the first line? I don't comprehend how you structure? or deduced it.

You do know the basic laws of exponents.
• Feb 24th 2009, 06:59 PM
Khorne
Indeed, but not with this type of proof setting.

format: plain


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Remark 37.40.3. In terms of topologies Lemmas 37.40.1 and 37.40.2 mean the following. Let $S$ be any scheme. Let $\{ f_ i : U_ i \to S\}$ be an étale covering of $S$. There exists a Zariski open covering $S = \bigcup V_ j$, for each $j$ a finite locally free, surjective morphism $W_ j \to V_ j$, and for each $j$ a Zariski open covering $\{ W_{j, k} \to W_ j\}$ such that the family $\{ W_{j, k} \to S\}$ refines the given étale covering $\{ f_ i : U_ i \to S\}$. What does this mean in practice? Well, for example, suppose we have a descent problem which we know how to solve for Zariski coverings and for fppf coverings of the form $\{ \pi : T \to S\}$ with $\pi$ finite locally free and surjective. Then this descent problem has an affirmative answer for étale coverings as well. This trick was used by Gabber in his proof that $\text{Br}(X) = \text{Br}'(X)$ for an affine scheme $X$, see [Hoobler].

Comment #418 by on

It's a question of English language (and I'm by no means an expert), but I think "descend problem" should be "descent problem".

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# zbMATH — the first resource for mathematics

New results on the periodic solutions for a kind of Rayleigh equation with two deviating arguments. (English) Zbl 1161.34345
Existence of periodic solutions for the class of equations with two deviating arguments of the form
$x''(t)+f(x'(t))+g_1(t,x(t-{\tau}_1(t)))+g_2(t,x(t-{\tau}_2(t)))=p(t)$
is investigated. The main tools used by the authors are: the continuation theorem of the coincidence degree, a priori estimates, and differential inequalities, thus improving and generalizing previous results.

##### MSC:
 34K13 Periodic solutions to functional-differential equations 47N20 Applications of operator theory to differential and integral equations
dde23
Full Text:
##### References:
 [1] Wang, G.; Cheng, S., A priori bounds for periodic solutions of a delay Rayleigh equation, Appl. math. lett., 12, 41-44, (1999) · Zbl 0980.34068 [2] Lu, S.; Ge, W., Some new results on the existence of periodic solutions to a kind of Rayleigh equation with a deviating argument, Nonlinear anal., 56, 501-514, (2004) · Zbl 1078.34048 [3] Lu, S.; Ge, W.; Zheng, Z., Periodic solutions for neutral differential equation with deviating arguments, Appl. math. comput., 152, 17-27, (2004) · Zbl 1070.34091 [4] Lu, S.; Ge, W.; Zheng, Z., A new result on the existence of periodic solutions for a kind of Rayleigh equation with a deviating argument, Acta math. sinica, 47, 299-304, (2004), (in Chinese) · Zbl 1293.34087 [5] Shampine, L.F.; Thompson, S., Solving DDEs in Matlab, Appl. numer. math., 37, 441-458, (2001) · Zbl 0983.65079 [6] Huang, X.; Xiang, Z., On existence of $$2 p$$-periodic solutions for delay Duffing equation $$x''(t) + g(t, x(t - \tau(t))) = p(t)$$, Chinese sci. bull., 39, 201-203, (1994) [7] Liu, B.; Huang, L., Periodic solutions for a kind of Rayleigh equation with a deviating argument, J. math. anal. appl., 321, 491-500, (2006) · Zbl 1103.34062 [8] Liu, B.; Huang, L., Periodic solutions for nonlinear $$n$$th order differential equations with delays, J. math. anal. appl., 313, 700-716, (2006) · Zbl 1105.34044 [9] Liu, B.; Huang, L., Periodic solutions for a class of forced Liénard-type equations, Acta math. appl. sin. engl. ser., 21, 81-92, (2005) · Zbl 1093.34020 [10] Peng, L., Periodic solutions for a kind of Rayleigh equation with two deviating arguments, J. franklin inst., 7, 676-687, (2006) · Zbl 1114.34051 [11] Gaines, R.; Mawhin, J., Coincidence degree and nonlinear differential equations, (1977), Springer-Verlag Berlin · Zbl 0339.47031 [12] Deimling, K., Nonlinear functional analysis, (1985), Springer New York · Zbl 0559.47040
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.


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# fraction with 2 variables, and radicals in numerator

I arrived at the following solution to a problem in pre-calculus:

$$\frac{2xh + h^2 + \sqrt{x+h} - \sqrt{x}}{h}$$

However, this can be simplified further to:

$$2x+h+\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

The steps to simplify were not provided. I substituted in $x = 9, h = 16$ to confirm and I've searched a few places, but am at a loss as to how the term

$$\frac{\sqrt{x+h} - \sqrt{x}}{h}$$

can be simplified to:

$$\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

What are the steps and relevant rules?

• difference of two squares Jun 26 '18 at 4:14

This is an example of the very useful "multiply by the conjugate" trick. The trick is based on the fact that $$(a+b)(a-b)=a^2-b^2,$$ which is easy to verify by direct multiplication — but it's so useful that it's worth remembering!
In your example, the conjugate of the numerator $\left(\sqrt{x+h}-\sqrt{x}\right)$ is the expression $\left(\sqrt{x+h}+\sqrt{x}\right)$, so we're going to multiply the numerator and denominator simultaneously by this conjugate: $$\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{\left(\sqrt{x+h}\right)^2-\left(\sqrt{x}\right)^2}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{x+h-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{1}{\sqrt{x+h}+\sqrt{x}}.$$
Note that $$h = x+h-x = (\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})$$
and therefore, $$\frac{h}{\sqrt{x+h}+\sqrt{x}} = \sqrt{x+h}-\sqrt{x}$$


text:
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Bug 35186 - MathJax: Several commands that work in texvc are not recognized
 Summary: MathJax: Several commands that work in texvc are not recognized
 Status: Product: ASSIGNED MediaWiki extensions Unclassified Component: Math (Other open bugs) unspecified All All Low normal with 1 vote (vote) --- Nobody - You can work on this! 70207 31406 make-mathjax-default Show dependency tree / graph

 Reported: 2012-03-13 08:56 UTC by Michael M. 2014-10-19 18:19 UTC (History) 11 users (show) chrkipping fred.wang gnosygnu hartman.wiki liangent mah mal.malego peter.krautzberger physik rich wikimedia.x.0x --- --- ---

Attachments

 Michael M. 2012-03-13 08:56:57 UTC I found these: \varcoppa, \sampi, \varstigma, \Digamma, \Sampi, \euro, \geneuro, \geneuronarrow, \geneurowide, \officialeuro, \arccot, \arcsec, \arccsc, \cancel, \bcancel, \xcancel, \cancelto Additionally, \color doesn't recognize all color names, at least not OliveGreen: { \color{OliveGreen} 1 } is black. Michael M. 2012-03-13 09:20:30 UTC For more colors, that don't work, see http://leuksman.com/mw/index.php/Help,_Formula#Color (at least one example where the help from en.wikipedia is better than that from de). Michael M. 2012-03-14 08:53:48 UTC The unrecognized colors are rendered black in Firefox 3.6, in IE 8 a [Math Processing Error] is displayed instead. Brion Vibber 2012-03-14 22:54:33 UTC Putting this on my list, these should be addable. Richard Morris 2012-03-15 00:50:04 UTC \arccot, \arcsec, \arccsc were solved in Nageh version by adding some macros. See http://en.wikipedia.org/wiki/User_talk:Nageh/mathJax#Missing_functions Michael M. 2012-03-17 08:30:07 UTC After a closer look at the sources: \color seems to be only a configuration issue, since there is MathJax/extensions/TeX/color.js. Probably you just have to mention it in MathJax/config/TeX-AMS-texvc_HTML.js next to the other extensions. The missing commands that can be emulated should go in MathJax/extensions/TeX/texvc.js. Richard Morris 2012-03-25 16:15:13 UTC $\dot \vec u$ works with tecvc but not with mathjax wihich only accepts the (correct) Latex $\dot{\vec u}$. There are a few similar bugs mentioned at http://en.wikipedia.org/w/index.php?title=User_talk:Nageh/mathJax including $$\hat \mathbf u$$, $$\underline \mathbb Z$$, $$\hat \boldsymbol \theta$$ Derk-Jan Hartman 2012-03-25 20:38:05 UTC Added the colors in https://gerrit.wikimedia.org/r/3723 Derk-Jan Hartman 2012-03-25 22:13:46 UTC Seems pagecolor is also not supported yet. Derk-Jan Hartman 2012-03-25 22:21:39 UTC \cancel, \bcancel, \xcancel, \cancelto can also be enabled by adding the extension to our config. Derk-Jan Hartman 2012-03-26 21:39:23 UTC \arccot, \arcsec, \arccsc: https://gerrit.wikimedia.org/r/3767 \cancel, \bcancel, \xcancel, \cancelto: https://gerrit.wikimedia.org/r/3768 Derk-Jan Hartman 2012-03-26 22:33:05 UTC Our defined symbols are smaller than the symbols from the AMS package due to different styling. Our symbols get font-family: STIXGeneral, 'Arial Unicode MS', serif; font-size: 85%; font-style: italic; font-weight: normal;, whereas AMS symbols only get font-family: MathJax_AMS; this will need fixing at some point. Michael M. 2012-03-28 08:58:20 UTC From source code I extracted lists of both the commands allowed in texvc and MathJax, and diffed them. I hope this is the complete list of commands still missing: AA Dagger Digamma emph euro geneuro geneuronarrow geneurowide H image officialeuro pagecolor Q sampi Sampi sen textsf texttt varcoppa varstigma vbox vline Derk-Jan Hartman 2012-04-01 14:16:03 UTC Added in https://gerrit.wikimedia.org/r/4060 \Dagger \Digamma \sampi \Sampi \varcoppa \varstigma \Q U+0051 doublestruck Q \H U+8461 doublestruck H \AA U+00C5 armstrong \sen spanish version of \sin (bug 18912) \image alias for \Im Still left: \euro \geneuro \geneuronarrow \geneurowide \officialeuro \pagecolor ignore Text instead of math formatting.. should we really include this ? \emph{} text emphasize (italics) \textsf{} text sans serif \texttt{} text typewriter \vbox{text} Encloses a paragraph's text to prevent it from running over a page break \vline draws a vertical line with the height of the row at the location where it appears nageh 2012-04-18 11:58:07 UTC \textsf{}, \texttt{} and \vline are implemented https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax%2Fconfig%2FTeX-AMS-texvc_HTML.js&diff=486528289&oldid=485202027 here. \emph{}, while implemented as \textit{}, really shouldn't be supported because it is a text mode command and not a math mode command. The case with the euro symbols is a different one. The \gen* commands are supposed to be generated from the 'C' letter, meant to support symbols that more closely match the font style. However, it turns out that Unicode and font designers, instead of adopting the official euro symbol, have preferred a symbol that matches the font style. Which means that the euro symbol in the font set most closely resembles \geneuro. \officialeuro is not available in the fonts, unless a specific euro font file will be added to MathJax. \euro is a link to one of the other symbols; while set to \officialeuro by default I have linked it to \geneuro. nageh 2012-04-18 12:07:03 UTC This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax%2Fconfig%2FTeX-AMS-texvc_HTML.js&diff=486653054&oldid=486652453, adds the RGB color space so \definecolor does not fail anymore. This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax%2Fconfig%2FTeX-AMS-texvc_HTML.js&diff=486652319&oldid=486638683, removes the noErrors.js extension, which, don't ask me why I have included it. This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax%2Fconfig%2FTeX-AMS-texvc_HTML.js&diff=486638683&oldid=486528289, implements pretty perfect \oiint support for both HTML-CSS and MathML output, and admittedly a (still) imperfect implementation of \oiiint. I possibly would be good to directly rely on the font symbols when STIX fonts are detected. This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax%2Fconfig%2FTeX-AMS-texvc_HTML.js&diff=487586044&oldid=486653054, resolves an issue where the \scriptstyle hack, which is so common on Wikipedia to effect a normal font size, is used in boxed display maths. I have updated my documentation accordingly (here: https://en.wikipedia.org/wiki/User:Nageh/mathJax#Usage_notes). nageh 2012-04-18 12:11:44 UTC This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax.js&diff=484537614&oldid=484068957, supports mtextFontInherit for both HTML-CSS and MathML output. Someone has complained that the vertical margin for display maths is too large. This change, https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax.js&diff=487684008&oldid=486653976, reduces that margin by a tiny bit. I think it pretty closely matches that of ordinary LaTeX. wikimedia.x.0x 2012-05-04 06:53:19 UTC So stuff like \texttt won't be supported? Should sites using these therefore be changed? Example: http://de.wikipedia.org/wiki/Lyndonwort nageh 2012-05-04 11:20:54 UTC Huh? \texttt is most likely to be supported as it can be trivially implemented. See comment 14. Richard Morris 2012-05-04 11:24:34 UTC As nageh has said \texttt \textsf and \emph can probably be fixed by adding MathJax.Hub.Config({ TeX: { Macros: { texttt: ['Macro','\\mathord{\\tt{\\text{#1}}}',1], textsf: ['Macro','\\mathord{\\sf{\\text{#1}}}',1], emph: ['Macro','\\textit{#1}',1], // doesn't switch to \textrm within italic text as it should } } } although I'm not sure if it works for doubly nested items (which don't actually work for \textbf{\textit{broken}} anyway). nageh 2012-05-04 20:48:50 UTC Oh? I didn't know that \text doesn't accept nested commands. I suppose this could be easily fixed by the MathJax devs. Do you want to file a bug report at https://github.com/mathjax/MathJax/ ? Should I file one on your behalf? Thanks. nageh 2012-05-05 11:44:23 UTC That should be https://github.com/mathjax/MathJax/issues . Richard Morris 2012-05-06 05:56:54 UTC Now not convinced \textbf{\textit{broken}} is broken I can't see anywhere in latex doc which say the \textfoo commands should nest. Only source which indicates it is seems to be http://en.wikibooks.org/wiki/LaTeX/Text_Formatting which isn't that reliable. nageh 2012-05-06 10:45:58 UTC Did you search on Google books? You could also try it yourself using your favorite LaTeX2e interpreter. \text* commands have been introduced in LaTeX(2e) over TeX for two reasons: old commands (\bf, \it, etc.) don't take a parameter so you always have to write {\bf text}; and they don't nest in contrast to \text*. nageh 2012-05-06 11:23:16 UTC CTAN is the definite web source for information on LaTeX and packages. You may want to read the top of page 124 of http://ctan.org/tex-archive/info/lshort/english/lshort.pdf nageh 2012-05-09 15:40:34 UTC Now reported at https://github.com/mathjax/MathJax/issues/243 . Liangent 2012-08-07 15:42:45 UTC \theta = 5^\operatorname{\omicron} This math code is problematic but I dunno which command is the cause. Richard Morris 2012-08-07 16:18:37 UTC An extra pair of curly brackets sorts it out. $\theta = 5^{\operatorname{\omicron}}$ works. MathJax is a bit more particular with precedence than texvc. Liangent 2012-08-07 16:28:33 UTC (In reply to comment #27) > An extra pair of curly brackets sorts it out. $\theta = > 5^{\operatorname{\omicron}}$ works. MathJax is a bit more particular > with precedence than texvc. This is still some inconsistency... Michael M. 2012-08-20 09:06:02 UTC (In reply to comment #27) > An extra pair of curly brackets sorts it out. $\theta = > 5^{\operatorname{\omicron}}$ works. MathJax is a bit more particular > with precedence than texvc. This is reported in bug 36534. Gerrit Notification Bot 2013-04-28 12:13:47 UTC Related URL: https://gerrit.wikimedia.org/r/61256 (Gerrit Change If6573901bafd8599a01979bcbff89e52b51e4ffd) Gerrit Notification Bot 2013-04-28 23:09:45 UTC Related URL: https://gerrit.wikimedia.org/r/61287 (Gerrit Change If38014bde908bb202dc63213fa21a72b52a6a42d) Derk-Jan Hartman 2013-04-28 23:18:14 UTC Still to do of things discussed in this ticket: Revive: https://gerrit.wikimedia.org/r/4060 the euro symbols \emph (only works if not in italics mode) \vbox https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax.js&diff=484537614&oldid=484068957 Would like to know if it was intentional here that Nageh has changed the CSS to only affect MathML mtext elements (and no longer the HTML-CSS class .mtext) https://en.wikipedia.org/w/index.php?title=User%3ANageh%2FmathJax.js&diff=484537614&oldid=484068957 nageh 2013-04-30 13:08:30 UTC DJ, you are right in your assumption that "mtext" targets MathML elements rather than the standard HTML/CSS output. For the latter, the "mtextFontInherit" property is being used, which has become available since MathJax version 2.0. nageh 2013-04-30 13:43:46 UTC I shall add that the mtext font scaling to 80% is almost certainly not correct, however, if omitted Firefox does not adapt the font size to that of the surrounding text (which is scaled to 80% by default). As MathML support has become available on Chrome, which correctly implements the scaling, I would suggest removing that part from the mtext CSS definition, i.e., simply use styles: { "mtext": { "font-family": "sans-serif ! important" } } Peter Krautzberger 2013-04-30 15:12:19 UTC Hi nageh, MathJax can be configured to scale differently per browser. But it shouldn't have to be scaled at all (except for personal preference), so if you can produce a bug report, we'd appreciate it. https://github.com/mathjax/mathjax/issues Oh, also: Chrome is currently removing the MathML code in Blink; the Chrome devs have hinted at a complete re-write. So MathML in Chrome is probably dead for the foreseeable future. Peter. nageh 2013-04-30 16:08:32 UTC Hi Peter, thank you for your input. Yes, scaling should not be needed at all, and I was actually just going to suggest to ignore that custom styling altogether (there is no point in introducing something that breaks on one or the other browser). I will file a bug report. Interesting note on Chrome's MathML support. What a pity. nageh 2013-05-02 11:36:19 UTC As for \textsf and \texttt, these commands are correctly being defined but are missing styling support in the HTML/CSS output jax of TeX-AMS-texvc HTML.js when mtextFontInherit is being enabled. This is fixed by amending the source with matches for "monospace" and "sans-serif" font variants. In explanation, you'll find code statements like if(!j.weight&&j.mathvariant.match(/bold/)){j.weight="bold"} Replicate these for the other font variants. I may do so in my user script sources so you can copy the changes. But possibly this would be a useful change for the official MathJax sources as well... Peter, what do you think? Derk-Jan Hartman 2013-05-02 13:07:01 UTC Nageh, ah, I was wondering why \texttt and \textsf weren't working for me indeed. Derk-Jan Hartman 2013-05-02 14:08:09 UTC @Nageh, sounds like a follow up of https://github.com/mathjax/MathJax/issues/226 right ? that \sf \tt fontinherit issue ? nageh 2013-05-02 15:21:02 UTC Yes, indeed, the same solution is needed to fix \sf and \tt. Peter Krautzberger 2013-05-02 15:43:40 UTC yes, this sounds like a bug in MathJax. Derk-Jan Hartman 2013-05-14 07:47:05 UTC upstream: https://github.com/mathjax/MathJax/issues/459 Gerrit Notification Bot 2013-05-19 17:15:17 UTC https://gerrit.wikimedia.org/r/4060 (Gerrit Change I51a72c378d3496b147eb87418c24718026306c13) | change RESTORED [by TheDJ] Gerrit Notification Bot 2013-05-19 18:06:53 UTC Related URL: https://gerrit.wikimedia.org/r/64530 (Gerrit Change I20ac8af2253835c20431585d9caebe4066606879) Gerrit Notification Bot 2013-05-19 18:13:41 UTC Related URL: https://gerrit.wikimedia.org/r/64531 (Gerrit Change If2fb219a1191caeecb0ed04bf0c0105621725dce) physikerwelt 2013-10-22 19:17:10 UTC https://gerrit.wikimedia.org/r/#/c/90748/ This solves all the issues since it passes only real TeX to MathJax. I home someone will review this one day. I even wrote a guide how to verify the correctness of the change. http://www.formulasearchengine.com/Verify%20texvc%20light Even that did not help... Maybe I should add some wrong spacing so that potential reviewers find something to comment. Richard Morris 2014-03-11 14:13:41 UTC The treatment of \color is very different between the two versions. Looking at http://docs.mathjax.org/en/latest/tex.html#color we see "The \color command in the core TeX input jax is not standard in that it takes the mathematics to be colored as one of its parameters, whereas the LaTeX \color command is a switch that changes the color of everything that follows it." This means in texvc we have $foo \color{Red} Red text \color{Black} bar$ which breaks mathjax. You can fix things so that $foo {\color{Red}{Red text}} bar$ which seems to work with both mathjax and also texvc. Note it needs the text to be colored in { } so mathjax recognises the first argument, and also { } around the whole thing so texvc does not get confused. A further issue is the colors supported. LaTeX's colors are Apricot Aquamarine Bittersweet Black Blue BlueGreen BlueViolet BrickRed etc But MathJax only support colors with CSS names. So no Apricot, Bittersweet, BlueGreen or BrickRed. You can see the difference by viewing https://en.wikipedia.org/wiki/Help:Formula#Color with the different renderers. There is MathJax color.js extension. Which makes the color handling like the texvc and I think define latex rather than CSS colors. It might be worth installing. Derk-Jan Hartman 2014-04-05 16:09:48 UTC @Richard The color problems were introduced in late january: Bug 63574. It's just that nobody noticed until today. Please try to open separate tickets for separate issues. Link them with dependencies if you want to group them. nageh 2014-04-06 11:34:46 UTC Oddly enough it seems that I was the first to notice despite I'm not being active anymore. Gerrit Notification Bot 2014-10-12 15:13:01 UTC Change 4060 abandoned by Physikerwelt: Math: Adding archaic greek symbols from teubner set Reason: conserved in https://www.mediawiki.org/wiki/Extension:Math/Roadmap#Bugs_and_Roadmap_and_random_thoughts_.28From_the_discussion_page.29 feel free to restore if any development on that is planned https://gerrit.wikimedia.org/r/4060 physikerwelt 2014-10-19 18:18:30 UTC This should be fixed within MathJax. I discussed with Peter that we don't want to have a WMF specific version of MathJax to reduce the maintenance effort. See https://github.com/mathjax/MathJax/pull/882

 Note You need to log in before you can comment on or make changes to this bug.


text:
How do you find the local extrema of a function?

First find the critical numbers. $c$ is a critical number for $f$ if and only if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.


text:
# 3 7 14 23 36 49 66 83 What number comes next?

Probably not easy:

3 7 14 23 36 49 66 83

What number comes next?

Last edited:

These are the smallest 8 roots of a 9th degree polynomial whose 9th root is 123456789

It's not quite that arbitrary.

Here's a hint: if I continued the sequence for 20 more terms then it would be much easier to figure out. In fact, from looking at the 10th and 20th terms you could get it almost immediately.

Last edited:
Are you sure 67 is correct for slot 7? I have a formula that gives all of them except it generates 66 for the 7th term.

EDIT: To check, my formula generates 104 for the 9th term.

Last edited:
I agree with Moo, and assume the next term to be 104 [in white].

You're right, I did make a mistake. 104 is the answer! Congratulations.

Last edited:
is someone going to show how to do it?

Explanation in white: It's the sum of the squares 1, 4, 9, ... with the corresponding primes 2, 3, 5, ...


text:
Signs of expressions challenge

Practice finding the signs of the answers to complex multiplication and division problems.

Problem

What is the sign of start fraction, s, start superscript, 67, end superscript, divided by, t, start superscript, 9, end superscript, end fraction, divided by, left parenthesis, 3, s, dot, start fraction, s, divided by, 4, t, end fraction, right parenthesis when s, is less than, 0 and t, is greater than, 0 ?
Please choose from one of the following options.


text:
# comm.MER System object

Package: comm

Measure modulation error ratio

## Description

The Modulation Error Ratio (MER) `MER` object measures the signal-to-noise ratio (SNR) in digital modulation applications. You can use these types of measurements to determine system performance in communications applications. For example, determining whether a DVB-T system conforms to applicable radio transmission standards requires accurate MER measurements. The block measures all outputs in decibels.

To measure modulation error ratio:

1. Define and set up your MER object. See Construction.

2. Call `step` to measure the modulation error ratio according to the properties of `comm.MER`. The behavior of `step` is specific to each object in the toolbox.

## Construction

`H = comm.MER` creates a modulation error ratio (MER) System object™, `H`. This object measures the signal-to-noise ratio (SNR) in digital modulation applications.

`H = comm.MER(Name,Value)` creates an MER object, `H`, with each specified property set to the specified value. You can specify additional name-value pair arguments in any order as (`Name1`,`Value1`,...,`NameN`,`ValueN`).

## Properties

 `MinimumMEROutputPort` Enable minimum MER measurement output When you set this property to `true`, the `step` method outputs minimum MER measurements. The default is `false`. The `step` method outputs the minimum MER output as the minimum MER value measured in the current input frame. `XPercentileMEROutputPort` Enable X-percentile MER measurement output When you set this property to `true`, the `step` method outputs X-percentile MER measurements. The default is `false`. The X-percentile MER measurements persist. These measurements are based on all the input frames since the last reset. `XPercentileValue` X-percentile value Specify the X-percentile value (as a percentage) that the object uses to calculate the $x$th percentile of the MER measurements. The default is `95`. Set this property to a real scalar value between `0` and `100`, inclusive. This property can have a data type of double, single, or integer. This property applies when you set the `XPercentileMEROutputPort` property to `true`. The $x$-th percentile is the MER value above which $x$% of all the computed MER values lie. `SymbolCountOutputPort` Enable symbol count output When you set this property to `true`, the `step` method outputs the number of accumulated symbols that have been used to calculate the $x$-Percentile MER measurements since the last reset. The default is `false`. This property applies when you set the `XPercentileMEROutputPort` property to `true`.

## Methods

 clone Create MER measurement object with same property values getNumInputs Number of expected inputs to step method getNumOutputs Number of outputs from step method isLocked Locked status for input attributes and nontunable properties release Allow property value and input characteristics changes reset Reset states of MER measurement object step Measure modulation error ratio

## Examples

collapse all

### Measure MER of a noisy 16-QAM modulated signal

```% Create a rectangular QAM modulator and an AWGN Channel hMod = comm.RectangularQAMModulator(16); hAWGN = comm.AWGNChannel('NoiseMethod',... 'Signal to noise ratio (SNR)',... 'SNR', 20, 'SignalPower', 10); % Create an MER object, output minimum and 90-percentile MER, and symbol % count hMER = comm.MER('MinimumMEROutputPort', true, ... 'XPercentileMEROutputPort', true,'XPercentileValue', 90,... 'SymbolCountOutputPort',true); % Generate modulated symbols and add noise refsym = step(hMod, randi([0 15], 1000, 1)); rxsym = step(hAWGN, refsym); % Calculate measurements [MERdB,MinMER,PercentileMER,NumSym] = step(hMER,refsym,rxsym) ```
```MERdB = 20.1071 MinMER = 11.4248 PercentileMER = 16.5850 NumSym = 1000 ```

## Algorithms

This object implements the algorithm, inputs, and outputs described on the MER Measurement block reference page. The object properties correspond to the block parameters.


text:
# Transpose of Matrix

## Transpose of the Matrix

The transpose of a matrix is an important phenomenon in the matrix theory. Let us understand the concept of the transpose of a matrix. If I say roughly then the process of taking transpose of a matrix is something equivalent to changing the rows of the matrix into columns and columns of the matrix into rows. If $A$ is a matrix then its transpose is denoted by $A^T$ or $A^t$ or $A'$.

Consider the following example of matrix and its transpose.

 \begin{aligned} A&=\left[\begin{matrix}1&7&2&9\\2&0&0&5\\1&9&8&3\end{matrix}\right]_{3\times4} \end{aligned}

The transpose of matrix $A$ is

 \begin{aligned} A^T&=\left[\begin{matrix}1&2&1\\7&0&9\\2&0&8\\9&5&3\end{matrix}\right]_{4\times 3} \end{aligned}

Here we can observe that first row of the matrix $A$ is first column of the matrix $A^T$; second row of the matrix $A$ is second column of the matrix $A^T$; and third row of the matrix $A$ is third column of the matrix $A^T$; second thing order of the matrix $A$ is $3\times 4$ while the order of the matrix $A^T$ get reversed and it is $4\times 3$.

Now let us give formal definition of the transpose of the matrix.

## Definition of Transpose of the Matrix

Let $A$ be a matrix then the transpose of the matrix $A$ is the matrix obtained on interchanging all the rows of $A$ into columns and vice versa.

More precisely, if

 \begin{aligned} A&=\left[\begin{matrix} a_{11}&a_{12}&\ldots&a_{1i}&\ldots&a_{1j}&\ldots&a_{1n}\\ a_{21}&a_{22}&\ldots&a_{2i}&\ldots&a_{2j}&\ldots&a_{2n}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{i1}&a_{i2}&\ldots&a_{ii}&\ldots&a_{ij}&\ldots&a_{in}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{j1}&a_{j2}&\ldots&a_{ji}&\ldots&a_{jj}&\ldots&a_{jn}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{m1}&a_{m2}&\ldots&a_{mi}&\ldots&a_{mj}&\ldots&a_{mn}\\ \end{matrix}\right]_{m\times n} \end{aligned}

then the transpose of matrix $A$ is

 \begin{aligned} A^T&=\left[\begin{matrix} a_{11}&a_{21}&\ldots&a_{i1}&\ldots&a_{j1}&\ldots&a_{n1}\\ a_{12}&a_{22}&\ldots&a_{i2}&\ldots&a_{j1}&\ldots&a_{n2}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1i}&a_{2i}&\ldots&a_{ii}&\ldots&a_{ji}&\ldots&a_{ni}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1j}&a_{2j}&\ldots&a_{ij}&\ldots&a_{jj}&\ldots&a_{nj}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1m}&a_{2m}&\ldots&a_{im}&\ldots&a_{jm}&\ldots&a_{nm}\\ \end{matrix}\right]_{n\times m} \end{aligned}

In other words if $A=[a_{ij}]_{m\times n}$, then the transpose of the matrix $A$ is $A^T=[a_{ji}]_{n\times m}$.

Thus on taking transpose of the matrix, we find two major changes; first one is $(i,j)^{th}$ entry of the matrix $A$ becomes $(j,i)^{th}$ entry of the matrix $A^T$ and second one is order of the transposed matrix get reversed than the order of the original matrix.

## Examples of Transpose of Matrix.

### Transpose of a square matrix is a square matrix

 \begin{aligned} A&=\left[\begin{matrix}1&2&9\\2&0&5\\1&9&8\end{matrix}\right]_{3\times 3} \end{aligned}

Transpose of matrix $A$ is

 \begin{aligned} A^T&=\left[\begin{matrix}1&2&1\\2&0&9\\9&5&8\end{matrix}\right]_{3\times 3} \end{aligned}

### Transpose of a row matrix is a column matrix

 \begin{aligned} B&=\left[\begin{matrix}3&2&7&8\end{matrix}\right]_{1\times 4} \end{aligned}

Transpose of matrix $B$ is

 \begin{aligned} B^T=\left[\begin{matrix}3\\2\\7\\8\end{matrix}\right]_{4\times 1} \end{aligned}

### Transpose of a column matrix is a row matrix

 \begin{aligned} C&=\left[\begin{matrix}1\\0\\3\\8\end{matrix}\right]_{4\times 1} \end{aligned}

Transpose of matrix $C$ is

 \begin{aligned} C^T&=\left[\begin{matrix}1&0&3&8\end{matrix}\right]_{1\times 4} \end{aligned}

### Transpose of a matrix is it self

 \begin{aligned} D&=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3} \end{aligned}

Transpose of matrix $D$ is

 \begin{aligned} D^T&=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3}\\ &=D \end{aligned}

## Transpose of a matrix is negative of the original matrix

 \begin{aligned} E&=\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right] \end{aligned}

Transpose of matrix $E$ is

 \begin{aligned} E^T&=\left[\begin{matrix}0&-2&5\\2&0&-3\\-5&3&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right]\\ &=-E \end{aligned}

The process of taking transpose produce a new matrix from the given matrix. Then there are some natural questions like; How the process of taking transpose behaves it applies more than one time? How the process of taking transpose behaves with sum and product of matrices? The following properties of transpose of matrices answer these questions.

## Properties of Transpose of Matrices

### Property 1. $(A^T)^T=A$

This property says that the transpose of the transpose is original. Let us verify this property by examples.

#### Verification of Property 1 by Examples

1. Example 1

 \begin{aligned} A&=\left[\begin{matrix}2&-7&0&6\\1&8&5&-2\\7&8&-1&6\end{matrix}\right] \end{aligned}

The transpose of matrix $A$ is

 \begin{aligned} A^T&=\left[\begin{matrix}2&1&7\\-7&8&8\\0&5&-1\\6&-2&6\end{matrix}\right]\\ \end{aligned}

Now, the transpose of $A^T$ is

 \begin{aligned} (A^T)^T&=\left[\begin{matrix}2&-7&0&6\\1&8&5&-2\\7&8&-1&6\end{matrix}\right]\\ &=A \end{aligned}

1. Example 2

Consider a matrix $B$ as follows:

 \begin{aligned} B&=\left[\begin{matrix}1&0&8\\3&1&7\\7&8&-1\end{matrix}\right] \end{aligned}

The transpose of matrix $B$ is

 \begin{aligned} B^T&=\left[\begin{matrix}1&3&7\\0&1&8\\8&7&-1\end{matrix}\right] \end{aligned}

Now, transpose of the $B^T$ is

 \begin{aligned} (B^T)^T&=\left[\begin{matrix}1&0&8\\3&1&7\\7&8&-1\end{matrix}\right]\\ &=B \end{aligned}

### Property 2. $(k\;A)^T=k\;A^T$

This property says that the process of taking the transpose does not affect the scalar which is multiplied to the matrix. Let us verify this property by examples.

#### Verification of Property 2 by Examples

Consider a $2\times 2$ matrix $A$ as

 \begin{aligned} A&=\left[\begin{matrix}2&-7\\1&8\end{matrix}\right]\\ \Longrightarrow\;2A&=2\left[\begin{matrix}2&-7\\1&8\end{matrix}\right]\\ &=\left[\begin{matrix}4&-14\\2&16\end{matrix}\right] \end{aligned}

Now

 \begin{aligned} (2A)^T&=\left[\begin{matrix}4&-14\\2&16\end{matrix}\right]^T\\ &=\left[\begin{matrix}4&2\\-14&16\end{matrix}\right]\\ &=2\left[\begin{matrix}2&1\\-7&8\end{matrix}\right]\\ &=2A^T\end{aligned}

1. Example 2

 \begin{aligned} B &=\left[\begin{matrix}1&5&-2\\7&-1&6\end{matrix}\right]\\ \end{aligned}

 \begin{aligned} 5B &=5\left[\begin{matrix}1&5&-2\\7&-1&6\end{matrix}\right]\\ &=\left[\begin{matrix}5&25&-10\\35&-5&30\end{matrix}\right] \end{aligned}

Now

 \begin{aligned} (5B)^T&=\left[\begin{matrix}5&25&-10\\35&-5&30\end{matrix}\right]^T\\ &=\left[\begin{matrix}5&35\\25&-5\\-10&30\end{matrix}\right]\\ &=5\left[\begin{matrix}1&7\\5&-1\\-2&6\end{matrix}\right]\\ &=5B^T \end{aligned}

### Property 3. $(A+B)^T=A^T+B^T$

This property says that the first you add two matrices and then take the transpose of its addition or first you take the transpose of the matrices and then take its addition these two things are same. i.e. the process of addition and taking transpose are commutative. Let us verify this property by an example.

#### Verification of Property 3 by an Example :

Let

 \begin{aligned} A&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right],\\ B&=\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right] \end{aligned}

 \begin{aligned} A^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]\\ B^T&=\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right] \end{aligned}

Then

 \begin{aligned} A^T+B^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]+\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\\ &=\left[\begin{matrix}5&3\\0&9\\7&5\end{matrix}\right] \end{aligned}

Also

 \begin{aligned} A+B&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right]+\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\\ &=\left[\begin{matrix}5&0&7\\3&9&5\end{matrix}\right] \end{aligned}

Thus

 \begin{aligned} (A+B)^T&=\left[\begin{matrix}5&3\\0&9\\7&5\end{matrix}\right]\\ &=A^T+B^T \end{aligned}

Property 3 can be generalized for finitely many matrices and it can be put mathematically as

$$(A_1+A_2+\cdots+A_n)^T=A_1^T+A_2^T+\cdots+A_n^T$$

$$i.e. \;\;\left(\sum\limits_{i=1}^nA_i\right)^T=\sum\limits_{i=1}^nA_i^T$$

### Property 4 $(A-B)^T=A^T-B^T$

This property is analogue to the property 3, it says that you can deal the process of taking transpose of the subtraction of two matrices is same as the case of addition. Let us verify this property by an example.

#### Verification of Property 4 by an Example :

Let

 \begin{aligned} & A=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right], B=\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\;\\ \Longrightarrow & A^T=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]\text{ and } B^T=\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\end{aligned}

Then

 \begin{aligned} A^T-B^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]-\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\\ &=\left[\begin{matrix}-1&-1\\-14&7\\1&-1\end{matrix}\right] \end{aligned}

Also

 \begin{aligned} A-B&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right]-\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\\ &=\left[\begin{matrix}-1&-14&1\\-1&7&-1\end{matrix}\right] \end{aligned}

Thus

 \begin{aligned} (A-B)^T &=\left[\begin{matrix}-1&-1\\-14&7\\1&-1\end{matrix}\right]\\ &=A^T-B^T. \end{aligned}

### Property 5. (Shock Shoes Property) $(A\cdot B)^T=B^T\cdot A^T$

This property says that the transpose of product of two matrices is product of transposed matrices in the reverse order.

This property is also know as shock shoes property. Why? Let us try to justify why. You think LHS i.e. $(A\cdot B)^T$ be a process of put on the shock and shoes and the RHS i.e. $B^T\cdot A^T$ be a process of take off shock and shoes. Then during the put on first we put on the shock and then shoes while during taking off it first we took off shoes and then shock. That is the process get reversed and exactly this happened in the case of takeing transpose of the product of two matrices.

#### Verification of Property 5 by an Example

Let

 \begin{aligned} A&=\left[\begin{matrix}2& 5& 3\\-1& 3& 6\end{matrix}\right] \end{aligned}

 \begin{aligned} B&=\left[\begin{matrix}2 &7& 9\\3& -2& 0\\ 1& 0 &8\end{matrix}\right] \end{aligned}

Now,

 \begin{aligned} A^T&=\left[\begin{matrix}2&-1\\5&3\\3&6 \end{matrix}\right] \end{aligned}

 \begin{aligned} B^T&=\left[\begin{matrix}2&3&1\\7&-2&0\\9&0&8\end{matrix}\right] \end{aligned}

Then

 \begin{aligned} B^T\cdot A^T&=\left[\begin{matrix}2&3&1\\7&-2&0\\9&0&8\end{matrix}\right]\left[\begin{matrix}2&-1\\5&3\\3&6 \end{matrix}\right]\\ &=\left[\begin{matrix}22&13\\4&-13\\42&39\end{matrix}\right] \end{aligned}.

Also

 \begin{aligned} A\cdot B &=\left[\begin{matrix}2& 5& 3\\-1& 3& 6\end{matrix}\right] \left[\begin{matrix}2 &7& 9\\3& -2& 0\\ 1& 0 &8\end{matrix}\right]\\ &=\left[\begin{matrix}22&4&42\\13&-13&39\end{matrix}\right] \end{aligned}

Thus

 \begin{aligned} (A\cdot B)^T &=\left[\begin{matrix}22&13\\4&-13\\42&39\end{matrix}\right]\\ &=B^T\cdot A^T \end{aligned}
Property 5 can be generalized for finitely many matrices and it can be put mathematically as

 $$(A_1\cdot A_2\cdot\cdots\cdot A_n)^T=A_1^T\cdot A_2^T\cdot\cdots\cdot A_n^T$$

 $$i.e.\;\;\left(\prod\limits_{i=1}^nA_i\right)^T=\prod\limits_{i=1}^nA_i^T$$

## Summary of Properties of Transpose of Matrices

• $(A^T)^T=A$
• $(k\;A)^T=k\;A^T$
• $(A+B)^T=A^T+B^T$
• $(A_1+A_2+\cdots+A_n)^T=A_1^T+A_2^T+\cdots+A_n^T$ i.e.$\left(\sum\limits_{i=1}^nA_i\right)^T=\sum\limits_{i=1}^nA_i^T$.
• $(A-B)^T=A^T-B^T$
• $(A\cdot B)^T=B^T\cdot A^T$
• $(A_1\cdot A_2\cdot\cdots\cdot A_n)^T=A_1^T\cdot A_2^T\cdot\cdots\cdot A_n^T$ i.e. $\left(\prod\limits_{i=1}^nA_i\right)^T=\prod\limits_{i=1}^nA_i^T$.

Milan holds Ph.D. Degree in Mathematics, his area of research is Banach Algebra and Harmonic Analysis, a branch of pure mathematics. He has cleared NET and GPSC (Gujarat Public Service Commission) with first rank. Milan enjoys teaching and problem solving in Mathematics. He has good hand on Real Analysis, Complex Analysis, Linear Algebra, Abstract Algebra and Differential Equation. Apart from teaching Milan spends his time to guide the students for various competitive examinations in Mathematics like JAM, CSIR NET-JRF, GSET, GATE, NBHM, TIFR etc. More than 29 students had cleared such type of examinations under his guidance. He has teaching experience of more than 13 years in the field of Mathematics at graduate level in Engineering and Science College.


text:
# For which energetic reasons are ions blocked by cell membranes?

The permeability of cell membranes is given by the following diagram from the book "Biochemistry":

The reason that ions get blocked can be explained chemically: The ions favor interactions with the polar parts, but cannot undergo favorable interactions with the non-polar parts of the membrane.

However, in my biophysics lecture, the reason for this phenomena was explained by the different relative permittivities $$\epsilon_r$$ in the Coloumb-Potential:

$$\Phi (r) = \frac{q}{4\pi\epsilon_0\epsilon_rr}$$

Outside of the cell, water has a relative permittivity of 80, inside the cell, $$\epsilon_r$$ is around 4 or 5. I'm not sure how to understand this energetic considerations, but I would formulate it in the following way:

I calculate the electric potentials of an ion with respect to the polar components outside of the cell membrane. Because of the difference in the relative permittivities, the potential outside of the membrane will be much less than the potential on the inside of the cell. The work the ion would have to do to overcome this energy barrier is too high, thus the ion gets blocked.

Do you think my formulation is correct?


text:
# Shortest Path in Layerwise Complete Graph

Consider a directed $$k$$ partite graph $$G$$ with a source node $$s$$ and a sink node $$t$$. Each vertex in the graph is labeled with a positive integer value. Both the source and sink are labeled with $$0$$. There are directed edges from the $$s$$ to all vertices in partition $$0$$. There are directed edges from all vertices in partition $$k$$,, ie, the last partition to $$t$$. $$G$$ is layerwise complete that is, in addition to the above, there is a directed edge from all vertices in partition $$i$$ to all vertices in partition $$i+1$$, ie, adjacent partitions form complete bipartite graphs. An edge between any two nodes is labeled with a negative integer value or with $$0$$.

There could also be directed edges between nodes in partitions $$m$$ and $$n$$ where $$1 \leq m, n \leq k, m \lt n, n \neq (m+1)$$, and labeled with a negative value. This is depicted in the following figure where there are two edges between partition $$1$$ and partition $$3$$ indicated with dashed lines. Non adjacent partitions do not form complete bipartite graphs. They only comprise edges as indicated. In the figure, the edge labels are shown with negative values, wherever edges are not labeled with values, they are labeled with $$0$$ by default.

Consider all paths from $$s$$ to $$t$$. The length of this path would be calculated as the addition of all the integer labels of the nodes on this path and the addition of the edge labels if there is an edge between any pair of vertices in this path. For example, the length of the path taking the first vertex from each partition would thus be $$0 + 50 + 30 + 50 + 0 -8$$, where the negative value of $$-8$$ corresponds to the edge between nodes labeled with $$50$$ and $$50$$. Similarly, if we took the first vertex from partition $$1$$, the last vertex from partition $$2$$ and the first vertex from partition $$3$$, the length of this path would be $$0 + 50 + 40 + 50 + 0 - 6$$, where the $$-6$$ value corresponds to the edge between the first vertex in partition $$1$$ and partition $$2$$. The $$0$$ values correspond to the source and sink vertices.

Find the shortest path from $$s$$ to $$t$$ taking exactly one vertex from each $$k$$-partition where the length would be calculated as described.

Attempt : I converted the node weights for a pair $$(a, b)$$ into edge weights where each edge weight would be the sum of the node value of $$b$$ deducting the edge cost between $$a$$ and $$b$$. This works for edges between successive partitions and produces the correct shortest path in cases where the edges are only between such. However, if I use the same approach for edges between different non successive layers it doesn't work. What would be an algorithm to find the shortest path and the transformation?

• What have you tried? Where did you get stuck? May 19, 2019 at 7:24
• Are you aiming for an algorithm or a hardness proof? May 19, 2019 at 7:30
• @YuvalFilmus I have updated what I have tried, I am looking for an algorithm if indeed this does not turn out to be a hard problem. If so, possible pointers towards an algorithm May 19, 2019 at 8:14

Your problem is NP-hard, by reduction from clique. Given a graph $$G=(V,E)$$ and a number $$k$$, we construct a graph with $$k$$ internal layers, each containing $$|V|$$ vertices. All vertices in a layer are connected to all vertices in the following layer. The weight of an edge is $$0$$ if the corresponding edge is not in $$G$$, and $$-1$$ otherwise. We also add shortcut edges corresponding to edges in $$G$$ with weight $$-1$$. The graph $$G$$ has a $$k$$-clique iff the shortest path in the new graph has length $$-k(k-1)/2$$.
• Does the number $k$ represent the number of layers in the original graph? May 30, 2019 at 9:58
• Okay, I get it. I get the intuition behind clique, essentially the original graph in the problem can be viewed as complete, with all non displayed edges having edge weight $0$. How does the shortest path length come up to $-k(k-1)$? Jun 1, 2019 at 5:37


text:
# Hotter than the sun

Which gives off more heat, you, or the Sun? If we're talking about net heat, it's obviously the Sun, but if we're talking about heat output per unit mass, it's not so clear. Calculate the ratio of your body's heat output per unit mass to that of the Sun.

Details and Assumptions

• The mass of the Sun is $$2\times10^{30}$$ kg
• The radius of the Sun is $$7\times 10^8$$ m
• The surface temperature of the sun is $$5780$$ K, and it is a perfect blackbody.
• Your daily consumption of food is 2,000 kilocalories, and you radiate it all as heat.
• You weigh 75 kg.
×


text:
# Beyond Billions: Threatened Bats are Worth Billions to Agriculture

Insect-eating bats provide pest-control services that save the U.S. agriculture industry over $3 billion per year, according to a study released today in the journal Science. However, scientists with the U.S. Geological Survey, University of Pretoria in South Africa, University of Tennessee, and Boston University who contributed to the study warn that these valuable animals are at risk: Bat populations are declining due to fatalities associated with White-Nose Syndrome and wind turbines, which could lead to significant economic losses on U.S. farms. Paul Cryan, USGS scientist and an author of the report, discusses these findings. ## Details Date Taken: Location Taken: US ## Transcript Marisa Lubeck: Welcome and thanks for listening to this episode of USGS CoreCast. I’m Marisa Lubeck. Insect-eating bats provide pest-control services that save the U.S. agriculture industry over$3 billion per year, according to a study released today. However, scientists with the U.S. Geological Survey, University of Pretoria in South Africa, University of Tennessee, and Boston University who contributed to the study warn that these valuable animals are at risk. Bat populations are declining due to fatalities associated with White-Nose Syndrome and wind turbines, which could lead to significant economic losses on U.S. farms.

I’m here with USGS scientist, Paul Cryan, an author of the report, to discuss these findings. Welcome, Paul.

Paul Cryan: Thank you for having me.

Marisa Lubeck:  Paul, you've estimated that the value of pest control services provided by bats in the US range from a low of $3.7 billion to a high of$53 billion. This is a huge range. What variables or pest control services make up these numbers and why is the range so large?

Paul Cryan:  Our analysis is a simple first approach to this question of how much cost savings do bats provide to agriculture in the United States. To do that we needed some information on the pest control services that bats provide, some monetary estimates.

So, the values we used to estimate the pest control services are based on some studies that have been done by my colleagues, Tom Kunz, Gary McCracken, and they're co-authors on this paper. They've spent about a decade or more working down in Texas with a species called the Brazilian Free-Tailed bat. They spent a lot of time to really get good numbers of how much those bats saved cotton farmers on a per acre basis under a number of different scenarios. This is the only system in the United System or anywhere in the world that we know about where these kinds of data have been gathered on bats in agriculture.

So those per-acre savings that we extrapolated from our analysis are based on three different scenarios. That's why we see such a wide range of values. One of them is $172 per acre that the bat saved farmers. That's in situations where the pest species they're quantifying, the cotton bollworm, is a major agriculture pest. That high dollar scenario area per acre is in years when the conditions favor growth of this crop pest, when the farmers lose more crops to it and they have to apply more pesticides. Those things are expensive. The low scenario that we extrapolated is in years when conditions don’t favor the pest species, the farmer loses less crops to the pest and there’s less pesticide application. That's why we came up with this final value of$3 to \$53 billion.

This is a very simple analysis that we think is an important conversation to start scientifically and we're hoping that other scientists follow up, check our work, add to this conversation and that we can learn a lot more about what we think is an under-appreciated value of bats to agriculture.

Marisa Lubeck:  How many bats are lost to White-nose Syndrome and wind turbines per year and how long will it take for population declines to begin having a noticeable effect on the agriculture economy?

For example, since its emergence in New York State, at what rate is White-nose Syndrome spreading across the country?

Paul Cryan:  White-nose Syndrome was first observed in Upstate New York during the winter of 2007 and 2008. Since then, it spread rapidly and we just heard about another state affected, Ohio. The fungus that now is known to cause these diseases moved to as far west as Oklahoma and at this point, after four winters, five winters, we're looking at 16 states and three Canadian provinces being affected.

The mortality caused by White-nose is unlike anything we've ever seen before in bats and in the northeastern United States where bat populations have been monitored for a long time. White-nose is a disease of hibernating bats and so we're seeing these hibernating colonies decline precipitously.

The number of bats dying at wind turbines in the United States, we don't have a good feel for that yet, although fatality rates we've seen at certain wind facilities are higher than any fatality rates we've seen in this particular species in the past. There's something strange happening with bats and wind turbines. We don't know why this particular species are so susceptible and why they're showing up in these surprising numbers.

The harder question that you asked is how do we know how long it will take to see these population declines having a noticeable effect, and that's where we run into a lot of uncertainty. You know, in our paper we suggest it could be as soon as four or five years in the heavy agricultural regions of the central continent. Perhaps in the next few years in the northeast where White-nose syndrome had a substantial influence on bat population

Marisa Lubeck:  What USGS-specific research contributed to the findings in this study?

Paul Cryan:  I was involved in this analysis because my expertise is with bat fatalities at wind turbines. I'm also involved in the response to White-nose syndrome in trying to tease apart the pathology of the disease, understand behaviors of bats with the disease and hopefully work towards solutions of minimizing the effects of some of these.

Marisa Lubeck:  What haven't you and your research partners accounted for in these economic estimates?

Paul Cryan: That's a good question. We only accounted for the actual savings to farmers from the reduced use of pesticides because the bats are there eating this pest insect, and the crop isn't damaged because the bats are there.

What we did not include is anything regarding forest agro-ecosystems. We know that bats suppress insects in forest systems as well.

We didn't account in any of these systems for the evolved resistance of insect pests to pesticides that are used against them or these genetically modified crops they can evolve around.

We didn't account for the downstream affects of pesticides on the environment. And we only extrapolated our values based on one insectivorous bat species feeding on one-type of agriculture pest in one region. The combined effects of multiple species feeding on many different agricultural pests particularly in agricultural landscapes with lots of different kinds of insects pests. There's a lot of things we couldn’t account for, so we feel that our estimates are likely conservative on that account.

Marisa Lubeck:  One of your partner research universities is the University of Pretoria in South Africa. Do these numbers vary across this country and do you think they’re comparable in other countries?

Paul Cryan:  In our analysis, the numbers suggest that the economic importance of bats are greater in the Midwest and Great Plains states where we have tremendous agricultural output and we know that there are a lot of bats occurring in those regions. Those are also the areas where White-nose syndrome is moving and also there’s a lot of wind energy development happening.

So the bottom line is that the economic value of bats varies across the country and we don't have good numbers for anywhere else, although there's over 1,100 species of bat that occur across the globe. Insectivorous bats are present in most temperate zones of the world and in the tropics. These are flying insect predators. They're at the top of these food chains that involve insects. So, pretty much wherever you are on Earth, I would say it's safe to reason that there's bats influencing the food web that those insects are in.

Marisa Lubeck:  What will it take to decrease these losses? Is it possible to stop or at least slow the spread of White-nose syndrome and is there such a thing as bat-friendly wind energy?

Paul Cryan:  That's what's on everyone's mind. What we really wanted to focus on is that these are a tremendously valuable asset we have in our world. They're out there eating bugs. It's a natural safe and free-pest control service and it's tremendously valuable to us, economically.

The losses that we're seeing due to White-nose Syndrome and wind turbines are unprecedented but there's still hope. We don't have a quick and easy solution to White-nose Syndrome but we're starting to understand the disease. We're looking for ways to at least minimize its impact and try to break the disease cycles if that's possible.

The same with wind energy, there's a lot of work going on. We work with the industry. We work with all kinds of other scientists to try and to find out why these bats are running into wind turbines. Is it an attraction? Is it some other factor that we can do something about at this point? So hopefully, our analysis will let people know that these are very valuable creatures and they deserve our help.

Marisa Lubeck:  Paul, what sort of value, ecological or otherwise, do bats have beyond their economic worth?'

Paul Cryan: That's a very good question and bats play key roles in all of earth's ecosystems. In tropical areas and arid deserts of the Southwest, they pollinate flowers. The large desert cacti that we have down in the Sonoran desert that characterize that landscape in people's minds, bats play a very important role in the pollination of those plants.

The agave, as you go farther into the tropics, bats disperse seeds.

So there's an ecological importance no matter where on earth you are except for the Polar ice caps.

The less tangible to most people value of bats is that these are unique creatures that have millions of years of evolution behind them. These are mammals just like us yet they can fly and find their way around in complete darkness using sound, and how cool is that?

Marisa Lubeck:  Certainly. Thanks for speaking with us today, Paul.

Paul Cryan:  Yes. Thanks. Bye-bye.

Marisa Lubeck:  This CoreCast is a product of the US Geological Survey, Department of the Interior.

I'm Marisa Lubeck. Thanks for listening.


text:
# How do I align my initial conditions?

I'm trying to align my initial conditions (bottom two lines) to match the top ODEs.

 \begin{empheq}[left=\empheqlbrace]{align} \nonumber
\begin{rcases}
\dv{x}{t} = v \\
\dv{v}{t} = \frac{-kx}{m}+Asin(\omega t) - bv
\end{rcases}
\\
\begin{rcases} \nonumber
x(0)=-2 \\
v(0)=1
\end{rcases}
\end{empheq}

• Welcome to TeX.SX! Please make your code compilable (if possible), or at least complete it with \documentclass{...}, the required \usepackage's, \begin{document}, and \end{document}. Please especially make clear where rcases and \dv are defined. – leandriis Feb 10 '18 at 16:26

edit: partly considered @marmot comment

like this?

it is not perfect ...

\documentclass{article}
\usepackage{empheq}

\begin{document}
\begin{empheq}[left=\empheqlbrace]{align}\nonumber
&  \begin{rcases}
\dfrac{\mathrm{d}x}{\mathrm{d}(t)}  = v                                 \\
\dfrac{\mathrm{d}v}{\mathrm{d}(t)}  = \dfrac{-kx}{m}+A\sin(\omega t) - bv \\
\end{rcases}          \\
&  \begin{rcases}\notag
x(0)   = -2\hphantom{,k+A\sin(\omega t) - bv } \\
v(0)   = 1
\end{rcases}
\end{empheq}
\end{document}

note: since mwe is not known, i anticipated the \dv{x}{t} can be own defined command, there fore i emulate it with suspected definition. however, such command is defined in physics package (not used in my mwe).

• The macro \dv{x}{t} very likely comes from the physics package, see my answer below, and it could be that you replaced by something that does not resemble the output of the macro. – user121799 Feb 10 '18 at 17:05
• @marmot, could be ... but this should tell as op with mwe, if it is from phisics package or it is defined by her/him :-) – Zarko Feb 10 '18 at 17:38
• I agree that the OP should provide an MWE. It's just that the output makes perfect sense if one uses the physics package (and becomes even better if one replaces sin by \sin ;-). Looks like I ws 2 seconds ahead of you ;-) – user121799 Feb 10 '18 at 17:44

I don't know if there is a better way than doing it "by hand".

\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{physics}
\begin{document}
\begin{empheq}[left=\empheqlbrace]{align} \nonumber
\begin{rcases}
\dv{x}{t} &= v \\
\dv{v}{t} &= \frac{-kx}{m}+A \sin(\omega t) - bv
\end{rcases}
\\
\begin{rcases} \nonumber
x(0)&=-2 \\
v(0)&=\makebox[0pt][l]{1}\hphantom{\frac{-kx}{m}+A \sin(\omega t) - bv}
\end{rcases}
\end{empheq}
\end{document}

With drcases, as suggested by Bernard

\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{physics}
\begin{document}
\begin{empheq}[left=\empheqlbrace]{align} \nonumber
\begin{drcases}
\dv{x}{t} = v \\
\dv{v}{t} = \frac{-kx}{m}+A \sin(\omega t) - bv
\end{drcases}
\\
\begin{drcases} \nonumber
x(0)=-2 \\
v(0)=\makebox[0pt][l]{1}\hphantom{\frac{-kx}{m}+A \sin(\omega t) - bv}
\end{drcases}
\end{empheq}
\end{document}

Or with TikZ.

\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{physics}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {\strut};}
\begin{document}
\begin{empheq}[left=\empheqlbrace]{align*} \nonumber
\dv{x}{t} &= v&\tikzmark{1} &\\
\dv{v}{t} &= \frac{-kx}{m}+A \sin(\omega t) - bv&\tikzmark{2} &
\\
x(0)&=-2 &\tikzmark{3} &\\
v(0)&=1&\tikzmark{4} &
\end{empheq}
\tikz[remember picture, overlay,decoration={brace}]{%
\draw[decorate,thick] (1.north) --
(2.south) node[right,midway] {(initial conditions on velocities)};
\draw[decorate,thick] (3.north) --
(4.south) node[right,midway] {(initial conditions on positions)};}
\end{document}

• Maybe with drcases? – Bernard Feb 10 '18 at 18:16
• @Bernard I agree that this will look better (d stands for \displaystyle, right?) but I am not sure if that will lead to a simpler solution to the alignment problem. – user121799 Feb 10 '18 at 18:21
• It was only for the general look. The derivatives are much smaller than the initial conditions, which looks odd, semantically. – Bernard Feb 10 '18 at 18:29
• @Bernard I fully agree, what is even worse is that the alignment &s cause bad spacing. But since I knew that Zarko will post an answer in 2 seconds, I had to hurry up ;-) . OK, more seriously, I think one may need to create a new environment for this, or, alternatively, draw the braces with TikZ (which introduces its own problems). Anyway, thanks for your input, I'll be happy to use drcases in a revision of my answer. – user121799 Feb 10 '18 at 18:33
• @Zarko Thanks for the edit. I do not understand how that happened. There was also a wrong figure, which I fixed. Really strange. – user121799 Feb 10 '18 at 19:32


text:
Measuring Activity

The formula for Activity of a radioactive substance is

$\frac{dN}{dt}=A=λN$.

If we have an initial number $N(0)$ of some Radionuclide, which has a halflife of, say, 12 hours, is there any particular reason why our time interval for measuring the number of remaining Radionuclides should be close to the halflife of the substance?

Is there any reason, for example, that I shouldn't take measurements of the number of remaining nuclei after every 10 seconds or 10 minutes? Is there a concrete reason that I should really measure the number of remaining nuclei every few hours or so?

The reason I ask is that as we experimentally can't measure the change in number of Nuclei after an infinitesimally small time interval dt, the above equation becomes:

$\frac{ΔN}{Δt} = λN = \frac {ln(2)}{T_{1/2}}N$

Where $Δt$ is the time interval after which each measurement of $N$ is taken.

Does this in some way say that $T_{1/2}$ should be close to $Δt$?

To my mind the obvious answer is no. I've been told, however, that the answer is yes. I don't understand why.

3 Answers

There is no reason why you can't measure the rate frequently. However, in order to estimate the half life, you need to see a change in the rate of decay. How long you need to measure for, and how far apart you need to change your measurements, depends on the number of decays per second that you observe as well as the required accuracy.

For example, if you have 10,000 decays per second, the standard deviation of a one second measurement will be 100 ($\sqrt{N}$) - so you get the answer to 1% accuracy. If you measure for just one second, and measure again for a second one hour later, you would see (for your huypothetical 12 hour half life) a change of about 6%. With an error of 1% for both the initial and final measurement, you would have an uncertainty of about 1.4% in a value of 6% - which is almost a 25% uncertainty in your estimate of the half life.

By waiting longer, and measuring the individual time points with greater precision, you can improve your estimate. It is a good rule of thumb, when making any measurement, to do a simple error propagation analysis like the above in order to understand the accuracy of your result.

If your activity measurement is more precise (you count more decays - either because the initial activity level is higher, or you count for longer) then it becomes possible to space your measurement intervals more closely. But whatever you do, your choices will affect your uncertainty. The rule of thumb to wait "roughly as long as the half life" takes some of the guess work out. It is not a substitute for a formal error analysis, but a shortcut.

you should resolve the differential equation of N. If you do that you'll get that $$A = A_0 e^{\lambda (t-t_0)},$$ where $A_0$ is the activity at time $t_0$. From there you can obtain the value of $\lambda$ from two measurements of the activity whatever the time interval.

Firstly, the activity formula is in fact:

$$-\frac{dN}{dt}=A=λN,$$

because $\frac{dN}{dt}<0$.

[...] is there any particular reason why our time interval for measuring the number of remaining Radionuclides should be close to the half-life of the substance?

No and that's not how it's done in practice. $\lambda$ and the half-life are determined by measuring activity a number of times using appropriate time intervals. Of course these intervals have to be in reasonable proportion to the expected half-life. For an expected half-life of about $12\:\mathrm{h}$, $12$ activity measurements at hourly intervals would be a good starting point.

Bear in mind that $\lambda$ and the half-life are determined by statistical analysis and as a general rule a larger number of data points will lead to higher confidence levels $P$ of the calculated observables.

The activity differential equation solves to:

$$\ln\frac{N(t)}{N_0}=-\lambda t,$$

Or:

$$\ln\frac{A(t)}{A_0}=-\lambda t.$$

So from a logarithmic plot of $\frac{A(t)}{A_0}$ versus $t$, $\lambda$ is derived from the gradient of the plot.


text:
Question

# The lines $$3x+y=0$$ may be perpendicular to one of the lines given by $$ax^2+2hxy+by^2=0$$.

Solution

## The line $$3x+y=0$$ may be perpendicular to one of the line given by $$a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$$Auxiliary equation$$=b{ m }^{ 2 }+2hm+a=0\longrightarrow (1)\\ 3x+y=0$$ slope$$({ m }_{ 1 })=-3$$ Hence slope  of the line perpendicular to given line.$$\therefore$$ Slope of one of the lines represented by the joint  equation is $$\cfrac { 1 }{ 3 } .$$ Now $$\cfrac { 1 }{ 3 }$$ must be one of the roots of equation$$b{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+2h\left( \cfrac { 1 }{ 3 } \right) +a=0\\ b\cfrac { 1 }{ 9 } +\cfrac { 2h }{ 3 } +0$$$$b+6h+9a=0\quad$$(The required condition)Maths

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text:
# \show for Lua commands

TeX (and all newer engines) has the primitive \show which helps me probe macros, and most of the time is enough to understand/remember the implementation (rather than having to go back to the source code).

Is there something similar for Lua commands? For instance, Heiko Oberdiek has written replacements for some pdfTeX primitives in LuaTeX, and I'd like to get more information about them using some hypothetical \showluatexcmd macro.

\input pdftexcmds.sty
\catcode@=11
\show\pdf@strcmp
%\showluatexcmd{oberdiek.pdftexcmds.strcmp}
\bye

-

I seriously doubt that there is something similar for Lua, but you can do the following to find out where the function is defined (to get to the source code).

\directlua{ require("showluatexcmd") }
\def\showluatexcmd#1{\directlua{showluatexcmd.show(#1)}}

\input pdftexcmds.sty
\catcode@=11
\show\pdf@strcmp
\showluatexcmd{oberdiek.pdftexcmds.strcmp}
\bye

and the file showluatexcmd.lua:

module(...,package.seeall)

function show( fun )
local debuginfo = { "source","linedefined","what" }
for _,v in ipairs(debuginfo) do
texio.write_nl(string.format("%20s=%q", v, debug.getinfo(fun)[v]))
end
end

One could extend this probably to directly load the function and display it. But in most cases this won't help as there are a lot of functions that are short and call other functions.

A sample output is

          source="@/opt/texlive2011/texmf-dist/scripts/oberdiek/pdftexcmds.lua"
linedefined="46"
what="Lua" )

-
there is also "lastlinedefined", so you could potentially print a portion of the source file .. –  Taco Hoekwater Feb 1 '12 at 8:36
I was hoping for something similar to Python's dis module. Actually, luac -l may help. Also, what does "what" mean in the debug info? –  Bruno Le Floch Feb 1 '12 at 14:10
@BrunoLeFloch a) what is Python's dis module? b) What describes where the function is defined: a "C" function or a "Lua" function (I remember seeing other things written there) –  topskip Feb 1 '12 at 14:29
@Patrick: sorry, the dis module is a disassembler: dis.dis(myfunction) will print (pseudo-)assembly code for myfunction. It seems luac -l does that too, but on compiled Lua files (see the well-hidden link in my previous comment), so I'm not sure whether it could be used for LuaTeX. –  Bruno Le Floch Feb 1 '12 at 14:39


text:
# KO   Tuesday July 9 - 19:00

## Poster Session

Claudio Arancibia-Ibarra - Spatial patterns in a diffusive modified Holling-Tanner predator-prey model [Abstract]

Swaroop Nandan Bora - Wave interaction with a semi-porous cylindrical storage tank [Abstract]

Gaëlle Brunet - Computation of Killing vector fields on compact manifolds [Abstract]

Stephen Coombes - Understanding sensory induced hallucinations: from neural fields to amplitude equations [Abstract]

Nelida Črnjarić-Žic - Data-driven predictions of dynamical systems in healthcare [Abstract]

Jaydev Dabas - Novel mesh free scheme for solving the inverse problem of heat conduction [Abstract]

Veronica Danesi - On the continuation of degenerate periodic orbits in nearly integrable Hamiltonian systems: normal form approach and possible applications [Abstract]

Ivan Dražić - 1-D flow of a $p$-th power viscous micropolar and heat-conducting fluid [Abstract]

Dominik Engl - An analytical discussion of polycrystalline structures in single-slip crystal plasticity [Abstract]

Timothy Faver - Singular perturbations and nanopteron traveling waves in nonlinear lattices [Abstract]

Valery Gaiko - Limit cycle bifurcations in multi-parameter polynomial dynamical systems [Abstract]

Rajib Haloi - Weighted norm inequality for maximal function [Abstract]

Christian Hamster - Noisy Patterns [Abstract]

Bastian Hilder - Stability of fast fronts in a Ginzburg-Landau equation with additional conservation law [Abstract]

Daniel Hill - Localised radial spots on the free surface of a ferrofluid [Abstract]

Victor Hissink Muller - A biofilm growth model - A two-fold degenerate reaction-diffusion system [Abstract]

Stefanie Hittmeyer - Existence of Blenders in a Hénon-Like Family: Geometric Insights from Invariant Manifold Computations [Abstract]

Sachiko Ishida - Boundedness, asymptotic behaviour vs. finite-time blow-up for quasilinear Keller-Segel model [Abstract]

Olfa Jaibi - Mathematical Tales of Fairy Circles [Abstract]

Mia Jukic - Dynamics on Planar Lattices [Abstract]

Young-Ho Kim - An Existence of the solution to neutral stochastic functional differential equations [Abstract]

David Kok - Stability and bifurcations of homoclinic stripe patterns near a saddle-node bifurcation [Abstract]

Sergey Kornev - On some versions of the coincidence degree theory [Abstract]

Chiharu Kosugi - Mathematical model for rotational motion of shape memory alloy ring partially within hot water [Abstract]

Sanjeev Kumar - A novel chaos based approach in conjunction with Pairing Function for Image Encryption [Abstract]

Matthias Lienert - Multi-time wave equations [Abstract]

Hueili Lin - Existence of solutions for the perturbed fourth-order Kirchhoff type elliptic equations [Abstract]

Alun Lloyd - After the Honeymoon, the Divorce: Unexpected Outcomes of Disease Control Measures [Abstract]

Emil Loevbak - An asymptotic preserving multilevel Monte Carlo method for particle based simulation of kinetic equations [Abstract]

Zhouqian Miao - Oscillations in a cAMP signaling model for cell aggregation -- a geometric analysis [Abstract]

Viktor Moskvin - Topology of saddle atoms-bifurcations for the billiards in non-convex domains [Abstract]

Wojciech Mydlarczyk - Nonuniqueness of solution for a sytem of nonlinear Volterra type integral equations [Abstract]

Hiroaki Niikuni - On the eigenvalues in the spectral bands for carbon nanotubes with impurities [Abstract]

Isamu Ohnishi - Hysterical Bifurcation Structure of modified Asakura-Honda model and its application to Circadian Rhythm and Biological Nitrogen fixation in a cell of Cyanobacteria [Abstract]

Zdeněk Opluštil - Oscillation and non-oscillation criteria for certain two-dimensional system of non-linear ordinary differential equations [Abstract]

Artur Prugger - Stability and Nonlinear Waves in Damped Driven Rotating Shallow Water Models [Abstract]

Giulio Romani - Unexpected behaviour of fundamental solutions of general higher-order elliptic operators [Abstract]

Benoit Sarels - Calcium in the egg is traveling, but is it a traveling wave? [Abstract]

Mustafa Sayli - Synchrony: From Franklin Bells to Brain Dynamics [Abstract]

Tomoharu Suda - Characterization of Filippov’s definition of the sliding vector field [Abstract]

Haruyoshi Tanaka - Perturbed finite-state Markov systems with holes and Perron complements of Ruelle operators [Abstract]

Petr Tomasek - Stability region analysis of a linear delay system with two parameters [Abstract]

Dennis Ulbrich - Dynamics and topological entropy of 1D-Greenberg-Hastings cellular automata [Abstract]

Takeo Ushijima - On traveling wave solution for a diffusive simple epidemic model with a free boundary [Abstract]

Vladimir Vasilyev - On some discrete elliptic boundary value problems [Abstract]

Viktoria Vedyushkina - The foliation on the Lens space as Liouville foliation of the integrable billiards. [Abstract]

Konstantin Vorushilov - Jordan-Kronecker invariants of semidirect sums of semisimple Lie algebras with a commutative ideal [Abstract]

Jichen Yang - Turing instability and energy estimates for reaction-subdiffusion equations [Abstract]

Alexander Zlotnik - Analysis of a finite element method for measure-valued optimal control problems governed by the 1D wave equation [Abstract]


text:
# Equation with multiplication of a matrix by a column vector

How do I solve this matrix equation?

$$\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 4\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix}.$$

I know each line represents an equation but what should I do with the zero in the first row, second column?

• Typesetting is your friend! – David H Apr 6 '14 at 12:19

First, multiply the left hand side. You can treat this as two matrices. One is 2x3 (2 rows one column), and one is 3x1 (3 rows one column). When multiplying these matrices the end results would be a matrix whose size is 2x1.

so, multiplying row 1 by column 1, and then row 2 by column 1 gives us

\begin{bmatrix} 1\cdot x_1 + 0\cdot x_2 + 3\cdot x_3\\ 0\cdot x_1 + 1\cdot x_2 +4\cdot x_3 \end{bmatrix}

and so

$$\begin{bmatrix}1\cdot x_1 + 3\cdot x_3\\ 1\cdot x_2 +4\cdot x_3\end{bmatrix}=\begin{bmatrix} 5\\ 7\end{bmatrix}$$

See more about multiplying: Multiplying a matrix by a column vector

You can then say that for the left hand side to be equal to the right hand side you need all the elements in the matrices to be equal. Both matrices are of size 2x1, and you get two simple equations.

$$x_1 + 3\cdot x_3 = 5$$ $$x_2 + 4\cdot x_3 = 7$$

You have a system of equations with infinite solutions. I assume you need to find the general solution:

$$(x_1 =5-3x_3, x_2=7-4x_3, x_3)$$

Hints :
1/ multiply out the left hand side, this should be a vector. Then equate this with the vector on the right hand side.
Each component has to be the same on LHS and RHS which gives three equations with two unknowns

2/ Compute the inverse of the matrix and left multiply by it. EDIT : I didn't see your equation is not square so forget this hint for your equation.

3/ Use echelon forms

4/ Expect to have a degree of freedom. If this is the case, write $x_2$ in terms of $x_1$ for example.


text:
Radiation conditions at the top of a rotational cusp in the theory of water-waves
ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Volume 45 (2011) no. 5, p. 947-979

We study the linearized water-wave problem in a bounded domain (e.g. a finite pond of water) of ${ℝ}^{3}$, having a cuspidal boundary irregularity created by a submerged body. In earlier publications the authors discovered that in this situation the spectrum of the problem may contain a continuous component in spite of the boundedness of the domain. Here, we proceed to impose and study radiation conditions at a point $𝒪$ of the water surface, where a submerged body touches the surface (see Fig. 1). The radiation conditions emerge from the requirement that the linear operator associated to the problem be Fredholm of index zero in relevant weighted function spaces with separated asymptotics. The classification of incoming and outgoing (seen from $𝒪$) waves and the unitary scattering matrix are introduced.

DOI : https://doi.org/10.1051/m2an/2011004
Classification:  76B15,  35J25,  35P99
Keywords: linear water-wave problem, cuspidal domain, radiation condition, scattering matrix
@article{M2AN_2011__45_5_947_0,
author = {Nazarov, Sergey A. and Taskinen, Jari},
title = {Radiation conditions at the top of a rotational cusp in the theory of water-waves},
journal = {ESAIM: Mathematical Modelling and Numerical Analysis - Mod\'elisation Math\'ematique et Analyse Num\'erique},
publisher = {EDP-Sciences},
volume = {45},
number = {5},
year = {2011},
pages = {947-979},
doi = {10.1051/m2an/2011004},
zbl = {1267.76013},
mrnumber = {2817552},
language = {en},
url = {http://www.numdam.org/item/M2AN_2011__45_5_947_0}
}

Nazarov, Sergey A.; Taskinen, Jari. Radiation conditions at the top of a rotational cusp in the theory of water-waves. ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Volume 45 (2011) no. 5, pp. 947-979. doi : 10.1051/m2an/2011004. http://www.numdam.org/item/M2AN_2011__45_5_947_0/

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