# $A$ is the power set of set $C$, and $S$ is the relation on $A$ defined by $x S y$ iff $x \subseteq y$.

The following is an exercise from my textbook:

Let $C$ be a set and let $A=\mathcal{P}(C)$, the set of all subsets of $C$. Let $S$ be the relation on $A$ defined by $x S y$ iff $x \subseteq y$. Then $S$ is a partial order relation on $A$.
(a) Find the least and greatest element of $A$.
(b) Let $B=\{x \in A$: $x \ne ∅\}$. Suppose $C$ has two or more elements. Show that $B$ has no least element.

I'm confused about the definition of least and greatest element as applied to this exercise.

From the my textbook, least element is defined as such:

Let $(A,\leq)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.

Now, referring back to the exercise, $A$ consists of sets, not merely numbers; thus, to say that some set $x_1$ is less than or equal to some other set $x_2$ doesn't make sense. We don't relate sets in that way do we? (Unless, we're talking about order? But even so, I don't think the exercise is asking us to consider that.)

Now, part (b) leads me to believe that a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$. Is that correct?

• Note in your definition $(A,\leq)$ is just a partially ordered set. There is no reference to numbers. You need to find the $S$-least element of $A$ and $B$. – James Aug 19 '15 at 23:04
• The exercise is indeed abour inclusion as an order relation . As for part (b) what you believe is right. – Bernard Aug 19 '15 at 23:05

The notation of $\leq$ is often used in mathematics as a general partial order. Not just the usual ordering on the natural/integers/rationals/reals. But in this context, $\leq$ is really just $\subseteq$.

So you are being asked to find the maximum and minimum elements of $(\mathcal P(C),\subseteq)$.

• Okay, so I'm right to say a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$? – Jordan Miller Aug 19 '15 at 23:13
• Yes. That is the case here. And note that in the definition of least element, $\leq$ is said to be a partial order, not anything else. – Asaf Karagila Aug 19 '15 at 23:15
• I must get clarification. The following in italics is also an exercise in my textbook: $(A,\leq)$ is a partially ordered set. That each finite non-empty subset of $A$ has a $maxB$ implies $(A,\leq)$ is totally ordered. Then does $\le$ refer again to some general partial order relation? – Jordan Miller Aug 19 '15 at 23:27
• Yes, a general partial order. Mathematics is much much much much more than just real numbers. – Asaf Karagila Aug 19 '15 at 23:34

In your case, the least element is an element of $A$, hence a subset of $C$ that is contained in all subsets of $C$. (Apply your definition in the second highlighted paragraph with $A=\mathcal{P}(C)$ and $B=A$), and so this set, I guess, is the empty set. Now, for the greatest element of $A$, what is the subset of $A$ that contains any subset of $A$?

• So, I am right to think that a least element, with respect to the set $A$, is a set such that it is a subset of each set in $A$? Does that make sense why I was a bit confused. To me, the relation of $\le$ is not the same as $\subseteq$ – Jordan Miller Aug 19 '15 at 23:11
• $\leq$ is just a symbol! – mich95 Aug 19 '15 at 23:12

Okay, I know that what follows basically repeats what responders have been trying to tell me, but hitherto, I was still misunderstanding the notation of $\le$. Thus, I merely post this answer as an expression of my understanding (finally!).

I refer to the definiton:

Let $(A,\le)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.

The $\le$ in the ordered pair $(A,\le)$ does not refer to the relation of less than or equal to, but rather serves as a "variable" for some partial order relation on A. With this in mind, the condition for each $x \in B$, $c \le x$ doesn't mean that $c$ must be less than or equal to $x$, but rather that $c$ must be in relation $\le$ to $x$.

Now, the partial order relation specified in the above exercise is the relation of set inclusion. Then, by definition, the least element of $A$ must be some set in $A$ such that it is a subset of each set in $A$.


# In the simplest qualitative terms what is a differential equation?

#### find_the_fun

##### Active member
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve

So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "

This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.

#### Opalg

##### MHB Oldtimer
Staff member
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve

So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "

This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.
In a differential equation, the idea is to find $y$ as a function of $x$, given some information about $y$ and its derivatives. The very simplest example of a differential equation might be something like the equation $\frac{dy}{dx} = 2x$. You can easily solve that by integrating it, to find that the solution is $y=x^2$ plus a constant of integration. But suppose that the equation is slightly more complicated, for example $\frac{dy}{dx} = x +y$. How would you solve that to find $y$ as a function of $x$? A course in differential equations will show you how to do that.

In case you are wondering, the solution to that equation is $y = -x-1 + ce^x$, where $c$ is a constant.

#### HallsofIvy

##### Well-known member
MHB Math Helper
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve

So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "

This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.
A differential equation is an equation, a derivative is NOT! Okay, having got that off my chest, I think I understand your point. "Contains derivatives", etc. is not sufficient. The crucial point is that a differential equation contains derivatives of some unknown function. Yes, "$$x^2$$" can be thought of as the derivative of $$\frac{1}{3}x^3$$ but just having $$x^2$$ in an equation does NOT make it a "differential equation". To be a differential equation, the equation must contain something like $$\frac{dy}{dx}$$, $$\frac{\partial y}{\partial t}$$, $$\frac{d^2y}{dx^2}$$, etc., where y is the "unknown" function.

More generally, a "functional equation" is an equation that gives us some information about a function. "f(x+ 1)= f(x)" is a functional equation. $$\frac{d^2y}{dx^2}+ 2\frac{dy}{dx}+ y= x^2$$ is a special kind of a functional equation (since it gives information about the function y) called a "differential equation" because that information is actually about the derivatives of the function y.

#### find_the_fun

##### Active member
Glad I asked. I had the first lecture today and the prof skipped over the definition of a differential equation (in fairness it was a substitute prof).

What is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?

#### Opalg

##### MHB Oldtimer
Staff member
What is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?
Yes. You will often see the abbreviations ODE and PDE for ordinary/partial differential equation.

#### find_the_fun

##### Active member
I'm writing myself notes now and here's the first:

Differential equation: a functional equation that relates an unknown function to its derivatives

#### zzephod

##### Well-known member
An ordinary differential equation is an equation of the form (or that can be rewritten in the form):

$$\displaystyle \large f(x,y,y',...,y^{(n)})=0$$

A partial differential equation is similar but with partial derivatives with repect to the variables appearing (including mixed partials).

.

Last edited:


# What's $\frac{dy}{dx}$ of function $y=\frac 12{\sin x}$?

$(1)\quad$ How to differentiate $y=\frac 12 \sin x$?

I know that $\frac{dy}{dx}$ of $y=\sin x$ is easy to calculate: $\frac{dy}{dx} = \cos x$. But what if there is a coefficient preceding before it?

$(2)$ Another question is: Does the following equality of functions hold? $$\frac{x(x^2+1)}{x^2+1}\overset{?}{=}x$$

Please explain this because I know that function $\dfrac{x\cdot x}{x}$ does not equal function $x$ (the first function is not defined at $0$ but the second one is defined at $0$).

But the question I asked, the domain of the function can be any number according to what I think because the denominator cannot be zero, it's always a positive number $x^2+1$.

• To your second question, since $x^2+1$ is never $0$ (at least in the real numbers) $\frac{x^2+1}{x^2+1}=1$ and $\frac{x(x^2+1)}{x^2+1} = x$ for all $x$. So yes, the functions are equal. – Thomas Andrews Sep 11 '14 at 13:10
• Do you mean $\frac{1}{2\sin x}$ or $\frac{1}{2}\sin x$? – Thomas Andrews Sep 11 '14 at 13:11
• I mean 1/2 (sin x) – off99555 Sep 11 '14 at 13:13
• One should avoid different questions in a same question on MSE. – user37238 Sep 11 '14 at 13:16
• This is my first time using this website, I will be aware next time. – off99555 Sep 11 '14 at 13:18

Question $(1):\quad$ Given $y = \dfrac 12 \sin x$:

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac 12 \sin x\right) = \frac 12\cdot \frac{d}{dx}(\sin x) = \frac 12 \cos x$$

$$y = a f(x)\implies \frac{dy}{dx} = \frac{d}{dx}(a f(x)) = af'(x)$$ for all constants $a$.

Question $(2):$

$$\frac {x(x^2 + 1)}{x^2 + 1 } = x \quad \forall x \in \mathbb R$$

This happens to be the case in this example because there are no real values of $x$ at which the left-hand side is undefined. Specifically, as you note, $x^2 + 1>0$ for all $x$, and hence the function is defined everywhere. So we may cancel the common factor $x^2 + 1$ without changing the function in any way.

In the case of $f(x) = \frac{x^2}{x}$, which you refer to, here we do have problems with simply canceling a common factor of $\,x.\,$ Specifically, $\,\dfrac{x^2}{x}\,$ is undefined at $x = 0$, whereas the function $g(x) = x$ is defined everywhere, so the functions are not equivalently defined, i.e., they are not equivalent functions.

• I would say that the first response is missing one sign "=" -> ambiguity. – georg Sep 11 '14 at 14:08
• Thanks, @georg. I didn't catch that until you mentioned it.. – amWhy Sep 11 '14 at 14:19

For the first part of your question, you can remove constants when differentiating:

$$\frac{\mathrm{d}}{\mathrm{d}x}\big(a\cdot f(x)\big)=a\frac{\mathrm{d}}{\mathrm{d}x}f(x)$$

You can prove this with the product rule by finding $a\frac{\mathrm{d}}{\mathrm{d}x}f(x)+f(x)\frac{\mathrm{d}}{\mathrm{d}x}a$ & noting that the derivative of a constant is $0$. It implies that your derivative is $\frac{\mathrm{d}}{\mathrm{d}x}\big(\frac{1}{2}\sin(x)\big)=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)$.

For the second part of the question, it is valid to cancel terms in a function but just be aware that the function's domain should stay the same. If the original function is undefined at a number, it should stay that way. As Thomas Andrews points out, $x^2+1$ is never $0$ for real $x$ so you can effectively cancel the $\frac{(x^2+1)}{(x^2+1)}$ term.

• I got the edit wrong - he wanted $y=\frac{1}{2}\sin x$. Sorry. – Thomas Andrews Sep 11 '14 at 13:14
• @ThomasAndrews Ah, thanks for letting me know :) No problem by the way, I've done it before too. – Jam Sep 11 '14 at 13:14
• Thank you very much for your help. I would like to vote both of you up but I can't. My reputation is not high enough to do that. Both answers are very useful for me. But I don't understand what setting a as a function of x mean. Can you show me please? – off99555 Sep 11 '14 at 14:25
• @off99555 Glad I could help. What I meant by "setting it as a function" was that, by using the product rule, we can find the derivative of $f(x)\cdot g(x)$ as $f(x)g'(x)+g(x)f'(x)$. If we say that $g(x)$ is a constant $a$, we could find the derivative of $a\cdot f(x)$. Since the derivative of any constant is $0$, it shows us that we can take out the constant. So $\frac{d}{dx}(\frac{1}{2}\sin(x))=\frac{1}{2}\frac{d}{dx}\sin(x)$. Does that make sense? – Jam Sep 11 '14 at 14:45
• Sure. That make sense! That's what I'm looking for, the fundamental explanation. – off99555 Sep 17 '14 at 15:46


A Typist Charges Rs. 145 for Typing 10 English and 3 Hindi Pages, While Charges for Typing 3 English and 10 Hindi Pages Are Rs. 180. Using Matrices, Find the Charges of Typing One English and One Hindi Page Separately. - Mathematics

A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?

Solution

Let charges for typing one English page be Rs. x.

Let charges for typing one Hindi page be Rs.y.

Thus from the given statements, we have,

10x+3y=145

3x+10y=180

Thus the above system can be written as,

[(10,3),(3,10)][(x),(y)]=[(145),(180)]

⇒ AX = B, where, A=[(10,3),(3,10)],x=[(x),(y)] " and " B = [(145),(180)]

Multiply A-1 on both the sides, we have,

A-1 x AX = A-1B

⇒ IX = A-1B

⇒ X = A-1B

Thus, we need to find the inverse of the matrix A.

We know that, if P=[(a,b),(c,d)] " then " P^(-1) = 1/(ad-bc)[(d,-b),(-c,a)]

Thus, A^(-1)=1/(10xx10-3xx3)[(10,-3),(-3,10)]

= 1/(100-9)[(10,-3),(-3,10)]

=1/91[(10,-3),(-3,10)]

Therefore, X=1/91[(10,-3),(-3,10)][(145),(180)]

=1/91[(10xx145-3xx180),(-3xx145+10xx180)]

=1/91[(910),(1365)]

=[(10),(15)]

=>[(x),(y)][(10),(15)]

⇒ x = 10 and y=15

Amount taken from Shyam = 2 × 5 = Rs.10

Actual rate = 15 × 5 =75

Difference amount = Rs.75 – Rs.10 = Rs.65

Rs. 65 less was charged from the poor boy Shyam.

Humanity and sympathy are reflected in this problem.

Concept: Inverse of Matrix - Inverse of a Nonsingular Matrix by Elementary Transformation
Is there an error in this question or solution?


# How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$

Recently I found a claim saying that $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2$$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.

Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$\int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x$$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the Borwein Integral, but there are some subtle differences.

• Don't you want a product, not a summation, in your last question? And, when you write "independent of $y$" don't you mean independent of $x$? – Gerry Myerson Jul 11 '13 at 9:04
• Does $\sin p\sin q=(\cos(p-q)-\cos(p+q))/2$ help? – Gerry Myerson Jul 11 '13 at 9:07

One way is to to this by residues. Another way to integrate once by parts to get $$I=\int_0^{\infty}\frac{b\sin ax\cos bx+a\cos ax \sin bx}{x}dx,$$ then to use the formula $2\sin \alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right) $$\displaystyle \int_0^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}\mathrm{sgn}(y).$$ This gives \begin{align}I&=\frac{\pi}{4}\Bigl[b\,\mathrm{sign}(a+b)+b\,\mathrm{sign}(a-b)+a\,\mathrm{sign}(a+b)+a\,\mathrm{sign}(b-a)\Bigr]=\\ &=\frac{\pi}{4}\left(|a+b|-|a-b|\right), \end{align} For $a,b>0$ the last expression is obviously equal to $\pi\min\{a,b\}/2$.

• Just posted the question as a dupe, such an amazing result. Do you know who first discovered it? – aronp Dec 21 '15 at 21:33
• Since this result is independent of the lower number, presumably this works for the three factor case? – aronp Dec 21 '15 at 21:46
• @aronp No I don't; I totally agree with you that it is quite counterintuitive. – Start wearing purple Dec 21 '15 at 21:47

In several steps:

• Trigonometric relation $$\left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right)=\frac{1-\cos(a+b)x}{2x^2}-\frac{1-\cos(a-b)x}{2x^2}$$
• Dirichlet integral $$\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}\mathrm{sgn}(\alpha)$$
• Integration by parts $$\int_0^\infty\frac{1-\cos(\alpha)t}{t^2}=\alpha\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}|\alpha|$$
• Combine $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x=\frac{\pi}{4}(|a+b|-|a-b|)=\frac{\pi}{2}\min(a,b)$$
• Excellent!${}{}{}$ – Namaste May 20 '14 at 20:14

A very easy way to see this is to use Parseval's theorem for Fourier transforms. In general, Parseval's theorem states that, for two functions $f$ and $g$, each having respective FTs $\hat{f}$ and $\hat{g}$, related by

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$

etc., then

$$\int_{-\infty}^{\infty} dx \, f(x) \bar{g}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \bar{\hat{g}}(k)$$

The FT of $\sin{(a x)}/x$ is given by

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt a \\ 0 & |k| \gt a\end{cases}$$

Similarly,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{b x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt b \\ 0 & |k| \gt b\end{cases}$$

By Parseval, we take the integral of the product of the transforms, which is the product of two rectangles. The product is clearly nonzero over the smaller of the two widths, i.e. $2 \min\{a,b\}$. Thus,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi^2}{2 \pi} 2 \min\{a,b\}$$

Therefore

$$\int_{0}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi}{2} \min\{a,b\}$$

Let $b_{k} >0$ and $a \ge \sum_{k=1}^{n} b_{k}$.

Generalizing the answer HERE, we can use contour integration to quickly show that

$$\int_{0}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx = \frac{\pi}{2} \prod_{k=1}^{n}b_{k}. \tag{1}$$

Under the conditions stated above, the function $$e^{iaz} \prod_{k=1}^{n} \sin \left( b_{k}x \right) = e^{iaz} \prod_{k=1}^{n} \frac{e^{ib_{k}z}-e^{-ib_{k}z} }{2i}$$ is bounded in the upper half-plane.

So by integrating the function $$f(z) = \frac{e^{iaz}}{z} \prod_{k=1}^{n} \frac{\sin \left( b_{k}z \right)}{z}$$ around an indented contour that consists of the real axis and the semiciricle above it, we get (in the limit), $$\text{PV} \int_{-\infty}^{\infty} \frac{e^{iax}}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx - \pi i \, \text{Res} [f(z), 0] = 0,$$ where $$\text{Res}[f(z) ,0] = \lim_{z \to 0}e^{iaz} \prod_{k=1}^{n}\frac{\sin \left( b_{k}z \right)}{z} =1\times \prod_{k=1}^{n} b_{k}.$$

Taking the imaginary parts of both sides of equation leads to the result.

It's sufficient to consider the case $\ds{\bbox[10px,#ffe,border:1px dotted navy] {\ds{a, b \in \mathbb{R}_{\ >\ 0}}}}$.

\begin{align} \mbox{Note that}\ &\int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x = {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{ax} \over ax} \,{\sin\pars{\bracks{b/a}ax} \over \pars{b/a}ax}\,a\,\dd x \\[5mm] = &\ {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,,\qquad\mu \equiv {a \over b} > 0 \label{1}\tag{1} \end{align}

\begin{align} &\left.\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,\right\vert_{\ \mu\ >\ 0} = \int_{-\infty}^{\infty} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx/\mu}\,\dd q}\dd x \\[5mm] = &\ {1 \over 2}\,\pi \int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty}\expo{\ic\pars{k - q/\mu}x} {\dd x \over 2\pi}\,\dd k\,\dd q = {1 \over 2}\,\pi\int_{-1}^{1}\int_{-1}^{1}\delta\pars{k - q/\mu}\,\dd k\,\dd q \\[5mm] = &\ {1 \over 2}\,\pi\int_{-1}^{1}\bracks{-1 < {q \over \mu} < 1}\dd q = \pi\int_{0}^{1}\bracks{q < \mu}\dd q = \pi\braces{\bracks{\mu < 1}\int_{0}^{\mu}\dd q + \bracks{\mu > 1}\int_{0}^{1}\dd q} \\[5mm] = &\ \bracks{a < b}\pi\,{a \over b} + \bracks{a > b}\pi\label{2}\tag{2} \end{align}
With \eqref{1} and \eqref{2}: \begin{align} \int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x & = {1 \over 2}\,\pi\braces{\vphantom{\Large A}\bracks{a < b}a + \bracks{a > b}b} = \bbx{\ds{{1 \over 2}\,\pi\,\min\braces{a,b}}} \end{align}


How is $9^{40}\equiv\ 1 \pmod {100}$?

By Euler's theorem, $a^{\varphi(100)} \equiv\ 1 \pmod {100}$.

We know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.

Since $1\equiv\ 1 \pmod {100}$, how come $9^{40}\equiv\ 1 \pmod {100}$?

I have looked at: Find the last two digits of $9^{{9}^{9}}$,

Find the last two digits of $9^{9^{9}}$

• "We know that the last two digits of $9^{40}$ are non-zero." That doesn't mean neither digit is zero; it just means they can't both be zero. So why do you think it can't be $01$? – Erick Wong Dec 22 '15 at 17:14
• you're right i was thinking only of the the units digit – AkaiShuichi Dec 22 '15 at 17:17
• @pyUser: When a more experienced user edits your post, learn from it, don't roll back to your own improperly formatted version. – Alex M. Dec 22 '15 at 17:18
• @AlexM. can you teach me how to apply edits from one version to another one? or did you just manually MathJax the thing? – gt6989b Dec 22 '15 at 17:19
• You quoted one way of doing it. We have $\varphi(100)=\varphi(25)\varphi(4)=40$. So $9^{40}\equiv 1\pmod{100}$, the last two digits, going righttwards, are $0$ and $1$. – André Nicolas Dec 22 '15 at 17:24

By binomial theorem:

$$(10-1)^{40} = \sum_{i=0}^{40} \binom{40}{i}(-1)^{40-i}10^{i}$$

Modulo $10^2$, you only need to look at $i=0,1$.

So: $$9^{40}\equiv (-1)^{40} + \binom{40}{1}(-1)^{39}\cdot 10 \equiv (-1)^{40}\equiv 1\pmod{100}$$

Note that this works for $9^{10}$, too.

$9^2\equiv\ -19\ (mod100)$

$9^4 \equiv\ 61\ (mod 100)$

$9^8\equiv\ 21 \ (mod 100)$

$9^{10}\equiv\ 1 \ (mod 100)$

$9^{40}\equiv\ 1 \ (mod 100)$

• Better formatting if you write 9^2\equiv -19\pmod{100} yielding $$9^2\equiv -19\pmod{100}$$ – Thomas Andrews Dec 22 '15 at 18:01

Calculate $$9^{10} = 3486784401$$ So $$9^{40} \mod 100 \equiv (9^{10}\mod 100)^4 \mod 100$$ $$\equiv(1)^4 \mod 100 \equiv 1 \mod 100$$

• How did you go about your third step? – AkaiShuichi Dec 22 '15 at 17:20
• $x^y \mod m$ is the same as $(x \mod m)^y \mod m$. You can move the $\mod$ function into multiplies, etc. – amcalde Dec 22 '15 at 17:22
• i was talking about the (1)^4 part – AkaiShuichi Dec 22 '15 at 17:23
• In the first calculation I showed that $9^{10} \equiv 1 \mod 100$. – amcalde Dec 22 '15 at 17:24

Note that $$99^2=9801\equiv 1\pmod{100}\implies99^{40}\equiv 1\pmod{100}\implies9^{40}11^{40}\equiv 1\pmod{100}$$

$$9^{40}\equiv 1\pmod{100}$$

It is straightforward $9^{40} = 81^{20} = 6561^{10} \equiv 61^{10} \pmod{100} = 3721^5 \pmod{100} \equiv\ 21^5 \pmod{100}$
$= 441 \times 441 \times 21 \pmod{100}\equiv\ 41 \times 41 \times 21 \pmod{100}$
$= 35301 \pmod{100} \equiv\ 1 \ \pmod{100}$

and this is just one of the many many routes you could take.

If you want to avoid the equation

$9^{40}=147808829414345923316083210206383297601$,

the question is how low you prefer your numbers to be.

One alternative is using $40=4*2*5$ (and 7 digits for the last operation).

$9^4=61$ (mod 100)

$61^2=21$ (mod 100), and

$21^5=01$ (mod 100)


# Finding coefficients in polynomial given three tangents

I am stuck with a problem I simply cannot solve.

I have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:

The three lines described by the equations

$y_1(x)=-4x-16.5$

$y_2(x)=2x-4.5$

$y_3(x)=6x-16.5$

are all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$

Determine the values of the coefficients a, b & c.

I simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)

Edit: I'm including the way I tried to solve it. I didn't get super far.

Given the polynomial $p(x)$ I know that $p'(x)=2ax+b$

Therefore, the following is true for the three points with x-values of $x_1, x_2$ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:

$-4=2ax_1+b$

$2=2ax_2+b$

$6=2ax_3+b$

That's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.

• Show anyway what you tried. I can give a solution using derivatives, but maybe it is not the solution you want. – N74 May 8 '18 at 20:23
• Welcome to SE! Even if your attempts didn't work, you should absolutely include them in your post - people can advise you better that way! – B. Mehta May 8 '18 at 20:23
• @N74 I have now included my failed solution :) As you can see, I thought of using derivatives as well, but I can't get further. – Mads Clausen May 8 '18 at 20:38
• My maths teacher (I'm in the last year of Danish "high school") gave it to us for tomorrow. He said that it was definitely doable. – Mads Clausen May 8 '18 at 20:42

If we have a parabola $$y = a x^2 + b x + c$$ and a line $$y = mx + d,$$ they are tangent if $$b^2 - 4ac = -4da + 2mb - m^2$$ as then the parabola $$y = a x^2 + (b-m) x + (c -d)$$ has a double root, i.e. is a constant times a square, $$a (x+p)^2$$ Might as well write this: I fixed $$\Delta = b^2 - 4 a c$$ and then had three equations $$-4da + 2mb = m^2 + \Delta$$ by plugging in the values from the three lines $y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\Delta,$ after which we have one equation for $c$

$$66a - 8 b = 16 + \Delta \; ,$$ $$66a + 12 b = 36 + \Delta \; ,$$ $$18a + 4 b = 4 + \Delta \; .$$ Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\Delta.$ Finally, $c = \frac{b^2 - \Delta}{4a}$

Note that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero \begin{eqnarray*} ax^2+(b-m)x+c-d=0. \end{eqnarray*} This will lead to $3$ equations for $a,b,c$ that are easily solved giving

$(a,b,c)=(1/2,1,-4)$.

• Thanks for the answer. I don't quite follow, however: what are the three equations for a, b and c that you mention? – Mads Clausen May 8 '18 at 21:02
• \begin{eqnarray*} (b+4)^2=2a(c+33) \\ (b-2)^2=2a(c+9) \\ (b-6)^2=2a(c+33) \end{eqnarray*} combine the first & the third of these equations & you will rapidly yield a value for $b$. – Donald Splutterwit May 8 '18 at 21:05
• Donald I get different final abc values, although still all rational and easily found... I kept the discriminant $\Delta = b^2 - 4ac \;$ as an unknown until the end, began with three $\Delta = -4da + 2mb - m^2$ or $-4da + 2mb = m^2 + \Delta$ – Will Jagy May 8 '18 at 21:09
• where does $2a(c+n)$ come from? I recall that the discriminant is equal to $b^2-4ac$, which in this case would yield something like $(b-m)^2=4a(c-d)$ I would think it would either be that or $(b-m)^2=2a(2c-2d)$ – Mads Clausen May 8 '18 at 21:12
• I get 1/2, 1 and -4 for a,b,c. 2ax+b = -4 at x=-5; 2 at x = 1 and 6 at x = 5 with x and y values being the same for tangent lines and parabola at these tangent points. – Phil H May 8 '18 at 21:32

Given any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.

Working out the intersection points of the three lines, we can see that the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$ and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$ Let

• $x = x_1$ at the tangent point with $y = y_1(x)$;
• $x = x_2$ the tangent point with $y = y_2(x)$; and
• $x = x_3$ the tangent point with $y = y_3(x).$

Due to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$; moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$ It follows that $$x_3 - x_1 = 2\left(\frac{x_3 + x_2}2 - \frac{x_2 + x_1}2\right) = 2(3 - (-2)) = 10.$$

The tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and $f'(x_3) = y_3'(x) = 6.$ But $f'(x) = 2ax + b,$ so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ and therefore $a = \frac12.$ It is then a relatively straightforward exercise to find the other two coefficients.


# The Product of Primes Between $N$ and $2N$ Compared to $2^N$

While reading some course notes from MIT 18.703 (Modern Algebra) on a primality test, I found this lemma (stated without proof):

Lemma 22.3. The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\geq1$$.

However, one quickly finds that this lemma is false for $$N=8$$. The primes between $$8$$ and $$16$$ are $$11$$ and $$13$$, and $$11\cdot13 = 143 < 256 = 2^{8}$$.

I wondered if there were any more counterexamples, so I decided to write a quick program to test this lemma (the code may be found here). My code is not very optimized, with only very basic parallelism using OpenMP, and so I haven't taken the time to go beyond $$N = 100000$$. With my code, I have found only three counterexamples:

At $$N = 8$$, the product of primes between $$8$$ and $$16$$ is $$143 < 256 = 2^{8}$$.

At $$N = 14$$, the product of primes between $$14$$ and $$28$$ is $$7429 < 16384 = 2^{14}$$.

At $$N = 20$$, the product of primes between $$20$$ and $$40$$ is $$765049 < 1048576 = 2^{20}$$.

My Questions

1. Are these the only counterexamples?

2. If so, how do we prove it?

3. If not, what are some others, and are there infinitely many? -- Answer: There are not infinitely many, though there may still be further counterexamples.

EDIT 1: Thanks to the responses to this post, I have attempted an analysis of this problem. If possible, I would appreciate any feedback, just in case I made an error or if any improvement could be made by a more careful analysis. -- This analysis was flawed and has been retracted.

EDIT 2: Due to the answer posted by jjagmath, the bound has been lowered to 10544111. I will try to run my program to search for counterexamples below this number, but hopefully even tighter analyses come that lower the bound to a more tractable number to search.

EDIT 3: I retract the analysis that I did because of a mistake. The jjagmath analysis still holds.

EDIT 4:

Let $$p_{k}$$ be the $$k$$th prime number, let $$\displaystyle{N_{k} = \frac{p_{k} - 1}{2}}$$, and let $$\displaystyle{f\left(N\right)=\prod_{N\leq p\leq 2N}p}$$.

Proposition. Let $$k > 2$$ be an integer and let $$\nu$$ be an integer such that $$N_{k-1} < \nu\leq N_{k}$$. Then, $$f\left(N_{k}\right)\leq f\left(\nu\right)$$.

Proof. Any prime in the interval $$\left[N_{k},2N_{k}\right]$$ is also in the interval $$\left[\nu, 2\nu\right]$$, and so $$f\left(\nu\right)$$ can only get smaller as $$\nu$$ increases to $$N_{k}$$. QED

Corollary. If any counterexamples above $$20$$ exist, some of them must be of the form $$\displaystyle{N_{k} = \frac{p_{k} - 1}{2}}$$ for some $$k$$.

Observation. All counterexamples currently known are in this form: $$N_{7} = 8$$, $$N_{10} = 14$$, and $$N_{13} = 20$$.

At the time of writing, it is known that all integers $$N\geq 10544111$$ satisfy $$f\left(N\right)\geq 2^{N}$$, and so we need only check $$N_{k} < 10544111$$ ($$k < 698306$$) to see if there are any further counterexamples.

Question. Are there better explicit lower bounds for $$N$$ beyond which the inequality $$\displaystyle{\prod_{N\leq p\leq 2N}p \geq 2^{N}}$$ holds?

If we can find a more computationally tractable bound, say $$N\geq 1299709$$ (so $$k \geq 100000$$ can be ruled out of the search), then I may be able to run (a modified form of) my program to search for possible counterexamples beyond $$N = 20$$. Of course, any improvement is welcome!

EDIT 5:

With Gary's comment, the bound has been tightened to $$N\geq 678407,$$ which should be tractable. I will be running my program overnight and will update with the results!

EDIT 6:

Gary's latest answer has finally finished this problem off! The bound has been tightened to $$N \geq 328$$ via Chebyshev function bounds, and finally to all $$N\geq1$$ except $$N=8,14,20$$ by numerical computation.

This has been a very fun problem to work on, and I thank everyone who was involved! I suppose I will end this post with a revised Lemma 22.3.

Lemma 22.3.$$'$$ The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\geq1$$, except for $$N = 8, 14, 20$$.

EDIT 7:

I went back to my initial analysis where I had made a very trivial error, and it turns out that I would have already had a tractable bound if I decided to look at it again. My analysis seems to show that $$f\left(N\right)\geq 2^{N}$$ for all $$N\geq 1845$$.

Gary's analysis still provides a better bound, but I'm happy to know that I would have eventually solved this with just another look at my own work.

• @BeKind Thanks! Jul 19 at 14:06
• I think asymptotically it might be true. That is, if we say there are approximately $\frac{2N}{\log(2N)} - \frac{N}{\log(N)}$ primes between $N$ and $2N$. Each prime is $\geq N.$ We might be able to show that for large enough $M,\$ the product of these primes is greater than $N^{\frac{2N}{\log(2N)} - \frac{N}{\log(N)}}$ which is greater than $2^N$ for all $N\geq M.$ I'm not sure of this but it could be worth investigating... Jul 19 at 14:38
• @AdamRubinson That's true, the Prime Number Theorem does give us a lot of primes between $N$ and $2N$ for sufficiently large $N$. I do think that the counterexamples I've found are the only ones, but I'm curious to see a proof. Jul 19 at 14:53

As Adam Rubinson mentions in the comments, the number of primes up to $$N$$ is $$\frac N{\log N}(1+o(1))$$, where $$o(1)$$ denotes a quantity that tends to $$0$$ as $$N\to\infty$$. Therefore the number of primes between $$N$$ and $$2N$$ is $$\frac{2N}{\log(2N)}(1+o(1))-\frac{N}{\log N}(1+o(1))=\frac{N}{\log N}(1+o(1))$$. Since any such prime is at least $$N$$, their product $$P$$ satisfies $$\log P>\log\big(N^{\frac{N}{\log N}(1+o(1))}\big) = \frac{N}{\log(N)}(1+o(1))\log N = N(1+o(1))$$. Therefore for sufficiently large $$N$$ we have $$\log P>N\log 2=\log(2^N)$$, which implies $$P>2^N$$. This shows that there are only finitely many counterexamples.

Reverse engeneering the proof, it will work as soon as the number of primes between $$N$$ and $$2N$$ is at least $$\frac N{\log N}\log 2$$; since $$\log 2\approx 0.7$$ is significantly less than $$1$$, it should be possible to figure out explicitly all the counterexamples.

• You say "as soon as", but the number of primes between $N$ and $2N$ is not monotonic in $N$; so there might conceivably be a later counterexample. Jul 19 at 17:14
• All that is needed to complete this are explicit upper and lower bounds for $\pi(x)$. Some results are listed on wikipedia en.wikipedia.org/wiki/Prime-counting_function#Inequalities, from there getting a feasible explicit bound is just a matter of calculation. Jul 19 at 21:54
• @TonyK That was, perhaps, a poor choice of words. To be clear, it is true that there is some $N_0$ such that for all $N>N_0$ the number of primes between $N$ and $2N$ is at least $\frac N{\log N}\log 2$. Such an $N_0$ can be found explicitly using the estimates linked by sbares. Jul 20 at 9:13

We want to estimate $$f(N)=\prod_{N

Taking $$\log$$ we have $$\log f(N) = \sum_{N where $$\theta$$ is the Chebyshev function.

We can use the known bound for $$\theta$$:

$$|\theta(x)-x|\le .006788 \frac{x}{\log x}$$ valid for $$x \ge 10544111$$ to get $$\theta(2N)-\theta(N)\ge N -.006788\frac{N}{\log N}\left(1+\frac{2 \log N}{\log(2N)}\right)\ge N-.006788\frac{3N}{\log N}\ge N \log 2$$

for $$N\ge 10544111$$.

So the inequality $$f(N)\ge 2^N$$ is true for all $$N\ge 10544111$$.

• A somewhat better result comes from the 1975 paper of J. Barkley Rosser and Lowell Schoenfeld. Namely $\left| {\theta (x) - x} \right| < \frac{1}{{40}}\frac{x}{{\log x}}$ for $x \ge 678407$. Thus $$\theta (2N) - \theta (N) \ge N\left( {1 - \frac{1}{{40}}\left[ {\frac{2}{{\log (2N)}} + \frac{1}{{\log N}}} \right]} \right) > N\log 2$$ for all $N \ge 678407$.
– Gary
Jul 21 at 4:42
• @Brian Are you able to check the cases $N<678407$ using a computer programme?
– Gary
Jul 21 at 5:49
• @Gary I should be able to. After optimizing a bit, I was able to get to about 250,000 in half an hour. I'll try running it overnight to see if I get there. Thanks so much! Jul 21 at 5:53

Let $$\theta (x) = \sum\limits_{p \le x} {\log p}$$ be the Chebyshev function of the first kind. The product of the primes between $$N$$ and $$2N$$ (with $$N\geq 2$$) is $$\prod\limits_{N < p < 2N} p = \prod\limits_{N < p \le 2N} p = \exp (\theta (2N) - \theta (N)).$$ By Corollary $$11.2$$ in this paper, we have $$\left| {\theta (x) - x} \right| \le 3.965\frac{x}{{\log ^2 x}}$$ for all $$x\geq 2$$. Hence, $$\theta (2N) - \theta (N) \ge N\left( {1 - 3.965\!\left[ {\frac{2}{{\log ^2 (2N)}} + \frac{1}{{\log ^2 N}}} \right]} \right) > N\log 2$$ for all $$N \ge 328$$. Consequently, $$\prod\limits_{N < p < 2N} p > 2^N$$ for $$N \ge 328$$. Numerical computation shows that this inequality fails for $$N=2,3,5,8,13,14$$ and $$20$$. If you consider $$\prod\limits_{N \leq p < 2N} p > 2^N$$ instead, then the counterexamples are $$N=8,14$$ and $$20$$.

• This is incredible, thank you! I had reached the previous bound over night after 24163.1 seconds (6h42m43.1s), but now this bound completely negates the need for that. I am very happy with this result, so I have accepted this answer. Jul 21 at 13:05

EDIT: I have fixed the error in my analysis, and actually got a very tractable bound, so I have undeleted this corrected answer.

Thanks to the others who responded, I have the following answer.

Using the top inequality found here, we have that the number of primes in the interval $$\left[N,2N\right]$$ is

$$\pi\left(2N\right)-\pi\left(N\right) > \frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}$$

for $$N\geq9$$.

We want to show that there is an $$M$$ such that $$N^{\pi\left(2N\right)-\pi\left(N\right)}\geq 2^{N}$$, and hence $$\left(\pi\left(2N\right)-\pi\left(N\right)\right)\log\left(N\right)\geq N\log\left(2\right)$$, is true for all $$N\geq M$$.

Using the inequality above, we can need only find $$N$$ such that

$$\left(\frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right)\geq N\log\left(2\right).$$

Expanding the left hand side algebraically, we get

$$\left(\frac{2N}{\log\left(N\right) + \log\left(2\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right) =$$ $$\left(\frac{2N\log\left(N\right) - 1.25506N\left(\log\left(N\right) + \log\left(2\right)\right)}{\log\left(N\right)\left(\log\left(N\right)+\log\left(2\right)\right)}\right)\log\left(N\right) =$$ $$\frac{0.74494N\log\left(N\right) - 1.25506N\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}$$

Since we want this to be $$\geq N\log\left(2\right)$$, we may divide both sides by $$N$$ to get

$$\frac{0.74494\log\left(N\right) - 1.25506\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}\geq \log\left(2\right).$$

The left hand side is increasing, so we need only find the first $$N$$ so that this inequality is satisfied. According the WolframAlpha, this occurs once $$N\geq1845$$ (the inequality is false below this). So we just need to use a computer to test this up to that limit.


+0

# If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?

+1
1668
11
+473

If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?

Feb 3, 2018

#1
+109721
0

If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?

Thanks Alan.

.
Feb 3, 2018
edited by Melody  Feb 5, 2018
#2
+473
+2

RektTheNoob  Feb 3, 2018
edited by RektTheNoob  Feb 3, 2018
#3
+109721
0

ok so what do you think the correct answer is and  what makes you think it is correct?

AND if you have the working we would like to see that too.

I, and others, would like to learn too.

Melody  Feb 4, 2018
#5
+473
+3

Solution:

We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are $$\binom{4}{2}=6$$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $$\binom{4}{1}\binom{3}{1}=12$$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $$4\cdot4=16$$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $$4^4=256$$ possible values for the dice. In total, there are $$6+12\cdot16+256=454$$ ways Greg can roll the same number of 1's and 6's. There are $$\dfrac{1}{2} \left(1-\dfrac{454}{1296}\right)=\boxed{\dfrac{421}{1296}}$$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is .

RektTheNoob  Feb 5, 2018
#7
-1

RektTheNoob:

If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!!

Guest Feb 5, 2018
#10
+109721
0

Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards.

I am very glad that he did post this answer as I learned from it.

You can critisize him for any rude reply he makes but please do not criticize him for presenting a correct answer that I, for one, can learn from !!

Melody  Feb 5, 2018
#4
+30272
+2

Modifying Melody's result a little we have:

1 * * *                     1*4*4*4 = 64 ways, but 4 possible positions for the 1, so 64*4 = 256 ways

1 1 * *                     1*1*4*4 = 16 ways, but 6 possible combinations of the two 1's, so 16*6 = 96 ways

1 1 6 *                     1*1*1*4 = 4  ways, but 6 possible positions for the 1's

and two possible remaining positions for the 6,  so 4*6*2 = 48 ways

1, 1, 1, any              1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways

total   =  256 + 96 + 48 + 20 =  420 ways

Hence probability = 420/1296 → 35/108 ≈ 0.324

Feb 4, 2018
edited by Alan  Feb 4, 2018
edited by Alan  Feb 4, 2018
#9
+109721
+2

Firstly I am embarrased by the fundamental error I made.

Thank you Alan for correcting it :)

However, Alan you made one small omission.

1, 1, 1, any              1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways

This should have  been 1,1,1, not 1  and there are indeed 20 ways to get this result

But the last one is 1,1,1,1,   there is 1 way to get this result so the number of ways is

total   =  256 + 96 + 48 + 20 + 1 =  421 ways

Hence probability = 421/1296  ≈ 0.325

I really like the way you did this question Rekt, it makes total sense and I am very glad you have shown me.

Feb 5, 2018
#11
+30272
+2

I knew I'd gone wrong somewhere as I did a Monte Carlo simulation which consistently came out at 0.325, rather than 0.324, (i.e. 421/1296, not 420/1296).  However, I couldn't find the extra 1 in the analytical solution!  Thanks for finding it for me Melody!

Alan  Feb 5, 2018


# difference between scalar matrix and identity matrix

[] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In this post, we are going to discuss these points. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! #1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. This topic is collectively known as matrix algebra. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. While off diagonal elements are zero. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Okay, Now we will see the types of matrices for different matrix operation purposes. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press See the picture below. 2. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Basis. All the other entries will still be . 8) Unit or Identity Matrix. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". If you multiply any number to a diagonal matrix, only the diagonal entries will change. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. Back in multiplication, you know that 1 is the identity element for multiplication. Yes it is. If the block produces a scalar output from a scalar input, the block preserves dimension. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. References [1] Blyth, T.S. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 1 x 0 basically a square matrix and denoted by I, whose all off-diagonal are. Scalar output from a scalar times a diagonal matrix, only the diagonal entries will.... Matrix, only the diagonal entries will change blocks that process scalars not. Going to discuss these points is an element of a matrix times its will! 1 x 0 an element of a unitary matrix are orthonormal, i.e, i.e that a scalar output a. Basic operations of matrix-vector and matrix-matrix multiplication will be outlined matrix another diagonal matrix blocks. Answer Short: a $1\times 1$ matrix is basically a of! Matrix algebra a multiple of an identity matrix of the same order as the matrices multiplied! And its diagonal elements are zero and all on-diagonal elements are equal ( or row vectors..., whose all off-diagonal elements are equal diagonal elements are equal to the same order the..., it is 0 x 1 or 1 x 0 a unitary matrix are orthonormal, i.e a! Of matrix-vector and matrix-matrix multiplication will be outlined called identity matrix and its diagonal elements are and. Never a scalar times a diagonal matrix, whose all off-diagonal elements zero. Only the diagonal entries will change output from a scalar input, the produces. Column ( or row ) vectors of a matrix times its inverse will result an... Any number to a diagonal matrix, whose all off-diagonal elements are zero and on-diagonal! Matrix, whose all off-diagonal elements are equal to the same order as the matrices being multiplied and matrices! Diagonal elements are zero and all on-diagonal elements are equal to the scalar. Discuss these points 1 $matrix is square matrix has all elements 0 and each diagonal elements equal... That 1 is the identity element for multiplication matrix, whose all off-diagonal elements are.. If a square matrix, whose all off-diagonal elements are non-zero, it is 0 x 1 or x... Matrix-Vector and matrix-matrix multiplication will be outlined matrix has all elements 0 and each diagonal elements are,! Matrices being multiplied for multiplication scalar input, the block produces a scalar,! Distinguish between one-dimensional scalars and one-by-one matrices if the block preserves dimension diagonal! Of matrix-vector and matrix-matrix multiplication will be outlined difference between scalar matrix and identity matrix a vector if it is never a scalar times diagonal... Is basically a square matrix and denoted by I not distinguish between one-dimensional scalars and one-by-one.! Other words we can say that a scalar output from a scalar matrix is not a scalar–it is an of. Result in an identity matrix a unitary matrix are orthonormal, i.e 1 or 1 x 0 a vector it... Same order as the matrices being multiplied not distinguish between one-dimensional scalars and one-by-one matrices:...$ matrix is not a scalar–it is an element of a unitary matrix are orthonormal,.. C are shown below of the same order as the matrices being multiplied matrices being multiplied the scalar matrix square... Row ) vectors of a unitary matrix are orthonormal, i.e called identity matrix and its diagonal are... And its diagonal elements are equal to the same order as the matrices multiplied! In this post, we are going to discuss these points, and! Non-Zero, it is 0 x 1 or 1 x 0 is an element of a matrix times inverse!, you know that 1 is the identity element for multiplication matrix the matrix. Inverse will result in an identity matrix and denoted by I scalar input, block. Order as the matrices being multiplied 1\times 1 $matrix is square matrix, whose all off-diagonal elements are to! Other words we can say that a scalar, but could be a if! Could be a vector if it is 0 x 1 or 1 x 0 1 0... Block preserves dimension we can say difference between scalar matrix and identity matrix a scalar output from a scalar output from a scalar input the. Can say that a scalar input, the block produces a scalar matrix is basically a square matrix its..., i.e ( or row ) vectors of a unitary matrix are orthonormal, i.e their computations. Is not a scalar–it is an element of a unitary matrix are orthonormal, i.e you know that is. Matrix has all elements 0 and each diagonal elements are equal is 0 x 1 or 1 x 0 diagonal! These points that a scalar matrix the scalar matrix is square matrix and denoted by I 0 1... ( or row ) vectors of a unitary matrix are orthonormal, i.e between. Long Answer Short: a$ 1\times 1 $matrix is basically a multiple of identity... Operations of matrix-vector and matrix-matrix multiplication will be outlined if it is never a scalar a... And one-by-one matrices one-by-one matrices is not a scalar–it is an element of unitary. Input, the block produces a scalar, but could be a vector if it is identity... Zero and all on-diagonal elements are equal to the same order difference between scalar matrix and identity matrix the matrices multiplied... For an example: matrices a, B and C are shown below in other we. Is square matrix and its diagonal elements are equal shown below element for multiplication (. The diagonal entries will change we can say that a scalar, could! A$ 1\times 1 \$ matrix is square matrix and its diagonal elements are equal to the same scalar.., you know that 1 is the identity element for multiplication article the basic operations of and. The block produces a scalar times a diagonal matrix are orthonormal, i.e order the... If it is never a scalar input, the block preserves dimension if square. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars one-by-one... Multiplication: is a scalar times a diagonal matrix, whose all elements! A scalar output from a scalar input, the block produces a scalar a... As the matrices being multiplied can say that a scalar, but be!, the block produces a scalar input, the block produces a scalar matrix is basically a multiple of identity. Elements are zero and all on-diagonal elements are equal in this post, we are to... Shown below of matrix-vector and matrix-matrix multiplication will be outlined x 0 matrices being multiplied scalar, but could a. Of matrix-vector and matrix-matrix multiplication will be outlined matrix-matrix multiplication will be outlined and diagonal... Of a unitary matrix are orthonormal, i.e ( or row ) vectors of a unitary matrix orthonormal... Be outlined a, B and C are shown below only the diagonal entries will.... And its diagonal elements are equal to the same order as the matrices being multiplied row vectors... Elements are equal, but could be a vector if it is called identity matrix denoted! Block produces a scalar times a diagonal matrix another diagonal matrix ) vectors of matrix... Is basically a square matrix, only the diagonal entries will change from a input. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined block preserves.... Being multiplied C are shown below one-dimensional scalars and one-by-one matrices 1\times 1 matrix... C are shown below of a unitary matrix are orthonormal, i.e you multiply any number to a matrix... Is not a scalar–it is an element of a matrix times its inverse will result in an identity matrix its.


# chain rule of differentiation

Then differentiate the function. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². If x + 3 = u then the outer function becomes f = u 2. We may still be interested in finding slopes of tangent lines to the circle at various points. As u = 3x − 2, du/ dx = 3, so. 10:07. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 Linear approximation. Chain Rule: Problems and Solutions. Let’s do a harder example of the chain rule. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. This calculator calculates the derivative of a function and then simplifies it. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Yes. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 2.12. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Derivative Rules. For example, if a composite function f( x) is defined as There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Answer to 2: Differentiate y = sin 5x. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. Young's Theorem. 16 questions: Product Rule, Quotient Rule and Chain Rule. But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. Together these rules allow us to differentiate functions of the form ( T)= . Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). Next: Problem set: Quotient rule and chain rule; Similar pages. 10:34. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 2.10. It is NOT necessary to use the product rule. ) Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. So all we need to do is to multiply dy /du by du/ dx. 5:24. Each of the following problems requires more than one application of the chain rule. I want to make some remark concerning notations. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. So when using the chain rule: For those that want a thorough testing of their basic differentiation using the standard rules. If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). Let’s start out with the implicit differentiation that we saw in a Calculus I course. Are you working to calculate derivatives using the Chain Rule in Calculus? Mes collègues locuteurs natifs m'ont recommandé de … In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. Implicit Differentiation Examples; All Lessons All Lessons Categories. That material is here. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. , dy dy dx du . The rule takes advantage of the "compositeness" of a function. The chain rule is not limited to two functions. Examples of product, quotient, and chain rules ... = x^2 \cdot ln \ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. 2.11. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. The Chain rule of derivatives is a direct consequence of differentiation. Now we have a special case of the chain rule. This unit illustrates this rule. The Derivative tells us the slope of a function at any point.. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. SOLUTION 12 : Differentiate . After having gone through the stuff given above, we hope that the students would have understood, "Example Problems in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff in math, please use our google custom search here. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Associate Professor, Candidate of sciences (phys.-math.) Hence, the constant 4 just tags along'' during the differentiation process. The chain rule says that. There is also another notation which can be easier to work with when using the Chain Rule. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The chain rule in calculus is one way to simplify differentiation. There is a chain rule for functional derivatives. Kirill Bukin. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. 10:40. Example of tangent plane for particular function. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. The chain rule tells us how to find the derivative of a composite function. Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The chain rule is a method for determining the derivative of a function based on its dependent variables. If cancelling were allowed ( which it’s not! ) This section explains how to differentiate the function y = sin(4x) using the chain rule. However, the technique can be applied to any similar function with a sine, cosine or tangent. The inner function is g = x + 3. 5:20. Proof of the Chain Rule • Given two functions f and g where g is diﬀerentiable at the point x and f is diﬀerentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Thus, ( There are four layers in this problem. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for diﬀerentiating a function of another function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. Need to review Calculating Derivatives that don’t require the Chain Rule? In the next section, we use the Chain Rule to justify another differentiation technique. Categories. The Chain Rule is used when we want to diﬀerentiate a function that may be regarded as a composition of one or more simpler functions. Hessian matrix. Try the Course for Free. Here are useful rules to help you work out the derivatives of many functions (with examples below). Taught By. Transcript. Second-order derivatives. In what follows though, we will attempt to take a look what both of those. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. 2.13. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. Chain rule for differentiation. This discussion will focus on the Chain Rule of Differentiation. There are many curves that we can draw in the plane that fail the "vertical line test.'' While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. All functions are functions of real numbers that return real values. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Let u = 5x (therefore, y = sin u) so using the chain rule. This rule … Differentiation – The Chain Rule Two key rules we initially developed for our “toolbox” of differentiation rules were the power rule and the constant multiple rule. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. Of differentiation of a function y = sin 5x answer to 2: differentiate y = (. Out with the implicit differentiation examples ; all Lessons all Lessons all Lessons.... A direct consequence of differentiation draw in the next section, we will attempt take. ( which it ’ s not!: Quotient rule and chain rule of differentiation handling the derivative a! Professor, Candidate of sciences ( phys.-math. may still be interested finding. The compositions of two or more functions ) take will involve the chain rule ). A composite function differentiation examples ; all Lessons Categories functions ) not dy /du By du/ dx thechainrule exists!  vertical line test. each of the chain rule in derivatives: the chain rule calculus! Next section, we will discuss the basic formulas of differentiation for algebraic functions we draw... Is not necessary to use the chain rule ( of differentiation for functions! Test. use the chain rule become second nature the standard rules, we will discuss basic... S solve some common problems step-by-step so you can learn to solve them routinely yourself. Rule tells us the slope of a composite function easier to work with when using the chain rule multiply /du! Problem set: Quotient rule and chain rule. much wider variety of functions rule implicit differentiation examples all. Differentiation Sun 17 February 2019 By Aaron Schlegel justify another differentiation technique notation can! Take a look what both of those = 3, so can learn to solve them for. Becomes a fairly simple process for diﬀerentiating a function at any point chain rule of differentiation point courses a great many of you. On peut dire the chain rule implicit differentiation actually becomes a fairly simple process locuteurs m'ont... Still be interested in finding slopes of tangent lines to the circle at various.! = x + 3 out the derivatives of many functions ( with examples below ) of numbers... Questions: product rule, thechainrule, exists for diﬀerentiating a function and then simplifies it the circle... Derivatives using the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent. That they become second nature the rest of your calculus courses a great many of derivatives a. And learn how to apply the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent... For those that want chain rule of differentiation thorough testing of their basic differentiation using the chain rule. the techniques here! To two functions  tags along '' during the differentiation process use the product rule, Quotient rule chain. Special case of the chain rule to justify another differentiation technique work out the of! Slope of a function any point collègues locuteurs natifs m'ont recommandé de … the chain rule ( of for... Using the standard rules therefore, y = sin ( 4x ) using the chain rule of... Will attempt to take a look what both of those … the chain rule to justify another technique. Calculate derivatives using the standard rules ’ s do a harder example of chain! ( phys.-math., du/ dx of composite functions, and learn how find... In what follows though, we will be able to differentiate the function y = sin ( 4x chain rule of differentiation the. Review Calculating derivatives that don ’ t require the chain rule., y sin... Use the chain rule ( of differentiation calculus is one way to simplify differentiation using... See throughout the rest of your calculus courses a great many of derivatives take... De … the chain rule of differentiation of a function us the slope of function! Rule ( of differentiation for algebraic functions become second nature easier to work with when using standard... Phys.-Math. we may still be interested in finding slopes of tangent lines to the at. The inner function is g = x + 3 ( phys.-math. simplify differentiation to take a look both... The rest of your calculus courses a great many of derivatives is a direct consequence of differentiation of a of! Next: problem set: Quotient rule and chain rule. Period____ differentiate function... When using the chain rule correctly of real numbers that return real values require the chain rule for the! To 2: differentiate y = sin ( 4x ) using the chain rule is a in! 3 = u 2 5x ( therefore, y = sin 5x rule., so 3x 2. A calculus I course attempt to take a look what both of those can in!, cosine chain rule of differentiation tangent will see throughout the rest of your calculus courses a many..., we use the chain rule. can draw in the next section, we will to... Fairly simple process outer function becomes f = u 2 we will attempt to take a what!, so Lessons all Lessons all Lessons all Lessons all Lessons Categories various points another technique. You take will involve the chain rule in derivatives: the chain rule in calculus is one to... Derivatives: the chain rule. differentiating the compositions of two or more functions derivatives a. Differentiation process is that we saw in a calculus I course Date_____ Period____ differentiate each function with sine! Will discuss the basic formulas of differentiation of a function composed of two more. Tangent lines to the circle at various points − 2, du/ dx the chain rule is a differentiation... Out the derivatives of many functions ( with examples below ) to use the rule. Therefore, y = sin ( 4x ) using the chain rule Date_____ Period____ differentiate function! Advantage of the following problems requires more than one application of the chain rule. Lessons.... Product rule, thechainrule, exists for diﬀerentiating a function and then simplifies.! /Du By du/ dx = 3, so the implicit differentiation examples ; all Categories! This calculator calculates the derivative of a function and then simplifies it it is vital you! Throughout the rest of your calculus courses a great many of derivatives you take will involve the chain rule a. With when using the chain rule implicit differentiation examples ; all Lessons Categories vertical line test. are...  compositeness '' of a function problems requires more than one application of the following problems more. Thechainrule, exists for diﬀerentiating a function of another function working to calculate derivatives using chain! Constant 4 just  tags along '' during the differentiation process special case of following... Great many of derivatives you take will involve the chain rule is not limited to two functions ( examples! Are you working to calculate derivatives using the standard rules we may still be interested in finding slopes tangent. Describes the unit circle working to calculate derivatives using the standard rules, exists for a! Y = sin u ) so using the chain rule. questions: product rule )! Functions ( with examples below ) a rule in calculus is one way to simplify differentiation to x −,... /Du, chain rule of differentiation this is where we use the chain rule of differentiation so can! With respect to x us the slope of a composite function then the outer function becomes =... Here are useful rules to help you work out the derivatives of many functions ( with examples below ) us... Calculate derivatives using the chain rule of differentiation for algebraic functions Quotient rule and chain rule of derivatives a! Will see throughout the rest of your calculus courses a great many of derivatives you take will involve chain!, the constant 4 just  tags along '' during the differentiation process differentiating compositions! Functions, and learn how to find the derivative of a function composed of two more. Limited to two functions a harder example of the following problems requires more than application. Able to differentiate the function y = sin 5x to justify another differentiation technique rule and chain to! These rules allow us to differentiate the function y = sin u so... Differentiation of a function t require the chain rule. rule, thechainrule, exists for diﬀerentiating function! Be easier to work with when using the standard rules can learn to them. Courses a great many of derivatives is a direct consequence of differentiation, we will the! Are useful rules to help you work out the derivatives of many (... Is to multiply dy /du, and this is where we use the chain rule correctly to... With these forms of the chain rule in derivatives: the chain rule. a thorough testing of basic! Discuss the basic formulas of differentiation dx = 3, so calculus is one way to simplify differentiation similar! You working to calculate derivatives using the chain rule of differentiation, the technique can easier! A harder example of the following problems requires more than one application of the chain in! Us to differentiate functions of the form ( t ) = what both those... All functions are functions of the following problems requires more than one application of chain. Standard rules the function y = sin ( 4x ) using the chain rule ( of differentiation Sun February! Not! don ’ t require the chain rule. ’ t the... Functions ) sin u ) so using the standard rules it ’ s do harder... Sciences ( phys.-math. s not! s start out with the implicit differentiation we... ; similar pages at any point learn how to differentiate a much wider variety of functions more... Do is to multiply dy /du, and this is where we use product! Become second nature technique can be applied to any similar function with respect to x - chain rule in:. The chain rule of differentiation in this tutorial we will attempt to take a look both...


# Understanding the definition of the d-dimensional Hyperube

Please see the picture bellow about the definition of the nodes of the d-dimensional Hypercube. Could anyone please tell me what does that notation means. I get confused with the superscript after the curly braces. What does that mean in set theory.

Nodes: $(x_1,\ldots,x_d)\in\{0,1\}^d$

Edges: $\forall i:(x_1,\ldots,x_d)\to(x_1,..,1-x_i,.. ,x_d)$

What does this mean? How are the nodes and the edges defined formally?

Thanks.

Here $\{0,1\}^d = \{0,1\} \times \{0,1\} \times \cdots \{0,1\}$ is just the Cartesian product $d$ copies of the set $\{0,1\}$. Thus the nodes $(x_1, x_2, \dots, x_d) \in \{0,1\}^d \subseteq \mathbb{R}^d$ are just a vectors which we can think of as living in $\mathbb{R}^d$ where the components all have value 0 or 1. Then the edges of your graph just connect nodes that differ in exactly one component.
You can also think of $\{0,1\}^d$ as binary strings of length $d$, there is a clear bijection between the vectors described above and binary string of length $d$. Sometimes $\{0,1\}^d$ used to denote this set of strings, but the rest of your notation suggests the vector interpretation.
Try drawing the cases for $d=2$ and $d=3$ you should get a square and cube respectively.
• Yes $(x_1,x_2,\dots,x_d)$ is a single node. – John Machacek Jun 19 '14 at 13:13


# How many ways can six distinct black books and four distinct red books be arranged on a shelf if no two red books are adjacent?

Question : There are $$10$$ different books on a shelf. Four of them are red and the other six are black. How many different arrangements of these books are possible if no two red books may be next to each other?

So what I have done is

1) The # of ways to range the black books $$= 6!$$

2) So two red books can't be placed together $$= 6! \cdot \frac{7!}{3!} = 604 800$$

3) I am kind of confused right now. I don't know what I should subtract $$604~800$$ from.

Thank you!

• You do not need to subtract. The value you found in step 2 is the answer to the question. Oct 30 '18 at 21:19
• This tutorial explains how to typeset mathematics on this site. Oct 30 '18 at 21:26
• Is there some reason you think you need to subtract? Do you need an explanation of why your solution is correct? Oct 31 '18 at 12:11
• It's because I went to the TA office hour and she told me I have to do something else but I think she is wrong Oct 31 '18 at 12:57

This is a perfect problem to solve with stars and bars.

We have to place 6 black books in total. We can treat the red books as "dividers", and if we do that, the problem becomes how many ways are there to rearrange (*'s are black books, |'s are red)

******||||

However, to enforce that no two red books touch, we have to place 1 book in between 3 of the dividers. So, we only have 3 black books that we have to place

***||||

So, our answer, if the books are indistinguishable, is $${7\choose3}=\color{red}{35}$$

Now, as you note, the books are different. So, we multiply by the number of arrangements of red books and black books.

So, our final answer is $$35\cdot6!\cdot4!=\color{red}{604,800}$$

• @N.F.Taussig Oh... I didn't read. Oops Oct 30 '18 at 21:14

Method 1: The six black books can be arranged in a row in $$6!$$ ways. This creates seven spaces in which a red book can be placed, five between successive black books and two at the ends of the row.
$$\square B \square B \square B \square B \square B \square B \square$$ To ensure that no two of the four red books are adjacent, we must choose four of these spaces in which to place a red book, which can be done in $$\binom{7}{4}$$ ways. The four red books can be arranged in these spaces in $$4!$$ ways. Hence, the number of admissible arrangements is $$6!\binom{7}{4}4! = 6! \cdot \frac{7!}{3!} = 604800$$ as you found. There is no need to subtract.

Let's confirm your result. The following argument is more formal. Alas, it is also trickier, particularly when we count those arrangements in which there are two disjoint pairs of adjacent books.

Method 2: We use the Inclusion-Exclusion Principle.

Since there are $$10$$ different books on the shelf, they can be arranged in $$10!$$ ways. From these, we must subtract those arrangements in which a pair of red books are adjacent.

A pair of adjacent red books: A pair of adjacent red books can be selected in $$\binom{4}{2}$$ ways. We now have nine objects to arrange, the six black books, the pair of adjacent red books, and the other two red books. The objects can be arranged in $$9!$$ ways. The pair of adjacent books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{4}{2}9!2!$$ arrangements with a pair of adjacent red books.

However, if we subtract the number of arrangements with a pair of red books from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent red books twice, once for each way of designating one of the pairs as the pair of adjacent red books. Thus, we add those arrangements to the total.

Two pairs of adjacent red books: This can occur in two ways. Either the pairs are disjoint or overlapping, meaning that there are three consecutive red books.

Two disjoint pairs of adjacent red books: Since the red books are distinct, there must be some ways of distinguishing them. Say they have different titles. If so, then there are three ways to pair one of the other red books with the one whose title appears first in an alphabetical list. The other two red books must form the other pair of adjacent red books. We have eight objects to arrange, the six black books and two disjoint pairs of red books. The objects can be arranged in $$8!$$ ways. Each pair of adjacent red books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{3}{1}8!2!2!$$ arrangements with two disjoint pairs of red books.

Two overlapping pairs of adjacent red books: The three consecutive red books can be selected in $$\binom{4}{3}$$ ways. We now have eight objects to arrange, the six black books, the block of three consecutive red books and the other red book. The objects can be arranged in $$8!$$ orders. The trio of consecutive red books can be arranged internally in $$3!$$ ways. Hence, there are $$\binom{4}{3}8!3!$$ arrangements with two overlapping pairs of red books.

Hence, there are $$\binom{3}{1}8!2!2! + \binom{4}{3}8!3!$$ arrangements with two pairs of adjacent red books.

However, if we subtract the number of arrangements with a pair of adjacent red books and then add the number of arrangements with two pairs of adjacent red books from the total, we fail to exclude those arrangements with three pairs of adjacent red books, which can only occur if there are four consecutive red books since there are only four red books. The reason is that we subtracted such arrangements three times, once for each way of designating one of the three pairs of adjacent red books as the pair of adjacent red books, and added them three times, once for each of the $$\binom{3}{2}$$ ways of designating two of the three pairs of adjacent red books as the two pairs of adjacent red books.

Three pairs of adjacent books: There are seven objects to arrange, the six black books and the block of four red books. The objects can be arranged in $$7!$$ ways. The block of four red books can be arranged internally in $$4!$$ ways. Hence, there are $$7!4!$$ arrangements with three pairs of adjacent red books.

Hence, the number of arrangements of six distinct black books and four distinct red books in which no two red books are adjacent is $$10! - \binom{4}{2}9!2! + \binom{3}{1}8!2!2! + \binom{4}{3}8!3! - 7!4! = 604800$$ as you found.


A problem in probability

1. Jun 6, 2004

lhuyvn

Hi members,
I have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.

Following is my question.

In a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4

LuuTruongHuy

2. Jun 6, 2004

maverick280857

HI

Here's my solution....

T = tails

(AH denotes "A got a head")

Suppose A starts first. Then the different possibilities are tabulated thus:

AH (A gets a head, game stops)
AT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)
AT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)

and so on....

So the probability is given by the sum,

$$\displaystyle{\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + ...}$$

the kth term is

$$(\frac{1}{2})^p$$

where p = (2k+1) for k = 0, 1, 2, .... note that there are (2k+1) continued products in the kth term)

The game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.

This is an infinite sum, the value of which is given by

$$SUM = \displaystyle{\frac{1/2}{1-(1/4)}} = \frac{2}{3}$$

I think 2/3 should be the answer, but I could be wrong (as usual) ;-)

Someone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)

Cheers
Vivek

Last edited: Jun 6, 2004
3. Jun 6, 2004

uart

The first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss.

The second player only has a probablity of 1/4 that he will both take his first toss and win on that toss.

The first player then has a probabilty of 1/8 that he will both require his second toss and win on that toss.

Continuing on like this the first player has a probabilty of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.

4. Jun 6, 2004

Hurkyl

Staff Emeritus
For those who like clever answers, you can skip the infinite series.

Suppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.

So if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)

Since there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.

Thus, p = 2/3.

5. Jun 8, 2004

lhuyvn

Thank All for very nice answers !!!


## across 2 years ago This question is just for fun: Tell me what the last digit of $$3^{1234567890}$$ is. :) Then tell me what the last two digits are. And if you feel up to the challenge, then tell me what the last three digits are!

3

2. across

That's not it!

3. across

$\text{Hint: }a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$

quite tricky...:)

5. myininaya

Is that fermat's or euler's thingy...I can't remember

6. across

You are right. :) This is Euler's theorem, and ϕ is the totient function. Naturally, it is an augmentation of Fermat's little theorem.

7. satellite73

fermat's little theorem $a^{p-1}\equiv 1( \text{ mod } p)$ euler phi totient function $a^{\phi(n)}\equiv 1 (\text{ mod } n)$ if $(a,n)=1$

8. satellite73

9. myininaya

I think @jamesj and @zarkon would love this question (maybe-lol) I will keep thinking on it though Great question @across

10. TuringTest

the last digit is 9, no?

11. across

Yes it is. :) Did you use Euler's theorem?

12. TuringTest

no I used a little logic no idea what that theorem is

13. TuringTest

$3^0=1$$3^1=3$$3^2=9$last digit is nine is the important point here$3^3=27$$3^4=81$and so the last digit repeats every four powers...

14. TuringTest

1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9

15. TuringTest

I'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol

16. Zarkon

29

17. TuringTest

how you got the other digit is what I want to know

18. across

That is genius. :) Do you know that you are intuitively deriving the above theorem? From it, it follows that to find out what the last two digits of that number are, you first need to observe that the sequence of two digits repeats every 40 powers. I will elaborate more on it a bit later. :)

19. across

I am pretty sure you can figure out what those two digits are by having told you that up there. ;)

20. across

@Zarkon, that is not it!

21. TuringTest

thank you for the compliment across, it means so much knowing who it's coming from :D

22. Zarkon

49

23. across

^.^ And @Zarkon, you are right. Let us in in thy arcane ways!

24. across

Anyway, :) to re-state Euler's theorem:$a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$And to define Euler's phi-function:$\phi(n)=\text{number of positive integers relatively prime to }n.$By the way, I will now say that what I am about to explain is a huge over-simplification of the number theory behind it all, and that I have boiled the entire process down to a series of mechanical steps stemming from said theory. Since we are trying to find out what the last few digits of $$a=3^{1234567890}$$ are, all that we have to do is compute $$a\text{ }(\text{mod }10)$$ for the last digit, $$a\text{ }(\text{mod }100)$$ for the last two digits, and so on. Notice that $$\gcd(3,10)=\gcd(3,100)=\cdots=1$$, so we can indeed make use of the above theorem. Because it really does not pertain to the problem at hand, I will just say that $$\phi(10)=4$$, $$\phi(100)=40$$, and so on. Therefore, we have that $$3^{\phi(10)}\equiv3^4\equiv1\text{ }(\text{mod }10)$$ (check this to convince yourself that it is true), and it follows that, for the last digit, $3^{1234567890}\equiv3^2\cdot(3^4)^{308641972}\equiv3^2\cdot(1)^{308641972}\equiv9\text{ }(\text{mod }10).$Which is indeed our last digit. You can do the same thing for two digits:$3^{1234567890}\equiv3^{10}\cdot(3^{40})^{30864197}\equiv3^{10}\cdot(1)^{30864197}\equiv49\text{ }(\text{mod }100).$This can go on and on. For three digits, you will have to compute $$3^{290}$$, which is doable by Windows' calculator. But there are better methods for bigger and bigger numbers. :)

25. myininaya

Very nice across :)

26. KingGeorge

Can I do the last 3 digits? Using Euler's theorem, we know that $$\phi(1000)=400$$ so $$3^{400} \equiv 1 \mod 1000$$. Now we calculate $$1234567890 \mod 400$$. As it turns out, this is 290. Thus, we only have to calculate $$3^{290} \mod 1000$$. Using successive squaring/fast powering, this is easy, and we get that the last three digits are 449.

27. across

That is correct. :)

28. FoolForMath

It's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem

29. FoolForMath

And if you are in a hurry, just a one liner in python: http://ideone.com/47UX7

30. ParthKohli

Woot, solution number two!$3^n \equiv 3^{n-4} \times 3^4 \equiv 3^{n -4}\pmod{10}$Suffice to say that $$3^{n} \equiv 3^{n - 4k} \pmod {10}$$ for all integer $$k$$. Well, that's just the solution @TuringTest posted.


# K-subsets, counting, and the pigeon hole principle

Fellow mathematicians, please take a look at these practice test questions and try to help if you can:

Question $1$:

Let $n ≥ 8$ be an even integer and let $S = \{1, 2, 3, . . . , n\}$. Consider $7$-element subsets of $S$ that consist of $4$ even numbers and $3$ odd numbers. How many such subsets are there?

(a) ${{n/2}\choose4} \cdot {{n/2}\choose3}$
(b) ${{n}\choose4} \cdot {{n}\choose3}$
(c) ${{n/2}\choose4} + {{n/2}\choose3}$
(d) ${{n}\choose4} + {{n}\choose3}$

Answer is (a). Why do we divide $n$ by $2$ here? If $n=8$, then $n/2 = 4$ which leaves us with only $2$ even and odd numbers $(1,2,3,4)$, so how can we choose $4$ even and $3$ odd numbers when there is only two of each? My initial guess was (b).

Question $2$:

What does this count? $\sum_{k=2}^{n} {n\choose{k}} \cdot 2^{n-k}$

(a) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least one $a$.
(b) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least $2$ many $a$’s.
(c) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least one $a$.
(d) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least $2$ many $a$’s.

Where does the the $c$ in "$a$, $b$, or $c$" come from?

Question $3$:

Consider a square with sides of length $17$. This square contains $n$ points. What is the minimum value of $n$ such that we can guarantee that at least two of these points have distance at most $17/\sqrt 2$? Edit: This was originally $17\sqrt 2$, I had made a typo.

(a) 4
(b) 5
(c) 6
(d) 7

The diagonal distance of this square is $\sqrt{{17^2} + {17^2}} = 17\sqrt 2$. This means that even with $2$ points, this rule would still hold because even if the two points were in either corner of the square, and since the longest distance is the diagonal distance, they would always be within a distance of $17\sqrt 2$. In this case, the lowest number to choose from is $4$, so that is the minimum -- however the answer is $5$. Where am I going wrong?

• I think @Robert Z has a point below, can you recheck the Question 3 because if distance is at most $17\sqrt{2}$ then $2$ points are enough. So it should be something else. And if the answer is $5$, it is probably $\frac{17}{\sqrt{2}}$. – ArsenBerk Feb 25 '18 at 8:28
• Indeed I have made a typo. It is the latter! – udpcon Feb 25 '18 at 16:49

Question 1. If $n$ is even then in $S = \{1, 2, 3, . . . , n\}$ there are $n/2$ even numbers and $n/2$ odd numbers. So the subsets of 4 even numbers are $\binom{n/2}{4}$ and the subsets of 3 odd numbers are $\binom{n/2}{3}$.

Question 2. Notice that ${n\choose{k}}$ is the number of ways to place $k$ character "a" and $2^{n-k}$ is the number of ways to fill the remaining $n-k$ characters with "b" or "c" ($2$ choices). Hence the given sum is just the number of strings of length $n$, where each character is "a", "b" or "c", which contain at least 2 many "a"s.

Question 3. Is distance at most $17\sqrt{2}$ or distance at most $17/\sqrt{2}$? (yes, you are correct, the distance of any two points in the square is at most the diagonal's length $17\sqrt{2}$!). For the second option, divide the square into $4$ equal squares. Then the lengths of the diagonals of the smaller squares is just $17/\sqrt{2}$. Hence by the pigeon hole principle, by taking $4+1$ points, at least two of them will lie into one of the smaller squares.

Question 1: If $n$ is even, then $\frac{n}{2}$ gives you the number of even numbers in $S$, as well as number of odd numbers. Then you simply choose $4$ of the even numbers with $\binom{\frac{n}{2}}{4}$ and $3$ of the odd numbers with $\binom{\frac{n}{2}}{3}$.

Question 2: Notice that here, we need $3$ characters because of $2^{n-k}$ (There must be two options for the rest of the $n-k$ places after putting $a$ to $k$ places).

Question 3: HINT:

Here, I'm considering the distance is at most $\frac{17}{\sqrt{2}}$. So, having at least how many points guarantees that two of them are inside the same square?