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Factor the expression and use the fundamental identities to simplify $7 \sin^2 x \csc^2 x − 7 \sin^2 x$ Factor the expression and use the fundamental identities to simplify. There is more than one correct form of the answer. $$7 \sin^2 x \csc^2 x − 7 \sin^2 x$$
I'm reviewing for a test and going over my old homework, is 7 a possible solution (I'm asking because the last time I did this on my homework I got $7\cos(x)$?
$7 \sin^2 x (csc^2 x − 1)$
$=7 \sin^2 x (1 + \cot^2x)$
$=7 \sin^2 x (1 + \frac {\cos^2x}{\sin^2x})$
$=7 \sin^2 x (\frac{\sin^2x}{\sin^2x} + \frac {\cos^2x}{\sin^2x})$
$=7\sin^2x (\frac{\sin^2x+\cos^2x}{\sin^2x})$
$=7\sin^2x (\frac{1}{\sin^2x})$
$=\frac{7\sin^2x}{\sin^2x}$
$=7$
| Notice, your mistake $\csc^2-1\ne 1+\cot^2 x$, one can easily simplify as follows $$7\sin^2 x\csc^2 x-7\sin^2 x$$ $$=7\sin^2 x(\csc^2 x-1)$$ $$=7\sin^2 x(\cot^2 x)$$ $$=7\sin^2 x\left(\frac{\cos^2 x}{\sin^2 x}\right)$$ $$=\color{red}{7\cos^2x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Am I right in calculating $\sin(2\arcsin(\frac{1}{3}))$ as $\frac{4\sqrt{2}}{9}$? I've been solving a problem in my textbook, and my result is at odds with the textbook's:
$$\sin\left(2\arcsin\left(\frac{1}{3}\right)\right)$$
My answer is
$$\frac{4\sqrt{2}}{9}$$
I've used the double-angle identity for sine.
$$\sin\left(2\arcsin\left(\frac{1}{3}\right)\right)=2\sin\left(\arcsin\left(\frac{1}{3}\right)\right)\cos\left(\arcsin\left(\frac{1}{3}\right)\right)$$
and this identity:
$$\cos(\arcsin(x))=\sqrt{1-x^2},$$
yielding
$$2*\frac{1}{3}*\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{8}}{3*3}=\frac{4\sqrt{2}}{9}$$
But the textbook's answer is
$$\frac{2\sqrt{2}}{3}$$
| Notice, $$\sin 2A=2\sin A\cos A$$ & $$|\cos A|=\sqrt{1-\sin^2 A}$$ Now, we have $$\sin\left(2\sin^{-1}\left(\frac{1}{3}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(2\frac{1}{3}\sqrt{1-\frac{1}{3^2}}\right)\right)$$
$$=\sin\left(\sin^{-1}\left(\frac{4\sqrt 2}{9}\right)\right)=\frac{4\sqrt 2}{9}$$
O.P. is right. There is some probable error in the answer in the textbook.
| {
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$f(x)=2x^4+x^4\sin(\dfrac 1x) , \forall x \ne 0 ; f(0):=0$ ; it's derivative has both positive and negative values in every neighbourhood of $0$? Let $f:\mathbb R \to \mathbb R$ a function defined as $f(x)=2x^4+x^4\sin\left(\dfrac 1x\right) , \forall x \ne 0 ; f(0):=0$ ; then how to show that it's derivative has both positive and negative values in every neighbourhood of $0$ ?
| Note that $f'(x) = 8x^3 + 4x^3 \sin \frac 1 x - x^2 \cos \frac 1 x$ for $x \ne 0$. Now $f' \Big( \frac 1 {2 n \pi} \Big) = \frac 1 {n^3 \pi^3} - \frac 1 {4 n^2 \pi^2} = \frac {4 - n \pi} {4 n^3 \pi^3} < 0$ for $n \ge 2$ and $f' \Big( \frac 1 {(2n+1) \pi} \Big) = \frac 8 {(2n+1)^3 \pi^3} + \frac 1 {(2n+1)^2 \pi^2} > 0$ for $n \ge 1$. Since the sequences $\frac 1 {2 n \pi}$ and $\frac 1 {(2n+1) \pi}$ both tend to $0$, this shows that in any neighbourhood of $0$ $f'$ will take infinitely many negative values and infinitely many positive values.
| {
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How do I prove that the $f(x)$ is positive for all real $x$? $$ \frac {f(x+y) - f(x)}{2}= \frac{f(y)-a}{2} +xy $$
for all real $x$ and $y$. If $f(x)$ is differentiable and $f'(0)$ exists for all real permisible values of $a$ and is equal to $\sqrt{5a-1-a^2}$. Prove that $f(x)$ is positive for all real $x$.
I differentiated the equation keeping $x$ constant and then put $y=0$ and then integrated and got $f(x)$ as
$$f(x)= x^2 +x\sqrt{5a-1-a^2}+c$$
by putting $x=y=0$ in
$$ \frac {f(x+y) - f(x)}{2}= \frac{f(y)-a}{2} +xy $$
I got $f(0) =a$ so I finally got the function as
$$f(x)= x^2 +x\sqrt{5a-1-a^2}+a.$$
Now how should I proceed, will $b^2 -4ac <0 $ help?
| Put $y=0$ giving $a=f(0)$. For non-$0$ $y$,divide the original equation by $y$ [with of course,$a=f(0)$], and let $y$ go to $0$,giving $f'(x)/2= f'(0)/2+x$. Integrating this,we have $f(x)=xf'(0)+x^2+c$. This satisfies the original equation because $c=a=f(0)$.
| {
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If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
| $$\sin^{-1}(x)+\sin^{-1}(y)=\frac{\pi}{2}\Longleftrightarrow$$
$$\sin^{-1}(y)=\frac{\pi}{2}-\sin^{-1}(x)\Longleftrightarrow$$
$$\sin(\sin^{-1}(y))=\sin\left(\frac{\pi}{2}-\sin^{-1}(x)\right)\Longleftrightarrow$$
$$y=\sqrt{1-x^2}$$
$$\sin^{-1}(x)+\sin^{-1}(y)=\frac{\pi}{2}\Longleftrightarrow$$
$$\sin^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(y)\Longleftrightarrow$$
$$\sin(\sin^{-1}(x))=\sin\left(\frac{\pi}{2}-\sin^{-1}(y)\right)\Longleftrightarrow$$
$$x=\sqrt{1-y^2}$$
$$x^2+y^2=1\Longrightarrow$$
$$\left(\sqrt{1-y^2}\right)^2+\left(\sqrt{1-x^2}\right)^2=1\Longleftrightarrow$$
$$1-y^2+1-x^2=1\Longleftrightarrow$$
$$1-y^2+1-\left(\sqrt{1-y^2}\right)^2=1\Longleftrightarrow$$
$$1-y^2+1-(1-y^2)=1\Longleftrightarrow$$
$$1-y^2-(1-y^2)+1=1\Longleftrightarrow$$
$$0+1=1\Longleftrightarrow$$
$$1=1$$
| {
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Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.
I can see two ways of proving it.
The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.
Another way is to solve it as a palindromic polynomial. It does not have roots.
But is there any way to directly manipulate the expression to show that it is always greater than zero?
| Another proof by contradiction.
Suppose $x^4+x^3+x^2+x+1$ has a real root. Necessarily, $x<0$. Set $x=-t,\enspace t>0$. Rewrite the equation as
$$t^4-t^3+t^2-t=t(t-1)(t^2+1)=-1$$
It shows $t-1<0$, hence $\;0<t<1$.
On another hand, $\;t=\dfrac1{(1-t)(1+t^2)}=\dfrac1{1-t+t^2-t^3}$, and since $0<t<1$, we have $0<1-t+t^2-t^3<1-t+t^2=1-t(1-t)<1$, hence
$$t=\dfrac1{1-t+t^2-t^3}>1,\quad\text{contradiction}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor Series of $\sin x/(1-x)$ Ιs there any fast way to calculate the first four non-zero terms a Taylor Series $\dfrac {\sin x}{1-x}$ at $x=0$ without making big derivatives calculations?
I know that $$\sin x = x- \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} \dots$$
and $$\frac{1}{1-x} = 1 + x + x^2 + x^3 \dots$$
Can we combine them together?
| Maybe
$$(1-x)(a_0+a_1x+a_2x^2+a_3x^3\cdots)=x-\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}\cdots$$
$$a_0=0\\a_1-a_0=1\\a_2-a_1=0\\a_3-a_2=-\frac1{3!}\\a_4-a_3=0\\a_5-a_4=\frac1{5!}\\\cdots$$
so that
$$a_{2k+2}=a_{2k+1}=\sum_{i=0}^k\frac{(-1)^i}{(2i+1)!}.$$
The coefficients are the successive Taylor approximations of $\sin(1)$.
| {
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Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad \quad \quad \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$
By observation if all the three value
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.
| Since $3$ can be written as
$$\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+a^2}{b^2+a^2}$$
we have$$\begin{align}\\&\frac{m-a^2}{b^2+c^2}+\frac{m-b^2}{a^2+c^2}+\frac{m-c^2}{b^2+a^2}=\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+a^2}{b^2+a^2}\\&\Rightarrow \frac{m-a^2-b^2-c^2}{b^2+c^2}+\frac{m-a^2-b^2-c^2}{a^2+c^2}+\frac{m-a^2-b^2-c^2}{b^2+a^2}=0\\&\Rightarrow (m-a^2-b^2-c^2)\left(\frac{1}{b^2+c^2}+\frac{1}{a^2+c^2}+\frac{1}{b^2+a^2}\right)=0\\&\Rightarrow m-a^2-b^2-c^2=0\\&\Rightarrow m=a^2+b^2+c^2\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Exponential Simultaneous Equations Solve the following simultaneous equations:
$$2^x + 2^y = 10$$
$$x + y = 4$$
Looking at it, it is obvious that the answers are $(3,1)$ and $(1,3)$, however, I was wondering if they could be solved algebraically. Here's my approach:
$$2^x + 2^{4-x} = 10$$
$$2^x + \frac{(2^4)}{(2^x)} = 10$$
$$2^x + \frac {16}{2^x} = 10$$
And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.
| Alternatively,
$$2^x+2^y=10\\2^x\cdot2^y=16.$$
You know the sum and the product, so by $\left(2^x-2^y\right)^2=\left(2^x+2^y\right)^2-4\cdot 2^x\cdot 2^y$, you get
$$2^x,2^y=2,8.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation in the following question Solve the equation $$\sqrt{x+1}+\sqrt{2-x}+2=x^2+2x$$
I tried with squaring both sides, but I got confused, any help will be thankful.
| You'll want just the terms with square roots on one side before you square both sides:
$$\sqrt{x+1}+\sqrt{2-x}=x^2+2x -2$$
Squaring, then simplifying so that again only the terms with square roots are on the left,
$$
(x+1) + 2\sqrt{x+1}\sqrt{2-x} + (2-x) = x^4+4x^3-8x+4\\
2\sqrt{x+1}\sqrt{2-x} = x^4+4x^3-8x+1\\
$$
Squaring again,
$$
4(x+1)(2-x) = x^8+8x^7+16x^6-16x^5-62x^4+8x^3+64x^2-16x+1\\
x^8+8x^7+16x^6-16x^5-62x^4+8x^3+68x^2-20x-7=0
$$
At this point I would throw my hands in the air and ask Wolfram Alpha, which tells me that there are four solutions; checking them in the original system, only one of them works: $x \approx 1.31295$.
What is the context of this problem? It would be extremely difficult to solve this by hand.
| {
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Integrating $\int\frac{xe^{2x}}{(1+2x)^2} dx$ I need help in integrating
$$\int \frac{x e^{2x}}{(1+2x)^2} dx$$
I used integration by parts to where $u=xe^{2x}$ and $dv=\frac{1}{(1+2x)^2}$ to obtain
$$=\frac{-1}{2(1+2x)}\left[e^{2x}+2xe^{2x}\right] + \frac{1}{2}\int \frac{1}{(1+2x)}\left(2e^{2x}+\frac{d}{dx}\left(2xe^{2x}\right)\right)dx$$
Which is pretty long, which made me question whether I'm doing it correctly or not. Are there any other ways to properly evaluate the integral?
Do note though that I already know the answer to this one since it was given in a textbook, I just don't know how to arrive at such an answer.
$$\int \frac{xe^{2x}}{(1+2x)^2}dx = \frac{e^{2x}}{4+8x}$$
| \begin{align}
& \int \frac{x e^{2x}}{(1+2x)^2} dx \\
= & -\frac{1}{2}\int xe^{2x}d\left(\frac{1}{1 + 2x}\right) \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int \frac{e^{2x} + 2xe^{2x}}{1 + 2x}dx \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int e^{2x}dx \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{4}e^{2x} \\
= & \frac{e^{2x}}{4 + 8x}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the definite integral Find the definite integral
$$\int_{1}^{2}\frac{(3x-1)(2x+3)}{x} dx$$
I have come to an answer of $16 - \ln(8)$ which I think is very wrong..
First used integration by parts
| If I'm interpreting your question correctly, your answer is right. A full derivation is as follows:
\begin{align*}
\int_1^2 \frac{(3x-1)(2x+3)}{x} \, dx &= \int_1^2 \frac{6x^2 + 7x -3}{x} \, dx \\
&= \int_1^2 6x \, dx + \int_1^2 7 \, dx - \int_1^2 \frac{3}{x} \, dx \\
&\phantom{=} \\
&= 6 \left. \frac{x^2}{2} \right|_1^2 + 7 x \biggr \rvert_1^2 - 3 \ln x \biggr \rvert_1^2 \\
&\phantom{=} \\
&= 6\left( \frac{4}{2} - \frac{1}{2} \right) + 7(2-1) - 3 (\ln 2 - \ln 1) \\
& \phantom{=} \\
& = 16 - 3 \ln 2 \\
& = 16 - \ln 2^3 \\
& = 16 - \ln 8
\end{align*}
| {
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Integration by u-substitution. Evaluate $$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) \mathrm{d}x$$
My attempt: Let $u = \dfrac{x^3}{3}+1$ so $\mathrm{d}u = x^2\mathrm{d}x$
$$\int_0^1\cos(u)\cos(u)\mathrm{d}u$$
Then I am stuck trying to integrate $\cos^2(x)$. Any help is good. Thanks.
| $$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) dx$$
Let $u=\sin\left(\frac{x^3}{3}+1\right)$, then
$$du=x^2\cos\left(\frac{x^3}{3}+1\right)dx$$
So now
$$\int_{\sin 1}^{\sin\frac43} \cos u\ du=\sin\left(\sin \frac43 \right)-\sin\left(\sin 1\right)$$
| {
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How can I complete this proof the cantor set does not meet $\left(\frac{3s+1}{3^k}, \frac{3s+2}{3^k}\right)$? We define the cantor set $K$ as the set of all ternary numbers $0.{a_1a_2a_3\cdots}$ such that $a_i = {0, 2}$ for all $i$, i.e. no digit is allowed to be 1, and the first one is zero.
Here is my proof in progress.
Theorem For all $x \in K$, either $x \leq \frac{3s + 1}{3^k}$ or $x \geq \frac{3s + 2}{3^k}$
Proof We proceed by induction on $k$.
(Base Case) For $k = 1$ we show that $x \in K$ implies $x \leq s + \frac{1}{3}$ or $x \geq s + \frac{2}{3}$ for all natural $s$. By case analysis on the first digit, either $x \leq \frac{1}{3}$ or $x \geq \frac{2}{3}$. Notice $x \leq \frac{1}{3}$ implies $x \leq \frac{1}{3} + s$ for all $s$. For $s \geq 1$, $s + \frac{2}{3} \geq \frac{5}{3} > 1$ and so there is no $x \in K$ such that $x \geq s + \frac{2}{3}$.
(Inductive case)
Suppose that $x = 0.a_1a_2\cdots$. If $a_1 = 0$, then $3x \in K$ so by the inductive hypothesis we have two cases. The first case is $3x \leq \frac{3s + 1} {3^k}$ for all $s$, so $x \leq \frac{3s+1}{3^{k + 1}}$ for all $s$. The second case is that $3x \geq \frac{3s + 2}{3^k}$ for all $s$, so $x \geq \frac{3s + 2}{3^{k + 1}}$ for all $s$.
Now here is where I get stuck. If $a_1 = 2$, then $3x - 2 \in K$, so we can apply the inductive hypothesis and similar reasoning to obtain that, for all $s$ either $x \leq \frac{3s + 1}{3^{k + 1}} + \frac{2}{3}$ or $x \geq \frac{3s+2}{3^{k + 1}} + \frac{2}{3}$. The problem is of course that rescaling the interval $[\frac{2}{3}, 1]$ to $[0, 1]$ "shifts" the $s$ value when compared to the intervals excluded in the previous step of the induction. I have an intuition that if $a_1 = 1$ it suffices to consider the case where $\frac{3s + 1}{3^{k + 1}} \geq \frac{2}{3}$, as the other cases are covered by the case where $a_1 = 0$, but I have no idea how to formalise this intuition. So could someone give me hints as to how to complete this proof, or at least how to formalise the aforementioned intuition?
| Here is a different approach. Consider an interval $I_k = \left(\frac{3s + 1}{3^k}, \frac{3s + 2}{3^k}\right)$. If $x \in [0, 1]$ is in this set, then $s < 3^{k - 1}$. So $s$ can be written in ternary with $k - 1$ digits $s_i$ as follows: $s_1, s_2,\dots, s_{k -1}$. Now, $\frac{3s + 1}{3^k} = \frac{s}{3^{k - 1}} + \frac{1}{3^k}$. Written in ternary, this is:
$$0.s_1,s_2,\dots, s_{k -1},1,0,0,\dots$$
Similarly,
$$\frac{3s + 2}{3^k} = \frac{s}{3^{k - 1}} + \frac{2}{3^k} = 0.s_1,s_2,\dots, s_{k -1},1,2,2,\dots$$
Now, any number strictly between* these numbers can only be written as
$$0.s_1,s_2,\dots, s_{k -1},1,x_1,x_2,x_3\dots$$ with some $x_i \neq 0$ and some $x_j \neq 2$. Clearly, this representation contains a $1$, and so it cannot be in the Cantor set.
*We do not include the extremes themselves because they are special cases, without unique representations. E.g. $0.s_1,s_2,\dots, s_{k -1},1,0,0,\dots = 0.s_1,s_2,\dots, s_{k -1},0,2,2,\dots$. This is a quirk of place-value number systems.
| {
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$\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$ $\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$
This is in the form of $\frac{1}{linear\sqrt{quadratic}}$.I put $x=\frac{1}{t}$
$\int_{\frac{2}{\sqrt2-1}}^{\infty}\frac{dt}{(2+t)\sqrt{t+1}}$Then put $t+1=p^2$
From now,it got complicated.Its answer is $\frac{\pi}{4}$.Answer is elusive.
| $$\int_{2/(\sqrt2-1)}^{\infty}\frac{dt}{(2+t)\sqrt{t+1}}$$
Let $t+1=u^2\implies dt=2udu$ $$\int_{\sqrt 2-1}^{\infty}\frac{2udu}{(1+u^2)u}$$
$$=2\int_{\sqrt 2-1}^{\infty}\frac{du}{1+u^2}$$
$$=2\left[\tan^{-1}(u)\right]_{\sqrt 2-1}^{\infty}$$
$$=2\left[\tan^{-1}(\infty)-\tan^{-1}\left(\sqrt 2-1\right)\right]$$
$$=2\left[\frac{\pi}{2}-\frac{\pi}{8}\right]$$
$$=2\frac{3\pi}{8}$$$$=\frac{3\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine whether the locus of the point P will intersect the straight line $y=-1$ A point $P(x,y)$ moves in such a way that its distance from the point $A(3,1)$ is always three times its distance from the straight line $x=-1$.
(a) Find the equation of the locus of the moving point P.
(b) Determine whether the locus of the point P will intersect the straight line $y=-1$.
I have solved (a) but was unable to solve (b). I have tried to substitute $y=-1$ into the locus equation.
| a) The distance of the point $P(x, y)$ from the point $A(3, 1)$ is such that $$\sqrt{(x-3)^2+(y-1)^2}=3\times \frac{|x+1|}{\sqrt{1^2+(0)^2}}$$
$$(x-3)^2+(y-1)^2=9(x+1)^2$$
$$x^2-6x+9+y^2-2y+1=9(x^2+2x+1)$$
$$8x^2-y^2+24x+2y-1=0$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the locus of point P:}\ 8x^2-y^2+24x+2y-1=0}}$$
Above is the locus of the point $P(x, y)$
b) setting $y=-1$ in the above equation, we get $$8x^2-(-1)^2+24x+2(-1)-1=0$$
$$2x^2+6x-1=0$$ Checking the nature of roots of above quadratic equation by using discriminant $\Delta$, as follows
$$\Delta=B^2-AC=(6)^2-4(2)(-1)=44>0$$ Above positive value of the discriminant shows that there are two distinct real roots i.e. the locus of the point P intersects the straight line: $y=-1$ at two different points .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Convergent/divergent series Is the following series divergent/convergent?
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks
| After placing brackets around the terms with the same sign the series becomes
$$1+\sum_{n=2}^\infty(-1)^{n+1}(H_{n(n+1)/2}-H_{n(n-1)/2}),$$
where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the $n$-th harmonic number. Now using well-known formula $$H_n=\log n+\gamma+O\left(\frac{1}{n}\right),$$
we get
$$a_n=H_{n(n+1)/2}-H_{n(n-1)/2}=\log\frac{n+1}{n-1}+O\left(\frac{1}{n^2}\right)=$$
$$=\log\left(1+\frac{2}{n-1}\right)+O\left(\frac{1}{n^2}\right)=\frac{2}{n}+O\left(\frac{1}{n^2}\right).$$
Therefore the main term of the series with braces is $$2\frac{(-1)^{n+1}}{n}+O\left(\frac{1}{n^2}\right),$$
which means that the series converges (by Leibniz's and comparison tests), and so does the original one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle If in a triangle $ABC$,$c$ is the longest side and $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle.
$a^2+b^2=2Rc\Rightarrow a^2+b^2=\frac{c^2}{\sin C}$
$\sin C=\frac{c^2}{a^2+b^2}$,how to proceed ahead?I am stuck.
| You are correct to use the sine law.
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \tag{1}$$
If you apply the sine law to sides $a,b$ the main equation becomes:
$$\begin{align}
&4R^2 \sin^2A + 4R^2 \sin^2B = 4R^2 \sin C \implies \\
&\sin^2A + \sin^2B = \sin C\tag{2}
\end{align}$$
For a triangle $C = 180^\circ - (A+B)$, so then
$$\sin C = \sin(A+B)$$
Substituting this into (2), we get:
$$\sin^2A + \sin^2B = \sin (A+B) \tag{3}$$
Can you show that a solution is $A+B=90^\circ$, so that $C=90^\circ$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve complex equation $5|z|^3+2+3 (\bar z) ^6=0$ I'm stuck in trying to solve this complex equation
$$ 5|z|^3+2+3 (\bar z)^6=0$$
where $\bar z$ is the complex conjugate.
Here's my reasoning: using $z= \rho e^{i \theta}$ I would write
$$ 5\rho^3+ 2 + 3 \rho^6 e^{-i6\theta} = 0 \\ 5\rho^3+ 2 + 3 \rho^6 (\cos(6 \theta) - i \cdot \sin(6 \theta)) = 0 \\$$
from where I would write the system
$$\begin{cases} 5\rho^3+ 2 + 3 \rho^6 \cos(6 \theta) = 0 \\ 3 \rho^6 \sin(6 \theta) = 0\end{cases}$$
But here I get an error, since, from the second equation, I would claim $ \theta = \frac{k \pi}{6}$ for $ k=0…5$, but $\theta = 0$ means the solution is real and the above equation doesn't have real solutions…where am I mistaken?
| Let $w = \overline{z}^3$. Then we have
$$
5|w|+2+3w^2 = 0
$$
As you point out, this constrains $w = k$ or $w = ki$ for real $k$.
Case 1. $w = k$
$$
3k^2+5|k|+2 = 0
$$
which yields no solutions since the left-hand-size is always positive.
Case 2. $w = ki$
$$
-3k^2+5|k|+2 = 0
$$
which yields $k = \pm 2$, so $w = \pm 2i$.
The rest is left as an exercise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction: for $n \ge 0$, $\frac{(2n)!}{n!2^n}$ is an integer Another prove by induction question:
for $n \ge 0$, $$\frac{(2n)!}{n!2^n}$$ is an integer
Base step:
$$n = 0$$
$$\frac{(2 \times 0)!}{0! \times 2^0} = \frac{0!}{1 \times 1} = 1$$
Induction step: please help
| Proof:: (Without Using induction::) Let $$Z=\frac{(2n)!}{n!\cdot 2^n} = \frac{(2n)\cdot (2n-1)\cdot (2n-2)\cdots4\cdot 3 \cdot 2 \cdot 1}{n!\cdot 2^n}$$
so $$Z = \frac{\underbrace{(2n)\cdot (2n-2)\cdot (2n-4)\cdots4\cdot 2}\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{n!\cdot 2^n}$$
So $$Z = \frac{2^n\cdot \left[n\cdot (n-1)\cdot (n-2)\cdots 4\cdot 2\right]\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{n!\cdot 2^n}$$
So $$Z = \frac{2^n\cdot n!\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{2^n\cdot n!}$$
So $$Z = \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}$$ is an Integer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Set up difference equation for the following recurrence. I have the following recurrence:
$t=0: 0$
$t=1: 0$
$t=2: 1$
$t=3: \beta+\alpha$
$t=4: (\beta+\alpha)\alpha+\beta^2$
$t=5: ((\beta+\alpha)\alpha+\beta^2)\alpha+\beta^3$
...
I was hoping to do something like:
$x_{t+1}=x_{t}f_{t}+g_{t}$ with $f_{t}=\alpha$ and $g_{t}=\beta^{t-1}$
But it does not fit with $t=0$ and $t=1$ both has to be $0$.
Is there anyway to set up this difference equation without using indicator functions or piecewiese functions for $g_{t}$?
| Call your value $a_t$, after $t = 3$ it is $a_{t + 1} = \alpha a_t + \beta^{t - 2}$, with $a_3 = \alpha + \beta$. Set up a generating function $A(z) = \sum_{t \ge 0} a_{t + 3} z^t$, multiply the recurrence by $z^t$ and sum over $t \ge 0$:
$\begin{align}
\sum_{t \ge 0} a_{t + 4} z^t
&= \alpha \sum_{t \ge 0} a_{t + 3} z^t
+ \sum_{t \ge 0} \beta^{t + 1} z^t \\
\frac{A(z) - a_3}{z}
&= \alpha A(z) + \frac{\beta}{1 - \beta z}
\end{align}$
Plugging in $a_3 = \alpha + \beta$ and solving for $A(z)$ gives, unless $\alpha = \beta$:
$\begin{align}
A(z)
&= \frac{(\alpha + \beta) - \beta(\alpha + \beta - 1) z}
{(1 - \alpha z) (1 - \beta z)} \\
&= \frac{\beta^2 - \beta - \alpha^2}{\beta - \alpha}
\cdot \frac{1}{1 - \alpha z}
+ \frac{\beta}{\beta - \alpha} \cdot \frac{1}{1 - \beta z}
\end{align}$
and your solution is:
$$
a_{t + 3}
= \frac{\beta^2 - \beta - \alpha^2}{\beta - \alpha} \cdot \alpha^t
+ \frac{\beta^{t + 1}}{\beta - \alpha}
$$
If $\alpha = \beta$:
$\begin{align}
A(z)
&= \frac{2 \alpha - \alpha(2 \alpha - 1) z}{(1 - \alpha z)^2} \\
&= \frac{2 \alpha - 1}{1 - \alpha z}
+ \frac{1}{(1 - \alpha z)^2}
\end{align}$
from which you have directly:
$$
a_{t + 3}
= (2 \alpha - 1) \cdot \alpha^t
+ (n + 1) \cdot \alpha^t
= (n + 2 \alpha) \cdot \alpha^t
$$
Earlier values don't follow the pattern.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to Translate two Equations for a "+/-" For a National Board Exam Review:
Find the Equation for the Asymptotes of a Hyperbola ${ (y-x)^2 - (x+5)^2 = 36 }$
Answer is ${ y-5 = \pm (x+5) }$
I've already solved the equations: here they are:
$${ y = x+10 }$$
$${ y = -x }$$
My problem is how to translate it into this " ${ y-5 = \pm (x+5) }$ " ? I know that if you reverse engineer the equation by doing seperate equations for each you could end up with my answer... But I want to know if there is a method for methodically translating my answer to the one with the ${\pm}$ sign on it...
| Using
$$b=\frac{b+c}{2}+\frac{b-c}{2}\quad\text{and}\quad c=\frac{b+c}{2}-\frac{b-c}{2},$$
we can see that
$$y=ax+b$$
$$y=-ax+c$$
can be written as
$$y=ax+\frac{b+c}{2}+\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=ax+\frac{b-c}{2}$$
$$y=-ax+\frac{b+c}{2}-\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=-ax-\frac{b-c}{2},$$
i.e.
$$y-\frac{b+c}{2}=\pm\left(ax+\frac{b-c}{2}\right)$$
Your case is $(a,b,c)=(1,10,0)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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mathematical induction proof of a square vs factorial So lets say I have $$ n^{2} \le n! $$
For what positive integers is this not true? $n=2$ and $ 3$
Base case? $$n=4 \implies 16 \le 24 $$
What is the inductive hypothesis and how do I show the inductive proof? Thanks
| For, $ x_0 = 4 $, $ n^2 \le n! $ is true.
Now, given, $n^2 \le n!$,
$(n+1)^2 = n^2 + 2n + 1$,
and, $ n^2 \le n! \implies n^2(n+1)^2 \le (n+1)^2n!$
$\implies (n+1)^2 \le (\frac{1}{n^2} + \frac{1}{n}) * (n+1)!$
$n > 4 => \frac{1}{n^2} < \frac{1}{16} $ and $\frac{1}{n} < \frac{1}{4} $, So,
$ (\frac{1}{n^2} + \frac{1}{n}) \lt \frac5{16} < 1$
As, $ (n+1)^2 \le \frac5{16} * (n+1)! \implies (n+1)^2 \le (n+1)!$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int\frac{x-1}{x^2-5x+6}dx$. Why my solution is different from book? I'm learning single variable calculus right now. Right now trying to understand integration with partial fraction. I'm confused in a problem from sometime. I think I'm doing right but answer in my book is something else.
Please have a look at the images.
The Solution given in my Book
I know there's is a difference in finding value of A and B. But in previous exercise I was applying the same method and was getting correct answer. Please help. thankyou in advance.
| HINT: we get $$\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}$$
can you show this?
now we calculate the zeros of $x^2-5x+6$, these are:$$x_{1,2}=\frac{5}{2}\pm\sqrt{\frac{25}{4}-\frac{24}{4}}$$ thus we get $$x_1=3$$ or $$x_2=2$$
and we get $$x^2-5x+6=(x-2)(x-3)$$ and we can make the ansatz:
$$\frac{5x-5}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}$$
multiplying by the denominators we obtain:
$$\frac{5x-5}{x^2-5x+6}=\frac{A(x-3]+B(x-2)}{x^2-5x+6}$$
and we get $$5x-5=x(A+B)-3A-2B$$
from here you will get $$5=A+B$$ and $$-5=-3A-2B$$ you must solve this system
| {
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"timestamp": "2023-03-29T00:00:00",
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How is that a rotation by an angle θ about the origin can be represented by this transformation matrix? $$
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
$$
How was this matrix derived? I know how to use it, but where did it come from? Can someone prove why this matrix represents a rotation about the origin by an angle theta?
| Consider what happens when we take a point $(x,y)$ and rotate it by an angle of $\theta$.
Using polar coordinates, we can write $(x,y) = (r\cos \phi,r\sin \phi)$ where $r$ is the distance from $(x,y)$ to the origin and $\phi$ is the angle made by the line from the origin to $(x,y)$ and the positive $x$-axis.
After rotating $(x,y)$ by an angle of $\theta$, the new point is still $r$ units away from the origin, but now, the angle made by the line from the origin to $(x',y')$ and the positive $x$-axis increases by $\theta$. Hence, $(x',y') = (r\cos(\phi+\theta),r\sin(\phi+\theta))$.
Now, let's use the cosine and sine sum of angle formulas to write $x'$ and $y'$ in terms of $x$ and $y$:
$$x' = r\cos(\phi+\theta) = r\cos\phi\cos\theta - r\sin\phi\sin\theta = x\cos\theta - y\sin\theta$$
$$y' = r\sin(\phi+\theta) = r\sin\phi\cos\theta + r\cos\phi\sin\theta = y\cos\theta + x\sin\theta$$
Putting this into matrix form gives us:
$$\begin{pmatrix}x' \\ y'\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$$
Hence, the matrix which rotates a point by an angle of $\theta$ counterclockwise is $\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$.
Remark: To add to what David C. Ullrich suggested, notice that the columns of this matrix $\begin{pmatrix}\cos\theta \\ \sin\theta \end{pmatrix}$ and $\begin{pmatrix}-\sin\theta\\ \cos\theta\end{pmatrix}$ are precisely the result of rotating the vectors $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ counterclockwise about the origin by an angle of $\theta$. This same property holds for other linear transformations. Specifically, the $j$-th column of the matrix of a transformation is the result of applying the transformation to the $j$-th basis vector.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-form of $\int_0^1 \operatorname{Li}_3\left(1-x^2\right) dx$ By using dilogarithm functional equations we can show that
$$
\int_0^1 \operatorname{Li}_2\left(1-x^2\right)\,dx = \frac{\pi^2}{2}-4,
$$
where $\operatorname{Li}_2$ is the dilogarithm function.
Could we evaluate in closed-form the following integral?
$$
I = \int_0^1 \operatorname{Li}_3\left(1-x^2\right)\,dx,
$$
where $\operatorname{Li}_3$ is the trilogarithm function.
A related integral with known closed-form is
$$\int_0^1 \operatorname{Li}_3\left(\frac{1}{x^2}\right)\,dx = \zeta(3)+\frac{\pi^2}{3}-8\ln2 - 4\pi\,i,$$
where $\zeta$ is the Riemann zeta function.
| \begin{align}
I&=\int_0^1\operatorname{Li}_3(1-x^2)\ dx\overset{IBP}{=}2\int_0^1\frac{x^2\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx\\
&=2\int_0^1\left(\frac1{1-x^2}-1\right)\operatorname{Li}_2(1-x^2)\ dx\\
&=2\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx-2\int_0^1\operatorname{Li}_2(1-x^2)\ dx\tag{1}
\end{align}
By the OP, the second integral is $\boxed{\frac{\pi^2}{2}-4}$.
To calculate the first integral, we are going to use the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.
$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=1$ and replacing $x$ with $1-x^2$ we get
$$\frac{\operatorname{Li}_{2}(1-x^2)}{1-x^2}=-\int_0^1\frac{\ln(u)}{1-ux+ux^2}\ du$$
Now we can write
$$\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx=-\int_0^1\ln u\left(\int_0^1\frac{dx}{1-ux+ux^2}\right)\ du$$
$$=-\int_0^1\ln u\left(\frac{\arctan\sqrt{\frac{u}{1-u}}}{\sqrt{u-u^2}}\right)\ du, \quad \color{red}{\arctan\sqrt{\frac{u}{1-u}}=\arcsin\sqrt{u}=x}$$
$$=-4\int_0^{\pi/2}x\ln(\sin x)\ dx=-4\left(\frac7{16}\zeta(3)-\frac{\pi^2}{8}\ln2\right)=\boxed{\frac{\pi^2}{2}\ln2-\frac74\zeta(3)}$$
where the last result follows from the Fourier series of $\ln(\sin x)=-\ln2-\sum_{n=1}^\infty \frac{(-1)^n \cos(2nx)}{n}$.
Plugging the boxed results of the two integrals in $(1)$, we get
$$I=\pi^2\left(\ln 2-1\right)-\frac72\zeta\left(3\right)+8$$
Note: Since $$\arctan x=-\frac{i}{2}\ln\left(\frac{1+ix}{1-ix}\right)$$
Then
\begin{align}
\arctan\frac{x}{\sqrt{1-x^2}}&=-\frac{i}{2}\ln\left(\frac{1+\frac{ix}{\sqrt{1-x^2}}}{1-\frac{ix}{\sqrt{1-x^2}}}\right)\\
&=-\frac{i}{2}\ln\left(\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}-ix}*\color{red}{\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}+ix}}\right)\\
&=-\frac{i}{2}\ln\left(\frac{(\sqrt{1-x^2}+ix)^2}{1}\right)\\
&=-i\ln\left(\sqrt{1-x^2}+ix\right)\\
&=\arcsin x
\end{align}
and if we replace $x$ with $\sqrt{x}$, we get
$$\arctan\sqrt{\frac{x}{1-x}}=\arcsin\sqrt{x}$$
Here is a different proof:
Since $$\frac{d}{dy}\arctan\frac{y}{\sqrt{1-y^2}}=\frac1{\sqrt{1-y^2}}$$
Then $$\left.\arctan\frac{y}{\sqrt{1-y^2}}\right|_0^x=\int_0^x\frac1{\sqrt{1-y^2}}\ dy$$
$$\arctan\frac{x}{\sqrt{1-x^2}}=\arcsin x$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the minimum of the $a^2+b^2+c^2+a+b+c+ab+bc$ (1): Let $a,b\in R$, Find this minimum of the value $a^2+b^2+ab+a+b$
I have prove $a^2+b^2+ab+a+b\ge-\dfrac{1}{3}$
(2): Let $a,b,c\in R$, find the minimum of the value
$$a^2+b^2+c^2+a+b+c+ab+bc$$
I think when $a=b=c=-\dfrac{1}{3}$ one gets a minimum.
(3): How to find the minimum $n>1,x_{i}\in R$
$$M=\sum_{i=1}^{n}(x^2_{i}+x_{i})+\sum_{i=1}^{n-1}x_{i}x_{i+1}$$For specific n, for example, n=2, I can get: $\min{M}=-\dfrac{1}{3}$
But in above generalized problem, how can I solve that ? Can I get the result by using sage or mathematica ?
| EDIT1: as Macavity pointed out the mistake, I correct solution.
hint :
$2M=(x_1+\dfrac{1}{2})^2+\sum_{i=1}^{n-1}(x_i+x_{i+1}+\dfrac{1}{2})^2+(x_n+\dfrac{1}{2})^2+ 2M_{min}$
but this is only work with $n=2k+1$ for $x_i$ can get reachable number.
for $n=2k$, I got answer but I am trying to find out a simple solution.
Edit2: following is the solution.
$2M=(x_1+\dfrac{k}{2k+1})^2+(x_1+x_2+\dfrac{k+1}{2k+1})^2+(x_2+x_3+\dfrac{k}{2k+1})^2+(x_3+x_4+\dfrac{k+1}{2k+1})^2+(x_4+x_5+\dfrac{k}{2k+1})^2+...+(x_{n-1}+x_n+\dfrac{k+1}{2k+1})^2+(x_n+\dfrac{k}{2k+1})^2+2M_{min}$
| {
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Which is the limit of the sequence $\sum_{i=1}^n \frac{\cos(i^2)}{n^2+i^2}$ I can't find the limit as n $\to$ infinity of the sequence:
$$\frac{\cos(1)}{n^2 + 1} + \frac{\cos(4)}{n^2 + 4} + \dots + \frac{\cos(n^2)}{n^2 + n^2}$$
I tried to use the inequality $\cos(n^2) < n$ which I could justify it by using Taylor series of $\cos x$ and replacing the $x$ with $x^2$ but my professor told me I was wrong.Plus I plotted $\cos(x^2)$ and $x$ and I saw that he was right. So I don't have any ideas left and I need some help here. Even though this is an exercise that my professor gave me it is not an exercise for the semester or anything like that it's more like exercise that he gave me for fun. (I asked for it.I should have been more careful :) )
| $$\lim\limits_{n\to\infty}\left(\frac{\cos 1}{n^2 + 1}+ \frac{\cos 4}{n^2 + 4}+ \dots + \frac{\cos n^2}{n^2 + n^2}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{\frac{\cos 1}{n^2}}{1 + \frac{1}{n^2}}+ \frac{\frac{\cos 4}{n^2}}{1 + \frac{4}{n^2}}+ \dots + \frac{\frac{\cos n^2}{n^2}}{1 + 1}\right)$$
Since the limit of each term as $n\to \infty$ yields a convergent limit, we have
$$\lim\limits_{n\to\infty} \frac{\frac{\cos 1}{n^2}}{1 + \frac{1}{n^2}}+\lim\limits_{n\to\infty} \frac{\frac{\cos 4}{n^2}}{1 + \frac{4}{n^2}}+ \dots + \lim\limits_{n\to\infty}\frac{\frac{\cos n^2}{n^2}}{1 + 1}$$
$$=\frac{0}{1 + 0}+\frac{0}{1 + 0}+ \dots + \frac{0}{1+1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the general method for solving $Ax = b$ when A is rectangular Suppose I am given that
$A = \begin{bmatrix} 2 &4& 6 &-2 \\ 1& 0& 1 &1 \\ 0& 2 &2& -2 \end{bmatrix}$
and
$b = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}$
What should I do to solve for $x$?
Should I attempt solving the simultaneous equation:
$2x_1 + 4x_2 + 6x_3 - 2x_4 = 4$
$x_1 + x_3 + x_4 = 2$
$2x_2 + 2x_3 -2x_4 = 1$
Is this the only way to approach this problem?
|
Is this the only way to approach this problem?
In fact, $Ax=b$ is equivalent to the given system of linear equations (just a different way of writing it).
But it would be easier to solve if we convert the augmented matrix to row echelon form first:
\begin{align*}
\left[\begin{array}{cccc|c} 2 & 4 & 6 & -2 & 4 \\ 1 & 0 & 1 & 1 & 2 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right]
& \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 2 & 4 & 6 & -2 & 4 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_2 \gets R_2-2R_1}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 4 & 4 & -4 & 0 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_2 \gets \tfrac{1}{4} R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_3 \gets R_3-2R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right].
\end{align*}
Since row operations preserve the set of solutions, the system of equations $Ax=b$ will have the same set of solutions as the system of equations
$$
\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 &
0 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ Struggling with this proof.
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.
| $$\left(\frac{a+b}{2}\right)^n = \frac{ \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k}{2^n}= \frac{ \sum_{k=0}^n \binom{n}{k} a^{k}b^{n-k}}{2^n}$$
Therefore
$$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n}$$
Now for each $k$ we have by AM-GM:
$$a^{n-k}b^k \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{(n-k)a^{n}+kb^n}{n} $$
and similarly
$$a^{k}b^{n-k} \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{ka^{n}+(n-k)b^n}{n} $$
Therefore, by adding them together we get
$$a^{n-k}b^k+a^kb^{n-k}\leq a^n+b^n$$
This yields
Therefore
$$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n} \leq\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n}+b^{n})}{2^n}\\=\frac{a^n+b^n}{2} \frac{ \sum_{k=0}^n \binom{n}{k} }{2^n} =\frac{a^n+b^n}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Give the equations that are a tangent to the parabola $y = x^2 + 5x + 6$ and pass through $(1,1)$ I have been given the question:
Give the equations that are a tangent to the parabola: $y = x^2 + 5x + 6$ and pass through the point $(1,1)$
I have tried two different methods for solving this. The first I don't know why am I incorrect (I would appreciate if someone would explain to me where I went wrong).
First Attempt:
$y = x^2 + 5x + 6$
Use $y - y_1 = m(x - x_1)$ to create lines that go through $(1,1)$
$y - 1 = m(x - 1)$
$y = mx + 1 - m$
Find when these lines intersect with the parabola:
$x^2 + 5x + 6 = mx + 1 - m$
$x^2 + x(5 - m) + (5 + m) = 0$
Find for what values of M these lines only intersect once using $b^2 - 4ac = 0$
$(5-m)^2 - 4(5+m) = 0 = m^2 - 14m + 5$
$(m - 7)^2 - 49= -5$
$m = \pm \sqrt{44} + 7$
Giving the equations:
$$
y = (\sqrt{44} + 7)x - \sqrt{44} + 1
$$
$$
y = (-\sqrt{44} + 7)x + \sqrt{44} + 1
$$
With the second attempt I am lost as to what that last equation is representing so I just got confused.
second:
$ y = x^2 + 5x + 6 $
$ y' = 2x + 5 $
All the lines that have the same gradient as a the parabola and pass through $(1,1)$
$ y - 1 = (2x + 5)(x - 1) $
$y = 2x^2 + 3x - 4$
Find when these intersect the parabola:
$x^2 + 5x + 6 = 2x^2 + 3x - 4$
which gave:
$x = \pm3 + 1$
From here I got lost and didn't know what was going on.
I would appreciate it if someone could explain why my first method is wrong, tell me if either of the methods are even on track to solving the issue and give an answer to the problem.
Thanks
-Kingpulse
Solution for method one
The first method was actually correct apart from at the very end.
The mistake was the incorrect substitution into $-m$, I was substituting $-\sqrt{44}$ into $-m$ not $-(\sqrt{44}+7)$.
Proof that it is a tangent:
Parabola : $y = x^2 + 5x + 6$
Find intersections of lines
$x^2 + 5x + 6 = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$
$x^2 + x(-2-\sqrt{44}) + (12 + \sqrt{44}) = 0$
$b^2 - 4ac = 0$ if it is a tangent (only one root)
$(-2-\sqrt{44})^2 - 4(12+\sqrt{44})$
$4+4\sqrt{44} + 44 - 48 -4\sqrt{44} = 0$
It is a tangent.
The equations are:
$$ y = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$$
and
$$ y = (-\sqrt{44} + 7)x - (-\sqrt{44} + 7) + 1$$
| Your computation for $b^2-4ac$ should have $5+m$ in the second term.
Also in the second method your notation is confusing you and you have $y=2x^2+3x-4$ as the equation of a straight line, which is wrong. You need to deal with the point you know, which is $(x_0,y_0)$ on the parabola, where the gradient is $2x_0+5$, work out the equation of the tangent line, and then, when you have it, find the condition that it passes through $(1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Another integral related to Fresnel integrals How would we prove this result by real methods ?
$$\int_0^{\infty } \frac{\sin \left(\pi x^2\right)}{x+2} \, dx=\frac{1}{4} \left(\pi-2 \pi C\left(2 \sqrt{2}\right)-2 \pi S\left(2 \sqrt{2}\right)+2 \text{Si}(4 \pi ) \right)$$
As you can easily see, Fresnel integrals are involved. What are your ideas on it?
| Integrating $f(z)=\dfrac{e^{iaz^2}}{z+b}$ along $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ gives
\begin{align}
\int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x
&=\int^\infty_0\frac{b\cos(ay^2)-y\sin(ay^2)}{y^2+b^2}\ {\rm d}y
\end{align}
To compute the first integral, we consider the function
$\displaystyle I(a)=\int^\infty_0\frac{e^{iay^2}}{y^2+b^2}\ {\rm d}y$
such that $\displaystyle-iI'(a)+b^2I(a)=\frac{\sqrt{\pi}}{2\sqrt{2a}}(1+i)$. Solving this ode while noting the initial value $I(0)=\dfrac{\pi}{2b}$,
\begin{align}
I(a)
&=-e^{-iab^2}\left(\frac{\sqrt{\pi}}{{2\sqrt{2}}}(1-i)\right)\int \frac{\cos(ab^2)+i\sin(ab^2)}{\sqrt{a}}\ {\rm d}a\\
&=-e^{-iab^2}\left(\frac{\sqrt{\pi}}{{2\sqrt{2}}}(1-i)\right)\left[\frac{\sqrt{2\pi}}{b}C\left(b\sqrt{\frac{2a}{\pi}}\right)+i\frac{\sqrt{2\pi}}{b}S\left(b\sqrt{\frac{2a}{\pi}}\right)-\frac{\sqrt{\pi}}{b\sqrt{2}}(1+i)\right]
\end{align}
Taking the real part of $bI(a)$,
\begin{align}
\int^\infty_0\frac{b\cos(ay^2)}{y^2+b^2}\ {\rm d}y
&=\frac{\pi}{2}\bigg{[}C\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)-\cos(ab^2)\right)-S\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)+\cos(ab^2)\right)\\
&\ \ \ \ +\cos(ab^2)\bigg{]}
\end{align}
The second integral readily reduces to sine and cosine integrals.
\begin{align}
\int^\infty_0\frac{y\sin(ay^2)}{y^2+b^2}\ {\rm d}y
&=\frac{1}{2}\int^\infty_{0}\frac{\sin(ay)}{y+b^2}\ {\rm d}y\\
&=\frac{1}{2}\int^\infty_{b^2}\frac{\sin(ay)\cos(ab^2)-\cos(ay)\sin(ab^2)}{y}\ {\rm d}y\\
&=\frac{1}{2}\bigg{[}\operatorname{Si}(ay)\cos(ab^2)-\operatorname{Ci}(ay)\sin(ab^2)\bigg{]}^\infty_{b^2}\\
&=\frac{\pi}{4}\cos(ab^2)-\frac{1}{2}\operatorname{Si}(ab^2)\cos(ab^2)+\frac{1}{2}\operatorname{Ci}(ab^2)\sin(ab^2)
\end{align}
Therefore, we have a generalised result that holds for positive, real $a,b$.
\begin{align}
\color{indigo}{\int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x}
&\color{indigo}{=\frac{\pi}{2}\bigg{[}C\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)-\cos(ab^2)\right)-S\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)+\cos(ab^2)\right)}\\
&\ \ \ \ \color{indigo}{+\frac{1}{2}\cos(ab^2)\bigg{]}+\frac{1}{2}\left(\operatorname{Si}(ab^2)\cos(ab^2)-\operatorname{Ci}(ab^2)\sin(ab^2)\right)}
\end{align}
Setting $a=\pi$, $b=2$ reproduces the identity stated in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Evaluation of $\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$ Evaluation of $\displaystyle\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$
$\bf{My\; Try::}$Let $$\displaystyle I = \int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx = \int\left(\frac{a+b\cos x}{b+a\cos x}\right)\cdot \frac{2\sin x}{(b+a\cos x)^2}dx$$
Now Let $$\displaystyle \left(\frac{a+b\cos x}{b+a\cos x}\right) = t\;,$$ Then $$\displaystyle \frac{\left[a^2-b^2\right]\sin x}{(b+a\cos x)^2}dx = dt\Rightarrow \frac{\sin x}{(b+a\cos x)^2}dx = \frac{1}{(a^2-b^2)}dt$$
So Integral $$\displaystyle I = \frac{2}{(a^2-b^2)}\int tdt = \frac{1}{(a^2-b^2)}\cdot \left(\frac{a+b\cos x}{b+a\cos x}\right)^2+\mathcal{C}$$
My Question is can we solve it any other Substution, If yes then plz explain here
Thanks
| Multiply with $\cos x$. Then $a2\cos x \sin x+b\cos x \sin 2x=\sin 2x(a+b\cos x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta $ Prove $$\sin^2(\theta)+\cos^4(\theta)=\cos^2(\theta)+\sin^4(\theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$\sin^2(\theta)-\cos^2(\theta)=\sin^4(\theta)-\cos^4(\theta)$
Then,
$\sin^2(\theta)-\cos^2(\theta)=(\sin^2(\theta)+\cos^2(\theta))(\sin^2(\theta)-\cos^2(\theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
| $$ \sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^4\theta - \cos^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = (\underbrace{\sin^2\theta + \cos^2\theta}_{1})(\sin^2\theta - \cos^2\theta) $$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^2\theta - \cos^2\theta $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 8
} |
Two straight lines one being a tangent to $y^2=4ax$ and the other to $x^2=4by$ are at right angles.Find the locus of their point of intersection. Two straight lines one being a tangent to $y^2=4ax$ and the other to $x^2=4by$ are at right angles.Find the locus of their point of intersection.
I tried but could not reach final answer.The tangent to $y^2=4ax$ is $y=m_1x+\frac{a}{m_1}$ and the tangent to $x^2=4by$ is $x=m_2y+\frac{b}{m_2}$.Let the point of intersection of these tangents be $(h,k)$.
$k=m_1h+\frac{a}{m_1}$ and $h=m_2k+\frac{b}{m_2}$.Now i need to eliminate $m_1$ and $m_2$ from these two equations using $m_1m_2=-1$.Final answer is $(ah+bk)(h^2+k^2)+(bh-ak)^2=0$.But i could not reach final answer.
Please help me.
| Note that the condition we have is not $m_1m_2=-1$.
Instead, since $y=\color{red}{m_1}x+\frac{a}{m_1}$ and $x=m_2y+\frac{b}{m_2}\iff y=\color{blue}{\frac{1}{m_2}}x-\frac{b}{m_2^2}$, we have
$$\color{red}{m_1}\cdot\color{blue}{\frac{1}{m_2}}=-1\iff m_2=-m_1$$
Thus,
$$\begin{align}\\&k=m_1h+\frac{a}{m_1},\qquad h=m_2k+\frac{b}{m_2}=-m_1k-\frac{b}{m_1}\\&\Rightarrow m_1k=m_1^2h+a,\qquad m_1h=-m_1^2k-b\\&\Rightarrow m_1^2h-m_1k+a=0,\qquad m_1^2k+m_1h+b=0\\&\Rightarrow m_1^2hk-m_1k^2+ak=0,\qquad m_1^2kh+m_1h^2+bh=0\\&\Rightarrow (m_1^2kh=)\ m_1k^2-ak=-m_1h^2-bh\\&\Rightarrow m_1=\frac{ak-bh}{k^2+h^2}\end{align}$$
Putting this into $k=m_1h+\frac{a}{m_1}$ gives
$$\begin{align}\\&k=\frac{ak-bh}{k^2+h^2}h+\frac{a(k^2+h^2)}{ak-bh}\\&\Rightarrow k(k^2+h^2)(ak-bh)=h(ak-bh)^2+a(k^2+h^2)^2\\&\Rightarrow h(ak-bh)^2-(k^2+h^2)(ak^2-bhk-ak^2-ah^2)=0\\&\Rightarrow (ak-bh)^2+(k^2+h^2)(ah+bk)=0\\&\Rightarrow (ah+bk)(h^2+k^2)+(bh-ak)^2=0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a Polynomial in $x-\frac1x$ Given that $x^n - (1/x^n)$ is expressible as a polynomial in $x - (1/x)$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P(x-(1/x)) = x^5 - (1/x)^5.$
To start, I factored, giving:
$P(x-\frac{1}{x}) = (x-\frac{1}{x})(x^4+\frac{1}{x^4}+x^2+\frac{1}{x^2}+1).$
However, I cannot find a way to connect this to $x-\frac{1}{x}$. What should I do?
| One has if we denote $Y_n=x^n-1/x^n$ and $Y_1=Y$
$$\begin{align}
\left(x-{1\over x}\right)^5
&=x^5-5x^3+10x-{10\over x}+{5\over x^3}-{1\over x^5}\\
&=Y_5-5Y_3+10Y
\end{align}$$
And
$$\begin{align}
\left(x-{1\over x}\right)^3
&=x^3-3x+{3\over x}-{1\over x^3}\\
&=Y_3-3Y
\end{align}$$
We then derive from the second identity that $5Y_3=5Y^3+15Y$ and therefore
$$\begin{align}x^5-{1\over x^5}&=Y_5\\&=Y^5+5Y_3-10Y\\&=Y^5+5Y^3+5Y\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Trouble with partial fractions and complex numbers $$f(z) = \frac{3}{(z+1)(z-i)} = \frac{A}{z+1} + \frac{B}{z-i}$$
$$z(A + B) + B - Ai = 3$$
$$A + B = 0$$
$$B-Ai=3$$
Somehow, I end up with $B = - \frac{3}{(1+i)}$
Is that the way to go?
| One practical way to do partial fractions which are of the type $\frac{1}{(x-a)(x-b)}$ is to simply let the answer be $\frac{1}{x-a}-\frac{1}{x-b}$ and adjust the constant term after calculation. In our case we have:
$$\frac{1}{z+1}-\frac{1}{z-i}=\frac{z-i-z-1}{(z+1)(z-i)}\\=-\frac{1+i}{(z+1)(z-i)}$$ from which one directly infers that we must have:
$$\frac{1}{(z+1)(z-i)}=-\frac{1}{1+i}\Big(\frac{1}{z+1}-\frac{1}{z-i}\Big)$$
or that
$$\frac{3}{(z+1)(z-i)}=-\frac{3}{1+i}\Big(\frac{1}{z+1}-\frac{1}{z-i}\Big)$$
This technique is usually much quicker than solving by forming linear equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A functional equation relating two harmonic sums. Introduction. I computed two Mellin transforms while browsing / working on the problem at this MSE link. No solution was found, but some interesting auxiliary results appeared. I am writing to ask for independent confirmation of these results, not necessarily using Mellin transforms.
Problem statement. Introduce
$$S(x) = \sum_{k\ge 1} \frac{1}{(2k-1)}
\frac{1}{\sinh((2k-1)x)}
\quad\text{and}\quad
T(x) = \sum_{k\ge 1} \frac{1}{k}
\frac{1}{\sinh(kx)}$$
Prove the functional equation
$$S(x) = \frac{1}{2} S(\pi^2/x)
- \frac{1}{16} x
+ \frac{1}{4} \log 2
+ \frac{3}{4} T(x).$$
Evaluate $T(x)$ at $x=\sqrt{2}\pi$ and prove that
$$T(\sqrt{2}\pi) = \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2.$$
Remark. It is hoped that these two problems might reward investigation, perhaps using several different methods. I do ask that possible details of the computations be included.
| Let $q = e^{-x}$ so that $$T(x) = \sum_{n = 1}^{\infty}\frac{1}{n\sinh nx} = 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{2n})} = f(q)\text{ (say)}\tag{1}$$ and then we have
\begin{align}
T(x) &= f(q)\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{2n})}\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{q^{n}}{n}\sum_{m = 0}^{\infty}q^{2mn}\notag\\
&= 2\sum_{m = 0}^{\infty}\sum_{n = 1}^{\infty}\frac{q^{(2m + 1)n}}{n}\notag\\
&= -2\sum_{m = 0}^{\infty}\log(1 - q^{2m + 1})\notag\\
&= -2\log\prod_{m = 1}^{\infty}(1 - q^{2m - 1})\notag\\
&= -2\log\left(2^{1/4}q^{1/24}\left(\frac{2k}{k'^{2}}\right)^{-1/12}\right)\tag{2}\\
&= -\frac{\log 2}{2} - \frac{\log q}{12} + \frac{1}{6}\log\frac{2k}{k'^{2}}\tag{3}
\end{align}
where $k$ is the elliptic modulus corresponding to nome $q$ and $k' = \sqrt{1 - k^{2}}$.
The product $\prod(1 - q^{2m - 1})$ is essentially Ramanujan's class invariant $g_{n}, g(q)$ and we have accordingly given its representation $(2)$ in terms of nome $q$ and modulus $k$. Also note that $\log q = -\pi K'/K$ where $K, K'$ are complete elliptic integrals of first kind with modulus $k$ and $k' = \sqrt{1 - k^{2}}$.
By the theory of modular equations (given in the link on Ramanujan's class invariant mentioned earlier) it is known that if $K'/K = \sqrt{r}$ where $r$ is a positive rational number then the value of modulus $k$ is an algebraic number and such value of $k$ is called a singular modulus and denoted by $k_{r}$. Here in the current notation we have $\log q = -x$ and hence if $x = \pi\sqrt{r}$ then the value of $k$ is an algebraic number and it is possible to have a closed form for $T(x)$ in terms of logarithm of an algebraic number plus $x/12$.
If we set $x = \pi\sqrt{2}$ then the corresponding value of singular modulus $k$ is $k = k_{2} = \sqrt{2} - 1$. And hence $2k/(1 - k^{2}) = 1$ and it follows from $(3)$ that $$T(\sqrt{2}\pi) = \frac{\sqrt{2}\pi}{12} - \frac{1}{2}\log 2\tag{4}$$ Next we discuss the sum $S(x)$ given by $$S(x) = \sum_{n\text{ odd}}\frac{1}{n\sinh nx} = \sum_{n = 1}^{\infty}\frac{1}{n\sinh nx} - \sum_{n\text{ even}}\frac{1}{n\sinh nx} = T(x) - \frac{T(2x)}{2}\tag{5}$$ Note that when $x$ is replaced by $2x$ then $q = e^{-x}$ is replaced by $q^{2}$ and hence $T(2x) = f(q^{2})$. By Landen's transformation replacing $q$ with $q^{2}$ leads to replacing $k$ with $(1 - k')/(1 + k')$ and hence
\begin{align}
T(2x) &= f(q^{2})\notag\\
&= -\frac{\log 2}{2} - \frac{\log q^{2}}{12} + \frac{1}{6}\log\dfrac{2\cdot\dfrac{1 - k'}{1 + k'}}{1 - \left(\dfrac{1 - k'}{1 + k'}\right)^{2}}\notag\\
&= -\frac{\log 2}{2} - \frac{\log q}{6} + \frac{1}{6}\log\frac{1 - k'^{2}}{2k'}\notag\\
&= -\frac{\log 2}{2} - \frac{\log q}{6} + \frac{1}{6}\log\frac{k^{2}}{2k'}\tag{6}
\end{align}
and hence it follows from $(3), (5)$ and $(6)$ that $$S(x) = -\frac{\log 2}{4} + \frac{1}{12}\log\frac{8}{k'^{3}}\tag{7}$$ Note that the above equation shows that for $x = \pi\sqrt{r}, r\in\mathbb{Q}^{+}$ the value of $S(x)$ is the logarithm of an algebraic number.
Next we deal with the transformation from $x$ to $\pi^{2}/x$. This changes $q = e^{-x}$ to $q' = e^{-\pi^{2}/x}$ and the effect of this is to swap $k$ and $k'$ so that $$S(\pi^{2}/x) = -\frac{\log 2}{4} + \frac{1}{12}\log\frac{8}{k^{3}}\tag{8}$$ and from $(7)$ and $(8)$ we get $$S(x) - (1/2)S(\pi^{2}/x) = -\frac{\log 2}{8} + \frac{1}{8}\log \frac{2k}{k'^{2}}\tag{9}$$ and looking at equations $(3)$ and $(9)$ we get $$S(x) = \frac{1}{2}S\left(\frac{\pi^{2}}{x}\right) - \frac{1}{16}x + \frac{1}{4}\log 2 + \frac{3}{4}T(x)\tag{10}$$ which is the desired functional equation connecting $S(x), S(\pi^{2}/x)$ and $T(x)$.
Like most of my answers dealing with sums involving hyperbolic functions this answer also requires a good understanding of the theory of elliptic integrals and their link with theta functions and other related topics.
Incidentally Ramanujan delved very deep into these topics and he had a very good understanding of both Mellin transform (favorite tool used by OP, see his answers) and elliptic/theta functions and moreover he somehow had the real-analysis equivalent of Mellin transform methods (Ramanujan's Master Theorem) so he could achieve his results without any use of complex analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Solution to the wave equation in $\mathbb{R}^{3}$ with certain initial data Suppose $f$ is a smooth function satisfying $f(0) = f'(0) = 0$. The question I am working on is to determine the solution $u$ to $u_{tt} - \Delta u = 0$ in $\mathbb{R}^{3}$ with $u(x, 0) = f(|x|)/|x|$ and $u_{t}(x, 0) = f'(|x|)/|x|$.
By (essentially) the Kirchhoff formula, I get
$$u(x, t) = \frac{\partial}{\partial t}\left(\frac{1}{4\pi t}\int_{\partial B(x, t)}\frac{f(|y|)}{|y|}\, d\sigma_{y}\right) + \frac{1}{4\pi t}\int_{\partial B(x, t)}\frac{f'(|y|)}{|y|}\, d\sigma_{y}.$$
However, this does not make use of $f(0) = f'(0) = 0$ which I think will simplify the above expression. I was wondering if one could simplify the above expression further?
| The initial conditions together with the (second order) differential equation determine $u(t)$ for all later $t$. One way of doing this is the following:
\begin{eqnarray*}
&&\partial _{t}^{2}u-\partial _{\mathbf{x}}^{2}u-0 \\
u_{1} &=&u,\;u_{2}=\partial _{t}u \\
\partial _{t}u_{1} &=&u_{2} \\
\partial _{t}u_{2} &=&\partial _{\mathbf{x}}^{2}u_{1}
\end{eqnarray*}
Set
\begin{eqnarray*}
\mathbf{u} &=&\left(
\begin{array}{c}
u_{1} \\
u_{2}
\end{array}
\right) \\
\partial _{t}\mathbf{u} &=&\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \mathbf{u} \\
\mathbf{u}(t) &=&\exp [\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t]\mathbf{u}(0)
\end{eqnarray*}
Then
\begin{eqnarray*}
\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ^{2} &=&\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) =\left(
\begin{array}{cc}
\partial _{\mathbf{x}}^{2} & 0 \\
0 & \partial _{\mathbf{x}}^{2}%
\end{array}
\right) =\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) \\
\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ^{3} &=&\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ,\;\mathrm{etc}
\end{eqnarray*}
Hence
\begin{eqnarray*}
\exp [\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t] &=&\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t+\frac{1}{2!}\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\frac{1}{3!}\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \cdots \\
&=&\cos (\partial _{\mathbf{x}}^{2}t)\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\frac{\sin (\partial _{\mathbf{x}}^{2}t)}{\partial _{\mathbf{x}}^{2}%
}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right)
\end{eqnarray*}
If you do not feel comfortable with ${∂_x}^2$ use the Fourier transform so ${∂_x}^2$ is replaced by $-k^2$.
Your initial conditions are tricky, since ${f_t}(x,0)$ need not vanish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show $\sum_0^\infty \frac{x\lambda^x} {x!} = \lambda e^\lambda$? I know that $\sum_0^\infty \frac{\lambda^x} {x!} = e^\lambda$, but I'm having a really difficult time dealing with the extra $x$.
| Very inelegantly and in schoolboy fashion we can write
\begin{align}
\sum_{x=0}^\infty x\frac{\lambda^x}{x!}
&= 0\frac{\lambda^0}{0!} + 1\frac{\lambda^1}{1!} + 2\frac{\lambda^2}{2!}
+ 3\frac{\lambda^3}{3!} + 4\frac{\lambda^4}{4!} + \cdots\\
&= 0 + \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda\left[1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!}
+ \frac{\lambda^3}{3!} + \cdots\right]\\
&= \lambda e^\lambda
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Probability of choosing a relatively prime number. Two integers m and n are called relatively prime if 1 is their only common positive divisor. Thus 8 and 5 are relatively prime, whereas 8 and 6 are not. A number is selected at random from the set {1, 2, 3, . . . , 63}. Find the probability that it is relatively prime to 63.
I'm not sure why this problem is giving me trouble. I know there are 36 numbers in that set that are relatively prime to 63, so that equals a 4/7 chance of picking a prime. I did this using the "brute force" method, but I'm not sure of how to set up a way of doing it using the $P(A\cup B) = P(A)+P(B) - P(A \cap B) $ formula.
| Since $63 = 3^2\cdot7$, an integer $n \in \{1,\ldots,63\}$ is relatively prime to $63$ if and only if $3 \nmid n$ and $7 \nmid n$.
Let $A$ be the event that $n$ is divisible by $3$, and let $B$ be the event that $n$ is divisible by $7$. Thus
\begin{align}
P(A \cup B) &= P(A) + P(B) - P(A \cap B)
\\ &= \frac{1}{3} + \frac{1}{7} - \frac{1}{21} = \frac{7+3-1}{21} = \frac{9}{21} = \frac{3}{7}
\end{align}
is the probability that a random integer in $\{1,\ldots,63\}$ shares a divisor with $63$, and therefore the probability that a random integer in $\{1,\ldots,63\}$ is relatively prime to $63$ is $1 - \frac{3}{7} = \frac{4}{7}$.
The calculation follows from observing that every third number in $\{1,\ldots,63\}$ is divisible by $3$, every seventh is divisible by $7$, and every "three times seven"-th number is divisible by $3$ and $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power Series with complicated Algebra I need to find the Power Series for $\frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$. I said let $\frac{b^2}{c^2}=x$. Therefore we have $\frac{ac^2}{\sqrt{1-x}}$. Well we know the Power Series for $\frac{1}{1-x}$. Therefore we substitute and get
$$ac^2(1+x+x^2+x^3+x^4+\cdots)^{0.5}$$
If I substitute back in for $x$ though I get stuck on the next step. Any tips or other ways would be much appreciated.
| Suppose
$f(x)
=\frac{1}{\sqrt{1-x}}
$.
Then
$f^2(x)
=\frac1{1-x}
=\sum_{n=0}^{\infty} x^{n}
$.
If
$f(x)
=\sum_{n=0}^{\infty} a_n x^n
$,
$\begin{array}\\
f^2(x)
&=\sum_{n=0}^{\infty} a_n x^n \sum_{m=0}^{\infty} a_m x^m\\
&=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} a_n a_m x^{n+m}\\
&=\sum_{k=0}^{\infty} \sum_{j=0}^{k} a_j a_{k-j} x^{k}\\
&=\sum_{k=0}^{\infty} x^k \sum_{j=0}^{k} a_j a_{k-j}\\
\end{array}
$
Therefore
$1
=\sum_{j=0}^{k} a_j a_{k-j}
$.
If $k=0$,
$1 = a_0^2$
so
$a_0 = 1$ or $-1$.
I will choose
$a_0 = 1$;
the other choice will
reverse all the signs.
If $k \ge 1$,
$1
=\sum_{j=0}^{k} a_j a_{k-j}
=2a_0a_k +\sum_{j=1}^{k-1} a_j a_{k-j}
=2a_k +\sum_{j=1}^{k-1} a_j a_{k-j}
$
or
$a_k
=\frac12(1-\sum_{j=1}^{k-1} a_j a_{k-j})
$.
For $k=1$,
$a_1 = \frac12
$.
For $k=2$,
$a_2
=\frac12(1-a_1^2)
=\frac12(1-\frac14)
=\frac{3}{8}
$.
As a check so far,
$(1+x/2+3x^2/8)^2
=1+x(1/2+1/2)+x^2(1/4+2(3/8)) + (...)x^3
=1+x+x^2+(...)x^3
$.
For $k=3$,
$a_3
=\frac12(1-a_1a_2-a_2a_1)
=\frac12(1-2a_1a_2)
=\frac12(1-2\frac12 \frac{3}{8})
=\frac{5}{16}
$.
And so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\sum_{n=0}^{\infty} {\frac{(2n+1)(n+1)}{3^n}}$ I had this problem on my recent exam and i got 0 points for my solution and I'd really like to know why :)
Calculate $$\sum_{n=0}^{\infty} {\frac{(2n+1)(n+1)}{3^n}}$$
My attempt:
Let us consider the following series:
$$\sum_{n=0}^{\infty} {\frac{(2n+1)(n+1)}{3^n}x^n}=\sum_{n=0}^{\infty} {\frac{(2n^2+3n+1)}{3^n}x^n}
$$
$$
\sum_{n=0}^{\infty} {\frac{(2n^2+3n+1)}{3^n}x^n}==\sum_{n=0}^{\infty} {\frac{2n^2}{3^n}x^n} + \sum_{n=0}^{\infty} {\frac{3n}{3^n}x^n}+\sum_{n=0}^{\infty} {\frac{1}{3^n}x^n}
$$
All three series have the radius of convergence $R=3$. Let us transform these series as follows:
$$
\sum_{n=0}^{\infty} {\frac{1}{3^n}x^n} = \sum_{n=0}^{\infty} {t^n}
$$
where $t=\frac{x}{3}$. This becomes:
$${\frac{1}{1-t}}$$
For $t=\frac{1}{3}$ this becomes $\frac{3}{2}$.
Using the same logic i transformed the remaining two series:
$$
\sum_{n=0}^{\infty} {\frac{3n}{3^n}x^n} = 3\sum_{n=0}^{\infty} n{t^n}= 3t\sum_{n=0}^{\infty} n{t^{n-1}}=3t \bigg( \sum_{n=0}^{\infty} {t^n} \bigg)'=3t \bigg( {\frac{1}{1-t}} \bigg)'={\frac{3t}{(1-t)^2}}
$$
Plugging in $t=\frac{1}{3}$ gives $\frac{9}{4}$
Doing the same for the remaining series gives me $3$ ( using the same method and derivating twice ) . Hence
$$\frac{3}{2} + \frac{9}{4}+3 = \frac{27}{4}$$
What is wrong with my method? I got no real explanation, even though the result is the same as the one Wolfram Alpha gave me. Is this mere coincidence that my result is correct ?
| Your solution is correct, and it seems wrong to give you $0$ points for it. However, there are two things to improve if you want. One is that introducing both $x$ and $t$ was wasteful; you could have just considered $\sum_n(2n+1)(n+1)x^n$ and substituted $1/3$ for $x$. The other is that you somewhat unnecessarily multiplied out the nice product and then had to juggle a lot of different terms with different orders of derivatives. The more elegant solution would be
$$
\sum_{n=0}^\infty\frac{(2n+1)(n+1)}{3^n}=\frac12\sum_{n=0}^\infty(2n+1)(2n+2)\left(3^{-1/2}\right)^{2n}
$$
with
\begin{align}
\frac12\sum_{n=0}^\infty(2n+1)(2n+2)x^{2n}&=\frac12\frac{\mathrm d^2}{\mathrm dx^2}\sum_{n=0}^\infty x^{2n+2}\\
&=\frac12\frac{\mathrm d^2}{\mathrm dx^2}\frac{x^2}{1-x^2}\\
&=\frac{1+3x^2}{(1-x^2)^3}\;,
\end{align}
whereupon substituting $x=3^{-1/2}$ yields your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$24$ divides $a^2+23$
Prove that $24$ divides $a^2 + 23$ if $a$ is not divisible by $2$ or $3$.
Well $a^2=8k+1$
for some $k$.
So, $$a^2+23 = (8k+1)+23=8(k+3)$$
So, $8$ divides $a^2+23$.
Now can $3$ divide $a^2+ 23$ so that in the end $24$ divides $a^2+23$?
| $a$ doesn't divide $3$ so:
$$
a\equiv\pm 1[3]\\
a^2 \equiv 1[3]\\
a^2+23 \equiv 24\equiv 0[3]
$$
So $3$ divides $a^2+23$ and since you correctly showed that $8|a^2+23$ and knowing that $\gcd(8,3)=1$ So :
$$
24|a^2+23
$$
I used the property that:
$$
a|b\quad \mathrm{and}\quad b|c \Longrightarrow \mathrm{lcm}(a,b)|c\\
ab=\gcd(a,b)\mathrm{lcm}(a,b)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} ) $ $$\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )
$$
My try:
$${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}} =
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 5{x^2} - {x^2} - 2x}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}
$$
And what's next...?
This task in first and second remarkable limits. I think i can replace variable, but how i will calculate it...
| $$(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} $$
Use Binomial theorem
$$[x^3+5x^2]^{1/3} - [x^2 - 2x]^{1/2}$$
$$[x^3(1+\frac{5x^2}{x^3}]^{1/3}-[x^2(1-\frac{2x}{x^2}]^{1/2}$$
$$\approx x(1+\frac{5}{3x}.....)-x(1-\frac{1}{x}.....)$$
$$\approx [x+\frac{5}{3}]-[x-1]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Check if an integral is convergent and calculate its value I have improper integral $$\int_0^\pi \frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx$$
I have to check if it is convergent. If yes, then i have to evaluate it.
I think it is convergent. I can make substitution $t=\sin(x)$. But what next?
| Formally, we can state
\begin{align}
\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{y=\pi/2}^{y=0}\frac{\cos (\pi-y)}{[1-2\sin (\pi-y)]^{1/3}}d(\pi-y)\\
\end{align}
Where $y=\pi-x$, then
\begin{align}
\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_{\pi/2}^{0}\frac{-\cos y}{(1-2\sin y)^{1/3}}dy\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^0\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_0^{\pi/2}\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\
&=0
\end{align}
This is true if $$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx$$ converges.
By using $t=\sin x,\;\;dt=\cos x dx$ we have
$$\int\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\int\frac{dt}{(1-2t)^{1/3}}=-\frac{3}{4}(1-2t)^{2/3}+C=-\frac{3}{4}(1-2\sin x)^{2/3}+C$$
Then
$$\int_0^{\pi/6}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}\int_0^{b}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}-\frac{3}{4}(1-2\sin b)^{2/3}+\frac{3}{4}=\frac{3}{4},$$
$$\int_{\pi/6}^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{a\to\frac{\pi}{6}^+}\int_a^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{4}+\lim_{a\to\frac{\pi}{6}^+}\frac{3}{4}(1-2\sin a)^{2/3}=\frac{3}{4}$$
Then, the integral converges, also
$$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{2}$$
Therefore
$$\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$ Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$. Then the left side is $2x^2+9x-5=\ln (1)+x^2+2x-15$, is that true? How should I proceed?
| Zero point's first step is correct. Alternatively, you could combine both natural logs to get
$$\ln\frac{2x^2+9x-5}{x^2+2x-15}=1$$
$$\frac{2x^2+9x-5}{x^2+2x-15}=e$$
It may help to factor next.
$$\frac{(2x-1)(x+5)}{(x+5)(x-3)}=e$$
Note that $x$ cannot equal $-5$ or else both natural logs in the original problem are not defined. So we have
$$\frac{2x-1}{x-3}=e$$
Now we can solve for $x$.
$$2x-1=e(x-3)$$
$$(e-2)x=3e-1$$
$$x=\frac{3e-1}{e-2}$$
We are not out of the woods yet. It still needs to be verified that both quadratics in the original problem are positive for the logarithms to be defined. Based on the factorizations, that requires $x>3$ or $x<-5$. We know $e>2$, so $x$ is positive. For the first condition, we can rewrite $x$ as $3+\frac5{e-2}>3$, so the solution checks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}$? How can I evaluate this?
$$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} = \frac{1}{1\cdot3\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{3\cdot5\cdot7}+ \frac{1}{4\cdot6\cdot8}+\cdots$$
I have tried:
$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \frac{1}{4\cdot6\cdot8}+\cdots
= \frac{1}{3\cdot5}\left(1+\frac{1}{7}\right)+\frac{1}{4\cdot6}\left(\frac{1}{2}+\frac{1}{8}\right)+\cdots$$
and so on... Been stuck for a while. Result should be $\dfrac{11}{96}$
| This might be overkill but,\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)}
&=\frac18\sum_{n=1}^{\infty}\left(\frac{1}{n+4}+\frac{1}{ n}
-\frac{2}{n+2}\right)\tag{1}\\
&=\frac18\sum_{n=1}^{\infty}\int_{0}^{1}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{2}\\
&=\frac18\int_{0}^{1}\sum_{n=1}^{\infty}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{3}\\
&=\frac{1}{8}\int_{0}^{1}\left(\frac{x^{4}}{1-x}+\frac{1}{1-x}-\frac{2x^2}{1-x}\right)\,\mathrm dx\tag{4}\\
&=\frac18\int_{0}^{1}\frac{(x^2-1)^2}{1-x}\,\mathrm dx\tag{5}\\
&=\frac18\int_{0}^{1}\left(1+x-x^2-x^3\right) \,\mathrm dx\tag{6}\\
&=\frac18\Bigg[x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}\Bigg]_{0}^{1}\tag{7}\\
&=\frac{11}{96}\tag{8}\\
\end{align}
$$\sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)}=\frac{11}{96}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Finding $z$ such that $\Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$ I wish to find $z$ such that $\displaystyle \Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$.
We may simplify $\displaystyle \frac{z}{-3+2i} = \frac{x+iy}{-3+2i} \frac{-3-2i}{-3-2i} = \frac{-3x+2y + i(-2x+3y)}{13}$.
Taking the real part of the fraction above gives us $\displaystyle \frac{-3x+2y}{13}$
Then, $-3x+2y = 10$
So now we move onto the other equation.
Recall that $|z|=z\overline{z}$ and $z=x+iy$. Then, $|3-2(x-iy)| = |2(x+iy)+3i|$, $|3-2x+2iy| = |2x+2iy+3i|$,
Then, $$(3-2x+2iy)(3-2x-2iy) = (2x+i(3+2y))(2x-i(3+2y))$$
which simplifies to $$4x^2-12x+4y^2+9 = 4x^2 - i(2y+3)^2$$.
But this is still messy and does not simplify further. Can anyone see where I messed up?
| again see last line $4x^2-12x+4y^2+9 = 4x^2 - i(2y+3)^2$, you have to wright
$4x^2-12x+4y^2+9 = 4x^2 +(2y+3)^2$
I sugesst you use complex modulus it will help you to reach answer in short way. $|z_1|=|z_2|$the $x_{2}^{2}+y_{2}^{2}=|z_1|^2=|z_2|^2=x_{1}^{2}+y_{1}^{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
nth derivative of: $F(x)=1-\sqrt{1-x^2}$ I've gotten this function from probability generating functions, and I want to calculate it's nth derivative (With respect to $x$).
This is:
$$F(x)=1-\sqrt{1-x^2}$$
Is there a practical way to do it?
Or for another approach, I just need the derivatives calculated in $x=0$, to calculate it's MacLaurin series. Any practical way to do it?
| let $g(x)=(1-x)^\frac 12$
$$g'(x)=-\frac 12 (1-x)^{-\frac 12} \\
g''(x)=-\frac 12 \frac 12 (1-x)^{-\frac 32} \\
g'''(x)=-\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 52} \\
g^{(4)}(x)=-\frac 52\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 72} \\
g^{(5)}(x)=-\frac 72\frac 52\frac 32 \frac 12 \frac 12 (1-x)^{-\frac 92} \\
$$
so
$$g^{(n)}(x)=-\frac {1}{2^n} (2n-3)(2n-5)(2n-7) ... (5)(3)(1) (1-x)^{-\frac{2n-1}{2} } \\
$$
$$g^{(n)}(0)=-\frac {1}{2^n} (2n-3)(2n-5)(2n-7) ... (5)(3)(1)
\\ = -\frac {1}{2^n} \frac{(2n-3)!}{(2n-2)(2n-4)(2n-6) ... (4)(2)}
\\ = -\frac {1}{2^n} \frac{(2n-3)!}{2^{n-1}(n-1)!}
\\ = -\frac {1}{2^{2n-1}} \frac{(2n-3)!}{(n-1)!}
$$
You should be able to complete this knowing that $F(x)=1-g(x^2)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$ For positive integers Numbers $a, b, c $ prove that $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$$
This inequality above take long time to prove it and still couldn't complete it. How I can prove this inequality? Any hint will help. Thanks
| Let $1+a^2=x$, $1+b^2=y$ and $1+c^2=z$.
Hence, by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{1+b+c^2}\geq\sum_{cyc}\frac{a^2+1}{1+\frac{1+b^2}{2}+c^2}=\sum_{cyc}\frac{2x}{y+2z}=\sum_{cyc}\frac{2x^2}{xy+2xz}\geq$$
$$\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(xy+2xz)}=\frac{2(x+y+z)^2}{3(xy+xz+yz)}=2+\frac{\sum\limits_{cyc}(x-y)^2}{3(xy+xz+yz)}\geq2.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$? If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?
At the end when you get $4ax=0$, can I divide by $4x$ to cancel out $4$ and $x$?
| $$(x-a)^2 = (x+a)^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$-2ax = 2ax$$
$$-a = a$$
Note that this statement is only true when $a=0$, which is thus your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
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How to calculate this integral via residues. I get into trouble in evaluating this integral:
$$
C(a)=\frac{1}{i\beta}\int_\Gamma \cot\frac{\pi z}{\beta}\frac{1}{\sin^2\frac{z}{2}}dz
$$
where the contour $\Gamma$ consists of two vertical lines, (−π − i∞, −π + i∞) and
(π + i∞,π − i∞).The result is:
$$
-\frac{2}{3}((\frac{2\pi}{\beta})^2-1)
$$
The integral can be evaluated via residues. Could show me how to do that integral?
Thanks a lot!
| We begin with the contour integral
$$C(a)=\frac{1}{i\beta}\oint_\Gamma \frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}dz$$
where for $|\beta|>\pi$, the only singularity within $\Gamma$ is at $z=0$ (a pole of order three). In order to calculate the residue at the origin, the following expansions are useful.
$$\begin{align}
\cos (\pi z/\beta)&=1-\frac12 (\pi z/\beta)^2+O(z^4) \tag 1\\\\
\sin^2(z/2)&=(z/2)^2\left(1-\frac{1}{12}z^2+O(z^4)\right) \tag 2\\\\
\sin(\pi z/\beta)&=(\pi z/\beta)\left(1-\frac16 (\pi z/\beta)^2+O(z^4)\right)\tag 3
\end{align}$$
Using $(1)-(3)$, we have the following expansion
$$\begin{align}
\frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}&=\frac{\cos (\pi z/\beta)}{\sin^2(z/2)\sin(\pi z/\beta)}\\\\
&=\frac{1-\frac12 (\pi z/\beta)^2+O(z^4)}{(z/2)^2(\pi z/\beta)\left(1-\frac{1}{12}z^2+O(z^4)\right)\left(1-\frac16 (\pi z/\beta)^2+O(z^4)\right)}\\\\
&=\frac{1-\frac12 (\pi z/\beta)^2+O(z^4)}{(z/2)^2(\pi z/\beta)\left(1-\left(\frac{1}{12}+\frac{\pi^2}{6\beta^2}\right)+O(z^4)\right)}\\\\
&=\frac{\left(1-\frac12 (\pi z/\beta)^2+O(z^4)\right)\left(1+\left(\frac{1}{12}+\frac{\pi^2}{6\beta^2}\right)+O(z^4)\right)}{(z/2)^2(\pi z/\beta)}\\\\
&=\frac{1+\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)z^2+O(z^4)}{(z/2)^2(\pi z/\beta)}\\\\
&=\frac{4\beta/\pi}{z^3}-\frac{(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)}{z^1}+O(z)
\end{align}$$
Therefore, the residue of $\frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}$ at $z=0$ is given by
$$\text{Res}\left(\frac{\cot(\pi z/\beta)}{\sin^2(z/2)},z=0\right)=(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)$$
Finally, using the Residue Theorem, we obtain
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{i\beta}\oint_\Gamma \frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}dz=2\pi i \frac{1}{i\beta}(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)=-\frac23\left(\left(\frac{2\pi}{\beta}\right)^2-1\right)}$$
as was to be shown!
NOTE:
While the use of expansions can often facilitate analyses such as the one herein, we can also have used Cauchy's Integral Formula or here to calculate the residue as
$$\text{Res}\left(\frac{\cot(\pi z/\beta)}{\sin^2(z/2)},z=0\right)=\frac1{2!} \lim_{z\to 0}\frac{d^2}{dz^2}\left(z^3\frac{\cot(\pi z/\beta)}{\sin^2(z/2)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the element with the highest order in a symmetric group? How does one find the element with the highest order in $S_9$ ?
I can only guess that the element may be (1 2 3 4)(5 6 7 8 9) whose order is 20.
| The conjugacy classes of $S_n$ are in one-to-one correspondence with the partitions of $n$, that is, a partition of $n$ is an ordered $k$-tuple ($k$ is arbitrary) of positive integers ordered increasingly whose sum is $n$. The partitions of $9$ are
$$
(1,1,1,1,1,1,1,1,1) \\ (1,1,1,1,1,1,1,2) \\ (1,1,1,1,1,1,3) \\ \vdots \\ (1,8), \\
(1,1,1,1,1,2,2) \\ (1,1,1,1,2,3) \\ (1,1,1,2,4) \\ (1,1,2,5) \\ (1,2,6) \\ (2,7), \\
(1,1,1,3,3) \\ (1,1,3,4) \\ (1,3,5) \\ (3,6) \\
(1,4,4) \\ (4,5) \\
(1,1,1,2,2,2) \\ (1,1,2,2,3) \\ (1,2,2,4) \\ (2,2,5) \\
(1,2,3,3) \\ (2,3,4) \\
(3,3,3)\\
(1,1,1,2,2,2)\\
(1,2,3,3) \\
(1,2,2,2,2) \\
(2,2,2,3) \\
(9)
$$
So by checking all their LCMs, you quickly notice that your 20 is indeed the correct one. I don't know if there is a general method for $S_n$ though.
Hope that helps,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Quadratic equation within quadratic Edited version:-
If both roots of equation $$4x^2 -20px+(25p^2 +15p- 66)=0 $$ are less than $2$, find the value of $p$. How to solve such type of equations?
| The roots are $$\frac{1}{2} \left(5 p\pm\sqrt{3} \sqrt{22-5 p}\right)$$ where clearly the biggest root is $\frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)$. Now if this root is smaller than $2$ the other root will also be smaller than $2$.
\begin{align}
\frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)&<2\\
5 p+\sqrt{3} \sqrt{22-5 p}&<4\\
\sqrt{3} \sqrt{22-5 p}&<4-5p\\
3 (22-5 p)&<(4-5p)^2\\
66-15 p&<16 - 40 p + 25 p^2\\
0&<-50 - 25 p + 25 p^2\\
0&<-2 - p + p^2\\
0&<(p+1)(p-2)\\
\end{align}
If $p<-1$ or $p>2$ will the inequality hold ... (why?) Also note that $\sqrt{3} \sqrt{22-5 p}<4-5p$ requires $0<4-5p$ i.e. $p<\frac45$. Hence only $p<-1$ is acceptable. On the other hand we require $p<\frac{22}{5}$ hence $$p<-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let x and y be integers, prove that if 3 doesn't divide x and 3 doesn't divide y then 3 divides $x^2 - y^2$ Let x and y be integers, prove that if $3 \nmid x$ and $3 \nmid y$ then
$3 \mid (x^2 - y^2)$
Attemmpt:
The only thing i get out of this is that there is a difference of squares: $$(x^2 - y^2) = (x-y)(x+y) $$
Other than that i am stuck. Perhaps a hint before a full blown solution?
| Note that $3$ does not divide $x$ and $y$ means that $3$ does not divide $x^2$ and $y^2$.
Hence, $x^2$ and $y^2$ leave remainder 1 when divided by $3$. This is a simple rule of perfect square numbers.
Let $x^2=3q+1$ and $y^2=3p+1$.
$\therefore x^2-y^2=3q+1-3p+1=3(q-p)\implies 3$ is a factor of $x^2-y^2$.
Thus, the answer follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why do I need to switch the sign of the natural log in this trig-sub integration problem? I'm attempting to integrate the following using trig substituion:
$$\int \frac{\sqrt{1+x^2}}{x}dx$$
and I am getting the result:
$$-\ln{\lvert \frac{ \sqrt{x^2 + 1} - 1}{x} \rvert} + \sqrt{x^2 + 1} + C$$
However, my answer guide and this yahoo answer both indicate that the $-\ln$ should now be positive, I can't seem to figure out why (although I'm sure it's painfully simple). I thought it might be related to the fact that $\ln{\frac1x} = -\ln{x}$ but I can't wrap my head around it.
I've provided my steps below:
I start with $x=\tan\theta$ and $dx=\sec^2\theta d\theta$
Which brings me to $$\int \frac{\sqrt{1+\tan^2\theta}}{\tan\theta}\sec^2\theta d\theta$$
and then: $$\int \frac{\sec^2\theta(\sec\theta)}{\tan\theta} d\theta$$
.... next I substitute $(1+\tan^2\theta)$ for $\sec^2\theta$ and multiply by the $\sec\theta$ already in the numerator, giving me:
$$\int \frac{\sec\theta+\sec\theta\tan^2\theta}{\tan\theta}d\theta$$
I separate this out into two separate fractions, canceling one power of $\tan\theta$ in its respective fraction:
$$\int \frac{\sec\theta}{\tan\theta} + \sec\theta\tan\theta d\theta$$
This can be split into two integrals, and the left side can be rewritten as $\csc\theta$:
$$\int \csc\theta d\theta + \int \sec\theta\tan\theta d\theta$$
Which using known integrals can be determined to be:
$$-\ln{\lvert \csc\theta - \cot\theta \rvert} + \sec\theta + C$$
Creating a triangle to determine $\csc\theta$, $\cot\theta$, and $\sec\theta$ in terms of x yields me:
$$-\ln{\lvert \frac{ \sqrt{x^2 + 1} - 1}{x} \rvert} + \sqrt{x^2 + 1} + C$$
Edit: So it turns out I had the formula for $\int\csc\theta d\theta$ wrong. As user84413 pointed out in the comments, the formula is:
$$\int\csc\theta d\theta=\ln|\csc\theta-\cot\theta|+C=-\ln|\csc\theta+\cot\theta|+C$$
Using this corrected formula for my second to last step, I get the correct answer.
| Your Second last line is $\displaystyle - \ln\left|\csc \theta-\cot \theta\right|+\sec\theta +\mathcal{C}$
Now Using $\displaystyle \csc^2 \theta-\cot^2 \theta = 1\Rightarrow (\csc \theta -\cot \theta) = \frac{1}{\csc \theta +\cot \theta}$
So we get $\displaystyle = -\ln\left|\frac{1}{\csc \theta+\cot \theta}\right|+\sec \theta+\mathcal{C}$
So we get $\displaystyle = \ln\left|\csc \theta+\cot \theta\right|+\sec \theta+\mathcal{C}$
So $\displaystyle =\ln\left|\frac{\sqrt{1+x^2}+1}{x}\right|+\sqrt{1+x^2}+\mathcal{C}$
$\bf{Solution \; without \; Trig. \; Substution.}$
Let $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{x}dx = \int\frac{1+x^2}{x\sqrt{1+x^2}}dx = \int\frac{1}{x\sqrt{1+x^2}}dx+\int\frac{x}{\sqrt{1+x^2}}dx$$
Now Let $$\displaystyle J = \int\frac{1}{x\sqrt{x^2+1}}dx\;,$$ Put $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So $$\displaystyle J = -\int\frac{1}{\sqrt{t^2+1}}dt = -\ln\left|t+\sqrt{t^2+1}\right|+C_{1} = -\ln\left|\frac{\sqrt{x^2+1}+1}{x}\right|+\mathcal{C_{1}}$$
And Let $$\displaystyle K=\int\frac{x}{\sqrt{x^2+1}}dx\;,$$ Now put $\displaystyle x^2+1=u^2\;,$ Then $xdx = udu$
So we get $$\displaystyle K = \int\frac{u}{u}du = u+\mathcal{C_{2}} = \sqrt{x^2+1}+\mathcal{C_{2}}$$
So we get $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{x}dx = -\ln\left|\frac{\sqrt{x^2+1}+1}{x}\right|+\sqrt{x^2+1}+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$ without using L' Hôpital
Compute $$\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$$ without using L'Hospital's rule.
By using L'Hospital's rule and
$$\tan'( \Diamond )=( \Diamond )'(1+\tan^{2}( \Diamond ))$$
I mean by $\Diamond $ a function
so I got
\begin{align}
\lim_{n\to +\infty}n\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)
&=\lim_{n\to +\infty}\dfrac{\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)}{\dfrac{1}{n}}\\
&=1+\tan^{2}\left(\dfrac{\pi}{3}\right)=1+\sqrt{3}^{2}=1+3=4
\end{align}
I'm interested in more ways of computing limit for this sequence.
| $$\lim_{n\to \infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3\right)$$
$$=\lim_{n\to \infty}\frac{\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{\frac{\tan\frac{\pi}{3}+\tan \frac{1}{n}}{1-\tan \frac{\pi}{3}\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{\frac{\sqrt 3+\tan \frac{1}{n}}{1-\sqrt 3\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{4\tan\frac{1}{n}}{\frac{1}{n}(1-\sqrt 3\tan\frac{1}{n})}$$
$$=4\lim_{n\to \infty}\frac{\tan\frac{1}{n}}{\frac{1}{n}}\times \lim_{n\to \infty}\frac{1}{1-\sqrt 3\tan \frac{1}{n}}$$
$$=4(1)\left(\frac{1}{1-0}\right)=\color{red}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving the given inequality. (Complex Analysis). We're given a function $P_n(x)$ for $-1\leq x\leq1$ as follows :
$$P_n(x) = \int \limits_0^\pi \dfrac{1}{\pi}(x+i\sqrt{1-x^2} \cos\theta)^n \, d\theta$$
for $n=(0,1,2,3,\ldots)$, we need to prove that $|P_n(x)| \leq 1$.
I tried the following :
Let $z=x+i\sqrt{1-x^2}\cos\theta$
I somehow want to prove that $|z|=|x+i\sqrt{1-x^2}\cos\theta|\leq1$, as that would imply that $|z^n|\leq1$, as $|z^n|=|z|^n$.
$$|z| = \sqrt{x^2+(1-x^2)\cos^2 \theta} \Longrightarrow \sqrt{x^2 \sin^2\theta + \cos^2\theta}.$$
Now, $x^2\leq1$ and $\sin^2\theta \leq1 \Longrightarrow x^2\sin^2\theta \leq1$.
Also, $\cos^2\theta \leq1 \Longrightarrow \sqrt{x^2\sin^2\theta+\cos^2\theta} \leq \sqrt{2}$,
But " $\leq1$" condition is required...
That's the point where I am stuck, could anyone help me with this?
| HINT: $(x^2+(1-x^2) \cos^2\theta) \leq (x^2+(1-x^2))=1$, because $\theta$ is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding an unknown linear transformation given that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$
Let $T\colon\mathbb{R}^2\to\mathbb{R}^3$ be the linear transformation such that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$.
Does such a linear transformation exist?
So far I've worked out that it cannot exist, as the first entry for $T(1,1)$ is 1, whereas the first entry for $T(2,3)$ is 1, which can only be obtained through 3-2, or half of 2.
Is there any way of presenting this formally?
| There is a very simple method to solve the problem described in "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely."
Consider the formula
$$
\vec{T}(\vec{p}) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{x} & \vec{y} \\
p_1 & a_1 & b_1 \\
p_2 & a_2 & b_2 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_1 & b_1 \\
a_2 & b_2 \\
\end{pmatrix}
},
$$
where $\vec{T}$ is linear transformation acting on arbitrary point $\vec{p}$. $\vec{T}$ has the property
$$
\vec{T}(\vec{a}) = \vec{x};\quad
\vec{T}(\vec{b}) = \vec{y}.
$$
Indices designate components of the corresponding vector.
Let's consider your case.
We need such $\vec{T}$ that
$$
\vec{T}: \begin{pmatrix}1\\ 1\end{pmatrix} \mapsto
\begin{pmatrix}1\\ 0\\ 2\end{pmatrix};~
\vec{T}: \begin{pmatrix}2\\ 3\end{pmatrix} \mapsto
\begin{pmatrix}1\\-1\\ 4\end{pmatrix}.
$$
Now I plug them into the general expression
$$
\vec{T}(\vec{p}) =
(-1)
\frac{
\det
\begin{pmatrix}
0 & (1,0,2)^T & (1,-1,4)^T\\
\begin{matrix}
p_{1} \\
p_{2}
\end{matrix} &
%
\begin{matrix}1\\ 1\end{matrix} &
%
\begin{matrix}2\\ 3\end{matrix}
\end{pmatrix}
}{
\det
\begin{pmatrix}
\begin{matrix}1\\ 1\end{matrix} &
%
\begin{matrix}2\\ 3\end{matrix}
\end{pmatrix}
}.
$$
Doing determinants I get
$$
\vec{T}(\vec{p})
= \left[
3\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} -
\begin{pmatrix}1\\-1\\ 4\end{pmatrix}
\right] p_1 -
\left[
2\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} -
\begin{pmatrix}1\\-1\\ 4\end{pmatrix}
\right] p_2
$$
or simplified
$$
\vec{T}(\vec{p}) =
\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} p_1 +
\begin{pmatrix} -1 \\ -1\\ 0 \end{pmatrix} p_2.
$$
Of course, you can write that in a vector form
$$
\vec{T}(\vec{p}) =
\begin{pmatrix}
2 &-1 \\
1 &-1 \\
2 & 0
\end{pmatrix}
\begin{pmatrix} p_1 \\ p_2 \end{pmatrix}.
$$
Now you can easily check
$$
\begin{pmatrix}
2 &-1 \\
1 &-1 \\
2 & 0
\end{pmatrix}
\begin{pmatrix}1\\ 1\end{pmatrix} =
\begin{pmatrix}1\\ 0\\ 2\end{pmatrix};~
\begin{pmatrix}
2 &-1 \\
1 &-1 \\
2 & 0
\end{pmatrix}
\begin{pmatrix}2\\ 3\end{pmatrix} =
\begin{pmatrix}1\\ -1\\ 4\end{pmatrix}.
$$
For more details on the methods used, you can always refer to "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely". The latter contains many problems similar to this one as explained by the authors of the method presented.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why do we need an integral to prove that $\frac{22}{7} > \pi$? We know this famous (and beautiful) integral which shows that $\dfrac{22}{7} > \pi$ as :
$$0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi$$
Now since the integrand is positive, hence:
$$\dfrac{22}{7}-\pi>0$$
$$\color{blue}{\dfrac{22}{7}>\pi}$$
Although I can see its beauty, why is it needed to show that $\dfrac{22}{7} > \pi$ ?
Can't we just say that :
$$\dfrac{22}{7}=\color{red}{3.142}857142857142857\cdots = 3. \overline{142857}$$
$$\pi =\color{red}{3.141}592653589793238\cdots$$
And hence it is greater ??
Thanks!!
| If you can derive
$\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
then you can extract $\pi<(22/7)$ via a comparison test. It does, however, take a little patience.
First separate the $n=1$ through $n=4$ terms:
$\dfrac{\pi^2}{6}-\dfrac{205}{144}=\displaystyle \sum_{n=5}^{\infty}\dfrac{1}{n^2}$
Then we have
$\dfrac{1}{n^2}<\dfrac{1}{(n-\frac{1}{2})(n+\frac{1}{2})}=\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}$
$\dfrac{\pi^2}{6}-\dfrac{205}{144}<\displaystyle \sum_{n=5}^{\infty}\left(\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}\right)$
The last sum telescopes to $2/9$. Therefore
$\dfrac{\pi^2}{6}<\dfrac{205}{144}+\dfrac{2}{9}$
$\pi^2<\dfrac{79}{8}$
And then
$\left(\dfrac{22}{7}\right)^2=\dfrac{484}{49}=\dfrac{\color{#0055ff}{3872}}{392}$
$\dfrac{79}{8}=\dfrac{\color{#0055ff}{3871}}{392}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Sum of series of real numbers. I am trying to find the sum of the series $$1+\frac{1}{3.4}+\frac{1}{5.4^{2}}+\frac{1}{7.4^{3}}\cdot\cdot\cdot$$ What is the idea behind to find the sum of this type of series? How i think to compare it to other known series like log series etc.? Please help. Thanks in advance.
| Recall the MacLaurin series for $\ln(1+x)$,
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$
and the series for $\ln(1-x)$,
$$x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots.$$
Add, divide by $2$. We get
$$\frac{1}{2}(\ln(1+x)+\ln(1-x))=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots.$$
Divide by $x$, and make the appropriate choice for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all primes $p$ and $q$ such that $p^2-2q^2=1.$
Find all primes $p$ and $q$ such that $p^2-2q^2=1.$
My idea so far was to observe that since $2q^2$ is even, then $q^2$ must be odd or even. If $q^2$ is even, then $q$ is even and the only even prime is $2.$ Thus one pair of primes $(p,q)=(3,2).$
But, if $q^2$ is odd, then $q$ is also odd and $q=2d+1$ for some integer $d.$ Then,
$p^2-2q^2=1$ turns into $p^2=4(2d^2+2d)+3.$
From here I do not know how to proceed.
Any ideas or suggestions would be greatly appreciated.
| From your last expression we have $$\tag1(p+1)(p-1)=2\cdot[\;2(2d^2+2d)+1\;].$$ If $p=2$ then you get from the original expression that $2q^2=3,$ which is impossible. Therefore $p>2$ and hence $p$ is odd. Then $p-1$ and $p+1$ are both even, which implies that their product $(p+1)(p-1)$ is divisible by $4$ which is clearly false by $(1).$ Therefore $q$ must be even and thus it must be equal $2,$ which implies that $p=3.$
Another approach: Since $p^2-2q^2=1$ then $(p+1)(p-1)=2q^2$ and the Unique Factorization Theorem implies that either $p+1=q^2\;\wedge\;p-1=2$ or $p+1=2q\;\wedge\;p-1=q$ but in either case $p=3$ and $q=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \sin ^{2} x+5 \cos ^{2} x} =& \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x \\
=& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{d(2 \tan x)}{(2 \tan x)^{2}+(\sqrt{5})^{2}} \\
=& \frac{1}{2 \sqrt{5}}\left[\tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)\right]_{0}^{\frac{\pi}{2}} \\
=& \frac{1}{2 \sqrt{5}} \cdot \frac{\pi}{2} \\
=& \frac{\pi}{4 \sqrt{5}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Find the postive integers such $xy+x+y\mid x^2+y^2-2$ Let $x,y$ be positive integers, and such
$$xy+x+y\mid x^2+y^2-2$$
Find the $x,y$
$$x^2+y^2-2=A(xy+x+y)\Longrightarrow x^2-(Ay+A)x+y^2-Ay-2=0$$
| Let $a=x+1$, $b=y+1$. We want $(ab-1)|(a-1)^2+(b-1)^2+2$, which is equivalent to: $(ab-1)$ dividing the numerator of (a-$1)^2+(\frac{1}{a}-1)^2+2$, which is $f(a)=a^4-2a^3-2a+1$. Now let's consider the case where $5 \leq a <b$, the others being solvable by hand. The key observation is that the other factor of $f(a)$ must be of the form $(ac-1)$ for $c<a$. [Otherwise $[a(a+1)-1][a^2+1]>f(a)]$. This means the pair $(a,c)$ is also a solution.
Thus if $(a,b)$ with $5<a<b$ is a solution, then there exists another solution $(c,a)$ with $c<a$. This chain of solutions must stop with the smaller value being at most 4, but for $a \leq 4$, we can check that the corresponding $b$s do not yield other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$ by induction So I have to prove the following using induction.
${\displaystyle \prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right)} = \frac{n+1}{2n}$
I showed the basis step that if $n=i=2$, then the two functions are equal $\frac{3}{4}$, and I know that the induction step involves simplifying the function where $n=n+1$. But I'm not sure how to do it with the following algorithm. I am quite confused. Can someone show me how I can prove this in the induction step?
| Without really using induction, you can use telescoping:
$$\prod_{i=2}^n \left(1-\frac{1}{i^2}\right)=\prod_{i=2}^n\frac{i^2-1}{i^2}=\prod_{i=2}^n\frac{(i-1)(i+1)}{i^2}$$
$$=\frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \not3}{2\cdot \not 2}\cdot \frac{\not 2\cdot 4}{\not 3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \Box}{2\cdot \Box}\cdot \frac{\Box\cdot \not 4}{\Box\cdot \not 3}\cdot \frac{\not 3\cdot 5}{\not 4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \Box}{2\cdot \Box}\cdot \frac{\Box\cdot \Box }{\Box\cdot \Box }\cdot \frac{\Box \cdot 5}{\Box \cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\cdots =\frac{1}{2}\cdot \frac{n+1}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the following using induction $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
I'm new to induction, but this is what I cam up with so far.
$$1 - \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{k+2+k}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2(k+1)}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2}{k(k+2)} = 1 - \frac{1}{k+2}$$
Where did I go wrong, or where do I go from here?
EDIT:
After some advice from Patrick (you're awesome!) I have...
$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{k + 2 + 1}{(k+1)(k+2)}$$
$$1 - \frac{k + 3}{(k+1)(k+2)}$$
$$\frac{(k+1)(k+2) - (k+3)}{(k+1)(k+2)}$$
$$\frac{k^2 + 2k + k + 2 - k - 3)}{(k+1)(k+2)}$$
$$\frac{k^2 + 2k - 1}{(k+1)(k+2)}$$
$$\frac{k^2 + 2k - 1}{k^2 + 3k + 2}$$
Did I goof up somewhere again?
Thanks.
| For every $n\in\mathbb{N}$, let $\mathcal{P}(n)$ the statement
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$
*
*The statement $\mathcal{P}(1)$ is $\frac{1}{1\times 2} = 1 - \frac{1}{2}$, which is clearly true.
*We prove that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$ for all $n\geq 1$. We use a direct proof. Suppose
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$
Then
$$
\begin{align*}
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)}
&= 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\
&= 1 - \frac{(n+2) - 1}{(n+1)(n+2)} \\
&= 1 - \frac{1}{n+2}.
\end{align*}
$$
This proves that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$.
It follows by induction that
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}
$$
for all $n\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to answer this kind of questions How can I answer this kind of questions?
I should prove the following
$$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \frac{1}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$
Sorry for my poor English
| $$\frac{\sqrt{a_1}-\sqrt{a_2}}{{a_1} + {a_2}}+\frac{\sqrt{a_2}-\sqrt{a_3}}{{a_2}-{a_3}}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-{a_n}}=\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+....+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{-d}\times\frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...
Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$
My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.
Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.
Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.
| Let $\theta/2=x$. Then the LHS is:
$$\sin^6 x+\cos^6 x=\underbrace{\left(\sin^2 x+\cos^2 x\right)}_{=1}\left(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x\right)$$
$$=\left(1-\cos^2 x\right)^2-\left(1-\cos^2 x\right)\cos^2 x+\cos^4 x\tag{1}$$
The RHS is:
$$\frac{1}{4}\left(1+3\cos^2 2x\right)=\frac{1}{4}\left(1+3\left(2\cos^2 x-1\right)^2\right)\tag{2}$$
Now prove $(1)=(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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A golden ratio series from a comic book The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:
That is,
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$
How might this be proven?
| This series appears in this place, referencing the derivation at "Golden Mean Series Derivation", both by Brian Roselle. That derivation starts with $$\phi = \frac{\sqrt{5}+1}{2}, $$ replaces $\sqrt{5}$ with $f(x) = \sqrt{x},$ expands $f$ in a Taylor series about $4$ (having a radius of convergence $> 1$, although this is not stated), then evaluating that series at $5$. Mr. Roselle describes finding the form of $f^{(a)}(x)$ via inspection.
We, however, can find these via induction. We wish to find $f^{(a)}(4)$:
*
*$f^{(0)}(4) = 4^\frac{1}{2} = 2$.
*$f'(4) = \frac{1}{2} 4^\frac{-1}{2} = \frac{1}{4}$.
*$f''(4) = \frac{1}{2} \frac{-1}{2} 4^\frac{-3}{2} = \frac{-1}{32}$.
*Suppose $f^{(n)}(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4}$, which is certainly true for $n=2$. Then $\left(f^{(n)}\right)'(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \left( \frac{3}{2}-n \right) \cdot \left( \frac{3}{2}-(n+1) \right) \right) x^{\frac{1}{2}-(n+1)} \right|_{x=4} = f^{(n+1)}(4)$.
So induction has given us, for $n \geq 2$,
$$\begin{align}
f^{(n)}(4) &= \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4} \\
&= 2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n},
\end{align}$$
where we have used $k!! = \begin{cases} k(k-2)(k-4) \cdots 4 \cdot 2, & k \text{ even}, \\ k(k-2)(k-4) \cdots 3 \cdot 1, & k \text{ odd} \end{cases}$. Then for $x$ close enough to $4$ (and a quick ratio test shows $5$ is close enough),
$$\begin{align}
f(x) &= \sum_{n=0}^\infty \frac{f^{(n)}(4)}{n!}(x-4)^n \\
&= \sum_{n=0}^\infty \frac{2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n}}{n!}(x-4)^n.
\end{align}$$
Replace $(2n-3)!!$ with $\frac{(2n-3)!}{2^{n-2}(n-2)!}$ and $x-4$ with $1$, pull out the first two summands and reindex $\sum_{n=2}^\infty \dots$ to start at $n=0$, plug into $\frac{f(5)+1}{2}$, and you get the given form of $\phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
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Show that $1^k+2^k+3^k+ \ldots +n^k$ is divisible by $ 1+2+3+ \ldots +n$ If $k$ is an odd positive integer, prove that for any integer $ n\geq 1$, $(1^k+2^k+3^k+ \ldots +n^k)$ is divisible by $ (1+2+3+ \ldots +n)$.
I didn't get how it is connected( or different) than this: suppose that $n$ is natural number and even, show that $ n∤1^n+2^n+3^n+…(n−1)^n$
I tried for small numbers $n=3,4$ then use induction on $k$ but how to show for any natural $n$? does the answer in the link provides an insight to the solution.
Hints would suffice. Thanks
| Let $S_k=1^k+2^k+\cdots +n^k$. Since $$S_1=1+2+\cdots+n=\dfrac{n(n+1)}2$$
and $gcd(n,n+1)=1$, it suffices to show that $2S_k$ is divisible by $n$ and by $n+1$ separately.
Now $$2S_k =(1+n^k)+(2^k+(n-1)^k)+\cdots+(n^k+1)$$
and also
$$2S_k=2n^k+(1+(n-1)^k)+(2^k+(n-2)^k)+\cdots+((n-1)^k+1).$$
And we are done (note that $k$ is odd).
| {
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"source": "stackexchange",
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Prove the lines of the orthocenter are concurrent BY Ceva's theorem Prove that the lines of the orthocenter are concurrent by Ceva's Theorem (or its converse).
Edit: Ceva's theorem is the theorem stating in a triangle $ABC$, if the lines $AX$, $BY$, and $CZ$ ($X$ being on $BC$, and so forth) are concurrent, then:
$(BX/XC) * (CY/YA) * (AZ/ZB) = 1.$
The converse would be that given ^^
| Let
*
*the altitude of $A$ intersect line $BC$ at $D$ (i.e. $AD \perp BC$)
*the altitude of $B$ intersect line $CA$ at $E$ (i.e. $BE \perp CA$)
*the altitude of $C$ intersect line $AB$ at $F$ (i.e. $CF \perp AB$)
*$a,b$ and $c$ be the lengths of segments $BC,CA$ and $AB$ respectively
$\triangle ABC$ has no obtuse angle
(source: mathwords.com)
$$\frac{AF}{FB} \frac{BD}{DC} \frac{CE}{EA} = \frac{b \cos A}{a \cos B} \frac{c \cos B}{b \cos C} \frac{a \cos C}{c \cos A} = 1
$$
By Ceva's Theorem, the orthocenter exists.
$\triangle ABC$ has an obtuse angle
Suppose $\angle B > 90^\circ$. From the diagram below, it can be seen that altitudes $AD$ and $CF$ lie outside $\triangle ABC$.
(source: mathwords.com)
\begin{align}
\frac{AF}{FB} \frac{BD}{DC} \frac{CE}{EA} =& -\frac{b \cos A}{a \cos (180^\circ - B)} \cdot -\frac{c \cos (180^\circ - B)}{b \cos C} \cdot \frac{a \cos C}{c \cos A} \\
=& \frac{b \cos A}{a \cos B} \frac{c \cos B}{b \cos C} \frac{a \cos C}{c \cos A} = 1
\end{align}
By Ceva's Theorem, the orthocenter also exists.
| {
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More than two solutions to $x^2 -1 = 0$ in $\Bbb{Z}/n\Bbb{Z}$ if $n$ is odd and not a power of a prime
Let $n$ be a positive odd integer such that it is not the power of a single prime, show that in $\Bbb{Z}/n\Bbb{Z}$ there are more than two solutions to $x^2 -1 = 0$
The requirement that $n$ is not the power of a single prime is stumping me. Does that mean that we can write n in the form of $p\times q^s\times\dots$ ?
| Let $n=ab$ with $\gcd(a,b)=1$ and $a,b>1$ each odd, By the Chinese remainder theorem we can choose $x$ in $\{1,...,n\}$ which is $-1$ mod $a$ and is $+1$ mod $b$. Since $a,b$ are odd, this $x$ will not be $\pm 1$ mod $n.$
However we have $x^2=1$ both mod $a$ and mod $b$, hence mod $n=ab.$ [ This uses that if $\gcd(a,b)=1$ and each of $a,b$ divides some $n$ then also $ab$ divides $n.$ ] So we have three solutions $-1,1,x$ to $x^2-1=0$
A detail: Since $x=-1$ mod $a,$ $a$ is a divisor of $x+1.$ Then using that $x+1$ is a divisor of $x^2-1,$ we have that $a$ divides $x^2-1$ since if $u$ divides $v$ and $v$ divides $w$ then $u$ divides $w.$ Similarly from $x=1$ mod $b$ we have $b$ divides $x-1$ and then since $x-1$ divides $x^2-1,$ we have $b$ divides $x^2-1.$ So at this point we have $\gcd(a,b)=1$ and each of $a,b$ divides $x^2-1,$ and it is from this that we can conclude $ab$ divides $x^2-1.$ [I didn't bring out before that $a,b$ each divide the same thing.] Finally to say $ab$ divides $x^2-1$ is the same, since $n=ab,$ as saying $x^2=1$ mod $n.$ I think it is clear that $x$ here is neither of $-1,1$ mod $n$ so we indeed have at least three solutions to $x^2=1$ mod $n.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $x+y \ge 2$ for $xy = 1$ and both positive? For every $x,y\in\mathbb{R}$ such that $x,y>0$ and that satisfy the condition xy=1, the following will hold:
$$x+y\ge 2$$
Steps I took and my thoughts on this:
$$xy=1\Rightarrow \frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } \quad and\quad xy=1\Rightarrow \frac { xy }{ y } =\frac { 1 }{ y } \Rightarrow x=\frac { 1 }{ y } $$
$$x\cdot \frac { 1 }{ x } =1\quad and\quad y\cdot \frac { 1 }{ y } =1$$
$$x+\frac { 1 }{ x } \ge 2\quad and\quad \frac { 1 }{ y } +y\ge 2$$
$$x(x+\frac { 1 }{ x } )\ge (2)x\quad and\quad y(\frac { 1 }{ y } +y)\ge (2)y$$
$$x^{ 2 }+1\ge 2x\quad and\quad y^2+1\ge 2y$$
$$x^{ 2 }-2x+1\ge 0\quad and\quad y^{ 2 }-2y+1\ge 0$$
$$(x-1)^{ 2 }\ge 0\quad and\quad (y-1)^{ 2 }\ge 0$$
$$x\in\mathbb{R} \quad and \quad y\in\mathbb{R} $$
I'm not sure where I am going wrong, but I don't know how to organize my thoughts about this in order to prove it with a formal mathematical proof. I'd like to be guided in the right direction. Hints are much appreciated.
| You actually have most of the pieces; you need to stop manipulating and think about putting them together.
If $xy=1$, then $y=\frac1x$, and
$$x+y=x+\frac1x=\frac{x^2+1}x\;.$$
Suppose that this is less than $2$; then since $x>0$, we can multiply by $x$ to find that $x^2+1<2x$, or $x^2-2x+1<0$. Now factor the lefthand side of that last inequality to get a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding combination using generating function Q. Find the no of ways of inserting r dolllars using 1 dollar, 2 dollar and 5 dollar tokens,when order doesn't matter and when order doesn't matter.
Ans.
When order doesn't matter..
$(1+x+x^2+x^3+..)(1+x^2+x^4+..)(1+x^5+x^{10}+..)$
Coefficient of $x^r$ in above generating function.
I clearly understand this. But I don't understand the process when order matters. Please somebody explain how (and why) to approach when order matters.
| When order does matter, the contents of the $k$-th bracket will be the possibilities for the $k$-th token to be inserted, which are $1,2$ or $5$ dollars regardless of $k$.
So a first guess for the generating function that respects order is
$$g(x)=(x+x^2+x^5)\times(x+x^2+x^5)\times(x+x^2+x^5)\times\cdots$$
but this has the problem that the degree of any term in the product is infinite.
But we know that at least $4$ and at most $19$ tokens can be used to make up the amount.
So
$$\begin{align}
g(x)&=\sum_{k=4}^{19}{(x+x^2+x^5)^k} \\
&=(x+x^2+x^5)^4\cdot\sum_{k=0}^{15}{(x+x^2+x^5)^k} \\[2ex]
&=(x+x^2+x^5)^4\cdot\frac{(x+x^2+x^5)^{16}-1}{x+x^2+x^5-1}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{dx}{3\sin(x)+2\cos(x)+3}$ $$\int \frac{1}{3\sin(x)+2\cos(x)+3}\ \text{d}x$$
This is one of the question which appeared in my exam today and I tried solving this but just couldn't solve it.
I tried converting sine and cosine in terms of $\tan\left(\frac{x}{2}\right)$ but didn't help very well..
Does anyone have idea how to evaluate this integral?
| HINT:
$$\int \frac{1}{3\sin(x)+2\cos(x)+3}\text{d}x =$$
Substitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)\text{d}x$. Than transform the integrand using the substitutions $\sin(x)=\frac{2u}{u^2+1}$, $\cos(x)=\frac{1-u^2}{u^2+1}$ and $\text{d}x=\frac{2}{u^2+1}\text{d}u$:
$$\int \frac{2}{(u^2+1)\cdot\left(\frac{6u}{u^2+1}+\frac{2(1-u^2)}{u^2+1}+3\right)}\text{d}u =$$
$$\int \frac{2}{u^2+6u+5}\text{d}u =$$
$$2\int \frac{1}{(u+3)^2-4}\text{d}u=$$
Substitute $s=u+3$ and $\text{d}s=\text{d}u$:
$$2\int \frac{1}{s^2-4}\text{d}s=$$
$$2\int -\frac{1}{4\left(1-\frac{s^2}{4}\right)}\text{d}s=$$
$$-\frac{1}{2}\int \frac{1}{1-\frac{s^2}{4}}\text{d}s=$$
Substitute $p=\frac{s}{2}$ and $\text{d}p=\frac{1}{2}\text{d}s$:
$$-\int \frac{1}{1-p^2}\text{d}p=$$
$$-\tanh^{-1}(p)+C$$
Substitute everything back and you'll get the final answer!
| {
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"timestamp": "2023-03-29T00:00:00",
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If (3,0) is a critical point, find if f(x,y) is a local maxima, local minima or a saddle point at (3,0) So i have this function:$f(x,y)=x^3+3xy^2-4y^3+x^2-33x$
Partials derivatives in $x$,$xx$,$y$,$yy$ and $xy$
$\frac{\partial }{\partial x}=3x^2+3y^2+2x-33$
$\frac{\partial^2 }{\partial x^2}=6x+2$
$\frac{\partial }{\partial y}=6xy-12y^2=6y(x-2y)$
$\frac{\partial^2 }{\partial y^2}=6x-24y$
$\frac{\partial^2 }{\partial x\partial y}=6y$
So i have this for the point $(3,0)$
$\Delta (3,0)=36x^2-144xy+12x-48y=360$
So the point $(3,0)$ is a saddle point.
Am i correct here? If not where did i go wrong?
Thank you.
| The solutions of the system
$$
\begin{cases}
3x^2+3y^2+2x-33=0,\\
6y(x-2y)=0
\end{cases}
$$
are:
$$(3,\,0),\;\;\left(-\frac{11}{3},\;0 \right) ,\\ \left(- \frac{4}{15} - \frac{2\cdot \sqrt{499}}{15}, \; - \frac{2}{15} - \frac{\sqrt{499}}{15}\right),\;\; \left(- \frac{4}{15} + \frac{2\cdot \sqrt{499}}{15}, \; - \frac{2}{15} + \frac{\sqrt{499}}{15}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic form from truth table with two outputs; simplifying boolean expression? I have a truth table like below:
x y z | a b
-----------
0 0 0 | 1 1
0 0 1 | 0 1
0 1 0 | 0 0
0 1 1 | 0 0
1 0 0 | 0 1
1 0 1 | 0 0
1 1 0 | 0 0
1 1 1 | 0 0
If the truth table had only one output, I would sum up the expressions that resulted in a true output, but I'm not entirely sure what to do for two outputs. Taking output $b$ for example:
$$b = \overline{xyz} + \overline{xy}z + \overline{x}yz$$
The complication arises with $a$ is introduced alongside $b$. Do I count only when both $a$ and $b$ are true?
$$ab = \overline{xyz}$$
Additionally, I'm not sure how to simplify an expression like this. Taking the expression for $b$ from earlier, is this how you do it?
$$\begin{align}b &= \overline{xyz} + \overline{xy}z + \overline{x}yz\\
&= \bar{x} + \bar{y} + \bar{z} + \bar{x} + \bar{y} + z + \bar{x} + y + z \text{ DeMorgan's Law}\\
&= \bar{x} + \bar{x} + \bar{x} + \bar{y} + \bar{y} + y + \bar{z} + z + z \text{ Commutation}\\
&= 3\bar{x} + \bar{y} + z\end{align}$$
The further I try to simplify, the less it makes sense in the context of circuitry.
| $$\begin{array}{ccc|cc} x & y & z & a & b \\ \hline
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0\end{array}$$ $$\implies \begin{cases}a= \overline x \cdot\overline y\cdot \overline z\\ b= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ a+b = \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z = b \\ a\cdot b = \overline x \cdot\overline y\cdot \overline z=a\end{cases}$$
Let's try to simplify $b$ a little:
$$\begin{align}b &= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x \cdot z + x \cdot \overline z) \\ &= \overline y \cdot [\overline x\cdot (\overline z + z) + x \cdot \overline z] \\ &= \overline y \cdot (\overline x + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot (\overline z+1) + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x + x\cdot \overline z) \\ &= \overline y \cdot ((\overline x +x)\cdot \overline z+\overline x) \\ &= \overline y \cdot (\overline z+\overline x) \\ &= \overline y \cdot \overline {z\cdot x}\end{align}$$
There's probably a slicker way of reaching that result, but it's been a while since I took electronics. But this way works! ;)
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I integrate $\sqrt{\frac{1+x}{1-x}}$? How do I integrate $\sqrt{\frac{1+x}{1-x}}$ using standard calculus techniques?
I tried trig substitution but it doesn't seems to work. is that some kind of u-substitution?
| To find $ \int \sqrt{\frac{1+x}{1-x}}dx $
You begin by multiplying $\frac{\sqrt{1+x}}{\sqrt{1+x}}$ to the integral.
then you get
$ \int \sqrt{\frac{(1+x)^2}{1-x^2}}dx $
now use trig substitution of
$x=sin \theta$
$dx = cos \theta d\theta$
we get $ \int \sqrt{\frac{(1+sin \theta)2}{1-sin^2 \theta}}cos \theta \, d\theta = \int 1+sin \theta \, d\theta = \theta - cos\theta + C = arcsin x - \sqrt{1-x^2} + C$
| {
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"timestamp": "2023-03-29T00:00:00",
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some functions, ploting them and defining positive and negative areas I have four following functions. I need to plot them and define positive and negative defined area. I have done most of them. But, i want you to check them, and correct them if false.
(1) $F(x,y)=x^2+2y^2$
for $x^2+2y^2=1$ , the graph is that
$C=\{(x,y)| -1\lt x \lt 1\ and -1/\sqrt{2} \lt y \lt 1/\sqrt {2}\} $
i.e the set C refers to inside of the ellipse
in this set C, the function is positive defined.
(2) $F(x,y)=xy$
for $xy=1$, the graph is that
this is hyperbol.
$C=\{(x,y)| x \lt 0 \ and\ y\lt 0,\ x\gt 0 \ and \ y\gt 0 \} $
in this set C, the function is positive defined.
(3) $F(x,y)=x+2y^2$
for $x+2y^2=0$ the graph is that
well, I cannot the set C which shows the positive and negative defined areas of the function.
please help me doing this.
(4) $F(x,y)=x^2+3xy+3y^2$
I can neither graph this function nor define the set C (positive and negatife defined areas).
Especailly, I cannot do part (3) and (4).
Hopefully help me doing these two parts. And please check other two parts (1) and(2). Thank you so much.
| (4)
The function $F(x,y)=x^2+3xy+3y^2$ can be thought as the quadratic form
$$F(x,y)=(x,y)\left(\begin{array}{cc}1&1\\2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right).\qquad (*)$$
Let us abbreviate $p=\left(\begin{array}{c}x\\y\end{array}\right)$ and
$Q=\left(\begin{array}{cc}1&1\\2&3\end{array}\right)$.
Now $Q$ isn't symmetric but can be decomposed as
$$Q=\frac{1}{2}(Q+Q^{\top})+\frac{1}{2}(Q-Q^{\top}).$$
This decomposition gives us the matrices
$$S=\frac{1}{2}(Q+Q^{\top})\quad \mbox{and}\quad A=\frac{1}{2}(Q-Q^{\top}),$$
which are $S$ symmetric and $A$ antisymmetric.
The matrix $S$ is
$$\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right).$$
Then ours expression for $F$ is
$$F(p)=p^{\top}Qp=p^{\top}(S+A)p=p^{\top}Sp,\qquad (**)$$
because it is easy to compute $p^{\top}Ap=0$.
Also note that
$$F(x,y)=(x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right)=x^2+3xy+3y^2.$$
Then the sign of $F$ will depend on the symmetric quadratic form $p^{\top}Sp$.
To decide the sign of this function, knowing that the level curve $x^2+3xy+3y^2=1$ is a tilted ellipse, one should try to calculate a new basis for the vectorspace $\Bbb R^2$.
This change gives a pair of axes that one gets from the eigenvectors of the matrix $S$.
The eigenvectors are
$$v_1=-\frac{2+\sqrt{13}}{3}e_1+e_2,$$
and
$$v_2=\frac{-2+\sqrt{13}}{2}e_1+e_2.$$
With these, one can take the matrix
$$B=
\left(\begin{array}{cc}
-\frac{2+\sqrt{13}}{3}&-\frac{2-\sqrt{13}}{3}\\
1&1
\end{array}\right),$$
and to get the interpretation of the quadratic form in $(**)$ with the basis $\{v_1,v_2\}$, this will be through the formula $B^{\top}QB$ (this is dubbed orthogonal diagonalization for $Q$).
So the quadratic form for $S$ is
$$p^{\top}Sp=({BB^{-1}p})^{\top}SBB^{-1}p=(B^{-1}p)^{\top}(B^{\top}SB)B^{-1}p$$
Hence any can compute that the resulting expression is
\begin{eqnarray*}
(x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right)
&=&
(s,t)B^{\top}SB
\left(\begin{array}{c}s\\t\end{array}\right)\\
&=&
(s,t)
\left(\begin{array}{cc}
\frac{26-5\sqrt{13}}{9}&0\\
0&\frac{26+5\sqrt{13}}{9}\end{array}\right)
\left(\begin{array}{c}s\\t\end{array}\right)\\
&=&
\frac{26-5\sqrt{13}}{9}s^2+\frac{26+5\sqrt{13}}{9}t^2
\end{eqnarray*}
Since both $\frac{26-5\sqrt{13}}{9},\frac{26+5\sqrt{13}}{9}$ are positive, then
$F(x,y)>0$ as well as $x,y$ aren't zero simultaneously.
We have
$\left(\begin{array}{c}s\\t\end{array}\right)=B^{-1}p$
to compute the new components of any vector $p=\left(\begin{array}{c}x\\y\end{array}\right)$.
| {
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How to find the desired Epsilon? Say for example I want to proof that:
$\forall 0<c \in \mathbb{Q} \; $ st. $\;c^2 < 2 \quad \exists 0<\varepsilon \in \mathbb{Q}\;$ st. $\; (c + \varepsilon)^2 < 2 $
And moreover, I want to actually find one.
Iv'e already been convinced it works with $\ \varepsilon = \frac{2-c^2}{c+2} \;$,
Now as I see it, it is just a way of dividing the distance between $c^2$ and $2$, but I'm wondering about the characteristic of this quotient.
My first question is how could i have found this, or any other $\ \varepsilon \; $ that satisfies ?
I was told that by solving $\; (c + \varepsilon)^2 < 2 $ for $\ \varepsilon \; $ I could have found one, but I can't really see how it works with the demand of it being a rational number (?). Now this is only one example, and finding a desired element is something I have seen repeats many times. Are there any main guidelines for this?
BTW,
I'm very new here and this is my very first post.
| Let $0<\varepsilon<1$ be a small number; then we have $$(c+\varepsilon)^2=c^2+2\varepsilon c+\varepsilon^2\le c^2+4\varepsilon+\varepsilon=c^2+5\varepsilon$$ since $c\le2$ and $\varepsilon^2\le\varepsilon$. Since $c^2<2$, we see that we can choose an $0<\varepsilon<1$ such that $c^2+5\varepsilon<2$, thus $(c+\varepsilon)^2<2$.
We can prove (by Archimedean property) that
Lemma 1. For any positive number $x>0$ there exists
a positive integer $N$ such that $x>1/N>0$.
Now, we want an $\varepsilon$ such that $c^2+5\varepsilon<2$, i.e., an $\varepsilon$ such that $\varepsilon<(2-c^2)/5$, which exists by Lemma 1.
In your case, since $0<(2-c^2)/(c+2)$, there is a $N$ such that $1/N<(2-c^2)/(c+2)$. Now, we can define $\varepsilon:=1/N$.
The key is, supposing that there is such $\varepsilon$, find some relation between $(x+\varepsilon)^2$ and $x^2+K\varepsilon$. Suppose $x^n<y$. There is many relations:
*
*First (Binomial formula): $$\sum_{k=0}^n\binom{n}{k}x^{n-k}\varepsilon^k.$$ Then use the expansion to obtain $(x+\varepsilon)^n\le x^n+x^{n-1}\varepsilon+\dotsb<y$.
*Second: $$(x+\varepsilon)^n\le x^n+\varepsilon((x+1)^n-x^n).$$
Then you can find $\varepsilon$ such that $(x+\varepsilon)^n<x$, just take a $\varepsilon$ such that $$0<\varepsilon<\min\left\{\frac{x-x^n}{(x+1)^n-x^n},1\right\},$$
which surely exists.
*Third: $$(x+\varepsilon)^n\le x^n+k\varepsilon$$
for some $k\in\Bbb R$. Then you obtain $(x+\varepsilon)^n\le x^n+k\varepsilon<y$, as desired.
All formulae are proved by induction on $m$.
A way to obtain this type of relations is see the behavior when $n$ increase.
For instance, let $x,y>0$ be rational numbers, and let $n\ge1$ be a integer number. We want to find some relation of the form $(x+\varepsilon)^n\le x^n+K\varepsilon$ for some $\varepsilon$ between $0$ and $1$.
Note that $\varepsilon^n\le\varepsilon$ for every $n$. Also, note that $\varepsilon$ always exists by Archimedean property.
Using some algebra, we can expand $(x+\varepsilon)^n$ when $n=1,2,3,4,5,\dots$
$$\begin{align}(x+\varepsilon)^2&=x^2+2x\varepsilon+\varepsilon^2\\&\le x^2+2x\varepsilon+\varepsilon\\&=x^2+\varepsilon(2x+1)\\\\(x+\varepsilon)^3&=x^3+3x^2\varepsilon+3x\varepsilon^2+\varepsilon^3\\&\le x^3+3x^2\varepsilon+3x\varepsilon+\varepsilon\\&=x^3+\varepsilon(3x^2+3x+1)\\\\(x+\varepsilon)^4&=x^4+4x^3\varepsilon+6x^2\varepsilon^2+4x\varepsilon^3+\varepsilon^4\\&\le x^4+4x^3\varepsilon+6x^2\varepsilon+4x\varepsilon+\varepsilon\\&=x^4+\varepsilon(4x^3+6x^2+4x+1)\\&\;\;\vdots\end{align}$$
So we have
$$\begin{align}(x+\varepsilon)^2&\le x^2+\varepsilon(2x+1)\\(x+\varepsilon)^3&\le x^3+\varepsilon(3x^2+3x+1)\\(x+\varepsilon)^4&\le x^4+\varepsilon(4x^3+6x^2+4x+1)\\(x+\varepsilon)^5&\le x^5+\varepsilon(5x^4+10x^3+10x^2+5x+1)\\(x+\varepsilon)^6&\le x^6+\varepsilon(6x^5+15x^4+20x^3+15x^2+6x+1)\\&\;\;\vdots\end{align}$$
Now, suppose that $x\le1$; so $x^n\le 1$ for every $n$. Thus
$$\begin{align}(x+\varepsilon)^2&\le x^2+3\varepsilon\\(x+\varepsilon)^3&\le x^3+7\varepsilon\\(x+\varepsilon)^4&\le x^4+15\varepsilon\\(x+\varepsilon)^5&\le x^4+31\varepsilon\\(x+\varepsilon)^6&\le x^4+63\varepsilon\\&\;\;\vdots\\\end{align}$$
Clearly, the relation from this is $$(x+\varepsilon)^n\le x^n+(2^n-1)\varepsilon.$$ (We can prove this by induction.)
Similarly, suppose $x>1$; so $x^n\ge x$ for every $n$. We can prove the relation $(x+\varepsilon)^n\le x^n+(2^n-1)x^{n-1}\varepsilon.$
But, if we don't want divide in cases, we can prove the relation $$(x+\varepsilon)^n\le x^n+n(2^n-1)(1+x)^n\varepsilon.$$
(To see this relation, try to prove the above relations and think how avoid the cases when some $k$ ensures the relation $(x+k)^n\ge x+k$.)
| {
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"url": "https://math.stackexchange.com/questions/1497483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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proving a hard inequality can someone help me to prove this inequality :
$\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } } } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } } } \right| \le \frac { 4 }{ { n }^{ 2 } } (2n+1)$
| First note that $1^2 + 2^2 + 3^2 + ... + n^2 =\frac{n(n + 1)(2n + 1)}{6}$
$\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } } } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } } } \right| \leq \sum _{ k=0 }^{ 2n }\left|\frac { k }{ k+{ n }^{ 2 }} -{ \frac { k }{ { n }^{ 2 } }}\right| = \sum _{ k=0 }^{ 2n }\left| \frac{k^2}{(k+n^2)n^2} \right| \leq \sum _{ k=0 }^{ 2n } \left|\frac{k^2}{n^4}\right| = \frac{1}{n^4}\sum _{ k=0 }^{ 2n } k^2 = \frac{1}{n^4}\frac{2n(2n + 1)(4n + 1)}{6} \leq \frac{1}{n^4}\frac{2n(2n + 1)(12n)}{6} = \frac{4(2n + 1)}{n^2} $
(through this estimation the bound can actually be smaller)
| {
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Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?
| Also, because for odd $n$ we have
$$\frac{(2n)!}{(n+1)!(n-1)!}\cdot\frac{(2n-2)!}{(n+1)!(n-3)!}\cdot...\cdot\frac{(n+1)!}{(n+1)!0!}\cdot(n-1)!(n-3)!...2!\geq1.$$
For the even $n$ it's the similar.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a recursive sequence with an independent term that goes to 0 I have a recursive sequence where the first element is $a_{1} = 1$ and then $a_{n+1}= \frac12a_{n} + \frac{1}{n+1}$.
The first three terms are $a_{1} = 1 > a_{2} = \frac56 > a_{3} = \frac23$ so it seems reasonable to try to prove that the sequence is decreasing and then use the monotonicity of the sequence to try to prove that it has a finite or infinite limit. (This is my usual approach to recursive sequences.)
However, I don't know how to deal with the term $\frac1{n+1}$. Clearly $\lim \frac1{n+1} = 0$, but I don't know if I can erase it and solve $a_{n+1} = \frac12 a_{n} < \frac12 a_{n-1} = a_{n}$ to prove the monotonicity.
| First prove a Lemma: $a_n > \frac{2}{n+1}$ for $n \geq2 $
When $n=2$, $a_2 = 5/6 > 2/3$
Now suppose $a_n > \frac{2}{n+1}$
then $a_{n+1} = \frac{a_n}{2}+ \frac{1}{n+1} > (\frac{1}{n+1})+\frac{1}{n+1} = \frac{2}{n+1} $
By induction, $a_n > \frac{2}{n+1}$ for all n
Now we can prove that $a_n$ is decresing.
$a_{n+1}-a_n = \frac{a_n}{2}+ \frac{1}{n+1}-a_n = \frac{-a_n}{2}+ \frac{1}{n+1} < (-\frac{1}{n+1}) + \frac{1}{n+1} = 0$
so $a_n$ is decresing. By our Lemma, we also have $a_n \geq 0$, so we have a decresing sequence bounded below, thus it converges.
To find the limit, let $lim_{n\to \infty} a_n = L$. By the recursive relation,
$L = L/2$ as $n\to \infty$
so $L = 0$
$lim_{n\to \infty} a_n = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the indefinite integral $\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $ Find the indefinite integral
$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$
Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
| Let $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\;,$$ Now Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$I = \int\frac{t^3}{(1+t^2)\sqrt{t^2-1}}\cdot -\frac{1}{t^2}dt = -\int\frac{t}{(1+t^2)\sqrt{t^2-1}}dt$$
Now Put $(t^2-1)=u^2\;,$ Then $2tdt = 2udu\Rightarrow tdt = udu$
So we get $$I = -\int\frac{u}{u^2+2}\cdot \frac{1}{u}du = -\int\frac{1}{u^2+2}du$$
So we get $$I = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathcal{C}=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt{2}}\right)+\mathcal{C}$$
So we get $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+\mathcal{C}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate: $\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}$. While solving a question that has to do with Darboux sums I reached the following sum:
$$\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}$$
and I didn't know how to do it. but I know its supposed to be $\frac{1}{12}$.
| Telescopic series
$$\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}
\sum_{i=1}^n=\frac{1-i}{3 (i+3 n-1)}+\frac{i}{3 (i+3 n)}=
\left(\frac{0}{3 (3 n)}+\frac{1}{3 (1+3 n)} \right)+
\left(\frac{-1}{3 (1+3 n)}+\frac{2}{3 (2+3 n)}\right)+
\left(\frac{-2}{3 (2+3 n)}+\frac{3}{3 (3+3 n)}\right)+
\left(\frac{-3}{3 (3+3 n)}+\frac{4}{3 (4+3 n)}\right)\dots=
\left( \frac{n}{3(4n)}\right)=\frac{1}{12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Number of squarefree positive integers less than $100$
An integer is called squarefree if it is not divisible by the square
of a positive integer greater than $1$. Find the number of squarefree
positive integers less than $100$.
My attempt:
I apply the inclusion-exclusion principle directly.
*
*Total number of integers = 99
*Number of integers divisible by $2^2$ = $24$
*Number of integers divisible by $3^2$ = $11$
*Number of integers divisible by $4^2$ = $6$
*Number of integers divisible by $5^2$ = $3$
*Number of integers divisible by $6^2$ = $2$
*Number of integers divisible by $7^2$ = $2$
*Number of integers divisible by $8^2$ = $1$
*Number of integers divisible by $9^2$ = $1$
*Number of integers divisible by $2^2$ and $3^2$ = $2$
*Number of integers divisible by $2^2$ and $4^2$ = $1$
Then the required solution would be $99-(24+11+6+3+2+2+1+1)+2+1=52$. But the solution is $61$. Where is my mistake?
| There are 6 integers divisible by 2^2 and 4^2, not 1. You are also not counting when integers are divisible by both 3^2 and 9^2, or 2^2, 4^2, and 8^2 for example. The other methods provided are easier to calculate, but your approach is correct except for these errors.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for $x$ :$|[x]-2x|=4$ Solve for $x$
$$|[x]-2x|=4$$ where $[.]$ denotes greatest integer function.
How to solve it graphically? can any one guide me step by step to get the solution ?
| Without Using Graph.
Given $\left|\lfloor x \rfloor -2x\right| = 4\Rightarrow \left|-\lfloor x \rfloor -2\left\{x\right\}\right|=4\Rightarrow \left|\lfloor x \rfloor +2\{x\}\right|=4$
Now Here Right hand Side is Integer. So Left hand Side must be integer.
So Using $0 \leq \{x\}<1\Rightarrow 0\leq 2\{x\}<2.$
So we get $2\{x\} =0\Rightarrow \{x\} =0$ or $\displaystyle 2\{x\} = 1\Rightarrow \{x\} = \frac{1}{2}$
$\bullet\; $ If $\{x\}=0\;,$ Then equation convert into $|\lfloor x \rfloor | = 4\Rightarrow |x|=4\Rightarrow x=\pm 4$
$\bullet\; $ If $\displaystyle \{x\}=\frac{1}{2}\;,$ Then equation convert into $|\lfloor x \rfloor +1| = 4\Rightarrow \lfloor x\rfloor =\pm 4-1\Rightarrow \lfloor x\rfloor =3,-5$
So we get $\displaystyle x = \lfloor x \rfloor +\{x\} = 3+\frac{1}{2}\;\;,-5+\frac{1}{2}=3.5\;\;,-4.5$
So we get $\displaystyle x = \left\{\pm 4\;, 3.5\;,-4.5\right\}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cauchy Product - Power Series Representation Question
Use Cauchy product to find a power series representation of
$$ (1+x^2+x^3+\cdots)(1-x^2+x^3-\cdots)$$
Solution
$$ (1+x^2+x^3+\cdots)=\frac{1}{1-x}= \sum_{n=0}^\infty x^n $$
$$ (1-x^2+x^3-\cdots)=\frac{1}{1+x}= \sum_{n=0}^\infty (-1)^nx^n $$
The Cauchy Product states:
$$
\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right)
=\sum_{n=0}^\infty\left(\sum_{j=0}^n a_jb_{n-j}\right)x^n\tag{2}
$$
and
$$\sum_{j=0}^n a_jb_{n-j}=\sum_{j=0}^n (-1)^{n-j}= \begin{cases} 0 & \text{even} \\ 1 & \text{odd} \end{cases}$$
But now I am stuck and not sure quite what to do even though I know the answer is just
$$\frac{1}{1+x}\cdot\frac{1}{1-x}=\frac{1}{1-x^2} = 1+x^2+x^4+\cdots=\sum_{n=0}^\infty x^{2n}$$
Any advice would be helpful thank you in advance
| Hint. Since
$$
\frac{1+(-1)^n}2= \left\{\begin{array}{ll}1 &\quad n = 2k\\ 0 &\quad n=2k+1, \end{array} \right.
$$ why not just write
$$
\frac{1}{1+x}\times\frac{1}{1-x}=\frac{1}{1-x^2} =\sum_{n=0}^{\infty}{x^{2n}}=\sum_{n=0}^{\infty}\frac{1+(-1)^n}2x^{n}, \quad |x|<1,
$$ the latter series being a power series representation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A Limit of Complex Trig Please help with a messy trig. I only want the process of evaluating the limits! Appreciate!
$$\lim_{x \to 0} \frac{(x-\sin x)\sin x}{(1-\cos x)^2}$$
or the alternative form
$$\lim_{x \to 0} \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(\sin x)}$$
Wolframalpha gave $2/3$
The original question(with image) is asking the area ratio. $\frac{ABD}{ADBC}$
colored segment ABD over the triangle ABC minus segment ABD
| First, write
\begin{equation*}
\frac{(x-\sin x)\sin x}{(1-\cos x)^{2}}=\left( \frac{x-\sin x}{x^{3}}\right)
\left( \frac{\cos (x)-1}{x^{2}}\right) ^{-2}\left( \frac{\sin x}{x}\right)
\end{equation*}
Next, using the standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}} &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(x-\sin x)\sin x}{(1-\cos x)^{2}}
&=&\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) \left( \frac{%
\cos (x)-1}{x^{2}}\right) ^{-2}\left( \frac{\sin x}{x}\right) \\
&=&\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) \cdot \left(
\lim_{x\rightarrow 0}\frac{\cos (x)-1}{x^{2}}\right) ^{-2}\cdot
\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right) \\
&=&\left( \frac{1}{6}\right) \cdot \left( -\frac{1}{2}\right) ^{-2}\cdot 1=%
\frac{2}{3}.\ \ \ \ \blacksquare
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Expected number of failures when tossing coins I'm trying to solve a problem but I suspect my solution is incorrect. I'm hoping someone can verify and perhaps give me an idea about how to solve it correctly.
We have three fair coins and each is tossed until the first head appears on each coin. (So once one head appears on one of the three coins, we stop tossing that particular coin.) When we are finished, a total of $6$ tails have been obtained. What is the expected value of the number of tails on the first coin?
This is how I have tried to solve it:
If a total of $6$ tails have appeared the possible number of tails obtained with the first coin is 0, 1, 2, 3, 4, 5 or 6. I define a random variable $X \sim \mathrm{Geometric}(1/2)$ so that the value of $X$ is the number of tails (failures) until the first head (success) is obtained.
I know that the PMF of a random variable with Geometric distribution is $(1-p)^k*p$ and in this case $p = 1/2$
From here I went on to compute the expected value as
$$\mathrm{E}(X) = \sum_{k=0}^6 k*\mathrm{P}(X=k) = \sum_{k=0}^6 k*(\frac{1}{2})^k*\frac{1}{2}$$
and reached the result $15/16.$
As I said, I'm pretty certain this is not the right answer, but I'm not sure why and how to solve it correctly, so any help is appreciated.
| If by "first coin" you mean "first coin to show a head", here is my answer.
The probability that the each coin shows $k$ tails before the first head is $\frac1{2^{k+1}}$, so the probability for each arrangement that the sum of the number of tails is $6$ is $\frac1{2^9}$. That is, each such arrangement has equal probability.
One Method to Compute the Number of Arrangements
The number of such arrangements where the lowest number of tails is $k$ or greater is the coefficient of $x^6$ in $\left(\frac{x^k}{1-x}\right)^3$ which is the coefficient of $x^{6-3k}$ in $\left(\frac1{1-x}\right)^3$, which is
$$
\begin{align}
(-1)^{6-3k}\binom{-3}{6-3k}
&=\binom{6-3k+2}{6-3k}\\
&=\binom{8-3k}2[k\le2]
\end{align}
$$
Another Method to Compute the Number of Arrangements
To count the number of ways to have $3$ non-negative integers that sum to $6$, we can use the Stars and Bars Method, which says the number is $\binom{6+3-1}{3-1}=\binom{8}{2}$ (the same as above).
To count the number of ways to have $3$ positive integers that sum to $6$, we can count the number of ways to have $3$ non-negative integers that sum to $3$ (by preloading each stars and bars partition with $1$). This gives $\binom{3+3-1}{3-1}=\binom{5}{2}$ (the same as above).
To count the number of ways to have $3$ integers, greater than or equal to $2$, that sum to $6$, we can count the number of ways to have $3$ non-negative integers that sum to $0$ (by preloading each stars and bars partition with $2$). This gives $\binom{0+3-1}{3-1}=\binom{2}{2}$ (the same as above).
Thus, the number of arrangements with exactly $0$ as the lowest number is $\binom{8}{2}-\binom{5}{2}=18$.
The number of arrangements with exactly $1$ as the lowest number is $\binom{5}{2}-\binom{2}{2}=9$.
The number of arrangements with exactly $2$ as the lowest number is $\binom{2}{2}-0=1$.
Thus, the expected value is
$$
0\cdot\frac{18}{28}+1\cdot\frac{9}{28}+2\cdot\frac1{28}=\frac{11}{28}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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find all complex solutions of $z^{12}-z^6-2z^3+2=0$ How could I find all the complex solutions to: $z^{12}-z^6-2z^3+2=0$
I tried substituting $y=z^3$ to get a simpler equation and I managed to get $y-1=0$ or $y^3+y^2-2=0$, but finding all complex solutions of the latter equation was also complicated.
| HINT:
$$z^{12}-z^6-2z^3+2=0\Longleftrightarrow$$
$$(z-1)^2(z^2+z+1)^2(z^6+2z^3+2)=0\Longleftrightarrow$$
$$(z-1)^2=0\Longleftrightarrow\space\space\vee\space\space(z^2+z+1)^2=0\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z-1=0\Longleftrightarrow\space\space\vee\space\space z^2+z+1=0\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z^2+z=-1\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z^2+z+\frac{1}{4}=-\frac{3}{4}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space \left(z+\frac{1}{2}\right)^2=-\frac{3}{4}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z+\frac{1}{2}=\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space z^6+2z^3+2=0\Longleftrightarrow$$
Substitute $x=z^3$:
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x^2+2x=-2\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space (x+1)^2=-1\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x+1=\pm i\Longleftrightarrow$$
$$z=1\Longleftrightarrow\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Longleftrightarrow\space\space\vee\space\space x=-1\pm i\Longleftrightarrow$$
$$z=1\space\space\vee\space\space z=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\space\space\vee\space\space z^3=-1\pm i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I was trying to solve a complex number to a power using de Moivre's Identity but got stuck. One of the assignments that I have to solve is:
$(7-i)^3$
I know that I can simply open the exponent since it is only 3 but I tried to use de Moivre's Identity.
$\theta=\tan^{-1}(\frac{-1}{7})$
$r=\sqrt{7^{2}+(-1)^{2}}=\sqrt{50}$
$z^{n}=r(\cos\theta+i\sin\theta)=50^{\frac{3}{2}}(\cos(3\cdot\tan^{-1}(\frac{-1}{7}))+i\sin(3\cdot\tan^{-1}(\frac{-1}{7})))$
How can I find out what the cosine of an arc-tangent is?
| Well, we have $\arctan(-x)=-\arctan(x)$, so $\arctan\left(\dfrac{-1}{7}\right)=-\arctan\left(\dfrac{1}{7}\right)$ and we also can simplify $\sqrt{50}=\sqrt{2\times25}=5\sqrt{2}$. Then your last expression simplifies to $$(7-i)^3=5\sqrt{2}\cos\left(3\arctan\left(\frac{1}{7}\right)\right)-5\sqrt{2}i\sin\left(3\arctan\left(\frac{1}{7}\right)\right)$$
Now, to really answer your question, first we need to get rid of that $3$ factor using the following trigonometric identities: $$\sin(3x)=3\sin(x)\cos(x)^2-\sin(x)^3,$$$$\cos(3x)=\cos(x)^3-3\cos(x)\sin(x)^2;$$ these can easily be derived for arbitrary integer factor once you learn about complex exponentiation and generally for arbitrary (symbolic) factor using the binomial theorem.
We have $$(7-i)^3=5\sqrt{2}\left(\cos\left(\arctan\left(\frac{1}{7}\right)\right)^3-3\cos\left(\arctan\left(\frac{1}{7}\right)\right)\sin\left(\arctan\left(\frac{1}{7}\right)\right)^2\right)-5\sqrt{2}i\left(3\sin\left(\arctan\left(\frac{1}{7}\right)\right)\cos\left(\arctan\left(\frac{1}{7}\right)\right)^2-\sin\left(\arctan\left(\frac{1}{7}\right)\right)^3\right);\label{eq1}\tag{1}$$
now we can evaluate $\sin\circ\arctan$ and $\cos\circ\arctan$: just recall
$$\sin(x)^2+\cos(x)^2=1;$$
we want to have $\sin$ and $\cos$ in terms of $\tan$ to cancel $\arctan$, therefore divide by $\sin(x)^2$:
$$1+1/\tan(x)^2=1/\sin(x)^2\implies$$
$$\implies \sin(x)^2=\frac{1}{1+1/\tan(x)^2}=\frac{\tan(x)^2}{\tan(x)^2+1}$$
$$\implies \sin(x)=\pm\frac{\tan(x)}{\sqrt{\tan(x)^2+1}},$$
$$\sin(\arctan(\alpha))=\frac{\tan(\arctan(\alpha))}{\sqrt{\tan(\arctan(\alpha))^2+1}}=\frac{\alpha}{\sqrt{\alpha^2+1}};$$
and $\cos(x)^2$:
$$\tan(x)^2+1=1/\cos(x)^2\implies$$
$$\implies\cos(x)^2=\frac{1}{\tan(x)^2+1}$$
$$\implies\cos(x)=\pm\frac{1}{\tan(x)^2+1},$$
$$\cos(\arctan(\alpha))=\frac{1}{\sqrt{\tan(\arctan(\alpha))^2+1}}=\frac{1}{\sqrt{\alpha^2+1}};$$
in particular then, we have
$$\sin\left(\arctan\left(\frac{1}{7}\right)\right)=\frac{1/7}{\sqrt{1/7^2+1}}=\frac{1}{7\sqrt{1/7^2+1}}=\frac{1}{\sqrt{7^2+1}}=50^{-1/2}=\frac{1}{5}2^{-1/2},$$
$$\cos\left(\arctan\left(\frac{1}{7}\right)\right)=\frac{1}{\sqrt{1/7^2+1}}=\frac{7}{\sqrt{7^2+1}}=\frac{7}{5}2^{-1/2};$$
substituting that in $\eqref{eq1}$ will yield the desired form after a bit of simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the points on the ellipse $x^2+2y^2=1$ where the tangent line has slope $1$. I did the following:
$$2x+4y\cdot y'=0\\y'=\frac{-2x}{4y}$$
Now I guess I should take all pairs $(x,y)$ such that:
$$\frac{-2x}{4y}=1$$
And I found: $\left(1,-\cfrac{1}{2}\right),\left(-1,\cfrac{1}{2}\right)$
And then I tried to do:
$$x^2+2y^2=1\\ y=\frac{\sqrt{1-x^2}}{\sqrt{2}}$$
And then derive $\frac{\sqrt{1-x^2}}{\sqrt{2}}$, this yielded the pairs: $\{(\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}),(-\sqrt{\frac{2}{3}},\frac{1}{\sqrt{6}})\}$
I don't see what I did wrong in the implicit differentiation method.
| we have $x=-2y$ and we get $$4y^2+2y^2=1$$ thus we get $$y=\pm\sqrt{\frac{1}{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\cos \frac{2\pi}{9} + \cos \frac{4\pi}{9} + \cos \frac{8\pi}{9}$=0 I tried letting $z=cis \frac{2\pi}{9}$
and using trigonometric identities i got $\frac{1}{2z^4} [\frac{z^9-1}{z-1}]-\frac{1}{2z^3} - \frac{1}{2}$
and since i assumed that z then $z^9=cis (2\pi) =1$
but i ended up with $\cos \frac{2\pi}{n} + \cos \frac{4\pi}{n} + \cos \frac{8\pi}{n} = \frac{(-1+i\sqrt(3))}{4}$
| $\cos \frac{2\pi}{9} + \cos \frac{4\pi}{9} + \cos \frac{8\pi}{9}=0$
Let $\frac{2\pi}{9}=t$, then
$\cos t + \cos 2t + \cos 4t=\cos t+\cos ^2t-\sin ^2t+\cos^2(2t)-\sin ^2(2t)=$
$=\cos t+\cos ^2t-\sin ^2t+(\cos ^2t-\sin ^2t)^2-(2 \sin t \cos t)^2=0$
$=\cos t+\cos ^2t-\sin ^2t+\cos ^4t+\sin ^4t-2 \cos ^2t \sin ^2t -4 \cos ^2t \sin ^2t =$
Try this. It should help you.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that the limit of a function doesn't exist using negation of epsilon delta $\lim \limits_{x \to \frac{1}{2}}\frac{1}{4x-2}$
I want to use the negation, $\exists \epsilon>0$ such that $ \forall \delta>0$ , $\lvert\frac{1}{4x-2}-L \rvert \ge \epsilon$, $\forall x$ with $0<\lvert x-\frac{1}{2} \rvert <\delta$
So can I say that because $\lvert \frac{1}{4x-2}\rvert =\lvert \frac{1}{4(x-\frac{1}{2})} \rvert = \frac{1}{4}\lvert \frac{1}{x-\frac{1}{2}} \rvert \ge \epsilon $
Then $\frac{1}{4} \ge \epsilon \lvert x-\frac{1}{2} \rvert$
Can I then let $\epsilon =\frac{1}{4 \delta}$
| No, behold! The choice of $\varepsilon$ should not depend on that of $\delta$.
In fact we can prove something stronger than necessary:
If $x \neq 1/2$, then
$$
\bigg| \frac{1}{4x-2} \bigg| = \frac{1}{4}\frac{1}{|x- \frac{1}{2}|};
$$
If $\varepsilon > 0$, then
$\frac{1}{4}\frac{1}{|x - \frac{1}{2}|} > \varepsilon$ if $|x-\frac{1}{2}| < \varepsilon/4$; hence $0 < |x-\frac{1}{2}| < \varepsilon/4$ only if
$$
\bigg| \frac{1}{4x-2} \bigg| > \varepsilon,
$$
which says that
$$
\bigg| \frac{1}{4x-2} \bigg| \to \infty
$$
as $x \to 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Simplifying using the rule of logs How would you simplify: $\left[\frac{16}{5}\ln(x + 2) - \frac{1}{5}\ln(x - 3) - \ln x \right]_4^6$ and put it in the form $\ln\frac{m}{n}$. Stating the values of m and n.
Note:
$a\ln b = \ln b^a$
$\ln a - \ln b = \ln \frac{a}{b}$
I just cant get the answer, my answer is always a decimal.
| The expression evaluated at $x=6$ is
$$
\frac{16}{5}\ln8-\frac{1}{5}\ln3-\ln6=
\frac{48}{5}\ln2-\frac{1}{5}\ln3-\ln2-\ln3=
\frac{43}{5}\ln2-\frac{6}{5}\ln3
$$
Evaluated at $x=4$ it is
$$
\frac{16}{5}\ln6-\frac{1}{5}\ln1-\ln4=
\frac{16}{5}\ln2+\frac{16}{5}\ln3-2\ln2=
\frac{6}{5}\ln2+\frac{16}{5}\ln3
$$
Thus the difference is
$$
\frac{43}{5}\ln2-\frac{6}{5}\ln3-\frac{6}{5}\ln2-\frac{16}{5}\ln3=
\frac{37}{5}\ln2-\frac{22}{5}\ln3=
\frac{1}{5}\ln\frac{2^{37}}{3^{22}}
$$
There's no way to represent it as $\ln(m/n)$, because this would mean
$$
\left(\frac{m}{n}\right)^{\!5}=\frac{2^{37}}{3^{22}}
$$
and this is impossible with integer $m$ and $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Analytic function on an open disc. Let $\mathbb{D}=\{ z\in\mathbb{C}:|z|<1\}$. Which of the following are correct?
*
*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(0)=0$ and $f'(0)=2$
*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=3/4$ and $f'(2/3)=3/4$
*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=-3/4$ and $f'(3/4)=-3/4$
*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(1/2)=-1/2$ and $f'(1/4)=1.$
Option one is not true by Schwartz-lemma. But i don't know how to think about other options. I don't know which theorem or result gives condition of existence of this type of holomorphic function. Please help me to solve this problem. If possible please solve this problem. Thanks in advance.
| For 1. Incorrect. Since Using Schwarz Lemma we must have $|f'(0)|\leq 1$.
For 2. Correct. consider $f(z) = \frac{3}{4}z + \frac{3}{16}$
For 3. Correct. consider $f(z) = -\frac{3}{4}z - \frac{3}{16}$
For 4. Incorrect. Using Schwarz Lemma (derivative version) i.e
$|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$. Here given that $f'(\frac{1}{4})=1$. So we have $|f(\frac{1}{4})| \leq \frac{1}{4}$.
Also using Schwarz Pick Lemma we have $$\frac{|f(z) -f(w)|}{|1-f(z)\overline{f(w)}|} \leq \frac{|z-w|}{|1-z\bar{w}|}$$
Using above inequality and the fact that $f(\frac{1}{2}) =-\frac{1}{2}$, one will get
\begin{align}
\frac{|f(\frac{1}{4}) + \frac{1}{2}|}{|1+ \frac{1}{2}f(\frac{1}{4})| } &\leq \frac{2}{7} \\
\implies |2 - \frac{3}{f(\frac{1}{4})+2}|&\leq \frac{2}{7} \\
\implies \frac{12}{7} \leq \frac{3}{|f(\frac{1}{4})+2|} &\leq \frac{16}{7}\\
\implies \frac{21}{16}\leq |f(\frac{1}{4})+2| &\leq \frac{21}{12}
\end{align}
So $f(\frac{1}{4})$ belong to Annulus with centre at $-2$ and with inner radius $\frac{21}{16}$ and outer radius $\frac{21}{12}$.
we also have $|f(\frac{1}{4})| \leq \frac{1}{4}$ which gives us $f(\frac{1}{4})$ lies in disk with centre $0$ and radius $\frac{1}{4}$. This two domain intersect at $-\frac{1}{4}$. So, one must have $f(\frac{1}{4}) = -\frac{1}{4}$.
Then we have $|f(\frac{1}{4})|= \frac{1}{4}$ and $f'(\frac{1}{4}) =1$. we have equality in the inequality $|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$ for $z = \frac{1}{4}$. Hence $f$ has to be scalar times identity function i.e $f(z) = cz$ for all $z$ in unit disk with $c$ is an unimodulur constant. But $f'(\frac{1}{2}) =1 $ gives us $c=1$. But that contradict $f(\frac{1}{2}) = - \frac{1}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/1521065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $xy'' + y' + xy = 0, y(0)=1, y'(0)=0$ The initial value problem $xy'' + y' + xy = 0; y(0)=1, y'(0) = 0$ has_____
Options:
a) a unique solution.
b) no solution
c) infinitely many solutions
d) two linearly independent solutions
(Is there any other way around to do it other than power series method?)
Ans) I started with y(x) = $x^r$ but fails to continue.
Then I choose $$y(x) =\sum_{n=0}^{\infty} a_n x^n$$
$$y'(x) =\sum_{n=0}^{\infty} na_nx^{n-1}$$
$$y''(x) = \sum_{n-0}^{\infty} n(n-1)a_nx^{n-2}$$
Substitution leads to
$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=0}^{\infty} a_n x^{n+1}=0$$
Change of index leads to
$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=2}^{\infty} a_{n-2} x^{n-1}=0$$
Rewrite as
$$ a_1 + \sum_{n=2}^{\infty} (n^2a_n + a_{n-2})x^{n-1}=0$$ which implies $a_1=0$.
Recursive relation is
$$n^2a_n = -a_{n-2}$$
Take $a_0 = a_0$ and $a_1 = a_1$
$a_2 = \displaystyle\frac{-a_0}{2^2}$ $a_3 = \displaystyle\frac{-a_1}{3^2}$
$a_4 = \displaystyle\frac{a_0}{2^2.4^2}$ $a_5 = \displaystyle\frac{a_1}{3^2.5^2}$
$a_6 = \displaystyle\frac{-a_0}{2^2.4^2.6^2}$ $a_7 = \displaystyle\frac{-a_1}{3^2.5^2.7^2}$
$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2} + a_1\sum_{k=0}^{\infty} \frac{(-1)^kx^{2k+1}}{\Pi_{n=0}^{k} (2n+1)^2}$$
already we got $a_1 = 0$ hence solution will be
$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$
Given condition $y(0) = 1$ gives $a_0 = 1$
Final solution will be $$y(x) = 1 + \sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$ and clearly $y'(0) = 0$ satisfies. So concludes a unique solution.
Is there any other way to do it smartly?
| Let $Y(s)$ be the Laplace transform of $y(x)$, then $y' = sY - y(0)$ and $y'' = s^2Y - sy(0) - y'(0) $
Taking the transform of the equation
$$ -(s^2Y - s)' + (sY-1) - Y' = 0 $$
$$ -(2sY + s^2Y' - 1) + sY - 1 - Y' = 0 $$
$$ (1+s^2)Y' + sY = 0 $$
Solving the above first-order ODE gives
$$ Y = \frac{C}{\sqrt{1+s^2}}$$
Using a table lookup, we find that
$$ y(x) = J_0(x) $$
As @Urgje pointed out, you can obtain the same solution by multiplying both sides by $x$
$$ x^2y'' + xy' + x^2y = 0 $$
which is the Bessel equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522357",
"timestamp": "2023-03-29T00:00:00",
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Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$.
Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto the set $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$.
My attempt:
Let $z= x+iy$ then we have \begin{align}f(z) &= \frac{1}{z+2}\\ &=\frac{\bar{z}-2}{(z+2)(\bar{z}-2)} \\ &= \frac{(x-2)+i(-y)}{(x+2)^2 +y^2} \\ &= \frac{(x-2)}{(x+2)^2+y^2} +i(-y)\end{align}
Now let \begin{align}u(x,y)= \frac{(x-2)}{(x+2)^2 +y^2}~~ \text{and}~~v(x,y) = -y\end{align}
then we have $$f(z)= u(x,y) + iv(x,y)$$
Now the part that I am stuck with, is showing that the circle of radius 2 centered at the origin gets mapped onto the vertical line through $\frac{1}{4}$ on the real axis.
What I've tried is \begin{align}x^2 + y^2 &= 4\end{align}
Then, since we're interested in the real part, we now only look at \begin{align}u(x,y) &= \frac{(x-2)}{(x+2)^2 + y^2} \\ &= \frac{(x-2)}{x^2 + 4x +4 +y^2} \\ &= \frac{(x-2)}{4x + 8} \\ &= \frac{(x-2)}{4(x+2)}\end{align}
Now this is where I see I have a problem - I cannot cancel those two out, did I make a mistake somewhere?
| z+2 shifts the given circle two units to the right that passes through the origin . Now 1/(z+2) takes this circle to a line that passes through the point 1/4 and perpenicular to the x-axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522942",
"timestamp": "2023-03-29T00:00:00",
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