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Fourier transform of squared exponential integral $\operatorname{Ei}^2(-|x|)$ Let $\operatorname{Ei}(x)$ denote the exponential integral:
$$\operatorname{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}tdt.$$
Now consider the function $\operatorname{Ei}(-|x|)$.
Its Fourier transform is
$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\operatorname{Ei}(-|x|)\,e^{i\hspace{.05em}k\hspace{.05em}x}\,dx=-\sqrt{\frac2\pi}\frac{\arctan|k|}{|k|}.$$
I'm interested in a Fourier transform of its square:
$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\operatorname{Ei}^2(-|x|)\,e^{i\hspace{.05em}k\hspace{.05em}x}\,dx=\,?$$
|
I changed my evaluation slightly, and I was able to get the result in a very simple form.
First notice that $$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx. $$
Then integrating by parts, and assuming for now that $k >0$,
$$ \begin{align}\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx &= 2 \left(\frac{ \text{Ei}^{2} (-x)\sin (kx)}{k} \Bigg|^{\infty}_{0} - \frac{2}{k} \int_{0}^{\infty}\frac{ \sin(kx) e^{-x} \text{Ei}(-x) }{x} \, dx \right) \\ &= - \frac{4}{k} \int_{0}^{\infty} \frac{ \sin(kx) e^{-x} \text{Ei}(-x) }{x} \, dx \\ &= \frac{4}{k} \int_{0}^{\infty}\frac{ \sin(kx) e^{-x} }{x} \int_{1}^{\infty} \frac{e^{-tx}}{t} \, dt \, dx \\ &= \frac{4}{k} \int_{1}^{\infty} \frac{1}{t} \int_{0}^{\infty} \frac{\sin(kx) e^{-(1+t)x} }{x} \, dx \, dt \\ &= \frac{4}{k} \int_{1}^{\infty} \frac{1}{t} \, \arctan \left(\frac{k}{1+t} \right) \, dt \tag{1} \\ &= \frac{4}{k} \left( \log(t) \arctan \left(\frac{k}{1+t} \right) \Bigg|^{\infty}_{1} + k \int_{1}^{\infty} \frac{\log(t)}{t^{2}+2t+1+k^{2}} \, dt \right) \\ &= 4 \int_{1}^{\infty} \frac{\log(t)}{t^{2}+2t+1+k^{2}} \, dt \\ &= - 4 \int_{0}^{1} \frac{\log(u)}{1+2u+(1+k^2)u^{2}} \, du \tag{2} \\ &= - \frac{4}{1+k^{2}} \int_{0}^{1} \frac{\log(u)}{u^{2}+ \frac{2}{1+k^{2}}u+ \frac{1}{1+k^{2}}} \, du \\& = \frac{2i}{k} \left( \int_{0}^{1} \frac{\log(u)}{u+ \frac{1}{1+ik}} \, du - \int_{0}^{1} \frac{\log(u)}{u+ \frac{1}{1-ik}} \, du\right) . \end{align}$$
In general, $$ \begin{align} \int \frac{\log(x)}{x+a} \, dx &= \frac{1}{a} \int \frac{\log(x)}{1+ \frac{x}{a}} \, dx \\ &= \log(x) \log \left(1+ \frac{x}{a} \right) - \int \frac{\log(1+ \frac{x}{a})}{x} \, dx \\ &= \log(x) \log \left(1+ \frac{x}{a} \right) + \text{Li}_{2} \left(- \frac{x}{a} \right) +C. \end{align}$$
Using this result we get
$$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = \frac{2i}{k} \Big( \text{Li}_{2} (-1-ik) - \text{Li}_{2} (-1+ik) \Big) . $$
This appears to be correct for various values of $k$. And the presence of the imaginary unit does not mean that the result is not real valued.
To get the result for the Fourier transform just replace $k$ with $|k|$ and multiply everything by $\frac{1}{\sqrt{2 \pi}}$.
For $k=0$ the right side of the equation should be interpreted as a limit.
You will indeed find that $$\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) \, dx= \frac{2 \sqrt{2} \log(2)}{\sqrt{\pi}}.$$
$ $
$(1)$ $\ \int_{0}^{\infty} \frac{\sin bx}{x} e^{-ax} \, dx= \arctan \left(\frac{b}{a} \right) , \, a>0$
$(2)$ Make the substitution $u = \frac{1}{t}$.
EDIT:
Using the fact that the dilogarithm is real-valued on the real axis for $x \le 1$, the answer can be simplified using the Schwarz reflection principle.
$$ \begin{align} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = \frac{2 \sqrt{2}}{ \sqrt{\pi} |k|} \, \text{Im} \, \text{Li}_{2} (-1+i|k|) \end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove inequality formula by induction my question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.
Page 35. Exercise 1. Prove the following formula by induction: $$1^3+2^3+3^3+\cdots+(n-1)^3<\frac{n^4}{4}<1^3+2^3+3^3+\cdots+n^3$$
The attempt at a solution: First I do the left side of the inequality, $$1^3+2^3+3^3+\cdots+(n-1)^3<\frac{n^4}{4}$$
For $n=2$, we have $$(2-1)^3<\frac{2^4}{4},$$ $$1<4$$ which is indeed true, now we do induction step, and add $n+1$ to both sides, we get: $$1^3+2^3+3^3+\cdots+(n-1)^3+n^3<\frac{n^4}{4}+n^3,$$ now I know from earlier exercise that $$1^3+2^3+3^3+\cdots+(n-1)^3+n^3=(1+2+3+\cdots+n)^2,$$ and $$(1+2+3+\cdots+n)^2=(\frac{n(n+1)}{2})^2=(\frac{n^2+n}{2})^2=\frac{n^4+2n^3+n^2}{4}$$so$$\frac{n^4+2n^3+n^2}{4}<n^4+n^3$$this is where I am stuck, I have continued on different paths and they all seem to give nothing, please help.
|
Assuming $\sum_{i=1}^{n-1} i^3 < \frac{n^4}{4}$, we see that
\begin{align*}
\sum_{i=i}^n i^3 &= \sum_{i=1}^{n-1} i^3 + n^3 \\
&< \frac{n^4}{4} + n^3 = \frac{n^4+4n^3}{4} \\
&< \frac{n^4 + 4n^3 + 6n^2 + 4n + 1}{4} \\
&= \frac{(n+1)^4}{4}.
\end{align*}
|
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|
Evaluating $\;\int x\cos(5x-1)\,\mathrm{d}x$ Integration by parts:
1) let $u = x$, $du = dx$, $v = \frac{1}{5} \sin (5x- 1)$, $dv = \cos (5x -1) dx$
$udv = x \frac{1}{5} \sin x ( 5x- 1) - \frac{1}{5} \int \sin (5x-1) dx$.
2) let $u = \sin (5x -1), du = \frac{1}{5} \cos(5x-1) dx , v = x, dv = dx$
Is my $du$ correct in part 2?
|
The expression should be $$I = \int u\,dv = uv -\int v\,du $$
which is what I suspect you meant, since you correctly obtained
$$I = \frac 15 x \sin(5x - 1) - \frac 15\int \sin(5x - 1)\,dx $$
From there, using substitution (see note below), we get
$$ I = \frac 15 x \sin(5x - 1) - \frac 1{25}( - \cos(5x - 1))+ C$$
$$= \frac 15 x \sin(5x - 1) + \frac 1{25}\cos(5x - 1)+ C$$
Note: Integrating $\frac 15 \int \sin(5x-1)\,dx$ doesn't require using integration by parts. Rather, you can use substitution putting $u = 5x-1\implies du = 5\,dx \iff \frac 15\,du = dx$ gives you
$$\frac 15 \int \sin u \left(\frac 15 \,du \right) = \frac 1{25} \int \sin u \,du$$ $$ = \frac 1{25}(-\cos u) + C = -\frac 1{25}\cos(5x-1) + C$$
|
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|
Integration Question: Completing the Square/Trig Sub yields a different answer than integral table. After the completing the square,
$$\int \frac{dx}{x^2 + 2x - 3}$$
becomes,
$$ \int \frac{dx}{(x+1)^2 - 4}$$
The integral table in my book says the antiderivative is,
$$\frac{1}{2a} ln \, \Biggl\lvert \frac{x-a}{x+a} \Biggr\rvert + \, C$$
or,
$$ \frac{1}{4} ln \, \Biggl\lvert \frac{x-1}{x+3} \Biggr\rvert + \, C$$
Using trig sub, and $\displaystyle u=a\sec\theta$ I obtained,
$$\frac{1}{2}\int csc \, \theta \, d\theta$$
which corresponds to
$$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
I am having a lot of trouble putting this result in the form presented in the table. Any help would be much appreciated.
|
$$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
You can take the square of the term in the absolut value signs. At the same time you take the square root. Remember the rule $log|x^a|=a\cdot log|x|$
$$-\frac{1}{2} ln \, \Biggl\lvert \left( \frac{(x+3)^2}{x^2+2x-3 } \right) ^{1/2}\Biggr\rvert + \, C=-\frac{1}{4} ln \, \Biggl\lvert \frac{(x+3)^2}{x^2+2x-3 } \Biggr\rvert + \, C $$
Linear factors of the denominator
$$=-\frac{1}{4} ln \, \Biggl\lvert \frac{(x+3)^2}{(x+3)\cdot(x-1) } \Biggr\rvert + \, C$$
Cancelling $(x+3)$. And taking away the negative sign. To neutralize this change you have to exchange the numerator and denominator.
$$=\frac{1}{4} ln \, \Biggl\lvert \frac{x-1}{x+3 } \Biggr\rvert + \, C$$
|
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|
Write it as an element of this ring? Since the degree of the irreducible polynomial $x^3+2x+2$ over $\mathbb{Q}[x]$ is odd, it has a real solution , let $a$. I am asked to express $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$.
I have done the following:
Since $a$ is a real solution of $x^3+2x+2$, we have that $a^3+2a+2=0$.
$\frac{1}{1-a}=y \Rightarrow a=\frac{y-1}{y}$
$$a^3+2a+2=0 \\ \Rightarrow \frac{(y-1)^3}{y^3}+2\frac{y-1}{y}+2=0 \\ \Rightarrow (y-1)^3+2(y-1)y^2+2y^3=0 \\ \Rightarrow 5y^3-5y^2+3y-1=0 \\ \Rightarrow 5\left ( \frac{a-1}{a} \right )^3-5\left ( \frac{a-1}{a} \right )^2+3\left ( \frac{a-1}{a} \right )-1=0 $$
That what I have found is not a polynomial of $a$, with coefficients at $\mathbb{Q}$, right??
Is there an other way to write $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$ ??
|
$$
a^3+2a+2=0\\
a^3+2a-3=-5\\
(a-1)(a^2+a+3)=-5\\
\frac{1}{5}(a^2+a+3)=\frac{1}{1-a}
$$
|
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|
Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$ Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.
My solution:
multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$
Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$
dividing by $\sqrt{x^2}$ gives:
$\displaystyle \frac{1}{\sqrt{1}-1}$
which equals $1/0$
However, I double checked my answers, and this does not seem to be correct, am I making a mistake (perhaps when I take the $\sqrt{1}$ in the denominator of the last step?
|
$\sqrt{x^2}$ is a positive number and since we are taking the limit as $x\to -\infty$, we have $\sqrt{x^2} = -x$. Your limit will be negative since $x^2>x^2+x-1$ as $x\to-\infty$.
|
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|
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
|
The mean value theorem gives $f(x+h) = f(x) + f'(x)h + {1 \over 2} f''(\xi) h^2$, where $\xi \in [x,x+h]$ (adjusted appropriately for sign of $h$).
This gives $|\sqrt{1+\theta} - (1+ {1 \over 2} \theta) | \le {1 \over 8} \theta^2$.
We have (if $x>0$),$\sqrt{x^2+2x} - \sqrt{x^2-7x} = x(\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over 7}})$.
Then $| \sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} - {1 \over 2}{9 \over x}| \le {1 \over 8} (({2 \over x})^2+({7 \over x})^2)$.
Multiplying through by $x$ and letting $x \to \infty$ yields the desired result.
Note:
The point of this answer is that you can guess the answer by
using $\sqrt{1+\theta} \approx 1+ {1 \over 2} \theta$ and
writing
$\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} \approx {1 \over 2}{9 \over x}$.
|
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|
Make x the subject given the formula for y I am given the following formula:
$$y=\frac{x}{a}+\sqrt{\frac{x}{b}}.$$
I want to make x the subject.
I rearranged the equation and got to:
$$y^{4}=x(\frac{y^{2}}{2}+2y^{2})-x^{2}.$$
and I don't know where to go from here. May be this is the wrong rearrangement.
The answer according to wolfram alpha is:
$$x=\pm \frac{a^{\frac{3}{2}}\sqrt{a+4by}+a^{2}+2aby}{2b}.$$
How can I get there?
|
Hint: Set $u=y^2$ and solve the quadratic equation for $u$.
Then your equation is rewritten as
$$
u^2=\frac{5x}{2}u-x^2,
$$
which by completing the square is written as
$$
\Bigg(u-\frac{5}{4}x\Bigg)^2 - \frac{9}{16}x^2=0,
$$
and thus you have
$$
u-\frac{5}{4}x = \pm\frac{3}{4}x,
$$
or
$$
u = \pm\frac{3}{4}x+\frac{5}{4}x = \frac{5\pm3}{4}x.
$$
Finally $y=\sqrt{u}=\sqrt{\frac{5\pm3}{4}x}$.
|
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|
Proving Identities. I tried to solve it but I cant get the answer. How to prove this by using a hand?
*
*$$ \sec^2x + \csc^2x = \sec^2x \csc^2x $$
*$$ \frac{\sec\theta + 1}{\sec\theta - 1} = \frac{1 + \cos\theta}{1 - \cos\theta}$$
*$$ \frac{1 - \cot^2\theta}{1 + \cot^2\theta} = \sin^2\theta - \cos^2\theta $$
Anyone can help? Thanks.
|
I think that #1 is an unfair question, because there is nothing to prove. For if we can assume that
$$\cos^2\theta+ \sin^2\theta=1$$
$$\tan^2\theta+ 1=\sec^2\theta$$
$$\text{and }\cot^2\theta+ 1=\csc^2\theta$$
Then it should be natural to assume that
$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$
Consider a right triangle with an acute angle $\theta$. Let the hypotenuse be of length $c$, the side adjacent to $\theta$ be of length $a$, and the side opposite angle $\theta$ be of length b.
By the Pythagorean theorem we have
$$\begin{array}{ll}a^2+b^2=c^2&(1)\end{array}$$
Dividing (1) by $c^2$ we have
$$\cos^2\theta+ \sin^2\theta=1$$
Dividing (1) by $a^2$ we have
$$1+\tan^2\theta = \sec^2\theta$$
Dividing (1) by $b^2$ we have
$$\cot^2\theta + 1= \csc^2\theta$$
Multiplying (1) by $\frac{c^2}{a^2b^2}$ we have
$$\frac{c^2}{b^2}+\frac{c^2}{a^2}=\frac{c^2}{a^2}\cdot\frac{c^2}{b^2}$$
$$\csc^2\theta+\sec^2\theta=\sec^2\theta\csc^2\theta$$
|
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|
study of two sequences I need to study wether thos two sequences converge or not.
1) $u_n=n\sum_{k=1}^{2n+1} \frac{1}{n^2+k}$
2) $v_n=\frac{1}{n}\sum_{k=0}^{n-1} \cos(\frac{1}{\sqrt{n+k}})$
For the first i get it converges to $0$ by just expressing the sum as a fraction (not very smart imo).
|
*
*Note that $(n^{2} +k) \leq (n^{2}+2n+1)$ whenever $k \in\left\{1,2,\cdots,(2n+1)\right\}$. So we have that $\frac{1}{n^{2}+k} \geq \frac{1}{n^{2}+2n+1} =\frac{1}{(n+1)^{2}}$. Similarly $n^{2}+1 \leq n^{2}+k$ and so we have that $\frac{1}{n^{2}+1}\geq \frac{1}{n^2+k}$. So we get that $$\sum_{k=1}^{2n+1} \frac{1}{n^{2}+k} \geq \sum_{k=1}^{2n+1} \frac{1}{n^{2}+2n+1} \ \ \mbox{and} \ \ \sum_{k=1}^{2n+1} \frac{1}{n^{2}+k} \leq \sum_{k=1}^{2n+1}\frac{1}{n^{2}+1}$$ Using this we have $$ \frac{2n^{2}}{n^{2}+2n+1} \leq u_{n} \leq \frac{2n^{2}}{n^{2}+1}$$ Now apply the Squeeze Theorem.
|
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Equation of line passing through point. The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$.
Find the equation of line passing through line $\overline{BC}$ if it also passes through point $(1,2)$.
I have found out slope of both the lines through $A$: $\frac{-3}{4}$ and $\frac{4}{3}$.
I can't figure out how to solve it.
|
inspection shows the lines are perpendicular and meet at $(-1,2)$.
define variables $(w,z)$ by:
$$
\begin{pmatrix} w \\ z \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}
\left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right)
$$
the equations of the lines are now $w=0$ and $z=0$, so the lines perpendicular to their bisector have equation $w+z=k$ for some $k$, i.e.
$$
\begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} w \\ z \end{pmatrix} = k
$$
in the original coordinates the line is
$$
\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}
\left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right) = k
$$
or
$$
7x+y =k'
$$
and since the line passes through $(1,2)$ we must have $k'=9$
|
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|
How to simplify $\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$ I in trouble simplifying this:
$$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$$
couldn't find a solution. Can you help?
|
Those expressions usually come from Cardan's formula
$$
\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}-\frac{q}{2}}-
\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}+\frac{q}{2}}
$$
(see Wikipedia).
So we need $45=q/2$ or $q=90$ and so
$$
\frac{p^3}{27}+\frac{90^2}{4}=2\cdot29^2,
$$
that is,
$$
\frac{p^3}{27}=-343=-7^3
$$
which gives $p=-21$. Thus the number is a root of the equation
$$
x^3-21x+90=0
$$
and some attempts with the rational root test give $x=-6$ as the only real root.
|
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|
What is the values of $a$ and $b$ without using the L'Hôpital's Rule Suppose that $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2}\to 1$$ as $x \to 1$. ThenWhat is the values of $a$ and $b$ without using the L'Hôpital's Rule?
Thanks for your help!
|
Factor an $x-1$ out of the denominator to get $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \frac{(2x)^x - 2}{x-1} \cdot \frac{1}{a + b(x-1)}$$ so that (as long as $a \not= 0$)$$\lim_{x \to 1} \frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \left.\frac{d}{dx} (2x)^x \right|_{x=1} \cdot \frac 1a.$$
But $\dfrac{d}{dx} (2x)^x = (2x)^x (1 + \log 2x)$, and at $x=1$ equals $2(1 + \log 2)$. Thus $$\lim_{x \to 1} \frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \frac{2(1 + \log 2)}{a}.$$
|
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|
Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$ I was given this proof for hw.
Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
So, far I've gotten this
Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)
Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
Inductive step:
LHS
$2^{k + 1 + 1} - 1
= 2^{k + 2}
= 2 \cdot 2^{k + 1}$
$2[ 2^{k + 1} - 1 + 1] - 1 \geqslant 2\cdot(2k^2 + 2k + 1 + 1) - 1$
RHS (Attempt to prove that the RHS is less than the LHS )
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant 2(k + 1)^2 + 2(k + 1) + 1$
Next step
$2(k + 1)^2 + 2(k + 1) + 1
= 2k^2 + 4k + 2 + 2k + 1 + 1
= (2k^2 + 2k + 1) + 4k + 3$
Now I'm stuck at how to prove that
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant (2k^2 + 2k + 1) + 4k + 3$
So, what can I do?
Thanks
|
Inductive step
$$2^{k+2}-1=2\times 2^{k+1}-1>2(2k^2+2k+2)-1>2(k+1)^2+2(k+1)+1$$
|
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|
I can't figure out this simplification in a differential equation I was watching PatrickJMT's video on first-order differential equations and while I think I should see what he's doing on the left side here from line one to line two, I just can't. I ran it past my roommate as well without any luck :(
Elaboration: I understand that he's multiplying everything by the integrating factor ($1+x^2$), but how does he compresses it to simply ($\frac{d}{dx}((1+x^2)y$)?
|
The second line is just the power rule $(uv)' = u'v + uv'$, but performed in reverse. If we expand it out starting from the second line you'll be able to see what he did:
$$
\begin{align}
\frac{d}{dx}\left[(1+x^2)y\right] & = y\frac{d}{dx}(1+x^2) + (1+x^2)\frac{dy}{dx} \\
& = 2xy + (1+x^2)\frac{dy}{dx} \\
& = \left(\frac{dy}{dx} + \frac{2x}{1+x^2}y\right)(1+x^2)
\end{align}
$$
or if you prefer moving from the first line to the second,
$$
\left(\frac{dy}{dx} + \frac{2x}{1+x^2}y\right)(1+x^2) = \frac{4}{(1+x^2)^2}(1+x^2) \\
\frac{dy}{dx}(1+x^2) + 2xy = \frac{4}{1+x^2} \\
y'(1+x^2) + y(1+x^2)' = \frac{4}{1+x^2} \\
\left[y(1+x^2)\right]' = \frac{4}{1+x^2} \\
\frac{d}{dx}\left[(1+x^2)y\right] = \frac{4}{1+x^2}
$$
|
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|
Proof modular equality by induction I'm trying to prove using induction that $5^{2^{x-2}} \equiv 1 + 2^x \pmod{2^{x+1}}$
So far, I have:
*
*Base case: $x = 2, 5 \equiv 5 \pmod{8}$, It is true. $x = 3, 25 \equiv 9 \pmod{16}$, It is true.
*Inductive step: let $x = n$ Assume $5^{2^{n-2}} \equiv 1 + 2^n \pmod{2^{n+1}}$ is true, so I need proof, that $5^{2^{n-1}} \equiv 1 + 2^{n+1} \pmod{2^{n+2}}$.
I tried express new equation through the old. But I am getting errors.
|
$$ \begin{align*}\begin{split}
5^{2^{n-1}} &= (5^{2^{n-2}})^2 \\
&= (1 + 2^{n}+2^{n+1}a)^2 \\
&= 1+ (2^n)^2+(2^{n+1}a)^2+
2\cdot2^n+2\cdot2^{n+2}a+2\cdot2^{n}\cdot2^{n+1}a\\
&=1+2^{n+1}+2^{n+2}b
\end{split}\end{align*}$$
|
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|
Composing Piecewise Functions I was wondering how to compose piecewise functions.
On a practice exam I was reading, a question asks what F(F(x)) will look like if F(x)= 2x if x<1/2 and = 2-2x if x>=1/2.
Would I just substitute the original parts into themselves? (like 2(2x)=4x and 2-2(2-2x)=4x-2?) On the solution to the question, there are 4 intervals instead of 2 for F(F(x)). How would I go about getting those intervals, since the method I listed only gives 2 intervals?
Thanks.
|
Our function F is given as:
$F(x)=\left\{\begin{matrix} 2x & x<1/2 \\ 2-2x & x \ge 1/2 \end{matrix}\right.$
So for inputs on both sides of the breakpoint $x=1/2$, it gives outputs in the range $\left(-\infty,1\right]$. Which means that when you compose it with itself, you have to consider four possible events:
*
*$x<1/2$ and $F(x)<1/2$
*$x<1/2$ and $F(x) \ge 1/2$
*$x \ge 1/2$ and $F(x)<1/2$
*$x \ge 1/2$ and $F(x) \ge 1/2$
Because in cases 1 and 3, the second time you plug it into the function you need to remember to double the value, while in cases 2 and 4 you have to double and then subtract the result from 2. The important thing to consider is whether all four cases exist - and from the preceding paragraph, we know that this is true.
So, let's see what happens in each case (and how we get to them):
*
*If $x<1/2$, then $F(x)=2x$ and so $F(x)<1/2\implies 2x<1/2 \implies x<1/4$. The intersection of those two is just $x<1/4$, so in this range $F(F(x))=2\times F(x)=2\times(2x)=4x$.
*Again, $x<1/2\implies F(x)=2x$, so $F(x)\ge 1/2 \implies x \ge 1/4$, so in the end we have $1/4 \le x < 1/2$. In this range, $F(F(x))=2-2\times F(x)=2-2\times 2x=2-4x$.
*This time, $x \ge 1/2 \implies F(x) = 2-2x$, so $F(x) < 1/2 \implies 2-2x < 1/2 \implies x > 3/4$. And in this range, $F(F(x))=2\times F(x) = 2(2-2x) = 4-4x$.
I'll leave case #4 for you as an exercise.
EDIT: To show it with some actual numbers.
$x = -1$: Clearly $x < 1/2$, so $F(x) = F(-1) = 2 \times -1 = -2$. So when we calculate $F(F(x))$, we want to take that value we just calculated and put it back in the function. So where does $F(x)$ lie? Clearly it's still less than 1/2, so $F(F(x)) = F(-2) = 2 \times -2 = -4$.
$x = 3/8$: We are still in $x < 1/2$ territory here, so $F(x) = F(3/8) = 2 \times 3/8 = 3/4$. Now when we feed that back into F, what range are we in? $3/4 \ge 1/2$, so this time $F(F(x)) = F(3/4) = 2 - 2 \times 3/4 = 1/2$.
$x = 1$: Here $x \ge 1/2$, so $F(x) = F(1) = 2 - 2 \times 1 = 0$. But since $F(x) = 0 < 1/2$, $F(F(x)) = F(0) = 2 \times 0 = 0$.
$x = 5/8$: Again $x \ge 1/2$, so $F(x) = F(5/8) = 2 - 2 \times 5/8 = 3/4$. So in this case $F(x) > 1/2$, so we have to again feed it in as $F(F(x)) = F(3/4) = 2 - 2 \times 3/4 = 1/2$.
To look at it another way, think of it as $F(F(x)) = F(y)$ where $y = F(x)$. Then ask yourself - how does $F(y)$ behave depending on the value of $y$? How does the value of $x$ affect the value of $y$? How do those two facts interact?
|
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|
Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6})= -\pi \sin(\pi \cdot \frac{1}{6})= \frac{-\pi}{2}$
And I put the $x$-value into $f(x)$ to find the $y$-value
$f(\displaystyle\frac{1}{6})=\cos(\pi \cdot \frac{1}{6})=\frac{\sqrt{3}}{2}$
Now I use the formula $y=m(x-x_0)+y_0$
$y= \displaystyle\frac{-\pi}{2}(x-\frac{1}{6})+\frac{\sqrt{3}}{2}$ = $\displaystyle\frac{-\pi \cdot x}{2} + \frac{\sqrt{3}}{2} + \frac{\pi}{12}$
Am I correct? Because my book got an different answer like $6 \pi x +12y =6\sqrt{3}+ \pi$
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You are correct, and so is your book. The two equations are equivalent.
|
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|
A inverse Trigonometric multiple Integrals How to calculate the closed form of the integral
$$\int\limits_0^1 {\frac{{\int\limits_0^x {{{\left( {\arctan t} \right)}^2}dt} }}{{x\left( {1 + {x^2}} \right)}}} dx$$
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Use integration by parts, taking $\displaystyle \int_{0}^{x} (\arctan t)^2 dt$ as the first function and $ \dfrac{1}{x(x^2+1)} $ as the second function to get:
$ \displaystyle \int \dfrac{dx}{x(x^2+1)} = \dfrac{1}{2} \ln\bigg(\dfrac{x^2}{x^2+1}\bigg)$
Hence,
$I = \bigg|\dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) \displaystyle \int_{0}^{x} (\arctan t)^2 dt \bigg|_{0}^{1} - \displaystyle \int_{0}^{1}
\dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) (\arctan x)^2 dx $
= $ \displaystyle \dfrac{1}{2}\int_{0}^{1} (\arctan x)^2 \ln\bigg(\dfrac{1+x^2}{2x^2}\bigg) dx $
I am not able to solve it further
|
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|
For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$
$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$
$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$
But I think this is wrong, please suggest how to proceed thanks..
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First we should note that $$1+2+3+...+n=\dfrac{n(n+1)}{2}$$ and $$1+3+5+...+(2n-1)=n^2.$$ Therefore general term in your series become to $$T_n=\dfrac{(1+2+3+..+n)^2}{1+3+5+...+(2n-1)}=\dfrac{(n+1)^2}{4}$$ $$S_n=\sum_{k=1}^nT_k\\=\sum_{k=1}^n\dfrac{(k+1)^2}{4}\\=\sum_{k=1}^{n+1}\dfrac{k^2}{4}-\dfrac{1}{4}\\=\dfrac{(n+1)(n+2)(2n+3)}{24}-\dfrac{1}{4}$$
|
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|
Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is
$$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$
Is this true? If so, how to prove it?
|
To get rid of both square root, I use the substitution$ \tan x = \sin^2\theta$. Then we rewrite the integral I as \begin{array}{l}
\displaystyle I=\int_{0}^{\frac{\pi}{2}} \sin \theta \sqrt{1-\sin ^{2} \theta} \cdot\frac{2 \sin \theta \cos \theta d \theta}{1+\sin ^{4} \theta} \\
=2 \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sin ^{4} \theta} d \theta.
\end{array}
Applying the identity $\cos^2x=1-\sin^2x$ yields
$$I= 2 \underbrace{\int_{0}^{\frac{\pi}{2}} \frac{1+\sin ^{2} \theta}{1+\sin ^{4} \theta} d \theta}_{J}-\pi $$
$$
J=2 \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta\left(\sec ^{2} \theta+\tan ^{2} \theta\right)}{\sec ^{4} \theta+\tan ^{4} \theta} d \theta$$
Letting $\tan \theta \mapsto t$, we have
$$J=2 \int_{0}^{\infty} \frac{1+2 t^{2}}{2 t^{4}+2 t^{2}+1} d t \\
=\int_{0}^{\infty} \frac{1+2 t^{2}}{t^{4}+t^{2}+\frac{1}{2}} d t\\
=\int_{0}^{\infty} \frac{2+\frac{1}{t^{2}}}{t^{2}+\frac{1}{2 t^{2}}+1} d t$$
Using the identity $\displaystyle 2+\frac{1}{t^{2}}=\frac{2+\sqrt{2}}{2}\left(1+\frac{1}{\sqrt{2} t^{2}}\right)+\frac{2-\sqrt{2}}{2}\left(1-\frac{1}{\sqrt{2} t^{2}}\right)$ yields
$$\displaystyle J=\frac{2+\sqrt{2}}{2} \int_{0}^{\infty} \frac{d\left(t-\frac{1}{\sqrt{2} t}\right)}{\left(t-\frac{1}{\sqrt{2}t}\right)^{2}+(1+\sqrt{2})}+\frac{2-\sqrt{2}}{2} \int_{0}^{\infty} \frac{d\left(t+\frac{1}{\sqrt{2} t}\right)}{\left(t+\frac{1}{\sqrt{2}t}\right) ^{2} -(\sqrt{2}-1)}$$
$$
=\frac{2 +\sqrt 2}{2 \sqrt{1+\sqrt{2}}} \left[\tan ^{-1}\left(\frac{t-\frac{1}{\sqrt{2} t}}{\sqrt{1+\sqrt{2}}}\right)\right]_{0}^{\infty}+\frac{2-\sqrt{2}}{4 \sqrt{\sqrt{2}-1}} \ln \left|\frac{t^{2}-t \sqrt{\sqrt{2}-1}+1}{t^{2}+t \sqrt{\sqrt{2}-1}+1}\right|_{0}^{\infty} \\ $$
$ \displaystyle \qquad\qquad =\frac{2+\sqrt{2}}{2 \sqrt{1+\sqrt{2}}} \pi \\
\displaystyle \qquad\qquad =\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} \pi
$
Finally, we can conclude that $$I=
\left(\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}}-1\right) \pi
$$
|
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|
Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$
2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$
I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???
|
Let:
$z_{1} = a + ib \\ z_2 =p+iq \\ z_{3} = x + iy$
Taking the square of modulus:
$a^ 2 +b^ 2 =p^ 2 +q^ 2 =x^ 2 +y^ 2 =1$
Equating the real and the imaginary part of $z_1 + z_2 + z_3$ to 0,
$a + p + x = b + q + y = 0$
$
|z_{1} - z_{2}|^ 2 \\
=(a - p) ^ 2 + (b - q) ^ 2 \\
=2(a^ 2 + p^ 2 )-(a+p)^ 2 + 2(b ^ 2 + q ^ 2) - (b + q) ^ 2 \\
=2(a^ 2 + b^ 2 + p ^ 2 + q ^ 2 ) - x ^ 2 - y ^ 2\\
=4 - 1
=3
$
Similarly, you can show all sides are equal.
|
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|
Row space of matrix
Hello! I am working on some differential equations homework and it is saying that my answer for this question is wrong, and i am not sure as to why. First I reduced the matrix A and then I read the first row of the reduced matrix into the given u=[], and the second row into v=[]. When that answer wasn't correct, I used my calculator to reduce A and got the same answer as when I did it by hand. Maybe I am not approaching this question correct, but if someone can help me solve it i would greatly appreciate it!
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$\begin{equation} \begin{aligned} \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 2 & 1 & -3 & -1 & 5 \\ 4 & 5 & 13 & -5 & -9} = & \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 2 & 1 & -3 & -1 & 5 \\ 0 & 3 & 19 & -3 & -19} \\ = & \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 0 & -1 & -\frac{19}{3} & 1 & \frac{19}{3} \\ 0 & 3 & 19 & -3 & -19} \\ = & \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 0 & -1 & -\frac{19}{3} & 1 & \frac{19}{3} \\ 0 & 0 & 0 & 0 & 0} \\ = & \pmatrix{ 3 & 0 & -14 & 0 & 17 \\ 0 & -1 & -\frac{19}{3} & 1 & \frac{19}{3} \\ 0 & 0 & 0 & 0 & 0} \\ = & \pmatrix{ 1 & 0 & -\frac{14}{3} & 0 & \frac{17}{3} \\ 0 & 1 & \frac{19}{3} & -1 & -\frac{19}{3} \\ 0 & 0 & 0 & 0 & 0} \end{aligned} \end{equation}$
Is this what you got when you row reduced? I feel like your work is correct.
|
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Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$
|
If you can avoid induction, you can use Gauss's trick:
$S=1+4+7+\cdots+(3n-8)+(3n-5)+(3n-2)$
$S=(3n-2)+(3n-5)+(3n-8)+\cdots+7+4+1$
$2S=(3n-1)+(3n-1)+\cdots+(3n-1)$
$2S=n(3n-1)$
$S = \frac{n}{2}(3n-1)$
|
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|
Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \frac{Gx+H}{x^2 + 4x + 16}$$
But I am told that the correct answer is
$$\frac{A}{x-4} + \frac{B}{(x-4)^2} + \frac{C}{(x-4)^3} + \frac{Dx+E}{x^2 + 4x + 16}$$
What exactly is the numerator of each fraction based on?
|
You may see the redundancy of the $Bx$ term as follows:
\begin{align}
\frac{Bx + C}{(x - 4)^2}
& = \frac{Bx}{(x - 4)^2} + \frac{C}{(x - 4)^2} \\
& = \frac{B(x - 4)}{(x - 4)^2} + \frac{4B}{(x - 4)^2} + \frac{C}{(x - 4)^2}\\
& = \frac{B}{x - 4} + \frac{4B + C}{(x - 4)^2}.
\end{align}
|
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Why does this method of solution for this system of equations yield an incorrect answer? We are required to solve the following system of equations:
$$x^3 + \frac{1}{3x^4} = 5 \tag1$$
$$x^4 + \frac{1}{3x^3} = 10 \tag2$$
We may multiply $(1)$ by $3x^4$ throughout and $(2)$ by $3x^3$ throughout (as $0$ is not a solution we may cancel the denominators) to yield.
$$3x^7 + 1 = 15x^4 \tag3$$
$$3x^7 + 1 = 30x^3 \tag4$$
Subtracting the two:
$$15x^4 - 30x^3 = 0$$
$$\implies x^3(x-2) = $$
As $0$ is not a solution we choose $x=2$.
But putting $x=2$ in the original equations does not satisfy them. How come?
|
Solutions to (3) and (4) are equivalent to solutions of $15x^4 - 30x^3 = 0$ and one of (3) or (4). But the solution $x = 2$ to $15 x^4 - 30x^3 = 0$ does not satisfy (3) [or equivalently (4)], as $3*2^7 + 1 \neq 15*2^4$.
You shouldn't be surprised that a system of two equations in one unknown may be inconsistent, i.e. not have any solutions.
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How integrate $ \iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0I'm trying to resolve this integral
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0<a<b
$$
I tried with polar coordinates:
$$
x = r\cos{\theta} \\
y = r\sin{\theta} \\
Jacobian = r
$$
But I'm confused about in how to calculate the domain for each integral
|
This is just annulus with inner radius $a$ and outer radius $b$, so $a \le r \le b$ and $0 \le \theta \le 2\pi$. The integral becomes
$$ \int_0^{2\pi} \int_a^b \frac{r^2 \cos^2 \theta}{r^2} r\,dr\,d\theta
\\= \int_0^{2\pi} \int_a^b r \cos^2 \theta \, dr \, d\theta
\\= \int_0^{2\pi} \frac{1}{2}(b^2 - a^2) \cos^2 \theta \, d\theta
\\ = \frac{1}{2}(b^2 - a^2) \int_0^{2\pi} \frac{1 + \cos 2\theta}{2} d\theta
\\ = \frac{\pi}{2}(b^2 - a^2) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
integral calculation is wrong. Why? $$\int \sqrt{1-x^2} dx = \int \sqrt{1-\sin^2t} \cdot dt= \int \sqrt {\cos^2 t} \cdot dt= \int \cos t \cdot dt = \sin t +C = x +C$$
The answer is wrong. Why?
|
Here's a way to do the integral without trigonometric substitution. All you need to know is the antiderivative $$\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin{x}+\color{grey}{constant},$$ as well as how to integrate by parts.
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}+x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,\mathrm{d}x\\
\implies 2\int\sqrt{1-x^2}\,\mathrm{d}x&=\arcsin{x}+x\sqrt{1-x^2}+\color{grey}{constant}\\
\implies \int\sqrt{1-x^2}\,\mathrm{d}x&=\frac{\arcsin{x}+x\sqrt{1-x^2}}{2}+\color{grey}{constant}.\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Questions about the computation of a limit. By L'Hôpital's rule, it is easy to see that
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 1/3.
$$
But if we use the following procedure to compute the limit, we get a wrong answer. The procedure is in the following.
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x}. \quad (2)
$$
Since $\sin x \sim x$ ($\sin x$ and $x$ are equivalent infinitesimals), we replace $\sin x$ by $x$ in (2). Then we obtain
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x} \\
= \lim_{x \to 0} \frac{\frac{x}{\cos x} - x}{x^3} \\
= \lim_{x \to 0} \frac{\frac{1}{\cos x} - 1}{x^2} \\
= \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \\
= 1/2.
$$
I don't konw where is the problem. Thank you very much.
|
What you did:
$${\sin x} \sim x \implies
\frac{\sin x}{\cos x} - x \sim \frac{x}{\cos x} - x
$$
is wrong, because you added equivalents: you can multiply by $1/\cos x$, to get
$$
{\sin x} \sim x \implies \frac{\sin x}{\cos x} \sim \frac x{\cos x}
$$
but you can't fo the manipulation with the sum:
$$
\frac{\sin x}{\cos x} \sim \frac x{\cos x} \nRightarrow
\frac{\sin x}{\cos x} - x \sim \frac{x}{\cos x} - x
$$
Even when $f\sim g$, you can be in the case for which
$$
g+h \nsim f+h
$$
For example, $f(n) = n^2 - n$, $g(n) = n^2$, $h(n) = 1-n^2$.
This is true in particular when:
*
*$f\sim a y$, $h\sim b y$ for a function $h$ and scalars $a,b$ with $a+b \neq 0$
*$h = o(g)$.
Otherwise, do not consider making the sum (or substitutions, which is the same) of equivalents.
|
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|
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
|
Consider the function for positive $x$:
$$f(x) = \frac{x}{\sqrt{1+x}}-\frac1{\sqrt 2}-\frac3{4\sqrt 2}\log x$$
Note that $f(x) \ge 0 \implies f(a)+f(b)+f(c) \ge 0 \implies $ the given inequality. Now
$$f'(x) = \frac{4x^2-3\sqrt2 (x+1)^{3/2}+8x}{8x(x+1)^{3/2}}$$
We need to check the sign of the numerator, $4(x+1)^2-3\sqrt2(x+1)^{3/2}-4$. Using $y = \sqrt{x+1}$, we get the numerator as
$$4y^4-3\sqrt2y^3-4 = (y-\sqrt2)(4y^3+\sqrt2y^2+2y+2\sqrt2)$$
As the second factor is positive, the numerator's sign is given by $y-\sqrt2$ which has the same sign as $x-1$, so $f'(x)< 0$ for $x < 1$ and $f'(x)> 0$ for $x> 1$. Hence $f(x)\ge f(1)=0$.
|
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|
What is the value of $\cos\left(\frac{2\pi}{7}\right)$? What is the value of $\cos\left(\frac{2\pi}{7}\right)$ ?
I don't know how to calculate it.
|
I suggest you have a look at http://mathworld.wolfram.com/TrigonometryAnglesPi7.html which clearly explains the problem.
As you will see, $\cos\left(\frac{2\pi}{7}\right)$ is the solution of $$8 x^3+4 x^2-4 x-1=0$$ Using Cardano, you will get $$\cos\left(\frac{2\pi}{7}\right)=\frac{1}{6} \left(-1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i
\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}\right)$$
If you want to approximate it, even very accurately, you could expand $\cos(x)$ as a Taylor series at $x=\frac{\pi}{3}$ which gives $$\cos(x)=\frac{1}{2}-\frac{1}{2} \sqrt{3} \left(x-\frac{\pi }{3}\right)-\frac{1}{4}
\left(x-\frac{\pi }{3}\right)^2+\frac{\left(x-\frac{\pi }{3}\right)^3}{4
\sqrt{3}}+\frac{1}{48} \left(x-\frac{\pi }{3}\right)^4+\cdots$$
Using the first term, you will get $0.6295570974$, and adding terms $0.6239620836$, $0.6234788344$, $0.6234892692$, $0.6234898099$, $0.6234898021$ for an exact value equal to $0.6234898019$.
I cannot resist to provide the approximation $\cos(x)=\frac{\pi^2-4x^2}{\pi^2+x^2}$ which would give an estimate of $\frac{33}{53}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Ellptic\Jacobi theta function and its residue integral The Ellptic\Jacobi theta function is given by
\begin{align}
\theta_1(\tau|z)&=\theta_1(q,y)=-iq^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^{k-1}) \\
&= -i\sum_{n\in \mathbb{Z}}(-1)^n e^{2\pi i z(n+\frac{1}{2})} e^{\pi i \tau(n+\frac{1}{2})^2}
\end{align}
Where $q=e^{2\pi i \tau}$, and $y=e^{2\pi i z}$
and dedkind eta function is given as
$\eta(\tau)=\eta(q)=q^{\frac{1}{24}} \prod_{n=1}^{\infty}(1-q^n) $
Now i want to evaluate following relation
$\theta'_1(\tau|0)=\theta'_1(q,1)=2\pi \eta(q)^3$
The results is well known (i guess...) but i can not find any papers or textbooks contain some procedure of this.
|
From Antonio DJC's comment i explicit compute
\begin{align}
&\theta_1(q, y)= -i y^{\frac{1}{2}}q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n-1} \right) \\
&\phantom{\Theta(q,y)}= -i y^{\frac{1}{2}}q^{\frac{1}{8}}(1-y^{-1}) \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right), \\
&\frac{\theta_1(q,y)}{1-y^{-1}}= -i y^{\frac{1}{2}} q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right) \\
& \lim_{z \rightarrow 0} \frac{\theta_1(\tau|z)}{1-y^{-1}} =\lim_{z \rightarrow 0} \frac{\theta_1'(\tau|z)}{y^{-2} 2\pi i} = \frac{\theta_1'(q,1)}{2\pi i}
= -i q^{\frac{1}{8}} \prod_{n=1}^{\infty} (1-q^n)^3 = -i \eta(q)^3
\end{align}
where we used $\frac{df(y)}{dz} = \frac{df}{dz} \frac{dy}{dz} = \frac{df}{dz} 2\pi i$. And Dedkind eta function is given as
$\begin{align}
\eta(q)=q^{\frac{1}{24}} \prod_{n=1}^\infty (1-q^n)
\end{align}$
|
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|
Examples where it is easier to prove more than less Especially (but not only) in the case of induction proofs, it happens that a stronger claim $B$ is easier to prove than the intended claim $A$ (e.g. since the induction hypothesis gives you more information). I am trying to come up with exercises for beginner students that help to demonstrate this point (and also interested in the general phenomenon). Do you know any good examples (preferably elementary ones) where strengthening a claim makes the proof easier?
Here is an example of what I mean (Problem 16 from chapter 7 of Engel's `Problem solving strategies'):
Show that $\frac{1}{2}\frac{3}{4}...\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n}}$ for $n\geq 1$. This is much harder than proving the stronger statement that $\frac{1}{2}\frac{3}{4}...\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$ for $n\geq 1$, which is a straightforward induction.
|
Here are two examples which might be useful
*
*Example 1: Generalize in order to use techniques from analysis resp. formal operator techniques
Problem: Evaluate
$$\sum_{k=1}^{n}\frac{k^2}{2^k}$$
This can be done by instead evaluating
$$\sum_{k=1}^{n}k^2x^k$$
For $x\neq 1$ we know that
$$S(x)=\sum_{k=1}^{n}x^k=\frac{1-x^{n+1 }}{1-x}$$
Now differentiating each side and multiplying with $x$ we get
\begin{align*}
(xD)S(x)&=\sum_{k=1}^{n}kx^{k}\\
&=(xD)\frac{1-x^{n+1 }}{1-x}\\
&=\left(1-(n+1)x^n+nx^{n+1}\right)\frac{x}{\left(1-x\right)^2}
\end{align*}
Differentiating again each side and multiplying with $x$ we get
\begin{align*}
{(xD)}^2S(x)&=\sum_{k=1}^{n}k^2x^{k}\\
&=(xD)^2\frac{1-x^{n+1 }}{1-x}\\
&=\left(1+x-(n+1)^2x^n-(2n^2+2n-1)x^{n+1}-n^2x^{n+2}\right)\frac{x}{\left(1-x\right)^3}
\end{align*}
We conclude
\begin{align*}
S\left(\frac{1}{2}\right)&=\sum_{k=1}^{n}\frac{k^2}{2^k}\\
&=\left(\frac{3}{2}-(n+1)^2\frac{1}{2^n}-(2n^2+2n-1)\frac{1}{2^{n+1}}-n^2\frac{1}{n+2}\right)\cdot 4\\
&=6-\frac{1}{2^n}\left(n^2+4n+6\right)
\end{align*}
Another example where generalization is useful
*
*Example 2: Generalize integrals by introducing a parameter and use the technique of parameter differentiation
Problem: Evaluate
$$\int_{0}^{\infty}\frac{\sin^2 x}{x^2}dx$$
given that $\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{1}{2}\pi$.
The idea is to introduce a parameter and evaluate the more general integral
\begin{align*}
I(a)=\int_{0}^{\infty}\frac{\sin^2 (ax)}{x^2}dx,\qquad a\geq 0\tag{1}
\end{align*}
and use parameter differentiation.
Differentiating each side of (1) with respect to $a$, we get
\begin{align*}
I^{\prime}(a)&=\int_{0}^{\infty}\frac{2\sin (ax)\cos (ax)\cdot x}{x^2}dx\\
&=\int_{0}^{\infty}\frac{2\sin (2ax)}{x}dx\\
\end{align*}
Now substituting $y=2ax$, we get $dy = 2a dx$, and
\begin{align*}
I^{\prime}(a)&=\int_{0}^{\infty}\frac{\sin y}{y}dx=\frac{1}{2}\pi\\
\end{align*}
Integrating each side gives
$$I(a) =\frac{1}{2}\pi a+C,\qquad C \text{ constant}$$
Since $I(0)=0$, we get $C=0$. Thus $I(a)=\frac{1}{2}\pi a, a\geq 0$. Setting $a=1$ yields
$$I(1)= \int_{0}^{\infty}\frac{\sin ^2 x}{x^2}dx = \frac{1}{2}\pi$$
Note: The examples above can be found in Problem-Solving Through Problems. They are stated as problems 1.12.1 and 1.12.3 in chapter 1.12 Generalize.
|
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|
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$1681\cos^2\theta-3280\cos\theta+1600=0$$
Solving the quadratic equation gives $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.
P.S: I've found the other method so I am self-answering the question.
|
This has already been mentioned multiple times, but for completeness:
Let $\sin \theta = x, \cos \theta = y$. Then we have to solve:
$$9x+40y=41 \quad \quad x^2 + y^2 = 1$$
Now from the second equation, $(9x)^2 + 81y^2 = 81$, or $(41 - 40y)^2 + 81y^2 = 81$. This gives:
$$(40^2+9^2)y^2 - 2 \cdot 41 \cdot 40y + (41^2-9^2)=0$$
$$41^2 y^2 - 2 \cdot 41 \cdot 40 + 40^2 = 0$$
using the Pythagorean triple $9^2 + 40^2 = 41^2$, which factors as a perfect square:
$$(41y - 40)(41y - 40) = 0$$
$$\implies \cos \theta = \frac{40}{41}$$
|
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|
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_{x\to\infty}\sqrt{x^2+x+1}-x=m
$$
I'm stuck.
How do i simplify this expression to find the limit?
I've tried manipulating the radical by converting it to the denominator:
$$\lim_{x\to\infty}\frac{x^2+x+1}{\sqrt{x^2+x+1}}-x.$$
Or by multiplying both terms with the conjugate:
$$\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}}$$
But in neither case do I know how to take the limit. Any help would be greatly appreciated.
|
$$\lim_{x->\infty}\sqrt{x^2+x+1}-x$$
$$=\lim_{x->\infty}\left(\sqrt{x^2+x+1}-x\right)\cdot\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}$$
$$=\lim_{x->\infty}\frac{1+x}{\sqrt{x^2+x+1}+x}$$
The limit of a sum is the sum of the limits
$$=\lim_{x->\infty}\frac{1}{\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$
The limit of a quotient is the quotient of the limits
The limit of a constant is the constant
$$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$
$$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
The limit of a sum is the sum of the limits
$$=\frac{1}{\infty+\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
Use the power law
$$=\frac{1}{\infty+\sqrt{\displaystyle\lim_{x->\infty}x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{\infty+\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=0 + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
The limit of a quotient is the quotient of the limits
The limit of a constant is the constant
The limit of a sum is the sum of the limits
$$=\frac{1}{1+\displaystyle\lim_{x->\infty}\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{1+\displaystyle\lim_{x->\infty}\sqrt{\frac{x^2+x+1}{x^2}}}$$
Use the power law
$$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}\frac{x^2+x+1}{x^2}}}$$
Use L'Hopital
$$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}1+\frac{1}{2x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot\lim_{x->\infty}\frac{1}{x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot0}}$$
$$=\color{lightgray}{\boxed{\color{black}{\dfrac{1}{2}}}}$$
|
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|
Examine limit of sequence I have to examine limit of following sequence $\{a_n\}_{ n \ge 1}$ $a_n = \sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}}$. We know that $\lim_{n \to \infty} \sqrt[n]{a} = 1$ for $a > 0$ and we know that $\sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}} > 0$ but sequence $a_n$ is increasing (I cannot prove it but I checked it in wolframalpha) so we cannot use the squeeze theorem. Any hints?
|
I think you can get your way around this one using limited expansion.
Here I use : $ln(1-u) = -u -\frac{u^2}{2} +o(u^2)$ ; and $\exp(u) = 1+ u +o(u)$ , when $ u \rightarrow 0$
Let : $u_k = (2-\frac{1}{k})^k = 2^k*(1-\frac{1}{2k})^k = 2^k*\exp{[k*ln(1-\frac{1}{2k})]} = 2^k*\exp[-\frac{1}{2} -\frac{1}{8k} + o(\frac{1}{k})] $
=> $ u_k = \frac{2^k}{\sqrt{e}}*[1 -\frac{1}{8k} + o(\frac{1}{k})] $
Hence, you get : $u_k$ ~ $\frac{2^k}{\sqrt{e}}$
$\sum 2^k$ is a diverging series, so we can say that partial sums are equivalent :
$ \sum_{k=1}^n v_k $ ~ $ \frac{1}{\sqrt{e}}*\sum_{k=1}^n 2^k $
$ \sum_{k=1}^n 2^k = 2*\frac{2^n -1}{2-1} $ ~ $2^{n+1}$
Hence : $ (a_n)^n = (\sum_{k=1}^n v_k) $ ~ $\frac{2^{n+1}}{\sqrt{e}}$
So : $ (a_n)^n = \frac{2^{n+1}}{\sqrt{e}} + o(2^{n+1}) $
=> $ (\frac{a_n}{2})^n = \frac{2}{\sqrt{e}} + o(1) $ => $ \frac{a_n}{2} = (\frac{2}{\sqrt{e}} + o(1) )^{\frac{1}{n}} \rightarrow 1$ , when $ n\rightarrow +\infty$
Hence $(a_n) \rightarrow 2$ , when $ n\rightarrow +\infty$
|
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|
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$
$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.
$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$
So Solutions of the above equation are $x=6$ and $x=8$
Can we solve it using Derivative or Algebraic way
Thanks
|
expanding and factorizing we obtain
$2\, \left( x-6 \right) \left( x-8 \right) \left( {x}^{2}-14\,x+56
\right)
=0$
from here you will get all solutions.
|
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Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.
I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
|
Hint: You are doing well. Now multiply top and bottom by $\sec x+1$, and note that $\sec^2 x-1=\tan^2 x$.
|
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|
If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
|
$$a^2+b^2=2+2\cos(A-B)\implies\cos(A-B)=?$$
$$ab=(\sin A+\sin B)(\cos A+\cos B)=\sin(A+B)+\frac{\sin2A+\sin2B}2$$
$$\implies ab=\sin(A+B)[1+\cos(A-B)]\implies \sin(A+B)=?$$
$$b^2-a^2=\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[\cos(A-B)]+1$$
$$\implies\cos(A+B)=?$$
|
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|
prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.
|
This is relatively easy to solve even without noticing some clever factorization, simply by checking several possible cases.
It suffices to check what are remainder of powers of $2$ modulo $7$.
You can see that:
$2^0 \equiv 1 \pmod 7$
$2^1 \equiv 2 \pmod 7$
$2^2 \equiv 4 \pmod 7$
$2^3 \equiv 1 \pmod 7$
$2^4 \equiv 2 \pmod 7$
$2^5 \equiv 4 \pmod 7$
We see, that this repeats with period $3$. We get that $2^k\equiv 1\pmod 7$ if $k$ is multiple of $3$, $2^k\equiv 2\pmod7$ if the remainder modulo $3$ is $1$ and $2^k\equiv 4\pmod 7$ in the remaining case.
Let us look at the case $n=3k+1$.
Then we get
$2^n = 2^{3k+1} \equiv 2 \pmod 7$
$2^{n-1} = 2^{3k} \equiv 1 \pmod 7$
$2^{n-2} = 2^{3(k-1)+1} \equiv 4 \pmod 7$
$8^n=2^{3n} \equiv 1 \pmod 7$
$8^{n-2} = 2^{3(n-2)} \equiv 1 \pmod 7$
(In fact, for the powers of $8$ it is easier to notice that $8^n \equiv 1^n \equiv 1 \pmod 7$.)
By summing the above congruences we get
$2^n+2^{n-1}+2^{n-2}+8^n - 8^{n-2} \equiv 2+1+4+1-1 \equiv 7 \equiv 0 \pmod 7$.
I leave remaining cases $n=3k$ and $n=3k+2$ for you. (And, as you have seen in another answer, if you think about this a bit, you should be able to come up with a solution where you don't need to try various cases.)
|
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|
Need help in figuring out what I am doing wrong when solving for n.. Here is the expression that I am trying to solve for n:
$$ \frac{4}{16+n} = \frac{10}{16+n} \frac{10}{16+n}$$
I am doing the following:
\begin{align}
\frac{4}{16+n} & = \frac{100}{(16 + n)^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{16^2 + 32n + n^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{256 + 32n + n^2} \\[8pt]
\frac{1}{16+n} & = \frac{100}{4(256 + 32n + n^2)} \\[8pt]
\frac{1}{16+n} & = \frac{20}{256 + 32n + n^2} \\[8pt]
16+n & = \frac{256 + 32n + n^2}{20} \\[8pt]
n & = \frac{256 + 32n + n^2 - 320}{20} \\[8pt]
n & = \frac{-64 + 32n + n^2}{20} \\[8pt]
20n & = -64 + 32n + n^2 \\[8pt]
20n -32n - n^2 & = -64 \\[8pt]
12n - n^2 & = -64 \\[8pt]
n(12 - n) & = -64
\end{align}
Not sure what to do next now... Book says that n = 9
Cannot get 9...
Where am I wrong?
Thank you
|
There's simple method
$$\frac{4}{16+n} = \frac{10}{16+n} \times \frac{10}{16+n}$$
$n\neq-16$
$${4} = \frac{10\times10}{16+n} $$
$$64+4n = 100 $$
$$36=4n$$
$$n=9$$
Well let's figure out your mstake,
$$\frac{4}{16+n} = \frac{100}{n^2+32n+256} $$
$n\neq-16$
$$n^2+32n+256=25(16+n)$$
$$n^2+7n-144=0$$
$$(n-9)(n+16)$$
since $n\neq-16$
solution is $$n=9$$
|
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|
Calculating limit-help needed (difference of cubes) I'm stuck trying to find this limit:
$$\lim_{x\to \infty} (\sqrt[3]{(x-2)^2(x-1)}-x)$$
Thanks in advance!
|
If the problem had a square root instead of a cube root, you probably know how to proceed. For example,
\begin{align*}
\lim_{x\to\infty} \big( \sqrt{(x-2)(x-1)} - x \big) &= \lim_{x\to\infty} \frac{(\sqrt{(x-2)(x-1)} - x)(\sqrt{(x-2)(x-1)} + x)}{\sqrt{(x-2)(x-1)} + x} \\
&= \lim_{x\to\infty} \frac{(x-2)(x-1) - x^2}{\sqrt{x^2-3x+2} + x} \\
&= \lim_{x\to\infty} \frac{-3x+2}{\sqrt{x^2-3x+2} + x} \\
&= \lim_{x\to\infty} \frac{-3+2/x}{\sqrt{1-3/x+2/x^2} + 1} = -\frac{3}{2}.
\end{align*}
That approach uses the identity $a^2-b^2=(a-b)(a+b)$ (with $a=\sqrt{(x-2)(x-1)}$ and $b=x$).
Given the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, can you use a similar method in your problem?
|
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|
Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
--
The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, and also the line $y = k$ (for $0 < k < 12$) at the points C and D. Determine $k$ in a way that the trapezoid $ABCD$ has a maximum area.
--
My solution was this
--
The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{4+2 \cdot \sqrt{\frac{12-k}{3}} \cdot k}{2}$$
$$A_{T}' = 0 \therefore \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$2(12-k)-k=0 \therefore k = 8$$
--
Where did I go wrong on?
Thank you guys!
|
The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{\left(4+2 \cdot \sqrt{\frac{12-k}{3}} \right)\cdot k}{2}$$
$$A_{T}' = 0 $$
$$ 2 + \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$4 \sqrt{3 \cdot (12-k)} + 24 - 3k = 0$$
$$4 \sqrt{3 \cdot (12-k)} = 3k - 24$$
Squaring both sides we get
$$16 \cdot 3 \cdot (12-k) = 9k^2-6 \cdot 24 \cdot k + 24^2 \therefore k = \frac{32}{3}$$
|
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|
Why is Implicit Differentiation needed for Derivative of $y = \arcsin (2x+1)$? my function is: $y = \arcsin (2x+1)$ and I want to find its derivative.
My approach was to apply the chain rule:
$$y' = \frac{dg}{du} \frac{du}{dx}$$
with $g = \arcsin(u)$ and $u = 2x+1$.
$$g' = \frac{1}{\sqrt{1-u^2}}.$$
${u}' = 2$.
My solution therefore was
$$\frac{1}{\sqrt{1-u^2}} \cdot 2 = \frac{2}{\sqrt{1-(2x+1)^2}}.$$
This seems to be wrong and the correct solution is given by:
$\frac{1}{\sqrt{-x^{2}-x}}$
I know that implicit differentiation should be used for this particular problem, but I do not really understand why. I appreciate any help!
|
$$\frac{2}{\sqrt{1-(1+2x)^2}}=\frac{2}{\sqrt{1-4x^2+
-4x-1}}=\frac{2}{\sqrt{-4x^2-4x}}=\frac{2}{\sqrt{4(-x^2-x)}}=\frac{2}{2\sqrt{-x^2-x}}=\frac{1}{\sqrt{-x^2-x}}$$
|
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|
How many zeros does $f(x)= 3x^4 + x + 2 $ have?
How many zeros does this function have?
$$f(x)= 3x^4 + x + 2 $$
|
For $x\ge -2$, we have $(x+2)\ge 0$ and $3x^4>0$, so $f(x)>0$.
For $x<-2$, we have $3x^4+x=x(3x^3+1)=(-x)(-1-3x^3)>(2)(-1+3\cdot 2^3)=46$, so $f(x)>0$.
|
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|
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.
Please Help!
|
I dont'have (yet) a complete solution for now.
Let $I=\displaystyle \int_0^1 (\arctan(x))^2 \ln\Big(\dfrac{1+x^2}{2x^2}\Big)dx$
$I=\displaystyle \int_0^1 (\arctan(x))^2\ln(1+x^2)dx-2\int_0^1 (\arctan(x))^2\ln(x)dx-\ln(2)\int_0^1 (\arctan(x))^2dx$
I know how to compute the last one ;)
Let $J=\displaystyle \int_0^{\frac{\pi}{2}} \log(\sin x)dx$
The value of $J$ is well known to be $-\dfrac{\pi}{2}\ln 2$
Perform integration by parts:
$J=\displaystyle \Big[x\log(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx=-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx$
Let $K=-J$
Perform the change of variable $u=\tan x$
$K=\displaystyle \int_0^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
$K\displaystyle =\int_0^{1}\dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
In the second integral, in the right member perform the change of variable $u=\dfrac{1}{x}$:
$K= \displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan \Big(\dfrac{1}{x}\Big) }{1+x^2}dx$
For $x>0$, $\arctan \Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
For $x\neq 0$ ,$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-2\int_0^1 \dfrac{x\arctan x }{1+x^2}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx$
The derivative of $(\arctan x)^2$ is $\dfrac{2\arctan x}{1+x^2}$
Hence:
$\displaystyle \int_0^1 \dfrac{2x\arctan x }{1+x^2}dx=\Big[x(\arctan x)^2\Big]_0^1-\int_0^1 (\arctan x)^2dx=\dfrac{\pi^2}{16}-\int_0^1 (\arctan x)^2dx$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1 (\arctan x)^2dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$
Recall $K=\dfrac{\pi}{2}\ln 2$
Therefore:
$\displaystyle\int_0^1 (\arctan x)^2 dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\ln 2$
Where $G=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx$ is the Catalan's constant.
|
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|
Find the domain of $\frac{x}{\sqrt{6x^2+3x+3/4}+x}$. My attempt:
Let's assume that $\sqrt{6x^2+3x+\frac{3}{4}}+x$ $>$ 0$$
\rightarrow\sqrt{6x^2+3x+\frac{3}{4}} > -x
\rightarrow {6x^2+3x+\frac{3}{4}} > x^2\\
\rightarrow {5x^2+3x+\frac{3}{4}} > 0\\$$
If x $<$ 0 then $x^2$ $>$ 0, and ${5x^2+3x+\frac{3}{4}} > 0$
the domain is $(-\infty,+\infty)\\$
But, I'm not totally convinced. Any help is appreciated.
Thank you
|
We have two cases to check.
First, $$6x^2 + 3x + \frac{3}{4} \ge 0$$
It is $\ge 0$ for all $x$.
Next, $$6x^2 + 3x + \frac{3}{4} - x\ne 0$$
$$6x^2 + 2x + \frac{3}{4} \ne 0$$
Solving we get complex values for $x$. Therefore, no real values exist for $x$.
Thus, you are correct. The domain is: $(-\infty, \infty)$
|
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|
Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks
|
$\forall a\in(0,1),n\in N^+$,let$x_n$ be the positive zero of :
$$x^2+x+1=(n+a)^2$$
then $x_{n+1}>x_n$,and $x_n>n+a-1$,so$\lim_{n\to+\infty}x_n=+\infty$,but:
$$\sqrt{x_n^2+x_n+1}-[x_n^2+x_n+1]\equiv a$$
|
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|
valid proof of series $\sum \limits_{v=1}^n v$ $$\sum \limits_{v=1}^n v=\frac{n^2+n}{2}$$
please don't downvote if this proof is stupid, it is my first proof, and i am only in grade 5, so i haven't a teacher for any of this 'big sums'
proof:
if we look at $\sum \limits_{v=1}^3 v=1+2+3,\sum \limits_{v=1}^4 v=1+2+3+4,\sum \limits_{v=1}^5 v=1+2+3+4+5$
i learnt rainbow numbers in class three years ago, so i use that knowlege here:
$n=3,1+3=4$ and $2$.
$n=4,1+4$ and $2+3$
$n=5,1+5$ and $2+4$ and $3$
and more that i have done on paper that i don't wanna type.
we can see from this for the odd case that we have $(n+1)$ added together moving in from the outside, so we get to add $(n+1)$ to the total $\frac{(n-1)}2$ times plus the center number, which is $\frac{n+1}2$.. giving $\frac{n-1}2(n+1)+\frac{n+1}2=\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}$ and i can get $\frac{n^2-1}2+\frac{n+1}2=\frac{n^2+n}2$ which is what we want.
so odd are proven.
for even we have a simplier problem: we have $n+1$ on each pair of numbers going in. since we are even numbers, we have $1+n=n+1$ , with $n$ even, $2+(n-1)=n+1$ and we can see this is good for all numbers since we increase one side by one and lower the other by 1. so we get $\frac{n}2$ times $n+1$ gives $\frac{n^2+n}{2}$
thus is proven for all cases. thus is is proven
|
$\displaystyle\sum_{v=1}^nv=\dfrac{1}{2}\displaystyle\sum_{v=1}^n2v=\dfrac{1}{2}\displaystyle\sum_{v=1}^n\left((v+1)^2-v^2-1\right)=\dfrac{1}{2}\left((n+1)^2-(n+1)\right)-=\dfrac{1}{2}n(n+1)$
|
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|
$sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$
let $x,y\in R$,prove or disprove
$$sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$$
I think we must show that
$$\cos{x^2}+\cos{y^2}-\cos{xy}\in (-3,3)$$
it is clear
$$|\cos{x^2}+\cos{y^2}-\cos{xy}|<|\cos{x^2}|+|\cos{y^2}|+|\cos{xy}|<1+1+1=3$$
But I think we must why
$$\cos{x^2}+\cos{y^2}-\cos{xy}\to 2.99999\cdots,$$
so I think this problem is key is prove this reslut?
on the other words: why
$$\cos{x^2}+\cos{y^2}-\cos{xy}$$ is dense in $(-3,3)$
|
Let $\displaystyle f(x,y)=\cos(x^2)+\cos(y^2)-\cos(xy)$. It is clear that $-3\leq f(x,y)\leq 3$ for all $x,y$.
a) Let $\displaystyle x_n=\sqrt{2\pi n}$, $\displaystyle y_n=\sqrt{2\pi(n+1)}$. We have $\cos(x_n^2)=\cos(y_n^2)=1$. Put $z_n=x_ny_n$. We have $\displaystyle z_n=2\pi \sqrt{n^2+n}=2\pi n\sqrt{1+\frac{1}{n}}$. As $\displaystyle \sqrt{1+x}=1+\frac{x}{2}+o(x)$, we get $\displaystyle z_n=2\pi n(1+\frac{1}{2n}+o(1/n))=(2n+1)\pi+\varepsilon _n$, with $\varepsilon_n\to 0$. Hence $\cos(z_n)=-\cos(\varepsilon_n)\to -1$, and $f(x_n,y_n)\to 3$. Hence ${\rm Sup}(f(x,y)=3$.
b) Let $\displaystyle a_n=\sqrt{(2n+1)\pi}$, $b_n=\sqrt{(2n+3)\pi}$. We have $\cos(a_n^2)=\cos(b_n^2)=-1$, and $\displaystyle c_n=a_nb_n=\pi(2n+1)\sqrt{1+\frac{2}{2n+1}}$. We have $\displaystyle \sqrt{1+\frac{2}{2n+1}}=1+\frac{1}{2n+1}+o(1/(2n+1))$. Hence $\displaystyle c_n=\pi (2n+1)+\pi+o(1/(2n+1)=2\pi(n+1)+\alpha_n$, with $\alpha_n\to 0$ if $n\to +\infty$. Hence $\cos(a_nb_n)=\cos(\alpha_n)\to 1$, as $n\to +\infty$. We get $f(a_n, b_n)\to -3$ as $n\to \infty$, and ${\rm Inf}\{f(x,y)\}=-3$. So the result $6$ is correct.
|
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|
Calculate the probability that the sum of 3 fair dice is at least 13 How would I approach this. I'm more concerned with method than answer.
|
Can be done using generating functions which reduces down to polynomial multiplication.
$G(x)=\left(\frac{x}{6}+\frac{x^2}{6}+\frac{x^3}{6}+\frac{x^4}{6}+\frac{x^5}
{6}+\frac{x^6}{6}\right)^3 $
$G(x)$ models throwing 3 fair dice.
When that is expanded.
$\frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 \
x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 \
x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 \
x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
The ones that count are all the x's whose exponent are greater than or equal to 13.
$\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
since the x's are just placeholders and the coefficients are what counts we substitute $x = 1$
$P(\geq 13)=\frac{7}{72}+\frac{5}{72}+\frac{5}{108}+\frac{1}{36}+\frac{1}{72}+\frac{1}{216}=\frac{7}{27}$
|
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|
Find the value of $x,y,z$ that minimize $3x^2+2y^2+z^2+4xy+4yz$ under constraint $x^2+y^2+z^2 = 1$ Find the value of $x,y,z$ that minimize $3x^2+2y^2+z^2+4xy+4yz$ under constraint $x^2+y^2+z^2 = 1 $
i am learning algebra and think a way solve this problem by matrix
|
by Lagrange Multiplier Method we get $$3x^2+2y^2+z^2+4xy+4yz\geq -1$$ and the equal sign holds if $$x=-\frac{1}{3},y=\frac{2}{3},z=-\frac{2}{3}$$
|
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|
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
|
Let $ y = \tan x + \cot x $.
Then $ y' = \sec^{2} x - \csc^{2} x $.
So when $ y' = 0 $, $ x = \pi/4 $, for $x$ acute.
You can prove this is a minimum by considering $ y' $ for $ x = \pi/6 $ and $ x = \pi/3 $.
Thus $ y_{min} = \tan \pi/4 + \cot \pi/4 = 1 + 1 = 2.$
|
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|
Proving that $x=\arccos(\sqrt{\sin\theta})$ is $\sin(x+iy)=\cos\theta+i\sin\theta$ Given:
$$\sin(x+iy)=\cos\theta+i\sin\theta$$
To prove:
$$x=\arccos (\sqrt{\sin\theta})$$
How I tried:
$$\begin{align*}
\sin x \cosh y &= \cos\theta \\
\cos x \sinh y &= \sin\theta
\end{align*}$$
Then tried to use logarithm of hyperbolic complex number.
Also various trignometric form manipulation but I can't get the answer.
|
given:
$sin(x+iy)=e^{i\theta}$
$sin(x)*cosh(y)+icos(x)*sinh(y)=cos(\theta)+isin(\theta)$
comparing the the real and imaginary part
$\implies sin(x)*cosh(y)=cos(\theta)$
&
$\implies cos(x)*sinh(y)=sin(\theta)$
consider hyperbolic form of $sin^2y +cos^2y=1 $
$\cosh^2y-sinh^2y=1$
put values of $sinh(y) \& cosh(y)$ in above equation
$ \frac {\cos^2\theta}{ \sin^2x} $-$\frac{sin^2\theta}{cos^2x}$=1
cross muliplying and arranging
$cos^2\theta*cos^2x-sin^2\theta*sin^2x=sin^2x*cos^2x$
$(1-sin^2\theta)*cos^2x-sin^2\theta*sin^2x=cos^2x*(1-cos^2x)$
$cos^2x-sin^2\theta(cos^2x+sin^2x)=cos^2x-cos^2x*cos^2x $
$-sin^2\theta=-cos^4x$
taking twice squre root
$cosx=\sqrt\sin\theta$
|
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|
showing that cos(y)-cos(x)=2*sin((x+y)/2)*sin((x-y)/2) how can one show that $\cos(y)-cos(x)=2sin(\frac{x-y}{2})sin(\frac{x+y}{2})$ by just using trigonometrical equalities like additional theorems?
I was trying to show it by rewriting $cos(y)=cos(0.5y+0.5y)$ and then using the additional theorems, but somehow I do not have the result, that is needed.
Starting on the righthandside wasn't successfull either.
Thanks for helping
|
I think the easier way to do it is by working with the right hand side, instead of the left hand side, making use of the addition formula for sine:
$$\begin{eqnarray}2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right) &=& 2\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{-y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{-y}{2}\right)\right)\\
&&\times\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)\end{eqnarray}$$
Since cosine is even and sine is odd, we can simplify this to
$$2\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)\\\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)$$
This is just the difference of two squares, so we get
$$2\left(\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{y}{2}\right)-\cos^2\left(\frac{x}{2}\right)\sin^2\left(\frac{y}{2}\right)\right).$$
Since $\sin^2x+\cos^2x=1$, we can rewrite this slightly as
$$2\left(\sin^2\left(\frac{x}{2}\right)\left(1-\sin^2\left(\frac{y}{2}\right)\right)-\cos^2\left(\frac{x}{2}\right)\sin^2\left(\frac{y}{2}\right)\right)$$
Expanding, we get
$$2\sin^2\left(\frac{x}{2}\right)-2\left(\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)\right)\sin^2\left(\frac{y}{2}\right)$$
We can see that the inner term has the Pythagorean identity so we can replace it with $1$ to get
$$2\sin^2\left(\frac{x}{2}\right)-2\sin^2\left(\frac{y}{2}\right)$$
Since $\cos(2x) = 1-2\sin^2(x)$, the above simplifies to exactly what we want.
|
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|
Compute the resultant of the polynomials $f(x)=x^2y+3xy-1$ and $g(x)=6x^2+y^2-4$ Consider these two polynomials:
$$f=x^2y+3xy-1$$
$$g=6x^2+y^2-4$$
I need to compute their resultant, denoted in my textbook as $h=Res(f,g,x)$.
Here's where I need help: setting up the Sylvester Matrix. I could be missing something obvious, but from the definitions I've read online, I kept getting a matrix like so:
$X =\begin{bmatrix} 1 & 0 & 0 & 6 & 0 & 0\\3 & 1 & 0 & 1 & 6 & 0\\-1 & 3 & 1 & -4 & 1 & 6\\0 & -1 & 3 & 0 & -4 & 1\\0 & 0 & -1 & 0 & 0 & -4\\\end{bmatrix}$
My understanding is that $X$ should be a square matrix- what am I missing?
Thanks in advance.
UPDATE:
Is this more like it?
$X =\begin{bmatrix} y & 0 & 6 \\-1 & y & 0 \\0 & -1 & y^2-4 \\\end{bmatrix}$
|
Write your polynomials as
$$
yx^2+ (3xy)-1
$$
and
$$
-6x^2+ (y^2-4)
$$
As polynomials in $x$, these have degree two, so your matrix should be a $4 \times 4$ matrix.
$$
\begin{pmatrix}
y & 3y & -1 & 0 \\
0 & y & 3y & -1 \\
-6 & 0 & y^2-4 & 0 \\
0 & -6 & 0 & y^2-4
\end{pmatrix}
$$
Taking the determinant gives
$$
y^{6}-62 y^{4}-12 y^{3}+232 y^{2}+48 y+36
$$
This can also be computed in Macaulay2 by the commands:
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : f1 = x^2*y+3*x*y-1
2
o2 = x y + 3x*y - 1
o2 : R
i3 : f2 = -6*x^2+y^2-4
2 2
o3 = - 6x + y - 4
o3 : R
i4 : resultant(f1,f2,x)
6 4 3 2
o4 = y - 62y - 12y + 232y + 48y + 36
o4 : R
|
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|
Principle of inclusion and exclusion How many integer solutions can we have for the equation $x_1 + x_2 + x_3 = 18$, if $0 \leq x_1 \leq 6$, $4 \leq x_2 \leq 9$ and $7 \leq x_3 \leq 14$?
Using the Principle of inclusion and Exclusion: I have found
$S_0$ = $\binom{9}{7} $
However, I am stuck as to how to go about calculating $S_1$, $S_2$ and $S_3$. I believe that there is no solution for $S_2$ and $S_3$.
|
This is the stars and bars problem. You can set up an equivalent question. Subtract out $4$ from both sides so that $0 \leq x_{2} \leq 5$. Similarly, subtract out $7$ so $0 \leq x_{3} \leq 7$. This leaves us with $x_{1} + x_{2} + x_{3} = 7$.
We can use a generating function to give us our inclusion-exclusion formula. Note that for $x_{1}$, we have the generating function: $f(x) = 1 + x + x^{2} + ... + x^{6} = \frac{1 - x^{7}}{1 - x}$.
Similarly, for $x_{2}, x_{3}$, we have the functions $g(x) = \sum_{i=0}^{5} x^{i} = \frac{1 - x^{6}}{1 - x}$ and $h(x) = \sum_{i=0}^{7} x^{i} = \frac{1 - x^{8}}{1 - x}$.
By rule of product, we multiply $f, g, h$ and are interested in the coefficient of $x^{7}$.
So our product is:
$$k(x) = \frac{1 - x^{5} - x^{6} - x^{7} + x^{11} + x^{12} + x^{13} - x^{18}}{(1 - x)^{3}}$$
Now $\frac{1}{(1-x)^{3}} = \sum_{i=0}^{\infty} \binom{i + 3 - 1}{i} x^{i}$
And so we multiply the expansion of the denominator by the numerator, keeping only the terms where we have $x^{7}$:
$$\binom{7 + 3 - 1}{7} - \binom{2 + 3 - 1}{2} - \binom{1 + 3 - 1}{1} - 1$$
And this gives you your inclusion-exclusion formula very methodically.
|
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|
variation of parameters for a Cauchy-Euler problem: where am I wrong? The problem is this one
$$x^2y'' - xy' -3y = 5x^4.$$
Then the complementary solution I get is
$$c_1x + c_2x\ln x.$$
Then after that the particular solution is
$$\ln x e^{4x} \left(\frac{x}{4} - \frac{1}{16}\right) - \frac{e^{4x}}{16} - \frac{1}{16} \int \frac{e^{4x}}{x}dx.$$
My doubt is how can integrate the last part or where am I doing it wrong
|
A.N.Other method is
$$
x^2y'' - xy' - 3y = 5x^4
$$
we can realise thath
$$
\dfrac{d}{dx}x^2y' = x^2y'' + 2xy'\implies x^2y'' = \dfrac{d}{dx}x^2y' - 2xy'
$$
so we can insert into the original equation to yeild
$$
\dfrac{d}{dx}x^2y' - 2xy' - xy' -3y = \dfrac{d}{dx}\left(x^2y'\right) - 3(xy' + y) = 5x^4\\
\dfrac{d}{dx}\left(x^2y'\right) -\dfrac{d}{dx}(xy) = 5x^4
$$
or
$$
y' -\frac{3}{x}y = x^3 + \frac{C_1}{x^2}
$$
integrating factor
$$
y\mathrm{e}^{-3\ln x} = \int x^3 \mathrm{e}^{-3\ln x} dx +C_1 \int \frac{1}{x^2}\mathrm{e}^{-3\ln x} dx +C_2\\
y\frac{1}{x^3} = \int x^3\frac{1}{x^3} dx + C_1 \int \frac{1}{x^2}\frac{1}{x^3}dx + C_2\\
\frac{y}{x^3} = \int 1 dx +C_1 \int x^{-5} + C_2 = \frac{1}{2}x^2 -\frac{C_1}{4}x^{-4} + C_2
$$
thus
$$
y = x^4 + C_2x^3 -\frac{C_1}{4}x^{-1}
$$
As obtained in the previous solutions.
|
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|
Evaluating $\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$ How to evaluate following integral
$$\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$$
I tried the feynman way
$$\begin{align}
I'(a)&=\int_0^1\frac{-x^{-a}\ln x-x^{a}\ln x}{1-x}\,\mathrm dx\\
&= \int_0^1\ln x\left(\frac{-x^{-a}-x^{a}}{1-x}\right)\,\mathrm dx\\
\end{align}$$
I don't have any idea about how to proceed!
Some help is appreciated
|
We can do it without using The Feynman Way
$$\begin{align}
\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx
&=\int_0^1 \left[\frac{x^{-a}}{1-x}-\frac{-x^a}{1-x}\right]\,\mathrm dx\tag{1}\\
&=\int_0^1\sum_{k=1}^\infty\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{2}\\
&=\sum_{k=1}^\infty\int_0^1\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{3}\\
&= \sum_{n=1}^\infty \left(\frac{1}{k-a} - \frac{1}{k+a}\right)\tag{4}\\
&= \sum_{n=1}^\infty \left(\frac{k+a-k+a}{(k-a)(k+a)}\right)\tag{5}\\
&= \sum_{n=1}^\infty \left(\frac{2a}{k^2-a^2}\right)\tag{6}\\
&= 2a\sum_{n=1}^\infty\frac{1}{k^2-a^2}\tag{7}\\
&= 2a\left[\frac{1}{2a} \left( \frac{1}{a} \, - \, \pi \cot(a\pi) \right)\right]\tag{8}\\
&= \frac{1}{a} \, - \, \pi \cot(a\pi)\tag{9}\\
\end{align}$$
Note that for convergence $a \in (-1,1)$
$$\large\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx= \frac{1}{a} \, - \, \pi \cot(a\pi)$$
|
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|
Another infinite series involving Gamma function It's not hard to see that Mathematica expresses this series in terms of hypergeometric function,
but how about finding a way of expressing the result in terms of elementary functions only?
Is that possible? How would you recommend me to proceed?
$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n\Gamma(2n-1)}{2^{2n}\Gamma\left(2n+\frac{1}{2}\right)}$$
|
$$\begin{eqnarray*}S=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{n\,\Gamma(2n-1)}{4^n\,\Gamma(2n+1/2)}&=&\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n\Gamma(3/2)}B(2n-1,3/2)\\&=&\frac{2}{\sqrt{\pi}}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n}\int_{0}^{1}x^{2n-2}(1-x)^{1/2}\,dx\\&=&\frac{1}{2\sqrt{\pi}}\int_{0}^{1}\sum_{n=0}^{+\infty}\frac{(-1)^n(n+1)}{4^n}x^{2n}(1-x)^{1/2}\,dx\\&=&\frac{8}{\sqrt{\pi}}\int_{0}^{1}\frac{(1-x)^{1/2}}{(4+x^2)^2}\,dx\\&=&\frac{8}{\sqrt{\pi}}\int_{0}^{1}\frac{x^{1/2}}{(5-2x+x^2)^2}\,dx\\&=&\frac{16}{\sqrt{\pi}}\int_{0}^{1}\frac{x^2\,dx}{(5-2x^2+x^4)^2}.\end{eqnarray*}$$
Now the last integral can be evaluated with standard techniques.
We get the terrifying identity:
$$ S =\frac{\phantom{}_3 F_2\left(\frac{1}{2},1,2;\frac{5}{4},\frac{7}{4};-\frac{1}{4}\right)}{3\sqrt{\pi}}=\\=\scriptstyle{\frac{1}{4\sqrt{5\pi }\, }\sqrt{2+ \sqrt{5}} \left(\pi -\arctan\left(2 \sqrt{2+\sqrt{5}}\right)+2\sqrt{5}\log 2-2 \sqrt{5} \log\left(1+\sqrt{5}+\sqrt{2 \left(1+\sqrt{5}\right)}\right)+2 \log\left(2+\sqrt{5}+2 \sqrt{2+\sqrt{5}}\right)\right)}\\=0.16981854\ldots.$$
|
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|
Understanding the limit $\lim_{t \to \infty} \int_0^\infty \frac{t}{t^2 + x} \sin(1/x)dx$ Prove :
$$\lim_{t\to +\infty}\int_{0}^{\infty}\frac{t}{t^2 +x} \sin\frac{1}{x} dx=0$$
Maybe dominated convergence theorem? Who can give a proof?
Thanks!
|
We can assume $t>1$ - we just throw out the beginning...
Since for $y<1$, $\sin y<y$, thus for $x>1$
$$\sin\frac{1}{x}<\frac{1}{x}$$
By AM-GM:
$$|\frac{t}{t^2+x}|<\frac{1}{t+\frac{x}{t}}<\frac{1}{2\sqrt{x}}$$
Also note that for $x\leq1$:
$$\frac{1}{1+x}-\frac{t}{t^2+x}=\frac{t^2-t+t(1-x)}{(1+x)(t^2+x)}>0$$
So we have a dominating function for $t>1$, as we used $1\geq \sin x$:
$$g(x)=\cases{ \frac{1}{1+x}& $0<x<1$\\ \frac{1}{2x\sqrt{x}}& $x>1$}$$
But
$$\int_0^\infty g(x)\text{d}x=\int_0^1 \frac{1}{1+x}\text{d}x+\int_1^\infty \frac{1}{2x\sqrt{x}}\text{d}x=1+\log 2$$
And apply dominated convergence.
|
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|
Using the half angle formula Find $\sin\frac { x }{ 2 }, \cos\frac { x }{ 2 }$, and $\tan\frac { x }{ 2 }$ from $\sin x=\frac { 3 }{ 5 } ,\quad 0°<x<90°$
What I did:
$$\sin\frac { x }{ 2 } =\sqrt { \frac { 1-\frac { 4 }{ 5 } }{ 2 } } =\sqrt { 10 } $$
$$\cos\frac { x }{ 2 } =\sqrt { \frac { 1+\frac { 4 }{ 5 } }{ 2 } } =\sqrt { \frac { 9 }{ 10 } } =\frac { 3 }{ \sqrt { 10 } } $$
$$\tan\frac { x }{ 2 } =\frac { \frac { 3 }{ 5 } }{ 1+\frac { 4 }{ 5 } } =\frac { \frac { 3 }{ 5 } }{ \frac { 9 }{ 5 } } =\frac { 1 }{ 3 } $$
The last one seems to be the only one that corresponds to the correct answer. The first two are wrong and I can't seem to find where I went wrong.
|
As $0<\dfrac x2<45^\circ\implies$ all the trigonometric ratios $>0$
$$\sin^2\frac x2=\frac{1-\cos x}2=\cdots=\frac1{10}\implies\sin\frac x2=+\frac1{\sqrt{10}}=\frac{\sqrt{10}}{10}$$
Similarly for $\cos\dfrac x2$
|
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|
Solving $\sin\theta -1=\cos\theta $
Solve$$\sin\theta -1=\cos\theta $$
Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi k, \theta =\frac { \pi }{ 2 } +2\pi k$$
Why is $\theta =0+\pi k$ wrong?
|
$$\begin{align}
\sin(\theta)-\cos(\theta)&=1\\
\sin(\theta)\frac{1}{\sqrt{2}}-\cos(\theta)\frac{1}{\sqrt{2}}&=\frac{1}{\sqrt{2}}\\
\sin(\theta)\sin(\pi/4)-\cos(\theta)\cos(\pi/4)&=\frac{1}{\sqrt{2}}\\
-\cos(\theta+\pi/4)&=\frac{1}{\sqrt{2}}\\
\cos(\theta+\pi/4)&=-{\frac{1}{\sqrt{2}}}\\
\theta+\pi/4&=(2k+1)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+3/4)\pi\pm\frac{\pi}{4}\\
\theta&=(2k+1/2)\pi\qquad\text{or}\qquad(2k+1)\pi
\end{align}$$
Thinking of the unit circle as a compass, these solutions are "north" and "west".
|
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|
evaluation of the integral of a certain logarithm I come across the following integral in my work
$$\int_a^\infty \log\left(\frac{x^2-1}{x^2+1}\right)\textrm{d}x,$$
with $a>1$.
Does this integral converge ? what is its value depending on $a$ ?
|
We have:
$$ I = \int_{\frac{a^2-1}{a^2+1}}^{1}\frac{\log x}{(1-x)^{3/2}(1+x)^{1/2}}\,dx $$
where the integrand function is integrable over $(0,1)$. Moreover:
$$ 0 > I > \int_{0}^{1}\frac{\log x}{(1-x)^{3/2}(1+x)^{1/2}} = -\frac{\pi}{2}-\log 2.$$
Integration by parts gives:
$$ I = x\,\left.\log\frac{x^2-1}{x^2+1}\right|_{a}^{+\infty}-\int_{a}^{+\infty}x\left(\frac{1}{x-1}+\frac{1}{x+1}-\frac{2x}{x^2+1}\right)\,dx,$$
or:
$$ I = -a\log\frac{a^2-1}{a^2+1}-\int_{a}^{+\infty}\frac{4x^2}{x^4-1}\,dx = -a\log\frac{a^2-1}{a^2+1}-4\int_{0}^{1/a}\frac{dx}{1-x^4}$$
so:
$$ I = -a\log\frac{a^2-1}{a^2+1}-2\left(\operatorname{arccot} a+\operatorname{arccoth} a\right),$$
or:
$$\color{red}{ I = -a\log\frac{a^2-1}{a^2+1}+\log\frac{a-1}{a+1}-\pi+2\arctan a}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1059759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Vector analysis - Curl of vector
How to prove it? I have tried several times to solve it, but I still get stuck everytime.
|
First Method (long)
Observe that $\operatorname{curl}(\pmb{a}\times\pmb{b})=\nabla\times (\pmb{a}\times\pmb{b})$ where $\nabla=\pmb{i}\frac{\partial}{\partial x}+\pmb{j}\frac{\partial}{\partial y}+\pmb{k}\frac{\partial}{\partial z}$.
Recalling that $\pmb{a}\times\pmb{b}=\left|\matrix{\pmb i & \pmb j & \pmb k\\
a_x & a_y & a_z\\
b_x & b_y & b_z}\right|=(a_yb_z-a_zb_y)\pmb{i}+(a_zb_x-a_xb_z)\pmb{j}+(a_xb_y-a_yb_x)\pmb{k}
$
we have
$$
\operatorname{curl}(\pmb{a}\times\pmb{b})=\nabla\times (\pmb{a}\times\pmb{b})=\left|\matrix{\pmb i & \pmb j & \pmb k\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
a_yb_z-a_zb_y & a_zb_x-a_xb_z & a_xb_y-a_yb_x}\right|
$$
The $x$ component is
$$\small\begin{align}
[\nabla\times (\pmb{a}\times\pmb{b})]_x &=\frac{\partial}{\partial y}(a_yb_z-a_zb_y)-\frac{\partial}{\partial z}(a_xb_z-a_zb_x)=\\
&=a_x\left(\frac{\partial b_y}{\partial y}+\frac{\partial b_z}{\partial z}\right)-b_x\left(\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z}\right)+
\left(b_y\frac{\partial }{\partial y}+b_z\frac{\partial }{\partial z}\right)a_x-
\left(a_y\frac{\partial }{\partial y}+a_z\frac{\partial }{\partial z}\right)b_x
\end{align}
$$
Adding $a_x\frac{\partial b_x}{\partial x}$ to the first of these four terms, and subtracting from the last, and doing the same with $b_x\frac{\partial a_x}{\partial x}$ to the other two terms, we find
$$
[\nabla\times (\pmb{a}\times\pmb{b})]_x=(\nabla\cdot\pmb{b})a_x-(\nabla\cdot\pmb{a})b_x+(\pmb{b}\cdot\nabla)a_x-(\pmb{a}\cdot\nabla)b_x
$$
and repeating for the $y$ and $z$ components, we find that
$$
\nabla\times (\pmb{a}\times\pmb{b})=(\nabla\cdot\pmb{b})\pmb{a}-(\nabla\cdot\pmb{a})\pmb{b}+(\pmb{b}\cdot\nabla)\pmb{a}-(\pmb{a}\cdot\nabla)\pmb{b}
$$
where the operator $\pmb{a}\cdot\nabla$ is defined as $\pmb{a}\cdot\nabla=a_x\frac{\partial }{\partial x}+a_y\frac{\partial }{\partial y}+a_z\frac{\partial }{\partial z}$ and $\nabla\cdot \pmb{a}=\operatorname{div}\pmb a$.
Now for $\pmb{a}=\pmb{v}$ and $\pmb{b}=\frac{\pmb{r}}{r^3}$ we obtain
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=
\left(\nabla\cdot\frac{\pmb{r}}{r^3}\right)\pmb{v}-(\nabla\cdot\pmb{v})\frac{\pmb{r}}{r^3}+\left(\frac{\pmb{r}}{r^3}\cdot\nabla\right)\pmb{v}-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}
$$
The second and third terms are null, because $\pmb v$ is a constant vector. The first term is the $\operatorname{div}\left(\frac{\pmb{r}}{r^3}\right)$ and is null; infact
$$
\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)=\frac{1}{r^3}\left(1-\frac{3x^2}{r^2}\right)
$$
Adding this to similar terms for $y$ and $z$ gives
$$
\nabla\cdot\frac{\pmb{r}}{r^3}=\frac{1}{r^3}\left(3-\frac{3(x^2+y^2+z^2)}{r^2}\right)=\frac{1}{r^3}\left(3-\frac{3r^2}{r^2}\right)=0
$$
So we have
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}.
$$
The $x$ component is
$$\begin{align}
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_x &=\left(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}+c\frac{\partial }{\partial z}\right)\frac{x}{\sqrt{x^2+y^2+z^2}}\\
&=a\left(1-\frac{3x^2}{r^2}\right)\frac{1}{r^3}-b\frac{3xy}{r^5}-c\frac{3xz}{r^5}
\end{align}
$$
In the same way, the other components are
$$\begin{align}
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_y &=
&=-a\frac{3xy}{r^5}+b\left(1-\frac{3y^2}{r^2}\right)\frac{1}{r^3}-c\frac{3yz}{r^5}\\
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_z &=
&=-a\frac{3xz}{r^5}-b\frac{3yz}{r^5}+a\left(1-\frac{3z^2}{r^2}\right)\frac{1}{r^3}
\end{align}
$$
It's easy to see that summing up the terms with $-\frac{3}{r^5}$
$$\small
-\frac{3}{r^5}\left[(by+cz)x\pmb{i}+(ax+cz)y\pmb{j}+(ax+by)z\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=
-\frac{3}{r^5}\left[(\underbrace{ax+by+cz}_{\pmb{v}\cdot\pmb{r}})x\pmb{i}-ax^2\pmb{i}+(ax+by+cz)y\pmb{j}-by^2\pmb{j}+(ax+by+cz)z\pmb{k}-cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=
-\frac{3}{r^5}\left[(\pmb{v}\cdot\pmb{r})(\underbrace{x\pmb{i}+y\pmb{j}+z\pmb{k}}_{\pmb{r}})-ax^2\pmb{i}-by^2\pmb{j}-cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=-\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}+\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=\color{blue}{-\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}}
$$
Summing up the terms with $\frac{1}{r^3}$ we find
$$
\frac{a}{r^3}\pmb{i}+\frac{b}{r^3}\pmb{j}+\frac{c}{r^3}\pmb{k}=\frac{1}{r^3}(a\pmb{i}+b\pmb{j}+c\pmb{k})=\color{blue}{\frac{1}{r^3}\pmb{v}}
$$
Putting all together, we have
$$\color{blue}{
\operatorname{curl}\left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}=\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}-\frac{1}{r^3}\pmb{v}}
$$
Second Method (shorter)
$$
\begin{align}
[\nabla\times(\pmb{a}\times\pmb{b})]_i
&=\epsilon_{ijk}\partial_j(\pmb{a}\times\pmb{b})_k\\
&=\epsilon_{ijk}(\partial_j\epsilon_{klm}a_lb_m)\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j(a_lb_m)\\
&=\partial_m(a_ib_m)-\partial_l(a_lb_i)\\
&=(\partial_m a_i)b_m+a_i(\partial_m b_m)-(\partial_l a_l)b_i-a_l(\partial_i b_l)\\
&=(\pmb{b}\cdot\nabla)b_i+(\nabla\cdot\pmb{b})a_i-(\nabla\cdot\pmb{a})b_i-(\pmb{a}\cdot\nabla)b_i
\end{align}$$
so we have finally
$$
\nabla\times (\pmb{a}\times\pmb{b})=(\nabla\cdot\pmb{b})\pmb{a}-(\nabla\cdot\pmb{a})\pmb{b}+(\pmb{b}\cdot\nabla)\pmb{a}-(\pmb{a}\cdot\nabla)\pmb{b}
$$
Now for $\pmb{a}=\pmb{v}$ and $\pmb{b}=\frac{\pmb{r}}{r^3}$ we obtain
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=
\left(\nabla\cdot\frac{\pmb{r}}{r^3}\right)\pmb{v}-(\nabla\cdot\pmb{v})\frac{\pmb{r}}{r^3}+\left(\frac{\pmb{r}}{r^3}\cdot\nabla\right)\pmb{v}-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}
$$
The second and third terms are null, because $\pmb v$ is a constant vector. The first term is the $\operatorname{div}\left(\frac{\pmb{r}}{r^3}\right)$:
$$
\nabla\cdot\frac{\pmb{r}}{r^3}=\partial_i\frac{x_i}{r^3}=\left(1-x_i^2\right)\frac{1}{r^3}\delta_{ii}=0
$$
So we have
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}.
$$
Observing that
$$
(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}=v_i\partial_i\frac{x_j}{r^3}=\frac{v_i\delta_{ij}}{r^3}-v_ix_jx_i\frac{3}{r^3}=\frac{v_i}{r^3}-(v_ix_i)x_j\frac{3}{r^5}=\frac{\pmb{r}}{r^3}-(\pmb{v}\cdot\pmb{r})\pmb{r}\frac{3}{r^5}
$$
we have finally
$$\color{red}{
\operatorname{curl}\left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{v}}{r^3}=\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}-\frac{1}{r^3}\pmb{v}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1060622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Extremely difficult log integral, real methods only $$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$
I tried this:
$$M_1 = \int_{0}^{1} \frac{1}{1-x} \cdot \left(\frac{x^2 + x\log(1-x) - \log(1-x) - x)}{x^2}\right) dx$$
$$M_1 = \int_{0}^{1} \frac{x^2 + x\log(1-x) - \log(1-x) - x}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x\left(x + \log(1-x) - 1\right) - \log(1-x)}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x + \log(1-x) - 1}{x(1-x)}dx - \int_{0}^{1} \frac{\log(1-x)}{(1-x)x^2} dx$$
but we cannot seperate as the integrals become divergent.
|
You can write the numerator as :
$$
x^2 +x\log(1-x) -\log(1-x)-x = (1 - \log(1-x))(1-x) +(x-1)(x+1)
$$
so that it remains to evaluate the integral:
$$
\int_0^1 \frac{1-\log(1-x)-1-x}{x^2} \mathrm dx = - \int_0^1 \frac{\log(1-x)+x}{x^2} \mathrm dx
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1060937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
Evaluation of $\int_0^{2\pi} \frac{1}{1+8\cos^2(\theta)}d\theta$ with Cauchy's residue Theorem I have to proof
$$\int_0^{2\pi} \frac{1}{1+8\cos^2(\theta)}d\theta = \frac{2\pi}{3}$$
with Cauchy's residue Theorem. I have showed it, but in my solution, there comes $-\frac{2\pi}{3}$. I Show you my solution:
I know that $\cos(\theta)=\frac{1}{2}(z+z^{-1})$, while $z=e^{i\theta}$, and $\frac{dz}{dt}=iz$, so $dt=\frac{dz}{iz}$. Let $C$ be an unit circle. Now, I Substitute and get:
\begin{align*}
\int_0^{2\pi} \frac{1}{1+8\cos^2(\theta)}d\theta &= \int_C \frac{1}{1+8(\frac{1}{2}(z+\frac{1}{z}))^2}\frac{dz}{iz} = \int_C \frac{1}{1+2(z+\frac{1}{z})^2}\frac{dz}{iz} \\
&= \int_C \frac{-i}{z+2z(z+\frac{1}{z})^2}dz = -i\int_C \frac{1}{z+2z(z+\frac{1}{z})^2}dz \\
&= -i\int_C \frac{1}{z+2z(z^2+2+z^{-2})}dz = -i\int_C \frac{1}{z+2z^3+4z+2z^{-1}}dz \\
&= -i\int_C \frac{1}{2z^3+5z+2z^{-1}}dz = -i\int_C \frac{z}{2z^4+5z^2+2}dz \\
&= -i\int_C \frac{z\cdot dz}{(z^2+2)(2z^2+1)} \\
&=-i\int_C \frac{z~dz}{(z+i\sqrt{2})(z-i\sqrt{2})(z+\frac{i}{\sqrt{2}})(z-\frac{i}{\sqrt{2}})}
\end{align*}
The singularities to be considered are at $2^{-1/2}i, -2^{1/2}i$. Let $C_1$ be a small circle about $2^{-1/2}i$, $C_2$ a small circle about $-2^{-1/2}i$.
Then I put it together:
\begin{align*}
&\Rightarrow -i\left[\oint_{C_1}\frac{z\left(\frac{1}{(z+i\sqrt{2})(z-i\sqrt{2})(z+\frac{i}{\sqrt{2}})}\right)}{z-\frac{i}{\sqrt{2}}} + \oint_{C_2}\frac{z\left(\frac{1}{(z+i\sqrt{2})(z-i\sqrt{2})(z-\frac{i}{\sqrt{2}})}\right)}{z+\frac{i}{\sqrt{2}}} \right] \\
&= -i\left.\left[2\pi i\left(\frac{z}{(z+\sqrt{2}i)(z-\sqrt{2}i)(z+\frac{i}{\sqrt{2}})} \right)\right|_{z=\frac{i}{\sqrt{2}}} +\left. 2\pi i\left(\frac{z}{(z+\sqrt{2}i)(z-\sqrt{2}i)(z-\frac{i}{\sqrt{2}})} \right)\right|_{z=-\frac{i}{\sqrt{2}}} \right] \\
&= 2\pi \left[ \frac{i/\sqrt{2}}{(\frac{3i}{\sqrt{2}})(\frac{-i}{\sqrt{2}})(\frac{2i}{\sqrt{2}})} + \frac{-i/\sqrt{2}}{(-\frac{3i}{\sqrt{2}})(\frac{i}{\sqrt{2}})(-\frac{2i}{\sqrt{2}})}\right] \\
&= 2\pi \left( \frac{1}{\sqrt{2}(+3)\sqrt{2}}+\frac{1}{\sqrt{2}(+3)\sqrt{2}}\right) \\
&= 2\pi \cdot \frac{+1}{3} = \frac{2\pi}{3}.
\end{align*}
|
It is better to notice that, by replacing $\theta$ with $\arctan t$:
$$ I = 4\int_{0}^{\pi/2}\frac{d\theta}{1+8\cos^2\theta} = 4\int_{0}^{+\infty}\frac{dt}{(1+t^2)(1+\frac{8}{1+t^2})}=2\int_{-\infty}^{+\infty}\frac{dt}{9+t^2}$$
and the last integral is very easy to compute with the residue theorem, or without:
$$2\int_{-\infty}^{+\infty}\frac{dt}{t^2+9}=\frac{2}{3}\int_{-\infty}^{+\infty}\frac{du}{u^2+1}=\frac{2\pi}{3}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
|
As Deepak wrote it's easier to start with the right hand side, but you could go ahead with your approach too, as shown by Arian.
\begin{equation*}
\sqrt{2}\sin 10{{}^\circ}
+\sqrt{3}\cos 35{{}^\circ}
=\sin 55{{}^\circ}+2\cos 65{{}^\circ}\tag{1}
\end{equation*}
Here is a variant that uses the cosine addition formula for $65{{}^\circ}=35{{}^\circ}+30{{}^\circ}$, the sine difference formula for $10{{}^\circ}=45{{}^\circ}-35{{}^\circ}$ and the sine/cosine complement formula.
*
*Use the complement formula $\sin \theta =\cos \left( 90{{}^\circ}
-\theta \right) $ for $\theta=55{{}^\circ}$, the addition formula $\cos \left( a+b\right) =\cos a\cos
b-\sin a\sin b,$ and the special values $\sin 30{{}^\circ}
=\frac{1}{2}$, $\cos 30{{}^\circ}
=\frac{\sqrt{3}}{2}$ to rewrite the right hand side of $(1)$ as [hover your mouse over the grey to see]
\begin{equation*}\sin 55{{}^\circ}+2\cos 65{{}^\circ}=\cos 35{{}^\circ}+\sqrt{3}\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{2}\end{equation*}
*Substitute $(2)$ in $(1)$ and simplify to obtain the equivalent identity
\begin{equation*}\sqrt{2}\sin 10{{}^\circ}=\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{3}\end{equation*}
*To show that $(3)$ holds, use the subtraction formula $\sin \left(
a-b\right) =\sin a\cos b-\cos a\sin b$ and the special values $\sin 45{{}^\circ}=\cos 45{{}^\circ}=\frac{\sqrt{2}}{2}$. Since we get the following trivial identity, we are
done.
\begin{equation*}\cos 35{{}^\circ}-\sin 35{{}^\circ}=\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{4}\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find maximum without calculus Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
|
$(f(x))^2 = 4x^2\left(1+\sqrt{1-x^2}\right)^2$. Let $y = \sqrt{1-x^2} \Rightarrow x^2 = 1 - y^2 \Rightarrow (f(x))^2 = 4(1-y^2)(1+y)^2 = 4(1-y)(1+y)^3$.
Apply AM-GM inequality we have:
$2 = (1-y) + \dfrac{1+y}{3} + \dfrac{1+y}{3} + \dfrac{1+y}{3} \geq 4\sqrt[4]{\dfrac{(1-y)(1+y)^3}{27}} \Rightarrow \dfrac{16\times 27}{4^4} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{16} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{4} \geq 4(1-y)(1+y)^3 = (f(x))^2 \Rightarrow f(x) \leq \sqrt{\dfrac{27}{4}} = \dfrac{3\sqrt{3}}{2} \Rightarrow f_{\text{max}} = \dfrac{3\sqrt{3}}{2}$.
$=$ occurs when $1-y = \dfrac{1+y}{3} \Rightarrow y = \dfrac{1}{2} \Rightarrow \sqrt{1-x^2} = \dfrac{1}{2} \Rightarrow x = \dfrac{\sqrt{3}}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1069438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was able to show using the following method that $c_n$ is bounded:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n} < \overbrace{\frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{2}}}{n}}^{n-1 \text{ times}} = \frac{\large {n - 1}}{n\sqrt{2}} < \frac{1}{\sqrt{2}}$$
So now we know that $\large {0 < c_n < \frac{1}{\sqrt{2}}}$.
I know from testing for large values of $n$ that $c_n \to 0$.
What's left is actually finding a way to show this.
Any hints?
|
We have $c_n>0$ and
$$\begin{align}c_{n+1}-c_n&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&<\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{n-1}{\sqrt{n}}\right]\\
&<0\end{align}$$
so $c_{n+1}<c_n$ and convergence follows from monotone convergence theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Compute the area of a parallelogram defined by a particular construction I got stuck with this mathematical task. Can someone help me how to solve this problem?
I need to find the F(area) value. It is kind of a thinking task
Context
The problem is extracted from a German book "Paul Eigenmann, Geometrische Wiederholungs- und Denkaufgaben. Ernst-Klett-Verlag, 1964."
I do not know where to start exactly. Can I assume that the base side is split on a half which gives us two sides with 25cm each?
There are no heights given. It is only given that what you see in the picture, the length of two parallel base lines $50$ and $57$.
|
If you don't want to rely on any geometric knowledge, or at least almost on none, then you can attack such a problem using coordinates.
$$
A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}50\\0\end{pmatrix}\qquad
C=\begin{pmatrix}57\\h\end{pmatrix}\qquad
D=\begin{pmatrix}7\\h\end{pmatrix}
$$
The circular arcs in the figure indicate $AD=AP=CQ=CB=r$ for some length $r$ which is the radius of these arcs. So you have coordinates
$$
P=\begin{pmatrix}r\\0\end{pmatrix}\qquad
Q=\begin{pmatrix}57-r\\h\end{pmatrix}
$$
and the condition
$$AD^2 = 7^2+h^2 = r^2\tag1$$
You also have the marked right angles. The slope of $DP$ is
$$\frac{7-r}{h}$$
while the slope of $AQ$ is
$$\frac{57-r}{h}$$
If they are to be orthogonal, the slopes must multiply to $-1$:
\begin{align*}
\frac{7-r}{h} \cdot \frac{57-r}{h} &= -1 \\
(7-r)(57-r) &= -h^2 \\
399 - 64r + r^2 + h^2 &= 0 \tag2
\end{align*}
Now subtract $(1)$ from $(2)$ and you obtain
\begin{align*}
399 - 64r + r^2 - 7^2 &= -r^2 \\
2r^2 - 64r + 350 &= 0 \\
r^2 - 32r + 175 &= 0 \\
r &= 16\pm 9 \\
r_1 &= 7 \quad r_2 = 25
\end{align*}
If you plug $r_1$ into $(1)$ you get $h=0$ which is not the solution you want. So take $r_2=25$ instead. This tells you that indeed the points $P$ and $Q$ are midpoints of their respective sides. Plugging them into $(1)$ you get
$$h = \sqrt{r_2^2-7^2} = \sqrt{25^2-7^2} = 24$$
I'll leave computing the area from this to you as an excercise.
|
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|
How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being between 1 and 2).
Any suggestion is welcomed.
|
Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$.
For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \frac{k}{n}}{1+0} = 1 + \dfrac{k}{n}$.
Therefore, $\displaystyle\dfrac{n}{n+1} \cdot \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right) \le \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$.
Now, use the fact that $\displaystyle\lim_{n \to \infty}\dfrac{n}{n+1} = 1$ and that $\displaystyle\dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$ is a Riemann sum for $\displaystyle\int_{0}^{1}(1+x)\,dx$ to get the result.
|
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|
3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?
|
I think using a projection into 2D might be the easiest way to actually calculate. If the points are $A, B, C$ then find
$$\begin{align}
\mathbf{u_1} & = B-A \\
\mathbf{w_1} &= (C-A) \times \mathbf{u_1} \\
\mathbf{u} & = \mathbf{u_1} / | \mathbf{u_1} | \\
\mathbf{w} & = \mathbf{w_1} / | \mathbf{w_1} | \\
\mathbf{v} & = \mathbf{w} \times \mathbf{u} \\
\end{align}$$
This gives three orthogonal unit vectors with $\mathbf{u}$ and $\mathbf{v}$ spanning the plane.
Get 2D coordinated by taking the dot products of $(B-A)$ and $(C-A)$ with $\mathbf{u}$ and $\mathbf{v}$. Let
$$\begin{align}
b & = (b_x,0) = ( (B-A) \cdot \mathbf{u} , 0 ) \\
c &= (c_x,c_y) = ( (C-A) \cdot \mathbf{u}, (C-A)\cdot \mathbf{v} ) \\
\end{align}$$
We know the center must lie on the line $x= b_x/2$. Let this point be $(b_x/2,h)$. The distance from c must be the same as the distance from the origin
$$(c_x-b_x/2)^2 + (c_y - h)^2 = (b_x/2)^2 + h^2$$
So
$$h = \frac{(c_x-b_x/2)^2 + c_y^2 - (b_x/2)^2}{ 2 c_y } $$
The actual center can then be recovered by taking $A + (b_x/2)\mathbf{u} + h \mathbf{v}$.
This is a nice explicit calculation.
|
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|
Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
|
The sequence given by:
$$ a_n = \left(b+\sqrt{b^2-4c}\right)^n + \left(b-\sqrt{b^2-4c}\right)^n $$
satisfies the recurrence relation:
$$ a_{n+2} = \color{red}{2}b\cdot a_{n+1} - \color{red}{4}c\cdot a_{n}.$$
Since $\nu_2(a_n)\geq n$ holds for $n=0$ and $n=1$, it holds for every $n$ by the previous relation.
Ok, this is still induction :D
|
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|
Finding $ \lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$ Finding
$$\lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$$
Suspecting that the limit is 0, but how do I show this? Was able to get an upper bound of $ \frac{1}{2\sqrt{e}} $ easily but it's not useful. Apparently this is a hard problem.
|
You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot \color{blue}2\cdot 3\cdot \color{blue}4 \cdot 5\cdot\color{blue}6\cdots(2n-1)\cdot \color{blue}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{blue}2}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 }\\
& =\frac{1}{\sqrt{\pi n}}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right), \quad \text{for} \, n \, \text{great}
\end{align}
$$
where we have used Stirling's approximation (approximating $(2n)!$ and $n!$), then you easily conclude for the limit (here you have more than the limit).
|
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|
How to express the equations as the Square root Like? $$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form that will be give us $\sqrt{2}$ like sin 45 degree?
|
Since, $\sin{\dfrac{\theta}{2}}=\sqrt{\dfrac{1-\cos\theta}{2}}$ and $\cos{\dfrac{\theta}
{2}}=\sqrt{\dfrac{1+\cos\theta}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\cos{\dfrac{\theta}{2}}}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\cos{\theta}}{2}}}{2}}$
Putting, $\theta=\dfrac{\pi}{4}$
$\sin{\dfrac{\pi}{16}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\dfrac{1}{\sqrt2}}{2}}}{2}}=\dfrac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2}$
Now, multiply both sides by $2$ to get the required equation.
$2\sin \dfrac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$
|
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|
Is there a lower bound for the following rational expression, in terms of $b^2$? In the following, $b$ and $r$ are both positive integers.
Is there a lower bound for the following rational expression, in terms of $b^2$?
$$L = \frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}$$
What I do know is that (a crude upper bound is) $L < b^2$. My question: Is it possible to get a lower bound for $L$ in terms of $b^2$ alone?
Edit: Here is an attempt to derive a (crude) lower bound for $L$, using no more than $r \geq 1$ and the AM-GM Inequality.
Since $r \geq 1$, the numerator of $L$ is
$$b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right) \geq 4b^4 + 4b^2 = 4{b^2}(b^2 + 1).$$
Now, for the denominator of $L$, we use the AM-GM Inequality to get
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \frac{\left(2^{r+1} + 2^r b^2\right)^2}{4}.$$
Simplifying, we obtain
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \left(2^r + 2^{r-1} b^2\right)^2 < 2^{2r}(b^2 + 1)^2.$$
Consequently, I finally have:
$$L > \frac{2^{2-2r}\cdot{b^2}}{b^2 + 1}$$
which is still expressed in terms of the extra variable $r$.
I was wondering if anybody out there has any bright ideas on how to eliminate $r$ from a (hopefully) nontrivial lower bound for $L$, in terms of $b^2$.
Thanks!
|
$$\frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}=\frac{2^{r+1}(2^r-1)b^4 + (3\cdot 2^r - 2)b^2}{\left(2^{r+1} - 1\right)2^r(b^2 + 2^{-r})}=\frac{b^4 + \frac{3\cdot 2^r - 2}{2^{r+1}(2^r-1)}b^2}{\left(1 +\frac{2^r}{2^r-1}\right)(\frac{1}{2}b^2 + 2^{-r-1})}\geq b^2\frac{b^2+3\cdot 2^{-r-1}- \frac{1}{2^{r}(2^r-1)}}{b^2+2^{-r}}\geq b^2\frac{b^2+2^{-r}+2^{-r-1}-2^{-2r+1}}{b^2+2^{-r}}\geq b^2$$
|
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|
Let $C$ be the curve of intersection of the plane $x+y-z=0$ with ellipsoid $\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$. Let $C$ be the curve of intersection of the plane $x+y-z=0$ and the ellipsoid $$\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$$ Find the points on $C$ which are farthest and nearest from the origin
When dealing with constraints I tried to consider the function $$F(x,y,z)=x^2+y^2+z^2-\lambda(x+y-z)-\mu\left(\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}-1\right)$$ However, I cannot solve this equation after differentiating respect to $x,y,z$ because it yields three equations with no common solution.
The system of equations are
$$2x=\lambda+\frac{\mu x}{2}$$
$$2y=\lambda+\frac{2\mu y}{5}$$
$$2z=-\lambda+\frac{2\mu z}{25}$$
How would I approach this problem, thanks.
|
You can simplify the problem by using $z=x+y$.
Then you minimize
$$x^2+y^2+(x+y)^2+\lambda\left(\frac{x^2}4+\frac{y^2}5+\frac{(x+y)^2}{25}-1\right).$$
by
$$\begin{cases}x+x+y+\lambda\left(\dfrac{x}4+\dfrac{x+y}{25}\right)=0,
\\y+x+y+\lambda\left(\dfrac{y}5+\dfrac{x+y}{25}\right)=0.\end{cases}$$
Eliminating $\lambda$, you end-up with
$$-\frac{xy}{10}+\frac{4y^2}{25}-\frac{21x^2}{100}=\frac{(7x+8y)(2y-3x)}{100}=0.$$
Now you intersect these two lines with the ellipse$$\frac{x^2}4+\frac{y^2}5+\frac{(x+y)^2}{25}-1=0$$ by eliminating $y$ and solving for $x$.
$$x=\pm\sqrt{\frac{800}{323}},y=-\frac78x,z=\frac18x,$$
or
$$x=\pm\sqrt{\frac{20}{19}},y=\frac32x,z=\frac52x.$$
The corresponding squared distances are
$$\frac{800}{323}\left(1+\frac{49}{64}+\frac1{64}\right)=\frac{75}{17}$$ and
$$\frac{20}{19}\left(1+\frac{9}{4}+\frac{25}{4}\right)=10.$$
|
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|
Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection? This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, not exceeding internal sum of 180°), for example: If one draws a right triangle with internal angles 90°, 60° and 30°, one can turn it in a way that the shadow triangle can be of angles 60°, 60°, 60° (equilateral) or 120°, 30° and 30° (isosceles), or any other triangle possibilities.
Is it true? How can we prove it? (if possible, showing not only that it's possible, but how to obtain the specific mapping that does it)
|
We'll take the special case of projection along the $z$-axis to a target triangle $\triangle A^\prime B^\prime C^\prime$ inscribed in the unit circle. Without loss of generality, we can take the triangle $\triangle ABC$ making the projection to be
$$A = (1,0,0) \qquad B = (\cos2C^\prime,\sin2C^\prime, p) \qquad C = (\cos2B^\prime, -\sin2B^\prime, q)$$
for some $p$ and $q$. (The counterpart $z$ coordinate of $A$ vanishes with an appropriate vertical translation.) In accordance with the Law of Sines, we have
$$\frac{|\overline{BC}|}{\sin A} = \frac{|\overline{CA}|}{\sin B} = \frac{|\overline{AB}|}{\sin C}$$
which gives a non-linear system in $p$ and $q$:
$$\begin{align}
p^2 \sin^2 A - ( p - q )^2 \sin^2 C + 4 \sin^2 A\sin^2C^\prime - 4 \sin^2A^\prime \sin^2 C &= 0 \\
q^2\sin^2 A - ( p - q )^2 \sin^2 B + 4 \sin^2 A \sin^2 B^\prime - 4 \sin^2 A^\prime \sin^2B &= 0
\end{align}$$
Eliminating $q$ from the system is straightforward enough, yielding a quadratic in $p^2$:
$$\begin{align}
0 \quad=\quad &p^4 \sin^2 A \sin^2 B \\
- &p^2 \left(\begin{array}{c}\left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2 \\
- 4 \sin^2A \sin^2B \sin^2C^\prime + 4 \sin^2A^\prime \sin^2B^\prime \sin^2 C\;) \end{array}\right) \\
- &4 \sin^2C^\prime \left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2
\end{align}$$
where
$$\alpha^{\pm} = A \pm A^\prime \qquad \beta^{\pm} = B \pm B^\prime \qquad \gamma^{\pm} = C\pm C^\prime$$
Because this quadratic has a non-negative leading term, and a non-positive constant term, Descartes' Rule of Signs guarantees that it has exactly one positive real root, so we can unambiguously write
$$p^2 = \frac{\left(\begin{array}{c}\left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2 \\
- 4 \sin^2A \sin^2B \sin^2C^\prime + 4 \sin^2A^\prime \sin^2B^\prime \sin^2 C\;) \end{array}\right) + \sqrt{\Delta}}{2\sin^2 A \sin^2 B}$$
where
$$\begin{align}
\Delta = \left(\;
\begin{array}{c}
\sin^2\alpha^{-} + \sin^2\beta^{-} + \sin^2\gamma^{-} \\
+ 2 \cos\alpha^{+} \sin\beta^{-} \sin\gamma^{-} \\
+ 2 \sin\alpha^{-} \cos\beta^{+} \sin\gamma^{-} \\
+ 2 \sin\alpha^{-} \sin\beta^{-} \cos\gamma^{+}
\end{array}
\;\right) \;\cdot \left(\;
\begin{array}{c}
\sin^2\alpha^{+} + \sin^2\beta^{+} + \sin^2\gamma^{+} \\
+ 2 \cos\alpha^{-} \sin\beta^{+} \sin\gamma^{+} \\
+ 2 \sin\alpha^{+} \cos\beta^{-} \sin\gamma^{+} \\
+ 2 \sin\alpha^{+} \sin\beta^{+} \cos\gamma^{-}
\end{array}
\;\right)
\end{align}$$
The corresponding equation for $q^2$ arises by exchanging the roles $B\leftrightarrow C$, $B^\prime \leftrightarrow C^\prime$, $\beta^{\pm}\leftrightarrow\gamma^{\pm}$. (Note that $\Delta$ remains unchanged.)
Determination of the signs of $p$ and $q$ themselves is left as an exercise to the reader.
|
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|
Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?
|
\begin{align}
\int_{-\infty}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx&=2\int_{0}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx\\[7pt]
&=2\int_{0}^{\infty} \frac{1}{\left(x-\frac{1}{x}\right)^2+3}\cdot\frac{\mathrm dx}{x^2}\\[7pt]
&=2\int_{0}^{\infty} \frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\tag1\\[7pt]
&=\int_{-\infty}^{\infty}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\\[7pt]
&=\int_{-\infty}^{0}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy+\int_{0}^{\infty}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\\[7pt]
&=\int_{-\infty}^{\infty}\frac{e^{z}}{\left(e^{z}-e^{-z}\right)^2+3}\,\mathrm dz+\int_{-\infty}^{\infty}\frac{e^{-z}}{\left(e^{-z}-e^{z}\right)^2+3}\,\mathrm dz\tag2\\[7pt]
&=\int_{-\infty}^{\infty}\frac{2\cosh z}{\left(2\sinh z\right)^2+3}\,\mathrm dz\tag3\\[7pt]
&=\int_{-\infty}^{\infty}\frac{1}{t^2+3}\,\mathrm dt\tag4\\[7pt]
&=\left.\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{\sqrt{3}}\right|_{-\infty}^{\infty}\\[7pt]
&=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{\sqrt{3}}}}\tag{$\color{red}{❤}$}
\end{align}
Explanation :
$(1)\;$ Use substitution $\;\displaystyle y=\frac{1}{x}\quad\implies\quad \mathrm dy=-\frac{\mathrm dx}{x^2}$
$(2)\;$ Use substitution $\;\displaystyle y=e^{z}\,$ for the left integral and $\;\displaystyle y=e^{-z}\,$ for the right integral
$(3)\;$ Adding both integrals then using the fact that $\;\displaystyle \cosh z=\frac{e^{z}+e^{-z}}{2}\,$ and $\;\displaystyle \sinh z=\frac{e^{z}-e^{-z}}{2}\,$
$(4)\;$ Use substitution $\;\displaystyle t=2\sinh z\quad\implies\quad \mathrm dt=2\cosh z\;\mathrm dz$
|
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|
limit of recursive function I've been struggling with this for a while now and can't seem to get it right.
Let $f(x)= \frac{x^2+3}{2x}$ from $0 $ to $\infty $. Show that $0\lt x \lt \sqrt{3}$ implies $\sqrt{3} \lt f(x)$
and show if we define $x_{n+1} = f(x_n)$ then
$$ \lim_{n\to\infty}x_n =\sqrt{3}$$ when $x_0 \gt 0$
It's easy to check that $f(\sqrt{3}) = \sqrt{3}$ but limits is not one of my strong points. I've also shown that $f(x) < 3/x$.
|
Note that $f(x)-\sqrt{3} = {(x-\sqrt{3})^2 \over 2x} \ge 0$. So we have $x_k \ge \sqrt{3}$ for $k \ge 1$.
Since $f(x)-\sqrt{3} = {x-\sqrt{3} \over 2x} (x-\sqrt{3})$, this suggests that we look at $\phi(x) = {x-\sqrt{3} \over 2x}$. It is straightforward to show that $\phi$ is strictly increasing for $x >0$, and $\lim_{x \to \infty} \phi(x) = {1 \over 2}$. Furthermore, $\phi(x) \ge 0$ for $x \ge \sqrt{3}$.
Hence we have $f(x) -\sqrt{3} \le {1 \over 2} (x-\sqrt{3})$ for $x \ge \sqrt{3}$.
In particular, this gives $0 \le x_{n+1}-\sqrt{3} \le {1 \over 2} (x_n-\sqrt{3})$ for $n \ge 1$ and so
$x_n-\sqrt{3} \le {1 \over 2^{k-1}} (x_1-\sqrt{3})$.
|
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|
Show that $(\ln a)^k \neq k \ln a $ I have a question that I am not sure how to answer:
Show that $(\ln a)^k \neq k \ln a $
|
I will show that
if $k > 1$
then $a \le e$.
In other words,
if $a > e$
there is no $k \ge 1$
that satisfies this.
(This is just playing around
with algebra and
elementary calculus.)
If
$(\log a)^k=k\log a$,
then
$(\log a)^{k-1}=k$
or
$\log a
=k^{1/(k-1)}
$.
Therefore,
for any given $k$,
there is at most one $a$.
Let
$f(k) = k^{1/(k-1)}$
and
$g(k) = \ln f(k) =\frac{\ln k}{k-1}$.
$g'(k)
=\frac{(k-1)/k-(\ln k)}{(k-1)^2}
=\frac{1-1/k-\ln k}{(k-1)^2}
$.
For small $c$,
$g'(1+c)
=\frac{1-1/(1+c)-\ln (1+c)}{c^2}
\approx \frac{1-(1-c+c^2)-(c-c^2/2)}{c^2}
= \frac{(c-c^2)-c+c^2/2}{c^2}
= \frac{(c-c^2)-c+c^2/2}{c^2}
=-\frac12
$.
Similarly,
for small $c$,
$g(1+c)
=\frac{\ln (1+c)}{c}
\approx \frac{c-c^2/2}{c}
=1-c/2
$
so $g(1) = 1$
and
$f(1) = e$.
If $h(k)
=1-1/k-\ln k
$,
the numerator of $g(k)$,
then
$h'(k)
=1/k^2-1/k
=\frac{1-k}{k^2}
< 0
$
for $k > 1$.
Therefore,
$g$, and therefore $f$
is decreasing for $k \ge 1$,
so $a$ can be at most
$e$ for
$k \ge 1$.
|
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}
|
Proving of $\cos (\frac{2}{3})>\frac{\pi }{4}$ How can I prove that $$\cos \left(\frac{2}{3}\right)>\frac{\pi }{4}$$
|
Use Carlson inequality
$$\arccos{x}>\dfrac{6\sqrt{1-x}}{2\sqrt{2}+\sqrt{1+x}},0\le x<1$$
then let
$x=\frac{\pi}{4}$,then we have
$$\arccos{\dfrac{\pi}{4}}>\dfrac{6\sqrt{1-\dfrac{\pi}{4}}}{2\sqrt{2}+\sqrt{1+\frac{\pi}{4}}}\approx 0.66741060\cdots>\dfrac{2}{3}$$
see wolf
so
$$\cos{\dfrac{2}{3}}>\dfrac{\pi}{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve for $y$: $x^2-3xy+y^2-x+4=0$ Please, I want to know the method to arrive at the solution $y=\dfrac{-3x \pm \sqrt{5x^2+4x-16}}{2}$ , step by step.
Thank you in advance.
|
$$\text{Hint: }$$
$$
\\x^2 - 3xy + y^2 - x + 4 = 0 \\ \Updownarrow \\ y^2 + (-3x)y +(x^2 -x+4) = 0
$$
$$$$
$$\text{Now solve the quadratic equation with: }$$
$$\begin{align}
a &= 1 \\ b &= -3x \\ c &= (x^2 - x + 4)
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Nasty Limit of sum at infinity $$\lim_{n \to \infty} \left (\sum_{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) $$
How would I even begin to approach this??
Edit:
So the hints helped... I ended up with $\space\ln\left(\frac{b}{a}\right)$
|
$$\begin{align}
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{a \left[ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \\ =
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \\= &\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{\left(\frac{b}{a}\right)^{\frac{i}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}} - 1 \right) \\ =&\lim_{n \to \infty} \sum_{i=1}^n \left[ \left(\frac{b}{a}\right)^{\frac1n} - 1 \right] \\ =&\lim_{n \to \infty} n\left(\frac{b}{a}\right)^{\frac{1}{n}} - n \\ =&\lim_{n \to \infty} \frac{\left(\frac{b}{a}\right)^\frac{1}{n} - 1}{\frac{1}{n}} \\ =&\lim_{n \to \infty} \frac{\ln\left(\frac{b}{a}\right)\cdot \frac{-1}{n^2} \cdot \left(\frac{b}{a}\right)^\frac{1}{n}}{\frac{-1}{n^2}} \\= &\lim_{n \to \infty} \ln\left(\frac{b}{a}\right) \cdot \left(\frac{b}{a}\right)^{\frac{1}{n}} \\ =& \space\ln\left(\frac{b}{a}\right)
\end{align}$$
|
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|
Need hint for $\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}$ Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it.
$$
\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}
$$
|
Easiest way is by series expansion:
$$\frac{1}{x}\log(1+x) = \frac{1}{x} \left( x - \frac{x^2}{2} + O(x^3) \right) =
\left( 1 - \frac{x}{2} + O(x^2) \right)
$$
$$
(x+1)^{1/x} = e^{\frac{1}{x}\log(1+x) } = e^{\left( 1 - \frac{x}{2} + O(x^2) \right)}
= \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!}
$$
$$
(x+1)^{1/x}-e = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!} - \sum_{n=0}^\infty \frac{1}{n!} = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n-1}{n!}
$$
$$
\frac{(x+1)^{1/x}-e}{x} = \sum_{n=0}^\infty \frac{\left( 1 - n\frac{x}{2} + O(x^2) \right)-1}{xn!} = \sum_{n=0}^\infty \frac{ - n\frac{x}{2} + O(x^2) }{xn!}
$$
$$
\frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n + O(x)}{n!} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n }{n!} +O(x)
$$
$$
\frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=1}^\infty \frac{n}{n!} +O(x) =
-\frac{1}{2} \sum_{m=0}^\infty \frac{1}{(m+1)!} +O(x) = -\frac{e}{2} + O(x)
$$
from which it follows that
$$
\lim_{x\rightarrow \infty}\frac{(x+1)^{1/x}-e}{x} = -\frac{e}{2} $$
|
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|
Integer root of a quadratic Determine the sum of all (distinct) positive integers $ n$ , such that for some integer $a$,
$$ n^2 -an + 6a = 0. $$
|
The equation is equivalent to
$$
a(n-6) = n^2.
$$
So we are looking for all $n$ such that $n-6$ divides $n^2$.
Polynomial long division of $n^2$ by $n-6$ gives $n^2 = (n-6)(n+6) + 36$, so equivalently, $n-6$ divides $36$.
Enumeration of the divisors of $36$ yields
$$
n-6\in\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18,\pm 36\}.
$$
Since $n$ is required to be positive, the answer is
$$
n\in\{2,3,4,5,7,8,9,10,12,15,18,24,42\}.
$$
|
{
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|
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.
$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!
|
If we solve $C=m^2+n^2$ for $n$, we get $n=\sqrt{C-m^2}$. We can then try values of $m$ to see which one(s) yield a positive integer for $n$, where
$$\text{To ensure }n<m:\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor:\text{ to ensure }n\in\mathbb{N}$$
$$\text{In the example of }65\qquad m_{min}=\lceil\sqrt{32.5}\space\rceil = 6\qquad m_{max}=\lfloor\sqrt{65}\rfloor = 8$$
If we test $n=\sqrt{C-m^2}$ using $m=6,7\text{, and }8$, we find positive integers for $n$ where $m=7\Rightarrow n=4$ and $m=8\Rightarrow n=1.$
$$\text{Using Euclid's formula }f(7,4)=(33,56,65)\quad f(8,1)=(63,16,65)$$
|
{
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|
Quadric and tangents planes Let $Q$ be the quadratic $x^2 + 4xy - 2y^2 + 6z^2 + 2y +2z = 0$
*
*Prove that $Q$ is a cone and find its vertex.
*Write the tangent plane $A$ to the cone in $(0,0,0)$ and say which kind of conic is the intersection of $Q$ with $A$.
*Find an ellipse and a parabola on the cone.
Am a bit ok for 1 and 2 but require some help for 3. For 1 we need the matrix representation of the cone and the 2 differentiate with respect to $x$ and then plug in $(0,0,0)$.
Anybody with any different ideas...
|
*
*With $A=\begin{pmatrix}1&2&0\\2&-2&0\\0&0&6\end{pmatrix}$ and $K=\begin{pmatrix}0&2&2\end{pmatrix}$, the equation is
${\bf x}^TA{\bf x}+K{\bf x}=0$. $A$ has eigenvalues $2,-3,6$ and the eigenvector matrix is $P=\begin{pmatrix}\frac{2\sqrt{5}}{5}&-\frac{\sqrt{5}}{5}&0\\\frac{\sqrt{5}}{5}&\frac{2\sqrt{5}}{5}&0\\0&0&1\end{pmatrix}$.
${\bf x}=P{\bf x}'$ gives ${\bf x}'^TP^TAP{\bf x}'+KP{\bf x}'=0$ where we complete the squares and that $x''=x'+\frac{\sqrt{5}}{10}, y''=y'-\frac{2\sqrt{5}}{15}, z''=z'+\frac{1}{6}$ transforms the equation into $2x''^2-3y''^2+6z''^2=0$. The vertex should be $P\begin{pmatrix}-\frac{\sqrt{5}}{10}\\\frac{2\sqrt{5}}{15}\\ -\frac{1}{6}\end{pmatrix}=\begin{pmatrix}-\frac{1}{3}\\\frac{1}{6}\\ -\frac{1}{6}\end{pmatrix}$.
*To get the tangent plane at $(0,0,0)$ we find the gradient which is $\begin{pmatrix}0&2&2\end{pmatrix}$ and is a normal vector. $2y+2z=0$. Intersecting is considering the two equations a system, so we put $z=-y$ into the first equation and get: $(x+2y)^2$ a double line.
*To get an ellipse we need a plane which only intersects one nappe of the cone. $y=0$ should do. Indeed putting it into the equation we get $6x^2+36(z+\frac{1}{6})^2=1$.
To get a parabola we need a plane parallel to a plane which is tangential to the cone's surface. We found one in 2. So 2y+2z=2 should do.
|
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|
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
c &
\end{array} ; 1\right] = \dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$
I give negative parameters to the $\Gamma$ function.
What other identity can I use?
I'm trying to find a closed form to this: $\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
Wolfram Mathematica answered this as a closed form: $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
But I would like to have a manual solution with proof.
This is how I got that hypergeometric series:
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
$\dfrac{t_{i+1}}{t_{i}} = \frac{\binom{a+i+1-1}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \frac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \frac{(a+i) (i-n)}{ (i-a-n+1)(i+1)}$
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
UPDATE
Thanks to David H, I got closer to the solution.
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
$\lim\limits_{\epsilon \to0} \frac{\Gamma (-2 a-2 \epsilon +1) \Gamma (-a-n-\epsilon +1)}{\Gamma (-a-\epsilon +1) \Gamma (-2 a-n-2 \epsilon +1)} = \frac{4^{-a} \Gamma \left(\frac{1}{2}-a\right) \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
As you can see this result is close to the expected $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$ formula. But the $\binom{a+n-1}{n}$ factor is still missing and I don't really understand why.
|
Using this identity to express the hypergeometric function as a Gegenbauer function, and this identity which gives the value of the Gegenbauer function evaluated at $1$, the hypergeometric function in question may then be expressed as a ratio of gamma functions whose arguments are each positive integers. These can be reorganized into binomial terms for a compact final expression:
$$\begin{align}
{_2F_1}{\left(a,-n;1-a-n;1\right)}
&=\frac{n!}{(a)_{n}}C_{n}^{a}{\left(1\right)}\\
&=\frac{\Gamma{\left(n+1\right)}\,\Gamma{\left(a\right)}}{\Gamma{\left(a+n\right)}}\cdot\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\
&=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(2a+n\right)}}{\Gamma{\left(a+n\right)}\,\Gamma{\left(2a\right)}}\\
&=\frac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}.\\
\end{align}$$
There's likely a much more direct way to derive this without this absurd detour into exotic special functions, but maybe this response will tide you over until someone more knowledgeable in combinatorial comes around. =)
Edit:
Here's a much nicer way of evaluating the sum using beta function machinery.
$$\begin{align}
s{(a,n)}
&=\sum_{k=0}^{n}\binom{a+k-1}{k}\binom{a+n-k-1}{n-k}\\
&=\sum_{k=0}^{n}\frac{\Gamma{\left(a+k\right)}}{\Gamma{\left(k+1\right)}\,\Gamma{\left(a\right)}}\cdot\frac{\Gamma{\left(a+n-k\right)}}{\Gamma{\left(n-k+1\right)}\,\Gamma{\left(a\right)}}\\
&=\frac{1}{\Gamma{\left(a\right)}^2}\sum_{k=0}^{n}\frac{\Gamma{\left(a+k\right)}}{\Gamma{\left(k+1\right)}}\cdot\frac{\Gamma{\left(a+n-k\right)}}{\Gamma{\left(n-k+1\right)}}\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\frac{\Gamma{\left(n+1\right)}}{\Gamma{\left(k+1\right)}\,\Gamma{\left(n-k+1\right)}}\cdot\frac{\Gamma{\left(a+k\right)}\,\Gamma{\left(a+n-k\right)}}{\Gamma{\left(2a+n\right)}}\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\binom{n}{k}\cdot\operatorname{B}{\left(a+k,a+n-k\right)}\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{1}t^{a+k-1}(1-t)^{a+n-k-1}\,\mathrm{d}t\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\int_{0}^{1}t^{a-1}(1-t)^{a-1}\sum_{k=0}^{n}\binom{n}{k}t^{k}(1-t)^{n-k}\,\mathrm{d}t\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\int_{0}^{1}t^{a-1}(1-t)^{a-1}\cdot1\,\mathrm{d}t\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\operatorname{B}{\left(a,a\right)}\\
&=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\
&=\binom{2a+n-1}{2a-1}.~~\blacksquare\\
\end{align}$$
|
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|
Mathematical induction involving inequalities and congruences I have the following two problems:
"Prove each of the following statements by induction for all positive integers $n$:"
*
*$2\cdot7^n \equiv 2^n\cdot(2+5n) \bmod 25 \quad$ <-- I have been going at this question for a couple of hours and can't seem to come up with an answer.
*$\frac{(2n)!}{(n!)^2} \leq 4^n \quad$ <-- This one I have an answer to, but I am curious if it's valid!
I'll attach a picture to show my work, but I would really like some help with question 1; I'm having a lot of trouble with that one.
|
$\underline{\text{Problem 1:}}$ For each positive integer $n$, let $M(n)$ be the statement that
$$
2\cdot 7^n \equiv 2^n\cdot(2+5n)\pmod{25}.
$$
Base step: $M(1)$ says that $14\equiv 14 \pmod{25}$, which is true.
Inductive step: Fix $k\geq 1$ and assume that
$$
M(k):\; 2\cdot 7^k \equiv 2^k\cdot(2+5k)\pmod{25}
$$
holds. It remains to show that
$$
M(k+1):\; 2\cdot 7^{k+1} \equiv 2^{k+1}\cdot(2+5(k+1))\pmod{25}
$$
follows. One can show this using just modular arithmetic; however, being a bit more pedantic, a few more steps may be added for clarity. Since $M(k)$ holds, we have the following:
$$
2\cdot 7^k \equiv 2^k\cdot(2+5k)\pmod{25} \Longleftrightarrow 2\cdot 7^k=2^k\cdot(2+5k) + 25\ell,
$$
where $\ell\in\mathbb{Z}$. Thus, we have the following:
\begin{align}
2\cdot 7^{k+1} &= (2\cdot 7^k)\cdot 7\\
&= (2^k\cdot(2+5k)+25\ell)\cdot 7\\
&= 7\cdot 2^k(2+5k)+7\cdot25\ell\\
&= 7\cdot 2^{k+1}+35k\cdot2^k+7\cdot 25\ell\\
&= 7\cdot 2^{k+1}+(25+2\cdot 5)k\cdot2^k+7\cdot 25\ell\\
&= 7\cdot 2^{k+1}+25k\cdot 2^k+5k\cdot2^{k+1}+7\cdot 25\ell\\
&= 2^{k+1}\cdot(7+5k)+25(7\ell+k\cdot2^k)\\
&\equiv 2^{k+1}\cdot(7+5k) \pmod{25}\\
&\equiv 2^{k+1}\cdot(2+5(k+1)) \pmod{25},
\end{align}
as desired, completing the proof of $M(k+1)$, and hence the inductive step. By mathematical induction, for each $n\geq 1$, the statement $M(n)$ holds.
$\underline{\text{Problem 2:}}$ Start by noting that we can use a strict inequality (i.e., $<$) here rather than $\leq$. Also note that
$$
\frac{(2n)!}{(n!)^2} < 4^n \Longleftrightarrow (2n)!<2^{2n}(n!)^2.
$$
There is nothing special about the equivalent expression on the right-hand side, but I find it to be a more natural expression to deal with; thus, I am going to prove this second problem with that expression in mind.
With that in mind, for each positive integer $n$, let $M(n)$ be the statement that
$$
(2n)!<2^{2n}(n!)^2.
$$
Base step: $M(1)$ says that $2<4$, which is true.
Inductive step: Fix $k\geq 1$ and assume that
$$
M(k):\;(2k)!<2^{2k}(k!)^2
$$
holds. It remains to show that
$$
M(k+1):\; [2(k+1)]!<2^{2(k+1)}[(k+1)!]^2
$$
follows. Beginning with the left side of $M(k+1)$,
\begin{align}
[2(k+1)]! &= (2k+2)!\\
&= (2k+2)(2k+1)(2k!)\\
&< (2k+2)(2k+1)[2^{2k}(k!)^2]\tag{ind. hyp.}\\
&< (2k+2)^2[2^{2k}(k!)^2]\tag{$2k+1 < 2k+2$}\\
&= [2(k+1)]^2\cdot[2^{2k}(k!)^2]\\
&= 2^2\cdot (k+1)^2\cdot 2^{2k}\cdot (k!)^2\\
&= 2^{2(k+1)}[(k+1)!]^2,
\end{align}
which is the right side of $M(k+1)$. By mathematical induction, for each $n\geq 1$, the statement $M(n)$ holds.
|
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|
How to integrate $\int \frac{dx}{\sqrt{ax^2-b}}$ So my problem is to integrate
$$\int \frac{dx}{\sqrt{ax^2-b}},$$
where $a,b$ are positive constants. What rule should I use here? Should substitution be used or trigonometric integrals?
The solution should be:
$$\frac{\log\left(\sqrt{a}\sqrt{ax^2-b}+ax\right)}{\sqrt{a}}+C$$
Thank you for any help =)
|
The question says "Should substitution be used or trigonometric integrals?". But a trigonometric substitution is a substitution, not an alternative to substitution.
$$
\int \frac{dx}{\sqrt{ax^2-b}} = \int\frac{dx/\sqrt{b}}{\sqrt{\frac a b x^2 - 1}} = \int \frac{\sec\theta\tan\theta\,d\theta/\sqrt{a}}{\sqrt{\sec^2\theta - 1}}
$$
Then use the fact that $\sec^2\theta-1=\tan^2\theta$. The substitution is $\sec\theta=x\sqrt{\frac a b}$ so that $\sec\theta\tan\theta\,d\theta=dx\sqrt{\frac a b}$ and thus $\sec\theta\tan\theta\,d\theta/\sqrt{a} = dx/\sqrt{b}$.
$$
\begin{array}{cc|c|c}
& \text{When you have} & \text{then use} & \text{so that} \\
\hline
& (\text{variable})^2 + \text{positive constant} & \text{variable}=\tan\theta & \tan^2\theta+1\text{ becomes }\sec^2\theta \\
& (\text{variable})^2 - \text{positive constant} & \text{variable}=\sec\theta & \sec^2\theta-1\text{ becomes }\tan^2\theta \\
& \text{positive constant} - (\text{variable})^2 & \text{variable}=\sin\theta & 1-\sin^2\theta\text{ becomes }\cos^2\theta \\
\end{array}
$$
In the last one, $\cos\theta$ rather than $\sin\theta$ will also work.
|
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|
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
|
$x^3-1 = (x-1)(x^2+x+1) = 0$, so $x^3=1$. That's all you need.
|
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"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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|
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?
Thanks.
|
i will use the maclaurin expansion for $\sqrt{1+x}, \ln(1+x), \tan(x)$
$\begin{align}
\ln[x + (1 + x^2)^{1/2}] &= \ln[x + 1 + \frac{1}{2}x^2 -\frac{1}{8}x^4 + \cdots]\\
&=\ln(1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots) \\
&=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{1}{2} \{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^2 \\
&+ \frac{1}{3}\{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^3 \cdots\\
&=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{x^2}{2} (1 + x + \cdots) + \frac{1}{3}(x^3 + \cdots) +\cdots\\
&=x-\frac{1}{2}x^3 + \frac{1}{3}x^3 + \cdots\\
&= x-\frac{1}{6}x^3 + \cdots
\end{align}$
the expansion for
$$\tan x = x + \cdots$$ putting the two together
$$\lim_{x \to 0}\dfrac{\ln[x + (1 + x^2)^{1/2}] - x}{\tan^3 x} = -\dfrac{1}{6}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1113338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Commutative matrices Knowing that $AB=BA$, find the matrices that commute with the matrix
\begin{pmatrix}
1 & 2 \\
3 & 4 \\
\end{pmatrix}
I have assumed that multiplying matrix
$\begin{pmatrix}
a & b \\
c & d \end{pmatrix}
$ by the first one should be equal to multiplying the first one by $\begin{pmatrix}
a & b \\
c & d \end{pmatrix}
$
Alright:
$\begin{pmatrix}
a+2c & b+2d\\
3a+4c & 3b+4d\end{pmatrix}
$
equals
$\begin{pmatrix}
a+3b & 2a+4b\\
c+3d & 2c+4d\end{pmatrix}
$
so:
$\begin{pmatrix}
a+2c = a+3b => 2c = 3b = 0\\
3a+4c = c +3d => a+c - d=0\\
b+2d = 2a+ 4b => -2a - 3b + 2d = 0\\
3b + 4d = 2c + 4d => 3b - 2c =0\end{pmatrix}
$
I was left with equations from which I have formed another matrix and used Gauss algorithm to evaluate it:
$\begin{pmatrix}
1 & 0 & 1 & -1 & 0 \\
-2& -3 &0 &2& 0\\
0 &3 &-2& 0 &0\\
0 &-3& 2& 0 &0\end{pmatrix}
$
I made some changes (r3:=r3+r4; r4:=r4+r3; r2:=r2+2r1) and the upper diagonal matrix is:
$\begin{pmatrix}
1 & 0 & 1 & -1 & 0 \\
0& -3 &2 &0& 0\\
0 &0 &0& 0 &0\\
0 &0& 0& 0 &0\end{pmatrix}
$
I was left with a set of equations
\begin{eqnarray}
a+ c + d &=& 0 \\
-3b + 2c &=& 0
\end{eqnarray}
and now I'm stuck. Any help would be highly appreciated!
EDIT: A is the given matrix.
|
You have figured it out. The equations you end up with:
2c=3b
and
a+c=d
are correct. Basically this means that you can specify any TWO values of the matrix arbitrarily and then the remaining two are fixed by the above equations. For example, suppose you want to specify a and b. The, rewrite the equations like this:
c=3b/2
and
d=a+3b/2
Given, for example, a=2.345 and b=3.14 then
c=1.5*3.14
and
d=2.345+1.5*3.14
Get it?
|
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"url": "https://math.stackexchange.com/questions/1113699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Chance of winning simple dice game Tossing two fair dice, if the sum is 7 or 11, then I win; if the sum is 2, 3 or 12, then I lose; if the sum is one of rest of numbers then I toss the two dices again. What is probability of winning?
What I have tried: every tossing, p(win)=$\frac{2}{9}$, and p(lose)=$\frac{1}{9}$, p(tossing again)=$\frac{2}{3}$, so the probability of winning is $\frac{2}{9}+\frac{2}{3}\frac{2}{9}+(\frac{2}{3})^2\frac{2}{9}+(\frac{2}{3})^3\frac{2}{9}+.....=\frac{2}{9}\sum^{\infty}_{n=0}(\frac{2}{3})^n=\frac{2}{3}$
Does anyone could tell me why my idea is wrong? What is the correct solution? Thanks!
|
With your edited solution i think you are now correct with 2/3.
Since $$\frac{\frac{2}{9}}{\frac{2}{9}+\frac{1}{9}}=\frac{2}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1118352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
|
Given the symmetry of the problem, wlog $x\ge y$. $x=y=0$ is a solution, from now on let us assume that $x>0$.
Let $f$ be the real-valued map $w\mapsto w+\sqrt{w^2+1}$. If $w>0$ then $f(w)>1$, which is enough to prove that $x>0>y$. Even better (without deriving it), $f$ is strictly increasing on the positive reals.
Define $z=-y>0$. Then $f(x)f(y)=1$ is equivalent to (just rationalize it)
$$f(x)=f(z),$$
which implies that $x=z$. This is equivalent to $x+y=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
}
|
How to integrate $(x^2 - y^2) / (x^2 + y^2)^2$ How do I integrate
$$\int \int \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dx dy?$$
The WolframAlpha page gives
$$
c_1 + c_2 + \tan^{-1}(x/y).
$$
And I kind of specifically need
$$
\int_{0}^{x} \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dy.
$$
Note
*
*I want to know integration technique to solve this without using $F' = f$.
*For the double integral above, what I'm interested is Lebesgue integral, but I guess what Wolfram gave is in the Riemann sense.
|
The main integral
$$I = \int\frac{x^2-y^2}{(x^2+y^2)^2}\,dx$$
can be solved using a tan substitution.
\begin{align*}
x &= y\tan\theta \\
dx &= y\sec^2\theta\,d\theta
\end{align*}
So that
\begin{align*}
I &= \int \frac{y^2(\tan^2\theta -1)}{y^4\sec^4\theta}\,y\sec^2\theta\,d\theta \\
&= \frac{1}{y}\int \frac{\tan^2\theta-1}{\sec^2\theta}\,d\theta \\
&= \frac{1}{y}\int \sin^2\theta - \cos^2\theta\,d\theta \\
&= - \frac{1}{y}\int \cos 2\theta\,d\theta \\
&= - \frac{1}{y}\frac{1}{2}\sin 2\theta + h(y) \\
&= - \frac{1}{2y} \frac{2\tan\theta}{1+\tan^2\theta} + h(y) \\
&= - \frac{1}{2y} \frac{2 (y/x)}{1+(y/x)^2} + h(y) \\
&= - \frac{x}{x^2+y^2} + h(y).
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1119784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Show that if $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$. Show that if $m$ and $n$ are integers such that $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$.
I am not sure where to begin.
|
Note that $n^2$ is even if and only if $n$ is even.
Since $m^2+n^2$ is even, you see that either both $m,n$ are even, or both are odd.
If both $m$, $n$ are even, then $mn$ is divisible by $4$.
If both $m$, $n$ are odd, you have $m=2k+1$, $n=2l+1$ so $m^2+n^2=4k^2+4k+4l^2+4l+2$ is not divisible by $4$. This gives the result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1120919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Proving that if $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.
Let $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.
My attempt:
If one or two numbers of $a,b,c$ are even then we're done, so we'll have to show that at least one of them is even.
Suppose $a,b$ are odd, then $a^2,b^2$ are odd, so $a^2+b^2$ must be even. So $c^2$ is even then $c$ is even. So $abc$ is even.
Is this enough to prove it?
|
Notice that a square of an integer is either $0$ or $1$ modulo $4$ therefore. If $$c^2\equiv0\mod{4}\Rightarrow c\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
If $$c^2\equiv1\mod{4}\Rightarrow a^2\equiv0, b^2\equiv1\mod{4}\Rightarrow a\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
or
$$a^2\equiv1, b^2\equiv0\mod{4}\Rightarrow b\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1121637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.