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Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?
|
Quite simply, $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int (x^2 + 1)^7 (x^2)(2x) \, dx,$$ and with the substitution $u = x^2 + 1$, $du = 2x \, dx$, we readily obtain $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int u^7 (u-1) \, du = \frac{1}{2} \int u^8 - u^7 \, du = \frac{1}{2} \left(\frac{u^9}{9} - \frac{u^8}{8}\right) + C = \frac{(x^2+1)^8}{18} \left( x^2 - \frac{1}{8} \right) + C.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Simplify $\tan^{-1}\Big[\frac{\cos x-\sin x}{\cos x+\sin x}\Big]$, where $x<\pi$ Write in the simplist form
$$
\tan^{-1}\bigg[\frac{\cos x-\sin x}{\cos x+\sin x}\bigg],\quad x<\pi
$$
My Attempt:
$$
x<\pi\implies -x>-\pi\implies \frac{\pi}{4}-x>\frac{-3\pi}{4}
$$
$$
\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-\sin^2 x}{(\cos x+\sin x)^2}=\frac{\cos 2x}{1+2\sin x\cos x}=\frac{\cos 2x}{1+\sin 2x}\\=\frac{\sin(\tfrac{\pi}{2}-2x)}{1+\cos(\tfrac{\pi}{2}-2x)}=\frac{2\sin(\tfrac{\pi}{4}-x)\cos(\tfrac{\pi}{4}-x)}{2\cos^2(\tfrac{\pi}{4}-x)}={\tan(\tfrac{\pi}{4}-x)}\\
\implies {\tan(\tfrac{\pi}{4}-x)}=\tan\big[\tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}\big]\\
\implies \tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}=n\pi+\frac{\pi}{4}-x
$$
If $-\pi/2<\frac{\pi}{4}-x<\pi/2$,
$$
\tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\pi}{4}-x
$$
this is what is given as answer.
But, since we have further restricted the domain to $\frac{-\pi}{2}<\frac{\pi}{4}-x<\frac{\pi}{2}\implies\frac{-\pi}{4}<x<\frac{3\pi}{4}$ from $x<\pi$, how can it be same as the original function ?
|
Indeed the given answer seems strange, as the usual range of the $\tan^{-1}$ function is $\left(-\frac\pi2,\frac\pi2\right)$, and if $\frac{3\pi}4 \leq x < \pi$ then $\frac\pi4-x$ is not in this range. I think you have the domain right.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Manipulating a functional equation
Consider a function $f$ such that $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ then find $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$$
And the options are as follows
A) $2f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
B) $2f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$
C) $f\left(\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )- f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
D) $f(3)+f\left (\frac {1}{3}\right )$
Through some algebraic manipulations I have reached until the following equations
1) $f(y)\,[ f(y^2)- 1]\, =f(y^3)$
2) $f(1)=0$ or $f(1)=2$
3) $ (f(y)\,)^2- f(1)= f(y^2)$
Also the question can be simplified as to finding $f((\sqrt 2-1)^3)-f((\sqrt 2+1)^3)$
|
By the properties of multiplication from this equation: $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ we obviously have: $$f\left (\frac {x}{y}\right )=f\left (\frac {y}{x}\right )$$ which means that for any $x$: $$f\left ( x \right )=f\left (\frac {1}{x}\right )$$
Since the arguments in:
$$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )
$$
are reciprocal of each other, the quantity in question is 0, which makes C) the correct answer
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why are we allowed to cancel fractions in limits? For example:
$$\lim_{x\to 1} \frac{x^4-1}{x-1}$$
We could expand and simplify like so:
$$\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} = \lim_{x\to 1} (x^3 + x^2 + x + 1) = (1^3 + 1^2 + 1^1 + 1) = 4$$
In this case we divided out $x-1$ on top and bottom even though technically, at $x=1$, we have $\frac{0}{0}$ that we're just tossing aside.
But what allows us to do this?
|
Algebraic Limit Theorem: Let the limits exist:
$$\lim_\limits{x\to a} f(x)=L \quad \text{and} \quad \lim_\limits{x\to a} g(x)=M.$$
Then:
$$\begin{align}&1) \ \lim_\limits{x\to a} (f(x)\pm g(x))=\lim_\limits{x\to a} f(x)\pm \lim_\limits{x\to a} g(x)=L\pm M;\\
&2) \ \lim_\limits{x\to a} (f(x)\cdot g(x))=\lim_\limits{x\to a} f(x)\cdot \lim_\limits{x\to a} g(x)=L\cdot M;\\
&3) \ \lim_\limits{x\to a} (f(x)/ g(x))=\lim_\limits{x\to a} f(x)/ \lim_\limits{x\to a} g(x)=L/M; \quad (\text{provided:} \lim_\limits{x\to a} g(x)=M\ne 0). \\ \end{align}$$
Note that:
$$\begin{align}\lim_{x\to 1} \frac{x-1}{x-1} = \lim_{x\to 1} 1&=1;\\
\lim_{x\to 1} (x^3 + x^2 + x + 1) &= 4;\\
\lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} &= \\
\lim_{x\to 1} \frac{x-1}{x-1}\cdot \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 1\cdot 4=4.\end{align}$$
See also: Limit Theorems.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Another summation identity with binomial coefficients Recently I have encountered on this page a rather nice identity:
$$
\sum_{k=0}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1},
$$
which however is valid only for $n\ge m$. This motivated me to try finding out a more symmetric expression with the aim to get $\frac{1}{2n+2m+1}$ on the RHS. Simple negation of $m$ did not help until the expression $\binom{2n}{2k}$ in the denominator was replaced with that one involving $m$:
$$
\sum_{k=0}^{n}(-4)^k\frac{\binom{n}{k}\binom{m+k}{k}}{\binom{2m+2k+1}{2k}\binom{2k}{k}}=\frac{2m+1}{2n+2m+1}.
$$
After swapping $n$ and $m$ the aim was achieved:
$$
\sum_{k=0}^{\infty}\frac{(-4)^k}{\binom{2k}{k}}\left[\frac{\binom{n}{k}\binom{m+k}{k}}{\binom{2m+2k+1}{2k}}+\frac{\binom{n+k}{k}\binom{m}{k}}{\binom{2n+2k+1}{2k}}\right]=1+\frac{1}{2n+2m+1}.
$$
But I am still not quite satisfied. Is there a suitable decomposition of 1 which could help to simplify the expression? Or maybe some different approach can better clarify the origin of the identity? I would appreciate any hint.
|
We prove a generalisation of OPs two binomial identities. The first being
\begin{align*}
\sum_{k=0}^m4^k\frac{\binom{\color{blue}{n}}{k}\binom{m}{k}}{\binom{\color{blue}{2n}}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1}\tag{1}
\end{align*}
In the second identity we exchange the variables $m$ and $n$ to get the identity more close to (1) and we also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$. We obtain
\begin{align*}
\sum_{k=0}^{m}&(-4)^k\frac{\binom{n+k}{k}\binom{m}{k}}{\binom{2n+2k+1}{2k}\binom{2k}{k}}
=\sum_{k=0}^m4^k\frac{\binom{\color{blue}{-n-1}}{k}\binom{m}{k}}{\binom{\color{blue}{-2n-2}}{2k}\binom{2k}{k}}=\frac{2n+1}{2n+2m+1}\tag{2}
\end{align*}
Comparison of (1) and (2) strongly indicates a generalisation where let's say $\alpha=n$ in the first case and $\alpha=-n-1$ in the second case.
The following is valid for non-negative integer values $m$ and $\alpha\in\mathbb{C}\setminus\left\{m-\frac{1}{2}\right\}$:
\begin{align*}
\color{blue}{\sum_{k=0}^m4^k\frac{\binom{\alpha}{k}\binom{m}{k}}{\binom{2\alpha}{2k}\binom{2k}{k}}
=\frac{2\alpha+1}{2\alpha-2m+1}}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^m}\color{blue}{4^k\frac{\binom{\alpha}{k}\binom{m}{k}}{\binom{2\alpha}{2k}\binom{2k}{k}}}
&=\sum_{k=0}^m\binom{m}{k}4^k\frac{\alpha^{\underline{k}}}{k!}\cdot\frac{(2k)!}{(2\alpha)^{\underline{2k}}}
\cdot\frac{k!k!}{(2k)!}\tag{1}\\
&=\sum_{k=0}^m\binom{m}{k}4^k\frac{\alpha^{\underline{k}}k!}{2^{2k}\alpha^{\underline{k}}\left(\alpha-\frac{1}{2}\right)^{\underline{k}}}\tag{2}\\
&=\sum_{k=0}^m\binom{m}{k}\binom{\alpha-\frac{1}{2}}{k}^{-1}\tag{3}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1\sum_{k=0}^m\binom{m}{k}z^k(1-z)^{\alpha-\frac{1}{2}-k}\,dz\tag{4}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-\frac{1}{2}}\sum_{k=0}^m\binom{m}{k}\left(\frac{z}{1-z}\right)^k\,dz\tag{5}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-\frac{1}{2}}\left(1+\frac{z}{1-z}\right)^m\,dz\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-m-\frac{1}{2}}\,dz\\
&=\frac{\alpha+\frac{1}{2}}{\alpha-m+\frac{1}{2}}\left[-(1-z)^{\alpha-m-\frac{1}{2}}\right]_0^1\\
&=\color{blue}{\frac{2\alpha+1}{2\alpha-2m+1}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we use the falling factorial
$\alpha^{\underline{k}}=\alpha(\alpha-1)\cdots(\alpha-k-1)$ notation and the identity
\begin{align*}
\binom{n}{k}=\frac{n^{\underline{k}}}{k!}
\end{align*}
*In (2) we do some simplifications and use the identity
\begin{align*}
(2\alpha)^{\underline{2k}}=2^{2k}\alpha^{\underline{k}}\left(\alpha-\frac{1}{2}\right)^{\underline{k}}
\end{align*}
*In (3) we cancel terms and use again the representation with binomial coefficients.
*In (4) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}
*In (5) we do some rearrangements to prepare for the usage of the binomial theorem in the next step.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim\limits_{t\to 0}(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4})$ It's from this answer.
https://math.stackexchange.com/a/368574/481435
I cannot evaluate the last limit.
I arranged that limit to slightly clearer form.
$$\lim\limits_{t\to 0}\left(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}\right)=\frac{1}{45}$$
I have no idea how to confirm that value. I know in theory I can use L'hospital's theorem but it would be endless derivatives for this problem. I tried Taylor expansion but I failed.
|
Note that
$$\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}=\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}$$
thus we need to expand in such way to have $8^{th}$ order terms, then
*
*$\cos2t=1-\frac{4t^2}{2}+\frac{16t^4}{24}+o(t^4)=1-2t^2+\frac{2t^4}{3}+o(t^4)$
*$\sin t=t-\frac{t^3}{6}+o(t^3)$
*$\sin^4
t=\left(t-\frac{t^3}{6}+o(t^3)\right)^4=t^4-\frac{2t^6}{3}+\frac{t^8}{5}+o(t^8)$
then
$$\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}=\frac{2t^4+t^4-2t^6+\frac{2t^8}{3}+o(t^8)-3t^4+2t^6-\frac{3t^8}{5}+o(t^8)}{3t^8+o(t^8)}=\frac{\frac{t^8}{15}+o(t^8)}{3t^8}=\frac1{45}+o(1)\to \frac1{45}$$
|
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"url": "https://math.stackexchange.com/questions/2631581",
"timestamp": "2023-03-29T00:00:00",
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|
Which of the two sums gives a better approximation to $\pi^2/6$? Which of the following two sums
$$\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{or} \quad 1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$
gives a better approximation to $\pi^2/6?$
I tested this on MATLAB and surprisingly obtained as a result the second sum. However, I do not know how to prove this rigorously.
|
Let: $$G_{10^6}=\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{and} \quad H_{10^3}+1=\sum_{n=1}^{1000}\frac1{n^2(n+1)}+1.$$
We can now compare the remainders of the two and see which is the smallest:
$$\begin{aligned}
&1) \int_{10^6+1}^\infty\frac1{x^2}dx\le\underbrace{\frac{\pi^2}6-G_{10^6}}_{\text{$R_G$}}\le\int_{10^6}^\infty\frac1{x^2}\,dx.\\
&2) \int_{10^3+1}^\infty\frac1{x^2(x+1)}dx\le\underbrace{\frac{\pi^2}6-1-H_{10^3}}_{\text{$R_H$}}\le\int_{10^3}^\infty\frac1{x^2(x+1)}\,dx.\end{aligned}$$
We can evaluate the improper integrals:
$$\begin{aligned}&1)\ \lim_{C\to\infty}\int_N^C\frac1{x^2}dx=\lim_{C\to\infty}\left[-\frac1x\right]^C_N=\frac1N.\\
&2)\ \lim_{S\to\infty}\int_K^S\frac1{x^2(x+1)}dx=\lim_{S\to\infty}\int_K^S\left(\frac1{x^2}-\frac1x+\frac1{x+1}\right)dx=\lim_{S\to\infty}\left[\ln\left(\frac{x+1}x\right)-\frac1x\right]^S_K=\frac1K-\ln\left(\frac{K+1}K\right).\end{aligned}$$
Thus: $$\begin{aligned}&1)\ \frac1{10^6+1}\le R_G\le\frac1{10^6}\Leftrightarrow 9.\overline{9}\cdot10^{-7}\le R_G\le 10^{-6}.\\ &2)\ \frac1{10^3+1}-\ln\left(\frac{10^3+2}{10^3+1}\right)\le R_H\le \frac1{10^3}-\ln\left(\frac{10^3+1}{10^3}\right)\end{aligned}$$ $$\approx4.98\cdot10^{-7}\le R_H \le4.99\cdot10^{-7}.$$
We can clearly see that $R_G>R_H$, therefore $$1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$ gives a better approximation to $\pi^2/6$.
|
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|
Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
|
Let $2^x=y$.
The equation becomes $$y+y^2=y^3\\
1+y=y^2\\
y^2-y-1=0\\
y=\frac{1\pm\sqrt{5}}{2}\\
2^x=\frac{1\pm\sqrt{5}}{2}\\
x=\log_2{\frac{1\pm\sqrt{5}}{2}}$$
Because $\frac{1-\sqrt{5}}{2}<0$ the logarithm of a negative number does not exist,
$$x=\log_2{\frac{1+\sqrt{5}}{2}}\text.$$
|
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"url": "https://math.stackexchange.com/questions/2633685",
"timestamp": "2023-03-29T00:00:00",
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|
Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta)-2\sin^3(\theta) $$ simply making the substitutions $$\sin(\theta)=\frac{y}{r},~\cos(\theta)=\frac{x}{r}$$ noting also that I can square both sides of the above to be substitutions we then can write down $$r=\dfrac{2yx}{r}\cdot \dfrac{x^2}{r^2}+\dfrac{y}{r}-\dfrac{2y^3}{r^3}$$ then multiplying through by $r^3$ we obtain $$r^4=2yx^3+yr^2-2y^3$$ then replacing $r^2$ with $x^2+y^2$ we get $$(x^2+y^2)^2=2xy^2+yx^2+y^3-2y^3 \\(x^2+y^2)^2=y(3x^2-y^2)$$
However I am unsure of where I have made a mistake as the true answer is$$(x^2+y^2)^2=4x^2y-(x^2+y^2)y$$ working backwards I've so far gotten to the point of asking how I would rearrange$$r=\sin (3\theta) \Rightarrow r=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ so that $$\sin(3\theta)=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ which I am afraid I'll have to ask help for the next steps. Thanks.
|
$$\begin{align}r&=\sin(3\theta)\\r&=\sin\theta\cos2\theta+\cos\theta\sin2\theta\\r^2&=r\sin\theta\cos2\theta+r\cos\theta\sin2\theta\\x^2+y^2&=y\cos2\theta+x\sin2\theta\\r^2(x^2+y^2)&=yr^2(1-2\sin^2\theta)+2xr^2\sin\theta\cos\theta\\(x^2+y^2)^2&=y(x^2+y^2)-2y^3+2x^2y\\(x^2+y^2)^2&=3x^2y-y^3\end{align}$$
I don't see a difference so far.
$$\begin{align}r&=\sin(3\theta)\\r&=3\sin\theta-4\sin^3\theta\\r^4&=3r^3\sin\theta-4r^3\sin^3\theta\\(x^2+y^2)^2&=3y(x^2+y^2)-4y^3\\(x^2+y^2)^2&=3x^2y-y^3\end{align}$$
No difference here either. Your answer looks good.
|
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|
continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$.
Where is the mistake? (I triple-checked and it seems fine to me)
|
Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 41} = 6 + \frac{ \sqrt {41} - 6 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {41} - 6 } = \frac{ \sqrt {41} + 6 }{5 } = 2 + \frac{ \sqrt {41} - 4 }{5 } $$
$$ \frac{ 5 }{ \sqrt {41} - 4 } = \frac{ \sqrt {41} + 4 }{5 } = 2 + \frac{ \sqrt {41} - 6 }{5 } $$
$$ \frac{ 5 }{ \sqrt {41} - 6 } = \frac{ \sqrt {41} + 6 }{1 } = 12 + \frac{ \sqrt {41} - 6 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccccccc}
& & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 6 }{ 1 } & & \frac{ 13 }{ 2 } & & \frac{ 32 }{ 5 } & & \frac{ 397 }{ 62 } & & \frac{ 826 }{ 129 } & & \frac{ 2049 }{ 320 } \\
\\
& 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1
\end{array}
$$
$$
\begin{array}{cccc}
\frac{ 1 }{ 0 } & 1^2 - 41 \cdot 0^2 = 1 & \mbox{digit} & 6 \\
\frac{ 6 }{ 1 } & 6^2 - 41 \cdot 1^2 = -5 & \mbox{digit} & 2 \\
\frac{ 13 }{ 2 } & 13^2 - 41 \cdot 2^2 = 5 & \mbox{digit} & 2 \\
\frac{ 32 }{ 5 } & 32^2 - 41 \cdot 5^2 = -1 & \mbox{digit} & 12 \\
\frac{ 397 }{ 62 } & 397^2 - 41 \cdot 62^2 = 5 & \mbox{digit} & 2 \\
\frac{ 826 }{ 129 } & 826^2 - 41 \cdot 129^2 = -5 & \mbox{digit} & 2 \\
\frac{ 2049 }{ 320 } & 2049^2 - 41 \cdot 320^2 = 1 & \mbox{digit} & 12 \\
\end{array}
$$
|
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|
Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$. Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.
I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have no prove.
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$f(x;y;z)=\frac{(xy+yz+zx)(y-z)^2 + (xy+xz-2yz)^2}{4(x+y)(z+x)(y+z)^2} +\frac{3}{4} \geq \frac{3}{4}$
Equality holds when $x=y=z$
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|
prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$$
using $\left( \begin{array}{c} n \\ 2\ \end{array} \right) = \dfrac{n(n-1)}{2}$?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = \frac{2n(2n-1)}{2} = 2n(n-1) = 2 \frac{n(n-1)}{2} + n^2 = 2\left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2 $$
Is this proof correct? if not how do I fix it?
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$$\binom{2n}{2} = \frac{2n(2n-1)}{2} = \frac{2n(n-1) + 2n^2}{2}
= 2 \binom{n}{2} + n^2$$
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Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$
I have attempted to do this but never seem to prove it.
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Proof by contradiction
Assumption: There are $n$ where $6$ doesn't divide $n(n+1)(n+2)$.
Let $x$ be the smallest $n$ where $6$ doesn't divide $n(n+1)(n+2)$.
Since $6|1\times 2\times 3$ and $6|2\times 3\times 4$ that $x$ has to be at least $3$.
If $6$ doesn't divide $x(x+1)(x+2)$ then it doesn't divide $x(x+1)(x+2)-6x^2 = ... = (x-2)(x-1)x = (x-2)(x-2+1)(x-2+2)$.
So the statement isn't true for $x-2$ and that is a contradiction to the fact that $x$ is smallest $n$ where the statement isn't true.
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Show the following inequality. Let $a,b,c \in \mathbb R^+$ and $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} =1$$
Show that $$(a^2 -3a +3)(b^2-3b+3)(c^2-3c+3) \ge 27$$
I tried using using the AM-GM inequality and some algebraic manipulation to who each of the quadratic terms cannot be smaller than $0.75$ and other little results but I am struggling to be able to solve this. Any help would be appreciated.
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For positive variables let $\frac{1}{a}=\frac{x}{3},$ $\frac{1}{b}=\frac{y}{3}$ and $\frac{1} {c}=\frac{z}{3}.$
Thus, $x+y+z=3$ and we need to prove that
$$\sum_{cyc}\left(\ln\left(\frac{9}{x^2}-\frac{9}{x}+3\right)-\ln3\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(x^2-3x+3)-2\ln{x}\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(x^2-3x+3)-2\ln{x}+3(x-1)\right)\geq0.$$
Let $f(x)=\ln(x^2-3x+3)-2\ln{x}+3(x-1).$
Thus, $$f'(x)=\frac{2x-3}{x^2-3x+3}-\frac{2}{x}+3=\frac{3(x-1)(x^2-2x+2)}{x(x^2-3x+3)},$$
which gives $x_{min}=1$ and we are done!
|
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U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
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Alternatively, change:
$$4x^2+9=u \Rightarrow x^2=\frac{u-9}{4}; xdx=\frac{du}{8};$$
$$\int_{9}^{36} \frac{\frac{u-9}{4}\cdot \frac{du}{8}}{u^{3/2}}=\frac{1}{32}\int_{9}^{36} \frac{1}{u^{1/2}}du-\frac{9}{32}\int_{9}^{36} \frac{1}{u^{3/2}}du=\cdots$$
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Removing middle fourths instead of middle thirds in Cantor Set If you removed middle fourths instead of middle thirds to form a generalized Cantor set G, then what would the Lebesgue measure $m(G)$?
The geometric series that I obtained from removing middle fourths was:
$\displaystyle\frac{1}{4}+(2\cdot\frac{1}{4}\cdot\frac{3}{8})+\big(4\cdot\frac{1}{4}\cdot\big(\frac{3}{8}\big)^2\big)+\big(8\cdot\frac{1}{4}\cdot\big(\frac{3}{8}\big)^3\big)+\dots$
$=\displaystyle\frac{1}{4}+(\frac{1}{2}).(\frac{3}{8})+\big(\frac{3}{8}\big)^2+2\cdot\big(\frac{3}{8}\big)^3+4\cdot\big(\frac{3}{8}\big)^4+\dots$
$=\displaystyle\frac{\frac{1}{4}}{1-\big(2\cdot\frac{3}{8}\big)}=\frac{\frac{1}{4}}{1-\frac{6}{8}}=\frac{\frac{1}{4}}{\frac{2}{8}}=1$
Did I arrive at the answer correctly? Is my geometric series correct? Any critiques are appreciated.
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Yes, your geometric series is correct and your new Cantor dust has measure zero.
You may try different middle intervals to remove and get different Cantor dusts with the same measure.
Note that while these Cantor sets have the same measure, their fractal dimensions are quite different.
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
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Use the distributive property.
$$(1 + x + x^2 + x^3 + x^4 + x^5)(1 + x + x^2 + x^3 + x^4 + x^5) $$
is equal to
$$
\begin{matrix}
(1)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^2)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^3)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^4)(1 + x + x^2 + x^3 + x^4 + x^5) \\
(x^5)(1 + x + x^2 + x^3 + x^4 + x^5) \\
\end{matrix}
$$
When the above is expanded, we get $36$ products. Arrange the results of these $36$ products into columns whose entries have the same degree.
$$
\begin{matrix}
1 & x & x^2 & x^3 & x^4 & x^5 \\
& x & x^2 & x^3 & x^4 & x^5 & x^6 \\
& & x^2 & x^3 & x^4 & x^5 & x^6 & x^7 \\
& & & x^3 & x^4 & x^5 & x^6 & x^7 & x^8 \\
& & & & x^4 & x^5 & x^6 & x^7 & x^8 & x^9 \\
& & & & & x^5 & x^6 & x^7 & x^8 & x^9 & x^{10} \\
\end{matrix}
$$
Adding like powers of $x$ shows how the pattern of coefficients arises, giving
$$= \;\;\; 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 5x^6 + 4x^7 + 3x^8 + 2x^9 + x^{10} $$
In the case of $\;(1 + x + x^2 + \ldots + x^{n-1} + x^n)^2,\;$ the same pattern of coefficients --- increasing by increments of $1$ from $1$ to some maximum value, and then decreasing by increments of $1$ from the maximum value to $1$ --- can be seen by thinking about the following.
$$
\begin{matrix}
1 & x & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n \\
& x & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} \\
& & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} \\
& & & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} & x^{n+3} \\
& & & & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} & x^{n+3} & x^{n+4} \\
\end{matrix}
$$
$$\cdot$$
$$\cdot$$
$$\cdot$$
|
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|
Maxima and minima of function with limiting condition (lagrange multiplier) problem
I have function:
$$ f(x,y,z)=x+2y+\frac{z^2}{2} $$ and i would want to find it's minima and maxima with condition $x^2+y^2+z^2=5$
Attempt to solve
I can form Lagrange function with the limiting condition and the function itself.
Lagrange function is defined as
$$ L(x,y,z,\lambda)=f(x)+\lambda(g(x)) $$
$g(x)$ would be limiting condition.
$$ L(x,y,z,\lambda)= x+2y+\frac{z^2}{2} + \lambda(x^2+y^2+z^2-5)$$
in order to find when this function has critical points i need to compute gradient.
$$ \nabla L(x,y,\lambda)=\begin{bmatrix} 1+2x\lambda \\ 2+2y\lambda \\ z+2z\lambda \\ x^2+y^2+z^2-5 \end{bmatrix} $$
$$ \begin{cases} 1+2x\lambda = 0 \\ 2+2y\lambda = 0 \\ z+2z\lambda =0 \\ x^2+y^2+z^2-5=0\end{cases} $$
$$ x=-\frac{1}{2\lambda} $$
$$ y=-\frac{2}{2\lambda} $$
$$ z=0 $$
$$ (-\frac{1}{2\lambda})^2+(-\frac{2}{2\lambda})^2+(0)^2-5=0 $$
$$ \frac{1}{4\lambda^2} + \frac{4}{4\lambda^2}-5=0 $$
$$ \frac{5}{4\lambda^2}-5=0 $$
$$ \frac{5}{4\lambda^2}=5 $$
$$ 20\lambda^2=5 $$
$$ \lambda = \sqrt{\frac{5}{20}}=\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}$$
since for all $n\ge 0$ $\sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}}$
Now i can compute values for $x$ & $y$
$$ 1+2(\cdot\frac{1}{2})x=0$$
$$ x=-1$$
$$ 2+2y\cdot (\frac{1}{2})=0 $$
$$ y=-2 $$
now for $z$
$$ z+2\cdot(\frac{1}{2})z=0$$
$$ z=0 $$
I found one such critical point $(-1,-2,0)$
now to compute what values this point has:
$$ f(-1,-2,0)=-1-2\cdot 2+ \frac{0^2}{2}=-5 $$
I don't know what tools i have for checking if this critical point is either maxima or minima but it certainly would look like a minima. I think there is version of hessian matrix called "bordered hessian" but this wasn't covered on our lecture so i am suppose to know only about the regular hessian which isn't sufficient way to tell on this if the point is minima or maxima. Also how do i find the another critical point (maxima) or how do i know if such point doesn't exist ?
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why don't you consider the function $$g(x,y)=x+2y+\frac{5-x^2-y^2}{2}$$ this simplifies the problem
and this is $$-\frac{1}{2}(x-1)^2-\frac{1}{2}(y-2)^2+5$$
and the minimum is given by $-5$
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|
Compute $\iint_D\frac{xy}{(1+y^2)^2}\,\mathrm{d}x\,\mathrm{d}y$ where $D=\{(x,y):x\geq 0,\ y\geq 0,\ x^2+y^2 \leq 1\}$
Compute $$I=\iint_D\frac{xy}{(1+y^2)^2}\,dx\,dy,$$ where $D = \{(x,y):x\geq
0, \ y\geq 0,\ x^2+y^2 \leq 1\}.$
In the $xy-$plane, this is just a quarter circle disk in the first quadrant. So choosing to integrate first w.r.t $x$ then $y$ I get that
$$I=\int_{0}^{1}\left(\int_0^{\sqrt{1-y^2}}\frac{xy}{(1+y^2)^2}dx\right)dy=\int_0^1\frac{y}{(1+y^2)^2}\left(\int_0^{\sqrt{1-y^2}}x \ dx\right)dy \\ = -\frac{1}{2}\int_0^1\frac{y^3-y}{(y^2+1)^2}dy.$$
Questions:
*
*This seems to work using partial fractions, I checked it with software. Howver I'm not sure how I should do it. Is this the correct ansatz: $$\frac{y^3-y}{(y^2+1)^2}=\frac{Ay+B}{y^2+1}+\frac{Cy+D}{(y^2+1)^2} \ ??$$
*Is there any other nicer way to solve this problem by making the integration simpler? Polar coordinates or something?
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Try this
\begin{eqnarray}
I&=&\int_{0}^{1}\left(\int_0^{\sqrt{1-x^2}}\frac{xy}{(1+y^2)^2}dy\right)dx\\&=&\int_0^1x\left(\int_0^{\sqrt{1-x^2}}\frac{y}{(1+y^2)^2} \ dy\right)dx \\ &=& \int_0^1x\left(-\frac{1}{2(1+y^2)^2} \right)\Large|_{0}^{\sqrt{1-x^2}} dx\\
&=& \int_0^1x\left(-\frac{1}{2(2-x^2)^2}+\frac{1}{2} \right)dx\\
&=& \int_0^1\left(-\frac{x}{2(2-x^2)^2}+\frac{x}{2} \right)dx\\
\end{eqnarray}
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Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}{2}$, because $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing and convergent to $e$, while $1+\frac{1}{1!}+\dots+\frac{1}{n!}$ is increasing and convergent to $e$, so $$\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>1+\frac{1}{1!}+\dots+\frac{1}{n!}$$
which means that $\frac{1}{2}>x_n$. I also know that for $a \in [0,\frac{1}{2}), \left(1+\frac{1}{n}\right)^{n+a}$ is eventually increasing, but I don't know how to get a lower bound for $x_n$ which goes to $\frac{1}{2}$
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Answer. $x_n\to \dfrac{1}{2}$
Explanation. Taylor series remainder
$$
\mathrm{e}=1+\frac{1}{1!}+\cdots+\frac{1}{n!}+\frac{\mathrm{e}^{\xi_n}}{(n+1)!}
$$
for some $\xi_n\in(0,1)$. Since
$$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!},$$
Then
$$
n+x_n=\frac{\log\left(\mathrm{e}-\dfrac{\mathrm{e}^{\xi_n}}{(n+1)!}\right)}{\log\left(1+\frac{1}{n}\right)}=\frac{1+\log\left(1-\dfrac{\mathrm{e}^{\xi_n-1}}{(n+1)!}\right)}{\dfrac{1}{n}-\dfrac{1}{2n^2}+{\mathcal O}(n^{-3})}=n\cdot\frac{1+{\mathcal O}\left(\frac{1}{(n+1)!}\right)}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})}\\=n+\frac{\dfrac{1}{2}+{\mathcal O}(n^{-1})}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})}
$$
and hence
$$
x_n\to \frac{1}{2}
$$
|
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Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem is can we prove the statement
"for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"
only using induction ?(without using the fact that convergence of the above series)
any ideas, thanks!
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The statement that we will prove with induction is that for every $K > k^*$,
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - \frac 1K $$
where $k^* = 4091641$.
This implies that, for every $K$,
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 $$
The idea is to find a function $f(K) > 0$ such that
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - f(K) $$
As @orlp states in his comment, it is clear that $f(K)$ cannot be a constant, because we know that the partial sums get arbitrarily close to $\frac{\pi^2}6$. So which function do we pick?
Well, just write the inductive step:
$$\sum_{n=1}^K \frac {1}{n^2} + \frac{1}{(K+1)^2} \le \frac {\pi ^2} 6 - f(K) + \frac{1}{(K+1)^2} \mathop{\le}^? \frac {\pi ^2} 6 - f(K+1)$$
If we can prove the last inequality, then the inductive step will work and we will have solved the problem.
The last inequality is equivalent to
$$f(K+1) \le f(K) - \frac 1{(K+1)^2}$$
which means that $f(K)$ must be decreasing. One can easily check that $f(K) = \frac 1K$, for example, works. The only thing that is left is the base case.
Turns out, the base case is not so easy to find. A quick numerical simulation, though, proved that for $n=4091641$ the case case is satisfied, and the rest follows. Note that since we know that the statement is true for a certain $K > k^*$, we also know that every partial sum up to $k^*$ is $\le \frac{\pi^2}6$, as the partial sums are increasing.
One could probably find a better $f(K)$ such that we don't need computers to verify the base case, but I'll leave that to someone else :)
|
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|
Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are the two solutions that I found:
I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?
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Your first way is right.
In the second solution $$A+\frac{\pi}{6}=B-\frac{\pi}{6},$$ which is very well or $$A+B=0,$$ which is impossible.
Another way:
$$2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\sqrt{\frac{3}{2}}$$ and
$$2\sin\frac{A+B}{2}\sin\frac{B-A}{2}=\sqrt{\frac{1}{2}}.$$
Thus,
$$\tan\frac{B-A}{2}=\frac{1}{\sqrt3}$$ and since $A$ and $B$ they are acute angles, we obtain $$B-A=60^{\circ},$$
$$\sin\frac{A+B}{2}=\frac{1}{\sqrt2}$$ and
$$A+B=90^{\circ}.$$
|
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|
Evaluate $\sum\limits_{i=1}^\infty a_i $ when $\sum\limits_{i=1}^na_i=\frac {n-1}{n+1}$ Question: $S_n=\frac {n-1}{n+1}$; Find $\sum_1^\infty a_i $
My answer: $a_n=S_n -S_{n-1}=\frac{2}{n^2+n}=\frac{2}{n}-\frac{2}{n+1}$
$\sum_1^\infty a_i=(\frac{2}{1}-\frac{2}{2})+(\frac{2}{2}-\frac{2}{3})+(\frac{2}{3}-\frac{2}{4})+......(\frac{2}{n}-\frac{2}{n+1}) = 2-\frac{2}{n+1}=2$
Am I correct in computing $a_n$ first and then finding $\sum_1^\infty a_i$? Or should I have done $\lim\limits_{n \to \infty}S_n =\lim\limits_{n \to \infty}\frac{n-1}{n+1}=1.$ Which is the correct method of evaluating this?
|
An option
$S_n:= \sum _{i=1}^{n}a_i = $
$\dfrac{n-1}{n+1}=1- 2\dfrac{1}{n+1}.$
$\lim_{ n \rightarrow \infty} S_n = 1- 2\lim_{n \rightarrow \infty}\dfrac{1}{n+1}.$
|
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Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
|
I start, like everyone else with $3\sin2x/(5-3\cos2x)$.
Consider a circle, radius 3, around 5+0i in the complex plane. The highest value for tan y is for the line through the origin, tangent to the circle. That forms a 3-4-5 right-angled triangle, so $\tan y=3/4$.
|
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|
Central Limit Theorem Proof using Logarithm Expansion I am trying to go over a proof of the CLT given by this site.
I understood everything up to the point where the expansion for the logarithm was used:
$$x=\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3).$$
This was used to expand:
$$n\ln\left(1+\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\ +\cdots\right).$$
How is this correct if the above is not in the form of $\ln(1+x)$? Am I missing something?
|
The logarithm of the moment generating function is
$$\ln m_u(t)=n\cdot \ln\left(1+\underbrace{\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)}_{\color{red}x} +\ldots\right)$$
Now you set $x$ equal to $\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3) \quad (*)$
Thus you can use
$$\ln m_u(t)=n\cdot \ln\left(1+x\right)$$
Using the series expansion of $\ln(x+1)$
$$n\cdot \ln m_u(t)=n\cdot \ln\left(1+x\right)=n\cdot \left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right)$$
Now you insert the expression for $x$, see (*). I do it for the first three summands only.
$n\cdot \ln m_u(t)$
$$=\color{blue}n\cdot \left(\frac{t^2}{2\color{blue}n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)-\frac{\left(\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\right)^2}{2}+\frac{\left(\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\right)^3}{3}+\ldots\right)$$
You see at the first summand $n$ is cancelling out only.
In the other cases the exponents of $n$ in the denominators are larger than 1 $\left(\frac{3}{2}, 2, 1+\frac{3}{2} , 3, ... \right)$.
Consequently for $n\to \infty$ the summands become $0$, exept the first one.
|
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|
general solution of the FODE $\frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x) -2$ What's the general solution of the first order differential equation ?
$$
\begin{align*}
\frac{1}{\sin(x)} \frac{dy}{dx} = y\sec(x)-2, \quad 0<x< \frac{\pi}{2}
\end{align*}
$$
my solution is as follows but i think it's wrong
$$
\frac{dy}{dx} = y\tan(x) -2\sin(x) \\
\frac{dy}{dx} - y\tan(x) = -2\sin(x) \\
e^{-\int \tan(x)dx} = \cos(x)\\
y\,cos(x) = \int(-2\cos(x)\sin(x))dx\\
y =\cos(x) + \frac{C}{\cos(x)}
$$
|
It seems correct to me...
$$y'-y\tan(x)=-2\sin(x)$$
$$y'-y\frac {\sin(x)}{\cos(x)}=-2\sin(x)$$
$$\frac {y'\cos(x)-y\sin(x)}{\cos(x)}=-2\sin(x)$$
$$({y}{\cos(x)})'=-\sin(2x)$$
Integrating ..
$${y(x)}{\cos(x)}=-\int \sin(2x) dx$$
$$y(x)=\frac 1 {\cos(x)}(\frac {\cos(2x)}2+K)$$
The same answer as yours.....with $\cos(2x)=2cos^2(x)-1$
$$\boxed {y(x)=\frac C {\cos(x)}+\cos(x)}$$
|
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|
Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}
Therefor, I thought that:
$$\lim_{n \to \infty}\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}}$$
$$ = \lim_{n \to \infty}\frac{1}{\left(n^2-n^2\right)\sqrt{1}} = \infty$$
I also tried a different way where I got to the result of $-\dfrac{1}{2}$.
I am not going to show that method here but it starts with using the 3rd binomial formula. Then, having the squareroot at the top of the fraction, I factorized $n^2$ and it all worked.
Why does this method like shown above not work?
I am very happy for any help.
P.S. This is not the only example where this kind of getting to a solution does not work for me. Are there cases where I am not allowed to factorize something?
|
$$ (n^2 + 2)^2 < n^4 + 4 n^2 + n < \left(n^2 + 2 + \frac{1}{2n} \right)^2 $$
$$ n^2 + 2 < \sqrt{n^4 + 4 n^2 + n \;} \; < \; n^2 + 2 + \frac{1}{2n} $$
$$ - 2 > n^2 -\sqrt{n^4 + 4 n^2 + n \;} \; > \; - 2 - \frac{1}{2n} $$
$$ - \frac{1}{2} < \frac{1}{n^2 -\sqrt{n^4 + 4 n^2 + n \;} } \; < \; - \frac{1}{2 + \frac{1}{2n}} $$
these are for $n > 4$
|
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|
Cauchy can't be applied on this inequality for this condition of a, b, c Given three numbers $a,b,c$ satisfy $0\le a,b,c\le 2$ and $a+b+c=3$.
Prove that $a^2+b^2+c^2\le \frac{1}{ab+bc+ca}+\frac{9}{2}$.
Attempt:
By trial and error, I know that the equality does not hold for $a=b=c=1$. However, the equality holds for all permutations of $(0;1;2)$. Because if the problem can be solved by Cauchy inequality, then the equality will usually (if not always) hold if $a=b=c$. My attempt is below:
This inequality is true:
$a^2+b^2+c^2\le \frac{1}{ab+bc+ca}+\frac{9}{2}$
$\Leftrightarrow \left(a+b+c\right)^2-2ab-2bc-2ca\le \frac{1}{ab+bc+ca}+\frac{9}{2}$
$\Leftrightarrow 3^2-\frac{9}{2}\le \frac{2\left(ab+bc+ca\right)^2+1}{ab+bc+ca}$
(after this I'll put $x=ab+bc+ca$)
$\Leftrightarrow 4.5\le \frac{2x^2+1}{x} \Leftrightarrow 2x^2+1 \ge 4.5x$
$\Leftrightarrow 2x^2-4.5x+1\ge 0$ $\Leftrightarrow x\le 0.25$ or $x\ge2$.
$\Leftrightarrow ab+bc+ca \le0.25$ or $ab+bc+ca\ge 2$.
At this point I'm stuck, is this the right way to solve?
|
Your way is right.
You need only to end it:
Let $a\geq b\geq c$.
Thus, $1\leq a\leq2$ and $$ab+ac+bc\geq a(b+c)=a(3-a)=2+(2-a)(a-1)\geq2.$$
|
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|
If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$
Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
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By Euler's theorem $O,G,H$ are collinear. By considering their trilinear coordinates it follows that
$$ \det\begin{pmatrix}\cos(A) & \cos(B) & \cos(C) \\ \frac{1}{\sin A}&\frac{1}{\sin B}&\frac{1}{\sin C}\\ \frac{1}{\cos A}&\frac{1}{\cos B}&\frac{1}{\cos C}\end{pmatrix}=0$$
and by multiplying the first column by $\cos(A)$, the second column by $\cos(B)$ and the third column by $\cos(C)$ we get that
$$ \det\begin{pmatrix}\cos^2(A) & \cos^2(B) & \cos^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$
and by replacing the first row with the difference between the third row and the first row
$$ \det\begin{pmatrix}\sin^2(A) & \sin^2(B) & \sin^2(C) \\ \cot(A)&\cot(B)&\cot(C)\\ 1&1&1\end{pmatrix}=0$$
readily follows. One may also notice that if $\theta\in\{A,B,C\}$ then
$$ 8R^2\cdot \sin^2\theta + 4[ABC]\cdot\cot\theta -(a^2+b^2+c^2)\cdot 1 = 0 $$
hence the nullspace of the given matrix has dimension $\geq 1$.
|
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|
Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If anyone could point me in the right direction I'd very much appreciate that.
Here was my working:
Let $z = x+iy$
$|z|$ = $\sqrt{x^2+y^2}$
$$|z+2-7i| = 2|z-10+2i| \\
\implies |(x+iy)+2-7i| = 2|(x+iy)-10+2i| \\
\implies |(x+2)(y-7y)i| = 2|(x-10)+(y+2)i| \\
\implies \sqrt{(x+2)^2+(y-7)^2i^2} = 2\sqrt{(x-10)^2+(y+2)^2i^2} \\
\implies (x+2)^2-((y-7^2) = 2((x-10)^2-(y+2)^2)$$
$$(x^2+4x+4)-(y^2-14y+49) = 2((x^2-20x+100)-(y^2+2y+4)) \\
\implies x^2+4x+4-y^2+14y-49 = 2x^2-40x+200-2y^2-4y-8 \\
\implies -237 = x^2-44x-y^2-18y\\
\implies x^2-44x-y^2-18 = -237$$
|
Consider $|z+a+bi|=2|z+c+di|$. Squaring gives
$$(x+a)^2+(y+b)^2=4(x+c)^2+4(y+d)^2$$
that is
$$3(x^2+y^2)+(8c-2a)x+(8d-2b)y+4c^2+4d^2-a^2-b^2=0.$$
In general this is a circle. Now put in your values of $a,\ldots,d$.
|
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|
Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2} = 3\cdot a+\frac{b}{2}$
Is this correct? How to prove it using resiudes?
Thank You.
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Note that $\gcd(n,6)=1 \implies n = 6 a + 1$ or $n=6a+5$, where $a \in \mathbb{Z}$.
For $n=6a+1$,
$$(n-1)(n+1) = 12a(3a+1).$$
Since $a(3a+1)$ is always even, i.e., $a(3a+1)=2k$, $(n-1)(n+1) = 24k$.
Similar is the case for $n = 6a+5$.
|
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|
Reducing a matrix to reduced row echelon form without introducing fractions I've been trying to figure out how to reduce this matrix without introducing fractions in the intermediate stages but can't figure out how to do it.
$$\begin{bmatrix}2&1&3\\
0&-2&-29\\
3&4&5\end{bmatrix}$$
|
Here's one way.
$\begin{bmatrix}2&1&3\\
0&-2&-29\\
3&4&5\end{bmatrix}
\xrightarrow {R3-R1}
\begin{bmatrix}2&1&3\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {R1-2R3}
\begin{bmatrix}0&-5&-1\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {-R1}
\begin{bmatrix}0&5&1\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {R1+2R2}
\begin{bmatrix}0&1&-57\\
0&-2&-29\\
1&3&2
\end{bmatrix}
\xrightarrow {R2+2R1}
\begin{bmatrix}0&1&-57\\
0&0&-143\\
1&3&2
\end{bmatrix}
\xrightarrow{-R2/143}
\begin{bmatrix}0&1&-57\\
0&0&1\\
1&3&2
\end{bmatrix}
\xrightarrow{R1+57R2}
\begin{bmatrix}0&1&0\\
0&0&1\\
1&3&2
\end{bmatrix}
\xrightarrow[R3-2R2]{R3-3R1}
\begin{bmatrix}0&1&0\\
0&0&1\\
1&0&0
\end{bmatrix}
\xrightarrow[R2\leftrightarrow R3]{R2\leftrightarrow R3}
\begin{bmatrix}1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
$
|
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|
Product of equation of two lines Why does the product of equations of straight lines represents 2 straight lines?
Why does a 2nd degree equation is so linear?
|
In the language of conics,
\begin{align}
0 &= (ax+by+c)(a'x+b'y+c') \\[10pt]
\iff 0 &=
\begin{pmatrix}
x & y & 1
\end{pmatrix}
\begin{pmatrix}
aa' & \frac{ab'+a'b}{2} & \frac{ac'+a'c}{2} \\
\frac{ab'+a'b}{2} & bb' & \frac{bc'+b'c}{2} \\
\frac{ac'+a'c}{2} & \frac{bc'+b'c}{2} & cc'
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ 1
\end{pmatrix}
\end{align}
Note that
$$\det
\begin{pmatrix}
aa' & \frac{ab'+a'b}{2} \\
\frac{ab'+a'b}{2} & bb'
\end{pmatrix}
=-\frac{(ab'-a'b)^2}{4}<0$$
providing non-parallel and
$$\det
\begin{pmatrix}
aa' & \frac{ab'+a'b}{2} & \frac{ac'+a'c}{2} \\
\frac{ab'+a'b}{2} & bb' & \frac{bc'+b'c}{2} \\
\frac{ac'+a'c}{2} & \frac{bc'+b'c}{2} & cc'
\end{pmatrix}
=0$$
implying a pair of straight lines.
It's a special case that the section plane cutting the radiant point of the cones.
See another answer here for your interest.
|
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|
Is $5^{1/5} - 3\cdot i$ algebraic? I am studying the book Complex Variables with Applications written by Herb Silverman. In this book, problem number 8 in Question 1.7 is as in the following.
Is $5^{1/5} - 3\cdot i$ algebraic? (i.e, Is the $5$th-root of $5$ minus $3\cdot i$ algebraic?)
Can you help me to solve this?
|
Multinomial Combiniations of Algebraic Integers
Given a set of monic polynomials defining a set of algebraic integers, how can we derive a polynomial for a multinomial combination of those algebraic integers?
For example, given $\alpha^2-3=0$ and $\beta^3-2=0$, how do we derive a polynomial for $\alpha+\beta$?
First, create the basis for all multinomial combinations of $\alpha$ and $\beta$:
$$
B=\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
$$
Next compute $(\alpha+\beta)B$:
$$
\begin{align}
(\alpha+\beta)B
&=(\alpha+\beta)
\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
=\begin{bmatrix}
\alpha+\beta\\
\alpha^2+\alpha\beta\\
\alpha\beta+\beta^2\\
\alpha^2\beta+\alpha\beta^2\\
\alpha\beta^2+\beta^3\\
\alpha^2\beta^2+\alpha\beta^3
\end{bmatrix}
=\begin{bmatrix}
\alpha+\beta\\
3+\alpha\beta\\
\alpha\beta+\beta^2\\
3\beta+\alpha\beta^2\\
\alpha\beta^2+2\\
3\beta^2+2\alpha
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
0&1&1&0&0&0\\
3&0&0&1&0&0\\
0&0&0&1&1&0\\
0&0&3&0&0&1\\
2&0&0&0&0&1\\
0&2&0&0&3&0\\
\end{bmatrix}
\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
=MB
\end{align}
$$
$M$ is an integer matrix since the polynomials for $\alpha$ and $\beta$ are monic.
Since multiplication by a scalar commutes with matrix multiplication, induction gives
$$
(\alpha+\beta)^kB=M^kB
$$
SInce $M$ satisfies its own characteristic polynomial,
$$
0=P(M)B=P(\alpha+\beta)B
$$
therefore, $\alpha+\beta$ also satisfies the characteristic polynomial of $M$:
$$
P(x)=x^6-9x^4-4x^3+27x^2-36x-23
$$
Since the characteristic polynomial, $P(x)=\det(M-Ix)$, for an integer matrix is monic, $\alpha+\beta$ is also an algebraic integer.
Application to the Question
The example above should be general enough to show how to apply the process to any set of monic polynomials, a cubic and quadratic were used to reduce the footprint of the example.
To show that $5^{1/5}-3i$ is an algebraic integer, apply the program above to $\alpha^5-5=0$ and $\beta^2+9=0$ to produce a monic polynomial for $\alpha-\beta$:
$$
x^{10}+45x^8+810x^6-10x^5+7290x^4+900x^3+32805x^2-4050x+59074
$$
Afterword
I had thought I had written something along these lines before. I finally came across this answer that I wrote a while back. Hopefully, between this post and that one, the ideas are sufficiently clear.
|
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|
Prove that every element in $V$ is on the form $\vec{v}$ Let $u_1=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}$, $u_2=\begin{pmatrix}2\\ 3\\ 4\end{pmatrix}$, $v_1=\begin{pmatrix}1\\ 1\\ 2\end{pmatrix}$, $v_2=\begin{pmatrix}2\\ 2\\ 3\end{pmatrix}$.
Let $U=span(\vec{u_1},\vec{u_2})$ and $V=span(\vec{v_1},\vec{v_2})$.
I've already shown that $(\vec{u_1}|\vec{u_2})$ and $(\vec{v_1}|\vec{v_2})$ are equivalent to $\begin{pmatrix}1 & 0\\ 0 & 1\\ 0 & 0\\ \end{pmatrix}$ and therefore that $U$ and $V$ are vector spaces of dimension 2.
Now I want to prove the following, but I could use some help (or at least some guidance on where to start):
*
*That any element in $V$ is of the form $\vec{v}=\begin{pmatrix}\alpha + 2\beta\\ \alpha + 2\beta\\ 2\alpha + 3\beta\end{pmatrix}$ where $\alpha$ and $\beta$ are real numbers.
*That the following system only has solutions if $\alpha+\beta=0$:
$\begin{pmatrix}1 & 2\\ 2 & 3\\ 3 & 4\end{pmatrix}\vec{x}=\begin{pmatrix}\alpha + 2\beta\\ \alpha + 2\beta\\ 2\alpha + 3\beta\end{pmatrix}$
*That $\begin{pmatrix}\alpha + 2\beta\\ \alpha + 2\beta\\ 2\alpha + 3\beta\end{pmatrix}=\beta\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}$ when $\alpha+\beta=0$.
*That $U\cap V=span(\vec{v})$, where $\vec{v}=\begin{pmatrix}1 \\ 1\\ 1\\ \end{pmatrix}$.
|
1) Since $V=span(\vec v_1,\vec v_2)$, there is nothing to prove, because by definition
$$
span(\vec v_1,\vec v_2)=\left\{\alpha\vec v_1+\beta\vec v_2~:~\alpha,\beta\in\mathbb R\right\}=\left\{\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}~:~\alpha,\beta\in\mathbb R\right\}.
$$
2) Write
$$
\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}
=(\alpha+\beta)\begin{pmatrix}1\\1\\2\end{pmatrix}+\beta\begin{pmatrix}1\\1\\1\end{pmatrix}
$$
and prove(!) $\begin{pmatrix}1\\1\\1\end{pmatrix}\in U$ and $\begin{pmatrix}1\\1\\2\end{pmatrix}\notin U$.
Not let us show: $\alpha+\beta=0\Rightarrow \begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}\in U$.
Let be $\alpha+\beta=0$, then
$$
\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}
=\beta\begin{pmatrix}1\\1\\1\end{pmatrix}\in U,
$$
because ... .
Next, we show: $\alpha+\beta\neq 0\Rightarrow \begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}\notin U$.
Assume $\alpha+\beta\neq 0$ and $\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}\in U$. Then
$$
(\alpha+\beta)\begin{pmatrix}1\\1\\2\end{pmatrix}=\begin{pmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{pmatrix}-\beta\begin{pmatrix}1\\1\\1\end{pmatrix}.
$$
The LHS is not in $U$ while the RHS ... .
3) Use the form from 2)
4) Combine the results 1)+2)+3).
|
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|
Squaring a Binomial How to find ab if:
$$
a - b = 3
$$
$$
a^2 - b^2 = 21
$$
What I already know is the following:
To find ab for:
$$
a - b = 3
$$
$$
a^2 + b^2 = 29
$$
ab is:
$$
2ab = (a^2 + b^2) - (a - b)^2
$$
|
From $a-b=3$ and $(a-b)(a+b)=21$ we get $a-b=3$ and $a+b=7$ and $a=5$, $b=2$.
|
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|
How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
|
We know that $\dfrac{99!}{100!}=\dfrac{1}{100}$
We rewrite $\dfrac{1}{100}=\dfrac{99!}{100!}=\dfrac{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{red}{98}\cdot\color{blue}{99}}{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{blue}{99}\cdot\color{red}{100}}=\dfrac{99!!\cdot 98!!}{100!!\cdot 99!!}$
We know trivially that both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are positive.
Now., there are four possibilities
*
*both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are less than $\frac{1}{10}$, which would lead to a contradiction as their product would be less than $\frac{1}{100}$
*both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are greater than or equal to $\frac{1}{10}$, which if at least one is strictly greater than $\frac{1}{10}$ would yield a contradiction as their product would be greater than $\frac{1}{100}$. If both are equal to $\frac{1}{10}$, then using a prime factorization argument yields a contradiction.
*$\frac{99!!}{100!!}$ is less than $\frac{1}{10}$ and $\frac{98!!}{99!!}$ is greater than $\frac{1}{10}$ (this is what we want to show must be the case)
*$\frac{99!!}{100!!}$ is greater than $\frac{1}{10}$ and $\frac{98!!}{99!!}$ is less than $\frac{1}{10}$. (we want to show this can't be the case)
So, the question has become proving whether or not $\frac{(n-1)!!}{n!!}$ must be less than $\frac{n!!}{(n+1)!!}$
|
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|
Evaluate $\lim_{(x,y,z)\to (0,0,0)}\frac{1}{x^2+y^2+z^2}e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}$
Evaluate $$\lim_{(x,y,z)\to (0,0,0)}\frac{1}{x^2+y^2+z^2}e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}$$
Let $t=\sqrt{x^2+y^2+z^2}$
$$\lim_{t\to 0}\frac{1}{t^2}e^{-\frac{1}{t}}$$
Let $r=\frac{1}{t}$
$$\lim_{r\to \infty}r^2 e^{-r}=\lim_{r\to \infty}\frac{r^2}{e^r}$$
using l'hopital
$$\lim_{r\to \infty}\frac{2r}{e^r}$$
using l'hopital again
$$\lim_{r\to \infty}\frac{2}{e^r}=0$$
Is the following valid? is there is a simpler way?
|
HINT
By spherical coordinates $\sqrt{x^2+y^2+z^2}=r\to 0$
$$\frac{1}{ x^2+y^2+z^2 }e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}=\frac1{r^2}e^{-\frac1r}$$
and for $y=\frac1r\to \infty$ since eventually $e^y\ge y^3$
$$\frac1{r^2}e^{-\frac1r}=\frac{y^2}{e^y}\le\frac{y^2}{y^3}=\frac1y\to 0$$
|
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|
How to determine whether an integral is convergent or divergent? For this question, I'm not sure if I'm doing it right, can anyone please help me out?
Determine whether the following integral is convergent or divergent.
$$\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$$
$$ \frac{1}{(x-3)\sqrt {x-5}}\le \frac{1}{x-3}$$
Since $\int_5^6 \frac 1 {x-3} \,dx$ converges, $\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$ must also converge.
|
The inequality is not valid. Rather,
\begin{align*}
0<\dfrac{1}{(x-3)\sqrt{x-5}}<\dfrac{1}{2\sqrt{x-5}},~~~~x\in(5,6],
\end{align*}
and
\begin{align*}
\int_{5}^{6}\dfrac{1}{\sqrt{x-5}}&=\lim_{\eta\rightarrow 5^{+}}\int_{\eta}^{6}\dfrac{1}{\sqrt{x-5}}dx\\
&=\lim_{\eta\rightarrow 5^{+}}2\sqrt{x-5}\bigg|_{\eta}^{6}\\
&=\lim_{\eta\rightarrow 5^{+}}\left(2-2\sqrt{\eta-5}\right)\\
&=2\\
&<\infty.
\end{align*}
|
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|
Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
|
If we allow $y$ and $z$ to be zero, this can be rephrased as:
For which positive integers $x$ is $2^{x} - 1$ is not divisible by any prime other than $3$ or $7$?
Call the set of such integers $S$.
Suppose $x \in S$. If $t$ divides $x$, then $2^{t} - 1$ divides $2^{x} - 1$, so $t \in S$ as well.
$3$ divides $2^{x} - 1$ iff $x$ is even, and $7$ divides $2^{x} - 1$ iff $x$ is divisible by 3. So we see that if $x \in S$, we must have either $x$ even, $x$ divisible by 3, or $2^{x} - 1 = 1$, i.e. $x = 1$.
Then if $x \in S$ has a prime factor $p$, $p \in S$, and so we must have $p = 2$ or $p = 3$.
Finally, we see $2^{4} - 1 = 15$ and $2^{9} - 1 = 511$, hence $4$ and $9$ are not in $S$, so no element of $S$ is divisible by $4$ or $9$. The only remaining integers are $1,2,3,6$. We can check $2^{1} - 1 = 1$, $2^{2} - 1 = 3$, $2^{3} - 1 = 7$, and $2^{6} - 1 = 63 = 3^{2} \cdot 7$. So $S = \{1,2,3,6\}$.
Finally, $x = 6$ is the only one which gives $y,z$ positive.
|
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|
guess 3 of 5 repeating non-ordered numbers from 1 to 6 I'm trying to calculate the probability of guessing 3 numbers on a set of 5.
The range is $[1, 6]$, the order is not relevant and numbers can repeat.
I did a small script to calculate it by brute force, and I got something unexpected.
Choosing 3 identical numbers (e.g. 1, 1, 1) has a lower probability of drawing than 2 identical numbers, and this is also lower than 3 different numbers.
I can't figure out why.
|
Guessing $1,1,1$ will match the sets that have $3$ or more $1$. There is $1$ set of $5$ ones, there are $5*5=25$ sets that have $4$ ones ($5$ positions for the non-one to be in and $5$ options for the value of the non-one), there are ${5\choose 2}*5 = 50$ ways to have $3$ ones and two of another value. And ${5\choose 2}5*4 = 200$ ways to have $3$ ones and two other values. This is $255$ ways.
Guessing $2,2,1$ will match sets that have $4$ twos and $1$ one: $5$ sets. $3$ twos and $2$ ones: ${5 \choose 2} = 10$ sets. $3$ twos, $1$ one and something else: ${5\choose 3}{2\choose 1}*4 = 80$ sets. $2$ twos, and $3$ ones: ${5\choose 2} = 10$ sets. $2$ twos, $2$ ones, and something else: ${5\choose 2}{3\choose 2}*4 = 120$ sets. $2$ twos, $1$ one and $2$ of something else: ${5\choose 2}*{3\choose 1}*4 = 120$ sets. $2$ twos, $1$ one and two other things: ${5\choose 2}{3\choose 1}4*3 = 360$. Or $360+120+120+10+80+10+5 = 705$ sets. Many more options!
Guessing $1,2,3$ can mean.... well, three of one of the numbers and one of the other two: $3*{5\choose 3}*2! = 60$( three choices for the triple, choices for where to put the triples, ways to place the remaining two). Or two of two of them and one of then third: $3{5\choose 2}{3\choose 2}= 90$ (Three choices for the single, where to place the first of te doubles, where to place the other doubles.) Two of one of them, one of the other two, and forth number. $3{5\choose 2}{3\choose 1}{2\choose 1}*3 = 540$. (3 chooses for the double, where to put them, where to put the singles, and what the fourth number is. Or exaclty one of each. ${5\choose 3}*3!*(3)^2 = 540$ (places to put the three, order to place the three within their posistions, 9 posible value of the remaining two numbers.) For $60 + 90 +540+540 = 1230$ possible ways.
|
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|
Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$
Then I got stucked, could anyone help me in solving it?
|
HINT:
We can write
$$\begin{align}
\frac{\tan(x)-\sin(x)}{x^3}&=\frac{\sin(x)(1-\cos(x))}{\cos(x)x^3}\\\\
&=\left(\frac{1}{\cos(x)}\right)\left(\frac{\sin(x)}{x}\right)\left(\frac{1-\cos(x)}{x^2}\right)
\end{align}$$
|
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|
Proving $\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2$ without calculus I would like to prove the following inequality without using calculus :
$$
\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2
$$
Any hint?
Thank you very much!
|
We have
\begin{align*}
\frac{1}{n\sqrt{n+2}}\leq\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}},\qquad n\geq 2,
\end{align*}
and here is why: First note that $\sqrt{n^2-1}\leq n$ for all $n\geq 1$. Moreover, $\sqrt{n+1}-\sqrt{n-1}=\frac{2}{\sqrt{n+1}+\sqrt{n-1}}\geq \frac{1}{\sqrt{n+2}}$. From this we get
\begin{align*}
\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n^2-1}}\geq\frac{1}{n\sqrt{n+2}},
\end{align*}
as claimed.
Now, we obtain
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n\sqrt{n+2}}&\leq\sum_{n=1}^6\frac{1}{n\sqrt{n+2}}+\sum_{n=7}^\infty\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\right) \\ &=\frac{1}{4}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{5}}+\frac{5}{4 \sqrt{6}}+\frac{6}{5 \sqrt{7}}+\frac{1}{12 \sqrt{2}}<2.
\end{align*}
|
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|
On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$. We wish to prove that $3-\epsilon$ is not an upper bound for $A$.
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{2\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
EDIT:
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
|
It is fine. Only a technical detail.
$$3-\frac1{2N^2}=3-\frac1{4n_0^2}>3-\frac{2\epsilon}4$$
This is certainly greater than $3-\epsilon$, so your proof is correct, but it would be more elegant if you simply say that $N>1/\sqrt\epsilon$ and even.
|
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|
Calculate $f(x)$ at a specific point Question:
Calculate $f(x) = \frac{49}{x^2} + x^2$ at points for which $\frac{7}{x}+x =3$
My attempt:-
I tried to find the value of $x$ and insert in $f(x)$
$$\frac{7}{x}+x =3$$
$$7+x^2 =3x$$
$$x^2 -3x + 7=0$$
$$x = \frac{3+\sqrt{9-7*4}}{2}$$
Now $x$ is coming out to be irrational and things get a lot more difficult from there. What should I do?
|
For $x \not =0$, $x$ real:
1) No real, negative $x$ satisfies above equation.
2) For $x >0$ :AM-GM:
$(7/x) +x \ge 2\sqrt{7} \gt 4 $.
Hence there is no real $x$ with: $(7/x) +x = 3$ .
Left to do, find a complex solution.
Note:
$(7/x+x)^2 =$
$ 49/x^2 +2(7) + x^2 =9.$
$49/x^2 +x^2 =-5.$
Hence $f(x)=-5.$
|
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|
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \sin \frac{x}{2} - \cos \frac{x}{2}$
So, $\frac{dy}{dx} = \frac{1}{2} \cdot (\cos\frac{x}{2} + \sin\frac{x}{2})$.
But apparently this is wrong. The correct solution is:
$\frac{dy}{dx} = -\frac{1}{2}\cdot(\cos\frac{x}{2} + \sin\frac{x}{2})$.
So I want to know what I have done wrongly here. Why is my answer not right?
|
An option:
$y^2 = 1-\sin x$; $0\lt x\lt π/2.$
Differentiate both sides with respect to x:
$2y\dfrac{dy}{dx} = -\cos x;$
Since $y \not =0:$
$\dfrac{dy}{dx} = -\dfrac{\cos x}{2\sqrt{1-\sin x}}$.
|
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|
Proving that an equation holds when we divide a square in triangles Consider the following square of side $1$ divided into right triangles:
Prove that the following equation holds:
I tried first using the Pythagorean theorem, getting 7 equations, one for each triangle, and after combining the equations, I got the following:
$$BI^2=-DE^2-FH^2-HI^2+2(GI)(EG)+2(GI)(CE)+2(EG)(CE).$$
Also, the sum of the areas of the triangles is $1$, and the area of each triangle is the product of the legs divided by $2$, but I know know if this can help (combined with the equation I got), or am I going on the wrong way?
|
First prove all the triangles are similar. This is easy as all angles are right or supplementary or congruent.
So
$AH = x$. $AB = 1$. $\frac {BI}{x} = \frac 1{\sqrt{1 + x^2}}; BI = \frac x{\sqrt{1+x^2}}$. $HI = \sqrt{ 1+x^2}- \frac x{\sqrt{1+x^2}}$.
$\frac {IG}{HI} = \frac x1; IG = x(\sqrt{ 1-x^2}- \frac x{\sqrt{1-x^2}});\frac {IG}{HG} = \frac {IB}1$. etc.
And so on.
By the end you will get $IG + GE +EC = \frac 1{\sqrt{1 - x^2}}$ and that $EC^2 + ED^2 = 1$.
One assumes putting all in terms of $x$ will yield the equation given.
====
Okay, let's put this together.
What is the scaling factor of each triangle:
$\triangle AHB$ is our basic the triangle with factor of $1$. It's dimensions are: $AH = x; AB= 1; BH = \sqrt{1 + x^2}$
$\triangle BIC$ is proportional ot $\triangle AHB$ and and it's sides are in proportion of $\frac {BI}{AH} = \frac {BC}{BH}=\frac {IC}{AB} = \frac{BI}x = \frac {1}{\sqrt {1 + x^2}} = {IC}$. So scaling factor is $\frac 1{\sqrt{1+x^2}}$. (Note: every scaling factor is the length of the long leg.)
$\triangle HIC$ has scaling factor of $HI= BH - BI = \sqrt{1+x^2} - AH*\frac 1{\sqrt{1+x^2}}= \frac {\sqrt{1+x^2}\sqrt{1+x^2}}{\sqrt{1+x^2}} - \frac x{\sqrt{1+x^2}}=\frac {1 - x +x^2}{\sqrt{1+x^2}}$
$\triangle HFG$ has scaling factor $HG =\frac {1 - x +x^2}{\sqrt{1+x^2}}*HB = 1-x + x^2$.
$\triangle FGE$ has scaling factor $FG = (1-x + x^2)HB = (1-x+x^2)\sqrt{1+x^2}$.
$\triangle FDE$ has scaling factor $FE = (1-x+x^2)\sqrt{1+x^2}*HB = (1-x+x^2)(1+x^2)$.
Finally $\triangle DEC$ has scaling factor $DE = (1-x+x^2)(1+x^2)*HB = (1-x+x^2)(1+x^2)\sqrt{1 + x^2}$.
So $1 = DC = HB* (1-x+x^2)(1+x^2)\sqrt{1 + x^2} = (1-x+x^2)(1+x^2)^2$.
So $(1+x^2)^2(1-x+x^2) - 1 = 0$
$(1 + 2x^2 + x^4)(1 - x +x^2)-1 = (1 + 2x^2 + x^4)- (x + 2x^3 + x^5)+(x^2 + 2x^4 + x^6)-1= x^6-x^5 +3x^4 -2x^3 +3x^2 - x + 1-1 =x(x^5-x^4 +3x^3 -2x^2 +3x - 1)= 0$
So $x^5-x^4 +3x^3 -2x^2 +3x - 1=0$.
|
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|
How to find the sum of all elements of inverse of a matrix without finding the inverse explicitly? Problem Statement:
Find all real $x$ so that the matrix is invertible. $$\begin{bmatrix}
2+x & 2 & 2 & 2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x & 2 \\ 2 & 2 & 2 &
2+x \end{bmatrix}$$
Assuming that $A^{-1}$ exists find the sum of all the elements of
$A^{-1}$ without explicitly computing $A^{-1}$.
For the first part I calculated the determinant (came out to be $x^4+8x^3$) and put it to $0$ to find that the matrix is not-invertible if $x=-8$ or $x=0$. But I've not idea how to approach the second part of the problem.
|
As suggested here, the sum of all elements in a matrix $M$ is given by $\langle Me, e\rangle$ where $e = \pmatrix{1 \\ 1 \\ 1 \\ 1}$.
In that case, if $v$ is a vector such that $Av = e$, then we have $$\langle A^{-1}e, e\rangle = \langle v, e\rangle$$
Therefore, we have to solve the system
$$\pmatrix{2+x & 2 & 2 &2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x &2 \\ 2 & 2 & 2 & 2+x} \pmatrix{v_1 \\ v_2 \\ v_3 \\ v_4} = \pmatrix{1 \\ 1 \\ 1 \\ 1}$$
Since $A$ is invertible, the solution is unique and it is given by $$v = \frac{1}{x+8}\pmatrix{1 \\ 1 \\ 1 \\ 1}$$
Therefore the sum is equal to
$$\langle v, e\rangle = \frac1{x+8}\|e\|^2 = \frac{4}{x+8}$$
Note that solving the linear system above is in general faster than inverting the matrix, especially in this case since your $A$ is symmetric.
|
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|
Some fundamental theory of calc 2 questions. I want to evaluate these integrals. I think I understand it, but I just wanted to check:
*
*$$\int_{-1}^1 x^{100} dx$$
$$\frac{1^{101}}{101} - \frac{-1^{101}}{101}$$
$$= \frac{1}{101} + \frac{1}{101} = \frac{2}{101}$$
*$$\int_{1}^9 \sqrt{x} dx$$
$$\frac{2}{3} \cdot 9^{\frac{3}{2}} - \frac{2}{3}$$
$$ = 18 - \frac{2}{3}$$
$$17 \frac{1}{3}$$
*$$\int_{1}^8 x^{\frac{-2}{3}} dx$$
So an antiderivative we can use is: $F(x) = \frac{x^{\frac{1}{3}}}{3}$
$$ \frac{8^{\frac{1}{3}}}{3} - \frac{1^{\frac{1}{3}}}{3}$$
$$ = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$$
*$$\int_{\frac{\pi}{6}}^\pi \sin \theta dx$$
$$ - \cos \pi - -\cos \frac{\pi}{6} $$
$$ = 1 + \frac{\sqrt{3}}{2}$$
$$= \frac{2 + \sqrt{3}}{2}$$
|
As @littleO points out, the antiderivative of part 3 is incorrect.
$$\int_{1}^8 x^{-2/3} dx = \left[ \frac{x^{1/3}}{1/3} \right]_1^8 = [3\sqrt[3]{x}]_1^8 = 3(2-1) = 3$$
The rest is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Asymptotic for the following series. $$S_N=\frac{\sum_{i=1}^{2N}\sqrt{i}+\sqrt{i-1}}{\sum_{i=1}^{N}\sqrt{i}+\sqrt{i-1}}$$
I want to find an equivalence/asymptotic for $S_N$ as $N$ become very large. I tried the following:
Edit
We know that $$\sum_{i=1}^{N}\sqrt{i}\approx\frac{2N^{3/2}}{3}$$ and so $$S_N\approx\frac{(2N)^{3/2}+(2N-1)^{3/2}}{N^{3/2}+(N-1)^{3/2}}\approx 2\sqrt{2}.$$ Is this estimation correct and also how can it be improved?
|
Here is a
reasonably elementary proof
for an arbitrary exponent,
not just $\frac12$.
It shows that
if the exponent is $a$
with $a > 0$,
then the limit is
$2^{a+1}$.
It also gives explicit bounds.
It is based on
this:
If $a > 0$ then
$\dfrac{n^{a+1}}{a+1}+n^a
\gt \sum_{1}^{n} k^a
\gt \dfrac{n^{a+1}}{a+1}
$.
(Similar bounds can be gotten
for $a < 0$.)
Proof:
$(k-1)^a
< \int_{k-1}^{k} x^a dx
\lt k^a
$
so
$\sum_{k=1}^n (k-1)^a
\lt \sum_{k=1}^n \int_{k-1}^{k} x^a dx
\lt \sum_{k=1}^nk^a
$
or
$\sum_{k=0}^{n-1} k^a
\lt \int_{0}^{n} x^adx
\lt \sum_{1}^{n} k^a
$
so that,
since
$\int_{0}^{n} x^adx
=\dfrac{n^{a+1}}{a+1}
$,
$-n^a
\lt \dfrac{n^{a+1}}{a+1}-\sum_{1}^{n} k^a
\lt 0
$.
or
$\dfrac{n^{a+1}}{a+1}+n^a
\gt \sum_{1}^{n} k^a
\gt \dfrac{n^{a+1}}{a+1}
$.
Let
$p(n)
=\sum_{k=1}^n (k^a+(k-1)^a)
$
and
$S(n)
=\dfrac{p(2n)}{p(n)}
$.
$\begin{array}\\
p(n)
&=\sum_{k=1}^n (k^a+(k-1)^a)\\
&=\sum_{k=1}^n k^a+\sum_{k=1}^n(k-1)^a\\
&=\sum_{k=1}^n k^a+\sum_{k=0}^{n-1}k^a\\
&=\sum_{k=1}^n k^a+\sum_{k=1}^{n-1}k^a\\
&=2\sum_{k=1}^n k^a-n^a\\
\end{array}
$
so that
$2\dfrac{n^{a+1}}{a+1}-n^a
\lt p(n)
\lt 2\dfrac{n^{a+1}}{a+1}+2n^a
$.
Therefore
$\begin{array}\\
S(n)
&\lt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}+2(2n)^a}{2\dfrac{n^{a+1}}{a+1}-n^a}\\
&= \dfrac{ 2^{a+2}+(a+1)2^{a+1}/n}{2-(a+1)/n}\\
&= \dfrac{ 2^{a+1}+(a+1)2^{a}/n}{1-(a+1)/(2n)}\\
\end{array}
$
and
$\begin{array}\\
S(n)
&\gt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}-(2n)^a}{2\dfrac{n^{a+1}}{a+1}+2n^a}\\
&= \dfrac{ 2^{a+2}-2(a+1)2^{a+1}/n}{2+2(a+1)/n}\\
&= \dfrac{ 2^{a+1}-2(a+1)2^{a}/n}{1+2(a+1)/n}\\
\end{array}
$
Therefore
$\lim_{n \to \infty} S(n)
=2^{a+1}
$.
|
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|
$ x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}> 0 $ Solve the equation
$$x^{3}+ x+ 6= 2\left ( x+ 1 \right )\sqrt{3+ 2x-x^{2}}$$
The only answer is $x= 1$ so
I have rewritten it to
$$\left ( x- 1 \right )\left ( x^{2}- x- 4 \right )= 2\left ( x+ 1 \right ) \left (\sqrt{3+ 2x-x^{2}}- x- 1 \right )= 2\left ( x+ 1 \right )\frac{3+ 2x- x^{2}- \left ( x+ 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}= \frac{-4\left ( x+ 1 \right )^{2}\left ( x- 1 \right )}{\sqrt{3+ 2x-x^{2}}+ x+ 1}\Leftrightarrow \left ( x- 1 \right ) \left ( x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1} \right ) = 0$$
I need to prove $$ x^{2}- x- 4+ \frac{4\left ( x- 1 \right )^{2}}{\sqrt{3+ 2x-x^{2}}+ x+ 1}> 0 $$
Who can help me? Thanks!
|
Let $t=x-1$ and rewrite the equation like this
$$ t^3+3t^2+4t+8 = 2(t+2)\sqrt{4-t^2}$$
We see that $t=0$ (and so $x=1$) is a solution. Let us show it is the only one.
We also see that $t\in[-2,2]$. Now it is easy to see that $y=4t+8$ is tangent at $f(t) = t^3+3t^2+4t+8$. Let's prove that for all $t\in[-2,2]$ we have:
$$ 2(t+2)\sqrt{4-t^2} \leq 4t+8$$
If $t=-2$ both sides are equal so let's $t>-2$. Then $t+2>0$ so we have
$$ \sqrt{4-t^2} \leq 2 $$
which is true.
Since $$f(t) > 4t+8\geq 2(t+2)\sqrt{4-t^2}$$ for each $t\in[-2,2]\setminus \{0\}$ we see that $t=0$ is the only solution.
|
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|
Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
|
Maybe there is a trick to it, but good ol' factoring can work here.
It is easy to check (using the identity $a^3=-a-1$) that
$$ (3a^2-2a+2)(3a^2+2a) = -5a-7a^2 $$
Let $\sqrt[3]{-5a-7a^2}+a^2 = y$. Given that $a\in \mathbb{R}$, we know that also $y\in \mathbb{R}$, and we must solve for it.
$$y-a^2 = \sqrt[3]{-5a-7a^2} \\ (y-a^2)^3 = -5a-7a^2 \\ y^3 - 3y^2a^2 + 3ya^4-a^6 = -5a-7a^2 \\ y^3-3y^2a^2+3ya(-a-1)-(-a-1)^2=-5a-7a^2 \\ y^3-3y^2a^2-3ya^2-3ya+6a^2+3a-1=0$$ So many threes, only the $6$ stands out. But if we use $6=3+3$: $$ (y^3-1)+(3a^2-3y^2a^2)+(3a^2+3a-3ya^2-3ya)=0 \\ (y^3-1)-3a^2(y^2-1)-(3a^2+3a)(y-1)=0$$ As one can see, it is possible to factor out $(y-1)$, thereby concluding that $y=1$ is a solution.
|
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|
How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$ First I looked the x that doesnt belong to this function.
$$|4x -6| - 21 \neq 0$$
$$ x \neq \frac{-15}{4}$$ and $$ x \neq \frac{27}{4}$$
Then I found the roots of the x
$$x = 0$$
$$x = \frac{3}{2}$$
After I found the roots I wrote the inequality like this:
$$ x + 14|x| - 10 > 3|4x-6| - 63$$
$$ x + 14|x| > 3|4x-6| - 53$$
to finish we may write differents function for each values of x based on roots we found.
for $$ x<0$$ the function is $$ x - 14x > 3(6-4x) - 53$$
for $$x<\frac{3}{2}$$ we have $$ x + 14x > 3(6-4x) - 53$$ for $$x \geq \frac{3}{2}$$ the function is $$ x + 14x > 3(4x-6) - 53$$
But using this inequalities I couldnt find the solutions for x!! If we calculate that on wolfram we can see the solutions for x are $$ x < \frac{-15}{4}$$ and $$ x > \frac{27}{4}$$. Can anyone explain me why?
|
Transform the inequality:
$$\frac{x+14|x|-10-3|4x-6|+63}{|4x-6|-21}>0.$$
Consider the $3$ cases:
$$1) \ \begin{cases} x<0 \\ \frac{x-35}{4x+15}>0\end{cases} \Rightarrow \begin{cases} x<0 \\ x<-\frac{15}{4} \ \ \text{or} \ \ x>35\end{cases} \Rightarrow \color{blue}{x<-\frac{15}{4}}.$$
$$2) \ \begin{cases} 0<x<\frac32 \\ \frac{27x+35}{-4x-15}>0\end{cases} \Rightarrow \begin{cases} 0<x<\frac32 \\ -\frac{15}{4}<x<-\frac{35}{27}\end{cases} \Rightarrow \emptyset.$$
$$3) \ \begin{cases} x>\frac32 \\ \frac{3x+71}{4x-27}>0 \end{cases} \Rightarrow \begin{cases} x>\frac32 \\ x<-\frac{71}{3} \ \ \text{or} \ \ x>\frac{27}{4}\end{cases} \Rightarrow \color{blue}{x>\frac{27}{4}}.$$
|
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|
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
|
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin x} d x
=& \int_{0}^{\frac{\pi}{2}} x^{2} d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\
=& {\left[x^{2} \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} x \ln \left(\tan \frac{x}{2}\right) d x } \\
=&-8 \int_{0}^{\frac{\pi}{4}} y \ln (\tan y) d y \textrm{, where }y=2x\\
=&-8\left[-\frac{\pi}{4} G+\frac{7}{16} \zeta(3)\right] \\
=& 2 \pi G-\frac{7}{2} \zeta(3)
\end{aligned}
$$
where $
\int_{0}^{\frac{\pi}{4}} x \ln (\tan x) d x= -\frac{\pi}{4} G+\frac{7}{16} \zeta(3)
$ from my post .
Can we go further with the integral
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{x^{n}}{\sin x} d x ?\tag*{} $
|
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|
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)^n(a-b)=n\pi+(-1)^n(\sin^{-1}x-\sin^{-1}\sqrt{x})
$$
Thus,
$$
y'=\frac{d}{dx}\big[n\pi+(-1)^n(a-b)\big]=\begin{cases}\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\text{ if }n\text{ is even}\\
-\bigg[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\bigg]\text{ if }n\text{ is odd}
\end{cases}
$$
Is it the right way to solve this problem and how do I check the solution is correct ?
Note: I think there got to be two cases for the derivative as the graph of the function is
|
Use that $$(\arcsin(x))'=\frac{1}{\sqrt{1-x^2}}$$
the whole derivative is given by $$f'(x)={\frac {1}{\sqrt {- \left( x\sqrt {1-x}+\sqrt {x}\sqrt {-{x}^{2}+1}
\right) ^{2}+1}} \left( \sqrt {1-x}-1/2\,{\frac {x}{\sqrt {1-x}}}+1/2
\,{\frac {\sqrt {-{x}^{2}+1}}{\sqrt {x}}}-{\frac {{x}^{3/2}}{\sqrt {-{
x}^{2}+1}}} \right) }
$$
after simplifying i got:
$$\frac {\sqrt {x} \left (3 x
\left (\sqrt {x} + \sqrt {x + 1} \right) - 2
\sqrt {x + 1} \right) - 1} {2 \sqrt {(-x + 1) x}
\sqrt {x + 1} \sqrt {2 (x - 1) \sqrt {x + 1} x^{3/2} + 2
x^3 - x^2 - x + 1}}$$
|
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|
Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e^{2x}+4}$
let $u = e^{2x} +4$
$\frac{dy}{dx}=2e^{2x}$
$dx=\frac{1}{2}e^{-2x}du$
$u = e^{2x} +4$
$e^{2x} = u-4$
$e^{-2x} = \frac{1}{u-4}$
$dx=\frac{1}{2}(\frac{1}{u-4})du$
Hence the integral is:
$$
\begin{aligned}
&\int\frac{2}{e^{2x}+4} \\
=& 2\int\frac{1}{u}\frac{1}{2}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u^2-4u}du\\
=&\int u^{-2}-\frac{1}{4}u^{-1}du\\
=&\frac{u^{-1}}{-1}-\frac{1}{4}\ln(u)+c\\
=&-\frac{1}{e^{2x}+4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
=&-e^{-2x}-\frac{1}{4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
\ne& \frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c\\
\end{aligned}
$$
???? help
|
You wrote
$$\int\frac{1}{u^2-4u}du=\int u^{-2}-\frac{1}{4}u^{-1}du$$
Which is incorrect
Try writing it as
$$\int\frac{1}{u^2-4u}du=\frac{1}{4}\int \frac{1}{u-4}-\frac{1}{u}du$$
Or
$$\int\frac{1}{u^2-4u}du=\int\frac{1}{(u-2)^2-4}du$$
And proceed through substitution
|
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|
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or?
|
Rather than using the constraint to find $x$ in terms of $y$ or vice versa, use implicit differentiation.
Differentiate the objective and set it equal to $0.$
$A = 4xy\\
\frac {dA}{dx} = 4y + 4x\frac {dy}{dx} = 0$
Differentiate the constraint.
$\frac {d}{dx} (4x^2 + 9y^2 = 3600)\\
8x + 18y\frac {dy}{dx} = 0\\
\frac {dy}{dx} = -\frac {4x}{9y}$
And now do your substitutions.
$-4x^2 + 9y^2 = 0\\
4x^2 + 9y^2 = 3600\\
y^2 = \frac {3600}{18}\\
|y| = \frac {20}{\sqrt 2}\\
|x| = \frac {30}{\sqrt 2}\\
4xy = 1200$
|
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|
Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be something easy I just can't see.
Thanks in advance.
|
We can rewrite the fraction as follows
\begin{align*}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}\times\frac{\sqrt{x^2+a^2}-x}{\sqrt{x^2+a^2}-x}\times \frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+b^2}-x}\\[4pt]
&=\frac{x^2+a^2-x^2}{x^2+b^2-x^2}\times\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x}
\end{align*}
So
\begin{align*}
\lim_{x\to-\infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x}\\[4pt]
&=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{\frac{x^2}{x^2}+\frac{b^2}{x^2}}-\frac x{|x|}}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2}}-\frac x{|x|}}\\[4pt]
&=\frac{a^2}{b^2}\times\frac{\sqrt{1}-(-1)}{\sqrt{1}-(-1)}\\[4pt]
&=\frac{a^2}{b^2}
\end{align*}
|
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|
Definite integral of $y=\sqrt{(16-x^2)}$ The integral is
$$\int^3_1\sqrt{16-x^2}dx$$
I've used the trig substitution method, replacing $x$ with $4\sin\theta$:
$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$
(I've excluded the intervals for the definite integral for now)
\begin{align}
I&=\int\sqrt{16-16\sin^2\theta} \ 4\cos\theta d\theta \\
&=\int\sqrt{16\cos^2\theta}\ 4\cos\theta d\theta\\
&=\int4\cos\theta 4\cos\theta d\theta \\
&=16\int\cos^2\theta d\theta \\
&=16\int\frac{\cos2\theta}{2}d\theta \\
&=8\int \cos2\theta d\theta\\
&=\left[4\sin2\theta\right]^{\arcsin(3/4)}_{\arcsin(1/4)}\\
&=4\sin\left(2\arcsin\frac 34\right)-4\sin\left(2\arcsin\frac 14\right)\\
\end{align}
At this last step, I'm not sure how to simplify the 2 arcsin part. It doesn't simplify because of the 2 in front of the arcsin, is there any other way? Or did I just do the wrong method to solve this integral in the first place?
|
$\sin(2\sin^{-1}(3/4))=2\sin(\sin^{-1}(3/4))\cos(\sin^{-1}(3/4))=2\cdot(3/4)\cdot(\sqrt{7}/4)$ because of the identity that $\cos(\sin^{-1}(3/4))=\sqrt{1-\sin^{2}(\sin^{-1}(3/4))}=\sqrt{1-9/16}=\sqrt{7}/4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using the Euclidean algorithm, deduce that $\gcd(x^3+2x^2+x +4;x^2+1)=1$ So, I've tried it but I keep getting to $$\frac{x^2+1} 2$$ and don't know how to proceed.
Question is Using the Euclidean algorithm, deduce that $$\gcd(x^3+2 x^2+x +4,\;x^2+1)=1.$$
|
$$ \left( x^{3} + 2 x^{2} + x + 4 \right) $$
$$ \left( x^{2} + 1 \right) $$
$$ \left( x^{3} + 2 x^{2} + x + 4 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x + 2 \right) } + \left( 2 \right) $$
$$ \left( x^{2} + 1 \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x + 2 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x + 2 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 2 x^{2} + x + 4 }{ 2 } \right) }{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } $$
$$ \left( x^{3} + 2 x^{2} + x + 4 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} + 1 \right) \left( \frac{ x + 2 }{ 2 } \right) = \left( 1 \right) $$
.......
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2724855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic progression, so
$$\begin{align}
2b&=a+c \\[4pt]
2\sin B &= \sin A+ \sin C \\[4pt]
2\sin(A+C) &=2\sin\frac {A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\sin\frac{A+C}{2}\;\cos\frac{A+C}{2}&=\sin\frac{A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\cos\frac{A+C}{2}&=\cos\frac{A-C}{2}
\end{align}$$
|
Expand your last line: $$2\left(\cos\frac A2\cos\frac C2 - \sin\frac A2\sin\frac C2\right)=\left(\cos\frac A2\cos\frac C2 +\sin\frac A2\sin\frac C2\right)$$
and your result is immediate after a cancellation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the area of circles in triangle In the figure there are infinitely many circles approaching the
vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of
length $1$, how can i find the total area occupied by the circles?
|
Denote the circumradius by $R$ and the radii of the circles by $r_1,r_2,r_3,...$.
We will find $R$ and $r_1$:
$$r_1=\frac{2S}{a+b+c}=\frac{2\cdot\frac{\sqrt{3}}{4}}{3}=\frac{1}{2\sqrt{3}} \\
R=\frac{abc}{4S}=\frac{1}{4\cdot \frac{\sqrt{3}}{4}}=\frac{1}{\sqrt{3}}.$$
From the similarity of the triangles:
$$\begin{align}\frac{r_2}{r_1}=\frac{R-r_1-r_2}{R} \Rightarrow r_2=&\frac{1}{2\cdot 3\sqrt{3}}, \\
\frac{r_3}{r_1}=\frac{R-r_1-2r_2-r_3}{R} \Rightarrow r_3=&\frac{1}{2\cdot 3^2\sqrt{3}}, \\
\frac{r_4}{r_1}=\frac{R-r_1-2r_2-2r_3-r_4}{R} \Rightarrow r_4=&\frac{1}{2\cdot 3^3\sqrt{3}}, \\
\vdots \\
r_n=&\frac{1}{2\cdot 3^{n-1}\sqrt{3}}
\end{align}$$
Hence:
$$\begin{align}S=&\pi r_1^2+3\cdot \pi \cdot \left(r_2^2+r_3^2+r_4^2+\cdots\right)=\\
&\frac{\pi}{12}+3\cdot \pi\cdot \left(\frac{1}{4\cdot 3^3}+\frac{1}{4\cdot 3^5}+\frac{1}{4\cdot 3^7}+\cdots\right)=\\
&\frac{\pi}{12}+\frac{3\pi}{4}\cdot \frac{\frac{1}{3^3}}{1-\frac{1}{3^2}}=\\
&\frac{\pi}{12}+\frac{\pi}{32}=\\
&\frac{11\pi}{96}.\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$
Given
$$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
and $a\in\Bbb{R}$, express $x$ in terms of $a$.
I rationalised the above expression and then again rationalised which gave me :
$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\sqrt{a-x}}$$
Now, what should I do?
If didn't rationalised the original Equation, or if the squared both sides twice then too bi-quadratic or higher degree polynomial will be obtained.
What should be done?
|
Squaring both sides, $$a\require{\cancel}\cancel{+\sqrt{a-x}}+a\cancel{-\sqrt{a-x}}+2\sqrt{\left(a+\sqrt{a-x}\right)\left(a-\sqrt{a-x}\right)}=2\left(a+\sqrt{a^2-a+x}\right)=4x^2.$$ since for all $m,n$ one has that $(m+n)(m-n)=m^2-n^2$. Now, dividing both sides by $2$,$$\begin{align} a+\sqrt{a^2-a+x}&=2x^2 \\ \Leftrightarrow (2x^2-a)^2 &= 4x^4 + a^2 - 4ax^2 \\ &= a^2 - a + x.\end{align}$$ Cancel out the term $a^2$ from both sides and expanding,$$4x^2(x^2-a) = x-a.$$ Now, notice that $$\frac{x^2-a}{x-a} = \frac{x^2-a+x-x}{x-a} = \frac{x^2-x}{x-a}+1.$$ Therefore, $$\begin{align}4x^2\left(\frac{x^2-a}{x-a}\right)= 4x^2\left(\frac{x^2-x}{x-a}+1\right)&=1 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a}+4x^2&=1.\end{align}$$ Now, since $4x^2 = (2x)^2$ then converting it to that and subtracting $1$ from both sides, $$\begin{align} \frac{4x^3(x-1)}{x-a} + (2x+1)(2x-1)&=0 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a} + x(2x+1) + (x-1)(2x+1) &= 0 \\ \Leftrightarrow (x-1)\left(\frac{4x^3}{x-a} + (2x+1)(x+1)\right)&=0.\end{align}$$ Solve for $x$ and $a$. To start you off, $x=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2)$. I want to prove that
$$ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2),$$
where $B(\cdot , \cdot)$ is the beta function.
My idea was to change it to something like
my previous question.
Edit 1.It follows from the absorption formula that
$$ \binom{2j+1}{k+j+1} = \frac{(2j+1)(2j) \ldots(j+2)(j+1)}{(k+j+1)(k+j) \ldots(k+2)(k+1)} \binom{j}{k}.$$
How can I go further with this binomial series?
Edit 2. This sum is going to diverge very fast as $j \to \infty,$ so I guess something like $\binom{2j}{j}$ inolves.
Edit 3. Due to $(-1)^k$, we have got lots of cancellations, which makes the series to be controlled.
Edit 4. The problem still open.
Edit 5. [Getting some progress] Consider two polynomials $$p_j(t):= \sum_{k=0}^{j} \binom{2j+1}{k+j+1} (-t)^k \ \, \text{and} \ \ q_j(t):=4^j \ (1-t)^{j}.$$
To prove our guess, it suffices to show that
$$ \color{red}{\int_0^1 t^{-1/2} \, p_j(t) \, dt = \int_0^1 t^{-1/2} \, q_j(t) \, dt} \tag{*}$$
since
$$ \begin{align} \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}
& = \sum_{k=0}^{j} \binom{2j+1}{k+j+1} (-1)^k \int_0^1 \frac{1}{2} t^{k-1/2} \, dt \\
& = \frac{1}{2} \int_0^1 t^{-1/2} \, p_j(t) \, dt \, ,
\end{align} $$
and
$$ \frac{1}{2} \int_0^1 t^{-1/2} \, q_j(t) \, dt= 2^{2j-1}\ B(j+1,\frac{1}{2}).$$
I guess we can use induction since for $j=1$, we have
$$ \int_0^1 t^{-1/2} \, (3-t) \, dx = \int_0^1 t^{-1/2} \, 4(1-t) \, dt = 16/3.$$
|
$S_n=\sum\limits_{k=0}^{n} \dbinom{2n+1}{k+n+1} \ \dfrac{(-1)^k}{2k+1}$
where $j=n$ (I am sorry I used $n$ instead of $j$ during the long proving).
1,
$\dbinom{2n+1}{k+n+1}=\dfrac{(2n+1)!}{(n+k+1)!(n-k)!}\dfrac{n!}{k!}\dfrac{k!}{n!}\dfrac{(n+1)!}{(n+1)!}=$ $\dbinom{n}{k}\dfrac{\dbinom{2n+1}{n}}{\dbinom{n+k+1}{k}}$
So we have:
$S_n=\sum\limits_{k=0}^{n} \dbinom{n}{k}\dfrac{\dbinom{2n+1}{n}}{\dbinom{n+k+1}{k}} \dfrac{(-1)^k}{2k+1}=\dbinom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \dbinom{n}{k}\dfrac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(n+k+2)}\dfrac{(-1)^k}{2k+1}$
2, Using that $\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
$S_n=\dbinom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \dbinom{n}{k}\beta(n+1,k+1)\dfrac{(-1)^k}{2k+1}$
3,
$(-1)^k\dfrac{1}{2k+1}=\int\limits_0^1 (-t^2)^kdt$
and
$\beta(k+1,n+1)=\int\limits_0^1 u^k (1-u)^n du$ by the definition of beta function. Put them back into the sum, and rearrange:
$S_n=\binom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \binom{n}{k}\int\limits_0^1u^k (1-u)^n du\int\limits_0^1 (-t^2)^kdt=$
$\binom{2n+1}{n}(n+1)\int\limits_0^1 (1-u)^n \int\limits_0^1\sum\limits_{k=0}^{n}\binom{n}{k}(-ut^2)^kdtdu$
4,
As $\sum\limits_{k=0}^{n}\binom{n}{k}(-ut^2)^k=(1-ut^2)^n$
$S_n=\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n\int\limits_0^1 (1-ut^2)^ndtdu$
Using the $ut^2=x$ substitution we get:
$S_n=\frac{1}{2}\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n u^{-\frac{1}{2}}\int\limits_0^u (1-x)^n x^{-\frac{1}{2}}dtdu$
5,
Applying the definitions of beta and incomplete beta function:
$S_n=\frac{1}{2}\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n u^{-\frac{1}{2}}\beta(u;n+1,\frac{1}{2})du$
There is the function and its derivate in the integral $(\int f'(x)f(x)dx=\frac{f^2(x)}{2})$ so
$S_n=\frac{1}{4}\binom{2n+1}{n}(n+1)\beta^2(n+1,\frac{1}{2})$
6,
Easy to prove that $\frac{1}{2}\binom{2n+1}{n}(n+1)\beta(n+1,\frac{1}{2})=2^{2n}$
Finally
$S_n=2^{2n-1}\beta(n+1,\frac{1}{2})$
|
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|
What is the limit of the average value of the first $n$ terms of $(1, 2, 1, 1, 1, 2, 1, 1, 2, 1, ...)$ as $n\to\infty$? There exists a sequence $a_n$ that begins $(1, 2, 1, 1, 1, 2, 1, 1, 2, 1, ...)$. It is fully defined on the OEIS at A293630, but I will give a simple explanation here. The sequence starts $1, 2$. The next part is generated by looking at the last term in the current sequence (currently $2$), and adding the rest of the sequence to the end that many times (resulting in $1, 2, 1, 1$). This continues $(1, 2, 1, 1, 1, 2, 1), (1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2)...$ $a_n$ represents the $n$th term of the final, infinite sequence. It appears that
$$\lim_{n\to\infty} \frac{\sum_{k=1}^{n} a_k}{n} = 1.2752618420911721359284772047801515149347600371...$$
While empirical evidence holds this to be true up to an absurd amount of terms, I have not been successful in even proving that the limit exists. I have found no relation of this number to other constants. Is there a way I could either prove the existence of the limit, prove this is the value of the limit, or find the significance of the number?
Edit: Now that a proof has been provided by Sangchul Lee, I am looking for what this value may represent and what causes this value in particular to appear. An acquaintance with the same interest has put a bounty up for answers that may present an explanation for this.
|
We establish the convergence of the limit.
Step 1. Designate $(1,2)$ as the $1$-st stage and let $L_n$ be the length of the sequence at the $n$-th stage. Also, let $S_n = \sum_{k=1}^{n} a_k$ and $T_n = S_{L_n}$. Then
$$
L_{n+1}
= \begin{cases}
2L_n - 1, & \text{if } a_{L_n} = 1 \\
3L_n - 2, & \text{if } a_{L_n} = 2
\end{cases},
\qquad
T_{n+1}
= \begin{cases}
2T_n - 1, & \text{if } a_{L_n} = 1 \\
3T_n - 4, & \text{if } a_{L_n} = 2
\end{cases}
$$
With this at our hands, we claim the following:
Claim. $T_n/L_n$ converges to some $\alpha \in [1, 2]$, and in fact, we have $\left|\frac{T_n}{L_n} - \alpha \right| \leq \frac{4}{L_n} $.
Notice that $L_{n+1} - 1 \geq 2(L_n - 1)$ for all $n \geq 1$. Inductively applying this inequality, it follows that
$$L_{n+k} \geq 1 + 2^k (L_n - 1) \qquad \forall n, k \geq 1.$$
Moreover, from the above recurrence relation we read out that
$$ \left| \frac{T_{n+1}}{L_{n+1}} - \frac{T_n}{L_n} \right|
= \left\{ \begin{array}{ll}
\dfrac{\left| T_n - L_n \right|}{L_n(2L_n - 1)}, & \text{if } a_{L_n} = 1 \\
\dfrac{2\left| 2L_n - T_n \right|}{L_n(3L_n - 2)}, & \text{if } a_{L_n} = 2
\end{array} \right\}
\leq \frac{1}{L_n}$$
So by the comparison test applied to $\frac{T_n}{L_n} = \frac{T_1}{L_1} + \sum_{k=1}^{n-1} \left( \frac{T_{k+1}}{L_{k+1}} - \frac{T_k}{L_k} \right)$, it follows that $T_n/L_n$ converges. Let $\alpha$ denote the limit. Then $1 \leq a_k \leq 2$ tells that $1 \leq \alpha \leq 2$ as well. Finally, applying both of the previous inequalities, we get
\begin{align*}
\left| \frac{T_n}{L_n} - \alpha \right|
&\leq \sum_{k=n}^{\infty} \left| \frac{T_{k+1}}{L_{k+1}} - \frac{T_k}{L_k} \right|
\leq \frac{1}{L_n} \sum_{k=0}^{\infty} \frac{L_n}{L_{n+k}} \\
&\leq \frac{1}{L_n} \sum_{k=0}^{\infty} \frac{L_n}{1 + 2^k (L_n - 1)}
\leq \frac{1}{L_n} \sum_{k=0}^{\infty} \frac{1}{2^{k-1}}
= \frac{4}{L_n}
\end{align*}
Therefore the claim follows. ////
Step 2. Now we want to show that $S_m/m$ converges to $\alpha$ as $m\to\infty$. To this end, we define
$$ \epsilon_{A} := \sup\left\{ \left| \frac{S_m}{m} - \alpha \right| : m \in A \cap \mathbb{Z} \right\} $$
(with the convention $\sup \varnothing = 0$, though this is irrelevant to the argument.) Our goal is to estimate $\epsilon_{[L_n, \infty)}$ and conclude that this converges to $0$ as $n\to\infty$, which is equivalent to the showing that $S_m/m \to \alpha$ as $m\to\infty$. It turns out that the following almost-repeating structure plays the key role:
Observation. Let $m \in (L_n, L_{n+1}]\cap\mathbb{Z}$. Then
*
*If $m \in (L_n, 2L_n - 1]$, then $S_m = T_n + S_{m-L_n}$.
*If $a_{L_n} = 2$ and $m \in [2L_n, 3L_n-2]$, then $S_m = 2T_n - 2 + S_{m-2L_n + 1}$.
Both are direct consequences of the construction. Now we temporarily fix $p \geq 1$ and let $n > p$. Then for $m \in (L_n, L_{n+1}] \cap \mathbb{Z}$,
*
*Case 1. Assume that $m \in [L_n, L_n+L_p) $. Using the fact that $\left| \frac{S_{m-L_n}}{m - L_n} - \alpha \right| \leq 1$, we obtain
\begin{align*}
\left| \frac{S_{m}}{m} - \alpha \right|
&\leq \frac{L_n}{m} \cdot\left| \frac{T_n}{L_n} - \alpha \right| + \frac{m-L_n}{m} \left| \frac{S_{m - L_n}}{m - L_n} - \alpha \right| \\
&\leq \frac{L_n}{m} \cdot \frac{4}{L_n} + \frac{m-L_n}{m} \\
&\leq \frac{4}{L_n} + \frac{L_p}{L_p + L_n}.
\end{align*}
*Case 2. Assume that $m \in [L_n + L_p, 2L_n - 1]$. Using the fact that $m - L_n \geq L_p$, it follows that
\begin{align*}
\left| \frac{S_{m}}{m} - \alpha \right|
&\leq \frac{L_n}{m} \cdot\left| \frac{T_n}{L_n} - \alpha \right| + \frac{m-L_n}{m} \left| \frac{S_{m - L_n}}{m - L_n} - \alpha \right| \\
&\leq \frac{L_n}{m} \cdot \frac{4}{L_n} + \frac{m-L_n}{m} \epsilon_{[L_p, \infty)} \\
&\leq \frac{4}{L_n} + \frac{1}{2} \epsilon_{[L_p, \infty)}.
\end{align*}
*Case 3. Let $m \in [2L_n, 2L_n+L_p) $. This case is possible only when $a_{L_n} = 2$ and hence we assume so. By mimicking the previous claim, we find that $\left| \frac{2T_n - 2}{2L_n - 1} - \alpha \right| \leq \frac{C}{2L_n - 1}$ holds for all $n \geq 1$ for some constant $C > 0$ which is independent of $n$. Using this and mimicking Claim 1,
\begin{align*}
\left| \frac{S_{m}}{m} - \alpha \right|
&\leq \frac{2L_n - 1}{m} \left| \frac{2T_n - 2}{2L_n - 1} - \alpha \right| + \frac{m - 2L_n + 1}{m} \left| \frac{S_{m-2L_n+1}}{m - 2L_n + 1} - \alpha \right| \\
&\leq \frac{C}{L_n} + \frac{L_p}{L_p + L_n}.
\end{align*}
*Case 4. Let $m \in [2L_n+L_p, L_{n-1}) $. Adopting the techniques in Case 2 and Case 3, we obtain
$$ \left| \frac{S_{m}}{m} - \alpha \right|
\leq \frac{C}{L_n} + \frac{2}{3}\epsilon_{[L_p, \infty)}. $$
Combining altogether, with $C' = 4+C$ the following holds:
$$ \epsilon_{(L_n, L_{n+1}]} \leq \frac{C'}{L_n} + \max\left\{ \frac{L_p}{L_p + L_n}, \frac{2}{3}\epsilon_{[L_p, \infty)} \right\}. $$
Then it follows that $ \limsup_{n\to\infty} \epsilon_{(L_n, L_{n+1}]} \leq \frac{2}{3}\epsilon_{[L_p, \infty)} $. Then $\limsup_{n\to\infty} \epsilon_{[L_n, \infty)} \leq \frac{2}{3}\epsilon_{[L_p, \infty)}$ holds as well, and then it is easy to conclude that
$$\limsup_{n\to\infty} \epsilon_{[L_n, \infty)} = 0.$$
This implies the desired convergence.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ become $$ 1^2 + 2^2 + \cdots + k^2 $$ and not $$ 1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$$
when adding $k^2$?
Thank you.
|
The reason why,$$1^2 + 2^2 + \cdots + (k-1)^2,$$ becomes, $$1^2 + 2^2 + \cdots + (k - 1)^2+ k^2,$$ is because you're adding the $k$th term squared on both sides. Thus, instead of having the sum up to the $k-1$ th term squared, now you have the sum up to the $k$th term squared.
|
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|
Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ 54abc(ab+bc+ca)^2(a+b+c)^2−27a^2b^2c^2(ab+bc+ca)(a+b+c) + (a+b+c)^9−9(ab+bc+ca)(a+b+c)^7+ 9(ab+bc+ca)^4(a+b+c)−30(ab+bc+ca)^3(a+b+c)^3+ 18a^2b^2c^2(a+b+c)^3+ 27(ab+bc+ca)^2(a+b+c)^5+ 9abc(a+b+c)^6−9abc(ab+bc+ca)^3$$
But I would like a solution without this because it's too ugly .
So could you help me ?
|
If you set $e_1=a+b+c,$ $e_2=ab+ac+bc,$ and $e_3=abc$ and you define a recurrence relation as follows:
\begin{eqnarray*}
x_0 &=& 3 \\
x_1 &=& e_1 \\
x_2 &=& e_1^2-2e_2 \\
x_{n+3} &=& e_1x_{n+2} - e_2x_{n+1} + e_3x_n
\end{eqnarray*}
then it can be shown that $x_n = a^n+b^n+c^n.$ In order to have $a^9+b^9+c^9=3,$ there must be $x_0,\ldots,x_9$ such that
$$\begin{pmatrix}
1 &&&&&&&&& \\
& 1 &&&&&&&& \\
&& 1 &&&&&&& \\
-e_3 & e_2 & -e_1 & 1 &&&&&& \\
& -e_3 & e_2 & -e_1 & 1 &&&&& \\
&& -e_3 & e_2 & -e_1 & 1 &&&& \\
&&& -e_3 & e_2 & -e_1 & 1 &&& \\
&&&& -e_3 & e_2 & -e_1 & 1 && \\
&&&&& -e_3 & e_2 & -e_1 & 1 & \\
&&&&&& -e_3 & e_2 & -e_1 & 1 \\
&&&&&&&&& 1\end{pmatrix}
\begin{pmatrix}
x_0 \\x_1 \\x_2 \\x_3 \\x_4 \\x_5 \\x_6 \\x_7 \\x_8 \\x_9
\end{pmatrix}
=
\begin{pmatrix}
3 \\ e_1 \\ e_1^2-2e_2 \\0\\0\\0\\0\\0\\0\\0\\3
\end{pmatrix}
$$
In other words
$$\det
\begin{pmatrix}
1 &&&&&&&&&& 3\\
& 1 &&&&&&&&& e_1\\
&& 1 &&&&&&&& e_1^2-2e_2\\
-e_3 & e_2 & -e_1 & 1 &&&&&&& \\
& -e_3 & e_2 & -e_1 & 1 &&&&&& \\
&& -e_3 & e_2 & -e_1 & 1 &&&&& \\
&&& -e_3 & e_2 & -e_1 & 1 &&&& \\
&&&& -e_3 & e_2 & -e_1 & 1 &&& \\
&&&&& -e_3 & e_2 & -e_1 & 1 && \\
&&&&&& -e_3 & e_2 & -e_1 & 1 & \\
&&&&&&&&& 1 & 3 \end{pmatrix} = 0
$$
|
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|
Proving that ${4a \choose 2a} - {2a \choose a}$ is divisible by 4 I'm trying to prove that ${4a \choose 2a} - {2a \choose a}$ is divisible by $4$, and I've currently tried using induction, though I have been unsuccessful. I'm not really sure where to go to for now, any help would be appreciated.
This is what I have currently done;
For $a=1$ we have that ${4a \choose 2a} - {2a \choose a} = {4 \choose 2} - {2 \choose 1} = 4$, which is divisible by $4$ and hence the statement is true for $a=1$.
Assume true for $a =n$ (where $n \in \mathbb{N}$).
Required to prove true for $a = n+1$. We have that;
$${4(a+1) \choose 2(a+1)} - {2(a+1) \choose a+1} = {4a+4 \choose 2a+2} - {2a+2 \choose a+1} $$
Though
$${2a+2 \choose a+1} = {2a+1 \choose a} + {2a+1 \choose a+1} = 2 {2a+1 \choose a}$$
However, applying the same binomial identity we get that ${2a+1 \choose a} = {2a \choose a - 1} + {2a \choose a}$. Hence our original expression can be written as;
$${4a+4 \choose 2a+2} - {2a+2 \choose a+1} = {4a+4 \choose 2a+2} - 2 {2a+1 \choose a} $$
After here I'm not sure where to go?
|
Beware: I am going for the overkill. By Legendre's theorem exponent of the largest power of $2$ dividing $n!$ is $\sum_{k\geq 1}\left\lfloor\frac{n}{2^k}\right\rfloor$, hence
$$ \nu_2\binom{2n}{n} = \sum_{k\geq 1}\left(\left\lfloor\frac{2n}{2^k}\right\rfloor-2\left\lfloor\frac{n}{2^k}\right\rfloor\right)\tag{1}$$
and the LHS is given by the number of "$1$"s in the binary representation of $n$. If we assume that $n$ has $k$ "1"s in its binary representation then both $\binom{2n}{n}$ and $\binom{4n}{2n}$ are numbers of the form $2^k\cdot\text{odd}$, hence their difference is a multiple of $2^{k+1}$. Since $k\geq 1$, we are done.
|
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|
$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ is convergent? Is $\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ convergent? If it is convergent find the sum of the series.
Gives series $$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....=\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x$$
$$\left|\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x\right|\le \sum _{n=1}^\infty \frac{1}{2^{n-1}}=2$$ so is it convergent?
|
Let $a:=\sin x$ then the series can be written as
$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin^nx}{2^{n-1}}=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^n}{2^{n-1}}=a\sum_{n=1}^{\infty}(-1)^{n-1}\Big(\frac{a}{2}\Big)^{n-1}=a\frac{1}{1-(-a/2)}=\frac{2a}{2+a}$$
by the geometric series whenever $|a|<2$ which is equivalent to $|\sin x|<2$ which is true for all $x\in \mathbb{R}$.
|
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|
How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
|
Note that
$$y=\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = \frac{3}{1-3^{x-2}} + \frac{3}{1-\frac1{3^{x-2}}}=\\= \frac{3}{1-3^{x-2}} - \frac{3^{x-1} }{1-3^{x-2}}=\frac{3-3^{x-1}}{1-3^{x-2}}=3\frac{1-3^{x-2}}{1-3^{x-2}}=3 $$
|
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|
An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi(2p)-\pi(p) \ge 2$ , a result by Ramanujan .
This means that $q<2p$ and $r<2p$ , so $\frac{1}{2p}<\frac{1}{q}$ and $\frac{1}{2p}<\frac{1}{r}$ which implies $\frac{1}{p} < \frac{1}{q} + \frac{1}{r}$ . If we square both sides of inequality we get $\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2}$ . Now , I don't know how to rule out term $\frac{2}{qr}$ .
|
This is a comment as opposed to an answer
All primes $p_{n+1} < 2p_n$. Let $p'$ be the prime before $p$, $q'$ be the prime before $q$, and $r'$ be the prime before $r$. Then, $$p+q+r< 2(p' + q' + r')$$ So $$\frac{1}{p^2} < \frac{1}{\big(2(p' + q' + r') - q - r\big)^2}$$ Also, $$\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2} = \left(\frac{1}{q} + \frac{1}{r}\right)^2.$$ Consider $$\frac{1}{\big(2(p' + q' + r') - q - r\big)^2} < \left(\frac{1}{q} + \frac{1}{r}\right)^2$$ then multiplying both sides by the denominator, subtracting $1$ from both sides, and then factoring, we get $$0 < \left(\left(\frac{1}{q} + \frac{1}{r}\right)\left(2(p'+q'+r')-q-r\right) + 1\right)\left(\left(\frac{1}{q} + \frac{1}{r}\right)\left(2(p'+q'+r')-q-r\right) - 1\right)$$ Which is true since primes are always positive, the entire inequality is literally just a bunch of multiplication, and if $p' = 2$, $q' = 3$ and $r' = 5$ then this inequality holds.
Your conjecture would be thus true if $$\frac{1}{\big(2(p'+q'+r')-q-r\big)^2} < \frac{1}{q^2} + \frac{1}{r^2}.$$
|
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|
Matrix Notation Form of Roots of a Quadratic Equation We know that the quadratic equation
$$f(x)=ax^2+bx+c=0$$
has roots
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$
Also, $f(x)$ can be written in matrix notation as follows:
$$f(x)=
\left(\begin{matrix}x&1\\\end{matrix}\right)
\left(\begin{matrix}a&\frac b2\\\frac b2&c\end{matrix}\right)
\left(\begin{matrix}x\\1\end{matrix}\right)=\mathbf{x^T Q x}$$
where the determinant of $\mathbf Q$ is $\left(ac-\frac {b^2}4\right)=-\frac 14\left(b^2-4ac\right)$, where coincidentally the familiar $(b^2-4ac)$ is the discriminant of the quadratic $f(x)$.
Hence the roots of the quadratic $f(x)=0$ may be written as
$$x=-\frac b{2a}\pm \frac 1a\sqrt{-\det(\mathbf Q)}$$
This is equivalent to
$$\left(x+\frac b{2a}\right)^2=\frac {-\det(\mathbf Q)}{a^2}$$
Or in neater form,
$$\left(ax+\frac b{2}\right)^2={-\det(\mathbf Q)}$$
Question
Can the roots of $f(x)=0$ be derived and written completely in matrix notation, given the link between the determinant and discriminant as shown above?
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Let us switch to homogeneous coordinates and use $2b$ instead of $b$ for convenience.
$$ax^2+2bxy+cy^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{p^TQp}=0.$$
Assuming $Q$ diagonalizable, we have (the transformation $P$ can be taken orthonormal)
$$\mathbf{p^TQp}=\mathbf{p^TP^TDPp}=\mathbf{q^TDq}=0.$$
The last expression is of the form
$$\lambda_0u^2+\lambda_1v^2=0$$ where the lambdas are the Eigenvalues of $\mathbf Q$. For the equation to have real solutions, the Eigenvalues must have opposite signs, or $\lambda_0\lambda_1\le0$, which is precisely $-\det(\mathbf Q)=b^2-ac\ge0$.
The conic factors as
$$\left(\sqrt{|\lambda_0|}u+\sqrt{|\lambda_1|}v\right)\left(\sqrt{|\lambda_0|}u-\sqrt{|\lambda_1|}v\right)=0$$
and the solutions are given by plugging $u=\mathbf{e_0p},v=\mathbf{e_1p}$ in the two factors, setting $y=1$ and solving the linear equations for $x$.
|
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|
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ to $A(y+1)^2$, $B(y+1)^1$ and $C$.
My question is why do we have to substitute $x-1=y$ and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting? Thanks in advance.
|
$$
(x+1)^n = Q(x)(x-1)^3 + ax^2+b x+c
$$
$$ 2^n = a+b+c $$
$$
n(x+1)^{n-1} = Q'(x)(x-1)^3+3Q(x)(x-1)^2 + 2ax+b \Rightarrow n2^{n-1}=2a+b
$$
In the same way
$$
n(n-1)(n-2)2^{n-2} = 2a
$$
etc.
|
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|
Show that the solutions from one quadratic equation are reciprocal to the solutions of another quadratic equation From Sullivan's Algebra & Trigonometry book: Chapter 1.2; Exercise 116:
Show that the real solutions of the equation $ax^2+bx+c=0$ are the
reciprocals of the real solutions of the equation $cx^2+bx+a=0$.
Assume that $b^2-4ac\geq0$.
The most I reached in my attempt to solve the problem was to prove that, as $ax^2+bx+c=0$ can be expressed as $a^2x^2+abx+b^2$, and that $cx^2+bx+a=0$ can be expressed as $b^2x^2+abx+a^2$; thus, the solutions from the one will be the reciprocal of the another...
$a^2x^2+abx+b^2=(ax+b)^2$
$(ax+b)^2= 0$
$x_1 = -\frac{b}{a}$ (root of multiplicity two)
$b^2x^2+abx+a^2=(bx+a)^2$
$(bx+a)^2=0$
$x_2 = -\frac{a}{b}$ (root of multiplicity two)
$x_1x_2=0$
The problem here is that the solution only works for perfect squares; I want to know, and that's what the problem is asking for, to prove it for the $ax^2+bx+c$ form.
Another attempts to solve the problem...
I've tried to solve it using the quadratic formula, I started from the equation $\frac{2a}{-b+\sqrt{b^2-4ac}}=0$ and then try to get to the expression $\frac{-b+\sqrt{b^2-4ac}}{2c}=0$, but I wasn't cappable, don't even know it it's possible to do or correct to try.
I also tried the same from above with the reciprocal of the actual equation: $\frac{1}{ax^2+bx+c}=0$. But for me it was also impossible to do anything with that expression.
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If $r$ is a root of $ax^2+bx+c$, then $ar^2+br+c=0$ and\begin{align}ar^2+br+c=0&\iff \frac{ar^2+br+c}{r^2}=0\\&\iff a++b\times\frac1r+c\times\frac1{r^2}=0\\&\iff\frac1r\text{ is a root of }cx^2+bx+c=0.\end{align}
|
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|
Convex quadrilateral; Area and parallel sides Consider a convex quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
Prove that the area of triangle $AOB$ = the area of the triangle $COD$ if an only if $AD$ and $BC$ are parallel.
Where should I start with this problem?
Thanks!
|
Good to have a diagram.
Now, if $AD || BC$ then $w = z$ and $x = y$, so we get that triangles $AOD$ and $BOC$ are similar. Hence, we get $\frac{OB}{OD} = \frac{OC}{OA}$, and therefore $OA \cdot OB = OC \cdot OD$. We know that $\angle AOB = \angle COD$, hence from the formula for area,
$$
\mbox{Area}(AOB) = \frac 12 OA \cdot OB \cdot \sin \angle AOB = \frac 12 OC \cdot OD \cdot \sin \angle COD = \mbox{Area} (COD)
$$
Conversely, if the areas are the same, then all steps reverse up till $OA \cdot OB = OC \cdot OD$. Let $k = \frac{OD}{OB} = \frac{OA}{OC}$. Then, we see that:
$$
AD^2 = AO^2 + OD^2 - 2AO \cdot OD \cdot \cos \angle AOD \\ = k^2(OC^2 + OB^2 + 2 OC \cdot OB \cdot \cos \angle COB) = k^2 BC^2 \\ \implies \frac{AD}{BC} = k
$$
Hence, the triangles $AOD$ and $BOC$ are similar, hence have the same angles, and from the way the sides map, we see that $w = z$ and $x = y$, and hence $AD || BC$.
I feel I am doing something too complicated, so I would still like suggestions.
|
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|
Prove that $\int_{\mathbb{R}} \frac{1}{(a^2+t^2)(b^2+t^2)} dt= \frac{\pi}{ab(a+b)}$
Prove that $\int_{\mathbb{R}} \frac{1}{(a^2+t^2)(b^2+t^2)} dt= \frac{\pi}{ab(a+b)}$
For $a,b>0$, the Fourier transform of $f(x) = e^{-a|x|}$ is $\widehat{f}(t) = \dfrac{2a}{a^2+t^2}$.
Otherwise, we have $\int_{\mathbb{R}}\dfrac{1}{(a^2+t^2)^2} dt= \dfrac{\pi}{2a^3} $.
I have to show that $\int_{\mathbb{R}} \dfrac{1}{(a^2+t^2)(b^2+t^2)} dt= \dfrac{\pi}{ab(a+b)}$ ?
I have no idea to do it. Could someone help me ?
|
Hint:
$$\frac{1}{(a^2+t^2)(b^2+t^2)}=\frac{1}{b^2-a^2}\left(\frac{1}{a^2+t^2}-\frac{1}{b^2+t^2}\right).$$
|
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|
Prove the given determinant Prove the given determinant:
$$\left|
\begin{matrix}
a&b&ax+by \\
b&c&bx+cy \\
ax+by&bx+cy&0 \\
\end{matrix}\right|=(b^2-ac)(ax^2+2bxy+cy^2)$$
I didn't get any idea. Please help me solve this.
|
A bit late in the day, here is the symmetric matrix version of this, writing $P^TAP = B,$ where $B$ is also symmetric with obvious determinant. We say that $A$ and $B$ are "congruent"
$$
\left(
\begin{array}{ccc}
1&0&0 \\
0&1&0 \\
-x&-y&0 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
a&b&ax+by \\
b&c&bx+cy \\
ax+by&bx+cy&0 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&0&-x \\
0&1&-y \\
0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
a&b&0 \\
b&c&0 \\
0&0& - \left(a x^2 + 2bxy+cy^2 \right) \\
\end{array}
\right)
$$
|
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|
Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
\left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
\frac{8!}{9!} < 1 \\
\frac{1}{9} < 1 \\
\end{align}
if it is not correct how it would be?
|
This step doesn't look right to me:
\begin{gather}
(8!)^{(9/72)} - (9!)^{8/72} < 0 \\[6px]
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) < 0
\end{gather}
When you divide $(8!)^{9/72} $ by $(9!)^{8/72}$, you should get
$$
\frac{(8!)^{9/72}}{(9!)^{8/72}} =
\frac{(8!)^{9/72}}{(9\cdot8!)^{8/72}} =
\frac{(8!)^{1/72}}{(9)^{8/72}} =
\frac{(8!)^{1/72}}{(9)^{1/9}} =
\left( \frac{(8!)^{(1/8)}}{9} \right)^{1/9}
$$
Also, note that your proof is basically of this form:
\begin{gather}
a < b \\[6px]
a - b < 0 \\[6px]
\left(\frac{a}{b} -1\right) < 0\\[6px]
\frac{a}{b} < 1
\end{gather}
You can skip several steps and just do
\begin{gather}
a < b \\[6px]
\frac{a}{b} < 1
\end{gather}
|
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|
A limit about $\left(1+\frac{1}{n}\right)^{n}$? Here is my question:
$$\displaystyle\lim_{n\rightarrow \infty}n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=?$$
Any hints will be fine. Thank you!
|
HINT
Note that (to be made more rigorous)
*
*$(1+\frac{1}{1+n}) ^{n+1}=e^{(n+1)\log\left(1+\frac{1}{1+n}\right)}\sim e^{(n+1)\left(\frac{1}{1+n}-\frac{1}{2(1+n)^2}\right) }=e^{1-\frac{1}{2(1+n)}}\sim e\left(1-\frac{1}{2(1+n)}\right)$
*$(1+\frac{1}{n}) ^{n}=e^{n\log\left(1+\frac{1}{n}\right)}\sim e^{n\left(\frac{1}{n}-\frac{1}{2n^2}\right) }=e^{1-\frac{1}{2n}}\sim e\left(1-\frac{1}{2n}\right)$
then
$$n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1} - \left(1+\frac{1}{n}\right)^{n}\right]\sim e\cdot n^2\left(\frac{1}{2n}-\frac{1}{2(1+n)}\right)=e\cdot n^2\left(\frac{2}{4n^2+4n}\right)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent series for $\frac{1}{z(z+3)(z-1)^2}$ Find the Laurent Series for
$$\frac{1}{z(z+3)(z-1)^2}$$ in $1 < |z-1| < 4$
So I did the partial fraction decomposition which yields:
$$\frac{1}{4(-1+z)^2} - \frac{5}{16(-1+z)} + \frac{1}{3z} - \frac{1}{48(3+z)}$$
Can anyone help me finish this problem?
|
You are on the right track. Partial fraction expansion reveals
$$\frac{1}{z(z+3)(z-1)^2}=\frac{1/3}{z}-\frac{1/48}{z+3}-\frac{5/16}{z-1}+\frac{1/4}{(z-1)^2}\tag1$$
The last two terms on the right-hand side of $(1)$ constitute part of the Laurent expansion in the annulus $1<|z-1|<4$.
We now expand the first term, $\frac{1/3}{z}$, for $|z-1|>1$ as
$$\begin{align}
\frac{1/3}{z}&=\frac{1/3}{z-1}\left(\frac{1}{1+\frac{1}{z-1}}\right)\\\\
&=\frac13\sum_{n=0}^\infty (-1)^n (z-1)^{-(n+1)}\tag2
\end{align}$$
Similarly, we expand the term $\frac{1/48}{z+3}$ for $|z-1|<4$ as
$$\begin{align}
\frac{1/48}{z+3}&=\frac{1/48}{4+(z-1)}\\\\
&=\frac1{192}\sum_{n=0}^\infty (-1)^n\left(\frac{z-1}{4}\right)^n\tag3
\end{align}$$
Now, assemble the expansion using $(2)$ and $(3)$ in $(1)$.
|
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|
If $a,b,c$ are the roots of $x^3+x^2-2x+1=0$ then what is the value of $\det\Delta$?
If $a,b,c$ are the roots of $ x^3+x^2-2x+1=0$, then what is the value of $\det\Delta$ where:
$$\Delta=\begin{bmatrix}
c^2 & b^2 & 2bc-a^2 \\ 2ac-b^2 & a^2 & c^2 \\ a^2 & 2ab-c^2 & b^2 \\ \end{bmatrix}$$
The roots are not simply resolvable. So my experience tells me that we need to convert $\Delta$ into a form that contains $a+b+c$ or $ab+bc+ca$ or $abc$.
Since expanding this determinant would be tedious, please suggest a method that doesn't require it.
All help will be appreciated
|
Hint: For your determinant we get $$ \left( b+a+c \right) ^{2} \left( {b}^{2}-ab-bc-ac+{a}^{2}+{c}^{2}
\right) ^{2}
$$
|
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|
Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common terms like this:
$$\frac{1}{3} x ^{-\frac{1}{3}} (6-x)^{\frac{-2}{3}} ( -x + (6-x) \cdot 2)$$
Is that right?
|
We've got to preserve the terms here, and we see that the ordering is mismatched, as in
$$\frac{d}{dx}(x^{\frac{2}{3}}(6-x)^{\frac{2}{3}})=\frac{2}{3}(\frac{6-x}{x})^{\frac{1}{3}}-\frac{1}{3}(\frac{x}{6-x})^{\frac{2}{3}}$$
Then we can factor out $(\frac{6-x}{x})^{\frac{1}{3}}$ to make the second term much nicer with no powers
Your answer and this one are ultimately the same
|
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|
Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $11$.
But the answer says it is $12$.
Where am I wrong and how to solve it correctly?
|
If $2^n$ divides $3^{1024}-1$, then $3$ has order at most $1024=2^{10}$, mod $2^n$.
Now, $3$ has order $2^{n-2}$ mod $2^n$. (*)
So, $n-2=10$ is the maximum possible, that is, $n=12$.
(*) By induction, $3^{2^{n-3}} \equiv 1+2^{n-1} \bmod 2^n$ and $3^{2^{n-2}} \equiv 1 \bmod 2^n$ for $n\ge 4$.
|
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|
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.
|
There is! You can use the identity $\binom{k+2}2 = \sum_{i=0}^{k+1}i$.
Our sum is $$\sum_{k=0}^n (-1)^k \binom{k+2}{2} = \sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i$$
For odd $n$: Letting $m=\frac {n-1}2$ and pairing up the terms we get
$$\begin{align}\sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i &= \sum_{j=0}^m [(-1)^{2j}\sum_{i=0}^{2j+1}i + (-1)^{2j+1}\sum_{i=0}^{2j+2}i] \\&=\sum_{j=0}^m [\sum_{i=0}^{2j+1}i -\sum_{i=0}^{2j+2}i] \\&= \sum_{j=0}^m-(2j+2) \\&= -2\sum_{j=0}^m(j+1) \\&= -2\sum_{j=1}^{m+1}j\\ &= -2[\binom{m+2}2-0] \\&= -(m+2)(m+1) \\&=-\frac{(n+3)(n+1)}4
\end{align}$$
For even $n$: $$\begin{align} \sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i &= \sum_{k=0}^{n-1} (-1)^k \sum_{i=0}^{k+1}i + \sum_{i=0}^{n+1}i
\\&= -\frac{((n-1)+3)((n-1)+1)}4 + \binom{n+2}2 \\
&=-\frac{n^2+2n}4 + \frac{n^2 + 3n + 2}{2}
\\&= \frac{ - n^2 - 2n + 2n^2 + 6n + 4}4 \\&=(\frac{n+2}2)^2
\end{align}$$
|
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|
How to transform arbitrary rectangle into specific parallelogram? I have a rectangle $S$ in $(u,v)$ coordinates. $S = [a,b] \times [c,d]$.
I want to find the transformation that yields a parallelogram given by $(0,0),(-2,1),(2,4),(4,3)$. I have deduced this is equivalent to
$$\bigg\lbrace (x,y) \bigg | \text{area bounded by } y = -\frac{1}{2}x, y = \frac{3}{4} x ,y = \frac{3}{4} x + \frac{5}{2}, y = -\frac{1}{2}x + 5\bigg\rbrace$$
I'm sure I'm doing this the hard way. But my intinct was there are four steps to do it.
*
*Translate the rectangle to the origin
*Resize the edges
*Shear it
*Rotate it
So if
$$\begin{bmatrix}
u \\ v \\
\end{bmatrix} \in S$$
Translation:
$$T = \begin{bmatrix}
-a \\ -c \\
\end{bmatrix}$$
Rescale:
$$A_1 = \begin{bmatrix}
\frac{\gamma}{b-a} & 0 \\
& \frac{5}{d-c}
\end{bmatrix}$$
Shear:
$$A_2 = \begin{bmatrix}
1 & -\frac{1}{2} \\
0 & 1
\end{bmatrix}$$
Rotation:
$$R = \begin{bmatrix}
4/5 & -3/5 \\
3/5 & 4/5
\end{bmatrix}$$
So then the transformation would look like
$$ \begin{bmatrix}
x \\ y \\
\end{bmatrix} = RA_2A_1\bigg(\begin{bmatrix}
u \\ v \\
\end{bmatrix} +T\bigg)$$
However, I can't figure out what $\gamma$ should be for the rescaling. Can anyone help?
|
After trial and error, I found that
Translation:
$$T = \begin{bmatrix}
-a \\ -c \\
\end{bmatrix}$$
Rescale:
$$A_1 = \begin{bmatrix}
\frac{5}{b-a} & 0 \\
& \frac{2}{d-c}
\end{bmatrix}$$
Shear:
$$A_2 = \begin{bmatrix}
1 & -\frac{1}{2} \\
0 & 1
\end{bmatrix}$$
Rotation:
$$R = \begin{bmatrix}
4/5 & -3/5 \\
3/5 & 4/5
\end{bmatrix}$$
Example
$S = [10,20] \times [0,2]$ and suppose we consider the point $ \begin{bmatrix}
20 \\ 2 \\
\end{bmatrix}$
$$\begin{bmatrix}
2 \\ 4 \\
\end{bmatrix} = R A_2 A_1 \bigg ( \begin{bmatrix}
20 \\ 2 \\
\end{bmatrix} + \begin{bmatrix}
-10 \\ 0 \\
\end{bmatrix}\bigg )$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find all rational solutions of $\ x^2 + 3y^2 = 7 $? I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at the right hand side ? Then how to deal with the $\ 3y^2 $ ? Thank you!
|
Using the method of pg 7 of this paper on this related equation
$$x^2+3y^2=7z^2 \quad \text{with initial solution} \quad (x,y,z)=(2,1,1)$$
A line $y=t(x-2)+1$, which will cut through the ellipse $x^2 + 3y^2 = 7$ at rational points if $t$ is rational.... when substituted into the ellipse yields:
$$\begin{align}
x^2+3\left[t(x-2)+1\right]^2&=7 \\
x^2+3\left[t^2(x^2-4x+4)+2t(x-2)+1\right]&=7 \\
(1+3t^2)x^2+(-12t^2+6t)x+(12-12t+3-7)&=0 \\
\text{vieta: the two roots,} \quad x_1, x_2 \quad \text{are such that}
\quad -(x_1+x_2)&=\frac{-12t^2+6t}{1+3t^2} \\
\text{since} \quad x_1=2, \quad \text{we have} \quad x_2(t)=\frac{12t^2-6t}{1+3t^2}-2 &=\frac{6t^2-6t-2}{1+3t^2} \\
\text{substituting this} \quad x(t) \quad \text {into the line:} \quad y&=t\left(\frac{6t^2-6t-2}{1+3t^2}-2\right)+1 \\
y(t)=t\left(\frac{-6t-4}{1+3t^2}\right) + \frac{1+3t^2}{1+3t^2}&=\frac{-3t^2-4t+1}{1+3t^2}
\end{align}$$
Letting $y(t)\to |y(t)|$, Solution set with one parameter:
$$\begin{cases}
x(t)&=\frac{6t^2-6t-2}{1+3t^2} \\
y(t)&=\frac{3t^2+4t-1}{1+3t^2}
\end{cases}$$
$t$ was rational, so let $t=\frac{m}{n}$
Solution set now with two parameters:
$$\begin{align}
x(m,n)&=\frac{6m^2-6mn-2n^2}{n^2+3m^2} \\
y(m,n)&=\frac{3m^2+4mn-n^2}{n^2+3m^2}
\end{align}$$
|
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|
Inhomogeneous Wave Equation Energy Method Let $u$ be a solution to the following equation:
$$
u_{tt}=u_{xx}−u^3
$$
Assume that $u(x,0) =u_t(x,0) = 0$ for all $x\in[a, b]$. Prove that $u(x, t) = 0$ if $a+t < x < b−t$.
I initially thought of using Duhamel's principle on this but realized that the cube in the integrand would make things a little tricky. I was also thinking of using an energy functional and showing that it was $0$ everywhere and nonincreasing. What would the functional be, and what would/should the domain of integration be?
|
Note that, thanks to the Leibniz integral rule,
\begin{align}
&\frac{\rm d}{{\rm d}t}\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u_xu_{xt}+u^3u_t\right){\rm d}x-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}.
\end{align}
Besides, the first term in above yields, following the usual integration by parts,
\begin{align}
&\int_{a+t}^{b-t}\left(u_tu_{tt}+u_xu_{xt}+u^3u_t\right){\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+\int_{a+t}^{b-t}u_xu_{xt}{\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+\int_{a+t}^{b-t}u_x{\rm d}u_t\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+u_xu_t|_{a+t}^{b-t}-\int_{a+t}^{b-t}u_tu_{xx}{\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_{tt}-u_{xx}+u^3\right)u_t{\rm d}x+u_xu_t|_{a+t}^{b-t}\\
&=u_xu_t|_{a+t}^{b-t},
\end{align}
where the last step is a gift from the governing equation.
Thanks to these results,
\begin{align}
&\frac{\rm d}{{\rm d}t}\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\\
&=u_xu_t|_{a+t}^{b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}\\
&=-\frac{1}{2}\left(u_t^2-2u_tu_x+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+2u_tu_x+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}\\
&=-\frac{1}{2}\left[\left(u_t-u_x\right)^2+\frac{1}{2}u^4\right]\Bigg|_{x=b-t}-\frac{1}{2}\left[\left(u_t+u_x\right)^2+\frac{1}{2}u^4\right]\Bigg|_{x=a+t}\\
&\le 0.
\end{align}
This immediately leads to
$$
\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\le\int_a^b\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right)\Bigg|_{t=0}{\rm d}x=0,
$$
where the last step is a gift from the initial conditions. Therefore,
$$
u=0
$$
holds for all $x\in\left(a+t,b-t\right)$ when this interval is non-degenerated.
|
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|
Values of $p$ such that the roots of $x^2+px+p$ are greater than $1$ I found this question in a book, the answer is given as $(-\infty,-5)$.
I know that for both roots of a quadratic polynomial to be greater than 1, $D \ge 0$, which implies that $p^2-4p \ge 0$, which means $p \ge 0$ and $p \ge 4$, or $p < 0$ and $p < 4$.
Also $f(1) >0$, which implies $2p>0$ or $p>0$.
Also $(-p/2)>1$ or $p<-2$.
This is not only contradictory but also doesn't match with the answer. What should I do? And if the answer is wrong what is the right answer?
|
Let $x=a+1$, then the equation is equivalent to
$(a+1)^2+p(a+1)+p=0$
$\Leftrightarrow a^2+2a+1+ap+p+p=0$
$\Leftrightarrow a^2+(p+2)a+2p+1=0$
We need to find $p$ so that $x>1$, or $a+1>1$, or $a>0$, which means we need to find $a$ so that the equation $a^2+(p+2)a+2p+1=0$ has two roots and both of them are positive roots.
This happens if and only if $\begin{cases}\Delta \ge 0 \\ a_1+a_2>0 \\a_1a_2>0\end{cases}\Leftrightarrow \begin{cases}(p+2)^2-4(2p+1) \ge 0 \\ -p-2>0 \\2p+1>0\end{cases}\Leftrightarrow\begin{cases}p^2-4p\ge0\\p<-2\\p>-0.5\end{cases}$
There are no real numbers $p$ satisfy the condition because $p<-2$ and $p>-0.5$.
|
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|
Showing $\lim_{x \to2^-}\frac{1}{x-2} = -\infty$ using the definition of the limit
Show
$$\lim_{x \to2^-}\frac{1}{x-2} = -\infty $$
My answer:
For all $B<0, \exists \delta >0 $ such that $\frac{1}{x-2} < B$ always that $-\delta < x-2 < \delta$, "in this part I don't understand because write $-\delta < x-2 < 0$, because approach for the left not necessary is for negatives values, not it sure??? I really need some explanation in this part"
So, I continue my answer with:
Looking in inequality between B, we have:
\begin{align}
\frac{1}{x-2} < B\\
x-2 > \frac{1}{B}\\
x> \frac{1}{B}+2
\end{align}
Then, We choose $\delta=\frac{1}{B}+2$
Like this, $\frac{1}{x-2} < B$ always that $-\delta < x-2 < 0.$
|
Remember we are assuming that $x < 2$ as $x\to 2^-$.
$-\delta < x - 2 < \delta;$ and $x < 2 \iff $
$-\delta < x-2 < 0 \iff$
$0 < 2-x < \delta \iff$
$\frac 1{x-2} > \frac 1{\delta} \iff$
$\frac 1{2-x} < -\frac 1{\delta}$
So if we set $-\frac 1{\delta} = B$ or in other words if we set $\delta = -\frac 1B$ we have our proof:
If $|x-2| < \delta = -\frac 1B$ then
$\frac 1B < x-2 < -\frac 1B$ and $x < 2$ so
$\frac 1B < x-2 < 0$ so
$0 < 2-x < -\frac 1B$ so
$ \frac 1{2-x} > -B > 0$ so
$\frac 1{x-2} < B < 0$.
|
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|
Evaluating $\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$ Greetings I want to evaluate $\displaystyle\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$.
Here is my try: We have that $x\in[0,\pi]$ so $$\cos(n\pi)\le \cos(nx) \le 1.$$ Here I am not sure, but if it's correct then it gives: $$\frac{1}{2}\le\frac{1}{1+\cos^2(nx)}\le \frac{1}{1+\cos^2(n\pi)},$$ giving $$\lim_{n\rightarrow\infty} \frac{1}{2}\int_0^{\pi} \sin x dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (n\pi)} dx.$$ Since $$\int_0^{\pi} \sin x dx =2$$ By squeeze theorem we may conclude that $\displaystyle \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx=1$. Could you help me evaluate this, if it's wrong?
|
One approach can be as follows:
\begin{align}
\int_0^{\pi}\frac{\sin x}{1+\cos^2nx}\,dx&\stackrel{(1)}{=}
\frac{1}{n}\int_0^{n\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy=
\frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy\stackrel{(2)}{=}\\
&\stackrel{(2)}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin\big(\frac{t+k\pi}{n}\big)}{1+\cos^2t}\,dt=
\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\Big[\frac{\pi}{n}\sum_{k=0}^{n-1}\sin\Big(\frac{t+k\pi}{n}\Big)\Big]\,dt\stackrel{(3)}{\to}\\
&\stackrel{(3)}{\to}\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\,dt\cdot\int_0^{\pi}\sin t\,dt=\sqrt{2}.
\end{align}
Explanations:
(1) change $y=nx$,
(2) change $t=y-k\pi$ and use that $\cos^2t$ is $\pi$-periodic,
(3) Riemann summa limit and dominated convergence theorem.
For the last integration: rewrite
$$
\cos^2x+1=\cos^2x+\cos^2x+\sin^2x=\cos^2x(2+\tan^2x),
$$
that gives
\begin{align}
\int_0^\pi\frac{1}{1+\cos^2x}\,dx&=2\int_0^{\pi/2}\frac{1}{1+\cos^2x}\,dx=2\int_0^{\pi/2}\frac{1}{2+\tan^2x}\,d\tan x=2\int_0^\infty\frac{1}{2+t^2}\,dt=\\
&=[t=s\sqrt{2}]=\sqrt{2}\int_0^\infty\frac{1}{1+s^2}\,ds=\sqrt{2}\frac{\pi}{2}=\frac{\pi}{\sqrt{2}}.
\end{align}
|
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|
Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}
$$
Following that log suggestion in the comments below:
\begin{align}
&\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\
&= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\
&= \int_1^2 \ln x \, dx \\
&= (x \ln x - x)\vert_1^2 \\
&= (2 \ln 2 - 2)-(1 \ln 1-1) \\
&= (\ln 4-2)-(0-1) \\
&= \ln 4-1 \\
&= \ln 4 - \ln e \\
&= \ln \left( \frac 4e \right)
\end{align}
but I read from somewhere that the answer should be $\frac 4e$.
|
If you know Sterling's Approximation:
$$
n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n},
$$
then you could approach it as follows:
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n}
= \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}}
= \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}}
= \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}}
= \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}.
$$
|
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|
Probability of flipping a coin an exact number of times A biased coin flips heads with probability 4/9 and tails with probability 5/9. The coin is flipped 90 times. What is the probability that heads is flipped exactly 40 times?
I believe this question requires the utilisation of the binomial distribution in which:
$Pr(H) = \frac{4}{9}$
$Pr(T) = \frac{5}{9}$
$n = 90$
$k = 40$
$Pr(X = 40) = {90\choose 40}(\frac{4}{9})^{40} (1 - \frac{4}{9}) ^{90-40} =
{90\choose 40} (\frac{4}{9})^{40} (\frac{5}{9}) ^{50}$
is approximately 0.084
Am I going abouts this correctly? Not feeling too confident
|
The standard deviation for this is $$\sqrt{np(1-p)} = \sqrt{90\cdot \frac{4}{9}\cdot\frac{5}{9}} = 4.714$$
Roughly $2/3$ of the outcomes for the number of heads would be between about $35$ and $45$. That's $11$ outcomes each with a probability of less than $0.09$ to ensure the total is less than $1$. So your answer of $.084$ is right in the ball park.
|
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|
Concurrency (I Think Using Menelaus' Theorem) Let $ABC$ be a triangle with incenter $I$, and let $B'$ and $C'$ be points on $BC$ such that $\angle{BIB'} = \angle{CIC'} = 90^\circ$. Let $AB'$ meet $CI$ at $P$, and let $AC'$ meet $BI$ at $Q$. Prove that $PQ$, $BC$, and the tangent to $\odot(BIC)$ at $I$ are concurrent.
So I have been trying to solve this problem by using Menelaus' theorem on $\triangle{BIC}$. So far, I have that if we the tangent at $I$ meet $BC$ at $R$, then by Menelaus theorem, we need $\frac{RB}{RC} \cdot \frac{BQ}{IQ} \cdot \frac{IP}{CP} = -1$. When trying to reduce this expression, the best I came up with was $\frac{RB}{RC} = \frac{IB^2}{IC^2}$. For those who are wondering, this result is not too hard to prove and can be done so using $\triangle{RBI} \sim \triangle{RIC}$. Any help on how to finish the problem would be very much appreciated. In particular, I am not sure how to deal with such seemingly nasty ratios.
Sincerely,
tworigami
|
Can be done with barycentric coordinates. Wlog, let $BC + AC + AB = a + b + c = 1$. The condition for the orthogonality of $P_1P_2$ and $P_3P_4$ is that the permanent of the matrix with the rows $(a^2, b^2, c^2), P_2 - P_1, P_4 - P_3$ is zero. For $IB$ and $IB'$, this becomes
$$\operatorname{perm} \begin{pmatrix}
a^2 & b^2 & c^2 \\ a & b - 1 & c \\ a & b - y_{B'} & c - z_{B'}
\end{pmatrix} = 0 \;\Rightarrow\;
B' = (0, 1 - 2 c, 2 c).$$
Similarly, $C' = (0, 2 b, 1 - 2 b)$.
Then the intersection of $CI$ and $AB'$ is given by
$$\lambda_1 (0, 0, 1) + \mu_1 (a, b, c) =
\lambda_2 (1, 0, 0) + \mu_2 (0, 1- 2 c, 2 c) \;\Rightarrow \\
\frac {IP} {PC} = \frac {\lambda_1} {\mu_1} =
\frac {c (1 - 2 a)} {1 - 2 c}.$$
Replacing $c$ with $b$ gives $IQ/QB = b (1 - 2 a)/(1 - 2 b)$.
It's easy to show that the tangent to the circumcircle of $IBC$ is orthogonal to $IA$. Writing the orthogonality condition for $IA$ and $IR$, we get
$$\operatorname{perm} \begin{pmatrix}
a^2 & b^2 & c^2 \\ a - 1 & b & c \\ a & b - y_{R} & c - z_{R}
\end{pmatrix} = 0 \;\Rightarrow\;
\frac {BR} {CR} = - \frac {c (1 - 2 b)} {b (1 - 2 c)}.$$
Now (segments aren't signed, but the ratios are)
$$\frac {BR} {CR} \frac {CP} {IP} \frac {IQ} {BQ} =
-\frac {c (1 - 2 b)} {b (1 - 2 c)}
\frac {1 - 2 c} {c (1 - 2 a)}
\frac {b (1 - 2 a)} {1 - 2 b} = -1,$$
therefore $PQ$ goes through $R$.
|
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|
How to apply the Chinese remainder theorem to $x^{3}+2x+1 \equiv 0 \bmod 15$? I want to apply the Chinese remainder theorem to the polynomial equation $ x^{3}+2x+1 \equiv 0 \bmod 15 $ which I have split into two equations $x^{3}+2x+1 \equiv 0 \bmod 3 $ and $x^{3}+2x+1 \equiv 0 \bmod 5$.
Looking at the Chinese remainder theorem the congruence equations are linear but I have seen examples which have applied the theorem to quadratic equations so is it possible to apply it this equation?
|
Yes, you may use the chinese remainder theorem.
$x^3+ 2x + 1 \equiv 0 \mod 3$
By Fermat $x^3 \equiv x \mod 3$ and we get the impossible $x^3 + 2x + 1 \equiv x + 2x + 1 \equiv 3x + 1 \equiv 1 \equiv 0\mod 3$. (Or we could have tried them each individually-- there's only three of them.)
so the chinese remainder theorem shows there is no solution.
====
As an example where it would have worked. Let's let $x\equiv 6 \mod 15$ and so $x^3 + 2x + 1\equiv 4\mod 15$ so suppose we we asked to solve $x^3 + 2x -3 \equiv 0 \mod 15$.
Then $x^3 + 2x -3 \equiv 0 \mod 3$ yields
$x^3 - x \equiv 0 \mod 3$
$x(x+1)(x-1) \equiv 0 \mod 3$ and any value mod 3 will solve this.
$x^3 + 2x -3 \equiv 0 \mod 5$ yields
$x^3 + 2x +2 \equiv 0 \mod 5$. $x = 0$ is not a solution. So if there is any solution it is relatively prime to $5$.
$x^4 + 2x^2 + 2x \equiv 2x^2 + 2x + 1 \equiv 0 \mod 5$.
$x \equiv \frac {-2 \pm \sqrt {4 - 4*2}}{4} = 2\pm \sqrt {-1+2}\equiv 2 \pm 1$.
So $x \equiv 1,3 \mod 5$.
So solutions are all $1,3,6,8,11,13\mod 15$. All in all, there was probably an easier way... but that was valid
|
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|
Definite integral concerning the greatest integer function Evaluate the integral $$I=\int_{0}^{x} \lfloor t+1 \rfloor^3 dt$$where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The answer is given as $$\Bigg[\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\Bigg]^2+(\lfloor x \rfloor+1)^3 \{x\}$$ where $\{x\}$ denotes the fractional part of $x$.
What I have done so far is this:
I used the property $\lfloor x+a \rfloor = a+\lfloor x \rfloor$ where $a$ is a positive integer and expanded the binomial.
$\int_{0}^{x}\lfloor t+1 \rfloor^3 dt = \int_{0}^{x} \lfloor t \rfloor^3 dt + 3\int_{0}^{x} \lfloor x \rfloor^2 dt + 3\int_{0}^{x} \lfloor x \rfloor dt + \int_{0}^{x} dt$
Where
$\int_{0}^{x}\lfloor t \rfloor^3 dt = \int_{1}^{2} 1 dt + \int_{2}^{3} 2^3 dt + \cdots +\int_{\lfloor x \rfloor -1}^{\lfloor x \rfloor}(\lfloor x \rfloor -1)^3+\int_{\lfloor x \rfloor}^{x}\lfloor x\rfloor^3 dt=\Big[\frac{\lfloor x \rfloor(\lfloor x \rfloor -1)}{2}\Big]^2 + \lfloor x \rfloor^3 \{x\} $
$\int_{0}^{x}\lfloor t \rfloor^2 dt = \int_{1}^{2} 1 dt + \int_{2}^{3} 2^2 dt + \cdots +\int_{\lfloor x \rfloor -1}^{\lfloor x \rfloor}(\lfloor x \rfloor -1)^2+\int_{\lfloor x \rfloor}^{x}\lfloor x\rfloor^2 dt=\lfloor x \rfloor^2(\lfloor x \rfloor -1) + \lfloor x \rfloor ^2 \{x\} $
And $\int_0^{x} \lfloor t \rfloor dt = \frac{\lfloor x \rfloor (\lfloor x\rfloor -1)}{2} + \lfloor x \rfloor\{x\}$
Therefore $$I=\Big[\frac{\lfloor x \rfloor(\lfloor x \rfloor -1)}{2}\Big]^2 + \lfloor x \rfloor^3 \{x\} + 3(\lfloor x \rfloor^2(\lfloor x \rfloor -1) + \lfloor x \rfloor ^2 \{x\}) + 3 \Big(\frac{\lfloor x \rfloor (\lfloor x\rfloor -1)}{2} + \lfloor x \rfloor\{x\}\Big)+x$$
But I am having trouble in reducing it further. I might be missing something, I am sorry about that. Please help me, thank you.
|
According to OP's stated answer we have
\begin{align*}
\color{blue}{\int_{0}^{x} \lfloor t+1 \rfloor^3\,dt=\left(\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\right)^2+(\lfloor x \rfloor+1)^3 \{x\}}\tag{1}
\end{align*}
This can be shown as follows
\begin{align*}
\int_{0}^{x}\lfloor t+1\rfloor ^3\,dt&=\int_{1}^{x+1}\lfloor u\rfloor ^3\,du\tag{2}\\
&=\sum_{j=1}^{\lfloor x\rfloor}j^3+\int_{\lfloor x\rfloor+1}^{x+1}\lfloor u\rfloor ^3\,du\tag{3}\\
&=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor+1\right)\right)^2+\left(\lfloor x\rfloor+1\right)^3\left\{x\right\}
\end{align*}
Comment:
*
*In (2) we substitute $t+1=u, dt=du$ and arrange the limits accordingly.
*In (3) we use the same representation as in OPs post regarding $\int_{0}^{x}\lfloor t \rfloor^3 dt$.
We can also show the result (1) using OPs calculation, with the correction regarding $\int_{0}^{x}\lfloor t \rfloor^2 dt$
\begin{align*}
\int_{0}^{x}\lfloor t \rfloor^3 dt
&=\sum_{j=1}^{\lfloor x\rfloor-1}j^3\int_{j}^{j+1}\,dt+\lfloor x\rfloor^3\{x\}\\
&=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)\right)^2+\lfloor x\rfloor^3\{x\}\tag{4}\\
\int_{0}^{x}\lfloor t \rfloor^2 dt
&=\sum_{j=1}^{\lfloor x\rfloor-1}j^2\int_{j}^{j+1}\,dt+\lfloor x\rfloor^2\{x\}\\
&=\frac{1}{6}\left(\lfloor x\rfloor-1\right)\lfloor x\rfloor\left(2\lfloor x\rfloor-1\right)+\lfloor x\rfloor^2\{x\}\tag{5}\\
\int_{0}^{x}\lfloor t \rfloor dt
&=\sum_{j=1}^{\lfloor x\rfloor-1}j\int_{j}^{j+1}\,dt+\lfloor x\rfloor\{x\}\\
&=\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)+\lfloor x\rfloor\{x\}\tag{6}
\end{align*}
With (4)-(6) we obtain according to OPs calculation
\begin{align*}
\color{blue}{\int_{0}^{x}\lfloor t+1 \rfloor^3 dt}
&= \int_{0}^{x} \lfloor t \rfloor^3 dt + 3\int_{0}^{x} \lfloor x \rfloor^2 dt + 3\int_{0}^{x} \lfloor x \rfloor dt + \int_{0}^{x}\,dt\\
&=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)\right)^2+\lfloor x\rfloor^3\{x\}\\
&\qquad +3\left(\frac{1}{6}\left(\lfloor x\rfloor-1\right)(\lfloor x\rfloor\left(2(\lfloor x\rfloor-1\right)+\lfloor x\rfloor^2\{x\}\right)\\
&\qquad +3\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)+\lfloor x\rfloor\{x\}\right) \\
&\qquad +\lfloor x\rfloor+\{x\}\\
&\,\,\color{blue}{=\frac{1}{4}\lfloor x\rfloor ^4+\frac{1}{2}\lfloor x\rfloor ^3+\frac{1}{4}\lfloor x\rfloor ^2+(\lfloor x \rfloor+1)^3 \{x\}}
\end{align*}
corresponding to the claim in (1).
|
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|
Verification of Computation of Galois group Let $f = x^8-10x^4 + 1$. We want the Galois group over $\mathbb{Q}$
The roots are $\pm \sqrt{ \pm \sqrt{5 \pm 2\sqrt6}}$. I'm not sure about the splitting field,but I think that it would just be $\mathbb{Q}(i, \sqrt[4]{5 + 2\sqrt6})$, but I can't figure out if this would cover the root $\sqrt[4]{5-2\sqrt6}$.
Then $|Gal_f{Q}| = [\mathbb{Q}(i, \sqrt[4]{5 + 2\sqrt6}): \mathbb{Q}] = [\mathbb{Q}(i, \sqrt[4]{5 + 2\sqrt6}) : \mathbb{Q}(\sqrt[4]{5 + 2\sqrt6})][\mathbb{Q}( \sqrt[4]{5 + 2\sqrt6}):\mathbb{Q}]$. I claim that $[\mathbb{Q}(\sqrt[4]{5 + 2\sqrt6}):\mathbb{Q}] = 8$ because of the minimal polynomial $f = x^8-10x^4 + 1$ and $[\mathbb{Q}(i, \sqrt[4]{5 + 2\sqrt6}) : \mathbb{Q}(\sqrt[4]{5 + 2\sqrt6})] = 2$ since the minimal polynomial x^2 + 1 is of degree 2. This means that $Gal_f(\mathbb{Q})| = 2\times8=16$ so the Galois group is isomorphic to some subgroup of $S_8$ of order $16$
Is this correct? Could someone please help me figure out what group of order $16$ it would be isomorphic to? Thanks
|
Note that the Galois group is non-abelian as complex conjugation will not commute with any automorphism that sends $\sqrt[4]{5 + 2\sqrt6} \to i\sqrt[4]{5 + 2\sqrt6}$. Now $\sqrt[4]{5 + 2\sqrt6} = \sqrt{\sqrt{2}+\sqrt{3}}=\alpha$, say, so we see that the splitting field contains at least 7 quadratic subfields, $\Bbb{Q}[\sqrt{6}],\Bbb{Q}[\sqrt{2}],\Bbb{Q}[\sqrt{3}],\Bbb{Q}[\sqrt{i}],\Bbb{Q}[\sqrt{-2}],\Bbb{Q}[\sqrt{-3}],\Bbb{Q}[\sqrt{-6}]$. I believe that limits us to only $D_4\times C_2$ as having the required 7 seven normal subgroups of order 8 (using the subgroup structure for groups of order 16 that can be found here).
If the Galois group is the direct product, we should also be able to show that using the result here. That is, we need to find 2 Galois subfields with intersection $\Bbb{Q}$ that generate the splitting field. For $D_4\times C_2$ we'll need to find a dihedral and a quadratic extension of $\Bbb{Q}$ contained in $\mathbb{Q}[i, \sqrt[4]{5 + 2\sqrt6}]=K$.
The dihedral extension we want will turn out to be the splitting field of $x^4-2x^2-2$, one root of which is $\sqrt{1+\sqrt{3}}$. It's a tedious but not too difficult check that
\begin{align}\frac{1}{4}\left(\alpha^7+\alpha^5-9\alpha^3-9\alpha\right)&= \frac{\alpha}{4}\left(\alpha^6+\alpha^4-9\alpha^2-9\right) \\
&= \frac{\alpha}{4}\left(11\sqrt{2} + 9\sqrt{3} +5 + 2\sqrt{6}-9\sqrt{2}-9\sqrt{3} -9\right) \\
&= \frac{\alpha}{4}\left(-4 + 2\sqrt{2}+ 2\sqrt{6} \right) = \frac{\alpha}{2}\left(-2 + \sqrt{2}+ \sqrt{6} \right) \\
&= \sqrt{\left(\sqrt{2}+\sqrt{3}\right)\left(3 - \sqrt{2} + \sqrt{3} - \sqrt{6}\right) } \\
&= \sqrt{1+\sqrt{3}}
\end{align}
$1+\sqrt{3}$ satisfies $g(x)=x^4-2x^2-2$ which is Eisenstein at 2, so it's clear that $1+\sqrt{3}$ isn't a square in $\Bbb{Q}[\sqrt{3}]$. Since $1-\sqrt{3}<0$ two of the roots of $g$ are complex and the degree of the splitting field, call it $F$, must then be 8. Thus the Galois group must be $D_4$. Now $F$ contains the quadratic subfields $\Bbb{Q}[\sqrt{3}],\Bbb{Q}[\sqrt{-2}]$ and $\Bbb{Q}[\sqrt{-6}]$, but any of the other 4 quadratic subfields of $K$ along with $F$ will generate $K$, so $\Bbb{Q}[i]$, say. Thus we get $\operatorname{Gal}(K/\Bbb{Q})\cong D_4\times C_2$ as desired.
|
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|
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help?
|
$$\int\sqrt{x^2-x}\; dx=\int\sqrt{x}\; \sqrt{x-1}\; dx\qquad\qquad x\rightarrow\; \cosh^2\theta\quad dx\rightarrow2\cosh\theta\sinh\theta\; d\theta$$
$$=2\int\cosh^2\theta\sinh^2\theta\; d\theta\ =2\int\cosh^2\theta(\cosh^2\theta-1)\; d\theta\ = 2\int\cosh^4\theta\; d\theta -2\int\cosh^2\theta\; d\theta\ $$
$$=\frac{1}{2}\int(1+\cosh(2\theta))^2\;d\theta\ -\int(1+cosh(2\theta))\; d\theta$$
$$\frac{1}{2}\int(1+2\cosh(2\theta)+\cosh^2(2\theta))\; d\theta\ -\theta-\frac{1}{2}\sinh(2\theta)+C$$
$$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)+C\ +\frac{1}{4}\int(1+\cosh(4\theta))\; d\theta$$
$$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)\ +\frac{\theta}{4}+\frac{1}{16}\sinh(4\theta)+C$$
$$\frac{1}{16}\sinh(4\theta)-\frac{\theta}{4}+C$$
$$\theta=\cosh^{-1}(\sqrt{x})$$
$$\frac{1}{16}\sinh(4\cosh^{-1}(\sqrt{x}))-\frac{1}{4}\cosh^{-1}(\sqrt{x})+C$$
This is the short answer if you don't want so much details. Of course if you have enough skills can try using Integration By Parts or another Trig. Subs.
|
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|
Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatrix}
-1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$c) \begin{pmatrix}
1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$d) \begin{pmatrix}
-1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
My attempt : I know that Determinant of A = product of eigenvalues of A, as option c and d is not correct because Here Determinant of A = 4
that is $ \det A = -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm in confusion about option a) and b).......how can I find the Jordan canonical form of A ?
PLiz help me.
Any hints/solution will be appreciated.
Thanks in advance
|
Another answer that requires only a very minimal amount of computations.
The trace of $A$ is preserved under similarity transformations. $\operatorname{Tr}(A) = 0+0+0+0 = \lambda_1+\lambda_2+\lambda_3+\lambda_4$ is enough to exclude (c) and (d). As others have noted, (b) isn't even a Jordan canonical form (and by the way the matrix (b) is similar to (a) anyway).
So (a) is the only remaining option and must be correct.
|
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|
Find $\int{\arctan x}\,\mathrm dx$ without substitution 2018-08-15: I'm still looking for an answer that does not rely on
$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$
I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to find $\int{\arctan x}\,\mathrm dx$.
I start with integration by parts:
$$\int{\arctan x}\,\mathrm dx
= \int{1\cdot\arctan x}\,\mathrm dx
= x\cdot\arctan x - \int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$
Next, it would be natural to use substitution, and I can do that to get the answer:
$$x\cdot\arctan x - \frac{\ln(x^2+1)}{2}+c$$
Which is correct, but it's clear from context that the book wants me to do it without using substituation. I feel like I've tried everything, looking at the book's own examples, but I must be missing some essential trick.
Here are some of the approaches I've tried:
1. Integration by parts, again
1.1. $u = x$ , $v' = \frac{1}{x^2+1}$
$$x\cdot\arctan x - \left(x\cdot\arctan x - \int{1\cdot\arctan x}\,\mathrm dx\right)
= \int{\arctan x}\,\mathrm dx$$
Back where I started.
1.2. $u = \frac{1}{x^2+1}$ , $v' = x$
$$x\cdot\arctan x - \left(\frac{x^2}{2} \cdot \frac{1}{x^2+1} - \int{\frac{x^2}{2} \cdot \frac{-2x}{(x^2+1)^2}}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2(x^2+1)} - \int{\frac{x^3}{(x^2+1)^2}}\,\mathrm dx$$
I've tried hacking away at this, but it doesn't look like it's getting any easier.
2. Adding and subtracting $x^2$ in the numerator
$$x\cdot\arctan x - \int{x\cdot \frac{(x^2+1)-x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x\cdot \left(1 - \frac{x^2}{x^2+1}\right)}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x - x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \int{x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
The book uses a similar trick (adding and subtracting 1 in the numerator) to solve $\int{x\cdot\arctan x}\,\mathrm dx$. In the process it finds that $\int{\frac{x^2}{x^2+1}}\,\mathrm dx = x - \arctan x$. Perhaps I need to use this result:
Partial integration with $u = x$ , $v´ = \frac{x^2}{x^2+1}$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x\cdot(x-\arctan x) - \int{1\cdot (x-\arctan x)}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx$$
$$= \int{\arctan x}\,\mathrm dx$$
Back where I started, again. I've been trying for days, different tricks and manipulations. The book doesn't even list it as a particularly tricky question, so I'm feeling a bit dumb.
3. Searching for the answer
I've looked at some other questions, including
*
*Integration of arctan(x) is itself?
*Indefinite integral of $\arctan(x)$, why consider $1\cdot dx$?
But they don't seem to answer my particular question.
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Observe that
$$I=\int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$
can be rewritten as
$$I=\frac{1}{2}\int{\frac{\mathrm d(x^2+1)}{x^2+1}}=\ln(x^2+1)$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2799115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Prove using calculus or otherwise the following inequality Prove that$$ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2$$where $$x\in\left(0,\frac{\pi}{2}\right]$$
My Attempt:$$ f'(x)=\frac{ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)}{(x-\sin x)^2}=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
After this not able to get breakthrough.
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If we can show that $f$ is decreasing on $(0,\frac{\pi}{2}]$, then it is easy to finish by checking the endpoints.
To show that $f$ is decreasing on this interval, it is enough to show that $f'$ is negative on this interval. You computed
$$f'(x)=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
which is negative if and only if the numerator is negative.
For convenience, we will rewrite the numerator as
$$x^2\sin x-(1-\cos x)(x+\sin x).$$
We can bound the numerator by using that fact that Taylor series for sine and cosine are alternating series, and therefore truncating it gives us bounds.
For $x\in(0,\frac{\pi}{2}]$, we have
$$x^2\sin x\le x^2(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!})$$
$$(1-\cos x)(x+\sin x)\ge(1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}))(x+(x-\frac{x^3}{3!}))$$
where for the second inequality we use that both factors on the right hand side are positive on this interval. Therefore,
\begin{align*}
&x^2\sin x-(1-\cos x)(x+\sin x)\\
\le &x^2(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}) - (1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}))(x+(x-\frac{x^3}{3!}))\\
=&-\frac{x^7((x^2-30)^2+108)}{725760}
\end{align*}
and we can see that the right hand side is always negative on the interval.
(How did I know how far to expand each Taylor series? Trial and error. There might be some way to see how far to expand each using the fact that they Taylor series for the numerator of $f'(x)$ is $O(x^7)$, but I don't see how at the moment.)
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2800023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
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