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Can we prove that $(2^{2^{n}}+1)+(2^{2^{n-1}}+1) -1$ have at least $n$ different prime divisors My point was to prove it using induction.
Then for $n=1$ it is true, for n = k it is also true so we assume that
$ \color{Red}{ 2^{2^{k}} + 2^{2^{k-1}} + 1}$ has at least k different prime divisors
When we consider $n=k+1$ we have:
$x^4+x^2+1$ where $x=2^{2^{k-1}}$
we also know that:
$x^4+x^2+1 = (x^2+x+1)(x^2-x+1) = \color{Red}{(2^{2^{k}} + 2^{2^{k-1}} + 1)}(2^{2^{k}}-2^{2^{k-1}}+1)$
And my question is: Is it sufficient to show that polynomial $x^2-x+1$ is not divisible by $x^2+x+1$ so $(2^{2^{k}}-2^{2^{k-1}}+1)$ gives us at least one more prime number which implies that this statement is true for all natural numbers?
|
Assuming $2^{2^k}+2^{2^{k-1}}+1$ has at least $k$ different prime divisors,
for the inductive step it suffices to show that $2^{2^k}-2^{2^{k-1}}+1$ has a different prime factor.
Well, if $p$ divides $2^{2^k}+2^{2^{k-1}}+1$ and $2^{2^k}-2^{2^{k-1}}+1$, then $p$ divides their difference, which is a power of $2$,
so $p$ is a power of $2$, but $2^{2^k}+2^{2^{k-1}}+1$ and $2^{2^k}-2^{2^{k-1}}+1$ are not powers of $2$, so $p=1$.
Since $2^{2^k}-2^{2^{k-1}}+1>1$ for $k\gt 0$, it follows that $2^{2^k}-2^{2^{k-1}}+1$ has different prime factor(s)
from those of $2^{2^k}+2^{2^{k-1}}+1$, so the inductive step works.
|
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Closed form for the integral $\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$
Somewhere, the integral
$$\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$$
has been found numerically to be real and finite, if $k \in (-\infty,3].$
Can one find a closed form for this integral?
|
Denoting the integral by $I(k)$, let us use $\int_{-a}^{a}f(x)dx, =\int_{0}^{a} [f(x)+f(-x)] dx,$ then
$$I(k)=\int_{-1}^{1} \frac{\ln[1+x^2-x\sqrt{k+x^2}]}{\sqrt{1-x^2}} dx= \int_{0}^{1} \frac{\ln[1+(2-k)x^2]}{\sqrt{1-x^2}}dx =\frac{1}{2} \int_{0}^{\pi} \ln \left (\frac{1+r^2}{2}+\frac{1-r^2}{2} \cos \theta \right) d \theta,$$ where we took $x=\sin(\theta/2)$ and $r^2=3-k$.
Next, we take $g(p)=\int_{0}^{\pi} \ln(p+q \cos\theta) d\theta$ , then $g'(p)=\int_{0}^{\pi} \frac{d\theta} {(p+q \cos \theta)}$. We can write
$$2g'(p)=\int_{0}^{\pi} \left( \frac{1}{p+q \cos \theta}+\frac{1}{p-q \cos \theta}\right) d\theta=$$
$$2p \int_{0}^{\pi} \frac{d\theta}{p^2-q^2 \cos^2 \theta}=2p\int_{0}^{\pi/2} \frac{\mbox{sec}^2\theta ~d\theta}{\tan^2\theta+(p^2-q^2)/p^2}.$$
$$\Rightarrow g'(p)=\frac{\pi}{\sqrt{p^2-q^2}} \Rightarrow g(p)=\pi \ln \frac{p+\sqrt{p^2-q^2}}{2}, ~p=\frac{1+r^2}{2},~ q=\frac{1-r^2}{2}.$$ Hence $$I(k)=\pi \ln \left(\frac{1+\sqrt{3-k}}{2} \right)$$
|
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Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
|
Because $$\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}=\frac{\left(\sqrt{a+b}\right)^2-\sqrt{(a-b)(a+b)}}{\sqrt{a+b}}=\sqrt{a+b}-\sqrt{a-b}$$
The domain is $a+b>0$ and $(a-b)(a+b)\geq0,$ which gives $a+b>0$ and $a-b\geq0.$
|
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Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tanh^{-1}\frac{2a\sin x}{1+a^2}dx$, with $I(0)=0$
$$ J’(a) = \int_0^{\frac\pi{3}}\frac{2(1-a^2)\sin x}{4a^2\cos^2x+(1-a^2)^2}dx
=\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a}
$$
Then
\begin{align}
I
& =J(1) =J(0)+\int_0^1 J’(a)da = \int_0^1\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a} da\\
&=\int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^3\to a}{\left.\frac{\tan^{-1}a^3}{a}\right)}da=\left(1-\frac13\right) \int_0^1\frac{\tan^{-1}a}{a}da=\frac23G
\end{align}
|
Let $I$ be the integral given by
$$\begin{align}
I&=\int_0^{\pi/3}\text{arctanh}(\sin(x))\,dx\\\\
&=\frac12\int_0^{\pi/3}\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\,dx\\\\
&=\frac12\int_{-\pi/3}^{\pi/3}\log\left(1+\sin(x))\right)\,dx\tag1
\end{align}$$
Then, enforcing the substitution $x\mapsto \pi/2-x$ in $(1)$, we find that
$$\begin{align}
I&=\frac12\int_{\pi/6}^{\pi/2}\log\left(\frac{1+\cos(x)}{1-\cos(x)}\right)\,dx\\\\
&=-\int_{\pi/6}^{\pi/2}\log\left(\tan(x/2)\right)\,dx\\\\
&=-2\int_{\pi/12}^{\pi/4}\log\left(\tan(x)\right)\,dx\tag2
\end{align}$$
Next, using the Fourier series for $\tan(x)=\sum_{k=1}^\infty \frac{(-1)^k-1}{k}\cos(2kx)$ in $(2)$ reveals
$$\begin{align}
I&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)-\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)-\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)}{(2k-1)^2}-2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{(-1)^k}{(2k-1)^2}-2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\left(G-\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\right)\tag3
\end{align}$$
Using judicious grouping, user @M.N.C.E. showed in THIS ANSWER that the right-hand side of $(3)$ was equal to $2G/3$. Hence we find that
$$I=2G/3$$
as was to be shown!
|
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compute the following integral in closed form : $\int_0^{\infty}\frac{\arctan x}{a^2x^2+1}\,dx$
Find :
$$\int_0^{\infty}\frac{\arctan x}{a^{2}x^2+1}\,dx,\qquad a > 0.$$
I think this integral related with polylogarithm function.
My attempt as follows:
Let $$I(b)=\int_0^{\infty}\frac{\arctan bx}{1+a^{2}x^{2}}\,dx.$$
Now differentiating with respect to $b$ we find:
$$I'(b)=\int_0^{\infty}\frac{x}{(1+a^{2}x^{2})(1+b^{2}x^2)}\,dx.$$
Then I use partial fraction:
$$
\begin{align*}
I'(b)&=\int_0^{\infty}\frac{1}{a^2-b^2}\left(\frac{a^{2}x}{1+a^{2}x^{2}}-\frac{xb^{2}}{1+b^{2}x^2}\right)\,dx\\
&=\frac{1}{2(a^2-b^2)}\left(a^{2}\ln(1+a^{2}x^2)-b^{2}\ln(1+b^{2}x^{2})\right)\bigg|_0^{\infty}
\end{align*}
$$
I don't know how I complete this work ?
|
Your final expression has extra $a^2$ and $b^2$. It should read $$I'(b)=\frac{1}{2(a^2-b^2)}\left.\ln\frac{1+a^2x^2}{1+b^2 x^2}\right|_{0}^{\infty}=\frac{\ln a-\ln b}{a^2-b^2}$$ which is integrated using $I(0)=0$: $$I(b)=\int_{0}^{b}\frac{\ln a-\ln x}{a^2-x^2}\,dx=\frac{1}{a}\int_{0}^{b/a}\frac{\ln(1/t)}{1-t^2}\,dt=\frac{1}{a}f\Big(\frac{b}{a}\Big),$$ where, integrating by parts and using the definition of $\mathrm{Li}_2$,\begin{align}f(x)&=\frac{1}{2}\left.\ln\frac{1}{t}\ln\frac{1+t}{1-t}\right|_{0}^{x}+\frac{1}{2}\int_{0}^{x}\frac{1}{t}\ln\frac{1+t}{1-t}\,dt\\&=\frac{1}{2}\left(\ln\frac{1}{x}\ln\frac{1+x}{1-x}+\mathrm{Li}_2(x)-\mathrm{Li}_2(-x)\right).\end{align}
|
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Inequality between $\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}}$ and $\frac{3}{\sqrt[n]{1+(abc)^{1/3}}}$ Earlier
Exploring an inequality between $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} $ and $\frac{3}{1+(abc)^{1/3}}$ if $a,b, c>0$
the expressions $$\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}}$$ and $$\frac{3}{\sqrt[n]{1+(abc)^{1/3}}}$$ have been found to display two inequalities for $n=1$ when $a,b,c \in [0,1]$ or in $[1,\infty)$.
Now the question is if there exist more general inequalities between them if $n>0$ ?
|
Take $f(x)=\frac{1}{(1+e^x)^{1/n}}, n>0$. then $f''(x)=\frac{e^x(e^x -n)}{n^2(1+e^x)^{1/n+2}}>0,~ \mbox{if}~ e^x>n.$ Then Jensen's inequality
$$\frac{f{x}+f(y)+f(z)}{3} \ge f\left(\frac{x+y+z)}{3}\right)~~~~(1)$$ holds. So we have $$\frac{1}{\sqrt[n]{1+e^x}}+\frac{1}{\sqrt[n]{1+e^y}}+
\frac{1}{\sqrt[n]{1+e^z}}\ge \frac{1}{\sqrt[n]{1+e^{(x+y+z)/3}}},~\mbox{if}~ e^x,e^y,e^z >n, n>0~~~~(2).$$ Hence, we can re-write it as
$$\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}} \ge \frac{3}{\sqrt[n]{1+(abc)^{1/3}}},~ \mbox{if}~ a,b,c>n,n>0.~~~~(3).$$
The sign of inequality(3) reverses if $0<a,b,c <n$. The equality holds when $a=b=c=n$.
|
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|
Find the sum of the even-degree coefficients I am doing practice problems from the AMC-10 math competition. I do not understand how to go about solving the problem
Let $(1+x+x^2)^n = a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$ be an identity of $x$. If we let $s = a_0 + a_2 + a_4 + ... + a_{2n}$, find $s$.
The choices are $A, 2^n$, $B, (2^n) + 1$, $C, \frac{3^n - 1}{2}$ , $D, \frac{3^n}{2}$ , and $E, \frac{3^n + 1}{2}$
Thanks in advance
|
As noted by lab in a comment, for $x=1,-1$ you get
$$3^n = a_o+a_1+\ldots+a_{2n} \\ 1 = a_o-a_1+\ldots+a_{2n}$$
so $3^n+1 = 2a_0+ 2a_2+\ldots+2a_{2n}$.
Another (lazy) idea if you already have some candidates: for $n=1, a_0+a_2=2$, for $n=2, (1+x+x^2)^2 = 1+2x+3x^2+2x^3+x^4$, so $a_0+a_2+a_4 = 5$. Then the only candidate possible is $\frac{3^n+1}{2}$.
|
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How do I take a fraction to a negative power? I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2
$$
So here were my steps:
*
*First step:
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\right)
$$
And that's as far as I got, because now I want to use the form $\ln(a/b) = \ln(a) - \ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.
How do I evaluate $\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}$ ?
Thanks
|
By definition $$a^{-k} = \frac 1{a^k}$$
So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$
$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$
It will help to realize that $(\frac ab)^{-1} = 1/(a/b) = \frac ba$ and that $(\frac ab)^k = \frac {a^k}{b^k}$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$
(Also $(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$ or that $(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$.)
In any event
$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$
|
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Prove that $\sum_{k=1}^{n} \frac{(-1)^{k}}{k}B(n,k) = 0 $ for all integer $n\geq 2$ For all $ n, k\geq 1$, define $B(n,k) = \sum_{a_{1}+a_{2} +\dots+a_{k} =n} \frac{1}{a_{1}!a_{2}! \cdots a_{k}! }$, where $ a_{1},a_{2},\dots,a_{k} $ are positive integers.
For example: $B(3,2) = \frac{1}{1!2!} + \frac{1}{2!1!} =1$ because we can write $3 $ as $2+1$ and $1+2$.
Another example:
$$B(5,3) = \frac{1}{1!1!3!} + \frac{1}{1!3!1!} + \frac{1}{3!1!1!}+\frac{1}{1!2!2!}+ \frac{1}{2!1!2!}+ \frac{1}{ 2!2!1!} =\frac{5}{4}. $$
Prove that $\sum_{k=1}^{n} \frac{(-1)^{k}}{k}B(n,k) = 0 $ for all integer $n\geq 2$.
I tried to approach this question by counting numbers of functions (onto), but someone said they did it by observing the coefficients of $f(x) = \ln(x+1) ,g(x) = e^{x}-1$. I don't really understand how this is helpful. Can anyone provide relevant pointers?
|
It is well-known that
$$
e^x = \sum_{a=0}^\infty \frac{x^a}{a!}.
$$
This implies that
$$
(e^x-1)^k = \sum_{n=0}^\infty B(n,k) x^n.
$$
It is also known that
$$
\ln (1+x) = -\sum_{k=1}^\infty (-1)^k \frac{x^k}{k}.
$$
Therefore
$$
\ln (1 + (e^x-1)) = -\sum_{k=1}^\infty (-1)^k \frac{(e^x-1)^k}{k} = -\sum_{k=1}^\infty \sum_{n=0}^\infty (-1)^k \frac{B(n,k) x^n}{k}=-\sum_{n=0}^\infty x^n \sum_{k=1}^\infty \frac{(-1)^k}{k} B(n,k).
$$
On the other hand,
$$
\ln(1 + (e^x-1)) = \ln(e^x) = x.
$$
The proof now follows by comparing coefficients.
|
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How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$? I'm trying to find such primes that are of the form $x^6+y^6$; $x,y\in\mathbb{Z}$.
If one of $x$ and $y$ is $0$, say $y=0$ then $x^6+y^6=x^6$ which is not a prime. So assume that $(x,y)\ne(0,0).$
I factored $x^6+y^6$ as follows:
$x^6+y^6=(x^2)^3+(y^2)^3=(x^2+y^2)(x^4+y^4-x^2y^2)=(x^2+y^2)[(x^2+y^2)^2-3x^2y^2]=(x^2+y^2)(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)$
If $x^4+y^4-x^2y^2=1$ then $(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)=1$ this implies that the two numbers $x^2+y^2+\sqrt{3}x^2y^2$ and $x^2+y^2-\sqrt{3}x^2y^2$ are reciprocal to each other. So if $x^2+y^2+\sqrt{3}x^2y^2\in\mathbb{Z}$ then $x^2+y^2-\sqrt{3}x^2y^2\not\in\mathbb{Z}$ or vise-versa.
But $x^2+y^2\pm\sqrt{3}x^2y^2\not\in\mathbb{Z}$ for $(x,y)\ne(0,0)$.
So that $x^4+y^4-x^2y^2$ can't be factored out in $\mathbb{Z}$. That is we can conclude that this is a prime.
To $x^6+y^6$ be a prime, we must have $x^2+y^2=1$, which is possible only for $(x,y)=(+1,0);(-1,0);(0,+1);(0,-1)$.
But in all the above cases, $x^6+y^6=1$, which is not a prime.
So I can conclude that there is no prime of the form $x^6+y^6.$
My question is my argument ok? And is there any other way to think about this problem?
|
If $x=0$ or $y=0$ then $x^6+y^6$ is $x^6$ or $y^6$ which cannot be a prime.
If $x\ne 0\ne y$ and $x^6+y^6$ is prime then $$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)=(x^2+y^2)(\,(x^2-y^2)^2+x^2y^2
)$$ and the factor $x^2+y^2\ge 2,$ so the factor $(x^2-y^2)^2+x^2y^2$ must be $1,$ which is not possible unless $|x|=|y|=1$ because $$(x^2-y^2)^2+x^2y^2\ge x^2y^2.$$ So $|x|=|y|=1$ and $x^6+y^6=2.$
|
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Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant, simpler method.
For ($i$) I tried cases like in part ($ii$) but since there are 2 variables that would lead to 4 cases.
An elegant and easy solution will be much appreciated
|
Part ii:
In general: $$\lfloor z\rfloor \leq z< \lfloor z\rfloor+1.$$
For ii, set $z=x/2$ then we get:
$$\left\lfloor\frac{x}{2}\right\rfloor\leq \frac{x}{2}<\left\lfloor\frac{x}{2}\right\rfloor +1$$
Double and you get:
$$2\left\lfloor\frac{x}{2}\right\rfloor\leq x<2\left\lfloor\frac{x}{2}\right\rfloor +2$$
Taking the floor gives:
$$2\left\lfloor\frac{x}{2}\right\rfloor\leq\lfloor x\rfloor<2\left\lfloor\frac{x}{2}\right\rfloor +2$$
Subtracting gets:
$$0\leq \left\lfloor x\right\rfloor-2\left\lfloor\frac{x}{2}\right\rfloor<2.$$
Which is the result you want.
Part i
You can actually prove the same way that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor$ is always either $0$ or $1.$ More specifically, you can show that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor=1$ if and only if $\{x\}+\{y\}\geq 1.$
In particular, when $y=x,$ $\lfloor 2x\rfloor -2\lfloor x\rfloor=1$ if and only if $\{x\}\geq \frac{1}{2}.$
Now, if $\{x\}+\{y\}\geq 1$ then one or both of $\{x\}$ and $\{y\}$ are $\geq \frac{1}{2}$, so one or both of $\lfloor 2x\rfloor -2\lfloor x\rfloor$ and $\lfloor 2y\rfloor-2\lfloor y\rfloor$ are one.
So that means, doe any $x,y:$
$$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor\leq \left(\lfloor 2x\rfloor -2\lfloor x\rfloor\right)+\left(\lfloor 2y\rfloor -2\lfloor y\rfloor\right)$$
This is because when the left side is $0,$ we know the right side is at least $0,$ and when the left side is $1$, then the right side is either $1$ or $2.$
Adding $2\lfloor x\rfloor + 2\lfloor y\rfloor$ to both sides gives you:
$$\lfloor x+y\rfloor +\lfloor x\rfloor +\lfloor y\rfloor\leq \lfloor 2x\rfloor +\lfloor 2y\rfloor$$
|
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"url": "https://math.stackexchange.com/questions/3294982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t X^t
\end{matrix}$$
Let $B$ be the standard basis for $\mathbb{R}^{2 \times 2}$ :
$$B =\left\{ \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
0 & 1\\
\end{pmatrix} \right\}$$
Calculate $\textsf{M}_B(\varphi)$ we come to $$\textsf{M}_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
We calculate a basis for the kernel like this:
If
$$X:= \begin{pmatrix}
a & b\\
c & d\\
\end{pmatrix}$$
then $$\varphi(X) = \begin{pmatrix}
a-b & -a+b\\
c-d & -c+d\\
\end{pmatrix}+\begin{pmatrix}
a-b & c-d\\
-a+b & -c+d\\
\end{pmatrix} = \begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
Now we have to look, for what values $$\begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
by definition, the kernel of a linear transformation is $\varphi(X) = 0$, therefore our basis for $\ker(\varphi)$ should be
$$\left\{ \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \right\}$$
Now here comes the part where I'm confused. How do I calculate a basis for $\operatorname{im}(\varphi)$ ?
$\textsf{M}_B(\varphi)$ is the transformation matrix. I've read that you'd just transpose the matrix $\textsf{M}_B(\varphi)$ and row reduce to calculate a basis. I just don't get it.
The solution according to the solution it should be the basis $$ \left\{\begin{pmatrix}
0 & 1 \\
1 & -2 \end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1 \end{pmatrix} \right\}$$
Also little side questions. I do know about
$$\dim(A) = \dim(\operatorname{im} A) + \dim(\ker A)$$
but how exactly do you know the dimension of the Kernel/Image?
|
Observe that $(XA+A^TX^T)$ is a symmetric matrix. So the image will be a subset of the set of symmetric matrices. But the question about a basis for the image set can be resolved as follows:
$$\varphi(X)=\begin{bmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{bmatrix}=(a-b)\begin{bmatrix}2&-1\\-1&0\end{bmatrix}+(c-d)\begin{bmatrix}0&1\\1&-2\end{bmatrix}.$$
This means that anything that belongs to the image is in the span of the two matrices on the RHS. Observe that both the matrices on the right side are LI. Hence they form a basis for the image set. Furthermore the dimension of the image set is $2$ (because we have $2$ vectors in the basis).
NOTE:
The answer given in your textbook is not different from what I gave above because $$\begin{bmatrix}2&-1\\-1&0\end{bmatrix}+\begin{bmatrix}0&1\\1&-2\end{bmatrix}=2\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$ Find:
$$\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$$
I don't know how I starte & evaluate this integral
Wolfram alpha give $=1,28553$
My problem whene I use $t=\cos x$ I get $\arccos x$
Same problem with $t=\sin x$
If any one have idea please help me
|
Let $x=\sin t$ and $a= \sin^{-1}\frac\pi4$
$$I=\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx = \int_0^a \frac{\cos ^2 t+\cos (\sin t)}{1+\sin t \sin (\sin t)} \, dt
$$
Use the shorthand $t_\pm=\frac 12 (t \pm {\sin t})$ to express the denominator and the numerator as
\begin{align}
1+ \sin t \sin (\sin t)&=1+\frac12[\cos(t-\sin t) - \cos(t+\sin t)]\\
&= \sin^2t_+ + \cos^2t_-\\
\cos ^2t+\cos (\sin t)&=\cos t\cos(t_+ + t_-) + \cos(t_+ - t_-)
\\ &= 2 \cos t_- (\sin t_+)’ - 2\sin t_+ (\cos t_-)’
\end{align}
Then
\begin{align}
I& =2\int_0^a \frac {\cos t_- d(\sin t_+)-\sin t_+ d(\cos t_-)}{ \sin^2t_+ + \cos^2t_-}\\
& =-2\int_0^a \frac {d\left(\frac{\cos t_-}{\sin t_+}\right) }{ 1 + \left(\frac{\cos t_-}{\sin t_+}\right)^2}
=2\cot ^{-1} \frac{\cos \left( \frac a2-\frac\pi8\right) }{\sin \left( \frac a2+\frac\pi8\right) }=1.28553
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\sin(x) - \sin(y))(\cos(x) - \cos(y)) = -\frac{1}{6} $$
$$ \sin(x)\cos(x) + \sin(y) \cos(y) - \sin(x+y) = \frac{1}{6}$$
thus
$$ \sin(x+y) =\sin(x)\cos(x) + \sin(y) \cos(y) - \frac{1}{6} $$
Next if I do a sum and organize and squaring we get:
$$ (\sin(x) + \cos(x))^{2} + (\sin(y) + \cos(y))^{2} - 2 (\sin(x) + \cos(x)) (\sin(y) + \cos(y)) = \frac{1}{36} $$
$$ 2 + 2 ( \sin(x) \cos(x) + \sin(y) \cos(y)) - 2 \left( \sin(x+y) + \cos(x-y) \right) = \frac{1}{36} $$
I have no idea after this.
Another method is I suspect that we have to solve for $\sin(x), \cos(x), \sin(y), \cos(y)$ instead of directly finding $\sin(x+y)$ algebraically.
|
HINT- $sinx-siny=2cos\frac{x+y}{2}sin\frac{x-y}{2}=-\frac{1}{3}$ ..........(1)
$cosx-cosy=-2sin\frac{x+y}{2}sin\frac{x-y}{2}=\frac{1}{2}$ ..........(2)
Divide to get $tan\frac{x+y}{2}=\frac{3}{2}$,Now use $sin\theta=2tan\frac{\theta}{2}/(1+tan^2\frac{\theta}{2})$
|
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|
Calculating $\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}$ I can't figure out how to calculate this limit (or prove it does not exist)
$$
\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}
$$
I've tried with restrictions on $y=mx$ and curves of the form $y=x^n$.
The limit should not exist but even with polar coordinates I can't figure it out
|
Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$.
Let $x^2+y^2=r^2$, with $0 < r \le 1$.
If $x\ne 0$, then
\begin{align*}
|f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt]
&\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt]
&=\left|\frac{y}{1+y^4}\right|\\[4pt]
&\le |y|\\[4pt]
&\le r\\[4pt]
\end{align*}
and if $x=0$, then $y\ne 0$, so
$$
f(x,y)=\frac{0}{y^4}=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$
In either case, we have $|f(x,y)|\le r$.
Letting $r$ approach zero from above, it follows that
$$
\lim_{(x,y)\to (0,0)}f(x,y)=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3297618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding values of $\tan^{-1} (2i)$. I am trying to find all solutions of $\tan^{-1} (2i)$. I don't see anything that I have done wrong, my answer doesn't match the one in the textbook. Here is what I have. (The convention in Brown and Churchill is to use $\log$ for a complex number and $\ln$ for a real number.)
\begin{align*}
\tan^{-1} (2i) & = \frac{i}{2} \log \frac{i + 2i}{i - 2i} \\
& = \frac{i}{2} \log \frac{3i}{-i} \\
& = \frac{i}{2} \log (-3) \\
& = \frac{i}{2} \left(\ln 3 + (2n + 1) \pi i \right) \\
& = \frac{i}{2} \ln 3 - \frac{2n + 1}{2} \pi \\
& = \frac{i}{2} \ln 3 - \left(n + \frac{1}{2}\right) \pi.
\end{align*}
The answer in the textbook, however, is:
\begin{align*}
\frac{i}{2} \ln 3 + \left(n + \frac{1}{2}\right)\pi.
\end{align*}
This leads me to believe that I have misplaced a sign somewhere, but I cannot see where. Might there just be a typo in the book?
Thanks in advance.
|
Both are correct.
If $n_1$ is the integer plugged into your solution, plugging $-n_1-1$ into the textbook solution gives the same answer. $$-\left(n_1+\frac{1}{2}\right) = k$$
$$(-n_1-1)+\frac{1}{2} = k$$
|
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|
How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
|
$ (\sqrt u + \sqrt v)^2 = u+v + 2\sqrt{u\,v }$
So at first bring in $2$ in front of the radical sign.
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}$$
Next find two factors for $3$ whose sum is $4$
They are easy to guess. They are $3$ and $1$;
(... or else you need to solve another quadratic equation)
$$ \sqrt{\frac{(\sqrt3 +1)^2}{2}}=\sqrt{ \frac{2(\sqrt3 +1)^2}{4}} $$
where the denominator is made to be $4$ so that while square rooting, $2$ would stay there.
Next manipulate numerator to make them squares of two terms and make denominator also a square
$$\sqrt{ \frac{(\sqrt{2})^2(\sqrt3 +1)^2}{2^2}} $$
that is now ( cancelling out squares and square roots) and multiplying two terms in the numerator:
$$ \dfrac{\sqrt 6 +\sqrt 2}{2}.$$
|
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"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding $P(X<2Y)$ given joint pdf $f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}$ for $x,y\in\mathbb R$
The joint pdf of $X$ and $Y$ is $$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}\,; \quad x,y\in \mathbb{R}$$ Find $P(X<2Y)$.
I have tried this:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{2y} f(x, y)\, dx\,dy$$
@StubbornAtom suggested polar transformations.
So here it goes,
Let $X = r \cos\theta, \, Y = r \sin\theta$.
Then The integral gets transformed as $\int \int re^{-r}d\theta$.
Could some one help find the new limits.
|
So this is my version that I have understood from @pre-kidney's answer please note that this is just an elaboration of his answer.
$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2+y^2}} \ ;\ -\infty<\ x \ <\infty\ ; -\infty < y < \infty$
$X=rcosθ,Y=rsinθ$
$f(r, \theta) = f(\theta).f(r) = (\frac {1}{2\pi} ).(re^{-r})\ \ ;\ \ 0 < r < \infty\ ;\ 0 < \theta < 2\pi$
Now, $P(X < 2Y) = P(rcos\theta < 2rsin\theta) = P(tan\theta > \frac {1}{2}) = P(\theta > tan^-\frac {1}{2}) = \frac {1}{2\pi} \int_{tan^-\frac {1}{2}}^{2 \pi}d\theta = \frac {1}{2\pi} [\theta]_{tan^-\frac {1}{2}}^{\pi + tan^-\frac {1}{2}} = \frac {1}{2\pi}[\pi + tan^-\frac {1}{2} - -tan^-\frac {1}{2}]) = \frac {1}{2\pi} . \pi = 0.5$
|
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|
Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\left(\frac{1+i(a+ib)}{1-i(a+ib)}\right)^n = a+ib$$
$$\left(\frac{(1-b)+ia}{(1+b)-ia}\right)^n = a+ib$$
$$\frac{(1-b)+ia}{(1+b)-ia}.\frac{(1+b)+ia}{(1+b)+ia} = \frac{1-b^2+2ia-a^2}{1+2b+b^2+a^2}$$
then
$$\left(\frac{1-b^2-a^2}{1+2b+b^2+a^2}+\frac{2ia}{1+2b+b^2+a^2}\right)^n = a+ib$$
where
$$a=0 \text{; & } 1+2b+b^2\neq0$$
|
Note that
$$
|1-iz|^2 - |1+iz|^2 = -2i(z -\bar z) = 4 \operatorname{Im}(z)
$$
so that
$$
z \in \Bbb R \iff \left | \frac{1+iz}{1-iz}\right|= 1 \, .
$$
It follows that
$$
\left(\frac{1+iz}{1-iz}\right)^n = a+ib
$$
has only real solutions $z$ if and only if $|a+ib|=1$, i.e. if $a^2+b^2=1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$? Question: What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$?
Source: James S. Rickards Invitational (Algebra II Individual)
I know how to calculate eccentricity, but I am having trouble on factoring it in to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ form.
This is what I did:
$4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$
$4\sqrt{5}x^2 − 17x + y=-\frac{1}{8}$
$4\sqrt{5}(x^2-\frac{17x}{4\sqrt{5}})+1(y+0)= -\frac{1}{8}$
$(x-\frac{17\sqrt{5}}{40})^2+\frac{y}{4\sqrt5}=\frac{-32\sqrt5}{5120}-\frac{289}{160}$
...
|
Note, as given, you have: $4\sqrt{5}x^2-17x+y+\frac{1}{8}=0$.
Just rewrite this as $y=-4\sqrt{5}x^2+17x-\frac{1}{8}$.
It is very easy to see now that this is just the equation of a parabola.
Now by definition, any parabola has an eccentricity of $1$.
No need to complete the square here.
|
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|
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
|
Let $3^a=4^b=k^{ab}\implies 3=k^b,4=k^a$
$$9^{a/b}+16^{a/b}$$
$$=(3)^{2a/b}+4^{2b/a}$$
$$=(k^b)^{2a/b}+(k^a)^{2b/a}$$
$$=k^{2b}+k^{2a}$$
$$=(k^b)^2+(k^a)^2=?$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that if n is divisible by $2$ and $3$, then n is divisible by $6$
For every integer n, if $2 | n$ and $3 | n$ then $6 | n$
! Note: x | y means y is divisible by x.
!! Note: I know that there are way better ways to prove it. However, I am just curious whether the proof below, admittedly peculiar, is correct.
Since 2 | n and 3 | n, we can write $\frac{n}{2} = x $ and $\frac{n}{3} = y$ where $x,y \in \mathbb Z$. Therefore
$$\tag1 \frac{n}{2} + \frac{n}{3} = x + y$$
$$\tag2 \frac{5n}{6} = x + y $$
$$\tag3 5\cdot\frac{n}{6} = x + y$$
Since $x, y \in \mathbb Z$, it follows that $x + y$ in integer and $5\cdot\frac{n}{6}$ is integer is as well. Need to prove that $\frac{n}{6} \in \mathbb Z$.
Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$. But then it will imply that $n = 6a + 1.2$, meaning that $n \notin \mathbb Z$, hence a contradiction. Therefore, $\frac{n}{6} \in \mathbb Z$
Is it correct?
|
Simpler: $\ \dfrac{n}2\in\Bbb Z,\, \dfrac{n}3\in \Bbb Z\,\Rightarrow\, \dfrac{n}6 = \dfrac{n}2-\dfrac{n}3\in\Bbb Z.\ $ Turning to your argument:
Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$.
This claim is unfounded.
Remark $ $ More generally $\,a,b\mid n\iff {\rm lcm}(a,b)\mid n\ $ and this can be proved as above.
|
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|
Matrix rearrangement I have a matrix A in this form:
$$
A=
\left[
\begin{array}{cccc}
x_1 & x_2 & 0 & 0 \\
0 & 0& x_1 & x_2
\end{array} \right]
$$
Here, $x_1$ and $x_2$ are variables, and A is a $2 \times 4$ matrix. I
would like to rearrange the matrix product so that
$$
A^TA=BC
$$
$A^T$ is the transpose of A. $B$ and $C$ are matrices to be found. $B$ is a $4 \times 2$ matrix that only contains constants, and C is a $2 \times 4$ matrix that contains constants and variables ($x_1,x_2$ etc). Is this possible to have such $B$ and $C$?
|
(Not really an answer but at least some development...)
Let's try it!
The left-hand side is easy to calculate:
$$
A^T A =
\left[
\begin{array}{cccc}
x_1^2 & x_1 x_2 & 0 & 0 \\
x_1 x_2 & x_2^2 & 0 & 0 \\
0 & 0 &x_1^2 & x_1 x_2 \\
0 & 0 &x_1 x_2 & x_2^2 \\
\end{array}
\right]
$$
The right-hand side should have this form also. Let's see what happens if we set
$$
B_0 =
\left[
\begin{array}{cc}
b_{00} & b_{01} \\
b_{10} & b_{11} \\
\end{array}
\right]
\qquad
B_1 =
\left[
\begin{array}{cc}
b_{20} & b_{21} \\
b_{30} & b_{31} \\
\end{array}
\right] \qquad
\Rightarrow \qquad
B_1 =
\left[
\begin{array}{c}
B_0 \\ B_1
\end{array}
\right]
$$
and
$$
C_0=
\left[
\begin{array}{cc}
c_{00} & c_{01} & \\
c_{10} & c_{11} & \\
\end{array}
\right]\qquad
C_1=
\left[
\begin{array}{cc}
c_{02} & c_{03} \\
c_{12} & c_{13} \\
\end{array}
\right]\qquad
\Rightarrow \qquad
C=
\left[
\begin{array}{cc} C_0 & C_1 \\
\end{array}
\right]
$$
Now their product is
$$
BC = \left[
\begin{array}{cc}
B_0 C_0 & B_0 C_1 \\
B_1 C_0 & B_1 C_1
\end{array}
\right]
$$
Now we get the equations
$$
\begin{array}{cl}
B_0 C_1 = B_1 C_0 &=
\left[
\begin{array}{cc}
0& 0 \\
0 & 0 \\
\end{array}
\right] \\
B_0 C_0 = B_1C_1
&=
\left[
\begin{array}{cc}
x_1^2 & x_1 x_ 2 \\
x_1 x_2 & x_2^2 \\
\end{array}
\right] \\
\end{array}
$$
Now, some insight would be needed to continue ...
|
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}
|
If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c}{2} = \sin \frac{b}{2} \cos \frac{b}{2}
$$
and
$$
\cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$
|
Let $a+b=2z$ etc.
$\implies2b=a+b+b+c-(c+a)=2(z+x-y)\implies b=z+x-y$
$$\sin(y+z-x)+\sin(x+y-z)=2\sin(z+x-y)$$
$$\sin(x+y-z)-\sin(z+x-y)=\sin(z+x-y)-\sin(y+z-x)$$
$$\sin(y-z)\cos x=\sin(x-y)\cos z$$
Expand using $\sin(A-B)$ formula and divide both sides by $\cos x\cos y\cos z$
|
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|
How to evaluate the following integral$\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$? Can anyone help me to evaluate the definite integral $\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$?I encountered this integral while doing a problem of particle dynamics in Ganguly Saha(Applied Mathematics).Can this integral be evaluated without the substitution of $x=asin^2\theta+pacos^2\theta$.Please anyone suggest some other method i.e. some direct method to calculate this integral.
|
Introducing integral by parts,
\begin{equation}
\begin{aligned}
I&=a \int^{pa}_a\frac{x}{\sqrt{(a-x)(p a-x)}}dx \\
&=\frac{a}{2}
\left\{
a(1+p)\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)}}dx
-\int^{pa}_a\frac{d[(a-x)(p a-x)]}{\sqrt{(a-x)(p a-x)}}
\right\} \\
&=\frac{a}{2}
\left\{
a(1+p)\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)}}dx
-2\left[\sqrt{(a-x)(p a-x)}\right]^{pa}_a
\right\} \\
&=\frac{a^2(1+p)}{2}\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)}}dx
\end{aligned}
\end{equation}
Transforming the variable $x$ to $t$ using the relation,
\begin{equation}
\begin{aligned}
t=\sqrt{\frac{pa-x}{x-a}}
\end{aligned}
\end{equation}
which gives,
\begin{equation}
\begin{aligned}
x=a\frac{p+t^2}{1+t^2}
\end{aligned}
\end{equation}
the integral $I$ is expressed as,
\begin{equation}
\begin{aligned}
I&=\frac{a^2(1+p)}{2}\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)}}dx \\
&=\frac{a^2(1+p)}{2}\int^{0}_{\infty}\frac{1}{t}\sqrt{\frac{(1+t^2)^2}{a^2(1-p)^2}}\frac{2a(1-p)t}{(1+t^2)^2}dt
\end{aligned}
\end{equation}
In the cases of (1)$a\geq 0, p\leq 1$ and (2)$a\leq 0, p\geq 1$, the integral can be expressed as
\begin{equation}
\begin{aligned}
I&=-a^2(1+p)\int^{\infty}_{0}\frac{dt}{1+t^2}=-\frac{a^2(1+p)\pi}{2}
\end{aligned}
\end{equation}
In the cases of (3)$a\leq 0, p\leq 1$ and (4)$a\geq 0, p\geq 1$, the integral can be expressed as
\begin{equation}
\begin{aligned}
I&=a^2(1+p)\int^{\infty}_{0}\frac{dt}{1+t^2}=\frac{a^2(1+p)\pi}{2}
\end{aligned}
\end{equation}
|
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|
Prove that $\int _{0}^1 \sqrt {(1+x)(1+x^3)} ≤ \sqrt {15/8}$
Prove that
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)}\ dx ≤ \sqrt {15/8}$$
Now in this question, the solution that was provided me was $\int _{0}^1 \sqrt {(1+x)(1+x^3)}dx ≤ \sqrt{\int _{0}^1(1+x)dx \int _{0}^1(1+x^3)dx}$
I have never even heard or seen this in inequality before. Can anyone explain me what they are doing or if it is a very standard result in mathematics what that result or theorem is. please help
|
Not an answer, just some thoughts.
In this case we can also try using the arithmetic mean - geometric mean inequality:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq \frac{1}{2} \int _{0}^1 (1+x) dx+\frac{1}{2} \int _{0}^1 (1+x^3) dx = \\ = \frac{3}{4}+\frac{5}{8}=\frac{11}{8}=1.375$$
Meanwhile:
$$\sqrt{\frac{15}{8}}=1.3693 \ldots < \frac{11}{8}$$
So AMGM inequality is not as good as Cauchy-Schwartz.
Among other famous means, we also have the logarithmic mean, which is between arithmetic and geometric means:
$$\sqrt{ab} \leq l(a,b)=\frac{a-b}{\log (a/b)} \leq \frac{a+b}{2}$$
So curiously, we have:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq \int_0^1 \frac{(x-x^3) dx}{\log \frac{1+x}{1+x^3}} \leq \frac{11}{8}$$
This last integral is clearly too complicated to evaluate exactly here, but numerically we have:
$$\int_0^1 \frac{(x-x^3) dx}{\log \frac{1+x}{1+x^3}}=1.3703117346 \ldots$$
Which is still worse estimate than $\sqrt{15/8}$. Such is the power of Cauchy-Schwartz.
As for the lower bound, the harmonic mean is a good way:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \geq 2 \int _{0}^1 \frac{(1+x)(1+x^3)}{2+x+x^3} dx= 3-\frac{12}{\sqrt{7}} \arctan \frac{1}{\sqrt{7}}=1.36099 \ldots$$
Which is a good lower bound, since now we know:
$$1.36099 \ldots \leq \int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq 1.36930 \ldots$$
And finally, the exact value:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx= \frac{3}{4}+ \frac{9}{16} \log 3=1.367969 \ldots$$
Which is very close to $\sqrt{15/8}$.
|
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|
Show that $7a^{2} = 13b^{2}$ Here is another question which states that -
The first and the third terms of an arithmetic sequence are
a and b respectively. The sum of the first n terms is denoted by $S_n$.
(a) Find $S_4$ in terms of a and b.
(b) Given that $S_4$, $S_5$,
$S_7$ are consecutive terms of a geometric sequence, show that
$$7a^{2} = 13b^{2}$$
My approach to part (a) -
$\Rightarrow\ U_1 = a$
$\Rightarrow\ U_3 = b$
$\Rightarrow\ a + 2d = b$
$\therefore\ d = \frac{b-a}{2}$
$\Rightarrow\ U_4 = b + \frac{b - a}{2}$
$\therefore\ U_4 = \frac{3b - a}{2}$
$\Rightarrow\ S_n = \frac{n}{2}(U_1 + U_4)$
$\Rightarrow\ S_4 = \frac{4}{2}(a + \frac{3b - a}{2})$
$\therefore\ S_4 = a+3b$ (Answer)
I do not know how to approach part (b) without the need for tedious calculations.
|
It is not hard to see that $S(n)=\frac{1}{4}\left(-a n^2+5 a n+b n^2-b n\right)$. So$$\frac{S(7)}{S(5)}=\frac{S(5)}{S(4)}\iff\frac{13 b^2-7 a^2}{10 b (a+3 b)}=0\iff13b^2=7a^2.$$
|
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|
totally differentiable function $\frac{x^3}{(x^2+y^2)}$ - check my proof Consider the function: $f: \mathbb{R^2} \to \mathbb{R}$ given by $f(x,y)=\begin{cases} \frac{x^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases}$
I want to check if it is totally differentiable at ${0 \choose 0}$. By definition this means that there must exists a linear transformation $A$ such that:
$lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert} = 0 $, where $\Vert \cdot \Vert$ is an arbitrary norm.
My proof:
I assume that $f$ is totally differentiable so that $A$ must consist of the partial derivatives. The partial derivative at point ${0 \choose 0}$ with respect to $x$ is $D_1f{0 \choose 0}=1$ and with respect to $y$: $D_2f{0 \choose 0}=0$. Let $A:= \left(1~~0\right)$. Plugging this into $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert}$ yields: $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-\left(1~~0\right){x \choose y}}{\Vert {x \choose y}\Vert}=lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}$. At this point, I am not sure how to continue. A first approach might be:
1.) As in $\mathbb{R^2}$ norms are equivalent it holds that $C \cdot \Vert {x \choose y} \Vert_{\infty} \geq \Vert {x \choose y} \Vert $ where $C >0$ and $\Vert {x \choose y} \Vert_{\infty}$ is the maximum norm. If I now assume that $x\leq y<0$, I can conclude: $lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}\geq lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{C(x^2+y^2) \vert y\vert} = lim_{{x \choose y} \to {0 \choose 0}} \frac{-x\vert y \vert}{C(x^2+y^2)}$. If $f$ were totally differentiable then we know that for an arbitrary null sequence $lim_{{x \choose y} \to {0 \choose 0}} \frac{-x\vert y \vert}{C(x^2+y^2)}$ must be $0$. However, if I plug in the null sequence ${\frac{-1}{n} \choose \frac{-1}{n}}$ the limit is $\frac{1}{2C}> 0$. Hence, $f$ is not totally differentiable at ${0 \choose 0}$.
A second approach could be:
2.) If I plug the null sequence ${\frac{-1}{n} \choose \frac{1}{n}}$ directly into $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert} $ then it should equal $0$ if $f$ was totally differentiable. However, rearrangeing yields: $lim_{{x \choose y} \to {0 \choose 0}} \frac{-(\frac{-1}{n})(\frac{1}{n})^2}{((\frac{-1}{n})^2+(\frac{1}{n})^2)\Vert {\frac{-1}{n} \choose \frac{1}{n}}\Vert}= lim_{{x \choose y} \to {0 \choose 0}} \frac{\frac{1}{n^3}}{\frac{2}{n^2}\cdot \frac{1}{n}\Vert {-1 \choose 1}\Vert}=lim_{{x \choose y} \to {0 \choose 0}} \frac{1}{2\cdot\Vert {-1 \choose 1}\Vert} > 0$. Hence, $f$ is not totally differentiable at ${0 \choose 0}$.
May be those approaches are not that straightforward but as I am currently learning multivariable calculus I am interested in wether the proves are correct or not. Any comments or suggestions are appreciated.
best regards Philipp
|
What you did is correct.
Maybe a simpler way would be to notice that for $x=y=t>0$
$$\frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}=-\frac{1}{2\sqrt 2}$$
and that consequently, above function can’t have $0$ for limit at $(0,0)$.
|
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|
solve the Lagrange multiplier equations the equations :
$$\left\{\begin{array}{l}{x+6 y+4 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
I've done many transformations,but I still can't get the answer.
the last I did this:
$$\lambda=\left(6 \frac{x}{y}+2\right)=\left(\frac{1}{4}+\frac{3}{2} \frac{y}{x}\right),y=tx \\=\left(6 \cdot \frac{1}{t}+2\right)=\left(\frac{1}{4}+\frac{3}{2} t\right)\\t^{2}-\frac{7}{6} t-4=0$$
the quadratic formula's roots have square number , but the answer doesn't have,so I think I might be wrong.
I don't know how to solve it
|
Restoring the objective function from the constraint (third) equation and partial derivatives:
$$Optimize \ \ z(x,y)=\frac12x^2+6xy+y^2 \ \ s.t. \ \ 4x^2+y^2-25=0.$$
If you take partial derivatives now:
$$\left\{\begin{array}{l}{x+6 y+\color{red}8 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
Note that if $x=0$, then from the first equation $y=0$, but it contradicts the last equation. So, $x\ne 0$. Similarly, if $y=0$, then from the second equation $x=0$, but it contradicts the last equation. So, $y\ne 0$ too. Thus:
$$\begin{cases}y=\frac{-(1+8\lambda)x}{6}\\ 6x+(2+\lambda)\frac{-(1+8\lambda)x}{6}=0 \Rightarrow x(36-(2+\lambda)(1+8\lambda))=0 \Rightarrow 36-(2+\lambda)(1+8\lambda)=0\end{cases} \Rightarrow \\
8\lambda^2+17\lambda-34=0 \Rightarrow \lambda_{1,2}=\frac{-17\pm 9\sqrt{17}}{16}.
$$
Can you do the rest?
|
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|
Solution verification on homework problem. Separable first order ODE IVP. The answer is supposedly $y^2 = 1 + \sqrt{x^2 - 16}$ I don't know where I went wrong cause I know for a fact that my substitution of $x = 4 \sec(\theta)$ is correct. I know for a fact that after substitution the integral becomes $\int \sec^2(\theta)= \tan(\theta)+C$. I am pretty sure that my substitution of $\tan(\theta)$ is correct. Am I missing something?
$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$
$\int 2y \space dy = \int \frac{4\sec(\theta)}{\sqrt{(4\sec(\theta))^2-16}}\space d(\theta)$
$y^2 = \int \frac{4\sec(\theta)}{\sqrt{(\sec^2(\theta)-1)16}}\sec(\theta) \ tan(\theta)$
$y^2 = \frac{4\sec(\theta)}{4\tan(\theta)}\sec(\theta)\tan(\theta)$
$y^2 = \int sec^2(\theta)$
$y^2 = \tan(\theta) + c$
Using the reference triangle: Tangent is equal to $\frac{\sqrt{x^2-16}}{4}$
$y^2= \frac{\sqrt{x^2-16}}{4} + C$
$4 = \frac{3}{4}+ C$
$C = \frac{15}{4}$
$y^2= \frac{\sqrt{x^2-16}}{4} + \frac{15}{4}$
|
You have to use another $4$ .
$x = 4\sec\theta \implies dx = \color{red}4\sec\theta\tan\theta
d\theta$
So, you'll have $y^2 = \sqrt{x^2-16} +c$
$2^2 = \sqrt{9} +c \implies c = 4-3= 1$
Thus,
$$y^2 =\sqrt{x^2+16} +1$$
|
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|
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
|
We can rewrite the equation as follows:
$$4x^2 - 4x + y^2 + 2y - 7 = 0$$
By completing the squares, we obtain:
$$4\left (x^2 - x + \color{red}{\frac 1 4} \right ) + (y^2 + 2y + \color{blue}1) - 7 - \color{red}1 - \color{blue}1 = 0$$
that is,
$$\frac {\left (x - \frac 1 2 \right )^2} {\left ( \frac 3 2 \right )^2} + \frac {(y + 1)^2} {3^2} = 1$$
which represents the ellipse with center $(\frac 1 2, -1)$ and semiaxes $\frac 3 2, 3$.
For a given $k \in \mathbb R$, the equation $5x + 6y = k$ represents a straight line parallel to the line of equation $5x + 6y = 0$. We want to find the greatest $k$ such that the line $5x + 6y = k$ intersects the ellipse at a point $(x, y)$.
As $k$ increases, the line will intersect the ellipse in either $0$, $1$ or $2$ points (see the picture below). It should be clear that the greatest $k$ is such that there is only one point of intersection. Therefore, we want to find $k$ such that the system
$$\begin{cases} 5x + 6y = k \\ 4x^2 - 4x + y^2 + 2y - 7 = 0 \end{cases}$$
has exactly one solution.
By substituting $y = -\frac 5 6 x + \frac 1 6 k$ in the second equation, we get
$$169 x^2 - (10k + 204) x + (k^2 + 12 k - 252) = 0$$
which has exactly one solution if and only if its discriminant with respect to $x$ is $0$:
$$\Delta = (10 k + 204)^2 - 4 \cdot 169 \cdot (k^2 + 12 k - 252) = 0$$
This last equation has solutions $k = -23$ and $k = 16$.
Therefore, the maximum value of $5x + 6y$ is $16$.
The solution is represented in the following picture:
|
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|
Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$?
An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?
I couldn't develop anything more than
$$N_b = \left(\;n\cdot (b-1)\;\right)_b = 6789_b$$
(with subscripts indicating "in base $b$").
|
It seems there are multiple answers.
Since $N$ can be written as $6789$ in base $b$ we know $$6b^3+7b^2+8b+9=N$$
If we divide this polynomial by $b-1$ we get $$6b^2+13b+21+\frac{30}{b-1}$$ but we know that $b-1$ divides $N$ so $\frac{30}{b-1}$ is an integer. $$b-1=1,2,3,5,6,10,15,30$$ $$b=2,3,4,5,7,11,16,31$$ but $b < 16$ and $b>9$ as noted by John Omielan hence $$b+1=12$$
|
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|
Solve for $\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$ It is known that $$\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x=\pi \ln \left|2 \cos \frac{a}{2}\right|+a \ln \left|\tan \frac{a}{2}\right|+2 \sum_{k=0}^{+\infty} \frac{\sin (2 k+1) a}{(2 k+1)^{2}}$$
In an attempt to derive the equation, I utilized
$$\sum_{n=0}^{\infty}\left(\frac{e^{i x}}{a}\right)^{n}=\frac{1}{1-\frac{e^{i x}}{a}}=\frac{a(a-\cos x)}{a^{2}-2 a \cos x+1}+i \frac{a \sin x}{a^{2}-2 a \cos x+1}$$, where a>1
and $$\sum_{n=0}^{\infty}\left(ae^{i x}\right)^{n}=\frac{1}{1-ae^{i x}}=\frac{a(a-\cos x)}{a^{2}-2 a \cos x+1}+i \frac{a \sin x}{a^{2}-2 a \cos x+1}$$, where a<1
The series expansion for $\ln \left(x^{2}+2 \sin a \cdot x+1\right)$ can be obtained by integrating the imaginary part of the equation and plugging in $x=\frac{\pi}{2}-a$, $a=x$
However, I still failed to get the desired result by using this series expansion. I wonder whether I can obtain the equation through such Fourier series or I just missed some important point. Any help would be appreciated.
|
Another series solution, which might be helpful.
$$I(a)=\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$$
$$x=\tan t$$
$$I(a)=\int_{0}^{\pi/2} \left(\ln \left(1+2 \sin a \sin t \cos t\right)-2\ln \cos t \right) d t$$
$$I(a)=\int_{0}^{\pi/2} \ln \left(1+2 \sin a \sin t \cos t\right)dt+ \pi \ln 2$$
Now substitute $t=y/2$:
$$I(a)=\pi \ln 2+\frac12 \int_{0}^{\pi} \ln \left(1+\sin a \sin y \right)dy $$
Expanding into a series:
$$I(a)=\pi \ln 2+\frac12 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin^n a \int_{0}^{\pi} \sin^n y dy $$
Using symmetry:
$$I(a)=\pi \ln 2+\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin^n a \int_{0}^{\pi/2} \sin^n y dy $$
The integrals can be evaluated easily enough:
$$\int_{0}^{\pi/2} \sin^n y dy=\int_0^1 \frac{u^n du}{\sqrt{1-u^2}}=\frac{1}{2} \int_0^1 \frac{v^{(n-1)/2} dv}{\sqrt{1-v}}= \frac{1}{2} B \left(\frac{n+1}{2}, \frac12 \right)$$
Which gives us:
$$I(a)=\pi \ln 2+\frac12 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} B \left(\frac{n+1}{2}, \frac12 \right) \sin^n a $$
This is not the same series the OP wanted to obtain, however it's nice and converges fast.
The OP series is Fourier series, which I think is harder to get.
|
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|
computing the Taylor series for $e^{1-\cos x}$ I was told to compute the Taylor series for $e^{1-\cos x}$ to the fourth degree.
I plugged the Taylor series for $1-\cos x$ into the Taylor series for $e$.
This got me: $$1-\frac{x^2}2+\frac{x^4}{24}+\frac{(\frac{-x^2}2)^2}{2}$$A bit of simplification results in: $1+\frac{x^2}2+\frac{x^4}6$.
Is this correct? The answer key to the question gave a different answer - it put the denominator of the last term as 12 instead of 6. However, I do not see the flaw in my calculations. Did I make the error or did they?
|
My answer agrees with the answer key:
$\cos x=1-\dfrac{x^2}2+\dfrac{x^4}{24}+O(x^6),$ so $1-\cos x=\dfrac{x^2}2-\dfrac{x^4}{24}+O(x^6),$
so $\exp(1-\cos x)=1 + \left(\dfrac{x^2}2 - \dfrac{x^4}{24} \right)+ \dfrac12\left(\dfrac{x^2}2\right)^2+O(x^6)=1 + \dfrac{x^2}2 + \dfrac{x^4}{12}+O(x^6).$
|
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Solving $x\sqrt{2}=1-x$ Solve for $x$ of the equation,
$$x\sqrt{2}=1-x$$
According to the solution, $x$ is supposed to be $\sqrt{2}-1$. The solutions to this test have proven wrong from case to case, so I wonder ...
If you solve it like this: $x\sqrt{2}+x = 1$ and pull out $x$ to get $x(\sqrt{2}+1) = 1$. Then, dividing by $\sqrt{2}+1$, we get
$$x=\frac1{\sqrt{2}+1}$$
right?
|
$\frac 1{\sqrt 2 + 1}=\frac 1{\sqrt 2+1}\frac {\sqrt 2 -1}{\sqrt 2 -1} = \frac {\sqrt 2 -1}{\sqrt 2^2 + \sqrt 2 - \sqrt 2 - 1} = \frac {\sqrt 2 -1}{2-1} =\frac {\sqrt 2-1}1 = \sqrt 2 -1$.
|
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|
What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$
What is the 94th term of the following sequence?
$$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
*
*8
*9
*10
*11
My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is written 2n: n is natural number. Here 94 = 2(47) so sum of first few natural numbers should be greater than or equal to 47. Since $$1+2+3+4+5+6+7+8+9 = 45 < 47$$ so options 1,2 are not possible and $$1+2+3+4+5+6+7+8+9+10 = 55 >47$$ But this is a lengthy process.
Please tell me easiest way to approach the answer.
|
Using your method,
$n(n+1) \lt 94$
Easy to see $9*10 = 90$ so the $90$th term value is $9$.
And $91$st term is the beginning of value $10$.
$a(a+1)$th term has the value $a$ and its the end of that value.
|
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|
Solve $\sin x + \cos x = \sqrt{1+k}$ for $\sin 2x$, $\sin x-\cos x$, and $\tan x$ in terms of $k$ Given that $\sin x + \cos x = \sqrt{1+k}$, $-1 \le k \le 1$
*
*Find the value of $\sin 2x$ in terms of k
*Given that $x \in (45^{\circ}, 90^{\circ})$ deduce that $\sin x - \cos x = \sqrt{1-k}$
*Hence, show that $\tan x = \dfrac{1 + \sqrt{1-k^2}}{k}$
Okay, I have figured out that
$$ \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1+k \implies \sin 2x = k $$
Now, I am not sure what approach I should use to deduce the second one.
And can somebody give me a hint on how to start with the third one?
|
Here’s two more very straightforward ways to go about it:
$${\cos x+\sin x\over\sqrt 2}=\sqrt{1+k\over 2}$$
$$\implies \sin\left(x+\frac\pi4\right) = \sqrt{1+k\over 2}$$ (note that this expression is valid for all $k$ in the range). Hence, $$x=\arcsin\left(\sqrt{1+k\over 2}\right)-\frac \pi4$$
For the first part, we have $$\sin2x=\sin\left(2\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi2\right)$$
$$=-\cos\left(2\arcsin\left(\sqrt{1+k\over 2}\right)\right)$$
$$={1+k\over 2}+{1+k\over 2}-1$$
$$=k$$
For the second, we have
$$\sin x-\cos x = \sin\left (\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)-\cos\left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$$
$$=\left(\sin\left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\sin \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)\right)\cdot\frac1{\sqrt 2} $$
$$=-\sqrt 2\cdot\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)$$
$$=-\sqrt2\cdot\mp\sqrt{1-k\over 2}$$
And since $x\in\left(\frac\pi4,\frac\pi2\right)$, $\sqrt{1+k\over 2}>0$ because $k\in(-1,1)$, and because $x = \left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$, $\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)<0$ and so we have $$\sin x-\cos x = \sqrt{1-k}$$
And finally, $$\tan x= \tan \left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$$ With this and $\tan\left(\alpha-\frac\pi4\right)={1-\tan\alpha\over1+\tan\alpha}$ we get our result that $$\tan x = {1+\sqrt{1-k^2}\over k}$$
Note that we have used $\cos(\arcsin\alpha) = \sqrt{1-\alpha^2}$ and $\tan(\arcsin\alpha) ={\alpha\over\sqrt{1-\alpha^2}}$ in our simplification. Also note that any part may be solved by steps from the other two similarly to the answer due to Quanto.
Yet another method picking off from the first line of the first method:
Writing $k$ in terms of $x$ we have:
$$k=2\sin^2\left(x+\frac\pi4\right)-1$$
$$\implies k=-\cos\left(2x+\frac\pi2\right)$$
$$\implies \boxed{k = \sin2x}$$
Next, we have
$$\sin x -\cos x = \sqrt{(\sin x +\cos x)^2-4\sin x\cos x}$$
$$=\sqrt{1+k-2k}$$
$$\implies \boxed{\sin x - \cos x = \sqrt{1-k}}$$
Also, I did not see the "Hence" in the last part of the question. So this last step is common to both parts of the solution
$$\tan x = \sin x + \cos x$$
By Componendo et Dividendo
$${\tan x - 1\over \tan x + 1} = {\sin x - \cos x \over \sin x + \cos x} = \sqrt{1-k\over 1+k}$$
$$\implies \sqrt{1+k}\tan x -\sqrt{1+k} = \sqrt{1-k}\tan x +\sqrt{1-k}$$
$$\implies \tan x(\sqrt{1+k}-\sqrt{1-k})=\sqrt{1+k}+\sqrt{1-k}$$
Multiplying both sides by $$\sqrt{1+k}+\sqrt{1-k}\over(\sqrt{1+k}+\sqrt{1-k})\cdot(\sqrt{1+k}-\sqrt{1-k})$$
we have
$$\tan x ={ (\sqrt{1+k}+\sqrt{1-k})^2\over \sqrt{1+k}^2-\sqrt{1-k}^2)}$$
$$ = {1+k+1-k-2\sqrt{1-k^2}\over 1+k-1+k}$$
$$\implies\boxed{\tan x = {{1+\sqrt{1-k^2}\over k}}}$$
The first method may look unwieldy due to the repeated appearance of $\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4$ but it is quite simple. Read it as some angle $\phi$ instead and it’ll make it better.
|
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|
Find the asymptotic form as $N \rightarrow \infty$ of $\sum_{a = 1}^{N} \sum_{u = 1}^{a - 2} \sum_{v = u + 1}^{a - 1} {\delta}_{N = u\, a + v}$ Here $N \ge 1$ is a positive integer and $a$, $u$, and $v$ are also integers. This triple sum arises from counting the number of reducible cubic polynomials. I am looking for a closed form solution if possible, or at least a reduction in one or two sums, but my main question is the asymptotic form as $N \rightarrow \infty$. Now through different computations (taking the better part of a day) I have a table of values of this triple sum, its difference from $N$ and the asymptotic correction that I have identified through these numerical tests. $$\left(\begin{array}{c c c c}
N &
\text{Triple Sum} &
\text{N-TSum} &
2 \sqrt{N} \\
1 & 0 & 1 & 2 \\
10 & 3 & 7 & 6 \\
10^2 & 77 & 23 & 20 \\
10^3 & 930 & 70 & 63 \\
10^4 & 9{,}789 & 211 & 200 \\
10^5 & 99{,}351 & 649 & 632 \\
10^6 & 997{,}977 & 2{,}023 & 2{,}000 \\
10^7 & 9{,}993{,}645 & 6{,}355 & 6{,}324 \\
10^8 & 99{,}979{,}961 & 20{,}039 & 20{,}000
\end{array}\right)$$
From this I see that the expected asymptotic form is now $$\begin{equation*}
\sum_{a = 1}^{N} \sum_{u = 1}^{a - 2} \sum_{v = u + 1}^{a - 1} {\delta}_{N = u\, a + v} \sim
N - 2 \sqrt{N}
+ \mathcal{O} \left({1}\right).
\end{equation*}$$
Where ${\delta}_{N = u\, a + v}$ is the KroneckerDelta function. I could also use the Inversion Brackets as $\left[N=a\,u+v\right]$. So how do I prove this?
Picking up after the comments, then to complete the asymptotic expansion as $N \rightarrow \infty$ for the number of divisors, Kevin A. Broughan,
"Restricted divisors sums" Acta Arithmetica 101(2), pp105-114, 2002 defines the restricted number of divisors
${d}_{\alpha} \left({n}\right) = \# \left\{{d : d \mid n \text{ and } 1 \le d \le \alpha}\right\}$ for real
$\alpha \ge 1$. Broughan furhter defines the sum of the restricted number of divisors as
$$D \left({x, \alpha}\right) =\sum_{1 \le n \le x} {d}_{\alpha} \left({n}\right)$$
with $1 \le \alpha \le x$.
From Broughan's Theorem 4.1 the asymptotic expansion as $x \rightarrow \infty$ of the sum of the restricted number of divisors is
$$D \left({x, \alpha}\right) \sim x\, \log \left({\alpha}\right) + \gamma\, x + O \left({\frac{x}{\alpha}}\right) + O \left({\alpha}\right)$$
Then as $N \rightarrow \infty$ we can now write the average number of divisors as
$$\sum_{u=2}^{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor} [n \mod u = 0] = {d}_{\left\lfloor{\left({\sqrt{4\, N + 5} - 3}\right)/2}\right\rfloor} \left({N}\right) - 1 = \frac{1}{N}\, D \left({N, \left\lfloor{\frac{1}{2} \left({ \sqrt{4\, N + 5} - 3}\right)}\right\rfloor}\right) \sim \frac{1}{2}\, \log \left({N}\right) + \gamma - 1 + O \left({\frac{1}{\sqrt{N}}}\right)$$
Now we can write
$$ S \sim N - 2 \sqrt{N} - \frac{1}{2}\, \log \left({N}\right) + O \left({1}\right)$$
|
We can simplify the triple sum by rephrasing it to $$S = \sum_{a = 1}^{n} \sum_{u = 1}^{a - 2} [u+1 \le n-ua \le a-1]$$
Now rearranging this, we get $$\sum_{a = 3}^{n} \sum_{u = 1}^{a - 2} \left[\frac{n-a+1}{a} \le u \le \frac{n-1}{1+a} \right]$$
Now solve for $u$ in terms of $a$. For each value of $u$ there will be $$\left \lfloor \frac{n-1-u}{u} \right\rfloor - \left\lceil \frac{n+1}{u+1} \right\rceil + 1$$ cases of $a$. After finding the bounds, this can be simplified to $$\sum_{u=1}^{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor} \left( \left \lfloor \frac{n-1-u}{u} \right\rfloor - \left\lceil \frac{n+1}{u+1} \right\rceil + 1 \right)$$
Of course, this can be simplified to $$\sum_{u=1}^{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor} \left( \left \lfloor \frac{n-1}{u} \right\rfloor - \left\lceil \frac{n+1}{u+1} \right\rceil \right)$$
This turns into something that is sort of a telescoping series, $$S = n-1- \left\lceil \frac{n+1}{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor+1} \right\rceil+\sum_{u=2}^{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor} \left( \left \lfloor \frac{n-1}{u} \right\rfloor - \left\lceil \frac{n+1}{u} \right\rceil \right)$$
Using the fact that $\left \lfloor \frac{n-1}{u} \right\rfloor - \left\lceil \frac{n+1}{u} \right\rceil = -2$ if $n \pmod u = 0$ and $-1$ otherwise, the sum can be simplified further to $$S = n- \left\lceil \frac{n+1}{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor+1} \right\rceil - \left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor -\sum_{u=2}^{\left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor} [n \mod u = 0]$$
That last part is basically the number of divisors $u$ of $n$, with $2 \le u \le \left \lfloor \frac{-3+\sqrt{4n+5}}{2} \right \rfloor$. After getting rid of the floors and ceilings, we get that $$S \approx n - \sqrt{n} - \sqrt{n} + O(1) = n - 2\sqrt{n} - O(1)$$
|
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|
Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$.
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inverse.
I have tried picking two arbitrary elements, $m,n$ in our set. Let $m= x+y\alpha+z\alpha^2$ and let $n=a+b\alpha+c\alpha^2$. Then setting our product $m\cdot n$ equal to 1 gets us:
$$(ax+2zb+2yc) + (bx+ay+2zc)\alpha + (cx + yb +za)\alpha^2 = 1.$$
I was wondering if this could lead me to my proof?
|
You have to solve
$$
\begin{cases}
ax+2bz+2yc=1\\
ay+bx+2cz=0\\
az+by+cx=0
\end{cases}
$$
in the variables $a,b,c$. The notation here is somewhat of counterintuitive with respect to the usual one.
The matrix associate to the non-homogeneous linear system is
$$
A(x,y,z)=\begin{pmatrix}
x & 2z & 2y\\
y & x & 2z\\
z & y & x
\end{pmatrix}
$$
and the determinant is $x^3+2y^3+4z^3-6xyz$. Assuming $A(x,y,z)$ invertible, you can determine $(a,b,c)$ as $\displaystyle A(x,y,z)^{-1}\cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}$.
It remains to show that $\det A(x,y,z)=x^3+2y^3+4z^3-6xyz=0$ iff $(x,y,z)=(0,0,0)$.
Set $X=x$, $Y=y\sqrt[3]{2}$ and $Z=z\sqrt[3]{4}$.
Hence the equality above becomes: $x^3+2y^3+4z^3-6xyz=X^3+Y^3+Z^3-3XYZ=0$
It is known that
$$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)$$
hence we get:
$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$.
There are two possible cases:
*
*$X+Y+Z=x+y\sqrt[3]{2}+z\sqrt[3]{4}=0$. Hence $x=y=z=0$ because $x,y,z\in \Bbb Q$. You can easily check that $1,\sqrt[3]{2},\sqrt[3]{4}$ are linearly independent over $\Bbb Q$.
*$\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$. Since this is a sum of three squares, the total amount is zero iff each summand in zero, that is $(X-Y)=(Y-Z)=(Z-X)=0$ iff $X=Y=Z$ iff $x=y\sqrt[3]{2}=z\sqrt[3]{4}=0$ iff $x=y=z=0$ because $\sqrt[3]{2},\sqrt[3]{4}\notin\Bbb Q$.
Therefore: $\det A(x,y,z)=0$ iff $x=y=z=0$, iff $m=x+y\alpha+z\alpha^2=0$. Hence any element different to zero has a multiplicative inverse.
|
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|
How to find eigenvalues of the matrix This is a question from our end-semester exam:
How to find the eigenvalues of the given matrix:
M=\begin{bmatrix}
5,1,1,1,1,1\\
1,5,1,1,1,1\\
1,1,5,1,1,1\\
1,1,1,5,1,1\\
1,1,1,1,4,0\\
1,1,1,1,0,4\\
\end{bmatrix}
I know that $4$ is an eigenvalue of $M$ with multiplicity atleast $3$ since $M-4I$ has $4$ identical rows.
Is there any way to find all eigenvalues of this matrix? I could find only $3$ out of $6$.
|
For block matrices, you can use Schur complement:
$$0=\det(M-I\lambda)=\begin{vmatrix}A&B\\ C&D\end{vmatrix}=\det(D)\cdot \det(A-B\cdot D^{-1}\cdot C)= \\
\begin{vmatrix}4-\lambda&0\\ 0&4-\lambda\end{vmatrix}\cdot \det\left(A-B\cdot \begin{pmatrix}\frac1{4-\lambda}&0\\ 0&\frac1{4-\lambda}\end{pmatrix}\cdot C\right)=\\
\small(4-\lambda)^2\det\left(\begin{pmatrix}5-\lambda&1&1&1\\ 1&5-\lambda&1&1\\ 1&1&5-\lambda&1\\ 1&1&1&5-\lambda\end{pmatrix}-\frac2{4-\lambda}\begin{pmatrix}1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\end{pmatrix}\right)=\\
\small\frac1{(4-\lambda)^2}\cdot \det\begin{pmatrix}\lambda^2-9\lambda+18&2-\lambda&2-\lambda&2-\lambda\\ 2-\lambda&\lambda^2-9\lambda+18&2-\lambda&2-\lambda\\ 2-\lambda&2-\lambda&\lambda^2-9\lambda+18&2-\lambda\\ 2-\lambda&2-\lambda&2-\lambda&\lambda^2-9\lambda+18\end{pmatrix}=\\
\small \frac{\lambda^2-12\lambda+24}{(4-\lambda)^2}\cdot \det\small{\begin{pmatrix}1&1&1&1\\ 2-\lambda&\lambda^2-9\lambda+18&2-\lambda&2-\lambda\\ 2-\lambda&2-\lambda&\lambda^2-9\lambda+18&2-\lambda\\ 2-\lambda&2-\lambda&2-\lambda&\lambda^2-9\lambda+18\lambda\end{pmatrix}}=\\
\small\frac{\lambda^2-12\lambda+24}{(4-\lambda)^2}\cdot\det\small{\begin{pmatrix}\lambda^2-8\lambda+16&0&0\\ 0&\lambda^2-8\lambda+16&0\\ 0&0&\lambda^2-8\lambda+16\end{pmatrix}}=\\
\frac{\lambda^2-12\lambda+24}{(4-\lambda)^2}\cdot (4-\lambda)^6=0 \Rightarrow \\
(4-\lambda)^4\cdot (\lambda^2-12\lambda+24)=0\Rightarrow \\
\lambda_{1,2,3,4}=4, \lambda_{5,6}=2(3\pm \sqrt{3}).$$
|
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
|
There are a few ways to do this problem : one of which is a more general method which turns out to be very elaborate in this case. Let $p(x) = x^3 - x$.
The first is to note that $x^3 \equiv x \mod p(x)$, therefore $x^{2n+1} \equiv x \mod p(x)$ for all $n \geq 0$(show this by induction). Thus, the given expression simplifies to $$
q(x) = x^{81} + x^{49} + x^{25} + x^9+x \equiv x+x+x+x+x \equiv 5x \mod p(x)
$$
which is the answer.
For the more general method, we have the Chinese remainder theorem for polynomials. We find the remainder when $q(x)$ is divided by $x, x+1$ and $x-1$, which are the relatively prime factors of $x^3 - x$. The CRT guarantees a unique polynomial modulo $x^3-x$ that leaves those remainders modulo $x,x+1$ and $x-1$ respectively, which then will be the remainder.
For example, it is clear that $q(x) \equiv 0 \pmod {x}$, and that $q(1) = 5$ and $q(-1) = -5$.
Thus, $q(x) \equiv -5 \pmod{x+1}$ and $q(x) \equiv 5 \pmod{x-1}$.
While the CRT provides us with an algorithm for computing the remainder in this case, we may also proceed by brute force : the remainder must be $ax^2+bx+c$, a degree two polynomial since $x^3-x$ is of degree $3$.
By substituting $x = 0$, $x= 1$ and $x=-1$ we get $c = 0 ; a+b = 5$ and $a-b = -5$. Thus, from here we get $a=c=0$ and $b=5$, thus the remainder is $5x$.
|
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|
If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is? If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is: A) $1$; B) $2$; C) $3$; D) more than $3$.
I started to try different values of $k$ in the first root ($13, 18, 25, 34$) and see if $\sqrt{k+36}$ is an integer. So far I found $k=13$ is a solution. Are there any other solutions? Also, I don't think I am doing it in the fastest way.
|
Let's first solve $a^2-b^2=45$ with $a:=\sqrt{k+36},\,b:=\sqrt{k-9}$ both integers $\ge0$. Since $(a-b)(a+b)=3^2\times 5$ must be a factorisation into two odd positive factors with $a-b\le a+b$, either $$a-b=1,\,a+b=45\implies a=23,\,b=22\implies k=b^2+9=493,$$ or $$a-b=3,\,a+b=15\implies b=6\implies k=45,$$ or $$a-b=5,\,a+b=9\implies b=2\implies k=13.$$So the answer is C).
|
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|
Find the image and inverse image of the following function $2x - x^2$ Where $f(E)$ denotes the image and $f^{-1}(E)$ denotes the inverse image.
$$E=[-2,2)$$
plugging in the values I get:
$f(-2)=-8$
$f(-1)= -3$
$f(0) = 0$
$f(1)= 1$
$f(2) = 0$
So all the numbers fall between the interval of $[-8,1)$ I think it has to be expressed differently than $[-8,1)$ because 0 appears twice. How do I find the inverse image? I cannot algebraically solve for the inverse function.
Edit: Looking at the graph on desmos APPEARS to verify my interval for $y$. Can I
|
We can complete the square to determine the image.
\begin{align*}
f(x) & = 2x - x^2\\
& = -x^2 + 2x\\
& = -(x^2 - 2x)\\
& = -(x^2 - 2x + 1) + 1\\
& = -(x - 1)^2 + 1
\end{align*}
Thus, the graph of $f$ is a parabola with vertex $(1, 1)$ that opens downwards. Therefore, its maximum value on the interval $E = [-2, 2)$ is $1$. The minimum value must occur at the endpoint $x = -2$ since $-2$ is farther from the axis of symmetry $x = 1$ of the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = 2x - x^2$ than $x = 2$ is. Since $f(-2) = -8$, the range of the continuous function $f: [-2, 2) \to \mathbb{R}$ defined by $f(x) = 2x - x^2 = -(x - 1)^2 + 1$ is $[-8, 1]$, as the graph below confirms.
The formal justification for the assertion that $f$ assumes every value between $-8$ and $1$ is the Intermediate Value Theorem, which states that if a function is continuous on an interval $[a, b]$, then it takes on every value between $f(a)$ and $f(b)$. In this case, we can apply the Intermediate Value Theorem to the interval $[-2, 1] \subseteq [-2, 2)$ to conclude that $f$ assumes every value between $f(-2) = -8$ and $f(1) = 1$.
As for $f^{-1}(E)$, we require that $-2 \leq f(x) < 2$.
\begin{align*}
-2 & \leq f(x) < 2\\
-2 & \leq -(x - 1)^2 + 1 < 2\\
2 & \geq (x - 1)^2 - 1 > -2\\
3 & \geq (x - 1)^2 > -3
\end{align*}
As Kavi Rama Murthy explained, the inequality $(x - 1)^2 > -3$ is automatically satisfied since $(x - 1)^2 \geq 0$ for any real number $x$. Hence,
\begin{align*}
(x - 1)^2 & \leq 3\\
|x - 1| & \leq \sqrt{3} && \text{since $\sqrt{u^2} = |u|$}
\end{align*}
Since $|x - 1| \leq \sqrt{3}$ means $x - 1$ is at most $\sqrt{3}$ units from $0$,
\begin{align*}
-\sqrt{3} \leq x - 1 \leq \sqrt{3}\\
1 - \sqrt{3} \leq x \leq 1 + \sqrt{3}
\end{align*}
Thus, $f^{-1}(E) = [-1 - \sqrt{3}, 1 + \sqrt{3}]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$ Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$
$$f(x) = \frac{\sin x}{x}, x \not= 0 $$
$$f(x) = 1, x = 0 $$
I calculated its derivative.
$$f'(x) = \frac{x\cos x-\sin x}{x^2}, x \not= 0$$
$$f'(x) = 0, x = 0$$
I can see that $f'(x)$ has a solution for x = 0.
I am not sure how to solve for $x\not= 0$ though. I have tried this:
$$ \frac{x\cos x - \sin x}{x^2} = 0 $$
Multiply both sides by $x^2$
$$ x\cos x - \sin x = 0 $$
Add $\sin x$ to both sides
$$ x\cos x = \sin x$$
Now divide both sides by $\cos x$
$$ x = \tan x $$
Did I do anything wrong? Can this equation be figured out without using any kind of graphing tool?
|
What you did is very correct and there is no explicit solution to the equation.
Discarding the trivial case $x=0$, it is better to consider that you look for the zero's of
$$f(x)=x \cos (x)-\sin (x)$$ Now, consider a Taylor expansion close to $\frac {3 \pi} 2$ (remember the symmetry) to get
$$f(x)=1+\frac{3}{2} \pi \left(x-\frac{3 \pi }{2}\right)+\frac{1}{2} \left(x-\frac{3 \pi
}{2}\right)^2-\frac{1}{4} \pi \left(x-\frac{3 \pi }{2}\right)^3+O\left(\left(x-\frac{3 \pi }{2}\right)^4\right)$$ and use series reversion to get
$$x=\frac{3 \pi }{2}-\frac{2}{3 \pi }-\frac{16}{81 \pi ^3}-\frac{16}{243 \pi ^5}- \cdots$$ This truncated expansion would give $x=4.4936$ while the "exact" solution would be $4.4934$.
Just compute the value of the function at this point.
|
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|
Minimal polynomial- fields Let $\zeta$ = $\cos(\frac{2\pi}{7} ) + i\sin(\frac{2\pi}{7} )$ , let $\alpha = \zeta +\zeta^{-1} $ note that $\zeta^{-1} =\zeta^6 $
I try to find the minimal polynomial of $\alpha$ over $\mathbb Q$.
I only managed to show that the degree of the minimal polynomial is 3.
My attempt so far:
$\alpha^3 = \zeta^3+ 3\zeta^{-1}\zeta^2+3\zeta\zeta^{-2}+\zeta^{18} = \zeta^3+\zeta^4+3\alpha $
And I don't know how to continue, Thank you for your help
|
Now add the equation
$$
α^4=ζ^4+ζ^{-4}+4(ζ^2+ζ^{-2})+6=ζ^4+ζ^3+4α^2-2
$$
to your consideration to find that
$$
α^4-α^3-4α^2+3α+2=0
$$
This has a root at $2=1+1^{-1}$, the remaining factor is
$$
α^3 + α^2 - 2α - 1=0
$$
You could of course also start at
$$
0=\frac{ζ^7-1}{ζ-1}=1+ζ+ζ^2+ζ^3+ζ^4+ζ^5+ζ^6=1+α+(α^2-2)+(α^3-3α)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
if the function such $f(m+nf(m))=mf(n+1)$ find the f if $f:N^{+}\to N^{+}$,and such for any postive integer $m,n$ have
$$f(m+nf(m))=mf(n+1)$$
find the $f$
I guess this function is $f(n)=n$,But How to solve it?Thanks.
|
Function $f(x)$ is defined on the countable set of points and can be considered as the countable sequence of values
$$\{f(1),f(2),f(3),\dots\},$$
with the system of the equations
$$f(nf(m)+m) = mf(n+1),\quad (m,n)\in \mathbb N^2. \tag1$$
Let us research the structure of this sequence.
The value $f(1)=p$ should satisfy the system of
$$f(np+1) = f(n+1),\quad n\in \mathbb N.\tag2$$
If $\mathbf{p=1}$ then the equation $(1)$ is satisfied for each value of $n$.
If $\mathbf{p>1}$ then $f(x)$ is a periodic function with the period $p$,
$$f(m+p) = f(m).\tag3$$
At the same time, the value $f(2)=q$ should satisfy the system of
$$f(f(m)+m) = mq,\quad m\in \mathbb N.\tag4$$
In particular, for the values $m\in\{1,2,1+q,1+q+q^2,1+q+q^2+q^3\dots\},$
$$\begin{cases}
f(p+1)=q\\
f(2+q) = 2q\\
f(2+q+q^2) = 4q\\
f(2+q+q^2+q^3) = 8q\\
\dots.
\end{cases} \tag5$$
If $f(1) = p>1,$ then the system $(4)$ contradicts with the periodiity condiion $(2).$
If $f(1) = p = 1,$ then for $m=1$ from $(1)$ should the identity
$$f(n+1) = f(n+1).\quad n\in\mathbb N,$$
then $f(1) = 1.$
Since $f(x)$ is a countable set of points, it can be presented in the polynomial form of
$$f(x) = 1+a_1(q-1)(x-1)+a_2(q-1)^2(x-1)^2+...,\quad a_i\in \mathbb R.\tag6$$
wherein from $(5)$ should
$$\begin{cases}
1+a_1(q-1)+a_2(q-1)^2+a_3(q-1)^3+\dots = q\\
1+a_1(q^2-1)+a_2(q^2-1)^2+a_3(q^2-1)^3+\dots = 2q\\
1+a_1(q^3-1)+a_2(q^3-1)^2+a_3(q^3-1)^3+\dots = 4q\\
1+a_1(q^4-1)+a_2(q^4-1)^2+a_3(q^4-1)^3+\dots = 8q\\
\dots\\
1+a_1(q^{k+1}-1)+a_2(q^{k+1}-1)^2+a_3(q^{k+1}-1)^3+\dots = 2^kq\\
\dots.
\end{cases}\tag7$$
Transitional limit $k\to \infty$ leads to the solution
$$q=2,\quad a_1=1,\quad a_2=a_3=\dots =0,$$
$$f(x)= x.\tag8$$
Substitution of $(8)$ to $(1)$ leads to the identity
$$nm+m=m(n+1).$$
Thus, $(8)$ is the single solution.
|
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|
What is the length of a side of square $DCEF$ where $AB=3, BC=4, AC=5$ in the following diagram?
$DCEF$ is a square and $\triangle ABC$ is an internal triangle. $AB=3,$
$BC=4, AC=5$. What is the length of the square?
I tried multiple times and ended up by converting variables again and again. Can someone give me a hint to solve this?
|
The key is $$3^2 + 4^2 = 5^2 \quad\implies\quad \angle ABC = 90^\circ$$
As a consequence,
$$\angle ABD = \angle BCF\quad\implies\quad \triangle ABD \simeq \triangle BCF$$
Let $s = CF$ be the side of the square, we have
$$BD : AB = CF : BC \iff BD = \frac{3s}{4} \implies BF = DF - BD = \frac{s}{4}$$
Apply Pythagoras theorem to $\triangle BCF$, we get
$$BF^2 + FC^2 = 4^2\quad\iff\quad\frac{s^2}{16} + s^2 = 4^2\quad\implies\quad s = \frac{16}{\sqrt{17}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\frac{1}{3}\frac{(x^2-1)^{3/2}}{x^3}+C$
$$\int\frac{\sqrt{x^2-1}}{x^4}dx$$
Let $x=\sec\theta$
Then $dx=\sec\theta\tan\theta d\theta$
$$\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4\theta}\sec\theta\tan\theta d\theta$$
$$=\int\frac{\sec\theta}{\sec^4\theta}\sqrt{\tan^2\theta}\tan\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta} \tan^2\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta}\frac{\sec^2\theta}{\csc^2\theta}d\theta$$
$$=\int\frac{1}{\sec\theta}\frac{1}{\csc^2\theta}d\theta$$
$$=\int \cos\theta\sin^2\theta d \theta$$
Using $u$-substition, let $u=\sin\theta$
Then $du=\cos\theta d\theta$ and $dx = \frac{1}{\cos\theta}du$
$$\int\cos\theta u^2 \frac{1}{\cos\theta}du$$
$$=\int u^2 du$$
$$=\frac{u^3}{3}+C$$
$$=\frac{\sin^3\theta}{3}+C$$
Since $x=\sec\theta$, $\sec^{-1}(x)=\theta$
$$=\frac{\sin^3(\sec^{-1}(x))}{3}+C$$
What am I doing wrong?
|
Taking your final answer we can sub back in the original subs
$$
\cos (\theta) =\frac{1}{x}
$$
We can then use the relationship
$$
\sin^2 \theta + \cos^2\theta = 1 = \sin^2 \theta +\frac{1}{x^2}
$$
The rest is straight forward.
|
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|
Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/smarter way to do this?
|
$$(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2= \frac {(x^2+1)^4(x^2-1)^2}{x^6}=$$
$$\frac {(x^4-1)^2(x^2+1)^2}{x^6} =\frac {(x^8-2x^4+1)(x^4+2x^2+1)}{x^6}=$$
$$\frac{x^{12} +2x^{10} -x^8-4x^6-x^4+2x^2+1}{x^6}=$$
$$x^{6} +2x^{4} -x^2-4-\frac {1}{x^2}+\frac {2}{x^4}+\frac {1}{x^6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}$ and generalize it In this post the following is proved
$$\small \sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}=\frac{1}{2} \pi J_0(a x)-\frac{\sin (a x)}{2 a},\ \ \sum _{n=1}^{\infty } \frac{\cos \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}=-\frac{1}{2} \pi Y_0(a x)-\frac{\cos (a x)}{2 a}$$
But how to established the harder one
$$\small\sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=-\frac{\sin (a x)}{2 a^3}+\frac{\pi x \coth (\pi a)}{2 a}+\frac{1}{4} \pi ^2 x^2 (\pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x))-\frac{1}{2} \pi x^2 J_0(a x)+\frac{\pi x J_1(a x)}{2 a}$$
Here $J, \pmb{H}$ denotes Bessel and Struve functions.
Update: By M-L theorem and repeated integration one have
$$\small \sum _{n=1}^{\infty } \frac{\cos \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^2}=-\frac{\cos (a x)}{2 a^4}+\frac{\pi \coth (\pi a)}{4 a^3}+\frac{1}{8} \pi ^2 x^3 \left(1-\frac{1}{a^2 x^2}\right) F(a x)+\frac{\pi ^2 \text{csch}^2(\pi a)}{4 a^2}+\frac{1}{4} \pi x^3 J_0(a x)-\frac{\pi x^2 \coth (\pi a)}{4 a}-\frac{\pi x^2 J_1(a x)}{4 a}$$
Where $F(t)=\pmb{H}_0(t) J_1(t)-\pmb{H}_1(t) J_0(t)$. Analytic continuation allows us to extend the range to $|a|<1$, $x\in (0,2\pi)$, for instance
$$\small\sum _{n=1}^{\infty } \frac{\cos \left(\sqrt{4 n^2-1}\right)}{\left(n^2-\frac{1}{4}\right)^2}=2 \pi ^2 \pmb{L}_1(1) I_0(1)-2 \pi ^2 \pmb{L}_0(1) I_1(1)+\pi ^2-8 \cosh (1)+2 \pi I_0(1)-2 \pi I_1(1)$$
Moreover, differentiating closed-form of $\sum _{n=1}^{\infty } \left(\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}-\frac{\sin (n x)}{n}\right)$ w.r.t $x$ yields:
$$\small\sum _{n=1}^{\infty } \left(\cos \left(\pi \sqrt{n^2+1}\right)-(-1)^n\right)=1-\frac{\pi J_1(\pi )}{2}$$
|
Using the representation
\begin{equation}
\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\int_0^1 \frac{\cos \left(tx \sqrt{a^2+n^2}\right)}{a^2+n^2}\,dt\\
\end{equation}
and integrating by parts,
\begin{equation}
\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}+x^2\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt
\end{equation}
We have then to evaluate
\begin{align}
S(x)&=\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}\\
&=xA(x)+x^2B(x)\\
A(x)&=\sum_{n\ge1}\frac{\cos \left(x \sqrt{a^2+n^2}\right)}{a^2+n^2}\\
B(x)&=\sum_{n\ge1}\int_0^1 t\frac{\sin \left(tx \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\,dt
\end{align}
We have
\begin{align}
A'(x)&=-\sum_{n\ge1}\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}\\
&=-\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}
\end{align}
And thus, considering that
\begin{equation}
A(0)=\sum_{n\ge1}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}
\end{equation}
we deduce
\begin{align}
&A(x)=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\int_0^x\left[\frac{\sin (a t)}{2 a}-\frac{1}{2} \pi J_0(a t)\right]\,dt\\
&=\frac{\pi}{2a}\coth \pi a-\frac{1}{2a^2}+\frac{x\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x\pi}{2}J_0(ax)+\frac{1-\cos ax}{2a^2}
\end{align}
Now,
\begin{align}
B(x)&=\int_0^1 \left[\frac{1}{2} \pi J_0(a xt)-\frac{\sin (a xt)}{2 a}\right]t\,dt\\
&=\frac{1}{x^2}\int_0^x\left[\frac{1}{2} \pi J_0(a u)-\frac{\sin (a u)}{2 a}\right]u\,du\\
&=\frac{\pi }{2ax}J_1(ax)+\frac{\cos ax}{2xa^2}-\frac{\sin xa}{2x^2a^3}\\
\end{align}
Finally, as expected,
\begin{align}
S(x)=&\frac{\pi x}{2a}\coth \pi a-\frac{x}{2a^2}+\frac{x^2\pi^2}{4}\left( \pmb{H}_1(a x) J_0(a x)-\pmb{H}_0(a x) J_1(a x) \right)-\frac{x^2\pi}{2}J_0(ax)\\
&+\frac{x}{2a^2}+
\frac{x\pi }{2a}J_1(ax)-\frac{\sin ax}{2a^3}
\end{align}
The series with cosines could be evaluated in the same way. In fact working directly with complex numbers $\exp\left(i x\sqrt{n^2+a^2} \right)$ and Hankel functions may simplify, but I didn't try. Indefinite integrals of $H_0^{1}(z)$ and $zH_0^{1}(z)$ are indeed tabulated DLMF.
|
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|
The sum of all digits of n for which $\sum^n_{r=1} r. 2^r=2+2^{n+10}$ If we start substituting values, we get
$$S_n= 2+8+24+64......n2^n$$
The nth term of the series will be
$$T_n = (n)(2^n)$$
So the sum would be $$\sum T_n=\sum n 2^n$$
Cause that’s how usually solve it. The exponent is throwing me off, and I don’t know what to do.
Thanks!
|
Note:
$$S_n=1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots +n\cdot 2^n\\
2S_n=1\cdot 2^2+2\cdot 2^3+3\cdot 2^4+\cdots +(n-1)\cdot 2^n+n\cdot 2^{n+1}\\
2S_n-S_n=-1\cdot 2^1-1\cdot 2^2-1\cdot 2^3-\cdots -1\cdot 2^n+n\cdot 2^{n+1} \Rightarrow\\
S_n=-(2+2^2+2^3+\cdots+2^n)+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-\frac{2(2^n-1)}{2-1}+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-2\cdot 2^n+2+n\cdot 2\cdot 2^n=2+2^{10}\cdot 2^n \Rightarrow \\
n\cdot 2\cdot 2^n=2^{10}\cdot 2^n+2\cdot 2^n \Rightarrow \\
2n=2^{10}+2 \Rightarrow \\
n=2^9+1=513 \Rightarrow \\
5+1+3=9.$$
|
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|
multiple choice question on group of matrices Consider the set of matrices $$G=\left\{ \left( \begin{array}{ll}s&b\\0&1 \end{array}\right) b \in \mathbb{Z}, s \in \{1,-1\} \right\}.$$Then which of the following are true
*
*G forms a group under addition
*G forms an abelian group under multiplication
*Every element of G is diagonolizable over $\mathbb{C}$
*G is finitely generated group under multiplication
I am getting
1) is false since not closed under addition
2)Forms a group under multiplication ( abelian or not i don't know)
3)Not true if $a=1$
4) dont know
please help me to complete
|
*
*$1$ is false:
Your approach is correct. Since, for example, $\begin{pmatrix}1&*\\0&1 \end{pmatrix}+\begin{pmatrix}1&*\\0&1 \end{pmatrix}=\begin{pmatrix}2&*\\*&* \end{pmatrix} \notin G$
*$2$ is false:
Take $b \neq 0$.$$\begin{pmatrix}1&b\\0&1 \end{pmatrix}\begin{pmatrix}-1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&2b\\0&1 \end{pmatrix}$$ whereas $$\begin{pmatrix}-1&b\\0&1 \end{pmatrix}\begin{pmatrix}1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&\color{red}{0}\\0&1 \end{pmatrix}$$
*$3$ is false too:
Since, for example, $\begin{pmatrix}1&b\\0&1 \end{pmatrix}$ is not diagonalizable when $b \neq 0$
*$4$ is true
The finite set $$\left\{\begin{pmatrix}1&1\\0&1 \end{pmatrix},\begin{pmatrix}1&-1\\0&1 \end{pmatrix},\begin{pmatrix}-1&0\\0&1 \end{pmatrix}\right\}$$ generates $G$(verify!)
|
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|
Prove $y = z$ if $y^p = z^p$ where $p \in \mathbb{N}$ and $y, z \in \mathbb{R}_{++}$ As above. I am beginner in analysis. I don't know how rigorously you have to prove things. To me it's obvious.
|
Are you familiar with induction and/or the well ordering principal?
If we assume $y \ne z$ and wolog we assume $0 < y < z$ then $y^1 < z^1$. Let's assume there is a $y^k \ge z^k$. By well ordering principal there most be so first $k$ where that is true.
But that would mean $y^{k-1} < z^{k-1}$. Byt then $y^k = y^{k-1}*y < z^{k-1}*y < z^{k-1}*z = z^k$ and thus we have a contradiction.
Alternatively by induction as $y^1 < z^1$ and $y^{k-1} < z^{k-1}\implies $y^k < z^k$ we can never have anything *but* $y^p < z^p$.
So if $y^p = z^p$ we can't have $y < z$ and by wolog (or symmetry) we can't have $y > z$ either. So if $y^p = z^p$ having $y = z$ is our only option.
....
or. Consider $y^p = z^p$ so $y^p - z^p = 0$ and notice (although this requires induction to actually state) that $(y-z)(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) = y^p - z^p=0$.
Now all $y^iz^k > 0$ so $(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) \ne 0$. So $(y-z)(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) = 0 \iff (y-z) = 0$. So $y-z = 0$ and $y= z$>
If $0 < y < z$ then $y^1 < z^1$ so we can't have $x^p =z^p$ if $p=1$.
Now if for some $p=k$ that $y^k < z^k$ then $y^{k+1} = y^k*y < z^k*y$ and $z^k *y < z^k*z = z^{k+1}$. So by induction it is true that $y^p < z^p$ for all $p$ and so if $y^p = z^p$ it
|
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|
Prove: $\sqrt{(n^{2}-1)}= [n-1,1,2n-2,1,2n-2,1,2n-2, \ldots ]$ I'm trying to show
$$
\sqrt{n^2-1}=n-1 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cdots}}}
$$
What I tried:
$$
\sqrt{n^2 - 1}- (n-1) = \frac{2n-2}{n^2-1+(n-1)}
$$
And here i got stuck on how to continue, would greatly appreciate any help
|
You want to show that
$n-1+\sqrt{n^2-1}=2n-2 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cfrac{1}{2n-2+\cdots}}}}$
Suppose
$y=x+\cfrac{1}{1+\cfrac{1}{x+\cfrac{1}{1+\cfrac{1}{x+\cdots}}}}$
then
$y = x + \cfrac{ 1}{1+\frac 1 y} \\\Rightarrow y=x+\frac {y}{y+1}\\\Rightarrow y^2+y = xy+x+y\\\Rightarrow y^2 - xy - x = 0 \\\Rightarrow y=\frac{x \pm \sqrt{x^2+4x}}{2}$
Now substitute $2n-2$ for $x$ ...
|
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|
Show that $3x^4+4y^4=19z^4$ has no integer solution If $x,y,z$ are integers such that $xyz\neq 0$, prove or disprove that
$$3x^4+4y^4=19z^4$$
has no solution.
Maybe this is can use quadratic residue to solve it, maybe this equation seem is famous? because I have solve paper Introduced this.
|
COMMENT.- Just for fun. Do you have for the equation $3x^4+7y^4=19z^4$ the following irrational parameterization
$$\begin{cases}x=\left(\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ There is a value of $t$ such that both $\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}$ and $\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}$ be rational $\ne0$?
If the equation $3x^4+7y^4=19z^4$ is not solvable, certainly NOT. But if the equation is solvable, certainly YES. For example for the equation $10x^4+6y^4=285z^4$ we have the parameterization $$\begin{cases}x=\left(\sqrt[4]{\dfrac{285}{10}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{285}{6}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ and for $t=\pm\dfrac15\sqrt{\dfrac{179\pm12\sqrt{114}}{5}}$ (which is root of $125t^4-358t^2+125=0$) we get the solution $(x,y,z)=(3,5,2)$.
|
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|
Indefinite integral of rational expression involving cubic polynomials I was given the following exercise:
$$\int \frac{x^3}{(x^2+1)^3}dx$$
As a tip, my professor suggested using the following substitution: $t=x^2+1$.
Notice that if $t=x^2+1$, then $x^2=t-1$ and therefore $x=(t-1)^\frac{1}{2}$. Then
$$x^3=x^2\cdot x=(t-1)(t-1)^\frac{1}{2}=(t-1)^\frac{3}{2}$$
and we have that
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}dt.$$
But now that I have applied the suggested substitution, I don't really see how to continue with this integral. Integrating by parts doesn't seem to get me nowhere, and I can't seem to find any way to apply the substitution method. Am I missing something, or perhaps I made a mistake changing my integration variable from $x$ to $t=x^2+1$?
Thanks in advance.
|
Partial integration may also work as follows:
\begin{eqnarray*} \int \frac{x^3}{(x^2+1)^3}dx
& = & \int x^2\frac{x}{(x^2+1)^3}dx \\
& = & x^2\left(-\frac{1}{4}\frac{1}{(x^2+1)^2}\right) + \frac{1}{4}\int \frac{2x}{(x^2+1)^2}dx \\
& = & -\frac{1}{4}\frac{x^2}{(x^2+1)^2} - \frac{1}{4}\frac{1}{x^2+1} (+C)
\end{eqnarray*}
|
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|
If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest possible value of $x+y$?
The book says "the greatest value of $4$ is divided by $6$, which produces a remainder of $4$. The greatest value of $y$ is when $2$ is divided by, which produces a remainder of $2$. Therefore, the greatest value of $x+y$ is $6$."
I think what's throwing me off is the phrase "multiple of $4$" bc it makes me think that any multiple of $4$ can be divisible by $6$ (i.e. $24/6 = 4$). The books answer doesn't use multiples, just the $4$ and $2$, respectively. I don't understand how this works. Can someone please clarify?
|
So let $N = 4M$ be a multiple of $4$. If you divide by $6$ and take the remainder $x$ so that $N=4M = 6k + x$ what are the possible values of $x$ if $x$ is not negative and $x < 6$.
Well, It's possible that $6$ divides into $N=4M$. If so $x = 0$.
And it's possible that $6$ doesn't divide into $N=4M$. What are the possible values of $x$.
Well, $4M = 6k+x$ so $2M =3k + \frac x2$. $2M$ and $k$ are is whole numbers, so $\frac x2$ is a whole number. And $x < 6$ so $x$ can be: $0, 2,4$.
Can $x =4$. Well of course. $4 = 0*4 + 4$ has a remainder of $4$. So does $28 = 6*4 + 4$.
So $x \le 4$.
....
Now let $N = 2K$ but a multiple of $2$. If you divide by $3$ and take the remainder so that $N =2k +y$ what are the possible values of $y$ if $y$ is not negative and $y < 3$.
Well, $y$ can only be $0, 1,$ or $2$. Which are possible.
All of them are $2=0*3 + 2$ so $y$ can be $2$. And $4=1*3 +1$ so $y$ can be $1$. And $6= 3*2 + 0$ so $y$ can be $0$.
So $y \le 2$. So $x + y \le 4 + 2 =6$.
That's it.
|
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|
Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\
&=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\
&=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx.
\end{align}
I can't go further from here, please help.
|
Whenever you have $\sqrt{a^2-x^2}$ in an integral, use the substitution $x=a \sin(\theta )$.
For information :
In case of having $\sqrt{x^2-a^2}$ use the substitution $x=a \sec(\theta )$.
In case of having $\sqrt{x^2+a^2}$ use the substitution $x=a \tan(\theta )$
|
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|
For continuous, monotonically-increasing $f$ with $f(0)=0$ and $f(1)=1$, prove $\sum_{k=1}^{10}f(k/10)+f^{-1}(k/10)\leq 99/10$ A question from Leningrad Mathematical Olympiad 1991:
Let $f$ be continuous and monotonically increasing, with $f(0)=0$ and $f(1)=1$.
Prove that:$$
\text{f}\left( \frac{1}{10} \right) +\text{f}\left( \frac{2}{10} \right) +...+\text{f}\left( \frac{9}{10} \right) +\text{f}^{-1}\left( \frac{1}{10} \right) +\text{f}^{-1}\left( \frac{2}{10} \right) +...+\text{f}^{-1}\left( \frac{9}{10} \right) \leqslant \frac{99}{10}
$$
I tried to express them in areas to find inequalities but failed.
|
Note: In the title of the question the upper limit of $k$ needs to be 9.
When $y=f(x)$ is monotonically increasing in domain $D$ then area under the curves $y=f(x)$ and $y=f^{-1}(x)$ is more that those of the resprctive rectangles below them.
Also the sum the area integrals
$$A=\int_{x_1}^{x_2} f(x)~ dx+\int_{y_1}^{y_2} f^{-1} (y)~ dy = x_2 y_2 -x_1 y_1$$
Here $f: [0,1]\rightarrow[0,1]$. Let the domain be divided into 10 strips of width $1/10$. Let $S$ denotes area under the rectangles of equal width. Then
$$S= \frac{1}{10}\sum_{k=1}^{9} [f(k/10)+ f^{-1}(k/10)] \le \sum_{k=1}^{9} \frac{(k+1)^2-(k)^2}{100}= \sum_{k=1}^{9}\frac{2k+1}{100}=\frac{99}{10}$$ $$\Rightarrow \sum_{k=1}^{10} [f(k/10)+ f^{-1}(k/10)] \le \frac{99}{10}. $$
As $S$ is the sum of area of rectangles below the curves $y=f(x)$ and $y=f^{-1}(x)$.
|
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|
Find the distribution of $X|Y =k$ given the distribution of $Y|X =x$ Good night, I'm trying to find the p.d.f of $Y$ and $X|Y=k$, and $\mathbb E (X|Y=k)$ in the following exercise:
Could you please verify if my attempt is correct or contains logical mistakes? Thank you so much for your help!
My attempt:
First, we have $$\begin{aligned} & \int_0^1 {n \choose k} x^k (1-x)^{n-k} \, \mathrm{d}x &&= {n \choose k} \int_0^1 x^{(k+1)-1} (1-x)^{(n-k+1)-1} \, \mathrm{d}x \\ &= {n \choose k} B(k+1,n-k+1) &&= \frac{n!}{(n-k)!k!} \frac{k!(n-k)!}{(n+1)!} \\ &= \frac{1}{n+1} \end{aligned}$$ and $$\begin{aligned} & \int_0^1 {n \choose k} x^{k+1} (1-x)^{n-k} \, \mathrm{d}x &&= {n \choose k} \int_0^1 x^{(k+2)-1} (1-x)^{(n-k+1)-1} \, \mathrm{d}x \\ &= {n \choose k} B(k+2,n-k+1) &&= \frac{n!}{(n-k)!k!} \frac{(k+1)!(n-k)!}{(n+2)!} \\ &= \frac{k+1}{(n+2)(n+1)} \end{aligned}$$
We have $X \sim \mathcal U([0,1])$ and $(Y|X=x) \sim \mathcal B(n,x)$. Then $f_{Y|X} (k|x) = {n \choose k} x^k (1-x)^{n-k}$ and so $$\begin{aligned} f_Y (k) &= \int_{\mathbb R} f_{Y|X} (k|x) f_X (x) \, \mathrm{d}x &&= \int_{\mathbb R} {n \choose k} x^k (1-x)^{n-k} \frac{1}{1-0} \textbf{1}_{[0,1]} (x) \, \mathrm{d}x \\&= \int_0^1 {n \choose k} x^k (1-x)^{n-k} \, \mathrm{d}x &&= \frac{1}{n+1} \end{aligned}$$
As such, $Y$ is uniformly distributed discrete random variable with $\operatorname{supp} (Y)= \{0,\ldots,n\}$.
We have $$\begin{aligned} f_{X|Y} (x,k) &= \frac{f_{Y|X} (k|x) f_X(x)}{f_Y (k)} \\ &= \frac{{n \choose k} x^k (1-x)^{n-k} \frac{1}{1-0} \textbf{1}_{[0,1]} (x)}{\frac{1}{n+1}}\\ &= (n+1) {n \choose k} x^k (1-x)^{n-k} \textbf{1}_{[0,1]} (x) \end{aligned}$$
As such, $$\begin{aligned} \mathbb E (X|Y=k) &= \int_\mathbb R x f_{X|Y} (x,k) \, \mathrm{d}x \\ &= \int_\mathbb R x (n+1) {n \choose k} x^k (1-x)^{n-k} \textbf{1}_{[0,1]} (x) \, \mathrm{d}x \\ &= (n+1) \int_0^1 {n \choose k} x^{k+1} (1-x)^{n-k} \, \mathrm{d}x \\ &= (n+1) \frac{k+1}{(n+2)(n+1)} = \frac{k+1}{n+2}\end{aligned}$$
|
I post my attempt in case $X \sim \operatorname{Beta}(a,b)$ here and is going to mark this answer as accepted to peacefully close this question.
Thank you again for your verification @Math1000 ^o^
We have $X \sim \operatorname{Beta}(a,b)$ and $(Y|X=x) \sim \mathcal B(n,x)$. Then $f_{Y|X} (k|x) = {n \choose k} x^k (1-x)^{n-k}$ and thus $$\begin{aligned} f_Y (k) &= \int_{\mathbb R} f_{Y|X} (k|x) f_X (x) \, \mathrm{d}x \\&= \int_{\mathbb R} {n \choose k} x^k (1-x)^{n-k} \frac{x^{a-1}(1-x)^{b-1}}{\mathrm{B}(a, b)} \textbf{1}_{[0,1]} (x) \, \mathrm{d}x \\&= {n \choose k} \frac{1}{\mathrm{B}(a, b)} \int_0^1 x^{(k+a)-1} (1-x)^{(n-k+b)-1} \, \mathrm{d}x \\&= {n \choose k} \frac{\mathrm{B}(k+a,n-k+ b)}{\mathrm{B}(a, b)} \end{aligned}$$
We have $$\begin{aligned} f_{X|Y} (x,k) &= \frac{f_{Y|X} (k|x) f_X(x)}{f_Y (k)} \\ &= \frac{{n \choose k} x^k (1-x)^{n-k} \frac{x^{a-1}(1-x)^{b-1}}{\mathrm{B}(a, b)} \textbf{1}_{[0,1]} (x)}{ {n \choose k} \frac{\mathrm{B}(k+a,n-k+ b)}{\mathrm{B}(a, b)}}\\ &= \frac{x^{k+a-1} (1-x)^{n-k+b-1}}{\mathrm{B}(k+a,n-k+ b)} \textbf{1}_{[0,1]} (x) \end{aligned}$$
As such, $(X|Y=k) \sim \operatorname{Beta}(k+a,n-k+b)$. Hence $\mathbb E(X|Y=k) = \frac{k+a}{n+a+b}$.
|
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|
In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time I've tried it so I'm stuck at a point...
So $$x_1+x_2+x_3 = 10$$
Subject to the condition that :
$$1\leq x_1 \leq8$$
$$1\leq x_2 \leq8$$
$$1\leq x_3 \leq8$$
As each beggar can get at maximum 8 blankets and at minimum, 1.
So the number of ways must correspond to the coefficient of $x^{10}$ in:
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)^3$
= coeff of $x^{10}$ in $x^3(1-x^7)^3(1-x)^{-3}$
= coeff of $x^{10}$ in $x^3(1-x^{21}-3x^7(1-x^7))(1-x)^{-3}$
= coeff of $x^{10}$ in $(x^3-3x^{10})(1+\binom{3}{1}x + \binom{4}{2}x^2+...+ \binom{12}{10}x^{10})$
= $\binom{9}{7} - 3 = 33$
Is this right? From here I get the answer as $\binom{9}{7} - 3 = 33$ but the answer is stated as $36$. I don't understand where I'm making a mistake
|
You may give a blanket to each one and and use stars and bars with remaining $7$ blanket.
You only need two bars with seven stars so the answer is $\binom {9}{2} =36$
|
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|
Solutions of $1+x+x^2+...+x^n$ I was thinking about the roots of unity, the solutions of the polynomial $P(x) = x^n - 1$ which are quite easy to find, they are of the form $ \cos \frac{2k\pi}{n} + i\sin \frac{2k\pi}{n} $. I was wondering if there are any other polynomials whose solutions we know. Particularly, is there any way to find the solutions of $ P(x) = 1 + x^2 + x^3 + \dots + x^n $?
|
Observe that
$x^{n + 1} - 1 = (x - 1)(x^n + x^{n - 1} + \ldots + 1); \tag 1$
therefore any root of
$x^{n + 1} - 1 = 0 \tag 2$
other than $x = 1$ is a root of
$x^n + x^{n - 1} + \ldots + 1 = 0; \tag 3$
thus the $n$ roots of (3) are
$x = e^{2k\pi i / (n + 1)}, \; 1 \le k \le n. \tag 4$
|
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Solve for $x$: $\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12}.$ Solve for $x$: $$\frac{1}{\log\big((x+2)^2\big)}+\frac{1}{\log\big((x-2)^2\big)} = \frac{5}{12}.$$ My Attempt: \begin{align*} & \frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12} \\ \implies &\> \frac{1}{2\log(x+2)}+\frac{1}{2\log(x-2)} = \frac{5}{12} \\ \implies & \> \frac{1}{\log(x+2)}+\frac{1}{\log(x-2)} = \frac{5}{6} \\ \implies & \> 6\log(x^2-4) = 5 \log(x-2) \log(x+2).\end{align*} Please help me how can I proceed from here?
|
We have that
$$f(x)=\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2}$$
is even and therefore we can assume $x> 0$ with $x\neq 1$ and $x\neq 2$ ($x=0$ is not a solution).
For $0<x<1$ and $1<x<2$ we have that
$$f(x)=\frac{1}{2\log(x+2)}+\frac{1}{2\log(2-x)}\\\implies f'(x)=\frac{1}{2(2-x)\log^2(2-x)}-\frac{1}{2(x+2)\log^2(x+2)}$$
from whic we can conlcude that for $0<x<1$
$$f(x)> \frac2{\log 4}$$
and for $1<x<2$
$$f(x)< \frac1{\log 16}$$
For $x>2$ we have
$$f(x)=\frac{1}{2\log(x+2)}+\frac{1}{2\log(x-2)}\\\implies f'(x)=-\frac{1}{2(x+2)\log^2(x+2)}-\frac{1}{2(x-2)\log^2(x-2)}<0$$
therefore on that interval $f(x)$ is strictly decreasing and since
*
*$\lim_{x\to 2^+} f(x)=\infty$
*$\lim_{x\to \infty} f(x)=0$
by IVT exactly a real solution exists which can be determined numerically and leads to $x\approx 11.3467$. For symmetry also $x\approx -11.3467$ is a solution.
|
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|
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c|
\geq |a+b|$ prove then $a+b+c = 0$
Solution:
$$
|a| + |b| + |c| \geq |b+c| + |c+a| + |a+b| \geq 0
$$
and
$$0 \leq |a+b+c|\leq |a+b| + |c| \leq |a| + |b| + |c|$$
so we have
$$0 \leq |a+b+c| \geq 0 $$
$$ |a+b+c|= 0 \iff a+b+c =0 $$
I need check my solution.
Many thanks!
|
We have:
$$a^2\geq(b+c)^2,$$
$$b^2\geq(a+c)^2$$ and $$c^2\geq(a+b)^2$$ or
$$(a-b-c)(a+b+c)\geq0,$$ $$(b-a-c)(a+b+c)\geq0$$ and $$(c-a-b)(a+b+c)\geq0.$$
Now, for $a+b+c>0$ we obtain:
$$a-b-c\geq0,$$ $$b-a-c\geq0$$ and $$c-a-b\geq0,$$ which gives
$$a-b-c+b-a-c\geq0$$ or
$$c\leq0.$$
Similarly, $b\leq0$ and $a\leq0,$ which gives
$$a+b+c\leq0,$$ which is a contradiction.
By the same way we'll obtain a contradiction for $a+b+c<0.$
Thus, $a+b+c=0$ and we are done because for $a=b=c=0$ we obtain $a+b+c=0.$
|
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|
$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.
My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0$$
First condition: If $\forall$ $x>0$, inequality is satisfied, it means $x=0$ must have been the root of the equation $\quad \{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}=0\quad$ at which given inequality would not be satisfied.
Second condition: If one root is real, then other root must be real as complex roots occur as conjugates if all quadratic equation coefficients are real. So $D>=0$
Third condition: $a>0$ as $\forall$ $x>0$, $\quad\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0\quad$
First condition:
$$x=0$$
$$a+\sqrt{2}=0$$
$$a=-\sqrt{2}$$
Second condition:
$$D>=0$$
$$4(a^2-2)^2-4(a-\sqrt{2})(a^2+a-3)(a+\sqrt{2})>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a^2-2-a^2-a+3)>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a-1)<=0$$
$$a\in \left(-\infty,-\sqrt{2}\right]\quad \cup \quad[1,\sqrt{2}]$$
Third condition:
$$a>0$$
$$(a-\sqrt{2})(a^2+a-3)>0$$
$$(a-\sqrt{2})\left(a-\left(\frac{-1+\sqrt{13}}{2}\right)\right)\left(a-\left(\frac{-1-\sqrt{13}}{2}\right)\right)>0$$
$$a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Taking intersection of all three conditions would give
$$a\in \{-\sqrt{2}\}\quad$$
But answer is $a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$
What am I missing here, I tried to think of it a lot but didn't any breakthroughs. Please help me in this.
|
For the first condition
($x = 0$),
you want
$a+\sqrt{2} > 0$
so
$a > -\sqrt{2}$.
|
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|
Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this without actually expanding everything out.
Additionally, does it provide a parametrisation of the surface
$$X^3+Y^3=Z^3+1$$ in $\mathbb{Q}^3$
with
$$X=\frac{x^2+7xy-9y^2}{2x^2+10y^2}$$
$$Y=\frac{2x^2-4xy+12y^2}{2x^2+10y^2}$$
$$Z=\frac{x^2-9xy-y^2}{2x^2+10y^2}$$
|
Observe that the discriminants of $x^2 + 7 x y -9 y^2$ and $x^2 - 9 xy - y^2$ are $85$ and the discriminants of $2x^2 - 4 x y + 12 y^2$ and $2x^2 + 10 y^2$ are $-80$. This suggests that these integral binary quadratic forms might be equivalent in pairs. Testing this, we find
$$ \left. x^2 + 7 x y -9 y^2 \right|_{\pmatrix{x \\ y} \mapsto \pmatrix{\frac{11}{\sqrt{85}}x - \frac{7}{\sqrt{85}}y \\ \frac{9}{\sqrt{85}}x + \frac{2}{\sqrt{85}}y}} = x^2 -9x y - y^2 $$
and
$$ \left. 2x^2 - 4 x y + 12 y^2 \right|_{\pmatrix{x \\ y} \mapsto \pmatrix{x + y \\ y}} = 2x^2 + 10y^2 $$
Now we know that the sets of integers generated by each member of the first pair are the same. (Generically, each member of that set is generated by different choices of $x$ and $y$ in the two forms.) Likewise, there is a set of integers generated by each member of the second pair.
The minimum positive member of the first set is $1$, from $x^2 - 9 x y - y^2$ with $x = 1, y = 0$. The minimum positive member of the second set is $2$ from $2x^2 + 10y^2$ with $x = 1$, $y = 0$. This suggests a method of choosing pairs from the class of the first pair and the class of the second pair:
*
*Pick random representatives from each class. Find the minimum positive representable value in each class and find $(x,y)$ pairs that realize that minimal value for the randomly chosen forms.
*If possible, find a change of variables that transforms one of the random class members to represent the minimal value at the same $(x,y)$ pair as the member from the other class. This gives the two forms on the right-hand sides of the two above display equations.
*Now perform a "random" transform on one. This gives the lower-left member in the above displays.
*Figure out where the lower transform moves the $(x,y)$ location of the minimal positive representable value and find a transform to the upper-right form moves its minimal point to the same place. This transform produces the upper-left form from the upper-right form.
(There's probably a slick way to get that upper transform from the lower transform, but I'm still working through my first cup of coffee, so I'll just leave the procedure above.)
|
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|
Why isn’t $f:\!R \rightarrow P : a \rightarrow ax^3 + a^2x^5$ a linear operator? Why isn’t $f:\!R \rightarrow P : a \rightarrow ax^3 + a^2x^5$ a linear operator?
Because $0 = 0$ and in my understanding it is closed in linear combinations. Isn’t it?
|
Simply because $f(a+b) \neq f(a)+f(b)$.
$$f(a+b)=(a+b)x^3+(a+b)^2x^5=(a+b)x^3+(a^2+2ab+b^2)x^5$$
while
$$f(a)+f(b)=ax^3+a^2x^5+bx^3+b^2x^5=(a+b)x^3+(a^2+b^2)x^5$$
And they are different unless $ab=0$ i.e. $a=0$ or $b=0$
|
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|
I have to solve this initial value problem and determine where the solution attains its minimum value. This is the differential equation : $y'= 4y^2 + xy^2 , y(0)=1.$
I was able to find the solution $y$ for this equation which is :
$$y=\frac{-2}{8 x + (x^2 - 2)},$$
but I don't know how to determine where the solution attains its minimum value.
So any help with it would be appreciated.
|
You can take the derivative of your solution (which is correct, by the way), and set it equal to zero to find the maximum value. Or you can work from the original DE:
\begin{align*}
y'&=0\\
4y^2+xy^2&=0\\
y^2(4+x)&=0.
\end{align*}
One solution is $y=0,$ and the other is $x=-4.$ But in looking at your solution, $y\not=0$ holds everywhere! So the only solution is $x=-4.$ But this doesn't tell us whether we have a max or a min here. Let us try the second derivative test:
\begin{align*}
y'&=y^2(4+x)\\
y''&=2yy'(4+x)+y^2\\
&=2yy^2(4+x)(4+x)+y^2\\
&=2y^3(4+x)^2+y^2.
\end{align*}
Now, at $x=-4,$ we have that $y''=y^2\ge 0,$ whereas the solution tells us that
$$y(-4)=\frac{-2}{-32 + (16 - 2)}=\frac{-2}{-18}=\frac19>0. $$
This tells us that $x=-4$ is a relative min, which is not what you are after.
So, moving on: are there any other critical points? Yes! Where the function $y$ blows up, which occurs at
\begin{align*}
x^2 +8x - 2&=0\\
x&=\frac{-8\pm\sqrt{64-4(-2)}}{2}\\
&=\frac{-8\pm\sqrt{72}}{2}\\
&=\frac{-8\pm 6\sqrt{2}}{2}\\
&=-4\pm 3\sqrt{2}.
\end{align*}
$y$ is not defined here, so these values are not in the domain. We conclude that there is no absolute maximum!
|
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|
The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$
I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?
|
Note that:
$2^2=1+3$, $3^2=4+5$, $5^2=16+9$, $9^2=64+17$, ...
Therefore
$$2=\sqrt{4^0+3}$$
$$2=\sqrt{4^0+\sqrt{4^1+ 5}}$$
$$...$$
$$2=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+17}}}}$$
$$...$$
$$...$$
Let $F_n=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+...}}}}$ where the sequence terminates after $n$ square roots.
For positive numbers $a$ and $b$, we have $\sqrt{a+b}<\sqrt{a}+\frac{b}{2\sqrt{a}}$ and therefore
$$F_n<2<F_n+\frac{2^n+1}{2^n(1+2+...+2^{n-1})}$$
Hence $F_n$ converges to 2.
|
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|
Possible Jordan canonical forms of identity matrix plus a nilpotent matrix I am working on the following Linear algebra problem:
Suppose that $N$ is a nilpotent $5 \times 5$ real matrix (so $N^5$ is the zero matrix). List all possible Jordan canonical forms of $I + N$.
Here is where my thinking is at: I know how to list all possible Jordan canonical forms of a matrix, given its characteristic polynomial. I also know that, in this case, the characteristic polynomial of $N$ will be given by $p_N(x) = x^5$. However, I'm struggling with this problem because I don't know how to deduce from this what the characteristic polynomial of $I + N$ is. Is there a nice way to see what this is?
I searched for if there is a nice relationship between a matrix's characteristic polynomial and the characteristic polynomial of the matrix plus the identity, but I couldn't find one. Does the fact that $N$ is nilpotent help at all to see what this is?
Thanks!
|
As you mentioned correctly- The characteristic polynomial of $N$ is $\chi_N(x)=x^5$, we know that not by computing, but by the nilpotent property and the dimension of $N$. To understand the characteristic polynomial of $N+I$ we need to go back to the definition: To any matrix $M$ it's characteristic polynomial is given by $\det(xI-M)$. Let's examine what happens when we add the identity matrix:
$$\det(xI-(N+I))=\det((x-1)I+N)$$
This is exactly the definition of the characteristic polynomial but instead of being a polynomial over of the variable $x$, it's shifted by $1$. It's a composition of the original characteristic polynomial with $x-1$, so if we had $\chi_N(x)=x^5$ then $\chi_{N+I}(x)=(x-1)^5$, and thus 1 is an eigenvalue of $N+I$.
Knowing all this we get these options of the jordan canonical form (up to rearranging etc.) (also, I'm using lower triangular matrices, I know some people use upper, these are equivalent):
$$
\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right]
$$
$$
\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right]
$$
|
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|
Then find the sum of all possible values of $abc$. Let $a, b, c$ be positive integers with $0 < a, b, c < 11$. If $a, b, $ and $c$ satisfy
\begin{align*}
3a+b+c&\equiv abc\pmod{11} \\
a+3b+c&\equiv 2abc\pmod{11} \\
a+b+3c&\equiv 4abc\pmod{11} \\
\end{align*}then find the sum of all possible values of $abc$.
What I tried:
I only got this far ...
If the equations from top to bottom were labeled $(1),(2),(3)$,
$(1)+3(2)+(3):$
$7a+11b+7c\equiv 11abc\pmod{11}$
$7(a+c)\equiv 0\pmod{11}$.
Since 7 and 11 are coprime, this means $a+c\equiv0\pmod{11}$ or rather just $a+c=11$ in the given range. Note that because 11 is prime, than modular division by any number not divisible by 11 is allowed henceforth.
Back in $(2)$:
$3b+11\equiv3b\equiv2abc\pmod{11}$
$3\equiv 2ac\pmod{11}$
$2ac=2a(11-a)=22a-2a^2\equiv-2a^2\equiv3\pmod{11}$
$2a^2\equiv-3\equiv8\pmod{11}$
$a^2\equiv 4\pmod{11}$
I just got this far ...
|
All equations below are understood to be modulo $11$.
Add the $3$ equations together
\begin{eqnarray*}
5(a+b+c)=7abc.
\end{eqnarray*}
Multiply the first equation by $5$
\begin{eqnarray*}
4a+5b+5c=5abc.
\end{eqnarray*}
Eliminate $b+c$ and divide by $2a$
\begin{eqnarray*}
10a=9abc. \\
bc=6.
\end{eqnarray*}
Similarly $ac=7$ and $ab=5$, now multiply these three equations gives $a^2b^2c^2=1$.
Now square root this gives $abc=1$ or $abc=10$.
|
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|
Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$ So,
I know that $\langle a\rangle$ = $aR$ and $\langle a,b\rangle$ = $aR + bR$ for any ring R.
I then multiplied each side by $\sqrt2$ and computed them.
These are the steps that I took:
$\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$
$(3+8\sqrt2)\sqrt2 + 7\sqrt2 = (3+\sqrt2)\sqrt2$
$3\sqrt2 + 16 + 7\sqrt2 = 3\sqrt2 + 2$
$16 + 10\sqrt2 = 2+3\sqrt2$
I'm not really sure where to go from here.
Any help would be much appreciated.
|
Note that since $1,-\sqrt{2} \in\mathbb{Z}[\sqrt{2}],$
$(3+8\sqrt{2})\cdot1-7\cdot\sqrt{2}=3+\sqrt{2}\in\langle 3+8\sqrt{2},7\rangle.$
Since $\forall r\in\mathbb{Z},1\cdot r\in\mathbb{Z}$ and $-\sqrt{2}\cdot r\in\mathbb{Z},$ we have that $\forall r\in\mathbb{Z},(3+\sqrt{2})r\in\langle 3+8\sqrt{2},7\rangle\\
\Rightarrow \langle 3+\sqrt{2}\rangle\in\langle3+8\sqrt{2},7\rangle.$
Divide $3+8\sqrt{2}$ by $3+\sqrt{2}$ to get a quotient of $-1+3\sqrt{2}\in\mathbb{Z}[\sqrt{2}].$
Similarly, the quotient of $\dfrac{7}{3+\sqrt{2}}$ is $3-\sqrt{2}\in\mathbb{Z}[\sqrt{2}].$ So we have that $\forall r\in \langle3+8\sqrt{2},7\rangle,r\in\langle3+\sqrt{2}\rangle\\
\Rightarrow \langle 3+8\sqrt{2},7\rangle\subseteq \langle 3+\sqrt{2}\rangle.$
This is because $3+8\sqrt{2}$ and $7$ are both multiples of $3+\sqrt{2}$ in $\mathbb{Z}[\sqrt{2}].$
Thus, since $\langle 3+\sqrt{2}\rangle\subseteq\langle 3+8\sqrt{2},7\rangle$ and $\langle 3+8\sqrt{2},7\rangle\subseteq \langle 3+\sqrt{2}\rangle,$ we have that $\langle 3+8\sqrt{2},7\rangle = \langle 3+\sqrt{2}\rangle.$
Both conditions must be satisfied, or else they may not be equal. For instance, the ideal
$\langle 6+2\sqrt{2}\rangle\subseteq\langle3+\sqrt{2}\rangle$ but
$\langle3+\sqrt{2}\rangle \not\subseteq\langle6+2\sqrt{2}\rangle,$ so the ideals are unequal.
As well, $\langle 3+\sqrt{2}\rangle \subseteq \langle 1+\sqrt{2}\rangle$ but $\langle 1+\sqrt{2}\rangle\not\subseteq \langle3+\sqrt{2}\rangle.$
|
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|
Find all matrices $A\in \mathbb{R}^{2\times2}$ such that $A^2=\bf{0}$ Attempt:
Let's consider $A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$.
$$\begin{align}
A^2 &=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\cdot
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \\&=
\begin{bmatrix}
a\cdot a+b\cdot c & a\cdot b + b\cdot d \\
c\cdot a + d\cdot c & c\cdot b + d\cdot d
\end{bmatrix}\\ &= \bf{0}
\end{align}$$
This gives us the system of equations:
$$\begin{align}
a\cdot a+b\cdot c &= 0 \tag{1}\\
a\cdot b + b\cdot d = b\cdot(a+d)&= 0 \tag{2}\\
c\cdot a + d\cdot c = c\cdot(a+d)&= 0 \tag{3}\\
c\cdot b + d\cdot d &= 0 \tag{4}
\end{align}$$
Now from equation $(2)$ and $(3)$, we have eight cases:
*
*$b = 0$
*$c = 0$
*$a+d = 0$
and 5 combinations of (1,2,3) which I won't bother listing.
Case 1 ($b=0$):
$b=0$ implies $a = 0$ in equation $(1)$ and $d = 0$ in equation $(4)$. This means that if $A = \begin{bmatrix}0&0\\c&0\end{bmatrix}$ then $A^2=\bf{0}$.
Case 2 ($c=0$):
From symmetry with $b$, $c=0 \implies A=\begin{bmatrix}0&b\\0&0\end{bmatrix}$.
So, we only consider cases where $b\neq0$ and $c\neq 0 $ which leaves us only with case 3 ($a+d=0$).
Case 3 ($a+d=0$):
In equation (1), $a+d=0 \implies a\cdot d - b\cdot c = 0$. So $A=\begin{bmatrix}a&b\\c&-a\end{bmatrix}$ is not invertible, $A^2 = \bf{0}$.
In summary, if $A$ has one of the following forms:
$$\begin{bmatrix}0&b\\0&0\end{bmatrix},
\begin{bmatrix}0&0\\c&0\end{bmatrix},
\begin{bmatrix}a&b\\c&-a\end{bmatrix} \text{ (and not invertible) }$$
then $A^2=\bf{0}$.
Questions:
*
*Is this a correct proof?
*What is the standard proof?
"Strange question":
*
*How can I know if there was only the 8 cases? As in, how do I know that only these 8 cases are relevant to $A^2 = \bf{0}$?
|
Your proof is correct, although rather naive---in the sense that you've converted the matrix equation into a system of simultaneous equations and solved it. There's no "standard" proof per se, but there are many more conceptual ways to prove it.
One example is the eigenvalue argument provided by Bungo in the comments. Another way is to note that if $A^2=0$, then the minimal polynomial of $A$ must be $x^2$ (except the trivial case $A=0$), and since the minimal polynomial of a matrix divides its characteristic polynomial, then the characteristic polynomial of $A$ must be $kx^2$ for a constant $k$. It is easy to deduce all the possible forms of $A$ from here on.
The advantage of these more "high-level" arguments is that they generalise easily, to e.g. higher dimensional vector spaces.
|
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|
The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because $\alpha=3 > 1$. Is this true?
*To calculate the integral, using the integration by parts where $u = \ln x$ and $dv = \frac{x \ln x}{(x^2+1)^2} dx$.
So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}- \frac{1}{4x^2} +\frac{\ln |x|}{2} ~\Big|_{0}^{\infty} $$
and this undefined while it should converge to $0$ ! what I missed?
I found an error in the calculation so the integral = So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}+ \frac{1}{8} \Big( \ln x^2 - \ln (x^2+1) \Big) ~\Big|_{0}^{\infty} $$
and it's still undefined!
|
Yes, the improper integral is convergent, but your computation is not correct. Note that
$$\begin{align}
\int \frac{x \ln(x)}{(x^2+1)^2} dx& = -\frac{\ln(x)}{2(x^2+1)}+ \int\frac{1}{2x(x^2+1)}dx\\
&=
-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx\\
&=-\frac{\ln(x)}{2(x^2+1)}+\frac{\ln(x)}{2}-\frac{\ln(1+x^2)}{4}+c\\
&=\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}+c.
\end{align}$$
which can be extended by continuity in $[0,+\infty)$.
Therefore
$$\int_0^{+\infty} \frac{x \ln(x)}{(x^2+1)^2} dx=
\left[\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}\right]_0^{+\infty}=0-0=0.$$
P.S. As regards the limit as $x\to +\infty$, note that
$$\begin{align}\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}&=-\frac{\ln(x)}{2(x^2+1)}
+\frac{ \ln(x^2) - \ln (x^2+1)}{4} \\&=-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\ln\left(\frac{x^2}{x^2+1}\right)\to 0+\ln(1)=0.\end{align}$$
|
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|
How to differentiate between external and internal angle bisectors of a triangle? I came up with this question when I was trying to figure out the coordinates of the incenter of a triangle with equations:
$4x-3y=0$, $3x-4y+12=0$, $3x+4y+2=0$.
I assumed the coordinates of the incenter to be $(h,k)$ and equated the perpendicular distances from all the sides and got:
$4h-3k=\pm(3h-4k+12)$, $3h-4k+12=\pm(3h+4k+2)$, $3h+4k+2=\pm(4h-3k) $
But I didn't know which sign to take. As different signs would mean different angle bisectors(external or internal).
Image showing intersection of Angle bisectors(posted as a link due to low reputation restrictions)
I found this Can we find incentre of a triangle by using equation of lines?, answering which bisector to take but I couldn't understand the method in the case when I do not have the vertices.
P.S. I want to know the incenter without knowing the vertices. Also I am curious why the above method works, intuitively. Thanks.
|
$4·x-3·y=0$
$3·x-4·y+12=0$
$3·x+4·y+2=0$
multiply the equations by the signs of the cofactors of the constants: 0, 12 and 2.
$M=\left [ \begin{array}{} 4 & -3 & 0 \\ 3 & -4 & 12 \\ 3 & 4 & 2\\ \end{array} \right ] $
$cofactor{M_{13}}=\left|\begin{array}{} 3 & -4 \\ 3 & 4 \\ \end{array} \right| =24$
$cofactor{M_{23}}=-\left|\begin{array}{} 4 & -3 \\ 3 & 4 \\ \end{array} \right| =-25$
$cofactor{M_{33}}=\left|\begin{array}{} 4 & -3 \\ 3 & -4 \\ \end{array} \right| =-7$
sign(24)=1, sign(-25)=-1 and sign(-7)=-1
$1·(4·x-3·y)=0⇒4·x-3·y=0$
$-1·(3·x-4·y+12)=0⇒-3·x+4·y-12=0$
$-1·(3·x+4·y+2)=0⇒-3·x-4·y-2=0$
with these "oriented" equations we obtain the bisectors
(the three generate internal bisectors)
$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$
$\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$
$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}$
$7·x+y+2=0$
$4·y-5=0$
$7·x-7·y+12=0$
bisectorInternal
|
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|
Prove that the number of different ways to write an integer X as a sum of positive integers is $2^z$, if the order matters. As an example for number 5, it is equal to $2^4$.
We can prove it by brute-forcing it, since it's a small number, as follows:
$1+1+1+1+1 $
$1+1+1+2 $
$1+1+2+1 $
$1+2+1+1 $
$2+1+1+1$
$1+1+3 $
$1+3+1$
$3+1+1$
$2+2+1$
$2+1+2$
$1+2+2$
$2+3$
$3+2$
$1+4$
$4+1$
$5$
What is a faster, more elegant way to do this?
|
Let $f(n)$ be the number you're looking for for a given $n$. Let's work by strong induction.
Consider $f(n+1)$. It could either have one term (which happens in one way) or more than one term. In this second case, the second term can be any number from $1$ through $n$. If the first term is $k$, then there are $f(n+1-k)$ ways to complete that sum. Adding all that together with our induction hypothesis that $f(k)=2^{k-1}$, we get
$$f(k+1)=1+2^0+2^1+...+2^{k-1}$$
which is obviously $f(k+1)=2^k$.
|
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how to find $\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$ how to find
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$$ without L'hopital or taylor/Laurent series
I tried but did not get any answer:
$$\frac{\tan^{2}\left(x\right)\tan\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\sin\left(x\right)\cos^{2}\left(x\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\left(\sin\left(x\right)\left(1-\sin^{2}\left(x\right)\right)\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2}{1}\frac{-3\left(\sin(x)-\sin^{2}\left(x\right)\right)+\sin^{3}\left(x\right)}{\left(1-\sin^{2}\left(x\right)\right)(\cos(x))\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}$$
|
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=$$
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan x (\tan x - \sqrt 3)(\tan x +\sqrt 3)}{\cos x \cos (\pi /6)-\sin x \sin (\pi /6)}=$$
$$\lim_{\large x \to \frac{\pi}{3}}\frac{(2\sec x)\tan x (\tan x - \sqrt 3)(\tan x +\sqrt 3)}{\sqrt 3 - \tan x}=-24$$
|
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|
Find all pairs of prime numbers $p$ and $q$ Find all pairs of prime numbers $p$ and $q$ such that $p^3 - q^5 = (p + q)^2$. Present your answers as an ordered pair $(p, q)$.
Is this resolution correct?
It suffices to show that one of $p$, $q$ is divisible by 3, then the rest of the problem collapses.
Assume otherwise. Observe that for any $n$, $n^3 \equiv n \pmod{3}$. So:
$$p^3-q^5 \equiv p-q \equiv p^2+2pq + q^2 \pmod{3}$$Since $p$, $q$ are not divisible by 3, they cannot have residue 0 when squares, so they must both have residue 1 when squared. We have:
$$p-q \equiv 2+2pq$$$$2p-2q \equiv 1+pq$$$$q-p \equiv 1+pq$$$$pq+p-q-1 \equiv 1$$$$(p-1)(q+1) \equiv 1$$Neither $p-1$, $q+1$ can be 0 mod 3, that would be absurd. They are either both congruent to $2$ mod 3, or both congruent to $1$ mod 3. Either case will lead to one of $p$, $q$ being divisible by 3, contradiction.
|
That looks correct to me. You can also establish the result by looking at $(p+q) \pmod{3}$. As you said, $p^3-q^5\equiv p-q\equiv (p+q)^2\pmod{3}$.
Case 1. $p+q\equiv 0\pmod{3}$. Then you have $3\mid p-q$ and $3\mid p+q$, which gives $3\mid p,q$.
Case 2. $p+q\equiv \pm1\pmod{3}$. Then $(p+q)^2\equiv 1\equiv p-q \pmod{3}$. Subtracting or adding (depending on $\pm$) the congruences $p+q\equiv \pm1\pmod{3}$ and $p-q\equiv 1 \pmod{3}$ will produce $3\mid p$ or $3\mid q$.
So the only solution is $(p,q)=(7,3)$.
|
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How to study this sequence $u_n=\sum_{k=1}^{n}\frac{1}{n+2k}$ Please is there any way to prove that sequence is increasing ?
I do: $u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}=\left[\frac{1}{n+3}+\frac{1}{n+5}+\ldots+\frac{1}{3n+1}+\frac{1}{3n+3}\right]-\left[\frac{1}{n+2}+\frac{1}{n+4}+\ldots+\frac{1}{3n}\right]$
i don't know how to continue
|
OP, a good start:
$$u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}$$
Let's continue. First,
$$ \sum_{k=1}^n
\frac 1{(n+2\!\cdot\! k-\frac 12)
\cdot(n+2\!\cdot\!k+\frac 32)}\,
\, =\,\, \frac 12\cdot\left(
\, \frac 1{n+2-\frac 12}\, -\, \frac 1{3\cdot n+\frac 32}
\right) $$
$$ =\, \frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3} $$
Next,
$$ u_{n+1}-u_n\quad =\quad \frac 1{3n+3}\,\, -\,
\, \sum_{k=1}^n\frac 1{(n+2\cdot k)
\cdot(n+2\!\cdot\!k+1)} $$
$$ =\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\, + $$
$$ \sum_{k=1}^n
\frac 1{(n+2\!\cdot\! k-\frac 12)
\cdot(n+2\!\cdot\!k+\frac 32)}
\, -\, \sum_{k=1}^n\frac 1{(n+2\cdot k)
\cdot(n+2\!\cdot\!k+1)} $$
$$ =\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\, + $$
$$ \frac 34\cdot\sum_{k=1}^n\frac 1
{(n+2\!\cdot\! k-\frac 12)\cdot(n+2\!\cdot\!k+\frac 32)
\cdot(n+2\cdot k)\cdot(n+2\!\cdot\!k+1)} $$
$$ >\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\ =
$$ $$ \frac{(2\!\cdot\! n+1)\!\cdot\!(2\!\cdot\! n+3)
+ (n+1)\!\cdot\!(2\!\cdot\! n+3) -
3\!\cdot\!(n+1)\!\cdot\!(2\!\cdot\! n+1)}
{3\cdot(n+1)\cdot(2\cdot n+1)\cdot(2\cdot n+3)}
$$ $$ =\, \frac{4\cdot n+3}
{3\cdot(n+1)\cdot(2\cdot n+1)\cdot(2\cdot n+3)} $$
This means that the following theorem holds,
Theorem
$$ \forall_{n=1\, 2\, \ldots}\quad u_n<u_{n+1} $$
Great!
|
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|
Find all the possible values of gcd$(n^2 +2 , n^4 +4)$ if $n$ is a natural number? Is this a complete “proof”? Im new to number theory so it might be wrong. So, $$n^4 +4 = (n^2 +2+2n)(n^2 +2 -2n)\ ,$$ so if $d$ divides $n^2 +2 +2n$ and $n^2 +2$, then $d | 2n$. , so $d|2$ or $d|n$.
Suppose $d| n$. We know $d | (n^2 +2)$ so $d| 2$ too, therefore $d=$gcd$(n , 2)$ , so $d=1$ (if $n$ odd) or $2$(if $n$ even) . (If we work with the $n^2 +2 -2n$ we get the same result).
If you have different solutions to this, please write them down! Thanks
|
We have $n^4+4 = (n^2+2)(n^2 - 2) + 8$. If $n$ is odd, then $gcd(n^2+2, 8) = 1$, and if $n$ is even then $n^2+2 = 4k^2 + 2$ is not divisible by $4$ or $8$ so $gcd(n^2+2,8) = 2$. This gives your original calculation (via a Euclidean Algorithm argument)
|
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|
Help Understanding Proof for Inequality Problem #3 from Problem Solving Strategies The problem is the following:
Prove for $a$, $b$, $c$, $d$ that
$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$
I understand the proof saying
$${\frac{abc+abd+acd+bcd}{4}}=\frac{(ab)(c+d)+(cd)(a+b)}{4}$$
Apply AM-GM to $ab$ and $cd$ to yield
$$\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}=\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}$$
However it then states the following:
$$\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}=\left(\frac{a+b+c+d}{4}\right)^3$$
I don't understand this step in the proof. Unless I'm missing something basic, the above expression does not factor as shown. It doesn't adduce a theorem to justify the equivalence either so I don't know how it was deduced.
|
Because by AM-GM and C-S we obtain:
$$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}=\sqrt[3]{\frac{ab(c+d)+cd(a+b)}{4}}\leq$$
$$\leq\sqrt[3]{\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}}=\sqrt[3]{\frac{a+b}{2}\cdot\frac{c+d}{2}\cdot\frac{a+b+c+d}{4}}\leq$$
$$\leq\sqrt[3]{\left(\frac{\frac{a+b}{2}+\frac{c+d}{2}+\frac{a+b+c+d}{4}}{3}\right)^3}=\frac{a+b+c+d}{4}\leq$$
$$\leq\frac{\sqrt{(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)}}{4}=\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$
|
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|
Integration based on beta function
I am getting the value of first integration using the beta function, but I am not able to find the second one. Please help
|
$\because$ $\int_0^1\dfrac{x^{a-1}(1-x)^{b-1}}{(z+x)^{a+b}}~dx=\dfrac{B(a,b)}{(z+1)^az^b}$ as https://dlmf.nist.gov/5.12 stated,
$\therefore$ $\int_0^1\dfrac{x^\frac{5}{2}(1-x)^\frac{7}{2}}{(3+x)^8}~dx=\int_0^1\dfrac{x^{\frac{7}{2}-1}(1-x)^{\frac{9}{2}-1}}{(3+x)^{\frac{7}{2}+\frac{9}{2}}}~dx=\dfrac{B\left(\frac{7}{2},\frac{9}{2}\right)}{4^\frac{7}{2}3^\frac{9}{2}}$
|
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How do I prove $1+x+x^2+\cdots+x^{n-1}=\frac{1-x^n}{1-x}$ using the induction method? I am having difficulties in solving the expression $1+x+x^2+\cdots+x^{n-1} =\frac{1-x^n}{1-x}$, $n\in \mathbb{N}$, using the induction method. How is the inductive step taken in this case?
|
Assume that $1+x + \dots + x^{n-1} = \frac{1-x^n}{1-x}$.
Then
$$1+x+ \dots + x^{n} = (1+ x + \dots + x^{n-1}) + x^n = \frac{1-x^n}{1-x} + x^n$$
$$= \frac{1-x^n}{1-x} + \frac{x^n(1-x)}{1-x} = \frac{1-x^n + x^n - x^{n+1}}{1-x}= \frac{1-x^{n+1}}{1-x}$$
|
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Which satisfy the system of inequalities below: For each nonnegative integer $n$, calculate the number of triples $(a, b, c)$ of nonnegative integers
which satisfy the system of inequalities below:
$ \begin{cases}
a + b \leq 2n \\
a + c \leq 2n \\
c + b \leq 2n \\
\end{cases}$
What I thought: We can solve this by plotting the inequalities with the bounds $x,y,z\geq0$ and getting that all such $(a,b,c)$ are lattice points bounded by the axis and $x+y+z=2n$.
|
Consider the triples $ (a, b, c) $ of nonnegative integers that satisfy the following relationships:
$$\begin{cases} \, a + b = x \\ a + c \leq 2n \\ b + c \leq 2n \end{cases} $$Let's divide the count of these triples into two steps. In the first we consider $ x = 2k $, while in the second we consider $ x = 2k + 1 $.
Thus, the $ (a, b) $ pairs that satisfy $ a + b = 2k $ are $ (2k, 0), ..., (k, k),..., (0.2k) $.
For each of these pairs, $ c $ will have to obey $ c \leq 2n - M $ where $ M = \max (a, b) $, which gives us $ 2n - M + 1 $ solutions.
Assuming that among the $ (a, b) $ pairs that satisfy the equation, the value of $ M $ ranges from $ k + 1 $ to $ 2k $ twice and then goes to $ k $, so the number of solutions for this case it will be:
$$\left(2\sum_{M=k+1}^{2k} (2n-M+1)\right)+ 2n - k + 1 $$$$= -3k^2+4nk+2n+1 $$For the second case, be $ n = 2k + 1 $. The difference is that we will not have the extra solution in which the components of pair $ (a, b) $ are equal. Like this:
$$\left(2 \sum_{M=k+1}^{2k+1}(2k+1-M+1)\right) $$$$= -3k^2+(4n-3)k+4n $$So just calculate the sum of all cases where $ a + b = 0 $, $ a + b = 1 $,… up to $ a + b = 2n $ and just get the sums for when $ 2k + 1 = 1, 3, 5, \dots, 2n-1 $ and sum with the sums for when $ 2k = 0, 2, 4, \dots, 2n $:
$$\sum_{k=0}^n 3k^2+4nk+2n+1 + \sum_{k=0}^{n-1} -3k^2+(4n-3)k+4n $$$$= 2n^3 + \frac{9n^2}{2} + \frac{11n}{2} + 1 $$
|
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|
Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} \equiv 9^{5} \bmod 11 = 9^{4} 9 \bmod 11, \:\: 9^{2} \equiv 4 \bmod 11 $$
$$ \implies 9^{5} \equiv 144 \bmod 11 \implies 20^{5} \equiv 1 \bmod 11 $$
Then the proof is done.
Now I will prove:
$$ 20^{15} \equiv 1 \bmod 31 $$
Notice $20^{2} \equiv 28 \bmod 31$, so
$$20 \times (20^{2})^{7} \equiv 20 \times (28)^{7} \bmod 31 \equiv 20 \times (-3)^{7} \bmod 31 \equiv -60 \times 16 \bmod 31\equiv 32 \bmod 31 $$
then the proof is done.
Also, in similar way to prove the $20^{15} \equiv 1 \bmod 61$.
Are there shorter or more efficient proof?
|
Alternatively:
$$\left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right)=(20^5-1)(20^{10}+20^5+1)\\
20^5\equiv (-2)^5\equiv -32\equiv 1 \pmod{11}\\
20^5\equiv 81^5\equiv 3^{20}\equiv (3^5)^4 \equiv1 \pmod{61}\\
20^5 \equiv (2^{5})^2\cdot (5^2)^2\cdot 5\equiv 1^2\cdot (-6)^2\cdot 5\equiv 25\pmod{31}\\
20^{10}\equiv (20^5)^2\equiv 25^2 \equiv (-6)^2\equiv 5\pmod{31}\\
20^{10}+20^5+1\equiv 5+25+1\equiv 0\pmod{31}$$
Note: The idea is to use the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ and decrease the exponent.
|
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|
Absolute difference between the first and second half of a random vector Suppose I have a zero vector of length $2n$, and I randomly choose $k$ of the entries of this vector to be $1$. The choice is taken uniformly from all $\binom{2n}{k}$ possibilities. Now let $S_1$ denote the sum of the first half of the vector, and $S_2$ denote the sum of the second half. I want to know the percentage of vectors such that the difference between $S_1$ and $S_2$ is nontrivial. More precisely, I want to compute the probability that $P(\frac{|S_1-S_2|}{n} > \epsilon)$ as $n$ and $k$ become large.
I'm almost sure that there is an established result somewhere, but since I'm not familiar with combinatorics, I don't know how to search for it.
|
TL;DR: I will show in this answer that
$$
\mathbb{P}\left(\frac{|S_1 - S_2|}{n} > \epsilon\right)
\le \frac{1}{\varepsilon^2 (2n-1)},
$$
independent of $k$ (better bounds can be given if $k$ is close to $1$ or $2n$).
In other words, the probability you ask for is bounded above by roughly $1 / (2 \varepsilon^2 n).$ In particular, it tends to $0$ as $n \to \infty$.
Markov's inequality is useful here. We consider the random variable $S_1 - S_2$ which is the sum of the first $n$ entries of the vector, minus the last $n$; that is, if the vector is $v$,
$$
S_1 - S_2 = (v_1 + v_2 + \cdots + v_n) - (v_{n+1} + v_{n+2} + \cdots + v_{2n}).
$$
We are interested in $\mathbb{P}\left(|S_1 - S_2| > n \epsilon\right)$.
But it will be more convenient to work with $(S_1 - S_2)^2$, so we note
$$
\mathbb{P}\left(|S_1 - S_2| > n \epsilon\right)
= \mathbb{P}\left((S_1 - S_2)^2 > n^2 \epsilon^2\right).
$$
We can then apply Markov's inequality as follows:
$$
\mathbb{P}\left((S_1 - S_2)^2 > n^2 \epsilon^2\right)
\le \frac{\mathbb{E}((S_1 - S_2)^2)}{n^2 \epsilon^2}.
$$
Now, observe that $S_1 + S_2 = k$, so $S_2 = k - S_1$. So we get
$$
= \frac{\mathbb{E}((2 S_1 - k)^2)}{n^2 \epsilon^2}.
$$
Now the expected value of $S_1$ is $\frac{k}{2}$, so the denominator is related to the variance. We get
$$
\mathbb{P}\left(|S_1 - S_2| > n \epsilon\right) \le \frac{4 \cdot \text{Var}(S_1)}{n^2 \varepsilon^2}. \tag{1}
$$
So the question becomes, what is the variance of $S_1$?
We can use this formula:
$$
\text{Var}(S_1)
= \text{Var}(v_1 + v_2 + \cdots + v_n)
= \sum_{i=1}^n \text{Var}(v_i) + \sum_{i \ne j} \text{Cov}(v_i, v_j).
$$
An individual $v_i$ is Bernoulli where the probability of being $1$ is $\frac{k}{2n}$, so the variance is $\frac{k}{2n} \left(1 - \frac{k}{2n}\right)$.
As for the covariance between $v_i$ and $v_j$, there are four possibilities for the pair of variables, and the covariance is calculated as
\begin{align*}
\text{Cov}(v_i, v_j)
= &\frac{k}{2n} \cdot \frac{k-1}{2n-1} \cdot \left(\frac{2n - k}{2n}\right)^2 \\
&+ 2 \cdot \frac{k}{2n} \cdot \frac{2n - k}{2n-1} \cdot \left( -\frac{k}{2n} \right) \left( \frac{2n - k}{2n} \right) \\
&+ \frac{2n - k}{2n} \cdot \frac{2n - k - 1}{2n-1} \cdot \left( -\frac{k}{2n} \right)^2 \\
&= \frac{k(2n-k)}{(2n)^3 (2n - 1)}
\left[ (k-1)(2n-k) - 2k(2n-k) + k(2n-k-1) \right] \\
&= \frac{k(2n-k)}{(2n)^3 (2n - 1)} \left[ -(2n- k) - k\right] \\
&= \frac{k(2n-k)}{(2n)^3 (2n - 1)} \cdot (-2n) \\
&= -\frac{k(2n-k)}{(2n)^2 (2n-1)}.
\end{align*}
Therefore,
\begin{align*}
\text{Var}(S_1)
&= n \cdot \frac{k}{2n} \left(1 - \frac{k}{2n}\right) + n(n-1) \cdot -\frac{k(2n-k)}{(2n)^2 (2n-1)} \\
&= \frac{n k (2n - k)}{(2n)^2} \left[ 1 - \frac{(n-1)}{2n-1}\right] \\
&= \frac{n^2 k (2n - k) }{(2n)^2 (2n-1)} \\
&= \frac{k(2n-k)}{4 (2n-1)}.
\end{align*}
Finally, we can return to (1). We get that
\begin{align*}
\mathbb{P}\left(|S_1 - S_2| > n \epsilon\right)
&\le \frac{4}{n^2 \varepsilon^2} \cdot \frac{k(2n-k)}{4 (2n-1)} \\
&= \frac{1}{\varepsilon^2} \cdot k (2n - k) \cdot \frac{1}{n^2 (2n-1)} \\
&\le \frac{1}{\varepsilon^2} \cdot n^2 \cdot \frac{1}{n^2 (2n-1)} \\
&= \frac{1}{\varepsilon^2} \cdot \frac{1}{2n-1}.
\end{align*}
Here, we used the fact that $k (2n - k)$ is maximized when $k = n$; better bounds can be given if $k$ is small (close to $1$) or large (close to $2n$).
In summary,
$$
\boxed{\mathbb{P}\left(|S_1 - S_2| > n \epsilon\right)
\le \frac{1}{\varepsilon^2 (2n-1)}.}
$$
We can make some concrete observations:
*
*This bound is independent of $k$.
*If $\varepsilon$ is fixed, the probability goes to $0$ as $n \to \infty$ (regardless of whether $k$ is held constant or also increasing).
|
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|
Evaluating $ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $
$$ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $$
I tried to do this with some kind of substitution but failed miserably. Any hints or help?
|
I nice approach using triangles. Use the following figure, where $\triangle ABC$ is right-angled with $\overline{BC} = x>0$ and $\overline{AB} = 1+x$. Take $D$ on $AB$ so that $\overline{AD} = 1$ and $E$ on $AC$ so that $DE\perp AC$.
By definition
$$\angle CAB = \arctan\left(\frac{x}{x+1}\right),$$
and by the External Angle Theorem
$$\angle ACD = \frac{\pi}4-\angle CAB.$$
Note that $\overline{AC} = \sqrt{2x^2+2x+1}$, and use the fact that $\triangle ADE \sim \triangle ABC$ to conclude that
$$\overline{DE} = \frac{x}{\sqrt{2x^2+2x+1}},$$
and
$$\overline{AE} = \frac{x+1}{\sqrt{2x^2+2x+1}},$$
so that
$$\overline{CE} = \frac{2x^2+x}{\sqrt{2x^2+2x+1}},$$
and
$$\angle ACD = \arctan\frac{\overline{DE}}{\overline{CE}}=\arctan\left(\frac{1}{2x+1}\right).$$
So, as per the other answers, your limit is equal to
$$\lim_{x\to +\infty} x\arctan\left(\frac1{2x+1}\right) = \frac12.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate
\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}
We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{x^2 + x}{\ln(1 + x + x^2) - x} \cdot \frac{1}{\sqrt{x^2 + x + 4} + 2} \quad \text{for all}\ x \in \mathbb{R},
\end{equation*}
so I think my exercise boils down to
\begin{equation*}
\lim_{x \to 0^{-}} \frac{x^2 + x}{\ln(1 + x + x^2) - x},
\end{equation*}
and this limit equals $-\infty$ by De L'Hôpital's Theorem.
Can I evaluate this limit without De L'Hôpital's Theorem?
|
For small $x$ so $x^2+x$ is also small,$$\sqrt{x^2+x+4}-2=2\left(\sqrt{1+\frac14(x^2+x)}-1\right)\approx\frac14(x^2+x)\approx\frac14 x$$while$$\ln(1+x+x^2)-x\approx x(1+x)-\frac12 x^2(1+x)^2-x\approx\frac12 x^2,$$so the ratio $\approx\frac{1}{2x}\to-\infty$ as $x\to0^-$.
|
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|
Calculate $\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$ $$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$
I know this limit must be equal to $\frac{1}{2}$ but I can't figure why. This is just one of the thing I tried to solve this limit:
$$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$
$$\lim_{x\to\infty}x\biggr(\sqrt{\frac{x}{x-1}}-1\biggr)$$
Now I try to evaluate the limit. I know that $\lim_{x\to\infty}\sqrt\frac{x}{x-1}$ is equal to 1 so that means the above limit evaluates to $\infty * 0$ which is indeterminate form. I do not know what to do next, would greatly appreciate some help.
|
Her is an elementary way:
\begin{eqnarray*} \biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)
& = & x \biggr(\sqrt{\frac{x}{x-1}}-1\biggr) \\
& = & x \biggr(\frac{\frac{x}{x-1}-1}{\sqrt{\frac{x}{x-1}}+1}\biggr) \\
& = & x \biggr(\frac{1}{(x-1)\sqrt{\frac{x}{x-1}}+(x-1)}\biggr) \\
& = & \frac{1}{(1-\frac{1}{x})\sqrt{1+\frac{1}{x-1}}+1-\frac{1}{x}} \\
& \stackrel{x \to \infty}{\longrightarrow} & \frac{1}{(1-0)\sqrt{1+0}+1-0} = \frac{1}{2}
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$
My attempt is as follows:-
$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$
Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$
$$f(x)=\sqrt{13}z$$
Let's find the range of $z^2$
$$z^2=\dfrac{4}{13}\cdot\sin^2x+\dfrac{9}{13}\cdot\cos^2x-\dfrac{6}
{13}|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot[0,1]$$
$$z^2=1-\left[0,\dfrac{6}{13}\right]$$
$$z^2=\left[\dfrac{7}{13},1\right]$$
As $0$ is not there in the range of $z^2$, so we can say $z\in \left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]$
Hence $y \in \sqrt{13}\left(\left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]\right)$
$$y\in[-\sqrt{13},-\sqrt{7}]\cup[\sqrt{7},\sqrt{13}]$$
But actual answer is $[-3,2]$
|
Your domain is 0 to $\frac{\pi}{2}$. That's where sine and cosine are positive. So at those points you are limited to the range of $[-3,2]$.
Unless you are given additional instructions, you are way over complicating it.
|
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|
Find the unit digit of $2124^{392}+3143^{394}*7177^{392}-8818^{394}$ What I do here is I find the remainder of each parcel in mod 10 and then do the maths and mod 10 again:
$2124^{392} \equiv 4^{392} \pmod{10}$
$$4^1 \equiv 4 \pmod{10} \\
4^2 \equiv 6 \\
4^3 \equiv 4 \\
4^4 \equiv 6 \\
(...)$$
Then I do $392 \pmod{2}$. Problem is that here, it's zero. If it were 1 or 2, the remainder would be either 4 or 6. How do I solve it this way?
|
Using Newton's binomial formula,
$$4^{392}=(10-6)^{392} $$
$$=6^{392} \mod 10 =6 \mod 10$$
Since $$6^2 = 6\mod 10$$
For the second and the third , $ 3$ and $ 7$ are coprime with $ 10$ so by Euler's Theorem
$$3^{\phi(10)}=1 \mod(10) =3^4$$
and
$$7^4 = 1 \mod 10$$
For example
$$3^{394}=3^{98.4+2}=9 \mod 10$$
For the last one, observe that
$$2^5 = 2 \mod 10$$
then
$$8^5= 2^{3.5}=2^3=8 \mod 10$$
thus
$$8^{394}=8^{78.5+4}=8^{78+4}=8^{16.5+2}=8^{16+2}=8^{3.5+3}=8^6$$
$$=8^{5+1}=8^2=4 \mod 10$$
|
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|
Tough integral from spherical coordinates One of my friend's students attempted to find the volume of a paraboloid using spherical coordinates. I'm trying to see if there's actually a way to finish it. I have been able to find the triple integral which describes the volume of the paraboloid in spherical coordinates and the part that I need help integrating is:
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx.$$
I know that it is equal to $4$ from Wolfram Alpha plus computing the same integral in cylindrical coordinates. Is there a way to finish this from here?
Someone asked for the original problem so here it is. Compute the volume of the region bounded by $z = 1 + x^2 + y^2$ and $z \leq 5$. This is the same as computing the region bounded by $z = 4 - x^2 - y^2$ and the $xy$-plane. Then from there I get the triple integral:
$$\int_{0}^{2\pi}\int_0^{\pi/2}\int_0^{\frac{-\cos{\phi} + \sqrt{\cos^2{(\phi)} + 16\sin^2{(\phi)}}}{2\sin^2{(\phi)}}} \rho^2\sin{\phi}d\rho d\phi d\theta.$$
Of course the integral up top that I care about I just substituted the $\phi$ for an $x$ when I originally stated it by itself.
|
Let $t=\cos x$, we have
\begin{align}
I&=\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx \\
&= \frac{1}3\int_0^1 dt \left(\frac{-t+\sqrt{t^2+16(1-t^2)}}{2(1-t^2)}\right)^3
\end{align}
Let $z$ be the larger root of the equation $(1-t^2)z^2+tz-4=0$, we have $z=2$ at $t=0$ and $z=4$ at $t=1$. Let $y=tz$, we have $z^2=y^2-y+4\Rightarrow zdz=(y-1/2)dy$, $y=0$ at $t=0$ and $y=4$ at $t=1$, then
\begin{align}
I&=\frac{1}3\int_0^1 dt z(t)^3 \\
&=\frac{1}3\left(z(t)^3t|_0^1-3\int_2^4t(z)z^2dz\right) \\
&=\frac{64}{3}-\int_0^4y\left(y-\frac{1}{2}\right)dy \\
&=\frac{64}{3}-\frac{64}{3}+\frac{16}{4}=4
\end{align}
Also a simple solution to your original problem:
area $A(z)=\pi r(z)^2=\pi(x^2+y^2)=\pi(z-1)$,
\begin{align}
V&=\int A(z) dz \\
&=\int_1^5\pi(z-1)dz \\
&=8\pi
\end{align}
|
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|
If $x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$ What is the value of $5x^2-5x-1?$ If $$x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$$ What is the value of $5x^2-5x-1?$
Efforts:
After rationalization, I got $x={\sqrt{5}+1\over 2}$ and $x^2={\sqrt{5}+1\over \sqrt{5}-1}$. Going by this method is very tedious and boring.
Is there a more clever and imaginative way to approach this problem?
Thanks.
|
You already have $x=\frac{\sqrt5+1}{2}$. Then $x^2=\frac{\sqrt5+3}{2}$ and so $x^2=x+1$.
Therefore $5x^2-5x-1=5x+5-5x-1=4.$
|
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|
Approximate \cos(x^2) using Composite Trapezoidal method Find an approximation to $$\int_2^3 \cos(x^2)dx$$ using the Composite Trapezoidal method with $r = 3$ and estimate the error.
Attempt: using the composite trapezoidal formula I calculated the approximation was equal to $$\frac 16 [\cos(4) + \cos(9) + \cos(16) + \cos(9) $$ which is approximately -0.5723. However when calculating the error I am not sure how to find the maximum of the second derivative, $ -4x^2\cos(x^2) - 2\sin(x^2) $ and I am stuck here. Have I gone about this correctly so far, and how do I find the error?
|
Assuming your "Composite Trapezoidal method" is my Trapezoidal rule, then you've made an error in applying the approximation.
With a uniform $$\Delta x = \frac{b-a}{n} = \frac{3-2}{3} = \frac{1}{3},$$
the trapezoidal rule gives
$$\begin{align*}\int_2^3 \cos\left(x^2\right)\,dx &\approx \frac{1}{6}\Big(\cos\left(2^2\right) + 2\cos\left((7/3)^2\right) + 2\cos\left((8/3)^2\right) + \cos\left(3^2\right)\Big) \\ &\approx 0.187472628\end{align*}$$
The error, as you've mentioned, is related to the second derivative
$$\frac{d^2}{dx^2} (\cos(x^2)) = \frac{d}{dx}(-2x\sin(x^2)) = -4x^2\cos(x^2) - 2\sin(x^2)$$
In particular, if we want to merely find an upper bound for the absolute value of the second derivative (which is what it sounds like you want), we can note
$$\left|-4x^2\cos(x^2) - 2\sin(x^2)\right| \leq 4x^2|\cos(x^2)| + 2|\sin(x^2)| \leq 4x^2 + 2 \leq 4(3^2)+2 = 38$$ (the actual maximum is $\approx 25.8$ at $x=3$)
Using our estimate, we then estimate the error as
$$\text{|Error|} \leq \frac{(b-a)^3}{12n^2}\cdot 38 = \frac{38}{108} \approx 0.351851852$$
(using the actual maximum of the second derivative updates this upper limit on the error to $0.239,$ and using the error formula on each subinterval could further refine the error estimate)
|
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|
prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
|
Use parity to deduce it is even. Or explicitly $\ \underbrace{105(\overbrace{2i\!+\!1}^{\textstyle 5^{N}}) - 15 (\overbrace{2j\!+\!1}^{\textstyle 3^{N}})}_{\textstyle 21\times 5^{N+1} - 5\times 3^{N+1}}\, =\, \overbrace{210 i - 30j + 90}^{\textstyle \color{#c00}{30}\,(7i -j + 3)}$
|
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|
Evaluating $\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$ for $p \in [0,1]$ I am trying to evaluate the following sum $$\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$$ for $p \in [0,1]$. This looks somewhat like the binomial theorem, but I don't know how I would go about applying it, as the index of summation is $y$ and it's at the top of the binomial coefficient.
I evaluated the sum using Mathematica, and I got $-\frac{(1-p)^{-a}}{p-1}$ which does make it seem like it's been obtained using the binomial theorem, but I am unable to find a way to use it.
Any help would be appreciated.
|
So we have
$$S = \binom{a}{a} + \binom{a+1}{a}p + \binom{a+2}{a}p^2 + \space ...$$
$$pS = \binom{a}{a}p + \binom{a+1}{a}p^2 + \binom{a+2}{a}p^3 + \space ...$$
subtracting, we obtain
$$(1-p)S = \binom{a}{a} + \binom{a}{a-1}p + \binom{a+1}{a-1}p^2 + \space ...$$
(I've used the identity $\binom{a}{a-1} + \binom{a}{a} = \binom{a+1}{a}$)
If we do the same with the above expression by taking $p(1-p)S$ and subtracting it, we obtain
$$(1-p)^2S = \binom{a}{a} + (\binom{a}{a-1} -\binom{a}{a})p + \binom{a}{a-2}p^2 + \space ...$$
notice that for $(1-p)^n$, the $(n+1)th$ binomial coefficient is reducing into $a$ at the top and the $nth$ term is reducing by the previous one's coefficient. If we extrapolate and do this a times, we get:
$$(1-p)^aS = \binom{a}{a} + (\binom{a}{a-1} - (a-1)\binom{a}{a})p + ( \binom{a}{a-2} - (a-2)\binom{a}{a-1} + \frac{(a-2)(a-1)}{2}\binom{a}{a})p^2 + \space ...$$
if we end up expanding the coefficients, they reduce to:
$$(1-p)^aS = 1 + p + p^2 + p^3 + \space ...$$
and vóila! The right hand side is now an infinite GP, which converges to $\frac{1}{1-p}$. Rearranging the terms, we obtain:
$$ S = \frac{1}{(1-p)\cdot(1-p)^a} = \frac{1}{(1-p)^{a+1}}$$
which is the final answer.
|
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|
Find the bisector in the given condition I need to find x which is bisectrix for that 60°. Given red segments 9 and 12 cm and a circle. It`s seems like school exercise, but i`m I am get stuck, please help (have tried several approaches). Please, give any advices or direction to solve it. Thanks.
↓ See the image below ↓
Image of the problem
|
Let
$|AB|=c=9$,
$|AC|=b=12$,
$|AD|=d=?$,
$\angle CAB=\alpha=60^\circ$,
$\angle DAB=\angle CAD=\tfrac12\,\alpha=30^\circ$.
Then by the cosine rule we have
\begin{align}
a^2&=b^2+c^2-2bc\cos\alpha=b^2+c^2-bc=117
,\\
a&=3\sqrt{13}
.
\end{align}
The area $S$ and the radius of the circle $R$
of $\triangle ABC$ are
\begin{align}
S&=\tfrac12bc\sin\alpha=27\sqrt3
,\\
R=|OA|=|OB|=|OC|&=\frac{abc}{4S}
=\sqrt{39}
.
\end{align}
\begin{align}
\triangle ABD:\quad
d^2&=c^2+R^2-2cR\,\cos(\beta+30^\circ)
=
120-27\sqrt{13}\,\cos\beta+9\,\sqrt{39}\,\sin\beta
,
\end{align}
\begin{align}
\triangle ABC:\quad
\cos\beta&=\frac{a^2+c^2-b^2}{2ac}
=\tfrac{\sqrt{13}}{13}
,\\
\sin\beta&=\sqrt{1-\cos^2\beta}=\tfrac{2\sqrt{39}}{13}
,\\
d^2&=147
,\\
d&=7\sqrt3
.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3442714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh(2x)}} + C
\end{align}
Since the answer contains $\cosh(2x)$
My first trial was
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = a^2 \cosh(2x) + (b^2 -a^2) \cosh^2(x)
\end{align}
Seems not good..
My second trial is
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = (a^2+b^2) \sinh^2(x) + b^2
\end{align}
and this also does not good for integrand....
How one can obtain this integral?
|
Using substitution: Let $u=\sinh(x)$. Then $\mathrm du$ corresponds to $\cosh(x)\,\mathrm dx$ and thus (since $a^2\sinh^2(x)+b^2\cosh^2(x)=(a^2+b^2)\sinh^2(x)+b^2$) $$I=\int \big(u^2\cdot(a^2+b^2)+b^2\big)^{-\frac32}\,\mathrm du.$$
Now we want to substitute $$u^2\cdot(a^2+b^2)=b^2\tan^2(s).$$
Then $$\mathrm du\sim \frac{b \sec^2(s)}{\sqrt{a^2+b^2}} \,\mathrm ds$$
and $$\big(u^2\cdot(a^2+b^2)+b^2\big)^{-\frac32}=\big(b^2\tan^2(s)+b^2\big)^{-\frac32}=\frac{1}{b^3\sec^3(s)}.$$
Hence, $$I=\frac{1}{b^2\sqrt{a^2+b^2}}\int \cos(s)\,\mathrm ds.$$
I'll let you do the rest.
You can use that $$s=\arctan\left(u\frac{\sqrt{a^2+b^2}}b\right)$$ and $$\sin(y)=\frac{\tan(y)}{\sqrt{1+\tan^2(y)}}$$ for all $y\in]-\frac\pi2,\frac\pi2[$.
Also, note that $$\cosh(2y)=\sinh^2(y)+\cosh^2(y).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3445092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
|
With $$b=2-a$$ we have to prove:
$$\sqrt{4a^2-a+3}\geq 4-\sqrt{2a^2-7a+9}$$ if $$4-\sqrt{2a^2-7a+9}\le 0$$ we have nothing to prove, in the other case we get
$$-3a+11\le 4\sqrt{2a^2-7a+9}$$ If $$-3a+11\le 0$$ then is all clear, in the other case we can square and collecting like terms we get
$$-21(a-1)^2\le 0$$ and this is true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3446873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
}
|
Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(\frac{x-1}{x})^{\frac{1}{x}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=0$
My answer is rather imprecise:$$\underset{x\rightarrow +\infty}\lim{\cos{\frac{1}{x^3}}}=1\implies\underset{x\rightarrow +\infty}\lim{\sqrt[6]{1-\cos{\frac{1}{x^3}}}}=0$$
$$\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}}=0?$$
I am aware of the mistake I have made by writting $0$ for an undefined term.
$$\underset{x\rightarrow +\infty}\lim{\Big(1-\frac{1}{x^3}\Big)}=1\implies \ln{\Big(1-\frac{1}{x}\Big)}<0\implies\underset{x\rightarrow +\infty}\lim{\Bigg(\frac{1}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}\Bigg)=-\infty}$$
The limit of the denumerator is $0$ $\&$ the limit of the whole expression is $0$.
How can I prove this concisely?
|
Let's rewrite the thing to $$L=\lim\limits_{x\to+0}\frac{\left( 1-\cos x^3\right)^\frac16\left(2^{-x}-3^{-x}\right)}{x\ln(1-x)}$$
Each of the things to first order will suffice, so let's start:
$$\lim\limits_{x\to0}\frac{e^x-1}{x}=1\Rightarrow
\lim\limits_{x\to0}\frac{a^x-1}{x}=\ln a,$$
$$1-\cos(x^3)=2sin^2\left(\frac{x^3}{2}\right),
\lim\limits_{x\to0}\frac{\sin(x)}{x}=1\Rightarrow
\lim\limits_{x\to0}\frac{2sin^2\left(\frac{x^3}{2}\right)}{\left(\frac{x^3}{2}\right)^2}=2,$$
$$\lim\limits_{x\to0}\frac{\ln(1+x)}{x}=1$$
Should I do the rest? )
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.