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Bhaskara approximation of cos(x) In my knowledge the best aproximation of $\sin$ and $\cos$ is Bhaskara approximation which is :
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
and
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$
what is the best approximation of $\cos(x)$ for $ -\pi\leq x\leq 0$ (by a regular function ) ?
| In terms of approximation, you can do a lot of things depending on how many terms you want to include in the expression. Most of the time, this is done using Padé approximants.
For example, for your question, using the simple $[3,2]$ Padé approximant built around $x=-\frac \pi 2$, we get
$$\cos(x)=\left(x+\frac{\pi }{2}\right)\frac {1-\frac{7}{60} \left(x+\frac{\pi
}{2}\right)^2 } {1+\frac{1}{20}\left(x+\frac{\pi
}{2}\right)^2 }$$ which leads to an absolute error of $0.0042$ at the bounds.
For sure, you could go further and use the $[5,4]$ Padé approximant
$$\cos(x)=\left(x+\frac{\pi }{2}\right)\frac {1-\frac{53}{396} \left(x+\frac{\pi
}{2}\right)^2+\frac{551 }{166320}\left(x+\frac{\pi
}{2}\right)^4 } {1+\frac{13}{396} \left(x+\frac{\pi }{2}\right)^2+\frac{5 }{11088}\left(x+\frac{\pi
}{2}\right)^4 }$$ which leads to an absolute error of $3.0\times 10^{-6}$ at the bounds. This is equivalent to the Taylor series to $O\left(\left(x+\frac{\pi }{2}\right)^{11}\right)$.
To give you an idea of the possible improvements; consider that $P_n$ is the $[2n+1,2n]$ Padé approximant of $\cos(x)$ built around $x=-\frac{\pi
}{2}$ and compute the norm
$$\Phi_n=\int_{-\pi}^0 \big[\cos(x)-P_n\big]^2\,dx$$
$$\left(
\begin{array}{cc}
n & \Phi_n \\
1 & 3.92 \times 10^{-6} \\
2 & 1.26 \times 10^{-12}\\
3 & 2.23 \times 10^{-20}
\end{array}
\right)$$
All of these have been done only based on the information (function and derivative values) at $x=-\frac{\pi}{2}$.
But we can try to cover the whole range and consider for example
$$\cos(x) \sim \left(x+\frac{\pi }{2}\right)\frac {1+a \left(x+\frac{\pi
}{2}\right)^2+b\left(x+\frac{\pi
}{2}\right)^4 } {1+c \left(x+\frac{\pi }{2}\right)^2+d\left(x+\frac{\pi
}{2}\right)^4 }$$ and obtain
$$a=\frac{-1952+672 \pi -192 \pi ^2+56 \pi ^3}{\pi ^3(\pi ^2-10)}$$
$$b=\frac{-3456+1056 \pi +768 \pi ^2-240 \pi ^3 } {\pi ^5 \left(\pi ^2-10\right) }$$
$$c=\frac{-192+72 \pi -104 \pi ^2+32 \pi ^3 } {\pi ^2 \left(\pi ^2-10\right) } $$
$$d=\frac{-4704+1504 \pi +400 \pi ^2-128 \pi ^3} {\pi ^4 \left(\pi ^2-10\right) }$$ which gives a maximum absolute error equal to $5.96\times 10^{-8}$ and a norm equal to $2.73\times 10^{-15}$ (which is $461$ times smaller than $\Phi_2$).
You could notice that the coefficients are extremely close to those of $P_2$ but these minor changes make a lot of difference.
All these approximations have been built for answering your question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit multivarible How do I solve this limit?
$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$
This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$
The first one is a standard limit with value one, but I'n not sure about the other term.
| $0 \leq | \frac {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac 1 2 |z| \to 0$ since $|xy| \leq \frac 1 2{(x^{2}+y^{2})}$. Hence the given limit is $1$.
| {
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"source": "stackexchange",
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Why substitution in irrational equation doesn't give equivalent equation? I have two examples of irrational equations:
The first example: $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$
In solution, they take cube of both sides and do following:
\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x}) +6+x=27\\
&\iff&
3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27 \iff \sqrt[3]{(3-x)(6+x)} = 2\\
&\iff&
x^2+3x-10=0\\
&\iff& x=2\quad \text{ or }\quad x= -5
\end{eqnarray*}
They conclude that both values are solutions, they satisfy the original equation.
The second example: $\sqrt[3]{x+1} + \sqrt[3]{3x+1} = \sqrt[3]{x-1}$
Here they do the following:
\begin{eqnarray*}
&\sqrt[3]{x+1}& + \sqrt[3]{3x+1} = \sqrt[3]{x-1}\\
&\iff&
x+1 + 3 \sqrt[3]{(x+1)(3x+1)} (\sqrt[3]{x+1} + \sqrt[3]{3x+1} ) + 3x+1 = x-1\\
&\implies&
3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3\\
&\iff&
(x^2-1)(3x+1)= -(x+1)^3\\
&\iff&
x= 0 \quad\text{ or }\quad x = -1
\end{eqnarray*}
but the only solutions is $x= -1$, because $0$ doesn't satisfy the equation.
What is difference between these examples, why do we have all equivalence signs in the first, and implication in the secondnd example?
The explanation in solution is that we have substituted $\sqrt[3]{3-x} + \sqrt[3]{6+x}$ by a number in the 1st example, and in the 2nd example we substituted by another expression which depends on x. I don't understand that, when does substitution give equivalent equation? Do we always have to verify if solutions satisfy the original equation in the end (is that the answer)?
Please help with this explanation, I need to understand better the irrational equations. Thanks a lot in advance.
| Very good question! The first example is misleading in writing equivalences everywhere; the second equivalence
$$3-x+ 9\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x})+6+x$$
$$\iff$$
$$3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27$$
should (at least conceptually) be an implication $(\implies)$. It is true that it is in fact an equivalence, but this is not yet clear at this point. Let me explain:
The argument starts from the assumption that if $x$ satisfies
$$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3,$$
then it also satisfies the expressions that follow. In both examples the first equivalence stems from the simple fact that
$$x^3=y^3\quad\iff\quad x=y,$$
but the second equivalence uses a substitution that need not be reversible; a number $x$ can satisfy
$$3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3,$$
but this does not imply that it should satisfy
$$\sqrt[3]{x+1} + \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$
This is illustrated by the solution $x=0$. In the first example it just so happens that we do not get any extra solutions, and hence, with hindsight, this implication turns out to be an equivalence. But that is not at all clear before checking whether all solutions to the last equation are also solutions to the original equation.
So to answer your question; yes, you should check whether all solutions to the last equation are also solutions to the original equation. If you are certain that every step along the way is unambiguously reversible, then you do not need to check. But if it is not clear whether a step is reversible (as in both examples here), you should check.
| {
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Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$. I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!
When $m$ is even, put $m=2k$.
$2^m-1=(2^k-1)(2^k+1)$
By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime.
So, these are integers to n-th power.
Put
$2^k-1=a^n,
2^k+1=b^n$ ($1≦a<b,a$ and $b$ are odd).
$2=(b-a)(b^{n-1}+...+a^{n-1})≧2n$
∴$n=1$
| From $p^n + 1 \ge p+1 \ge 4$ we have $m \ge 2$.
Hence $p^n = 2^m - 1 \equiv -1\pmod 4$, and thus $p^n$ cannot be a square.
This forces $n$ to be odd.
From $p^n + 1 = (p+1)(p^{n-1} - p^{n-2} + \dots + 1) = 2^m$, we see that
$$(p^{n-1} - p^{n-2} + \dots + 1) \mid 2^m$$
Since $p$ is odd, this factor, which is a sum of $n$ odd numbers, is odd as well.
This forces $p^{n-1} - p^{n-2} + \dots + 1 = 1$, and thus $p^n+1 = (p+1)(1)$.
| {
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Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$
Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$
I dont know what to do from here, can you tell me what is the next step?
| The key is the generalized binomial theorem
in the handy special case
$(1-x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k
$.
This implies that
$(1+x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}(-1)^kx^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}(-1)^kx^k
$
and,
for any $m$,
$(1-x^m)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^{km}
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^{km}
$.
For $m=2$ this is
$(1-x^2)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^{2k}
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^{2k}
$.
Your problem is
$\begin{array}\\
g(x)
&= \dfrac{x^3} {(1+x)^5 (1−x)^6}\\
&= \dfrac{x^3(1+x)} {(1+x)^6 (1−x)^6}\\
&= \dfrac{x^3(1+x)} {(1−x^2)^6}\\
&= \dfrac{x^3} {(1−x^2)^6}+\dfrac{x^4} {(1−x^2)^6}\\
\end{array}
$
The results above
should make this
straightforward.
Actually,
it's not that simple.
Here's the rest.
$\begin{array}\\
\dfrac{x^3} {(1−x^2)^6}
&=x^3\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k+3}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2(k+1)+1}\\
&=\sum_{k=1}^{\infty} \binom{k+4}{5}x^{2k+1}\\
\\
\dfrac{x^4} {(1−x^2)^6}
&=x^4\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2k+4}\\
&=\sum_{k=0}^{\infty} \binom{k+5}{5}x^{2(k+2)}\\
&=\sum_{k=2}^{\infty} \binom{k+3}{5}x^{2k}\\
\end{array}
$
This gives the coefficients
of the even and odd terms.
To map these cases
into a single one,
use
$h(k)
=k-2\lfloor k/2 \rfloor
$
where
$h(k) = 0$ for even $k$
and
$h(k) = 1$ for odd $k$.
Since we want
even $k \to \lfloor k/2 \rfloor+3$
and
odd $k \to \lfloor k/2 \rfloor+4$,
we want
$\begin{array}\\
k
&\to \lfloor k/2 \rfloor+3+h(k)\\
&\to \lfloor k/2 \rfloor+3+k-2\lfloor k/2 \rfloor\\
&\to k+3-\lfloor k/2 \rfloor\\
\end{array}
$
Therefore
$g(x)
=\sum_{k=3}^{\infty} x^k\binom{k+3-\lfloor k/2 \rfloor}{5}
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$
Original image
| $$\sin3A-\cos3A$$
$$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$
$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$
$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$
Similarly, $\sin A+\cos A=\sqrt2\sin(?)$
Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$
| {
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Unable to represent the function as a power series Assuming I have this function below :
$$f(x) = \frac{(x-1)}{x^2+1}$$
How do I represent it as a power series ?
The only solution that I know is when using the geometric series :
$\sum_{k=1}^\infty q^k = \frac{1}{1-q}$ but this is not possible in our case .
| $$\frac{x-1}{x^2+1}=\frac{x}{x^2+1}-\frac{1}{x^2+1}$$
Integrate
$$\int\left(\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)\,dx=\frac{1}{2}\log(1+x^2)+\arctan x+C$$
Series for these functions are known, we get
$$\frac{1}{2}\log(1+x^2)-\arctan x+C=\frac{x^{10}}{10}-\frac{x^9}{9}-\frac{x^8}{8}+\frac{x^7}{7}+\frac{x^6}{6}-\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}-x+O(x^{11})+C$$
Now differentiate all terms
$$\frac{x-1}{x^2+1}=-1 + x + x^2 - x^3 - x^4 + x^5 + x^6 - x^7 - x^8 + x^9+O(x^{10})$$
bonus
$$\frac{x-1}{x^2+1}=\sum _{n=0}^{\infty } a_n x^n;\quad a_n=3 n^2+n+3-4 \left\lfloor \frac{1}{4} \left(3 n^2+n+4\right)\right\rfloor $$
| {
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Expression of sum of squares as a sum of a specific form This question is a follow-up of this one.
Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$, and that $x,y$ are not both odd.
Do there exist $a,b,c,d \in \mathbb{Z}$ such that
$
(a+d)^2+(b-c)^2=x^2+y^2
$
and $ad-bc=1$?
This question is motivated by an attempt to characterise the norms of matrices in $SL_2(\mathbb{Z})$: If $A =\begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, then
$$
(a+d)^2+(b-c)^2=\|A\|^2+2,
$$
so I wonder whether $\|A\|^2+2$ can be any sum of squares $x^2+y^2$ where $x,y$ are not both odd.
This parity restriction on $x,y$ is certainly necessary for the existence of such a representation:
If $x,y$ are both odd, then $a+d,b-c$ must be odd (by looking at equality $\mod 4$). Thus one of $a$ and $d$, and one of $b$ and $c$ must be even, hence $ad-bc$ is even which is incompatible with $ad-bc=1$.
| Let $x=0$ and $y=3$ so that $x^2+y^2=9$. Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that
$$(a+d)^2+(b-c)^2=x^2+y^2=9,\tag{1}$$
then $\{a+d,b-c\}=\{0,\pm 3\}$ for some for some choices of the signs. Then of course the matrices
$$\begin{pmatrix}-a&\hphantom{-}b\\\hphantom{-}c&-d\end{pmatrix},
\begin{pmatrix}\hphantom{-}a&-b\\-c&\hphantom{-}d\end{pmatrix},
\begin{pmatrix}\hphantom{-}b&\hphantom{-}a\\-d&-c\end{pmatrix}
\in\operatorname{SL}_2(\Bbb{Z}),$$
also satisfy $(1)$, so without loss of generality $a+d=0$ and $b-c=3$. Then
$$1=ad-bc=a(-a)-b(b-3),$$
which shows that $b$ is an integral root of a quadratic with discriminant
$$\Delta=-4a^2+77.$$
But this is never a perfect square; a contadiction! Hence no such matrix exists.
| {
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solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$. I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \frac{3}{z+3}+\frac{2}{z-2}$. Now based on the region to where the function $f$ should be valid on is that $1< |z-1|< 4$, I'm not sure if I did it right when I assumed that $\left|\frac{2}{z}\right|<1$ and $\left|\frac{3}{z}\right|<1$. (I just did it with brute force to arrive on the two inequalities). Can you help me with that?
Assuming my two inequalities are correct, I have now
$\frac{3}{z+3} = \frac{3/z}{1+3/z} = 3/z \sum_{n=0}^{\infty} \left(- \frac{3}{z}\right)^n = \sum_{n=0}^{\infty} (-1)^{n-1}3^nz^{-n} = \sum_{n=1}^{\infty} (-1)^n3^{n+1}z^{-(n+1)}$,
and
$\frac{2}{z-2} = \frac{2/z}{1-2/z} = 2/z \sum_{n=0}^{\infty} \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}}.$
Then after that, I combined which I have the result as
$f(z) = \frac{z-12}{z^2+z-6} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}} + \sum_{n=1}^{\infty} (-1)^n(3)^{n+1}z^{-(n+1)}$.
Now, did I do it correctly? or I did wrong on the inequalities of the region to where the laurent should be valud to? Thanks for those who can help.
| $$
\begin{align}
\frac{z-12}{z^2+z-6}
&=\frac3{z+3}-\frac2{z-2}\\
&=\frac3{(z-1)+4}-\frac2{(z-1)-1}\\
&=\frac34\underbrace{\sum_{k=0}^\infty\left(\frac{1-z}4\right)^k}_\text{converges for $|z-1|\lt4$}-2\underbrace{\sum_{k=1}^\infty\left(\frac1{z-1}\right)^k}_\text{converges for $|z-1|\gt1$}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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GCD of $n^2-3n-1$ and $n-4$ $n$ is a natural number and after trying the division algorithm
$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$
For the last part I'm not sure whether it does equal to $\gcd(n-1,3)$.
If yes, then should I take the cases where $n$ is $3k+i, 0\le i\le2$?
| Using this implementation of the Extended Euclidean Algorithm, rotated $90^{\large\circ}$ and applied to polynomials, gives
$$
\begin{array}{c|c|c}
\color{#00F}{n^2-3n-1}&1&0\\
\color{#00F}{n-4}&0&1\\
\color{#090}{n-1}&\color{#C00}{1}&\color{#C00}{-n}&n\\
\color{#090}{-3}&\color{#C00}{-1}&\color{#C00}{n+1}&1
\end{array}
$$
which says that
$$
\overbrace{\begin{bmatrix}
\color{#C00}{1}&\color{#C00}{-n}\\\color{#C00}{-1}&\color{#C00}{n+1}
\end{bmatrix}}^{\det=1}
\begin{bmatrix}
\color{#00F}{n^2-3n-1}\\\color{#00F}{n-4}
\end{bmatrix}
=\begin{bmatrix}
\color{#090}{n-1}\\\color{#090}{-3}
\end{bmatrix}
$$
which means any integer linear combination of $n-1$ and $3$ is an integer linear combination of $n^2-3n-1$ and $n-4$ and vice-versa (since the determinant of the matrix is $1$). This means that
$$
\gcd\left(n^2-3n-1,n-4\right)=\gcd(n-1,3)
$$
| {
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find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$ find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$
I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$.
But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.
| By plugging in $x=16$, we find $P(16)=0$. So $P$ has some positive roots. If $P$ is not $\equiv 0$, it has a minimal positive root $x_0$.
Plug in $\frac 12x_0$ to find $0=16(\frac12x_0-1)P(\frac12x_0)$ and so $x_0=2$ as $P(\frac12x_0)\ne 0$. By plugging in $x=2$, we then find $P(4)=0$, next $P(8)=0$ and $P(16)=0$. Similar to above, we find that the maximal positive root is $16$.
So
$$P(x)=(x-2)(x-4)(x-8)(x-16)Q(x) $$
and therefore
$$ (x-16)P(2x)=(x-16)(2x-2)(2x-4)(2x-8)(2x-16)Q(2x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(2x)$$
whereas
$$ 16(x-1)P(x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(x).$$
Therefore $Q(2x)=Q(x)$ for all $x$ (except possibly for $x=1,2,4,8,16$). Hence $Q$ is bounded, hence constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simultaneous algebra question Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$
Find $$x^3+3y^3$$
Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$
I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$
| Being lazy, I changed a little the notations
$$ax+by=v \tag 1$$ $$ax^2+by^2=w \tag 2$$
From $(1)$, you have $y=\frac{v-a x}{b}$. Plug in $(2)$ to face
$$\left(v^2-b w\right)-2 a vx+a\left(a+ b\right)x^2=0$$
$$x_\pm=\frac{av\pm\sqrt{a b \left(w (a+b)-v^2\right)}}{a (a+b)}\qquad \qquad y_\pm=\frac{v-a x_\pm}{b}$$
Now, just a bit of patience for the simplifications and
$$a x_+^3+b y_+^3=\frac{a^2 \left(bv+\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3+b^2 \left(a
v-\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3}{a^2 b^2 (a+b)^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other, but got nothing. No idea what to do.
| $$\overline{xyz138}=1000\overline{xyz}+138\equiv 6\overline{xyz}-2\pmod 7\\ \overline{138xyz}=138000+\overline{xyz}\equiv 5+\overline{xyz}\pmod{13}\\ \overline{x1y3z8}\equiv 10x+1+10y+3+10z+8\equiv 1-x-y-z\pmod{11}$$
Denote $a:=\overline{xyz}$, then you have the linear system
$$\begin{cases}6a-2\equiv 0\pmod 7\\5+a\equiv 6\pmod{13}\end{cases}$$
along with $x+y+z\equiv 7\pmod{11}$
Apply CRT to solve modulo $7\times 13$ and then check the values of $a$ that satisfy the congruence modulo $7\times 13$ against the final constraint. There are only $11$ values of $a$ to check.
| {
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For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$ For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$
I couldn't solve this question and hence looked at the solution. The solution substitutes $x=\sqrt{a}$, $y=\sqrt{b}$, $z=\sqrt{c}$, hence $x^3+y^3+z^3=3$. We also have that $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le24$
which is also written as $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le \frac{8}{3}(x^3+y^3+z^3)^2$ which holds true, after expanding etc.
My question is, is there a simpler and more intuitive approach? And if not how was I supposed to think of the substitution used above?
| Here I will show the thoughts that could help you to do all the steps above.
*
*The first substitution is unnecessary, but it simplifies the algebra, since you have $(\sqrt{a})^3+(\sqrt{b})^3+(\sqrt{c})^3=3$
*Now, for the inequality we want to prove we have $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3$ on the left side. The general trick is to make inequality Homogeneous. It means that if you imagine $x,y,z$ to be lengths in meters, you will get $meters^6$ on the left side, which means that we want to get $meters^6$ on the right side as well. Since $(x^3+y^3+z^3)$ is $meters^3$, we want to square it to get $meters^6$.
*The last step is just to find the coefficient before $(x^3+y^3+z^3)^2$. Here it helps to check the equality point $(x^3+y^3+z^3)^2=9$, but we want to have 24. Therefore we put $\frac{8}{3}$ as a coefficient and get $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3 \leq \frac{8}{3}(x^3+y^3+z^3)^2$.
| {
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.
| $$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$
$$
= \sqrt{
9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
2\sum_{cyc}
{\sqrt{
\left(3 +
\frac{1}{a^{2}}\right)
\left(3 +\frac{1}{b^{2}}\right)}}
}
$$
$$\tag{By C-S}\geqslant \sqrt{9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
2\sum_{cyc}{3 + \frac{1}{ab}}}$$
$$= \sqrt{9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
54}$$
$$\geqslant \sqrt{9 +18 + 54} =\sqrt{81} = 9$$
Which uses $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 18$ and $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geqslant \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer
I said that since $a+b+c=0$ then $c=-a-b$ and hence factoring $K=a^3b+b^3c+c^3a$ we have that $K=-(a^2+ab+b^2)^2$
This is where I got stuck. I can't work out how to finish off the solution. Could you please explain to me the full solution and how you intuitively thought of each step?
| you have $$ 2017 + \left( a^2 + ab + b^2 \right)^2 = w^2 $$ or
$$ w^2 - \left( a^2 + ab + b^2 \right)^2 = 2017 $$
Difference of squares on the left, factors, meanwhile 2017 is prime. Thus
$$ w + a^2 + ab + b^2 = 2017 , $$
$$ w - a^2 - ab - b^2 = 1 , $$
or
$$ w = 1009$$
$$ a^2 + ab + b^2 = 1008 = 2^4 3^2 \cdot 7 = 7 \cdot 144 $$
You can check, this means $a,b$ are both divisible by $12.$ Let $\alpha = \frac{a}{12}$ and $\beta = \frac{b}{12},$ then
$$ \alpha^2 + \alpha \beta + \beta^2 = 7$$
which has just a few integer solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$ The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?
| $$\sin(x)\cos(x)=\sin(x)+\cos(x)$$
$$\frac12 (\sin(2x))=\sin(x)+\cos(x)$$ $$\frac{\sin(2x)}{2}=\sin(x)+\cos(x)$$ $$\frac12 (\sin(2x))=\frac{2}{\sqrt2}(\sin(x+\frac{π}{4}))$$ $$\sin(2x)=\frac{4}{\sqrt2}\left(\sin(x+\frac{\pi}{4}) \right)$$ $$\sin(2x)=2\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ $$2\sin^2\left(x+\frac{π}{4}\right)-1=2\sqrt{2}\sin\left(x+\frac{π}{4}\right)$$ $$2\sin^2\left(x+\frac{π}{4}\right)-1-2\sqrt{2}\sin\left(x+\frac{π}{4}\right)=0$$ $$2{\color{green}{\sin^2\left(x+\frac{π}{4}\right)}}-2\sqrt2 {\color{green}{\sin\left(x+\frac{π}{4}\right)}}-1=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions? When factoring the quadratic $x^2 + bx + c $, where $b$ and $c$ are integers, why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions?
We wish to write this as $(x + u) (x + v)$. Where $u = \frac{p}{q}$ and $v =\frac{m}{n}$.
\begin{align}
\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right)&= x\left(x+\frac{m}{n}\right)+\frac{p}{q}\left(x + \frac{m}{n}\right)\\
&=x^2 + \frac{mx}{n} + \frac{px}{q} + \frac{pm}{qn}\\
&=qnx^2 + (mq + pn)x + pm.
\end{align}
This is quite different from the quadratic $x^2 + bx + c$.
Am I along the right path? What's the complete reasoning?
| Notice that if $u = \frac{p}{q}$ and $v = \frac{m}{n}$ for some integers $p,q,m,n$ then you obtain the equation $$qn\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right) = qnx^2 + (mq + pn)x + pm.$$ In the expression $qnx^2 + (mq + pn)x + pm$ we can conclude that $qn, mq+pn$, and $pm$ are all integers. If we use the labels
\begin{align}a &= qn\\
b&=mq+pn\\
c&=pm\\
u &= \frac{p}{q}\\
v &= \frac{m}{n}\end{align} then we again end up with $$(x-u)(x-v) = ax^2 + bx + c,$$ so in reality this form covers your case as well. Generally, though, we don't require $a,b,c,u,v$ to be integers.
| {
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Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
| Well, in the worst case when we are stuck we can do cases:
Case 1: $x+y\ge 0$ and $x-y \ge 0$ then
$x\ge -y$ and $x \ge y$ so $x \ge \max(y,-y) = |y|$.
And $|x+y| + |x-y| = (x+y)+(x-y) = 2x = 2$ so $x = 1$ and $|y| \le x=1$ so $-1 \le y \le 1$.
$x^2 - 6x +y^2 = -5+y^2$ and the max that $y^2$ can be is $1$ so the max of $x^2 -6x +y^2$ is $-4$ when $y = \pm 1$ if was assume $x+y\ge 0$ and $x-y \ge 0$.
(Likewise the max of $x^2 + y^2 + 10y = 1+y^2 + 10y$ is $12$ if $y=1$.)
Case 2:
$x +y < 0$ and $x-y \ge 0$ then
$x < -y$ and $x \ge y$ so $y \le x < -y$ so $y < 0$.
And $|x+y| + |x-y| = -x -y + x-y = -2y = 2$ so $y = -1$ and $-1\le x < 1$.
$x^2 -6x + y^2 = x^2 -6x +1$ and $x^2 \le 1$ and $x \ge -1$ so $-6x \le 6$ so the very max that $x^2-6x +y^2$ is $1+6 +1 = 8$ if $x=-1$.
(Likewise $x^2 +y^2 + 10y = x^2 + 11$ an the max that can be is $12$ if $x = -1$.)
Case 3: $x +y \ge 0$ and $x -y < 0$ then
$x \ge -y$ and $x < y$ so $-y \le x < y$ so $y > 0$.
$|x+y| + |x-y| = x+y+y-x = 2y =2$ so $y =1$ and $-1 \le x < 1$ and
the max value of $x^2-6x +y^2 = x^2-6x+1$ and $x^2 \le 1$ and $-6x \le 6$ so $x^2 -6x+1 \le 8$ and the max value is $8$ when $x=-1$.
case 4: $x+y < 0$ and $x-y < 0$ so $x< -y$ and $x < y$ so $x< \min (y,-y) = -|y|$
And $|x+y| + |x-y| = -x -y +y -2 =-2x = 2$ so $x = -1$ and $-|y| > x=-1$ so $|y|< 1$.
so $x^2 -6x +y^2 = 7 + y^2< 8$. but $x^2-6x +y^2 = 8$ never occurs; it's an unacheived least upper bound.
Of those four cases the max value of $x^2 -6x +y^2 = 8$.
....
that's probably not an efficient or clever way but it is a complete hit the ground running systematic way.
| {
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Probability -- Proving Cantelli’s inequality by setting up a $E[φ(X)]$ and using Markov's inequality I'm working on proving cantellis inequality. We're told to use this setup:
$P(X >= a) ≤ E[φ(X)]/ φ(α)$ where $var[X] = σ^2$ and $E[X] = 0$ and $φ(x) = (x+c)^2$
An intermediate step was proving that $P(X ≥ α ) ≤ (σ^2+c^2)/(α +c)^2$ which I did successfully.
The next part was finding a value of c that minimizes the right hand side of the equation, and therefore gets the best bound. We're to find the c that gets this bound:
$P(X ≥ α ) ≤ (σ^2)/(α^2 +σ^2)$
The answer is supposed to be $c = σ^2/α$ but I'm getting, by partial differentiation on c and setting that RHS to 0, $c = ασ^2/(α- σ^2) $
$d/dc ((σ^2 + c^2)/(α + c)^2) = (2c(α+c)^2 - 2(α + c)(σ^2+c^2))/(α + c)^4 = 0$
$2c(α+c)^2 - 2(α + c)(σ^2+c^2) = 0$
$2c(α+2αc+c^2) - 2(ασ^2 + αc^2+ cσ^2+c^3 = 0 $
$2αc+2αc^2+2c^3 - 2ασ^2 - 2αc^2-2cσ^2-2c^3 = 0$
$2αc - 2ασ^2 -2cσ^2 = 0$
$αc - ασ^2 -cσ^2 = 0$
$c(α- σ^2) = ασ^2 $
$c = ασ^2/(α- σ^2) $
What did I do wrong?
| Given that $$\mathbb P(X\geqslant\alpha)\leqslant\frac{\sigma^2+c^2}{(\alpha+c)^2},$$
we consider the map $f:[0,\infty)\to\mathbb R$ defined by $f(c) = \frac{\sigma^2+c^2}{(\alpha+c)^2}$. Now, $f(0) = \left(\frac\sigma\alpha\right)^2$, and $\lim_{c\to\infty}f(c)=+\infty$. It is clear that $f$ is differentiable on $(0,\infty)$, as a rational function whose denominator is always positive, and so differentiating we find that
$$
f'(c) = \frac{2 \left(\alpha c-\sigma ^2\right)}{(\alpha +c)^3}.
$$
It follows that $f$ is decreasing on $\left(0, \frac{\sigma^2}\alpha\right)$ and increasing on $\left(\frac{\sigma^2}\alpha,\infty\right)$, and thus attains a global minimum at $c^* = \frac{\sigma^2}\alpha$. We conclude that
$$
\mathbb P(X\geqslant\alpha)\leqslant\inf_{c\geqslant 0} \frac{\sigma^2+c^2}{(\alpha+c)^2} = \frac{\sigma ^2}{\alpha ^2+\sigma ^2}.
$$
(Note that
| {
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Compute $a+b+c+d$ where $(ab+c+d =13),\ (bc+d+a = 27),\ (cd+a+b = 30),\ (da+b+c = 17)$ Compute $(a+b+c+d)\ $ if $\ a,b,c,d\ $ satisfy the system of equations below:
$$\begin{cases}
ab+c+d = 13\\\\
bc+d+a = 27\\\\
cd+a+b = 30\\\\
da+b+c = 17
\end{cases}$$
.................................................................................My method:
I added all equations to get:
$(ab+bc+cd+da)+2(a+b+c+d)=87$
$(a+c)(b+d)+2(a+c)+2(b+d)=87$
$(a+c)(b+d+2)+2(b+d)=87$
$(a+c+2)(b+d+2)=91$
$(a+c+2)(b+d+2)=13\times 7$
So, $a+c=11$ and $b+d=5$
or, $a+c =5$ and $b+d =11$
Hence, $a+b+c+d = 16$
But I am not sure if this is the correct method since it was not stated that $a,b,c,d$ were positive integers.
I would be thankful if someone provided a solution to this question.
| Let $u = a+b+c+d$, the set of equations can be rewritten as
$$
\begin{cases}
(a-1)(b-1) + a + b + c + d &= 14\\
(b-1)(c-1) + a + b + c + d &= 28\\
(c-1)(d-1) + a + b + c + d &= 31\\
(d-1)(a-1) + a + b + c + d &= 18
\end{cases}
\iff
\begin{cases}
(a-1)(b-1) &= 14- u\\
(b-1)(c-1) &= 28 -u\\
(c-1)(d-1) &= 31 -u\\
(d-1)(a-1) &= 18 -u
\end{cases}$$
This leads to
$$(14-u)(31-u) = (a-1)(b-1)(c-1)(d-1) = (28-u)(18-u)$$
which can be simplified to $u = 70$. So $a + b + c + d = 70$.
| {
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| For $a=x^2+5x+4$ we get to factor $a(a+2)-48=a^2+2a-48=(a+8)(a-6)$.
| {
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the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$
Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$
Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$
Now I can use the squeeze theorem and get:
*
*$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$
*$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$
And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?
Unfortunately, I can not use integrals while doing that exercise.
| The Stolz–Cesàro theorem indeed suffices, with a derivative-like limit (the last one below).
Let $a_n=\sum\limits_{k=1}^n\sqrt{k}$ and $b_n=\sum\limits_{k=1}^n k/a_k$; we're computing
\begin{align*}
\lim_{n\to\infty}\frac{b_n}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{b_n-b_{n-1}}{\sqrt{n}-\sqrt{n-1}}
\\&=\lim_{n\to\infty}\frac{n}{a_n}(\sqrt{n}+\sqrt{n-1})
\\&=2\lim_{n\to\infty}\frac{n^{3/2}}{a_n}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{a_n-a_{n-1}}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{\sqrt{n}}
\\&=\color{gray}{2\lim_{n\to\infty}\frac{1-(1-1/n)^{3/2}}{1/n}}=\mathbf{3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that the unique solution of $P_n(X) = 0$ is $x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)}$ I need to how that when $ k \neq 0$, the unique solution of $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i} = 0$ is $ x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)} $ with $\theta=\frac{k\pi}{2n+1} \in ]-\frac{\pi}{2},\frac{\pi}{2}[ \backslash {0}$.
In the previous questions I've shown that there is no solutions when $k=0$, that if $P_n(x) = 0$, then $x$ verify $(\frac{x+i}{x-i})^{2n+1} = 1$, and that $P_n(x) = 0$ when $\frac{x+i}{x-i} = \exp(\frac{2ik\pi}{2n+1})$ with $k$ between $-n$ and $n$.
I don't really know where to start with this question. Do i just need to replace x in the polynomial with what is said to be the unique solution ?
| We want the solutions to $$(x+i)^{2n+1} - (x-i)^{2n+1} =0.$$
Assume $x$ is real and $z=x+i = r e^{i\theta}$ so that $z^*=x-i=re^{-i\theta}$ is the conjugate of $z$.
Substituting and rearranging:
$$z^{2n+1} = {z^\ast}^{2n+1}$$
$$e^{i(2n+1)\theta}=e^{-i(2n+1)\theta}$$
$$e^{i2(2n+1)\theta}=1=e^{2\pi k i}$$
so $$\theta = \frac{k\pi}{2n+1}$$
but $z=x+i$ thus $\Im z \ne 0$, so $k\ne 0$ nor is $k$ a multiple of $2n+1.$
$$x=\Re \left(re^{i\theta}\right)=r\cos \theta, \quad { \text{and} }\quad
1=\Im \left(re^{i\theta}\right)=r\sin \theta,$$
so $r = 1/\sin \theta$ and $$x=\frac{\cos \theta}{\sin \theta}=\cot \theta=\frac{1}{\tan \theta} =i \frac{e^{i\theta} + e^{-i\theta}}{e^{i\theta} - e^{-i\theta}}=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}.$$
One indexing is $k\in \{1,2,\cdots, 2n\}.$
The solution is not unique. $P_n$ is a polynomial of order $2n$, and $P_n(x)=0$ has $2n$ solutions.
$$\begin{aligned} P_n(x)&= \frac{1}{2i}\left[(x+i)^{2n+1}-(x-i)^{2n+1}\right]\\&=\frac{1}{2i} \left[ \sum_{k=0}^{2n+1} \pmatrix{2n+1\\k}x^k i^{2n+1-k}-\sum_{k=0}^{2n+1} \pmatrix{2n+1\\k}x^k (-i)^{2n+1-k}\right]\\ &=\sum_{k=0}^{n} \pmatrix{2n+1\\2k}x^{2k} (-1)^{k+n}\\&=(2n+1)x^{2n}-\pmatrix{2n+1\\2n-2}x^{2n-2} + \cdots +(-1)^n.\end{aligned}$$
Please see a continuation of the discussion for further developments.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx} = \_\_\_\_\_\_\_$.
My approach is as follow
$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
y=h
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$
How will I proceed from here
| If you define $g(x) = f(x) + e^{-x}$, then the equation becomes $g(x + y) = g(x) + g(y)$.
It is well-known that all continuous solutions to that equation are linear, i.e. there exists $c$ such that $f(x) = cx$.
The rest should be easy.
| {
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"source": "stackexchange",
"question_score": "3",
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} |
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$
We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$
However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?
| The given answer is larger than $r$, which clearly isn't true, as mirroring $B$ across $AC$ shows $|AB|<|AO|$, where $O$ is the center of the circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.
Any help would be appreciated.
| Because $\sin x$ is convex on $x \in (0, \pi)$ then the sum of $\sin (kx)$ inside that interval will be not less than $n$ times the linear average on the secant $(\sin x+sin (nx))/2$ and not greater than $n$ times the linear average on the tangent $n \sin((n+1)/2 x)$, so in this case
$$
\eqalign{
& n\left( {{{\sin \left( {1/n^{\,2} } \right) + \sin \left( {n/n^{\,2} } \right)} \over 2}} \right) \le x_{\,n} = \cr
& = \sin \left( {{1 \over {n^{\,2} }}} \right) + \sin \left( {{2 \over {n^{\,2} }}} \right) + \cdots + \sin \left( {{n \over {n^{\,2} }}} \right)
\le n\sin \left( {{{\left( {n + 1} \right)/2} \over {n^{\,2} }}} \right) \cr}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Brick wall with maximum height 3 Given n same-sized rectangular bricks. We want to build a wall with these constraints:
*
*All bricks should be horizontal.
*We can put a brick on two other bricks, such that the middle of the top brick is on the border of two other bricks.
*The maximum height of the wall should be 3.
*Bricks on the first row (the ground) should stick together.
For example, if we have 4 bricks, we can build the wall in 3 ways. If we have 6 bricks, we can make the wall in 9 ways.
Some examples of this recursive function are:
$f(1) = {0 \choose 0} = 1$
$f(2) = {1 \choose 1} = 1$
$f(3) = {2 \choose 0} + {1 \choose 1} = 1 + 1 = 2$
$f(4) = {3 \choose 0} + {2 \choose 1} = 1 + 2 = 3$
$f(5) = {4 \choose 0} + {3 \choose 1} + {2 \choose 2} = 1 + 3 + 1 = 5$
$f(6) = {5 \choose 0} + {4 \choose 1} + {3 \choose 2} + {2 \choose 2}{1 \choose 1} = 1 + 4 + 3 + 1 = 9$
This is the illustration of how we can build the wall using 6 bricks:
I tried to find a recurrence relation for this problem, but I didn't succeed.
| This is only a partial answer, deriving a sixth-order linear recurrence.
A wall with $k$ bricks in the bottom row can be decomposed into $k$ diagonals with slope $-1$. The first diagonal contains only the leftmost brick in the bottom row. After that each diagonal can be at most one brick longer than the previous one, up to a maximum length of $3$ bricks. Thus, we are in effect counting the sequences $\langle a_1,\ldots,a_k\rangle$ such that
*
*$a_i\in\{1,2,3\}$ for $i=1,\ldots,k$,
*$a_1=1$,
*$a_{i+1}-a_i\le 1$ for $i=1,\ldots,k-1$, and
*$a_1+\ldots+a_k=n$.
Let $b_n$ be the number of such sequences, and for $i=1,2,3$ let $b_{n,i}$ be the number of such sequences ending in $i$. Then
$$\begin{align*}
b_{n,3}&=b_{n-3,2}+b_{n-3,3}=b_{n-3}-b_{n-3,1}\\
b_{n,2}&=b_{n-2}\\
b_{n,1}&=b_{n-1}\,,
\end{align*}$$
so
$$\begin{align*}
b_{n}&=b_{n,1}+b_{n,2}+b_{n,3}\\
&=b_{n-1}+b_{n-2}+b_{n-3}-b_{n-3,1}\\
&=b_{n-1}+b_{n-2}+b_{n-3}-b_{n-4}\,,
\end{align*}$$
with initial values $b_0=1$, $b_1=1$, $b_2=1$, $b_3=2$, $b_4=3$, and $b_5=5$. The next few values are easily computed:
$$\begin{array}{rcc}
n:&0&1&2&3&4&5&6&7&8&9&10\\
b_n:&1&1&1&2&3&5&9&15&26&45&77
\end{array}$$
By the way, if the height were limited to $2$, we would simply be getting the Fibonacci numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Small angle approximation on cosine The problem is
Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant
I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{x\to 0}{\approx}1-\frac{x^2}{2}$ to get $$3-2(1-\frac{x^2}{2})+4(1-x^2)=5-3x^2$$ hence $k=3$. But why does using $\cos^2(x)\overset{x\to 0}{\approx}(1-\frac{x^2}{2})^2$ doesn't work out? I originally tried plugging that into the $\cos^2(x)$ but got another complete thing. why?
| By Taylor, as the function is even the development to second order is $f(0)+\dfrac{f''(0)}2x^2$ or
$$(3-2\cos(0)+4\cos^2(0))+(2\cos(0)+8\sin^2(0)-8\cos^2(0))\frac{x^2}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$ Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.
| Using the binomial theorem,
\begin{align*}
\left( 1 - \frac{x}{3} \right)^5 &= \sum_{i=0}^{5} \binom{5}{i} 1^{5-i} \left(\frac{-x}{3} \right)^i \\
&= \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i \text{ and } \\
\left( 1 + \frac{2}{x} \right)^3 &= \sum_{j=0}^{3} \binom{3}{j}1^{3-j} \left(\frac{2}{x} \right)^j \\
&= \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j}
\end{align*}
So \begin{align*}
\left( 1 - \frac{x}{3} \right)^5 \left( 1 + \frac{2}{x} \right)^3
&= \left( \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i \right) \left( \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j} \right) \\
&= \sum_{i=0}^5 \sum_{j=0}^{3} \binom{5}{i} \binom{3}{j} (-3)^{-i} 2^j x^{i-j}
\end{align*}
The only contributions to $x^0$ have $i - j = 0$, so $i = j$. The range of $i$ is $[0,5]$ and the range of $j$ is $[0,3]$. The intersection of these two intervals is $[0,3]$, so we have four terms contributing to the coefficient of $x^0$:
$$ \sum_{j=0}^{3} \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j $$
(It seems that we have done a lot of work to get here. With practice, you can jump straight to this expression.)
Call the four terms $a_j = \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j$ for $j \in [0,3]$. Notice
$$ a_0 = \binom{5}{0} \binom{3}{0} (-3)^{-0} 2^0 = 1 \text{.} $$
Then
\begin{align*}
a_1 &= a_0 \cdot \frac{5}{1} \cdot \frac{3}{1} \cdot \frac{-1}{3} \cdot 2 = -10 \text{,} \\
a_2 &= a_1 \cdot \frac{4}{2} \cdot \frac{2}{2} \cdot \frac{-1}{3} \cdot 2 = \frac{40}{3} \text{, and} \\
a_3 &= a_2 \cdot \frac{3}{3} \cdot \frac{1}{3} \cdot \frac{-1}{3} \cdot 2 = \frac{-80}{27}
\end{align*}
This says we can compute the next term from the previous one. The second and third factors are the old mental arithmetic method for calculating rows of Pascal's triangle by multiplying by successive fractions having decreasing numerators and increasing denominators.
Then the coefficient of $x^0$ is
$$ 1 - 10 + \frac{40}{3} - \frac{80}{27} = \frac{37}{27} \text{.} $$
| {
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Prove the inequality $4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$ $$4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$$, $a,b,c\ge0$
I tried to work out this inequality the way I did
So we have to show $$ 4(a + b + c) ^ 3 \ge 27 (a ^ 2b + b ^ 2c + c ^ 2a + abc) $$
One way is to use cyclic symmetry, and WLOG assumes $ a $ is $ a, b, c $ minute. Then we can write $ b = a + x, c = a + y $, where $ x, y \ge 0 $. Now the inequality reduces to 9 $$ (x ^ 2-xy + y ^ 2) + (x-2y) ^ 2 (4x + y) \ge 0 $$ which is obvious. Also from the above, we get then equality is possible iff $ x = y = 0 $ or when $ a = 0, x = 2y $, i.e. when $ (a, b, c) = (1, 1, 1) $ or permutation $ (0, 2, 1) $.
I think it's a mistake please help
| The last inequality should be
$$9(x^2-xy+y^2)a+4x^3-15x^2y+12xy^2+4y^3 \geqslant 0,$$
or
$$9(x^2-xy+y^2)a+(4x+y)(x-2y)^2 \geqslant 0.$$
It's called BW method.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$ If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$
Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that this is true, but I don't seem to be able to prove it.
| Let $x = a^2+b^2, y = c^2+d^2 \implies x+y \le 1+xy \implies (x-1)(1-y) \le 0$, which is true since $ 0 \le x,y \le 1$.
| {
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Solving $x^2 + py^2 = 4z^3$ with a fixed $z$ Given an odd prime $p$ and a positive integer $z$, I'd like to find all positive integers $x$, $y$ such that
$$
x^2 + p y^2 = 4 z^3.
$$
Because the left is positive definite there is only a finite number of solutions.
I could solve by incrementally guessing values of $y$ and checking if $4z^3-py^2$ is a square, until $py^2> 4 z^3$. But this is slow, and many times the answer just turns out to be there are no solutions, or wasted time after finding the first solution which turns out to be unique.
I've learned the left hand side is called a binary quadratic form. Which has a class size $h(-4p)$. If it somehow helps, assume I know this value.
*
*Is there an easy way to determine that, if a solution exists, it is unique?
*Is there an easy way to find all the solutions?
| I recommend Lehman as an introduction, simultaneous to binary quadratic forms and quadratic number fields.
I'm going to ignore the $4.$ It is there for a reason, but things begin more simply without it.
$$\left( x^2 + p y^2 \right)^3 = \left( x^3 -3 px y^2 \right)^2 + p \left( 3x^2 y - p y^3 \right)^2 $$
When $p \equiv 3 \pmod 8$ ( and $p>3$) we have
class number divisible by 3. Sometimes we can use the easiest description (available if we can solve $a^3 - t^2 = p$):
$$\left( ax^2 +2txy + a^2 y^2 \right)^3 = \left(tx^3 + 3a^2x^2 y + 3atxy^2 +(t^2 - p) y^3 \right)^2 + p \left( x^3 - 3axy^2 - 2t y^3 \right)^2 $$
The smallest primes for which this works are $11, 19, 67, 83, 107.$ When there is no such expression, for example primes $43, 59,$ there is a less formulaic way to write it out, but there are still infinitely many cubes where your $z$ has no expression as $x^2 + p y^2$
Example $p=11, a=3,t=4.$
$$\left( 3x^2 +8xy + 9 y^2 \right)^3 = \left(4x^3 + 27x^2 y + 36xy^2 +5 y^3 \right)^2 + 11 \left( x^3 - 9xy^2 - 8 y^3 \right)^2 $$
Example $p=19, a=7,t=18.$
$$\left( 7x^2 +36xy + 49 y^2 \right)^3 = \left(18x^3 + 147x^2 y + 378xy^2 +305 y^3 \right)^2 + 19 \left( x^3 - 21xy^2 - 36 y^3 \right)^2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.
Can you help me?
I have not seen determinats yet
| Suppose that we have $$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
and we want $Q$ such that $AQ =B$. Let
$$ Q = \begin{pmatrix} a & b &c \\ d & e & f \\ g & h & i \end{pmatrix}$$
Then using matrix multiplication you'd see that
$$ AQ = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix} \begin{pmatrix} a & b &c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a + \frac{g}{3} & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} $$
now set it equal to $B$
$$ \begin{pmatrix} a + \frac{g}{3} & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
Multiply by $3$ on both sides and we get
$$ \begin{pmatrix} 3a + g & 3b + h & 3c + i \\ -g & -h & -i \\ 3a + 3d + 2g & 3b + 3e 2h & 3c + 3f + 2i \end{pmatrix} = \begin{pmatrix} 4 & 1 & 3 \\ -1 & -1 & 0 \\ 5 & 8 & 6\end{pmatrix}$$
Then you directly get $g,h,i$
$$ -g = -1 \implies g = 1 \\ -h = -1 \implies h = 1 \\ -i = 0 \implies i=0$$
Now substitute those into the top row
$$ 3a +1 = 4 \implies a =1 \\ 3b + 1 = 1 \implies b = 0 \\ 3c = 3 \implies c =1 $$
then continue and get the last three
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing a double integral over $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$
Compute the integral $\int_{D} f \ dx \ dy$, when $f : D \to \mathbf{R}, f(x,y)= y^2x+x^3$ and $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$
It seems that $D$ defines a semicircle with radius $5$ in the $xy$-plane. So converting to polar coordinates I have $x=r\cos\theta, y=r\sin\theta$ and the limits for integration would be $r \in [0,5]$ and $\theta \in [0, \pi]$, thus the integral would become $$\int_{0}^{\pi} \int_{0}^{5} r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta \ dr \ d\theta.$$
However I doubt that this is correct since seems to evaluate to zero?
| Note that
$$
\begin{split}
I
&= \int_0^\pi \int_0^5
\left(r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta\right) r\ dr \ d\theta \\
&= \int_0^5 r^4 dr
\int_0^\pi \left(\sin^2\theta +\cos^2 \theta\right) \cos\theta d\theta \\
&= \frac{5^5}{5} \int_0^\pi \cos\theta d\theta
\end{split}
$$
which does evaluate to zero since cosine is symmetric around $\pi/2$...
A different way to see it without going polar is the boundary curve (since $y>0$) is given by $y < \sqrt{25-x^2}$ and $f(x,y) = x(x^2+y^2)$, which is odd in $x$, and since we will integrate an odd function over a symmetric interval, the integral will be zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Laurent Series Expansion of $f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$ at $z=0$ For my function $$f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$$ I have determined that it contains $3$ poles at $z=0,z=-a,z=-\frac{1}{b}$
I believe these to be the only singularities.
I found them by factoring the denominator
$$f(z) = \frac{bz^2+2z+b}{z(bz^2+(1+b^2)z+b)}$$
Then clearly there is a pole at $z=0$ and the remaining singularities are found by solving the quadratic
$$bz^2+(1+b^2)z+b = 0$$
$z = \frac{-(1+b^2) + \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$ and $z =\frac{ -(1+b^2) - \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$
This simplifies to
$z = \frac{-1-b^2 -b^2 +1}{2b}$ and $z=\frac{-1-b^2+b^2-1}{2b}$
$z = \frac{-2b^2}{2b}$ and $z= \frac{-2}{2b}$
So,
$z=-b$ and $z=-\frac{1}{b}$
I want to Laurent Expand these singularities and determine the radius of convergence for each singular point.
All help is appreciated. I want to focus on the $z=0$, once I have clarified the process I will be able to expand the remaining myself.
| Before factoring take notice of the similarity of the coefficients in the numerator and denominator. This suggests that there is a short cut that can be taken before attempting to partial fraction that would make our lives easier.
$$\frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz} = \frac{bz^2+(1+b^2)z+b+(1-b^2)z}{bz^3+(1+b^2)z^2+bz} = \frac{1}{z}+\frac{1-b^2}{bz^2+(1+b^2)z+b}$$
Now partial fractions is automatic
$$\frac{1}{z}+\frac{1-b^2}{(z+b)(bz+1)} = \frac{1}{z}+\frac{1}{z+b}-\frac{1}{bz+1}$$
$z=0$ won't actually provide illumination on what to do for the other singularities, so instead I will show the procedure for $z=-b$. Every term needs to be centered at $z=-b$ like so
$$\frac{1}{(z+b)-b}+ \frac{1}{z+b} + \frac{1}{b(z+b)+1-b^2} = -\frac{1}{b}\frac{1}{1-\frac{(z+b)}{b}}+ \frac{1}{z+b} + \frac{1}{1-b^2}\frac{1}{\frac{b(z+b)}{1-b^2}+1}$$
The scaling is done because it is important to have the $+1$ to apply geometric series. Now we assume the quantity $z+b$ is small so we can use the Taylor expansion
$$-\frac{1}{b}\sum_{n=0}^\infty \frac{(z+b)^n}{b^n} + \frac{1}{z+b} + \frac{1}{1-b^2}\sum_{n=0}^\infty (-1)^n\left(\frac{b}{1-b^2}\right)^n(z+b)^n$$
The case $b=1$ needs to be handled separately since in that case $b=\frac{1}{b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to simplify the fraction? How to simplify the fraction $ \displaystyle \frac{\sqrt{3}+1-\sqrt{6}}{2\sqrt{2}-\sqrt{6}+\sqrt{3}+1} $ to $ (\sqrt{2}-1)(2-\sqrt{3}) $?
I've checked it in the calculator and both give the same value.
I was able to simplify it until $\displaystyle \frac{1+\sqrt2-\sqrt3}{3+\sqrt2+\sqrt3} $ which also gives the same value.
For anyone who is interested, $ (\sqrt{2}-1)(2-\sqrt{3}) =tan\frac{\pi}{8}tan\frac{\pi}{12} $ and I need this as I was trying to use coordinate geometry to solve the following
question.
| \begin{align}
\frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} &= \frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} \times \frac {3 - \sqrt 2 - \sqrt 3}{3 - \sqrt 2 - \sqrt 3}\\
&=\frac {3 - \sqrt 2 -\sqrt 3+3\sqrt 2-2-\sqrt 6-3\sqrt 3+\sqrt 6+3}{9 - (2 + 2\sqrt 6 + 3)}\\
&=\frac {4+2\sqrt 2-4\sqrt 3}{4 -2\sqrt 6}\\
&=\frac {2+\sqrt 2-2\sqrt 3}{2 -\sqrt 6} \times \frac {2 +\sqrt 6}{2 +\sqrt 6}\\
&=\frac {4 + 2\sqrt 2 - 4\sqrt 3 + 2\sqrt 6+2\sqrt 3 - 6\sqrt 2}{4-6}\\
&=\frac {4 -4\sqrt 2 - 2\sqrt 3 + 2\sqrt 6}{-2}\\
&=-2+2\sqrt 2 +\sqrt 3-\sqrt 6\\&=-2(1-\sqrt 2)+\sqrt 3(1-\sqrt 2)\\
&=(2-\sqrt 3)(\sqrt 2-1)
\end{align}
Just bash away.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$
*$\left(1- \frac {x+1}{4x+3}\right)^x$
*$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$
*$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$
*$e^{\infty} = \infty$
answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!
| First of all let investigate terms of inside the parentheses. When $x$ approaches to infinity $\displaystyle \frac{3x+2}{4x+3}=\frac{3}{4}$ then the limit transformed into $\displaystyle \lim_{x \to \infty} \left(\frac{3}{4}\right)^{x}$. Rewritre $\frac{3^x}{4^x}$. Obviously denomiter is greater than numerator. So $\lim$ approaches to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Calculating series expansions within a matrix: matrix exponential I have a $(3 \times 3)$ matrix $$ Y = \begin{pmatrix} 0 & - e^{-i \theta} & 0 \\ e^{i \theta} & 0 & - e^{-i \theta} \\ 0 & e^{i \theta} & 0 \end{pmatrix} $$ for which I would like to calculate the matrix exponential $\exp(t Y) = I + t Y + \frac{t^2 Y^2}{2!} + \ldots $ If I let $z : = e^{i \theta}$, I have $$ Y^2 = \begin{pmatrix} - |z|^2 & 0 & |z|^2 \\ 0 & -2 |z|^2 & 0 \\ |z|^2 & 0 & - |z|^2 \end{pmatrix} \\ Y^3 = \begin{pmatrix} 0 & 2 \overline{z} |z|^2 & 0 \\ |z|^2 (-z - \overline{z}) & 0 & |z|^2 (z + \overline{z}) \\ 0 & -2z |z|^2 & 0 \end{pmatrix} $$ and $$ Y^4 = \begin{pmatrix} - \overline{z} |z|^2 (-z - \overline{z}) & 0 & - \overline{z} |z|^2 (z + \overline{z}) \\ 0 & 4 |z|^4 & 0 \\ z |z|^2 (-z- \overline{z}) & 0 & z |z|^2 (z+ \overline{z}) \end{pmatrix}. $$ Setting $|z| = 1$ and calculating the matrix exponential above up the fifth power $Y^5$, I got $$ \begin{pmatrix} 1 - \frac{t^2}{2!} + \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots & - t \overline{z} + \frac{t^3}{3!} (2 \overline{z}) - \frac{t^5}{5!} 4 \overline{z} + \ldots & \frac{t^2}{2!} - \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots \\ tz - \frac{t^3}{3!} (z + \overline{z}) + \frac{t^5}{5!} 2 (z + \overline{z}) + \ldots & 1 - \frac{2 t^2}{2!} + \frac{t^4}{4!} 4 + \ldots & - t \overline{z} + \frac{t^3}{3!} (z + \overline{z}) - \frac{t^5}{5!} 2 ( z+ \overline{z}) + \ldots \\ \frac{t^2}{2!} - \frac{t^4}{4!} z (z + \overline{z}) + \ldots & tz - \frac{t^3}{3!} 2 z + \frac{t^5}{5!} 4 z + \ldots & 1 - \frac{t^2}{2!} + \frac{t^4}{4!} z (z + \overline{z}) + \ldots
\end{pmatrix} $$
I think I must be able to rewrite this with the help of $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots$.
For example, if I look at the $a_{22}$ term above, I see that it is almost $\cos(t)$, except the numerical factors that don't work out. Also, the $a_{11}$ term is almost $\cos(t)$, except there appears a term $\overline{z} (z+ z)$ from the fourth power onwards, and the same happens for the $a_{33}$ term with $z$ and $\overline{z}$ switched. The $a_{32}$ term seems to be $z \sin(t)$, but again the numerical coefficients don't work out.
Question: Does anyone recognize the pattern in these entries (i.e. the series) and is able to calculate the matrix exponential $e^{tY}$ in closed form?
Also, what would be matrix exponential $\exp(tZ)$ of the 'generalization' $$Z = \begin{pmatrix} 0 & - \overline{z} & - \overline{z} \\ z & 0 & - \overline{z} \\ z & z & 0 \end{pmatrix} $$ with $z = e^{i \theta}$ again?
| Setting $z = e^{i \theta}$ is a good idea. It becomes a bit clearer if $(- e^{-i \theta})$ is replaced by $-1/z$ instead of $-\overline z$ (and it makes the result correct even for complex $\theta$).
So we have
$$
Y = \begin{pmatrix}
0 & -1/z & 0 \\
z & 0 & -1/z \\
0 & z & 0
\end{pmatrix}
$$
and the first powers are
$$
Y^2 = \begin{pmatrix}
-1 & 0 & 1/z^2 \\
0 & -2 & 0 \\
z^2 & 0 & -1
\end{pmatrix}\, , \,
Y^3 = \begin{pmatrix}
0 & 2/z & 0 \\
-2z & 0 & 2/z \\
0 & -2z & 0
\end{pmatrix}\,. \\
$$
One can see that $\boxed{Y^3 = -2Y}$, which allows to compute all powers $Y^n$ in terms of $Y$ or $Y^2$:
$$
Y^{2k+1} = (-2)^{k} Y \\
Y^{2k+2} = (-2)^{k} Y^2
$$
for $k \ge 1$. Therefore
$$
\begin{align}
\exp(tY) &= I + \left(t-\frac{2t^3}{3!} + \frac{2^2t^5}{5!} - \frac{2^3t^7}{7!} + \ldots\right)Y
\\ &\quad + \left(\frac{t^2}{2!} - \frac{2t^4}{4!} + \frac{2^2t^6}{6!} - \frac{2^3t^8}{8!} + \ldots \right)Y^2 \\
&= I + \frac{\sin(\sqrt 2 t)}{\sqrt 2}Y + \frac 12 \left(1- \cos(\sqrt 2 t)\right)Y^2 \, .
\end{align}
$$
The general case is described in Computing the Matrix Exponential The Cayley-Hamilton Method: If $A$ is an $n$-dimensional square matrix and $\lambda_1, \ldots, \lambda_n$ the zeros of the characteristic equation $\det(\lambda I - A) = 0$, then
$$
\exp(tA) = \sum_{k_0}^{n-1} \alpha_k A^k
$$
where $\alpha_0, \ldots, \alpha_{n-1}$ are the solutions of the linear equation system
$$
e^{\lambda_i t} = \sum_{k_0}^{n-1} \alpha_k \lambda_i^k \, , \,
1 \le i \le n \, .
$$
In our case $\det(\lambda I - Y) = \lambda^3 + 2 = 0$ has the zeros $\lambda_1 = 0$, $\lambda_2 = i\sqrt 2$, $\lambda_3 = -i \sqrt 2$. The linear equation system is
$$
\begin{align}
1 &= \alpha_0 \\
e^{i\sqrt 2 t} &= \alpha_0 + i \sqrt 2 \alpha_1 - 2 \alpha_2 \\
e^{-i\sqrt 2 t} &= \alpha_0 - i \sqrt 2 \alpha_1 - 2 \alpha_2
\end{align} \, .
$$
The solution is
$$
\alpha_0 = 1, \, \alpha_1 = \frac{\sin(\sqrt 2 t)}{\sqrt 2}, \,
\alpha_2 = \frac 12 \left(1- \cos(\sqrt 2 t)\right)
$$
confirming the result for $\exp(tY)$ that we obtained above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Frenet frame field of a curve $\vec{C}(t)$ = $(\frac{t^3}{3}, 2t-1,t^2+2)$ Working out the unit tangent $T$
$\vec{r} = \frac{t^3}{3}i + (2t-1)j + (t^2+2)k$
$\frac{dr}{dt} = t^2 +2j+2tk$
$\frac{ds}{dt} = |\frac{dr}{dt}| = \sqrt{t^4+4+2t^2} = t^2+2$
$T = \frac{dr}{ds} = \frac{dr}{dt}/\frac{ds}{dt} = \frac{t^2i+2j+2tk}{t^2+2}$ = $\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k$
Now working out the Principal Normal N,
$\frac{dT}{dt} = \frac{4t}{t^4+4}i-\frac{4t}{t^4+4}j-\frac{2t^2+4}{t^4+4}$ By the Quotient Rule
$\frac{dT}{ds} = \frac{dT}{dt}/\frac{ds}{dt} = \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$
Now $N$ is $\frac{1}{k}$$\frac{dT}{ds}$ so in order to find $N$, I must first find $k$.
$k$ = $|\frac{dT}{ds}$| and this is where is becomes problematic. Since $|\frac{dT}{ds}$| = $\sqrt{(\frac{4t}{t^6+8})^2i-(\frac{4t}{t^6+8})^2j-(\frac{2t^2+4}{t^6+8})^2k}$
=$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}$
So $N$ = $\frac{1}{$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}} \times \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$
Which I don't know what it equals to.
I wonder how I would be able to work out the Binormal B, knowing that $B = T \times N$.
Maybe I'm making a mistake... Any tips?
| You've missed an $i$ in your calculations for $\frac{d\vec{r}}{dt}$.
$$\frac{d\vec{r}}{dt} = t^2i +2j+2tk$$
Even though the final result for $\frac{ds}{dt}$ is good, the rightmost term in the radical sign is incorrect.
$$\frac{ds}{dt} = \left| \frac{dr}{dt} \right| = \sqrt{t^4+4+(2t)^2} = t^2+2$$
Luckily, you put back the $i$ in the expression for $\vec{T}$.
However, you've incorrectly squared the denominator $t^2 +2$ as $t^4 + 4$.
The correct $\frac{d\vec{T}}{dt}$ should be
$$\frac{d\vec{T}}{dt} = \frac{4t}{(t^2+2)^2}i-\frac{4t}{(t^2+2)^2}j-\frac{2(t^2-2)}{(t^2+2)^2}k$$
because $$\frac{d}{dt} \frac{2t}{t^2+2} = \frac{2(t^2+2)-2t(2t)}{(t^2+2)^2} = \frac{4-2t^2}{(t^2+2)^2}.$$
You've got the denominator wrong in the same way.
$$\frac{d\vec{T}}{ds} = \frac{d\vec{T}}{dt}/\frac{ds}{dt} = \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k$$
\begin{aligned}
\kappa &= \left| \frac{d\vec{T}}{ds} \right| \\
&= \frac{\sqrt{(4t)^2+(4t)^2 + 4(t^2-2)^2}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(4t^2+4t^2+t^4-4t^2+4)}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(t^4+4t^2+4)}}{(t^2+2)^3} \\
&= \frac{2}{(t^2+2)^2}
\end{aligned}
So
\begin{aligned}
\vec{N} &= \frac1\kappa \frac{d\vec{T}}{ds} \\
&= \frac{(t^2+2)^2}{2} \left[ \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k \right] \\
&= \frac{1}{2(t^2+2)} [4ti-4tj-2(t^2-2)k] \\
&= \frac{2ti-2tj-(t^2-2)k}{t^2+2}
\end{aligned}
Finish the exercise with $\vec{B} = \vec{T} \times \vec{N}$.
\begin{aligned}
\vec{B} &= \vec{T} \times \vec{N} \\
&= \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2-2)k}{t^2+2} \\
&= \frac{1}{(t^2+2)^2} [(-2(t^2-2)-(-2t)(2t))i - ((t^2)(-(t^2-2))-(2t)(2t))j + ((t^2)(-2t)-(2t)(2))k] \\
&= \frac{2(t^2+2)i+t^2(t^2+2)j-2t(t^2+2)k}{(t^2+2)^2} \\
&= \frac{2i+t^2j-2tk}{t^2+2}
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\lim\limits_{x \to 0}(\frac{\sin x}{x})^{\cot ^2x}$
Calculate
$$\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}$$
I tried to use the natural logarithm and L'hopital rule but it didn't help me.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\lim\limits_{x \to 0}\frac{\ln\frac{\sin x}{x}}{\tan ^2x}=\lim\limits_{x \to 0}\frac{x}{2\sin x\tan x \sec^2x}$
| Composing Taylor series one piece at the time
$$y=\Bigg[\frac{\sin (x)}{x}\Bigg]^{\cot ^2(x)} \implies \log(y)={\cot ^2(x)}\log\Bigg[\frac{\sin (x)}{x}\Bigg]$$
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\Bigg[\frac{\sin (x)}{x}\Bigg]=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^6\right)$$
$$\cot^2(x)=\frac{1}{x^2}-\frac{2}{3}+\frac{x^2}{15}+\frac{2 x^4}{189}+O\left(x^5\right)$$
$$\log(y)=-\frac{1}{6}+\frac{19 x^2}{180}-\frac{22 x^4}{2835}+O\left(x^6\right)$$
$$y=e^{\log(y)}=\frac{1}{\sqrt[6]{e}}+\frac{19 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bound on subsequent terms of inductively defined sequence Let $(a_n)$ be the sequence defined by $a_0=1$ and $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$. Show that $|a_{n+1}-a_n|\leq(\frac{1}{2})^{n+1}$.
Using the induction hypothesis I can show that
$$|a_{n+1}-a_n|\leq \frac{|a_n-a_{n-1}|}{2} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|\leq (\frac{1}{2})^{n+1} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|.$$
I cannot get rid off the extra term and I have tried different approaches, but this one seems like the most promissing one because the bound appears naturally. I think I am missing something obvious.
Thank you.
| Assume you want $|a_{n+1}-a_n| \le \frac{1}{2^{n+1}}$
First $a_n\ge 0$. If $n\ge 1$ then $a_n \ge 2\sqrt{\frac{a_{n-1}}{2} \cdot \frac{1}{a_{n-1}}} = 2\sqrt{\frac 12}=\sqrt 2$ via AM-GM.
$a_{n+1} - a_n = \frac{1}{a_n} - \frac{a_n}{2}=\frac{2-a_n^2}{2a_n} \le 0$
Lastly,
$$a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}, a_n=\frac{a_{n-1}}{2} + \frac{1}{a_{n-1}}\\
\implies a_{n+1} - a_n = (a_n-a_{n-1}) \left( \frac 12 - \frac{1}{a_n a_{n-1}}\right) \\
\implies |a_{n+1} - a_n | \le \frac 12|a_n-a_{n-1} | \le \frac{1}{2^n}|a_1-a_0|=\frac{1}{2^{n+1}}$$
Note: we have $0 \le \frac 12 - \frac{1}{a_n a_{n-1}} \le \frac 12$. LHS because $a_n\ge \sqrt 2 $, RHS because $a_n>0$
| {
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"url": "https://math.stackexchange.com/questions/3960989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.
What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?
| The equation's defined for $\;x>1\;\;\text{or}\;\;x\le-1\;$ , so now square the whole thing:
$$x^2-1=(x+5)^2\,\frac{x+1}{x-1}\iff x^3-x^2-x+1=x^3+11x^2+35x+25\iff$$
$$12x^2+36x+24=0\iff12(x+1)(x+2)=0\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
| Hint:
Multiply for $12\sqrt{x^2-1}$ your initial equation,
you have $$12(\sqrt{x^2-1})x+12x=35\sqrt{x^2-1} \iff 12x(\sqrt{x^2-1}+1)=35\sqrt{x^2-1}$$
and square (two times to delete the sqrt) to the left and the right.
$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$
$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$
This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$
I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.
| sometimes its easier to break it up into other expressions first:
$$I=\int x^3\sqrt{x^2-9}$$
$x=3u\Rightarrow dx=3du$
$$I=\int(3u)^3\sqrt{3^2(u^2-1)}(3du)$$
$$I=3^5\int u^3\sqrt{u^2-1}du$$
now we want to look for an identity that matches this, we know:
*
*$\cosh^2\theta-\sinh^2\theta=1$
*$\cos^2\theta+\sin^2\theta=1$
so we want to choose one of these that can be rearranged to our form, the nicest would be: $\cosh^2\theta-1=\sinh^2\theta$ so:
$u=\cosh(v)\Rightarrow du=\sinh(v)dv$ now replace everything:
$$I=3^5\int\cosh^3v\sqrt{\cosh^2v-1}\sinh v dv$$
$$I=3^5\int\cosh^3v\sinh^2vdv$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dirichlet's Integral to find Volume with gamma function values Evaluate volume $\iiint_D dx dy dz$ over a domain $D$ where $D$ is the region bounded by $x\ge 0$, $y\ge 0$ and $z\ge 0$ and ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$.
My approach:
I used the RMS$\ge$AM inequality (since $x,y,z\ge 0$ and $a,b,c>0$ without loss of generality) to arrive at $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\le\sqrt{3}$
and then did a substitution $X=\frac{x}{a}$, $Y=\frac{y}{b}$ and $Z=\frac{z}{c}$
And I got answer to be
$\frac{abc(\cdot\Gamma(1)\Gamma(1)\Gamma(1)\cdot(\sqrt{3})^3)}{\Gamma(4)}$ ie $abc\sqrt{3}/2$
My friend's approach:
Direct substitution of $X=\frac{x^2}{a^2}$, $Y=\frac{y^2}{b^2}$ and $Z=\frac{z^2}{c^2}$ and his answer was
$(\frac{abc}{8})\frac{(\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}))}{\Gamma(\frac{5}{2})}$ ie $\frac{abc\cdot\pi}{6}$
Professor says mine is wrong and friend's method is right but I am not satisfied without the reason.
| Did you just replace the ellipsoid $(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2\le 1$ with the octahedron $|\frac{x}{a}|+|\frac{y}{b}|+|\frac{z}{c}|\le\sqrt{3}$ ?
No wonder you are getting a different result for its volume. The RMS$\ge$AM
inequality guarantees that the ellipsoid is inscribed in the octahedron, i.e. fully contained in it - but of course, they are not the same bodies.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation using $f(x)=\frac{\sin\pi x}{x^2} $ If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$
Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.
Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.
Then which of the options is(are) correct.
(A) $x_1<y_1$
(B) $x_{n+1}-x_n>2$ for all $n$
(C) $x_n \in (2n,2n+\frac{1}{2})$ for every $n$
(D) $|x_n-y_n|>1$ for every $n$
The official answer is $B,C,D$
My approach is as follow
$$y' = \frac{x^2 \times \pi \times \cos \pi x - 2x\sin \pi x}{x^4}$$
$$y' = \frac{x^2 \times \pi \times \cos \pi x - 2x\sin \pi x}{x^4} = \frac{x \times \pi \times \cos \pi x - 2\sin \pi x}{x^3} = \frac{2\cos \pi x \left( \frac{\pi x}{2} - \tan \pi x \right)}{x^3}$$
We should consider $x\in\frac{1}{2},\frac{3}{2},\ldots.$ as inflection point even though $\tan\pi x$ is not defined because we need to consider the numerator part $\pi x\cos \pi x - 2\sin \pi x$
| $$f'(x)=\frac{\pi x \cos (\pi x)-2 \sin (\pi x)}{x^3}$$
$$f'(x)=0\to \sin(\pi x)=\pi x\cos(\pi x)\tag{1}$$
Second derivative is
$$f''(x)=-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi x)+4 \pi x \cos (\pi x)}{x^4}$$
To understand which are max/min we use second derivative test using relation $(1)$ in the second derivative
$$f''(x^*)-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi x)+4 \sin(\pi x)}{x^4}=-\frac{\left(\pi ^2 x^2+4 \pi -6\right) \sin (\pi x)}{x^4}$$
As the factor $\left(\pi ^2 x^2+4 \pi -6\right) $ is positive for any $x$ we have that second derivative is negative when $\sin(\pi x)>0$ that is $2k\pi <\pi x<(2k+1)\pi$ $\forall k\in\mathbb{N}$ and finally $2k<x<2k+1$.
In other words the solutions of $(1)$
are max when
$ 2k<x<2k+1$ and are minimum when $k+1<x<2k$
So we can see that statement (A) is false.
| {
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"timestamp": "2023-03-29T00:00:00",
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show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$
| You can convert the recurrence equation into a differential equation by multiplying it with $x^n$, summing over $n=1,2,3,...$ and defining $f(x)=\sum_{n=0}^\infty a_{n+1} x^n$, which gives the first order DE $$\sum_{n=1}^\infty a_{n+1}x^n = 2\sum_{n=1}^\infty a_n x^n + \sum_{n=1}^\infty \frac{a_n}{n} x^n - \sum_{n=1}^\infty \frac{x^n}{n} \\
\Rightarrow \quad f(x)-a_1=2x f(x) + \sum_{n=1}^\infty \frac{a_n}{n} x^n + \log(1-x) \\
\Rightarrow \quad (1-2x)f'(x) - 3f(x) + \frac{1}{1-x} = 0$$
that can be solved in full generality. You can use that formula or also just check that $$f(x)=-\frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} + \frac{2}{1-2x} + \frac{\pi-c}{(1-2x)^{3/2}}$$
solves the ODE where $c$ is an arbitrary constant. The initial condition $$-\frac{\pi}{2}+2 + \pi -c=f(0)=a_1=2-\frac{\pi}{2}$$ implies $c=\pi$ and it is easy to see that this very special case removes the singularity at $x=1/2$. This singularity would lead to the violation of the inequality $a_n \leq \frac{1}{n+1}$ as the coefficients $a_{n+1}$ would then obey $\sim (-2)^n\binom{-3/2}{n}$ asymptotically. Now we express $f(x)$ as an integral and use the geometric series $$\frac{2}{1-2x} - \frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} = \int_0^1 \frac{2t^2}{1+(1-2x)t^2} \, {\rm d}t = \sum_{k=0}^\infty x^k \, \underbrace{2^{k+1} \int_0^1 {\rm d}t \, \frac{t^{2k+2}}{(1+t^2)^{k+1}}}_{a_{k+1}}$$
where $0<x<1/2$ was assumed. Interchanging summation and integration is allowed by dom. convergence. Hence it needs to be shown that $$0\leq \frac{1}{n+1} - 2^n \int_0^1 {\rm d}x \, \frac{x^{2n}}{(1+x^2)^{n}} = \int_0^1 {\rm d}x \left( x^n - \frac{2^n x^{2n}}{(1+x^2)^{n}} \right) \\ = \int_0^1 {\rm d}x \, \frac{x^n}{(1+x^2)^n} \left\{ (1+x^2)^n - (1+1)^n x^n \right\} \, .$$ From this it suffices to show
$$0\leq (1+x^2)^n - (1+1)^n x^n \\
= \sum_{k=0}^n \binom{n}{k} \left(x^{2k} - x^n\right)
= \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{n/2 < k \leq n} \binom{n}{k} (x^{2k} - x^n) \\
= \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{0 \leq k < n/2} \binom{n}{n-k} (x^{2(n-k)} - x^n)
=\sum_{0 \leq k < n/2} \binom{n}{k} \left( x^k - x^{n-k} \right)^2 \, .$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| For $|x| \le 1$ the Taylor series
$$
\cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
is an alternating series with terms that decrease in absolute value. It follows that for these $x$
$$
\cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{x^2}{4!} = 1 - \frac{11}{24} x^2
$$
and therefore
$$
1 - \cos(x) \ge \frac{11}{24} x^2 \, .
$$
That is an even better estimate since $\frac{11}{24} > \frac 13$.
The same approach can be used to show that $1 - \cos(x) \ge \frac{x^2}{3}$ holds on the larger interval $[-2, 2]$:
$$
\cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{4 x^2}{4!} = 1 - \frac 13 x^2
$$
because the $x^{2n}/({2n})!$ terms decrease in absolute value for $n \ge 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many rectangles can be found in this shape?
Question: How many rectangles can be found in the following shape?
My first solution:
Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1$ to the shape with side length $n$ is, $n^2$. Similarly, the number of the newly added $2 \times k$ rectangles while expanding is, $(n-2)^2$, the number of the $3 \times k$ ones is $(n-4)^2$, and so on. So, we have, $$a_n = a_{n-1} + 1^2+3^2+\ldots +n^2$$ or, $$a_n = a_{n-1} + 2^2+4^2+\ldots +n^2$$ depending on the parity of $n$. In either case, we have $$a_n=a_{n-2}+1^2+2^2+ \ldots n^2$$ So, $a_{n+1} = a_{n-1} + 1^2+2^2+ \ldots (n+1)^2$. Subtracting both sides, we have, $a_{n+1}-a_n = a_{n-1} - a_{n-2} + (n+1)^2$. Repeating the same method, we obtain $$a_{n+1}-4a_n+5a_{n-1} = 5a_{n-3} - 4a_{n-4}+a_{n-5}$$ This recurrence relation has characteristic equation $r^6-4r^5+5r^4-5r^2+4r-1=0 \iff (r-1)^5 \cdot (r+1) = 0$. So, using the usual ways to solve recurrence relations, we can obtain $a_n=\frac{n^4}{24} + \frac{n^3}{4} + \frac{11n^2}{24} + \frac{n}{4} = \binom{n+3}{4}$.
However, since the answer is so simple, I tried to solve it with a method which can give the result in a more 'direct' way:
Let us find these rectangles by choosing 2 rows and 2 columns and looking at their intersection. If we choose the right most column, we have $n$ choices for the other column and only $1$ choice for the rows. If we choose column which is the second from right, we have $n-1$ choices for the other column (in order not to over count), and, $\binom{2+2-1}{2}$ (number of ways to choose 2 rows out of 2 rows including repetitions) choices for the row. Using this argument, we can obtain, $$a_n = \sum_{k=0}^{n-1} (n-k) \cdot \binom{(k+1)+2-1}{2} = \sum_{k=0}^{n-1} (n-k) \cdot \binom{k+2}{2}=\sum_{k=0}^{n-1} \frac{(n-k) \cdot (k+2) \cdot (k+1)}{2}$$ After some computation, we can see from here that $a_n=\binom{n+3}{4}$.
However, I think this solution is also overcomplicated and I think that there is a really elegant and/or 'direct' solution for this problem. So, my question is, how can this problem be solved with a 'direct' method (for example, a combinatorial method which immediately gives the number as $\binom{n+3}{4}$)? It would be nice to see your 'nice' solutions too, thank you.
| Label the lines starting from $1$ on the left to $n+1$ on right/above. Extend the lines by $1$ unit to create tetherings. There are $n+3$ of them, label these from $1$ to $n+3$. The diagram is for $n=5$.
There is a bijection between the coordinates of four vertices of any rectangle and the coordinates at end of tetherings which determine the four edges of the given rectangle.
For example, the rectangle with vertices at $(1,1),(5,1),(5,3),(1,3)$ corresponds to the unique quadruplet $(1,3,4,8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?
| To get things going you can calculate the inverse of $[3]_{100}$ and then try relating it to powers of $[3]_{100}$, keeping in mind that $10^2 \equiv 0 \pmod{100}$.
In $\text{modulo-}100$ you know that the inverse has the form $[10n +7]_{100}$ and solving you can write out
$\tag 1 3 \cdot 67 \equiv 1 \pmod{100}$
We can form a relation right away, suspending further ${[3^n]_{100}}$ calculations,
$\quad 3^3 \equiv 67 - 40$
and so
$\tag 2 3^4 \equiv 1 - 20 \pmod{100}$
We can now use the binomial theorem.
$\quad \displaystyle \large 3^{1234} \equiv (3^4)^{308} \cdot 3^2 \equiv$
$\quad \displaystyle \large 9(1-20)^{308} \equiv$
$\quad \displaystyle \large 9(1-308 \cdot 20^1 + \Large 0 \large ) \equiv \quad \quad \normalsize \; (20^k \text{ usually has residue } 0)$
$\quad \displaystyle \large 9 \cdot 41 \equiv 69 \pmod{100}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 & 4 \\ \hline
1 & 6 & 6 & 0 \\ \hline
2 & 26 & 30 & -4 \\ \hline
3 & 94 & 102 & -8 \\ \hline
4 & 306 & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.
| $$d(n+1)=3d(n)+8n-12,\\d(0)=4.$$
The homogeneous solution is
$$d_h(n)=3^nd_h(0).$$
Plugging the initial condition,
$$3^0d_h(0)=3\cdot4+8\cdot0-12=0.$$
Then with the ansatz $d_a(n)=an+b,$
$$an+a+b=3an+3b+8n-12$$ or by identification,$$d_a(n)=4-4n=d(n).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find the volume of the set in $\mathbb{R}^3$ defined by the equation $x^2 + y^2 \leq z \leq 1 + x + y$. Let $D=\left\{\left(x,y,z\right) \in \mathbb{R}^3: x^2 + y^2 \leq z \leq 1 + x + y \right\}$. I tried to integrate with cartesian coordinates, so that
$$\iiint_Ddxdydz = \int_{x=\frac{1-\sqrt{5}}{2}}^{{\frac{1+\sqrt{5}}{2}}} \int_{z=x^2}^{1+x} \left( \int_{y=z-(1+x)}^{\sqrt{z-x^2}}dy\right)dxdz$$
but the calculation seems to be quite laborious. Is there a way to solve it easily, maybe with another coordinate system?
Edit: After your suggestions I finally solved the problem. If someone is interested in the calculation itself I'll report the main steps below:
$$\int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}}\left(\int_{z=r^2+r(\sin\theta+\cos\theta)+\frac{1}{2}}^{r(\sin\theta+\cos\theta)+2}dz\right)rdrd\theta = \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}} \left( \frac{3}{2}r-r^3\right)drd\theta = \pi\int_{0}^{\frac{3}{2}}\left(\frac{3}{2}-t\right)dt = \frac{\pi}{2} \left[3t-t^2\right]_{0}^{\frac{3}{2}}=\frac{9\pi}{8}$$
| $\textbf{Hint}$: Use this modified cylindrical coordinate system:
$$\begin{cases}x = r\cos\left(\theta-\frac{\pi}{4}\right)\\ y = r\sin\left(\theta-\frac{\pi}{4}\right)\\ z = z \\\end{cases}$$
What does the change do? Does it alter the Jacobian (which measures how much the volume is distorted between coordinate systems) ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root? this might sound like a stupid question,
bear with me it probably is.
I know the solutions for $x=x^6$ are 1 and 0.
Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?
Thank you so much in advance.
| You find $x=0$ by using the property that if $A\cdot B=0$, and $A,B$ are real numbers (or elements of a field), then $A=0$ or $B=0$.
Write $x=x^6\iff x^6-x=0\iff x(x^5-1)=0$
Either $x^5-1=0$ or $x=0$. The last thing shows that $x=0$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$ Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$
where $a$ and $b$ are real numbers and not equal. Find the value of $\frac{a+b}{a\cdot b}$.
At first I applied integration by parts but then it got even more complicated and I couldn't solve it. How can I solve this problem?
| To be zero the integral, $h(x)=\log \left(\frac{1+a x}{1+b x}\right)$ must be odd, thus
$$\log \left(\frac{1+a x}{1+b x}\right)=-\log \left(\frac{1-a x}{1-b x}\right)$$
$$\log \left(\frac{1+a x}{1+b x}\right)+\log \left(\frac{1-a x}{1-b x}\right)=0$$
$$\log\frac{1-a^2x^2}{1-b^2x^2}=0$$
$$1-a^2x^2=1-b^2x^2\to b=-a$$
since the problem says $a\ne b$.
$h(x)=-h(-x)\to b=-a$
Must be added that $h(x)$ exists on $[-1/2,1/2]$ only if
$$\frac{1+a x}{1-a x}>0\to -2<a<2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3977522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$
Evaluate the series $$S=\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$$
I have tried many values of $(a,b,c)$ and plugged into Wolframalpha, it always converges. I know that for particular values of $a,b,c$, we solve it by forming a telescoping series by using the fact that $\displaystyle \arctan x-\arctan y=\arctan\left(\dfrac{x-y}{1+xy}\right)$ and converting it into a form $f(r+1)-f(r)$.
But I think that we cannot convert all types into this form. Even if this was possible, what is the way for us to know what $f$ to use? In particuar, I was evaluating $\displaystyle \sum_{r=1}^{\infty} \cot^{-1}\left(3r^2-r-\frac13\right)$, but couldn't convert it into telescoping series. So, how then, do we solve this? and for what values of $(a,b,c)$ is the sum convergent?
| Given quadratic $ar^2 +br+c$, suppose that we can find some function $f(r)$ such that
$$
(\dagger)\quad\quad\frac{1+f(r+1)f(r)}{f(r+1)-f(r)} = ar^2 + br + c.
$$
Then as $$\cot^{-1}(ar^2+br+c)=\arctan\frac{1}{ar^2+br+c}\\
=\arctan \frac{f(r+1)-f(r)}{1+f(r+1)f(r)}=\arctan(f(r+1))-\arctan(f(r)),
$$
we can proceed to find $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)$ telescopically as mentioned. Indeed we would have $$\sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r)),$$
provide convergence.
Now to find $f(r)$, we propose to seek it in the form $$f(r)=\frac{Ar+B}{Cr+D},$$ a linear fraction. If so, then the LHS of the $(\dagger)$ equation will be a quadratic in $r$. Indeed, substituting this linear fractional into the LHS yields
$$\frac{A^2+C^2}{AD-BC}r^2 + \frac{A^2+C^2+2AB+2CD}{AD-BC} r + \frac{B^2+D^2+AB+CD}{AD-BC}.$$
So the task becomes to find $A,B,C,D$ such that
$$ \begin{eqnarray}a &=& \frac{A^2+C^2}{AD-BC} \\
b &=& \frac{A^2+C^2+2AB+2CD}{AD-BC}\\
c &=& \frac{B^2+D^2+AB+CD}{AD-BC}\end{eqnarray}$$
Since here we overparametrize $a,b,c$ with $A,B,C,D$, we can in principle (see remark below) find these values. For instance you could set $A=0$ to simplify your search. Also note that as $\displaystyle \lim_{r\to\infty}f(r) = \frac AC$, we will have $ \displaystyle \lim_{r\to\infty}\arctan(f(r)) = \arctan\left(\frac AC\right)$.
An example, find $\displaystyle \sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac 13\right)$. We seek $A,B,C,D$ as above that works. Take $A=0$, we have by wolfram alpha $B = 1,C = -3 ,D = 2$ (among many other possible solutions). So $f(r) = \dfrac{1}{-3r+2}$ and
$$
\sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac13\right)\\
=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r))\\
=-\arctan(-1)= \arctan(1) = \frac{\pi}4.
$$
Remark. There is a limitation to this, for instance say $a\neq 0 $ for this $f(r)$ to take this form. Indeed, if $a =0$, then we see that $A=C=0$, which will give a contradiction if $b\neq 0 $. So if $f(r) = \dfrac{Ar + B}{Cr+D}$, then $(a,b,c)$ needs to be in the range of the function
$$(A,B,C,D)\mapsto \left(\frac{A^2+C^2}{AD-BC},\frac{A^2+C^2+2AB+2CD}{AD-BC},\frac{B^2+D^2+AB+CD}{AD-BC}\right).$$
Remark 2. Despite this limitation, you can use this the other way: Pick your favorite four numbers $A,B,C,D$ and write down $f(r) = \dfrac{Ar + B}{Cr+D}$. This generates a quadratic $ar^2 + br + c$, and with this you will have the value of $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c) = -\arctan\left(\dfrac{A + B}{C+D}\right)+\arctan\left(\dfrac{A}{C}\right)$.
| {
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"url": "https://math.stackexchange.com/questions/3980216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
\left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$
I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.
Please, advice.
| if $x\ge 0$ by AM-GM $$x^4+3=x^4+1+1+1\ge 4x$$ and the inequality $$\color{blue}{x^4+3-4x}+4x^3+4x^2>0$$ is obvious as equality of am-gm and $x=0$ is simultaneously is not achieved.
If $x\le 0$ then replace $x$ by $-t $ and we have to prove $$t^4-4t^3+4t^2+4t+3>0$$which is true as $$t^4+4t^2\ge 4t^3$$ by AM-GM and as $t\ge 0$ and the rest of terms are postive (note :again equality of both AM-GM and t=0 is not achieved)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to get from $x^2 + y^2 = e^2(d + x)^2$ to $(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$ I'm working through a problem in my pre-calc book and I don't know how to get the equation into the same form as the book. We start with $x^2 + y^2 = e^2(d + x)^2$ and the book gets to the following form:
$(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$
When I try I get stuck. Here are my steps:
1.) $x^2 + y^2 = e^2(d + x)^2$
2.) $x^2+y^2 = e^2(d^2 + dx + x^2)$
3.) $x^2+y^2 = e^2d^2 + e^2dx + e^2x^2$
4.) $x^2 - e^2x^2 - e^2dx = e^2d^2 -y^2$ (Grouping like terms)
5.) $(1-e^2)x^2 - e^2dx= e^2d^2 - y^2$
6.) $x^2 - \frac{e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$
7.) $x^2 - \frac{e^2dx}{1-e^2} + (\frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$ Completing the square.
8.) $(x^2 - \frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$
But from here I don't know where to go. Any help is Much appreciated.
| To be completed
*
*given equation
*you missed a term $dx$ inside the bracket on the RHS
*$x^2+y^2 = e^2d^2 + 2e^2dx + e^2x^2$
*$x^2 - e^2x^2 - 2e^2dx = e^2d^2 -y^2$ (same logic with coefficient corrected)
*$(1-e^2)x^2 - 2e^2dx= e^2d^2 - y^2$
*$$x^2 - \frac{2e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$$
*$$x^2 - \frac{2e^2dx}{1-e^2} + (\frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$ (same logic here)
*The first term inside the bracket on the LHS should be $x$. $$(x - \frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$
Can you continue?
| {
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"url": "https://math.stackexchange.com/questions/3984919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the Maximum Trigonometric polynomial coefficient $A_{k}$
Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.
I don't know if this question has been studied
If $n=2$ it is easy to solve it.
| My second answer:
Edit 2021/03/11
According to NTstrucker@AoPS's result (https://artofproblemsolving.com/community/c6h2477208), I give the following conjecture.
Conjecture 2: $\max A_1 = 2\cos \frac{\pi}{n+2}$.
When $n = 2$, $\max A_1 = \sqrt{2} = 2 \cos \frac{\pi}{4}$.
When $n = 3$, $\max A_1 = \frac{\sqrt{5} + 1}{2} = 2\cos \frac{\pi}{5}$.
When $n = 4$, $\max A_1 = \sqrt{3} = 2\cos \frac{\pi}{6}$.
When $n = 5$, $\max A_1 = 2\cos \frac{\pi}{7}$.
When $n = 6$, $\max A_1 \approx 1.8478$ (I only get numerical result) should be $2\cos \frac{\pi}{8}
\approx 1.847759065$.
When $n = 7$, $\max A_1 \approx 1.8794$ (I only get numerical result) should be $2\cos \frac{\pi}{9}
\approx 1.879385242$.
When $n = 8$, $\max A_1 \approx 1.9021$ (I only get numerical result) should be $2\cos \frac{\pi}{10}
\approx 1.902113033$.
When $n = 9$, $\max A_1 \approx 1.9190$ (I only get numerical result) should be $2\cos \frac{\pi}{11}
\approx 1.918985947$.
$\phantom{2}$
Let us prove that $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
We need Theorem 1. The proof is given at the end.
Theorem 1: Let $n\ge 2$ be a given integer. Let $A_i, i=1, 2, \cdots, n$ be given real numbers such
that $A_1\cos x + A_2 \cos 2x + \cdots + A_n \cos n x \le 1, \ \forall x\in \mathbb{R}$. Then
$A_m \le 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1, \cdots, n$.
By Theorem 1, we have immediately $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$
(simply let $A_m = 1$ and $A_k = 0$ for all $k\ne m$
and the condition $A_m \cos m x \le 1, \forall x\in \mathbb{R}$ is clearly satisfied.).
$\phantom{2}$
Proof of Theorem 1: First, we have Fact 1. The proof of Fact 1 is easy and thus omitted.
Fact 1: Let $M\ge 2$ be an integer. Let $K$ be a positive integer. Then
$$\frac{1}{M} + \frac{\cos K\pi}{M}\cdot \frac{1 + (-1)^M}{2}
+ \frac{2}{M}\sum_{j=1}^{\lfloor \frac{M+1}{2}\rfloor - 1} \cos \frac{2jK\pi}{M} = \left\{\begin{array}{cc}
0 & M \nmid K \\[5pt]
1 & M \mid K.
\end{array}
\right.$$
Second, we have Fact 2.
Fact 2: Let $n\ge 2$ be a given integer. Let $m$ be an integer with $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
Let $A_i, i= 1, 2, \cdots, n$ be given real numbers.
Let $f(x) = \sum_{k=1}^n A_k\cos (kx)$. Then
$$\frac{f(0)}{m} + \frac{f(\pi)}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} f\left(\frac{2j\pi}{m}\right) = A_m. $$
(Proof of Fact 2: It suffices to prove that
$$\frac{1}{m} + \frac{\cos k\pi}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} \cos \frac{2j k \pi}{m} = \left\{\begin{array}{cc}
0 & k \ne m \\[5pt]
1 & k = m.
\end{array}
\right. $$
This is true by Fact 1.)
By Fact 2, Theorem 1 is proved. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove by induction that $9^{n+1}+2^{6n+1}$ is divisible by 11 I need to prove that $9^{n+1}+2^{6n+1} $ is divisible by $11$ $\forall n\in N$
Steps I did:
*
*basis step
$$\\ P(0)\ 11\mid(9+2 ) \ (True)$$
*inductive step
$$\\ P(n)\implies P(n+1) \ where: \\ P(n): 9^{n+1}+2^{6n+1} = 11k\\P(n+1): 9^{n+2}+2^{6n+7} = 11k$$
so $$9^{n+2}+2^{6n+7}=11k\\9^n\times9^2+2^{6n}\times2^7=11k$$
Then I don't know how to proceed
| Hypothesis gives that if $9^{n+1} \equiv r \pmod{11}$ then $2^{6n+1} \equiv -r \pmod {11}$
The induction step follows:
$$\begin{align}P(n+1) &=9^{n+2}+2^{6n+7} \\
&= 9\cdot 9^{n+1}+2^6\cdot2^{6n+1} \\
&\equiv 9 r + 2^6(-r) \\
&= - 55r \\
&\equiv 0 \pmod{11}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
|
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$,$x+y=23$
The obvious way to solve it is by substituting for one variable.
No, cart before the horse. The first step is to move one of the radicals to the RHS and square both sides. This gives:
$$(x^2 + 12y) = 1089 + (y^2 + 12x) - 66\sqrt{y^2 + 12x}.\tag1$$
At this point, there are two key factors:
*
*When you square both sides, you will generate an equation such that some of its solutions may not satisfy the original equation.
*You are looking for some elegant way of utilizing that $(x + y) = 23.$
I experimented with various tries and then decided on the following:
*
*$(x^2 + 12y) = x^2 + 12[23 - x] = x^2 - 12x + 276.$
*$(y^2 + 12x) = [23-x]^2 + 12x = 529 - 46x + x^2 + 12x = x^2 - 34x + 529.$
Therefore, equation (1) above simplifies to
$$(x^2 - 12x + 276) = 1089 + (x^2 - 34x + 529) - 66\sqrt{x^2 - 34x + 529} \implies $$
$$(1342 - 22x) = 66\sqrt{x^2 - 34x + 529} \implies $$
$$(61 - x) = 3\sqrt{x^2 - 34x + 529} \implies $$
$$x^2 - 122x + 3721 = 9[x^2 - 34x + 529] \implies $$
$$(8x^2 - 184x + 1040) = 0 \implies$$
$$x = \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{33856 - \left(4 \times 8 \times 1040\right)}\right]$$
$$= \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{576}\right]$$
$$= \left(\frac{1}{16}\right) \times
\left[184 \pm 24\right] \implies x \in \{10, 13\}.$$
Here, you have to check each candidate value of $x$ against the original equation. Doing so, you realize that there are in fact two solutions: $(x=10, y=13)$ and $(x = 13, y = 10).$
| {
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Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.
| A nice trick to know.
$\newcommand{\d}{\frac{\mathbb Z_2[x]}{\langle x^4+x^2+x\rangle}}$In $\mathbb Z_2[x]$, we have the nice identity that $(a+b)^2 = a^2+2ab+b^2 = a^2+b^2$, because $2ab \equiv 0$ in $\mathbb Z_2[x]$! This naturally becomes $(a+b+c)^2 = a^2+b^2+c^2$ in $\mathbb Z_2[x]$ as well.
We can use this to great advantage as follows : let $p$ be an element satisfying $p^2 = x$ in the ring $\d$. Then $(p^3)^2 = x^3$ and $1^2 = 1$ therefore $(p^3+p+1)^2 = p^6+p^2+1 = x^3+x+1$ in $\d$.
So let's search for such a $p$ : $p^2 = x + h(x)(x^4+x^2+x)$ for some polynomial $h \in \mathbb Z_2[x]$. We start with the simplest $h$ (note $h = 0$ is not useful) , i.e. $h \equiv 1$. For this , we get the RHS as $$x^4+x^2+2x = x^4+x^2 = x^2(x^2+1) = x^2(x+1)^2 = [x(x+1)]^2$$
(The step $x^2+1 = (x+1)^2$ follows from the same $\mathbb Z_2$ identity I discussed earlier)
Et voilà! So $p = x(x+1)$, and $p^3+p+1$ provides the required square root.
Confirmation :
$$
p^3+p+1 = x^3(x+1)^3 + x(x+1)+1 = x^6 + x^5+x^4+x^3+x^2+x+1
$$
modulo $x^4+x^2 + x$ this reduces to $x^3+x+1$, whose square is in fact $(x^3+x+1)^2 = x^3+x+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you
| In fact you are very close to the right answer. I would suggest you check Euler - Maclaurin formula first - plus, not minus in the second term, though it does not change the result that you got $(+\frac{1}{2})$: $$\sum\limits_{k=1}^nf(\frac{k}{n})=n\int_1^nf(t)dt+\frac{1}{2}\left(f(\frac{1}{n})+f(1)\right)+\sum\limits_{k=2}^{\infty}(\frac{1}{n})^{k-1}\frac{B_k}{k!}\left(f^{(k-1)}(1)-f^{(k-1)}(\frac{1}{n})\right)$$
Next, the integral in fact is $$n\int_0^1\frac{t^2}{1+\frac{1}{n}t}dt=n\int_0^1t^2dt-\int_0^1t^3dt+O(\frac{1}{n})$$ the second term here gives you additionally $-\frac{1}{4}$.
All together,
$$ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$
| {
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Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.
| Note that
$$
(a^2-a^3)^6-1=(a^6 - 2a^5 + a^4 - a^3 + a^2 + 1)(a^4 - a^3 - a^2 + a + 1)(a^3 - a^2 + 1)(a^3 - a^2 - 1)(a^2 - a + 1),
$$
so that the claim of the title follows.
| {
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Evaluate the I? This is question is inspired from this
find the value of $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}\sqrt{ \frac{u^4}{a^2} +u^2} du dv$$
My attempt : $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du dv$$
after that im not able to solved this
| Before I worked out a complete solution, @TitoEliatron has left a key step in his comment.
\begin{align}
I&=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} \,du \, dv\\
&=\int_0^{2\pi}dv \; \frac12 \int_0^{\sqrt2 a} \sqrt{ \frac{u^2}{a^2} +1} \, d(u^2) \\
&= \pi \int_0^{2a^2} \sqrt{x/a^2 + 1} \, dx \tag{$x = u^2$} \\
&= \frac{\pi}{a} \int_0^{2a^2} \sqrt{x + a^2} \, dx \\
&= \frac{2\pi}{3a} \left[ (x+a^2)^{3/2} \right]_0^{2a^2} \\
&= \frac{2\pi (3\sqrt3 - 1) a^2}{3}
\end{align}
Edit : The $a^3$ in the numerator cancelled with $a$ in the denominator to give $a^2$.
| {
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Logarithm to trigonometry conversion I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$
I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.
As an attempt to solve the problem, I was trying to play around the relation
$$\frac{1}{2} i \log \left(\frac{a+i c}{-a+i c}\right)=\tan^{-1}\left(\frac{a}{c}\right)$$ to change into arctan function but does not look like this alone is going to help in this case.
Any suggestions?
| $\frac {i}{2}\ln \frac {a+b+ic}{a-b-ic} = x\\
\ln \frac {a+b+ic}{a-b-ic} = -2ix\\
\frac {a+b+ic}{a-b-ic} = e^{-2ix}\\
a+b+ic = e^{-2ix}(a-b-ic)\\
a+b+ic = e^{-2ix}(a-b-ic)\\
(ic-b)(1+e^{-2ix}) = a(e^{-2ix} - 1)\\
\frac {ic - b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}$
At this point the right hand side equals very nearly $\tan x.$ That is:
$i\frac {c}{a} - \frac {b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}\\
\frac {ic - b}{a} = \frac {(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {i(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {e^{ix} - e^{-ix}}{i(e^{-ix} + e^{ix})}\\
\frac {-c - bi}{a} = \tan x\\
x = \tan^{-1} (-\frac {c}{a} - i\frac {b}{a})$
| {
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Estimate a value with square roots squared without calculating If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.
This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do that?
| Use the difference of two squares to get:
$$6+\sqrt{29} = \frac{(6+\sqrt{29})(6-\sqrt{29})}{6-\sqrt{29}} = \frac{7}{6 - \sqrt{29}}$$
and so:
$$n<\frac{49}{(6-\sqrt{29})^2}<n+1 \Rightarrow \frac{1}{n+1} < \frac{(6 - \sqrt{29})^2}{49} < \frac{1}{n}.$$
Adding $49$ times this inequality with the original gives:
$$n + \frac{49}{n+1} < (6 + \sqrt{29})^2 + (6 - \sqrt{29})^2 < n+1 + \frac{49}{n}$$
$$\Rightarrow n + \frac{49}{n+1} < 130 < n+1 + \frac{49}{n}$$
From the right-hand side, the minimum value of $n$ which satisfies the inequality is $129$. But from the left-hand side, $n$ cannot be $130$ or greater. Therefore $n = 129$ which is the unique solution.
| {
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(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola I searched but couldn't find the proof.
Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$
And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this point $P(x, y)$ to the foci of the isosceles hyperbola is equal to the square of the distance from point $P$ to the center of the hyperbola.
Proof?
I took this question from my analytical geometry project assignment. I tried various ways (I found the foci $F(x,y)$ and $F^{'}(x,y)$ terms of $x$, $y$ and chose any point on the hyperbola...) but I couldn't prove.
I request your help.
| For any point on the hyperbola, $x^2 - y^2 = a^2$
Foci of the hyperbola are $(\pm a\sqrt2,0)$ and the center is $(0, 0)$.
So product of distance of point $P(x,y)$ on the hyperbola to foci is
$\sqrt{(x-a\sqrt2)^2 + y^2} \times \sqrt{(x+a\sqrt2)^2 + y^2}$
$\sqrt{x^2 + 2a^2 + y^2 - 2 \sqrt2 a x} \times \sqrt{x^2 + 2a^2 + y^2 + 2 \sqrt2 a x}$
Using $y^2 = x^2 - a^2$,
$ = \sqrt{((2x^2+a^2) - 2\sqrt2 a x) ((2x^2+a^2) + 2\sqrt2 a x)}$
$ = \sqrt{ 4x^4 + a^4+4a^2x^2 - 8a^2x^2}$
$ = 2x^2 - a^2 = x^2 + y^2$
Which is the square of distance of point $P$ to the center of the hyperbola $(0, 0)$
| {
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Inverse of fourth root function I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$
but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!
| \begin{align}
y=a\sqrt[^4]{b^2-(x-c)^2} &\implies y^{4} = a^{4}(b^{2} - (x-c)^2)\\
&\implies \frac{y^{4}}{a^{4}} = b^{2} - x^{2}+2cx-c^{2}\\
&\implies x^{2} - 2cx + \left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right) = 0\\
&\implies x = \frac{2c \pm \sqrt{4c^2-4\left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right)}}{2}\\
&\implies\boxed{x = c\pm\sqrt{b^2-\frac{y^{4}}{a^{4}}}}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify conditional probability tree There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".
This question is similar to my last question except the desired outcome is rolling one of each number, the probability being:
$$\frac{6!}{6^6} = \frac{720}{6^6}$$
An example of this kind of roll would be "123456" or "352461".
I need to determine the breakdown of the amount of rolls by number of gold sides. Using the method of the accepted answer I could get the number of rolls with 6, 4, and 3 gold sides (exactly 5 gold sides is impossible as the last side must also be gold in order for the roll to contain one of each number).
$$G_6 = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 1$$
$$G_4 = {6 \choose 4} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 15$$
$$G_3 = {6 \choose 3} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 40$$
Now for $G_2$ the probability of the 4th dice being gold became conditional on the position of the 3rd dice, and the 5th conditional on the 3rd and 4th. I drew a probability tree to find there are 9 positions out of 24 for the last 4 dice not to be the number of their gold side.
And then plug that in to find the answer:
$$G_2 = {6 \choose 2} \times \frac{1}{6} \times \frac{1}{5} \times \frac{9}{24} \times 6! = 135$$
But I am wondering what is the better way to find the answer (without having to draw the tree, as a tree for $G_1$ and $G_0$ would be even larger and more tedious).
| There is no easy way to do it with a probability tree. For any $k\in \{0,\dots,6\}$, the number of rolls with $k$ gold faces is equal to the number of permutations of $\{1,\dots,6\}$ with $k$ fixed points, which is in turn equal to $\binom{6}k$ times the number of derangements of a set of size $6-k$. The number of derangements of a set of size $n$ is well known to be
$$
n!\sum_{i=0}^n\frac{(-1)^i}{i!}
$$
which can be found using the principle of inclusion-exclusion.
| {
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John Wallis and calculating ratios I was asked the following question:
Using Wallis's method, calculate the values $n=0,\frac12,1,\frac32,2,\frac52$ in row $p=\frac32$ of his ratio table.
The Wallis rereferred to here is John Wallis of England (1616-1703). The ratio table looks like this: $$\begin{matrix}p/n&0&1&2&3&4&5&6&7&...\\0&1&1&1&1&1&1&1&1&...\\1&1&2&3&4&5&6&7&8&...\\2&1&3&6&10&15&21&28&36&...\\3&1&4&10&20&35&56&84&120&...\\...&...&...&...&...&...&...&...&...&...&\\\end{matrix}$$
I don't understand how to read this table or how to actually go about solving the question. The table's goal is to find the ratio of the area of the unit square to the area aenclosed in the first quadrant by the curve $y=(1-x^{\frac1p})^n$.
| The ratio table represents the ratio of the unit square over the area under the stated curve, $y = (1-x^{1/p})^n$. We shall assume $p, n \geqslant 0$ and adopt the convention that the area is $1$ when $p=0$.The area of the unit square is always $1$, while the area under the curve is
\begin{align}
A_{p,n} = \int_0^1 (1-x^{1/p})^n ~ dx
\end{align}
and the relevant Wallis number in the table is simply the reciprocal, $W_{p,n} = 1 / A_{p,n}$. It would seem by convention the area $A_{p,n}$ is taken to be $1$ when $p=0$.
To evaluate the integral, you can begin by making the substitution $x = \sin^{2p} \theta$ for $0 \leqslant \theta \leqslant \pi/2$, so that $dx = 2p\sin^{2p-1}\theta \cos \theta ~d\theta$
\begin{align}
A_{p,n} &= \int_0^{\pi/2} (1-\sin^2\theta)^n \cdot 2p \sin^{2p-1}\theta \cos \theta d\theta \\
&= 2p\int_0^{\pi/2} \cos^{2n+1} \theta \sin^{2p-1}\theta ~d\theta.
\end{align}
In this case you are only interested in a particular instance, $p=3/2, n=0,\frac{1}{2},1, \cdots$, giving
$$A_{p,n} = 2p \int_0^{\pi/2}\sin^2\theta \cos^{2n+1}\theta ~d\theta$$ where $m=2n+1=1,2,3,\cdots$. Integrate once by parts,
\begin{align}
A_{p,n} = 2p \int_0^{\pi/2} \cos^{2n+1} \theta - \cos^{2n+3}\theta ~ d\theta .
\end{align}
These integrals are standard, see the Wikipedia entry for Wallis's integral, giving,
\begin{align}
\begin{array}
~A_{3/2,0} = 1 & A_{3/2,1/2} = \frac{3\pi}{16} \\
A_{3/2,1} = \frac{6}{15} & A_{3/2,3/2} = \frac{3\pi}{32} \\
\cdots & \cdots\\
n=k, k=0,1,\cdots & n=k+\frac{1}{2},k=0,1,\cdots\\
A_{3/2,n}=3\frac{2^{2k}k!^2}{(2k+1)!}\frac{1}{2k+3} &
A_{3/2,n}=3\frac{(2k+2)!}{2^{2k+3}(k+1)!^2}\frac{1}{k+2} \frac{\pi}{2}
\end{array}
\end{align}
and the corresponding Wallis ratios are the reciprocals of these numbers.
More generally, you can evaluate $A_{p,n}$ for $p$ and $n$ when either is a multiple of $\frac{1}{2}$ by considering the integral,
$$I_{r,s} = \int_0^{\pi/2} \sin^r\theta \cos^s \theta~d\theta.$$
in which $r, s$ are non-negative integers; first notice $I_{r,s}=I_{s,r}$. Evaluate the special cases when $r=0,1, s \geqslant 1$ and $s=0,1, r \geqslant 1$. Then use integration by parts to obtain the recurrence relation,
$$I_{r,s}=\frac{s-1}{r+1} I_{r+2,s-2}$$
where by symmetry we can take $r \geqslant s \geqslant 2$. Using the special cases this should give all values for $I_{r,s}$. Then $A_{p,n} = 2p I_{2p-1,2n+1}$.
| {
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Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.
| We may observe that our polynomial has the form
$$ (x^2+x+1)\left(x^3 - \frac{5}{2}\right)^2 + \frac{1}{4} (3x^2+3x-1) $$
The first term is always non-negative (and positive for $x \neq \sqrt[3]{\frac{5}{2}}$, that case we can check separately), and the second one is non-negative outside interval $(-2,1)$, so we just need to check the values inside this interval. But $x^2+x+1 = (x+ \frac{1}{2})^2) + \frac{3}{4} \geqslant \frac{3}{4}$, and $(x^3-\frac{5}{2})^2$ has on this interval values not less than $\frac{9}{4}$. It suffices to show that minimum of $\frac{1}{4}(3x^2+3x-1)$ is greater than $\frac{27}{4}$, which is easy to check with elementary methods.
| {
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Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.
There are $2$ cases to consider:
Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:
\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) + abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}
Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?
| Unfortunately, I wasn't able to pick up from where you left off.
Starting from the following inequality:
$$a(b+c-a) \leq b(a+c-b) \leq c(a+b-c) $$
which is true given your initial WLOG $a\geq b\geq c$.
And note that $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$.
You can use Rearrangement Inequality so that:
\begin{align}\frac{1}{a} a(b+c-a) + \frac{1}{b} b(a+c-b) + \frac{1}{c} c(a+b-c) & \geq \frac{1}{c} a(b+c-a) + \frac{1}{a} b(a+c-b) + \frac{1}{b} c(a+b-c) \\ a+b+c &\geq \frac{a(b-a)}{c} + a + \frac{b(c-b)}{a} + b + \frac{c(a-c)}{b} + c \\ 0 &\geq \frac{a(b-a)}{c} + \frac{b(c-b)}{a} + \frac{c(a-c)}{b} \\ \frac{a(a-b)}{c} + \frac{b(b-c)}{a} + \frac{c(c-a)}{b} & \geq 0 \\ a^2b(a-b) + b^2c(b-c)+c^2a(c-a) &\geq 0\end{align}
Hope it's correct.
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"answer_id": 2
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If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $ If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $
Prove $ abc \geq 8 $
I have tried AM-GM and substituting the 1, without results.
| This simplifies to $$a+b+c+2=abc$$
$$⇒\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}=1$$
Applying AM-GM inequality we get
$$\frac{1}{4}=\frac{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}}{4}≥(\frac{2}{a^3b^3c^3})^\frac{1}{4}$$
Riasing both sides to the power of 4 we get
$$\frac{1}{256}≥(\frac{2}{a^3b^3c^3})$$
$$⇒abc≥8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $ax^2+3x+b$ that goes through the given points using the least squares The points are $ (2,1.5),(−1,1.7),(1,1.9)$.
I did:
$A = \left [ \begin{matrix}
1 & 2 & 4 \\
1 & -1 & 1 \\
1 & 1 & 1
\end{matrix} \right ] \\
x = \left [ \begin{matrix}
b \\
3 \\
a
\end{matrix} \right ]\\
b = \left [ \begin{matrix}
1.5 \\
1.7 \\
1.9
\end{matrix} \right ]$
I solved for $$A^TAx = A^Tb$$
I got the Cholesky decomposition of $A^TA$
Then I solved $Ly = A^Tb$ and got $$
\left [ \begin{matrix}
51/10 \\
16/5 \\
48/5
\end{matrix} \right ]$$$
The next part is where I am having a problem. Now I have to solve $L^tx = y$. I already have some values for x, so I multiplied $L^tx$ and made it equal to y and got a system of equations
How do I solve this?
| $3$ is not an unknown variable, so your solution is incorrect.
We have
$$
\left\{\begin{array}{rrrrr}
2^2\cdot a &+3\cdot 2&+b&= &1.5 \\
(-1)^2\cdot a&+3\cdot(-1)&+b&= &1.7 \\
1^2\cdot a&+3\cdot 1&+b&= &1.9 \\
\end{array}\right.
$$
or
$$
\left\{\begin{array}{rrrrr}
4a&+&b&= &-4.5 \\
a&+&b&= &4.7 \\
a&+&b&= &-1.1 \\
\end{array}\right.
$$
or
$$
Ay=d,
$$
where
$$
A=\left(\begin{array}{ll}
4 & 1\\
1 & 1\\
1 & 1\\
\end{array}\right),\quad
d=\left(\begin{array}{r}
-4.5 \\
4.7 \\
1.9 \\
\end{array}\right),\quad
y=\left(\begin{array}{r}
a \\
b \\
\end{array}\right).
$$
This system has no solutions, thus, there is no curve of the form $ax^2+3x+b$ which goes exactly through the given points. It is only possible to find a least squares solution. This solution won't make a lot of practical sense since $a+b$ can't be equal to $4.7$ and $-1.1$ at the same time even approximately.
$$
y_{ls}=A^{+}d=(A^TA)^{-1}A^Td=\left(\begin{array}{r}
-2.1 \\
3.9 \\
\end{array}\right).
$$
Hence, the answer is $-2.1x^2+3x+3.9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| Divisibility by $7$ is congruence to zero modulo $7.$
So we might get some insights by looking at the numbers' congruences mod $7$.
Note that $x^{2^{k+1}} = x^{2^k\cdot 2} = \left(x^{2^k}\right)^2.$
That is, every time we add $1$ to the exponent $k$ in
$2^{2^k}$ or $4^{2^k}$, we square the number.
So let's try applying these two ideas: do the arithmetic modulo $7$; and try the first few values of $n$ and see what happens.
So what happens is
\begin{align}
2^2 \equiv 4 \pmod7,\\
4^2 \equiv 2 \pmod7.\\
\end{align}
That is,
\begin{align}
\text{for } n &= 1,
& 4^{2^1} + 2^{2^1} + 1 = 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\text{for } n &= 2,
& 4^{2^2} + 2^{2^2} + 1 \equiv 2^2 + 4^2 + 1 &\equiv 4 + 2 + 1 \pmod7,\\
\text{for } n &= 3,
& 4^{2^3} + 2^{2^3} + 1 \equiv 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\end{align}
and so forth.
You should now know the congruence classes of $4^{2^n}$ and $2^{2^n}$ for every $n$ and be able to prove the obvious pattern of congruences for alternating odd and even $n$ by induction.
This is just a little less elegant that some of the other proofs in that you may find it necessary to have two cases in the inductive step, one for odd $k$ and one for even $k.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 0
} |
Prove that $\sqrt{13}$ is irrational in 3 ways. I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.
However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.
| \begin{align}
\sqrt{13} & = 3 + \text{a fractional part, since } 3^2<13<4^2 \\[8pt]
& = 3 + (\sqrt{13}-3) \\[8pt]
& = 3 + \frac{4}{\sqrt{13}+3} \text{ by rationalizing the numerator} \\[8pt]
& = 3 + \frac 1 {(\sqrt{13}+3)/4}
\end{align}
Since $3<\sqrt{13}<4,$ we have $1 < (\sqrt{13}+3)/4 < 2,$ so $(\sqrt{13}+3)/4$ is between $1$ and $2.$
So we have
\begin{align}
\sqrt{13} & = 3 + \frac 1 {1 + \left( \frac {\sqrt{13}-1} 4 \right)} \\[12pt]
& = 3 + \frac 1 {1 + \cfrac 3 {\sqrt{13} + 1}} \text{ by rationalizing the numerator as before}
\end{align}
Continuing in this way, we get
$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}.
$$
In the steps before this, the expression in the denominator was $\dfrac{\sqrt{13} + (\text{an integer})}{\text{an integer}}$ and the denominator in this last fraction was more than $1.$ But this time, the denominator is $1,$, i.e. we have $\dfrac{\sqrt{13}+3} 1.$ This means that now in place of $\sqrt{13}$ at the end of this long expression one can put this entire continued fraction, and we get
$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{6 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}}}}}}.
$$
And then we can do that again. And again. We get this repeating sequence:
$$
3, \, \overbrace{1,1,1,1,\,6,}\, 1,1,1,1,\,6,\, 1,1,1,1,\,6,\, \ldots
$$
The part under the $\overbrace{\text{overbrace}}$ must keep repeating. This could not happen if this number were rational, as we will argue below. To see, why, suppose we start with a ration approximation to $\sqrt{13}$:
\begin{align}
\sqrt{13}\approx\frac{119}{33} & = 3 + \frac{20}{33} = 3 + \cfrac 1 {1 + \cfrac{13}{20}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 6 7}}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 6}}}}.
\end{align}
But we cannot keep getting smaller positive integers forever. So if we start with a rational number, this process cannot go on forever; it must end. But as shown above, if we start with $\sqrt{13},$ it does not end; it just keeps repeating.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem in proving Cauchy sequence - how to tackle inequalities and changing indices [EDITED] We are given this sequence:
$a_1 =1, a_{n+1} = 1 + \frac{1}{a_{n}} \forall \; n\geq 1$
Now before marking this question as "duplicate", consider that I have already referred these sources:
How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$
Convergence of sequences and limits
Now, I am relatively new to Real Analysis and for this particular question, I don't need to determine the limit, only to prove that it is Cauchy and hence convergent.
My approach:
Observe that $a_1 =1, a_2 =2, a_3 = \frac{3}{2}, a_4= \frac{5}{3}$ and so on.
Also observe,
$a_{n+1} = 1 + \frac{1}{a_n} \forall n\ge 1$
Multiplying throughout by $a_n$ we get,
$a_{n+1}a_{n} = a_{n} + 1 \; \forall n\ge 1$
Claim: $a_n \; \ge 1 \; \forall \; n\ge 1$
Proof by Induction:
Base Case: $n=1:\; a_1 =1 \ge 1$ which is true.
Induction Hypothesis: $a_k \ge 1 \;$ for some $k \in \mathbb{N}, k \ge 1$
Induction Step: To prove: $a_{k+1} \ge 1$ for some $k \in \mathbb{N}, k\ge 1$
Now,
$a_k \ge 1$
$\therefore 0 \le \frac{1}{a_{k}} \le 1$
$\therefore 1 \le 1 + \frac{1}{a_{k}} \le 2$
$\therefore 1 \le a_{k+1} \le 2 $
Therefore by Priniciple of Mathematical Induction,
$a_n \; \ge 1 \; \forall \; n\ge 1$
Now Consider,
$a_{n+1}a_{n} = a_{n} + 1 \ge 1+1 =2 \; \; \forall \; n\ge 1$
Changing indices, we get
$a_{n}a_{n-1} = a_{n-1} + 1 \ge 2 \; \; \forall \; n\ge 2$
Now Consider:
\begin{aligned}
&\left|a_{n+1}-a_{n}\right| \\
=&\left|1+\frac{1}{a_{n}}-\left(1+\frac{1}{a_{n-1}}\right)\right| \\
=&\left|\frac{1}{a_{n}}-\frac{1}{a_{n-1}}\right| \\
=&\left|\frac{a_{n}-1-a_{n}}{a_{n} \cdot a_{n-1}}\right| \\
=&\left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right|
\end{aligned}
Since we have that
$a_{n}a_{n-1} \ge 2 \; \; \forall \; n\ge 2$
We get,
$\left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$
$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$
$\therefore$
$$\left|a_{3}-a_{2}\right| = \left|\frac{a_{2}-a_{1}}{a_{2} \cdot a_{1}}\right| \leqslant \frac{\left|a_{2}-a_{1}\right|}{2} = \frac{1}{2}$$
$\therefore$
$$\left|a_{4}-a_{3}\right| = \left|\frac{a_{3}-a_{2}}{a_{3} \cdot a_{2}}\right| \leqslant \frac{\left|a_{3}-a_{2}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^2} $$
$\therefore$
$$\left|a_{5}-a_{4}\right| = \left|\frac{a_{4}-a_{3}}{a_{4} \cdot a_{3}}\right| \leqslant \frac{\left|a_{4}-a_{3}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^3} $$
$\therefore$
$$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^{(n+1)-2}}= \frac{\left|a_{2}-a_{1}\right|}{2^{n-1}} \forall n \ge 2 $$
Hence,
for some $m \ge n$
$$\left|a_m-a_n\right| \leqslant \left|a_{m}- a_{m-1}\right| + \left|a_{m-1}-a_{m-2}\right| + ... \left|a_{n+1}-a_{n}\right| $$
$\;$
$$\leqslant \frac{\left|a_2-a_1\right|}{2^{m-2}} + \frac{\left|a_2-a_1\right|}{2^{m-3}} + ... \frac{\left|a_2-a_1\right|}{2^{n-1}}$$
$$ = \left| a_{2}-a_{1} \right| \cdot \frac{1 - {\frac{1}{2}}^{n-m}}{1 - \frac{1}{2}}$$
Setting $$n = N, m= 2N$$
We get for given $ \epsilon > 0 $
For $N> log_{2}{\frac{2-\epsilon}{\epsilon}}$
We have, $m>n \ge N$
$\left|a_m - a_n\right| < \epsilon $
Hence, this sequence is Cauchy.
Is my approach right?
| Step 1. By induction show that $$\alpha:=\frac32 \le a_n \le \frac{15}8.$$
(The upper-bound is the main trick!)
Step 2. We have
$$
|a_{n+1}-a_n| =
|\frac{1}{a_n}-\frac{1}{a_{n-1}}| =
\frac{|a_n-a_{n-1}|}{a_na_{n-1}} \le
\frac{|a_n-a_{n-1}|}{\alpha^2}.
$$
So by induction we have
$$|a_{n+1}-a_n|\le \alpha^{-2(n-1)}|a_2-a_1| = \frac12\alpha^{-2(n-1)}.$$
Step 3. For $n>m$, we have
$$|a_n-a_m| \le \sum_{k=m}^{n-1} |a_{k+1}-a_k|
\le
\frac12\sum_{k=m}^{n-1}\alpha^{-2(k-1)}
\le
\frac12\sum_{k=m}^{\infty}\alpha^{-2(k-1)} = \frac{\alpha^2}{2(1-\alpha^{-2})}\alpha^{-2m}.
$$
This shows that the sequence is Cauchy and convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $? How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$
I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to :
$$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
has anyone a idea to how to find the closed-form of the last series or how to start out differently with the integral?
| Substitute $t=\tan\frac x2$
\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\
=&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt
=\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5
\end{align}
where, per the results
\begin{align}
I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt
= -G \ln 2-K+\frac{3 \pi ^3}{128}+\frac{3\pi}{32} \ln ^22\\
I_2 &= \int_0^1 \frac{\ln^2(1+t)}{1+t^2} dt
= -2 G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{16} \ln ^22 \\
I_3 &= \int_0^1 \frac{\ln (1+t^2)\ln(1-t)}{1+t^2} dt
= -\frac{1}{2} G \ln 2+4 K -\frac{5 \pi ^3}{64}+\frac{\pi}{8} \ln ^22 \\
I_4 &= \int_0^1 \frac{\ln (1+t^2)\ln(1+t)}{1+t^2} dt
= -\frac{5}{2} G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{8} \ln ^22\\
I_5 &= \int_0^1 \frac{\ln^2(1+t^2)}{1+t^2} dt
= -2 G \ln 2+4 K-\frac{7 \pi ^3}{96}+\frac{7\pi}{8} \ln ^22
\end{align}
with $K= \Im\text{Li}_3\left(\frac{1+i}{2}\right)$. Together
$$ \int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx
=4\Im\text{Li}_3\left(\frac{1+i}{2}\right)-\frac{11\pi^3}{96}+\frac{3\pi}8\ln^22
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$
I can use Werner and prostaferesis formulas and I find the identity $(1)$.
But if we suppose of not to use these formulas I have done a try writing:
$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$
$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$
$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$
Then I have abandoned because I should do long computations and I think that is not the right way.
Is there any trick without to use the prostapheresis or Werner's formulas?
| $$\sin^22\theta=4\sin^2\theta\ (1-\sin^2\theta)$$
$$=\sin\theta\ (\sin\theta+(3\sin\theta-4\sin^3\theta))$$
$$=\sin\theta\ (\sin\theta+\sin 3\theta)=\sin^2\theta+\sin 3\theta\sin \theta.$$
Therefore, $\cos^22\theta=\cos^2\theta-\sin 3\theta\sin\theta.$ Finally,
replace $\theta$ by $2\alpha.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Optimizing a Conical Frustum Using Partial Differentiation I am working on minimizing the surface area of a frustum with a known volume, however, the equations for the volume and surface area (listed below) have three variables. I am trying to use partial differentiation to optimize surface area with volume as a constraint, yet, differentiating with respect to any variable results in very messy work that can't be solved for a variable. Is there a better way to do it, or can anyone please show me how to do it using partial differentiation?
$
V=1/3(πh)(r^2+rR+R^2)
$
$
S.A=π(r+R)(\sqrt{(r-R)^2+h^2}+πr^2+πR^2
$
Thank you.
| Define the Lagrangian function
$$\Lambda(r,R,h;\lambda):=\pi(r+R) \sqrt{(r-R)^2+h^2}+\pi r^2+\pi R^2 -\lambda(V_0-(\pi h) (r^2+rR+R^2)/3), $$
with $V_0$ denoting the allocated volume.
The partial derivatives are
$$\begin{align}\Lambda_r&=\pi \sqrt{h^2+(r-R)^2}+\frac{\pi (r-R) (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi h \lambda (2 r+R)+2 \pi r, \\ \Lambda_R&=\frac{\pi \left(R^2-r^2\right)}{\sqrt{h^2+(r-R)^2}}+\pi \sqrt{h^2+(r-R)^2}+\frac{1}{3} \pi h \lambda (r+2 R)+2 \pi R, \\ \Lambda_h &=\frac{\pi h (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi \lambda \left(r^2+r R+R^2\right), \\ \Lambda_\lambda&=\frac{1}{3} \pi h \left(r^2+r R+R^2\right)-V_0. \end{align}$$
Proceed with a change of variables, motivated by the square root in the above
$$
\begin{align}
h &= \rho \cos \theta, \\
r &= \frac{s+\rho \sin \theta}{2}, \\
R &= \frac{s-\rho \sin \theta}{2}.
\end{align}
$$
Setting the partial derivates above equal to zero, and applying the substitution gives
$$
\begin{align}
0&=\frac{1}{6} \pi \lambda \rho \cos (\theta ) (\rho \sin (\theta )+3 s)+\pi (\sin (\theta )+1) (\rho +s),\\
0&=-\frac{1}{6} \pi \lambda \rho \cos (\theta ) (\rho \sin (\theta )-3 s)-\pi (\sin (\theta )-1) (\rho +s), \\
0&=\frac{1}{24} \pi \lambda \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right)+\pi s \cos (\theta ),\\
0 &=\frac{1}{12} \pi \rho \cos (\theta ) \left(\rho ^2 \sin ^2(\theta )+3 s^2\right)-V_0.
\end{align}
$$
The first three equations can be written in linear form as
$$
\left(
\begin{array}{ccc}
\pi (\sin (\theta )+1) & \pi (\sin (\theta )+1) & \frac{1}{6} \pi \rho \cos (\theta ) (\rho \sin (\theta )+3 s) \\
\pi (1-\sin (\theta )) & \pi (1-\sin (\theta )) & \frac{1}{6} \pi \rho \cos (\theta ) (3 s-\rho \sin (\theta )) \\
0 & \pi \cos (\theta ) & \frac{1}{24} \pi \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right) \\
\end{array}
\right) \begin{pmatrix} \rho \\ s \\ \lambda \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}.
$$
If the matrix in the above is invertible, we must have $\rho=0$, which implies a planar frustum ($h=0$). Hence, we can assume that the matrix is singular, meaning its determinant
$$\frac{1}{3} \pi ^3 \rho \sin (\theta ) \cos ^2(\theta ) (\rho -3 s)=0. $$
This vanishing determinant condition reduces the search to two seemingly admissible cases:
*
*$\rho=3s \implies h =2 \sqrt{(2 r+R) (r+2 R)}$
*$\rho \sin \theta =0 \implies r=R$
Case 1:
The first three equations, upon divisions allowed by $s>0, \cos \theta \neq 0$ become
$$\frac{1}{2} \pi (3 \lambda s \cos (\theta )+8) =0, \\\frac{1}{2} \pi (3 \lambda s \cos (\theta )+8)=0, \\\pi \cos (\theta )+\frac{1}{8} \pi \lambda s (5-3 \cos (2 \theta ))=0. $$
It takes a little work to deduce from this that $\sin^2 \theta = \frac{1}{9}$, which gives $r=0$ or $R=0$. This renders Case 1 inadmissible.
Case 2:
Let us return to the Cartesian form of the system, assuming that $r=R$:
$$\pi (h \lambda r+h+2 r)=0, \\ \pi (h \lambda r+h+2 r)=0, \\\pi r (\lambda r+2)=0, \\\pi h r^2-V_0 .$$
Since $r>0$ the third equation immediately gives $\lambda = -2/r$. It follows from there that $h=2r$ and finally that $r=\sqrt[3]{\frac{V_0}{2 \pi }}.$
Since there is a unique critical point, it has to be the minimum. That is, the frustum of volume $V_0$, of least surface area is a cylinder with radii $R=r=\sqrt[3]{\frac{V_0}{2 \pi }}$ and height $h=2\sqrt[3]{\frac{V_0}{2 \pi }}$.
Remark: This is somewhat expected based on the iso-perimetric inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
compute $ \iiint_Kxyz\ dxdydz$ The question is:
$$ \iiint_Kxyz\ dxdydz\quad k:=\{(x,y,z):x^2+y^2+z^2\leq1, \ \ x^2+y^2\leq z^2\leq 3(x^2+y^2), \ x,y,z\geq 0\} $$
Here how i have tried to solve this:
$$\iint_{x^2+y^2\leq1}\int_{\sqrt{x^2+y^2}}^{\sqrt{3(x^2+y^2)}}xyz \ dz\ dxdy=\frac{1}{2}\iint xy\left(3(x^2+y^2)-(x^2+y^2)\right)=...=0$$
But the answer that i got is zero which is obviously wrong what is wrong with my solution?
any suggestion would be great, Thanks
| Basically we are trying to find the volume of the sphere $x^2+y^2+z^2 \leq 1$ between two cones $z^2 = x^2 + y^2$ and $z^2 = 3(x^2+y^2)$.
So the bounds are much easier to set up in spherical coordinates.
$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$
$0 \leq \rho \leq 1, 0 \leq \theta \leq \frac{\pi}{2}$ are obvious as we are in first octant and radius of the sphere is $1$.
Now to find the bounds of $\phi$, we observe that
$x^2 + y^2 \leq z^2 \leq 3(x^2+y^2)$
plugging in $x,y,z$, $ \frac{1}{\sqrt3} \leq \tan \phi \leq 1 \implies \frac{\pi}{6} \leq \phi \leq \frac{\pi}{4}$
$xyz = \rho^3 \cos\theta \sin \theta \sin^2\phi \cos\phi$
So the integral becomes,
$\displaystyle \int_{0}^{\pi/2} \int_{\pi/6}^{\pi/4} \int_0^1 \rho^5 \cos\theta \sin \theta \sin^3\phi \cos\phi \ d\rho \ d\phi \ d\theta$
I did not do the integral by hand but WolframAlpha shows the result as $\displaystyle \frac{1}{256}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
By Using the Binomial expansion as follow that the expression for the first three term is mathematically correct or not? Let the expression is defined as $(1-\frac{1}{i})^{2i}$.
For example: Using the Binomial expansion as follow $$(1-\frac{1}{i})^{2i}=\sum_{n=0}^{n}\frac{(2i)!}{n!(2i-n)!}(-1)^n(\frac{1}{i})^n$$
$$\approx1-\frac{1}{i}+O((-\frac{1}{i})^3)$$.
Please need your help that the above expression for the first three term is mathematically correct or not, where the big $O$ notation means that the first neglected or unknown term is of the order $(-\frac{1}{i})^3$.
| That's fairly straight forward. When n= 0, n!= 0!= 1 and 2i- n= 2i so (2i- n)!= 2i!. The coefficient is $\frac{(2i)!}{n!(2i- n)!}= \frac{(2i)!}{(2i)!}= 1$. Of course, $(-1)^0= 1$ and $\left(\frac{1}{i}\right)^01= 1$ so the first term is 1
When n= 1, n!= 1!= 1 and $\frac{(2i)!}{(2i- n)!}$$= \frac{2i(2i-1)!}{(2i-1)!}= 2i$, $(-1)^1= -1$ and $\left(\frac{1}{i}\right)^1= \frac{1}{i}$ so the second term is $-2$, not $-\frac{1}{i}$.
When n= 2, n!= 2!= 2 and $\frac{(2i)!}{(2i-2)!}= $$\frac{2i(2i-1)(2i-2)!}{(2i-2)!}= $$2i(2i-1)$, $(-1)^2= 1$, and $\left(\frac{1}{i}\right)= \frac{1}{i^2}$ so the third term is $\frac{2(2i)(2i-1)}{i^2}= \frac{4(2i- 1)}{i}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Taylor and Laurent series with center at $z_{o}= 1$ of the function $\frac{\sinh(z)}{(z-1)^4}$ Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$
First I do the following:
$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$
The expansions of $\sinh(z)$ and $\cosh(z)$ are
$$\begin{align*}
\sinh(z) &= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \\
\cosh(z) &= 1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots
\end{align*}$$
Then
$$\begin{align*}
\sinh(z-1) &= (z-1)+\frac{(z-1)^3}{3!}+\frac{(z-1)^5}{5!}+\cdots \\
\cosh(z-1) &= 1+\frac{(z-1)^2}{2!}+\frac{(z-1)^4}{4!}+\cdots
\end{align*}$$
Now let's see what
$$\begin{align*}
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4}
&= \cosh(1) \biggl( \frac{1}{(z-1)^3}+\frac{1}{3!(z-1)}+\frac{(z-1)}{5!}+\cdots \biggr) \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4}
&= \sinh(1) \biggl( \frac{1}{(z-1)^4}+\frac{1}{2!(z-1)^2}+\frac{1}{4!}+\cdots \biggr)
\end{align*}$$
Rewriting these last two equalities we have
$$\begin{align*}
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4}
&= \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4}
&= \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
\end{align*}$$
Hence the Taylor and Laurent series of $f(z)$ is
$$
f(z) = \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} + \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
$$
Where we have
$$
a_{n} = \frac{\cosh(1)}{(2n-1)!}
\quad \text{and} \quad
b_{n} = \frac{\sinh(1)}{(2n)!}.
$$
But I don't know if it's okay. I feel a bit confused regarding these series. Could you tell me if I'm okay? or in another case, could you give me any suggestions?
| Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did)
$$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did
$$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then have
$$\frac{\sinh (t+1)}{t^4}=\sum_{n=0}^\infty a_n\, t^{n-4}$$ where
$$a_{2n+1}=\frac {\cosh(1)}{(2n+1)!}\qquad \text{and} \qquad a_{2n}=\frac {\sinh(1)}{(2n)!}$$ which is your result (just make $t=z-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030024",
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"source": "stackexchange",
"question_score": "1",
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} |
Conditional expectation of sum of numbers on 2 throws of a dice given their difference Let$ X$ and $Y$ be the numbers obtained on first and second throws of a fair die .
Calculate $E(X+Y|(X-Y)^2=1)$
I am doing basically $$E(X|(X-Y)^2=1)=E(Y|(X-Y)^2=1)$$
So $E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
$E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
implies $E(X|(X-Y)^2=1)=1/6(1+2+3+4+5+6)$
So I am getting it $7$ please help
| The result is correct.
To "see" the solution observe that throwing 2 fair dice the events that correspond to
$$(X-Y)^2=1$$
are 10
$$\{(1,2);(2,1);(2,3);(3,2);(3,4);(4,3);(4,5);(5,4);(5,6);(6,5)\}$$
given this sample space, the sum $X+Y$ is a rv taking values in
$$ Z=
\begin{cases}
\frac{1}{5}, & \text{if $z=3$} \\
\frac{1}{5}, & \text{if $z=5$} \\
\frac{1}{5}, & \text{if $z=7$} \\
\frac{1}{5}, & \text{if $z=9$} \\
\frac{1}{5}, & \text{if $z=11$}
\end{cases}$$
Thus
$$\mathbb{E}[Z]=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$
$$=\int \frac {1}{(25\tan^2\theta + 25)} $$
$$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$
$$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$
$$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$
$$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$
To generalize for $x$, $\theta = \arctan(7x/5)$
$$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$
$$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$
But taking the derivative of this gets me:
$$ \frac{35}{(49x^2 +25)^2}$$
Where is my mistake?
| Welcome to MSE. The issue is that you left off $dx$:
$$\int \frac1{49x^2 + 25} dx = \int \frac1{49(5/7\tan(\theta))^2 + 25} d(5/7 \tan(\theta) = \frac57\int \frac1{25\tan^2(\theta) + 25} \sec^2(\theta)d\theta = \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
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"source": "stackexchange",
"question_score": "3",
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System of differential equations: why does it solve it the example? consider the system with conditions: $\quad y_1(0) = 1,\quad y_2(0) = -1,\quad y_3(0) = 0,\quad y_4(0) = 2$
$\begin{align}
&{y_1}'= y_1 + 6\,y_2\\
&{y_2}'= y_2 + 6\,y_3\\
&{y_3}'= y_3 + 6\,y_4\\
&{y_4}'= y_4
\end{align}$
what can be written as matrix: $A = \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right)$
Thus the solution: $\exp(A\,x)\,y_0 = e^{x}\,\underbrace{\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)}_{B} \,\left(\begin{array}{c}y_1(0)\\y_2(0)\\y_3(0)\\y_4(0)\end{array}\right) = B\,\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) $
Now to my question: how does it solve the system above? May sound stupid, because I understand solving something without knowing what. For example: what's $y_1, y_2,y_3,y_4$ in the solution matrix? The 1st, 2nd, 3rd, 4th row or entry? But even then I can't put it together.
Also: How is the starting condition vector built? What would be the difference between $y_1(0) = 1$ and $y_1(1) = 1$
May be essential questions but these Systems are really kicking me off.
| We are given the system
$$\begin{align}
&{y_1}'= y_1 + 6\,y_2 = 1\, y_1 + 6\, y_2 + 0 \,y_3 + 0\, y_4\\
&{y_2}'= y_2 + 6\,y_3 = 0 \,y_1 +1\, y_2 + 6\, y_3 + 0\, y_4\\
&{y_3}'= y_3 + 6\,y_4 = 0\, y_1 + 0\, y_2 + 1\, y_3 + 6\, y_4\\
&{y_4}'= y_4 = 0 \,y_1 + 0 \,y_2 + 0\, y_3 + 1\, y_4
\end{align}$$
This can be compactly written as
$$y' = A y = \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right) \begin{pmatrix} y_1\\y_2\\y_3\\y_4 \end{pmatrix}$$
We now solve for the matrix exponential or matrix exponential (note that there are many approaches to solve a system, like eigenvalues/eigenvectors, diagonalization (when possible), Putzer's Algorithm...)
$$e^{Ax} = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)$$
Once we have the matrix exponential and are given initial conditions (they could have chosen anything), we can write the solution as
$$y(x) = e^{Ax} y_0 = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) = \left(
\begin{array}{c}
72 e^x x^3-6 e^x x+e^x \\
36 e^x x^2-e^x \\
12 e^x x \\
2 e^x \\
\end{array}
\right)$$
This means
$$\begin{align} y_1(x) &= 72 e^x x^3-6 e^x x+e^x\\ y_2(x) &= 36 e^x x^2-e^x \\ y_3(x) &= 12 e^x x \\y_4(x) &= 2 e^x \end{align}$$
We can verify this solution, for example
$$y_1' = y_1 + 6 y_2$$
Calculating the LHS
$$y_1' = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$
Calculating the RHS
$$y_1 + 6 y_2 = 72 e^x x^3-6 e^x x+e^x + 6(36 e^x x^2-e^x) = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$
The LHS $= $ RHS, that checks out and this looks like a valid solition - as will the three other results.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$ Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) = \frac{1}{2}\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)$$
$$\alpha = \cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$$
$$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$
$$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$
Hmm where I was wrong ?
| You are not wrong because
$$
\frac32\sqrt{3}\ge \sqrt3.
$$
You just have proved a stronger inequality than required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Cross product and determinants The determinant of a $3 \times 3$ matrix gives the volume of the parallepiped formed by 3 vectors. With the cross product, say to find the torque, we find the $3 \times 3$ determinant value of displacement vector $\mathbf{r}$ and force $\mathbf{F}$,
\begin{align}
\boldsymbol{\tau} &= \mathbf{r} \times \mathbf{F}
\\
&=
\begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
r_1&r_2&r_3\\
F_1&F_2&F_3\\
\end{vmatrix},
\end{align}
and the definition of the cross product yields the area vector.
But the determinant by definition is volume. How is the definition of the determinant changed to find the cross product ?
| When we write a cross-product as a determinant, we are really abusing notation. If you look at the expression,
\begin{align}
\mathbf{a\times b}
&= \begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
a_1&a_2&a_3\\
b_1&b_2&b_3\\
\end{vmatrix}
\\
&=
(a_2b_3 - a_3b_2)\mathbf{i} -(a_1b_3 - a_3b_1)\mathbf{j} +(a_1b_2 - a_2b_1)\mathbf{k}
\\
&=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix},
\end{align}
you can see that in each term we have a product of two dimensionful `length' coordinates, giving an interpretation as area.
When we talk about it being numerically the area of a parallogram, we are talking about the norm of this vector,
\begin{align}
\left\lVert
\mathbf{a\times b}
\right\rVert
&=
\sqrt{(a_2b_3 - a_3b_2)^2 + (a_1b_3 - a_3b_1)^2 + (a_1b_2 - a_2b_1)^2}.
\end{align}
Counting the dimensions is easiest if we just consider
\begin{align}
&\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix}
&\mathbf{b} = \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix},
\end{align}
then
\begin{align}
\left\lVert
\mathbf{a\times b}
\right\rVert
&=
\sqrt{(a_1b_2)^2}
\\
&=
a_1 b_2,
\end{align}
ie an area.
Why do we understand a determinant as corresponding to volume? Consider instead the scalar triple product,
\begin{align}
\left(\mathbf{a\times b}\right) \cdot \mathbf{c}
&=
\left( a_2b_3-a_3b_2 \right) c_1 +
\left( a_3b_1-a_1b_3 \right) c_2 +
\left( a_1b_2-a_2b_1 \right) c_3
\\
&=
\begin{vmatrix}
c_1&c_2&c_3\\
a_1&a_2&a_3\\
b_1&b_2&b_3\\
\end{vmatrix}.
\end{align}
Then if we count the dimensions, eg by choosing
\begin{align}
&\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix}
&\mathbf{b} &= \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix}
&\mathbf{c} = \begin{pmatrix} 0 \\ 0 \\ c_3\end{pmatrix},
\end{align}
to simplify the algebra,
we get
$$
\left(\mathbf{a\times b}\right) \cdot \mathbf{c}
=
a_1 b_2 c_3,
$$
a product of three lengths and hence a volume.
Final comment. Why do we abuse notation to write a cross-product as a determinant?
Really, we are interested in the only totally-antisymmetric tensor in three dimensions, called the Levi-Civita, or alternating, symbol $\varepsilon_{ijk}$, defined as
$$
\varepsilon_{ijk} =
\begin{cases}
+1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1), \text{ or } (3,1,2), \\
-1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2), \text{ or } (2,1,3), \\
\;\;\,0 & \text{if } i = j, \text{ or } j = k, \text{ or } k = i
\end{cases}.
$$
Both the cross-product and determinant are properly defined using this, with
$$
\mathbf{a}\times\mathbf{b} = \sum_{i,j,k=1}^3 \mathbf{e}_i \;\varepsilon_{ijk} \, a_j b_k
$$
and
$$
\det A
=
\begin{vmatrix}
A_{11}&A_{12}&A_{13}\\
A_{21}&A_{22}&A_{23}\\
A_{31}&A_{32}&A_{33}\\
\end{vmatrix}
=
\sum_{i,j,k=1}^3
\varepsilon_{ijk} \,
A_{1i}
A_{2j}
A_{3k}
$$
which allows us to use our mnemonic for calculating the latter (an expansion in the minors, the $2\times 2$ submatrices) to remember how to calculate the former.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{ This is what we need to prove}$$
$$
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing and is bounded sufficient?
| Assuming the nested radical is infinite we have $x=\sqrt{2+\sqrt{2+...+\sqrt{2}}}\Rightarrow x^2-2=x \iff x^2-x-2=0\Rightarrow x=2$
| {
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Find the probability of placing $5$ dots on an $8 \times 8$ grid. $5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.
What is the probability that no row or column has more than $1$ dot?
I thought about this in the following way, the number of ways to place $5$ dots on $64$ cells is $$\begin{pmatrix} 64 \\ 5 \end{pmatrix} $$
Now if I want the $8 \times 8$ to have no row or column with more than $1$ dot then I reasoned that the dots must be placed in the diagonal which has $8$ cells, so the number of ways I can place $5$ dots into $8$ cells is $$\begin{pmatrix} 8 \\ 5 \end{pmatrix} $$
So the desired probability would be $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
However I know that this is not the answer, the correct answer is $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
but I can't see what the term $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ counts, can you explain how this term came?
| Pick $5$ of $8$ rows to occupy. Sort those in ascending order. Hence the order they were picked in will be forgotten. This gives:
$$
8C5=\binom 85
$$
Pick $5$ of $8$ columns to occupy. Pair those with the ordered set of chosen rows in the order you pick them. Hence order counts this time, and we have:
$$
8P5=8!/(8-5)!=8\cdot 7\cdot 6\cdot 5\cdot 4
$$
which is where the extra product comes from.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| Since
$$f(x,y)= \sqrt{(2y-3)^2+1}+\dfrac{\sqrt2}2\sqrt{(6x-3)^2+1}+\dfrac12\sqrt{(6x-4y+2)^2+(6x)^2},$$
then the substitutions
$$2y-3 = \sinh s,\quad 6x-3=\sinh t,\quad s,t\in\mathbb R,\tag1$$
transform $\;f(x,y)\;$ to
$$g(s,t)= \cosh s + \dfrac{\sqrt2}{2} \cosh t+\dfrac12\,r(s,t),\tag2$$
where
$$r(s,t)=\sqrt{\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2}.\tag3$$
The stationary points of $\;g(s,t)\;$ correspond to solutions of the system $\;g'_s=g'_t=0,\;$ or
\begin{cases}
\sinh s-\dfrac1{r(s,t)}\,\cosh s\,(\sinh t-2\sinh s-1)=0\\[4pt]
\dfrac{\sqrt2}2\sinh t-\dfrac1{r(s,t)}\,\cosh t\,(\sinh t-\sinh s+1)=0,\\[4pt]
\end{cases}
\begin{cases}
r(s,t) = \coth s\,(\sinh t-2\sinh s-1)\\[4pt]
r(s,t) = \sqrt2\,\coth t\,(\sinh t-\sinh s+1),\\[4pt]
\end{cases}
\begin{cases}
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2s\\[4pt]
=(\sinh t-2\sinh s-1)^2(1+\sinh^2s)\\[4pt]
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2t\\[4pt]
=2(\sinh t-\sinh s+1)^2(1+\sinh^2t)\\[4pt]
\coth s\,(\sinh t-2\sinh s-1)\ge0\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\left(\sinh t+3\right)\sinh s=\sinh t-2\sinh s-1\\[4pt]
\big(2\sinh^2t-4\sinh t \sinh s +4\sinh^2s+4\sinh t+4\sinh s+10\big)\sinh ^2t\\[4pt]
=2(\sinh^2t-2\sinh t \sinh s+\sinh^2s+2\sinh t-2\sinh s+1)(1+\sinh^2t)\\[4pt]
\sinh t+3\ge0,\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t+3 = \dfrac{2\sinh s+4}{1-\sinh s}\\[4pt]
(\sinh s+2)^2\sinh ^2t =(\sinh t-\sinh s+1)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(5\sinh s+1+(\sinh s-1)^2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(\sin s+1)^2(\sinh s+2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}
$$\begin{pmatrix}\sinh s\\ \sinh t\\ g(s,t)\end{pmatrix} =
\left\{\begin{pmatrix}-2\\ -3\\ 2\sqrt5\end{pmatrix},
\begin{pmatrix}-\frac13\\ -\frac12\\ \sqrt10 \end{pmatrix},
\begin{pmatrix}0 \\ 1\\ 4\end{pmatrix}\right\}.$$
Therefore, the least value of $\;f(x,y)\;$ is $\;\color{brown}{\mathbf{\sqrt{10}}}\;$ at $\;\color{brown}{\mathbf{(x,y)=\left(\dfrac5{12},\dfrac43\right)}}.\;$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Interesting logarithmic inequality Recently the following question was published on the site and soon after this deleted:
Prove the inequality
$$
\left(\log\frac{a^2+b^2}{2ab}\right)^n\le \left(\log\frac{a^2+c^2}{2ac}\right)^n+\left(\log\frac{b^2+c^2}{2bc}\right)^n\tag1
$$
for all positive $a,b,c$ and $n=\frac12$.
The inequality has many nice features. Particularly $\frac12$ seems to be the largest possible value of $n$ such that (1) generally holds.
I have succeeded to prove the inequality in a rather awkward way but I believe there should be a simple and nice way to prove it. Possibly it is just a particular case of a more general inequality.
Any hint is appreciated.
| My attempt :
$\sqrt{ln (\frac{1+(\frac{b}{a}) ^2}{2\frac{b} {a} })}<\sqrt{ln (\frac{1+(\frac{c}{a}) ^2}{2\frac{c} {a} })}+
\sqrt{ln (\frac{1+(\frac{c}{b}) ^2}{2\frac{c} {b} })} $
Suppose $x=\frac{b}{a} $ and $y=\frac{c} {a} $
$\sqrt{ln (\frac{1+x^2}{2x})}<\sqrt{ln (\frac{1+y^2}{2y})}+\sqrt{ln (\frac{x^2+y^2}{2xy})}$
$\Leftrightarrow$$ln (\frac{1+x^2}{2x})<ln (\frac{1+y^2}{2y})+ln (\frac{x^2+y^2}{2xy})+2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})} $
$\Leftrightarrow$$ln (\frac{1+x^2}{2x})-ln (\frac{1+y^2}{2y})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y(1+x^2)}{2x(1+y^2})-ln (\frac{x^2+y^2}{2xy})-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y(1+x^2)(2xy) }{2x(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{2y^2(1+x^2) }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
$\Leftrightarrow$$ln (\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}-2\sqrt{ln (\frac{1+y^2}{2y})×ln (\frac{x^2+y^2}{2xy})}<0 $
So that is always correct if $ ln(\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}$<0
$\Rightarrow$$\frac{(1+x^2)2y^2 }{(1+y^2)(x^2+y^2)}<1$
$\Rightarrow$$y^2x^2+y^2-x^2-y^4<0$$\Rightarrow$$ (x^2-y^2)(y^2-1)<0$$\Rightarrow$$ x<y<1$ or $x>y>1$
that is correct if
$b<c<1$ or $b>c>1 $
So I am showed this inequality for $b<c<1 $ and $b>c>1 $
| {
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Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different? Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?
Probably a really dumb question but I just don't see it :(
| The difference is:
We can have $4 \ge x(x-4)$ if $x(x-4) \le 0$.
But we can not have $\frac 4{x(x-4)} \ge 1$ if $x(x-4) \le 0$.
(if $x(x-4) = 0$ then $\frac 4{x(x-4)}$ doesn't exist. And if $x(x-4) < 0$ then $\frac 4{x(x-4)} < 0 < 1$.)
(Otherwise, when $x(x-4) > 0$ then $\frac {4}{x(x-4)} \ge 1$ and $4 \ge x(x+4)$ are the same thing.)
So $4\ge x(x-4)$ is true more often and will have more solutions than $\frac 4{x(x-4)}\ge 1$.
(Although, come to think of it, $4 \ge x(x-4) > 0$ and $\frac 4{x(x-4)} \ge 1$ will have the exact same solutions.....)
======
To solve $4 \ge x(x-4)$ we have
$x(x-4) = x^2 - 4x \le 4$ so
$x^2 - 4x + 4 \le 8$ so
$0 \le (x -2)^2 \le 8$ so
$0 \le x - 2 \le \sqrt 8$ so
$2-\sqrt 8 \le x \le \sqrt 8+2$
But notice these solutions will include cases where $x(x-4) \le 0$.
.....
TO solve $\frac 4{x(x-4)}\ge 1$ we must consider three cases:
Case 1: $x(x-4) < 0$.
Then $\frac 4{x(x-4)} \ge 1 \implies$
$\frac 4{x(x-4)} x(x-4) \le 1\cdot x(x-4)\implies$
$4 \le x(x-4) < 0$. But that is impossible.
Case 2: $x = 0$
Then $\frac 4{x(x-4)} = \frac 4 0$ and that is impossible.
Case 3: $x(x-4) > 0$.
Then we have either $x > 0$ and $x-4 > 0$; so $x> 4$. Or $x < 0$ and $x-4 < 0$; so $x < 0$. So we are restricted to requiring $x > 4$ or $x < $. In other words $0 \le x \le 4$ is impossible.
But if we make that restriction then
$\frac 4{x(x-4)} \ge 1 \implies $
$\frac 4{x(x-4)} x(x-4) \ge 1\cdot x(x-4) \implies$
$4 \ge x(x-4)$.
But note !!!VERY IMPORTANT!!!! we can only do this because we are assuming $x(x-4) > 0$. If we didn't know $x(x-4) > 0$ we could not say this.
Now we solved $4 \ge x(x-4)$ above and got $2-\sqrt 8 \le x \le 2 +\sqrt 8$.
But we MUST combine it with $x \not \in [0, 4]$.
So the solution here is $2-\sqrt 8 \le x < 0$ or $4 < x < 2 + \sqrt 8$.
.....
These two solutions are the same EXCEPT $4 \ge x(x-4)$ includes cases where $x(x-4) \le 0$ (which occures if and only if $0\le x \le 4$) while $\frac 4{x(x-4)} \ge 1$ does not include those.
[SO as the solutions to $4\ge x(x-4)$ is $[2-\sqrt 8, 2 + \sqrt 8]$ then the solutions to $\frac 4{x(x-4)} \ge 1$ is $[2-\sqrt 8, 2+\sqrt 8] \setminus [0, 4] = [2-\sqrt 8, 0) \cup (4, 2+\sqrt 8]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two maximum values for $|iz+3-4i|$ what went wrong? I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$
$|iz+3-4i|\leq |iz|+|3-4i|$
$|iz+3-4i|\leq 4+5$
But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.
What went wrong here? , thanks in advance.
| Put $z = a + bi\implies |iz+3-4i|= |i(a+bi)+3-4i|=|3-b+(a-4)i|=\sqrt{(a-4)^2+(b-3)^2}$. The problem translates to: Maximize $f(a,b) = \sqrt{(a-4)^2+(b-3)^2}$ given that $a^2+b^2 \le 16$. First observe that if the max occurs at an interior point inside the disk $D: a^2+b^2 \le 16$ at $M$, then consider the point of intersection $B$ of the line joining $A = (4,3)$ to $M$ and the circle $C: a^2+b^2 = 16$, we have: $AM \le AB$ and $AB$ is the max value. This can be proven rigorously but is quite easy. Thus the constraint is reduced to the circle $a^2+b^2 = 16$, and $f(a,b) = \sqrt{a^2-8a+16+b^2-6b+9}=\sqrt{41-8a-6b}$. Again observe that $f(a,b)$ is maximized iff $g(a,b) = 41 - 8a-6b$ is maximized under the condition $a^2+b^2=16$. This can be done with Lagrange Multiplier from calculus. Indeed, $\nabla g(a,b) = \lambda\nabla h(a,b), h(a,b) = a^2+b^2-16\implies (-8,-6)=(2\lambda a, 2\lambda b)\implies 2\lambda a= -8, 2\lambda b = -6\implies \lambda a = -4, \lambda b = -3\implies \lambda^2(a^2+b^2)= (-4)^2+(-3)^2=25\implies 16\lambda^2=25\implies \lambda = \pm \dfrac{5}{4}$. We can see that only $\lambda = \dfrac{5}{4}$ yields the max value, and it corresponds to $a= -\dfrac{16}{5}, b = -\dfrac{12}{5}\implies g_{\text{max}} = 41-8\cdot \dfrac{-16}{5}-6\cdot \dfrac{-12}{5}=81\implies f_{\text{max}}= \sqrt{g_{\text{max}}}=\sqrt{81} = 9$, and this is the maximum value of $|iz+3-4i|$.
| {
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Antiderivative of $\frac{1}{x^2-1}$ for $x>1$ or $x<-1$ We know that the antiderivative of $\frac{1}{x^{2} - 1}$ on $(-1,1)$ is $\operatorname{artanh}(x)$. Is there a nice way of writing the antiderivative on the intervals $(-\infty,-1)$ and $(1,\infty)$?
I realize we could write $$\int_{a}^{x}\frac{1}{y^{2} - 1}\,dy$$ for $a>1$, for the antiderivative on $(1,\infty)$, for example, but I'm curious if there's an expression that doesn't include an integral, and WolframAlpha chokes when I ask it to compute this.
| For the sake of completeness, here are the two suggested ways of writing the antiderivative:
First Way:
$$\boxed{\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.}$$
This holds because:
\begin{align}
\frac{d}{dx}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| &=\frac{1}{2}\frac{d}{dx}\bigl(\ln|x-1| - \ln|x+1|\bigr)\\[5pt]
&=\frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\\[5pt]
&=\frac{1}{x^2-1}.
\end{align}
Second Way:
$$\boxed{\int \frac{dx}{x^2-1} = \begin{cases}
-\operatorname{artanh}(x)+C, &|x|<1\\
-\operatorname{arcoth}(x)+C, &|x|>1.
\end{cases}}$$
This holds because:
$$-\operatorname{artahn}(x) = \frac{1}{2}\ln\left(\frac{1-x}{x+1}\right) = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for } |x|<1$$ and
\begin{align}
-\operatorname{arcoth}(x) &= -\operatorname{artanh}\left(\frac{1}{x}\right)\\[5pt]
&= \frac{1}{2}\ln\left(\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\right)\\[5pt]
&= \frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)\\[5pt]
&=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for }|x|>1.
\end{align}
| {
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$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$ Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdot(a+b+c)(ab+bc+ca)(a+bc)(b+ca)(c+ab)}$$
$$P \ge3\sqrt[3]{\frac{8}{3}\cdot(a+b+c)(a+bc)(b+ca)(c+ab)} $$
I can't prove $(a+bc)(b+ca)(c+ab)\ge 8$
Could you help me ?
| Hints :
We have :
$$P=(a+b)(b+bc) + (b+c)(c+ca) + (c+a)(a+ab) \ge 12$$
Using :$$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$$
Now we use Cauchy-Schwarz inequality to get :
$$P\geq \sum_{cyc}^{}(a+\sqrt{abc})^{2}\geq 12$$
Now expand and use the substitution $3u=a+b+c$,$3v^2=ab+bc+ca$ and $w^3=abc$
Conclude using uvw's method .
| {
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Functional Expectation Problem: Let $X$ be distributed over the set $\Bbb N$ of non-negative integers, with pmf: $$P(X=i)=\frac{\alpha}{2^i}$$
*
*$\alpha$
*$E[X]$
For $Y=X$ mod $3$, find:
*
*$P(Y=1)$
*$E[Y]$
*
*I will assume $\alpha=1$ for now to set the stage althought is there a way to find this value rather than assume?
*Assuming this is simlar to an infinite expectation that starts with Basel problem: $$\sum_{i=0}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}$$
$$\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i^2}=1$$
$$p_i = \frac{6}{\pi^2}\cdot\frac{1}{i^2}$$
$$E[X]=\sum_{i=0}^\infty i \cdot p_i=\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i}=\infty$$
However, just not certain.
For the "$Y=X$ mod $3$" part I don't even know what that means. Does that mean $Y=X+3$?
Thus I don't know how to answer the rest of them without understanding what "mod $3$" means.
|
*
In order to guarantee that $p_i=\frac{\alpha}{2^i}$ is a distribution, we have that:
$$\sum_{i=0}^{+\infty} p_i = 1 \Rightarrow \\\alpha \sum_{i=0}^{+\infty}\frac{1}{2^i} = \alpha \frac{1}{1 - \frac{1}{2}} = 2 \alpha = 1 \Rightarrow \alpha = \frac{1}{2}.$$
1.
$$P(Y=1) = P(X ~\text{mod}~ 3 = 1) = \sum_{i : i ~\text{mod}~ 3 = 1}\frac{\alpha}{2^i}.$$
But all $i$ such that $i ~\text{mod}~ 3 = 1$ can be rewritten as:
$$i = 3j + 1,$$
for $j \geq 0.$ Hence:
$$P(Y=1) = \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \frac{\alpha}{2}\sum_{j=0}^{+\infty}\frac{1}{8^{j} } = \frac{\alpha}{2}\frac{1}{1 - \frac{1}{8}} = \frac{4}{7}\alpha = \frac{2}{7} .$$
2.
$$\mathbb{E}[Y=1] = \sum_{j=0}^{+\infty} (3j + 1)\frac{\alpha}{2^{3j + 1}} = \sum_{j=0}^{+\infty} \frac{3j\alpha}{2^{3j + 1}} + \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \\
=\frac{3\alpha}{2}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^j + \frac{4}{7}\alpha = \frac{3\alpha}{16}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^{j-1} + \frac{4}{7}\alpha =\\
= \frac{3\alpha}{16}\frac{1}{\left(1 - \frac{1}{8}\right)^2} + \frac{4}{7}\alpha = \frac{3\alpha}{16}\frac{64}{49} + \frac{4}{7}\alpha = \frac{12\alpha}{49} + \frac{28 \alpha}{49} = \frac{40}{49}\alpha = \frac{20}{49}.
$$
| {
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
| First Solution
By AM-GM inequality we have
$$
\frac{x^2+y^2}{2} \geq |xy| \geq xy \\
\frac{x^2+1}{2} \geq |x| \geq x \\
\frac{1+y^2}{2} \geq |y| \geq y \\
$$
Add them together.
Second Solution: By Cauchy-Schwarz we have
$$
\left( xy+1\cdot x+y\cdot 1 \right)^2 \leq ( x^2+1^2+y^2)(y^2+x^2+1^2)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Exact differential equation $\frac{y}{(y+x)^2}dx+ ( \frac{1}{y} - \frac{x}{(x+y)^2})dy=0$ I have to solve the differential equation $\dfrac{y}{(y+x)^2}dx+ ( \dfrac{1}{y} - \dfrac{x}{(x+y)^2})dy=0$
Let $ M= \dfrac{y}{(y+x)^2}$ and $ N=\dfrac{1}{y} - \dfrac{x}{(x+y)^2}$
Since it is an exact differential equation, the solution is a function U(x,y)
However, I started from $U_x(x,y) = M$
When performing the procedure I obtained that the solution $U(x,y) = - \dfrac{y}{x+y} + ln|y|+C$
On the other hand, if I start from $U_y(x,y) = N$
When performing the procedure I obtained that the solution $U(x,y) = \dfrac{x}{x+y} + ln|y|+C$
So are these solutions different?
| Both solutions are correct. They differ by a constant, which you can see by pulling out $-1$ from the general constant:
$$\frac x{x+y}+C=\frac x{x+y}-1+C=\frac{x-(x+y)}{x+y}+C=-\frac y{x+y}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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