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Integrate $\int\frac{x+1}{(x^2-x-5)^3} dx$ can you help with this math problem? I dont know how to start? There is any good method? $$\int\frac{x+1}{(x^2-x-5)^3} dx$$
Thank you
|
As in my other answer,
$$\int \frac{x+1}{\left(x^2-x-5\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x-5\right)}{\left(x^2-x-5\right)^3}+\int\frac{3}{\left(x^2-x-5\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x-5\right)^2}+3\cdot 2^5\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}\right)$$
Let $2x-1=\sqrt{21}\sin u$. Then $d(2x-1)=\sqrt{21}\cos u\, du$.
$$\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}=\frac{1}{21^2\sqrt{21}}\int \frac{1}{-\cos^5 u}\, du$$
Now integrate by parts. Here are some hints:
$$\int \sec^5 u\, du=\int \sec^3 u\, d\left(\tan u\right)=\sec^3 u\tan u-\int \left(3\sec^2 u\cdot \sec u\tan u\right)\tan u\, du$$
$$\int \sec^3 u\tan^2 u\, du=\int \sec^3 u\left(\sec^2 u-1\right)\, du=\int \sec^5 u\, du-\int \sec^3 u\, du$$
$$\int \sec^3 u\, du=\int \sec u\, d(\tan u)=\sec u\tan u-\int (\sec u\tan u)(\tan u)\, du$$
$$\int \sec u\tan^2 u\, du=\int \sec\left(\sec^2 u-1\right)\, du=\int \sec^3 u\, du-\int \sec u\, du$$
Now, there are many ways to solve $\int \sec u\, du$. See Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ (including the question body). You can always use the Tangent Half-Angle substitution on rational $\sin x,\cos x$ integrals, including $\int \sec u\, du$.
|
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|
$(a+b)^2+4ab$ and $a^2+b^2$ are both squares I cannot find a complete answer to the following problem (this is the source):
Q. Find all positive integers $(a,b)$ for which $(a+b)^2+4ab$ and $a^2+b^2$ are both squares.
Just something: clearly if $(a,b)$ works then $(a/c,b/c)$ works as well, where $c$ is the greatest common divisor of $(a,b)$. Hence we can assume they are coprime.
List of known primitive solutions: As remarked by Michael below, this is equivalent to solve the equation $(x^2-y^2)^2+12xy(x^2-y^2)+4x^2y^2=z^2$ with positive integers $x,y$. Adding the constraint $\mathrm{gcd}(x,y)=1$, by computer calculations we can see that all solutions $(x,y)$ with $x,y \le 30000$ are only the following ones: $(3,2)$, $(5,1)$, $(7,85)$, $(39,46)$, $(2717,1380)$, $(4097, 1337)$. This leads to the primitive solutions
$$
(a,b)=(5,12),
$$
$$
(a,b)=(5477689,7498920).
$$
Edit (05 Dec 2015): the ones above are the unique primitive solutions also with the (last new) constraint $x,y \le 80000$..
|
Let $q^2 = (a+b)^2$, $p^2 = a^2 + b^2$ and $r^2 = (a+b)^2 + 4ab$. Note that $r^2 + p^2 = 2q^2$. Now let $X = r+p$, $Y = r - p$ and $Z = 2q$.
You can easily show that $r^2 + p^2 = 2q^2$ if and only if $X^2 + Y^2 = Z^2$.
So, each solution can be found as follows:
*
*Pick a Pythagorean triple $(X, Y, Z)$ in which $Z$ is even.
*Let $q = \frac{Z}{2}$, $r = \frac{X+Y}{2}$ and $p = \frac{X-Y}{2}$.
*Check if there is any Pythagorean triple $(a', b', p)$ for all $a', b'$ that $a'+b' = q$.
Hope this helps :)
|
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Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
|
Notice, $$2^x+2^{x+1}=3^{x+2}+3^{x+3}$$
$$2^x+2\cdot 2^x=3^2\cdot 3^x+3^3\cdot 3^x$$
$$2^x+2\cdot 2^x=9\cdot 3^x+27\cdot 3^x$$
$$3\cdot 2^x=36\cdot 3^x$$ $$\frac{3^x}{2^x}=\frac{1}{12}$$
$$\left(\frac{3}{2}\right)^x=\frac{1}{12}$$
$$x\ln\frac{3}{2}=-\ln 12$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=-\frac{\ln 12}{\ln(3/2)}}}$$
$$
|
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|
Limit of function $x^2-x\cot\left(\frac{1}{x}\right)$ How To compute $\lim_{x \to \infty} x^2 -x \cot(1/x)$? Wolfram says it is $\frac{1}{3}$ and I know it is supposed to converge to a number other than 0 but I keep getting infinity.
|
Near $x=0$,
$$
\begin{align}
\tan(x)
&=x+\frac{x^3}3+O\left(x^5\right)\\
&=x\left(1+\frac{x^2}3+O\left(x^4\right)\right)
\end{align}
$$
therefore,
$$
\begin{align}
\cot(x)
&=\frac1x\left(1-\frac{x^2}3+O\left(x^4\right)\right)\\
&=\frac1x-\frac x3+O\left(x^3\right)
\end{align}
$$
So as $x\to\infty$,
$$
\cot\left(\frac1x\right)=x-\frac1{3x}+O\left(\frac1{x^3}\right)
$$
Now we see that
$$
x^2-x\cot\left(\frac1x\right)=\frac13+O\left(\frac1{x^2}\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to Evaluate $\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$? How to find $$\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$$
In fact,
\begin{align*}
&\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} \\
=& \int_0^1 {\frac{{u\arcsin u}}{{\left( {{u^2} - \sqrt {2\sqrt {13} - 2} u + \sqrt {13} } \right)\left( {{u^2} + \sqrt {2\sqrt {13} - 2} u + \sqrt {13} } \right)}}du} \\
= &\frac{1}{{2\sqrt {2\sqrt {13} - 2} }}\left[ {\int_0^1 {\frac{{\arcsin u}}{{{u^2} - \sqrt {2\sqrt {13} - 2} u + \sqrt {13} }}du} - \int_0^1 {\frac{{\arcsin u}}{{{u^2} + \sqrt {2\sqrt {13} - 2} u + \sqrt {13} }}du} } \right].
\end{align*}
But how can we continue?
|
This is probably not an answer
Being too lazy to attack such monsters, I should use Taylor series $$\sin^{-1}(u)=\sum_ 0^\infty\frac{(2n)!}{4^n(n!)^2(2n+1)}u^{2n+1}\qquad (|u|\leq 1)$$ and perform the long division by the denominator. This would lead to $$\frac{{u\sin^{-1} u}}{{{u^4} + 2{u^2} + 13}}=\frac{u^2}{13}+\frac{u^4}{1014}-\frac{79 u^6}{263640}+O\left(u^7\right)$$ If you are brave enough, take more terms !
|
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|
Let $X$ ~ Geometric with $p=\frac{1} {4}$ and $Y$ ~ Uniform on $(1,2,3,4)$. Given X and Y are independent, I need to find $VAR(2X + 2Y)$.
Does independence mean that $VAR(2X + 2Y) = VAR(2X) + VAR(2Y)$? If so, then $VAR(X)=\dfrac{1-p} {p^2} = \dfrac{\dfrac {3} {4}} {\dfrac{1} {16}}=12$. And $VAR(Y)= \dfrac {1} {4}(1^2 +2^2 +3^2 +4^2)-[\dfrac{1} {4}(1+2+3+4)]^2=\dfrac{5} {4}.$ Then $VAR(2X+2Y)=2VAR(X)+2VAR(Y)=24+10/2=\dfrac{53} {2}$?
Did I do this right or am I off somehow?
Thanks!
|
$$Var(2X+2Y) =Var(2(X+Y)) = 4Var(X+Y) = 4[Var(X) + Var(Y)]$$
Or:
If X and Y are independent, then 2X and 2Y are independent.
$$Var(2X+2Y)=Var(2X)+Var(2Y)=4Var(X)+4Var(Y)$$
Assuming your $Var(X)$ and $Var(Y)$ are correct, we have $Var(2X+2Y)=53.$
|
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|
Finding $\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$ Find $$\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$$
I think this has to be evaluated with riemann sums. I tried using $\sin x\approx x$ as n is big, but that got me to $\int_0^1 (1+1/x)^{-1/2}$ but that gives me the wrong answer. Any suggestions?
EDIT: I would like to see a solution with Riemann sums.
|
Use the inequality
$$x - \frac{1}{6}x^3 \leq \sin x \leq x, \quad x \in [0, \pi/2).$$
For sufficiently large $n$, the summand satisfies:
$$\frac{\pi}{\sqrt{n^2 + k}} - \frac{1}{6}\frac{\pi^3}{(n^2 + k)^{3/2}} \leq \sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) \leq \frac{\pi}{\sqrt{n^2 + k}} \tag{1}.$$
For $k \in \{1, \ldots, n\}$, the right side bound in $(1)$ is bounded above by $\frac{\pi}{n}$, while the left side bound in $(1)$ is bounded below by
$$\frac{\pi}{\sqrt{n^2 + n}} - \frac{1}{6}\frac{\pi^3}{n^3}.$$
It thus follows that
$$\frac{n\pi}{\sqrt{n^2 + n}} - \frac{1}{6}\frac{\pi^3}{n^2} \leq \sum_{k = 1}^n\sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) \leq \pi.$$
By the squeeze principle, we conclude that
$$\lim_{n \to \infty} \sum_{k = 1}^n\sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) = \pi.$$
|
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$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
Options are: A) $1:2$, B) $3:4$, C) $5:4$, D) $4:5$, E) None of these
So my first observation was that $|a|$ has to be greater than $|b|$. So only options that suffice this condition are $C)$ and $E)$. Also, if $C)$ was correct, then we get $a=5k$; $b=4k$, $25k^2-16k^2=9k^2$. Which is giving us a perfect square. But isn't there any rigorous way also?
|
In a multiple choice question it can often be quicker to answer it via trial and error as you know one of the choices must be correct. If we ignore the multiple choice part then are looking at: $a^2-b^2=x^2$. This is simplify Pythagoras' Theorem rewritten so if you know solutions to that you have additional solutions. A general solution to Pythagoras is for any $a$ and $b$ with $a>b>0$ then you can construct the following three numbers which will satisfy Pythagoras' Theorem:
$$(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$$
Each number could also be multiplied by a constant.
Update: To use express these as a ratio rewrite it to the form you needed:
$$(a^2+b^2)^2-(a^2-b^2)^2=(2ab)^2$$
or
$$(a^2+b^2)^2-(2ab)^2=(a^2-b^2)^2$$
Hence you want a ratio of:
$$a^2+b^2\,:\,a^2-b^2$$
or
$$a^2+b^2\,:\,2ab$$
for any integers $a>b>0$.
Note: Your solution has $a=2$ and $b=1$ as is the second form of the ratio.
|
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|
Find the limit of the recursive sequence Definition of the sequence :
$$a_1=a;\\a_2=b;\\$$ and $$\ \ \ a_{n+2}={{a_n+a_{n+1}}\over2}$$ for $n\geq 1$.
Find the limit of this sequence in terms of $a$ and $b$.
Now in this case , taking $\lim_{n\rightarrow \infty}$ on both sides does not help at all . So what I did was try to find the $n$-th term in therms of $a$ and $b$ and then take $n$ to $\infty.$
So , using the given recursion formula I wrote down like $7$-$8$ terms, grouped them in various ways but still could not get the pattern of the $n$-th term.
Terribly sorry for the lack of work/context in this post of mine but I tried for like an hour to figure things out but can't really put those scribbles down here.
Please give me some hints as to how to find the answer . Thank you.
|
If you would like to have a closed form for the individual terms of the recurrence, you can use the method of generating functions. This first step places your sequence as the terms of a power series $G(x) = \sum_{n=1}^\infty a_n x^n$, and then you generate a separate representation using the recursion. It's a long process but very powerful. It's real strength comes from combinatorics, but it can solve linear recurrences well.
$$G(x) = \sum_{n=1}^\infty a_n x^n = ax + bx^2 + \sum_{n=3}^\infty a_n x^n$$
$$= ax + bx^2 + \sum_{n=1}^\infty a_{n+2} x^{n+2}$$
$$= ax + bx^2 + \frac x2 \sum_{n=1}^\infty a_{n+1} x^{n+1} + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}$$
$$= ax + bx^2 + \frac x2 \left(\sum_{n=1}^\infty a_{n} x^{n} - ax\right) + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}.$$
Thus recognizing $G(x)$ as the series on the right, we have
$$G(x)= ax + bx^2 + \frac x2 \left(G(x) - ax\right) + \frac{x^2}2 G(x).$$ Thus solving for $G(x)$ we find:
$$G(x)(1-\frac{x}{2}-\frac{x^2}{2})=ax + (b-\frac{a}{2})x^2$$
and
$$G(x) = \frac{2ax}{2-x-x^2} + \frac{(2b-a)x^2}{2-x-x^2}.$$
Using partial fraction decomposition we get:
$$G(x) = (2ax + (2b-a)x^2) \left( \frac{1}{6} \frac{1}{1-\left(-\frac x 2\right)} + \frac13 \frac{1}{1-x} \right)$$
Now applying geometric series identities:
$$G(x) = (2ax + (2b-a)x^2) \sum_{n=0}^\infty\left( \frac{1}{6} \frac{(-1)^n}{2^n} + \frac13\right) x^n $$
Manipulating series:
$$G(x) = \sum_{n=0}^\infty 2a \left( \frac{1}{6} \frac{(-1)^n}{2^n} + \frac13\right) x^{n+1} + \sum_{n=0}^\infty (2b-a)\left( \frac{1}{6} \frac{(-1)^n}{2^n} + \frac13\right) x^{n+2}$$
$$=\sum_{n=1}^\infty 2a \left( \frac{1}{6} \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) x^{n} + \sum_{n=2}^\infty (2b-a)\left( \frac{1}{6} \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) x^{n}$$
$$=a + \sum_{n=2}^\infty \left[ 2a \left( \frac{1}{6} \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6} \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right] x^n$$
Thus since $G$ is a power series centered about zero, its coefficients are uniquely determined. For $n \ge 2$ we have: $$a_n = \left[ 2a \left( \frac{1}{6} \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6} \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right]$$
now as $n \to \infty$ we are left with $L = \lim_{n\to\infty} a_n = 2a \cdot \frac13 + (2b-a) \cdot \frac13 = \frac{1}{3}a + \frac{2}{3} b$.
|
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Is it possible to estimate the sign of real part of eigenvalues of a 10 by 10 matrix only by observing all the entries? I have a symbolic 10 by 10 matrix. It is not difficult to get the eigenvalue expressions by using Matlab. But the expressions of some eigenvalues are too long to be analyzed. I was wondering if there is a way to analyze the sign of real part of eigenvalues only by observing the matrix since all the entry expressions are pretty simple?
The matrix is like this:
\begin{bmatrix} -k_1-p_1\lambda & 0 &0 &0 & 0 &0 &0 &-\lambda^2p_1 &0 &0\\
0 & -k_2-p_2\lambda & a_{13}\omega &0 & 0 &0 &0 &0 &-\lambda^2p_2 &0 \\
0 &a_{12}\omega & -k_3-p_3\lambda &0 & 0 &0 &0 &0 &0 &-\lambda^2p_3\\
-\frac{\sin{\theta}}{2} &0 &0 &0 &-\frac{\omega}{2} &0 &0 &0 &0 &0\\
\frac{\cos{\theta}}{2} &0 &0 &\frac{\omega}{2} &0 &0 &0 &0 &0 &0\\
0 &\frac{\cos{\theta}}{2} &-\frac{\sin{\theta}}{2} &0 &0 &0 &\frac{\omega}{2}&0 &0 &0\\
0 &\frac{\sin{\theta}}{2} &\frac{\cos{\theta}}{2} &0 &0 &-\frac{\omega}{2}&0 &0 &0 &0\\
-1 &0 &0 &0 &0 &0 &0 &-\lambda &0 &0\\
0 &-1 &0 &0 &0 &0 &0 &0 &-\lambda &0 \\
0 &0 &-1 &0 &0 &0 &0 &0 &0 &-\lambda
\end{bmatrix}.
Thanks.
|
Arrange the rows and columns in the order $1,8,4,5,2,3,9,10,6,7$. Your matrix is permutation-similar to the direct sum of
$$
A=\left[\begin{array}{cc|cc}
-k_1-p_1\lambda &-\lambda^2p_1 &0 &0\\
-1 &-\lambda &0 &0\\
\hline
-\frac{\sin{\theta}}{2} &0 &0 &-\frac{\omega}{2}\\
\frac{\cos{\theta}}{2} &0 &\frac{\omega}{2} &0\\
\end{array}\right]
$$
and
$$
B=\left[\begin{array}{cccc|cc}
-k_2-p_2\lambda & a_{13}\omega &-\lambda^2p_2 &0 &0 &0\\
a_{12}\omega & -k_3-p_3\lambda &0 &-\lambda^2p_3 &0 &0\\
-1 &0 &-\lambda &0 &0 &0\\
0 &-1 &0 &-\lambda &0 &0\\
\hline
\frac{\cos{\theta}}{2} &-\frac{\sin{\theta}}{2} &0 &0 &0 &\frac{\omega}{2}\\
\frac{\sin{\theta}}{2} &\frac{\cos{\theta}}{2} &0 &0 &-\frac{\omega}{2}&0
\end{array}\right]
$$
The eigenvalues of the three $2\times2$ subblocks should be easy to analyse. It remains to find the eigenvalues of the $4\times4$ subblock of $B$.
|
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There is no square number that is $3 \mod 4$ Prove that the following cannot be true:
There is no square number that is $3 \mod 4$
$x^2 \equiv 3 \mod 4$, I started with examples:
$1^2 \mod 4=1$
$2^2 \mod 4=0$
$3^2 \mod 4=1$
I am more interested in the proof but Im not seeing how to do it.
|
If $x$ is even then $x=2k$ so the square of $x$ is equal $4k^2$ and leaves no remainder when divided by $4$.
If $x$ is odd then $x=2k-1$ so the square of $x$ is equal $4k^2-4k+1$ and leaves a remainder $1$ when divided by $4$.
So we have either $x^2 \equiv 0 \mod 4$ or $x^2 \equiv 1 \mod 4$.
|
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Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
|
By C-S $$3(x^2+y^2+z^2)=(1+1+1)(x^2+y^2+z^2)\geq(x+y+z)^2=1.$$
Thus, $x^2+y^2+z^2\geq\frac{1}{3}$.
The equality occurs for $x=y=z=\frac{1}{3}$, which is also the answer.
|
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|
8-puzzle which has the numbers in order but has gap in between If the numbers of the 8-puzzle are all in order, but the blank tile is somewhere in between, is this puzzle solvable? (for example, $\begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix}$, where 0 represents the blank tile)
|
Count how far the gap tile is displaced. In your case it has been moved up 2 and over 1. So that is a displacement of 3. Call that "odd" or 1.
Now do that for every square.
1 is in place. So that's displaced 0.
2 is over 1 so that's displaced 1.
3 is down one and over 3 so that's 3. odd. 1.
4 is over 1. 1.
5 is over 1. 1.
6 is displaced 3. odd. 1.
7 is displace 1.
8 is displaced 1.
Add those all up. 8. even. 0.
Even parity is solvable.
Odd parity is not.
$\begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 & 5 \\ 0 & 3 & 4 \\ 6 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 2 & 3 & 5 \\ 1 & 0 & 4 \\ 6 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 0 \\ 6 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 8 \\ 0 & 6 & 7 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 5 & 0 & 8 \\ 4 & 6 & 7 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 5 & 6 & 8 \\ 4 & 0 & 7 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 5 & 6 & 0 \\ 4 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \\ 0 & 7 & 8 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \\ 7 & 8 & 0 \end{bmatrix}$
|
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|
On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the "congruent number problem" (Much revised for brevity.) An integer $n$ is a congruent number if there are rationals $a,b,c$ such that,
$$a^2+b^2 = c^2\\
\tfrac{1}{2}ab = n$$
or, alternatively, the elliptic curve,
$$y = x^3-n^2x = x(x-n)(x+n)\tag1$$
is solvable in the rationals. Assume $x=(p/q)^2$. Then $(1)$ becomes,
$$\frac{p^2}{q^6}(p^4-n^2q^4) = y^2$$
or simply,
$$(p^2+nq^2)(p^2-nq^2) = w^2\tag2$$
Assuming $w=z\,t$ and equating factors, then $(2)$ becomes,
$$p^2 + nq^2 = z^2\\
p^2 - nq^2 = t^2\tag3$$
This is the implication given by Mathworld and OEIS.
Questions:
*
*Is it true that if $p^4-n^2q^4 = w^2$ is solvable, then is
$$p^2 + nq^2 = m z_1^2\\
p^2 - nq^2 = m z_2^2\tag4$$
necessarily also solvable for $m=1$?
*Is the solution to $(4)$ for $m=1$ the smallest for $p^4-n^2q^4 = w^2$?
P.S. This post made me re-visit congruent numbers.
|
(Edited 2018): After 2 years finally found the answer.
I. For Question $1$, the answer is yes.
If one has a solution to $x^4-n^2y^4 = z^2$, then,
$$p^2-nq^2 = (z^2-2nx^2y^2)^2\\
p^2+nq^2 = (z^2+2nx^2y^2)^2$$
where $p=x^4+n^2y^2$ and $q=2xyz$.
II. For Question $2$, the answer is no.
For example, in this Mathworld link, we see that the smallest solution to,
$$p^2+101q^2 = \color{brown}mz_1^2\\p^2-101q^2 = \color{brown}mz_2^2$$
for $ \color{brown}{m=1}$ is the large,
$$p_1,\,q_1 =2015242462949760001961,\, 118171431852779451900\tag1$$
but for $\color{brown}{m=101}$ is the smaller,
$$p_2,\,q_2 =2125141,\, 63050\tag2$$
Thus, $(2)$ is a smaller solution to,
$$p^4-101^2q^4 = w^2$$
|
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Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.
Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.
I tried a few different values for $p$, namely $2,3,5,7,$ and $11$. The prime $p =2,7,11$ all worked but $p =13$ didn't so it makes me wonder. How can I find all primes such that it is a perfect square?
EDIT: it turns out this is problem P25 in post number #63 at http://artofproblemsolving.com/community/c3h1171106p5665470
|
If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.
By choosing the appropriate sign we can write
$$
\begin{align}
n^2=(-n)^2 &= (kp+3)^2 \\
p^3-4p+9 &= k^2p^2+6kp+9 \\
0 &= p^2-k^2p-(6k+4) \\
p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2}
\end{align}
$$
where we used $p\ne 0$ since it's a prime. This can only have an integer solution for $p$ when $k^4+24k+16$ is a square (as @tkr suggested in the comments).
But
$$
\begin{array}{ll}
(k^2-1)^2 = k^4-2k^2+1 < k^4+24k+16<(k^2)^2 & \text{if }k<-11 \\
(k^2)^2 < k^4+24k+16< k^4+2k^2+1 =(k^2+1)^2 & \text{if }k>12
\end{array}
$$
so if $k<-11$ or $k>12$ then $k^4+24k+16$ lies between two squares and hence is not a square. This leaves 24 values to check, which you could do by hand. It leads to solutions when $k=-3,0,3$ which give integral values for $p$ of $-2,2,7,11$, of which only $2,7,11$ would normally be considered prime.
|
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What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$
Part II. Strangely, for the similar concordant forms/numbers $n$ such that,
$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$
is not even in the OEIS,
$$n = 1,7,10,11,17,20,22,23,24,27,30,31,34,\dots$$
The list of $104$ prime $n<10^3$ is known (by Kevin Brown, David Einstein (hm?), and Allan MacLeod) though several troublesome primes were excluded assuming the Birch/Swinnerton-Dyer conjecture.
Question: Anybody knows how to generate the list of all concordant forms $n<1000$? (Elkies describes a method here.)
P.S. Incidentally, the special case $n=52$ appears in equal sums of like powers. Let,
$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$
Then for $k=1,2,4,6,8,10,$
$$(8b)^k + (5a-4b)^k + (-a-2d)^k + (a-2d)^k + (-5a-4b)^k + (-12b+4c)^k + (12b+4c)^k =\\
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$
found by J. Wroblewski and yours truly. An initial primitive solution is $a,b,c,d = 3,4,5,29$ and an infinite more.
|
For the system of equations:
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.$$
Solution 1:
$$a=p-s,\quad b=2t,\quad c=p+s$$
$$w=\mp2q+p+s\pm2$$
$$q=(p\pm1)(s\pm1)$$
$$ps=t^2$$
Solution 2:
$$a=t^2-1,\quad b=2t,\quad c=t^2+1$$
$$w=3t^2-1$$
$$q=2t^2-1$$
|
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Integrate: $\int \frac{2x^2-3x+8}{x^3+4x} \, dx$ $$\int \frac{2x^2-3x+8}{x^3+4x}\,dx$$
My main problem is calculating the $B$ and $C$. This is the algebra part. Thus, what is a technique I can use that is in line with what I did to calculate $A$?
|
Multiplying both sides of $$\frac{2x^2-3x+8}{x^3+4x}=\frac Ax+\frac{Bx+C}{x^2+4}$$
with $x(x^2+4)$, you can remove denominators, and get
$$2x^2-3x+8=A(x^2+4)+(Bx+C)x. $$
Then set $x=\;$ the poles of the fraction:
*
*$x=0$ yields $8=4A+0$, whence $A=2$.
*$x=2\mathrm i$ yields $-8-6\mathrm i+8=-6\mathrm i=0+(2B\mathrm i+C)2\mathrm i=-4B+2C\mathrm i$. Identifying the real and imaginary parts, we get
$$B=0,\enspace C=-3.$$
|
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How to find the exponent of a prime in $n!$ Let the positive integer $n$ be written as powers of prime $p$ so that we have $n=a_kp^k+....+a_2p^2+a_1p+a_0,$ where $0\leq a_i<p.$ Show that the exponent of the highest power of $p$ appearing in the prime factorization of $n!$ is
$\frac{n-(a_k+....+a_1+a_0)}{p-1}$.
I know that the exponent of $p$ in $n!$ is $\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^k}\right\rfloor$. But I got stuck on how to use the given expression of $n$. Any suggestions?
|
$$
\begin{align}
\sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor
&= \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \dotsb + \left\lfloor \frac{n}{p^k} \right\rfloor + \sum_{i=k+1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor \\
&= \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \dotsb + \left\lfloor \frac{n}{p^k} \right\rfloor \\
&= \left\lfloor \frac{a_kp^k+\dotsb+a_2p^2+a_1p+a_0}{p} \right\rfloor + \left\lfloor \frac{a_kp^k+\dotsb+a_2p^2+a_1p+a_0}{p^2} \right\rfloor + \dotsb \\
&\phantom{= \lfloor} + \left\lfloor \frac{a_kp^k+\dotsb+a_2p^2+a_1p+a_0}{p^k} \right\rfloor \\
&= \left\lfloor \frac{n - a_0}{p} + \frac{a_0}{p} \right\rfloor + \left\lfloor \frac{n - (a_1p + a_0)}{p^2} + \frac{a_1p + a_0}{p^2} \right\rfloor + \dotsb \\
&\phantom{= \lfloor} + \left\lfloor \frac{n - (a_{k-1}p^{k-1}+\dotsb+a_2p^2+a_1p+a_0)}{p^k} + \frac{a_{k-1}p^{k-1}+\dotsb+a_2p^2+a_1p+a_0}{p^k} \right\rfloor \\
&= (a_kp^{k-1} + \dotsb + a_2p + a_1) + (a_kp^{k-2} + \dotsb + a_3p + a_2) + \dotsb + (a_k) \\
&\phantom{= \lfloor} + \left\lfloor \frac{a_0}{p} \right\rfloor + \left\lfloor \frac{a_1p + a_0}{p^2} \right\rfloor + \dotsb \left\lfloor \frac{a_{k-1}p^{k-1}+\dotsb+a_2p^2+a_1p+a_0}{p^k} \right\rfloor \\
&= (a_kp^{k-1} + \dotsb + a_2p + a_1) + (a_kp^{k-2} + \dotsb + a_3p + a_2) + \dotsb + (a_k) \\
&= p^{k-1}a_k + p^{k-2}(a_{k-1} + a_k) + \dotsb + p(a_2 + a_3 + \dotsb + a_k) + (a_1 + a_2 + \dotsb + a_k) \\
&= \frac{(p - 1)(p^{k-1}a_k + p^{k-2}(a_{k-1} + a_k) + \dotsb + p(a_2 + a_3 + \dotsb + a_k) + (a_1 + a_2 + \dotsb + a_k))}{p - 1} \\
&= \frac{a_kp^k - a_kp^{k-1} + a_{k-1}p^{k-1} + a_kp^{k-1} - \dotsb + p(a_1 + a_2 + \dotsb + a_k) - (a_1 + a_2 + \dotsb + a_k))}{p - 1} \\
&= \frac{a_kp^k + a_{k-1}p^{k-1} + \dotsb + a_0 - (a_1 + a_2 + \dotsb + a_k)}{p - 1} \\
&= \frac{n - (a_1 + a_2 + \dotsb + a_k)}{p - 1}
\end{align}
$$
*
*Line 2: The remainder of the sum is zero.
*Line 5: The first part in each floor bracket is an integer, and hence comes outside the brackets.
*Line 6: The fractions inside the brackets are less than one due to the way $n$ and $a_0, a_1, \dotsc, a_k$ are defined.
*Line 10: The cancellation works somewhat like a telescopic sum, cancelling all terms except those on the next line. (This can be verified.)
|
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If $a,b,c>0$ and $a+b+c=1$, prove $\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$ If $a,b,c>0$ and $a+b+c=1$, then prove
$$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$$
|
HINT: set $a=xy,b=yz,c=zx$ and after this use the $\tan(\alpha/2)$ etc substitution4}
after simplification we get
$$\frac{1}{1+z^2}+ \frac {1}{1+x^2}+\frac {y}{1+y^2}\le 1+\frac{3\sqrt{3}}{4}$$
now set $$x=\tan(\alpha/2),y=\tan(\beta/2),z=\tan(\gamma/2)$$
after simplification we get
$$\cos(\gamma)+\cos(\alpha)+\sin(\beta)\le \frac{3\sqrt{3}}{2}$$
|
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|
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}$ by trigonometric function ?
The signs : + - - + - - + - - ... .
Thanks for helping.
|
I can´t solve in detail this question, however I want to write a remark about because I think it might be of interest to some people (we disregard the convergence whose existence is implicit by the statement assuming to be true).
(1) Let $E$ be equal to the RHS; we have the challenging equality $$E^3+E^2-6E-7=0$$ (you can calculate the numerical value of $E$ and verify the corresponding approximation; I give below the source of this relation).
(2) The answer given by H.R. is quite “natural” but without solving the equation this appears almost as the question itself (where $E$ must be a root of the equation).
(3) The equation in (2), of 8-degree, factorizes as $$(x^2+x-4)(x^3-2x^2-3x+5)(x^3+x^2-6x-7)=0$$ so $E$ must be root of one of the two cubic factors (why?).
(4) One finds that the third factor is what suits. We add here, JUST FOR THE BEGINNERS, the trigonometrical solution of this equation with some pertinent discussion. $$x^3+x^2-6x-7=0\iff X^3+ax^2+bx+c=0 \space\text{where} \space a=1;\space b=-6;\space c=-7$$
$$p={3b-a^2\over 9}={-19\over 9}$$ $$q={9ab-27c-2a^3\over 54}=\frac{133}{54}$$ $$\Delta=p^3+q^2<0$$ There are three distinct real roots because the discriminant $\Delta$ is negative. Making $$\cos(\theta)=\frac{q}{\sqrt{-p^3}}=\frac{7\sqrt{171}}{114}$$
the solutions are given by $$\begin{cases}x_1=2\sqrt{-p}\cos (\frac{\theta}{3})-\frac a3\\ x_2=2\sqrt{-p}\cos (\frac{\theta+2\pi}{3})-\frac a3\\x_3=2\sqrt{-p}\cos (\frac{\theta+4\pi}{3})-\frac a3\end{cases}$$
These three roots are $$x_1\approx 2,5077;\space x_2\approx -1,2219;\space x_3\approx -2,2851$$ We get an approximation for the angle $\theta\approx 36,5867$ degrees so $\frac{\theta}{3}\approx 12,1955$ degrees. This calculation allows us to say that
$$\color{red}{E=x_1}$$
|
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Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$
*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
|
Observe $$(1 - \cos x + \sin x)(1 + \cos x) = (1 - \cos^2 x) + (1 + \cos x)\sin x = \sin^2 x + (1 + \cos x)\sin x = (1 + \cos x + \sin x)\sin x,$$ from which the result immediately follows.
|
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|
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polylogarithms)?
|
As can be checked by differentiation, there is an antiderivative continuous on $(0,1)$:
$$\begin{align}&\int\arcsin x\cdot\ln^3x\,dx=\\
&\hspace{1cm}\frac32\left[\operatorname{Li}_3\left(\frac{\alpha}2\right)-\operatorname{Li}_3\left(\frac\beta2\right)\right]+3\,(2-\ln x)\cdot\operatorname{Li}_2\left(\frac\alpha2\right)-\frac{\ln^3\alpha}2+24\,\beta\\
&\hspace{1cm}+3\,(\ln x-1)\cdot\ln^2\alpha-\left(\pi^2+12\ln^2x-6\ln^22+24\ln2-72\right)\cdot\frac{\ln\alpha}4\\
&\hspace{1cm}-\beta\,\ln^3x+3\,(2\,\beta+\ln2)\cdot\ln^2x+\left(\frac{\pi^2}2-18\,\beta-3\ln^22\right)\cdot\ln x\\
&\hspace{1cm}+x\left(\ln^3x-3\ln^2x+6\ln x-6\right)\cdot\arcsin(x)\color{gray}{+C},\end{align}$$
where
$$\alpha=1+\sqrt{1-x^2},\quad\beta=1-\sqrt{1-x^2}.$$
Here is an outline of an approach leading to this result:
*
*Integrate by parts to get rid of $\arcsin$.
*Change variable $y=\sqrt{1-x^2}$ to get rid of $\sqrt{1-x^2}$ in the denominator.
*Use identity $\ln(1-y^2)=\ln(1+y)+\ln(1-y)$ and expand parentheses, this will result in a sum of integrals with powers and products of $\ln(1+y),\, \ln(1-y)$ terms.
*Evaluate those integrals in terms of polylogarithms using CAS, WolframAlpha or integral tables.
*Simplify dilogarithm terms.
Bonus:
$$\begin{align}&\int\arcsin x\cdot\ln^4x\,dx=\\
&\hspace{1cm}120\alpha-\ln^4\alpha-6\operatorname{Li}_4\left(\frac\alpha2\right)
+6\operatorname{Li}_4\left(\frac\beta2\right)-3\operatorname{Li}_4\left(-\frac{\alpha^2}{x^2}\right)\\
&\hspace{1cm}+6\,(\ln x-2)\cdot\left[\operatorname{Li}_3\left(\frac\alpha2\right)-\operatorname{Li}_3\left(\frac\beta2\right)\right]
-6\,\left(\ln^2x-4\ln x+6\right)\cdot\operatorname{Li}_2\left(\frac\alpha2\right)\\
&\hspace{1cm}+(\alpha-4)\cdot\ln^4x+(4\ln x+4-2\ln2)\cdot\ln^3\alpha\\
&\hspace{1cm}-\left[\vphantom{\Large|}6\,(\ln x+8-2\ln2)\cdot\ln x+\pi^2-36\right]\cdot\frac{\ln^2\alpha}2\\
&\hspace{1cm}+\left[\vphantom{\Large|}(\ln x+2-\ln2)\cdot\pi^2+6\,(4-\ln2)\cdot\ln^2x\right]\cdot\ln\alpha\\
&\hspace{1cm}+\left[\vphantom{\Large|}6\zeta(3)-96+2\ln^32-12\ln^22+36\ln2\right]\cdot\ln\alpha\\
&\hspace{1cm}+\left[\vphantom{\Large|}96\beta-6\zeta(3)-2\ln^32+24\ln^22+(\ln2-4)\cdot\pi^2\right]\cdot\ln x\\
&\hspace{1cm}-3\left(12\beta+\ln^22+8\ln2\right)\cdot\ln^2x+8\,(\beta+\ln2)\cdot\ln^3x\\
&\hspace{1cm}+x\left(\ln^4x-4\ln^3x+12\ln^2x-24\ln x+24\right)\cdot\arcsin(x)\color{gray}{+C},\end{align}$$
where
$$\alpha=1+\sqrt{1-x^2},\quad\beta=1-\sqrt{1-x^2}.$$
|
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|
Groups with Complex Numbers I'm working on a quick proof related to the complex numbers.
Take $G$ to be the group of all nonzero complex numbers under
multiplication ($\circ$). Let $H$ be the set of of complex
numbers such that the sum of squares of the two real parts of the
complex number is equal to 1. I want to show that $H$ under
multiplication is a subgroup of $G$. I can see how to do the
closure part of the proof, but I am having issues finding an
accurate inverse for each element, likely due to the fact that
I am uncertain about the identity element in the complex case.
|
Remember that the identity under multiplication is 1. So the inverse of $z \in \mathbb C$ is the reciprocal $\frac{1}{z}$. Since $|z| = 1$, we know $\left|\frac{1}{z}\right| = \frac{1}{1} = 1$.
We can compute this explicitly if you are not comfortable with the absolute value of a complex number. Let $z = x+iy$ such that $x^2+y^2=1$ and $x, y \in \mathbb R$. We now have $$\frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}.$$
Of course, we need to check that sum of squares of real and complex parts is 1. So we get
$$\left(\frac{x}{x^2+y^2}\right)^2 + \left(\frac{-y}{x^2+y^2}\right)^2= \frac{x^2+y^2}{(x^2+y^2)^2},$$ which is 1 because we assumed that $x^2+y^2=1$.
|
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|
Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
|
We have:
$$(1+2+...+k+(k+1))^2=(1+2+...+k)^2+(k+1)^2+2(k+1)(1+2+...+k)$$
So by induction hypothesis:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+2(k+1)(1+2+...+k)$$
We know $1+2+...+k = \frac{k(k+1)}{2}$.(You can prove it by induction easily). Therefore:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+k(k+1)^2=1^3+2^3+...+k^3+(k+1)^3$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to compute $\lim_{x \to 0} (\frac{x^5 e^{-1/x^2}+x/2 - \sin(x/2))}{x^3})$? I have a problem with this limit. I have no idea where is the problem.
Can you correct my mistake? Thanks
$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$
I used the developments of McLaurin $e^x$ and $\sin x$
$$\lim\limits_{x \to 0} \left(\frac{x^5 (1-\frac{1}{x^2}+\frac{1}{2x^4})+\frac{x}{2} - ((\frac{x}{2})-(\frac{x^3}{48}))}{x^3}\right) =
\lim\limits_{x \to 0} \left(\frac{x^5-x^3+ \frac{x}{2} +\frac{x}{2} - \frac{x}{2}+\frac{x^3}{48}}{x^3}\right)=$$
$$\lim\limits_{x \to 0} \left(\frac{-x^3+\frac{x^3}{48}}{x^3}\right)=
\lim\limits_{x \to 0} \left(\frac{-\frac{47x^3}{48}}{x^3}\right)=\lim\limits_{x \to 0} \left(-\frac{47x^3}{48x^3}\right)= -\frac{47}{48}$$
but the result is wrong.
|
One may write, as $x \to 0$,
$$
\begin{align}
\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}&=\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \left(\frac{x}2-\frac{x^3}{48}+O(x^5)\right)}{x^3}\\\\
&=\frac{x^5 e^{-\frac1{x^2}} + \frac{x^3}{48}+O(x^5)}{x^3}\\\\
&=\frac{x^2 e^{-\frac1{x^2}} + \frac{1}{48}+O(x^2)}{1}\\\\
&= \frac{1}{48}+O(x^2)\\\\
\end{align}
$$ giving $\dfrac{1}{48}$ for the sought limit, where we have used $\displaystyle x^2 e^{-\frac1{x^2}}=O(x^2)$ (to say the least).
|
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|
Find $a,b$ for which $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has unique solution Find all values of $a,b$ for which the system of equations $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has only one real solution.
$xyz+z=a$
$xyz^2+z=b$
So,$\frac{xy+1}{xyz+1}=\frac{a}{b}$
I can think no method by so as this system of equations has only one real solution. What should I do? Please help me.
|
If $(x, y, z)$ is a solution, then $(y, x, z)$ is a solution. Hence if there is a unique solution, we need $x=y$. This reduces the problem to $2$ variables.
$$\begin{align*}
x^2z+z&=a\\
x^2z^2+z&=b\\
2x^2+z^2&=4\\
\end{align*}$$
If $(x, x, z)$ is a solution, note that $(-x, -x, z)$ is also a solution. Hence, $x=0$ if the solution is unique.
$$\begin{align*}
z&=a\\
z&=b\\
z^2&=4\\
\end{align*}$$
Hence $a=b$. For there to be a solution, we need $a=b=2$ or $a=b=-2$.
We now check the $2$ solutions of $(a, b)$ by solving the equations.
$$\begin{align*}
xyz+z&=a\\
xyz^2+z&=b\\
x^2+y^2+z^2&=4\\
\end{align*}$$
Since $a=b$, we subtract the first equation from the second equation to get $xy(z^2-z)=0$. There are a few cases:
If $z=0$, then $xyz+z=0\neq a$.
Otherwise, if $x=0$ (or $y=0$ symmetrically), we have $z=a$ and $z^2=4$, which, substituting into the third equation, means that $y^2=0$ (or $x^2=0$), which means that $(x, y, z)=(0, 0, a)$ is one solution.
The final case $z=1$ should have no solutions. Let's check the $2$ cases $a=b=2$ and $a=b=-2$.
If $a=b=2$, the system of equations reduces to:
$$\begin{align*}
xy&=1\\
x^2+y^2&=3\\
\end{align*}$$
Solving, we find another solution, $(x, y)=\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$, so $a=b=2$ does not admit a unique solution.
If $a=b=-2$, the system of equations reduces to:
$$\begin{align*}
xy&=-3\\
x^2+y^2&=3\\
\end{align*}$$
Note that $(x+y)^2=x^2+y^2+2xy=-3$. As such $x+y$ is not real and there is no extra solution here.
Hence the only $(a, b)$ for the system to have a unique solution is $(-2, -2)$.
|
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|
Evaluate the following integral I got stuck with the integral below. I have tried to make it look like the derivative of arctan.
$$\int \frac{2-x}{x^2-x+1}\,dx$$
Thank you!
|
HINT: $$\int \frac{2-x}{x^2-x+1}\ dx=\frac{1}{2}\int \frac{3-(2x-1)}{x^2-x+1}$$
$$=\frac{1}{2}\int \frac{3}{x^2-x+1}\ dx-\frac 12\int \frac{(2x-1)}{x^2-x+1}\ dx$$
$$=\frac{3}{2}\int \frac{1}{\left(x-\frac 12\right)^2+\frac{3}{4}} dx-\frac 12\int \frac{d(x^2-x+1)}{x^2-x+1}$$
|
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|
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
|
You are all wrong. The determinant of this matrix is 380.
You can check it here:
https://www.wolframalpha.com/input/?i=matrix+determinant&assumption=%7B%22F%22,+%22Determinant%22,+%22detmatrix%22%7D+-%3E%22%7B+%7B+5,-7,2,2%7D,%7B0,3,0,-4%7D,+%7B+-5,-8,0,3%7D,%7B+0,5,0,6%7D%7D%22&assumption=%7B%22C%22,+%22matrix+determinant%22%7D+-%3E+%7B%22Calculator%22,+%22dflt%22%7D
|
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|
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
|
Yet another way in which this can be shown using Schur's inequality in tandem with the AM-GM inequality is as follows:
$$
a^2+b^2+c^2+3(a^2b^2c^2)^{1/3}\geqslant a^{2/3}b^{4/3} + a^{4/3}b^{2/3} +b^{2/3}c^{4/3} + b^{4/3}c^{2/3} + a^{2/3}c^{4/3} + a^{4/3}c^{2/3} \\[2ex]= 2\left({a^{2/3}b^{4/3} + a^{4/3}b^{2/3}\over 2} + {b^{2/3}c^{4/3} + b^{4/3}c^{2/3}\over 2} + {a^{2/3}c^{4/3} + a^{4/3}c^{2/3}\over 2}\right)\\[2ex] \geqslant 2(ab + bc + ac)
$$
|
{
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|
Not Understanding a specific substitution rule I was given the question,
If $f(3x+5) = x^2-1$, what is $f(2)$? I am trying to understand the reasoning why $3x+5$ is set equal to $2$.
|
Let's say $u=3x+5$. Then, $x^2-1=(\frac{u-5}{3})^2-1$. $\Rightarrow$ $f(u)=(\frac{u-5}{3})^2-1$ $\Rightarrow$ $f(x)=(\frac{x-5}{3})^2-1$. Now we can easily see that $f(2)=0$ . However, instead of doing all these complicated stuff, we could have made the following deduction: $3x+5=2$ $\Rightarrow$ $x=-1$ $\Rightarrow$ $x^2-1=0$. To summarize, $3x+5=2 \Rightarrow x^2-1=0$ $\:\:\:\equiv\:\:\:$ $f(2)=0.$
SIMILAR EXAMPLE
$f(x-1)=x^2\Rightarrow f(0)=?$
Say $u=x-1 \Rightarrow (u+1)^2=x^2 \Rightarrow f(u)=(u+1)^2 \Rightarrow f(x)=(x+1)^2 \Rightarrow f(0)=1.$
OR
Set $$x-1=0 \Rightarrow x=1 \Rightarrow x^2=1$$
$$\Downarrow$$
$$x-1=0 \: and \: x^2=1 \:\:\equiv\:\: f(0)=1$$
|
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|
Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some trick or other technique to get the result quicker . Can anybody help please $?$
Thank you .
|
You could consider carrying out polynomial long division, dividing the given polynomial by the minimal polynomial of $-5+4i$, which is $x^2+10x+41$. Viz:
$x^4+9x^3+35x^2-x+4=(x^2-x-4)(x^2+10x+41)+160$
We know that if $x=-5+4i$, then $x^2+10x+41$ vanishes, so the given polynomial evaluates to $160$ there.
|
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|
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } = \frac{65}{8}; a \neq 0
$$
|
factorizing the left minus the right-hand side we get $$-\frac{20 x \left(x^2+5\right)}{(x+1) (x+2) (x+3) (x+4)}=0$$
can you proceed from here?
making the same with the second equation we get $$-\frac{x (x+3) \left(x^2+3 x-16\right)}{(x-1) (x+1) (x+2) (x+4)}=0$$
|
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|
Pythagorean triples with the same c value $a^2 + b^2 = c^2$
There are, Primitive Pythagorean Triples, that share the same c value. For example,
$63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.
I have been trying to figure out why the following theorem for finding such triples works.
Take any set of primes. Ex: $5,13,17$.
Now take their product, $1052$, this is the new $c^2$ value. You can express that $c^2$ value as a triple by factoring the product of primes as Gaussian Integers:
$(2-i)(2+i)(3+2i)(3-2i)(4+i)(4-i)$
Now if you take three of those Gaussian Integers and solve for their product:
ex, $(2+i)(3+2i)(4+i) = 9+32i$ and you have found an $a$ value $9$ and a $b$ value $32$ that works in the theorem. $9^2 + 32^2 = 1105$
You can continue with this:
$(2-i)(3+2i)(4+i) = a + bi$
$(2+i)(3-2i)(4+i) = a + bi$
$(2+i)(3+2i)(4-i) = a + bi$
In fact, the number of triples with this method is $2^{||p|| - 1}$ where $||p||$ is the number of primes used.
Can someone please explain why this method of finding these "c stuck triples" works the way it does?
EDIT: It appears that you cannot take any prime, $p$ but rather must use primes congruent to $1mod4$ according to Fermat's theorem on sums.
|
1) $(a + bi)(a - bi) = a^2 + b^2$ always.
2) $(a + bi)(c + di)(e + fi) = g + hi \ne (a + bi)(c + di)(e - fi) = j + ki$
yet $(a - bi)(c - di)(e - fi) = g - hi \ne (a - bi)(c - di)(e + fi) = j - ki$
so
3) $(a + bi)(c + di)(e + fi)(a - bi)(c - di)(e - fi) = (g+hi)(g - hi) = g^2 + h^2 =K$
and
4) $(a + bi)(c + di)(e - fi)(a - bi)(c - di)(e + fi) = (j+ki)(j - ki) = j^2 + k^2 = pqr$
So $pqr = g^2 + h^2 = j^2 + k^2$.
You do have to choose primes that a gaussian factorable which seems to me like begging the question.
|
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|
Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$
Consider the set containing $A$ all positive integers with no prime
factor larger than $3$, and define $B$ as
$$
B= \sum_{k\in A} \frac{1}{k}
$$
Thus, the first few terms of the sum are:
$$
\frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{6} +\frac{1}{8} +\frac{1}{9} +\frac{1}{12} +\frac{1}{16} +\frac{1}{18} + \cdots
$$
a) Write a
closed-form expression for $X$ that makes the equation below true. In
other words what expression should $X$ be so that the following
equation is true, i.e. writing $B$ in terms of $X$.
$$
B = \sum_{i=0}^\infty \sum_{j=0}^\infty X
$$
b) Write a closed-form expression for .
|
So there are obviously only two primes, $2$ and $3$. Thus, we have to consider the sum
$$\sum_{j\ge0}\sum_{i\ge0}\frac{1}{2^i3^j}$$
I would break this down further into the following:
$$\sum_{i\ge0}\frac{1}{3^0\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^1\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^2\cdot2^i}+\cdots$$
Now the first term is a geometric series with sum $\frac{1}{1-\frac{1}{2}}=2$. The second term is simply one third of that, or $\frac{2}{3}$. Continuing, the entire sum is equal to
$$\frac{2}{3^0}+\frac{2}{3^1}+\frac{2}{3^2}+\cdots=\frac{2}{1-\frac{1}{3}}=\boxed{3}$$
|
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|
Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$ As said in the title, I have to find the minimum of the following:
$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$
In my very last attempt, I tried to work it out using AM-GM:
Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\frac {a}{\sqrt{a(b+c)}}\ge \frac {\sqrt{2}a}{a+b+c}$$
$$\Rightarrow\sum_{cyc} {\sqrt{\frac a{2(b+c)}}}\ge \frac {\sqrt{2}(a+b+c)}{a+b+c}=\sqrt{2}$$
The problem is, to get the minimum value, $$a=b+c$$
$$b=c+a$$
$$c+a+b$$
or $a=b=c=0$. This is wrong since $a,b,c>0$
And after some calculus I found the minimum value is $\frac 32$ when $a=b=c$.
So what mistake I have made?
|
Stronger inequality:
With $a, b, c> 0$: $$\sqrt{\frac{a}{b+ c}}+ \sqrt{\frac{b}{c+ a}}+ \sqrt{\frac{c}{a+ b}}\geq \sqrt{2+ 2.\frac{abc}{\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )}}$$
Proof:
Thus, we have:
$$\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )+ abc= \left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$
From the assumed inequality, we obtain the equivalent inequality:
$$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}\geq 2\sqrt{\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )}$$
Other way, we have:
$$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}= \sqrt{a^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{b^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{c^{2}\left ( a+ b+ c \right )+ abc}\geq \sqrt{\left ( a\sqrt{a+ b+ c}+ b\sqrt{a+ b+ c}+ c\sqrt{a+ b+ c} \right )^{2}+ 9abc}= \sqrt{\left ( a+ b+ c \right )^{3}+ 9abc}\geq 4\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$
Done!
|
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|
Find the value of the expectation $E(X)$ of the following distribution.
Let $X$ be a random variable having the distribution function
$$F(x)=\begin{cases}
0 &\text{if } x<0\\
\frac{1}{4} & \text{if } 0\leq x<1\\
\frac{1}{3} & \text{if } 1\leq x<2\\
\frac{1}{2} & \text{if } 2\leq x<11/3\\
1& \text{if } x\geq 11/3\end{cases}$$
Find the value of $E(X)$.
Now we know $E(X)=\int_{-\infty}^{\infty}xf(x)dx=\int_{0}^{1}\frac{1}{4}xdx+\int_{1}^{2}\frac{1}{3}xdx+\int_{2}^{11/3}\frac{1}{2}xdx$,
now I am stuck as i don't know what to do with $x\geq 11/3$? Please help me out. Thank you.
|
This thing isn't a continuous random variable, so the formula you're using doesn't work.
I recommend drawing this thing out.
The expectation is
$$E[X] = (1-0)\left(1-\frac{1}{4}\right)+(2-1)\left(1-\frac{1}{3}\right) +\left(\frac{11}{3}-2\right)\left(1-\frac{1}{2}\right)=2.25$$
Alternatively, the distribution of $X$ is
\begin{array}{r|ccccc}
x & 0 & 1 & 2 & \frac{11}{3}\\\hline
P(X = x)& \frac{1}{4} & \frac{1}{12} & \frac{1}{6} & \frac{1}{2}
\end{array}
Then,
$$E[X] = 0(1/4)+1(1/12)+2(1/6)+(11/3)(1/2) = 2.25$$
|
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|
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
|
$1=x^3+y^3\implies 0=3 x^2+3 y^2 y',$ so $$0=x^2+y^2 y'.$$ Differentiate this to get $$0=2 x +2 y y'^2+y^2 y''.$$ Therefore for $y\ne 0$ we have $$y''=-y^{-2}(2 x+2 y y'^2).$$ From the first differentiation we have $$y'=-x^2/y^2.$$ Therefore for $y\ne 0$ (equivalently, for $x\ne 1$),$$y''=-y^{-2}(2 x +2 y y'^2)=- y^{-2}(2 x +2 y (x^4/y^4))=-2 y^{-5} x (y^3+x^3)=-2y^{-5}x$$ because $y^3+x^3=1.$ This can also be written as $y''=-2 y x y^{-6}=-2 y x (1-x^3)^{-2}.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$. Let $x,y,z>1$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that
$$\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$$.
I took $x=\sec^2{a}$, $y=\sec^2{b}$, $z=\sec^2{c}$ but it was not useful. Then I took $x-1=a$ and similarly $b$ and $c$. On simplifying I got
$$b+1\ge{2\sqrt{ac}}$$
|
Applying Cauchy the following is true. $$(x+y+z)(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})\ge (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$$
However, $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1$ from the condition in the problem.
$\therefore$ $\sqrt{x+y+z}\ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$.
However $\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\ge \sqrt{x-1}+\sqrt{z-1}$. Our proof is done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating the Fourier coefficients of $abs(x)$ Let's get started:
$$\hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} |x|e^{-inx} dx$$
since $|x|$ is an even function:
$$= \frac{1}{\pi}\int_0^{\pi} xe^{-inx} dx$$
Integration by parts yields:
$$e^{-inx}\Big|_0^{\pi} + \frac{1}{in} \int_0^\pi e^{-inx} dx = (-1)^n - 1 + \frac{1}{in} \left( \frac{(-1)^n}{-in} + \frac{1}{in} \right) \\ = (-1)^n - 1 + \frac{(-1)^n - 1}{n^2}$$
So if $n$ is even then $\hat f(n) = 0$. Otherwise:
$$\hat f(n) = \frac{1}{\pi} \left( -2 -\frac{2}{n^2} \right)$$
but that doesn't make sense since we know that $\hat f(n) \to 0$.
Where is my mistake?
EDIT
it should be
$$x e^{-inx}\Big|_0^{\pi} + \frac{1}{in} \int_0^\pi e^{-inx} dx = \frac{\pi e^{-in\pi}}{-in} + \frac{(-1)^n - 1}{n^2}$$
So $$\hat f(n) = \frac{1}{\pi} \left( \frac{(-1)^n}{-in} + \frac{(-1)^n - 1}{n^2} \right)$$
|
Since $|x|$ is positive in $[0, 2 \pi]$, then
\begin{align}
\hat f(n) &= \frac{1}{2\pi} \, \int_0^{2\pi} |x|e^{-inx} \, dx \\
&= \frac{1}{2\pi} \, \int_{0}^{2\pi} x \, e^{-i n x} \, dx \\
&= \frac{1}{2\pi} \, \left[ \int_{0}^{\pi} x \, e^{-i n x} \, dx + \int_{\pi}^{2\pi} x \, e^{-i n x} \, dx \right] \\
&= \frac{1}{2\pi} \, \left[ \int_{0}^{\pi} x \, e^{-i n x} \, dx + \int_{0}^{\pi} (x+\pi) \, e^{-i n (x+\pi)} \, dx \right] \\
&= \frac{1}{2\pi} \, \int_{0}^{\pi} \left[ x + (-1)^{n} (x+\pi) \right] \, e^{-i n x} \, dx \\
&= \frac{1}{2\pi} \, \left\{ \left[\frac{i}{n} \left(x + (-1)^{n} (x+\pi)\right) + \frac{1+ (-1)^{n}}{n^{2}}\right] \, e^{-i n x} \right\}_{0}^{\pi} \\
&= \frac{i \, (1 + (-1)^{n})}{2 \, n}
\end{align}
This leads to
\begin{align}
\hat f(2n) &= \frac{i}{2n} \\
\hat f(2n+1) &= 0.
\end{align}
|
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|
Induction proof without summation I have to prove this induction:
$\dfrac{1}{(n+1)}+\dfrac{1}{(n+2)}+\dots+\dfrac{1}{2n} = \dfrac{1}{(1\times2)}+\dfrac{1}{(3\times4)}+\dots+\dfrac{1}{(2n-1)\times2n}$
Can someone help me with it?
|
If you want to use the summation symbol, note that
$$
a_n=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=
\sum_{k=1}^n\frac{1}{n+k}
$$
Therefore
$$
a_{n+1}=\sum_{k=1}^{n+1}\frac{1}{n+1+k}
$$
The right-hand side can be written
$$
b_n=\sum_{k=1}^{n}\frac{1}{2k(2k-1)}
$$
So your task is to prove $a_n=b_n$. The case $n=1$ is trivial. Suppose the assertion holds for $n$. Then
\begin{align}
a_{n+1}&=\sum_{k=1}^{n+1}\frac{1}{n+1+k}\\[6px]
{\scriptsize\text{(add and subtract, detach two terms)}\quad}
&=-\frac{1}{n+1}+\biggl(\,\sum_{k=0}^{n-1}\frac{1}{n+1+k}\biggr)+
\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(set $l=k+1$)}\quad}
&=-\frac{1}{n+1}+\biggl(\,\sum_{l=1}^{n}\frac{1}{n+l}\biggr)+
\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(induction hypothesis)}\quad}
&=-\frac{1}{n+1}+b_n+\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(rearrange)}\quad}
&=b_n+\frac{1}{2n+1}-\frac{1}{2n+2}\\[6px]
&=b_n+\frac{2n+2-2n-1}{(2n+1)(2n+2)}\\[6px]
&=b_n+\frac{1}{(2n+1)(2n+2)}\\[6px]
&=b_{n+1}
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculte indefinte integral of $\int \frac{dx}{\sqrt{(x+2)(3-x)}}$ I have to calculate $$\int \frac{dx}{\sqrt{(x+2)(3-x)}}$$.
I tried to use -
$\int u'v = uv - \int v'u$, but im pretty stuck.
Thanks.
|
\begin{align}
\int \frac{dx}{\sqrt{(x+2)(3-x)}}
=& \ \text{sgn} (\frac12-x)\int \frac{d\left(\sqrt{(x+2)(3-x)}\right)}{\sqrt{\frac{25}4- (x+2)(3-x)}}\\
=& \ \text{sgn} (\frac12-x) \sin^{-1}\bigg(\frac25 \sqrt{(x+2)(3-x)}\bigg)+C
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Linear algebra: Solving a system of equation matrix with a variable as coefficient. Let's consider this augmented matrix
$$\left(\begin{array}{ccc|c}
3 &-6 &6 &15\\
-2 &7 &a &-25\\
2 &-6 &6 & 20
\end{array}\right)$$
I'm trying to figure how to solve a matrix like this when there is $a$ as one of the coefficients.
|
Procede with the usual row reduction, trying to avoid pivoting on the element with $a$ as long as possible:
$$\left(\begin{array}{ccc|c}
3 &-6 &6 &15\\
-2 &7 &a &-25\\
2 &-6 &6 & 20
\end{array}\right)
\xrightarrow{\begin{matrix}R1~/~3\\R3~/~2\end{matrix}}
\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
-2 &7 &a &-25\\
1 &-3 &3 & 10
\end{array}\right)
\xrightarrow{\begin{matrix}R2~+~2R1\\R3~-~R1\end{matrix}}\\
\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
0 &3 &a+4 &-15\\
0 &-1 &1 & 5
\end{array}\right)
\xrightarrow{\begin{matrix}R_3~\times~-1\\R3~\leftrightarrow~R2\end{matrix}}
\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
0 &1 &-1 & -5\\
0 &3 &a+4 &-15
\end{array}\right)
\xrightarrow{\begin{matrix}R_3~-~3R2\end{matrix}}\\
\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
0 &1 &-1 & -5\\
0 &0 &a+7 &0
\end{array}\right)
$$
What happens next depends on whether $a+7=0$:
*
*Case $a=-7$: then the bottom row is $0~0~0~|~0$, the set of equations is consistent and has an infinite number of solutions.
*Case $a\ne-7$: then we can divide $R3$ by $a+7$ to get $$\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
0 &1 &-1 & -5\\
0 &0 &1 &\dfrac{1}{a+7}
\end{array}\right)
$$ and the set of equations has a unique solution.
|
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|
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $$\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$$
I am having a hard time relating the square roots in the inequality to the given condition. I was thinking that maybe there is some substitution I could do, but I couldn't think of any.
|
My attempt (not a full solution). The best I was able to achieve is this inequality:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}>\sqrt{2+\sqrt{3}}=1.93185\dots$$
We will use the mean inequalities:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
By squaring the first inequality and using $xy+xz+yz=1$ and $x,y,z>0$ we obtain:
$$x^2+y^2+z^2 \geq 1$$
By adding $2=2(xy+xz+yz)$ we get:
$$x+y+z \geq \sqrt{3}$$
Additionally from the GM-HM inequality and using $xy+xz+yz=1$ we obtain:
$$\sqrt[3]{xyz} \geq 3 xyz$$
$$xyz \leq \frac{1}{\sqrt{27}}=\frac{1}{3\sqrt{3}}$$
I don't know how to use it here though.
Now let's denote:
$$a=\sqrt{x}+\sqrt{y}+\sqrt{z}$$
$$a^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}) \geq \sqrt{3}+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$
$$(a^2-\sqrt{3})^2 \geq 4 (1+2(z \sqrt{xy}+x \sqrt{yz}+y \sqrt{zx}))$$
Notice that:
$$z \sqrt{xy}+x \sqrt{yz}+y \sqrt{zx}=a \sqrt{xyz}$$
Then we obtain a polynomial inequality:
$$(a^2-\sqrt{3})^2 \geq 4 (1+2a \sqrt{xyz})$$
We need a lower bound for $xyz$, but the best I could come up with:
$$xyz>0$$
Using this trivial bound we obtain:
$$(a^2-\sqrt{3})^2 > 4$$
The positive solution is:
$$a>\sqrt{2+\sqrt{3}}$$
I'll update if I manage to solve the original problem.
This may be useful (I got it by applying GM-HM inequality to three pairs of numbers):
$$\sqrt{xy}+\sqrt{yz}+\sqrt{zx} \geq 2\frac{1+(x+y+z)xyz}{x+y+z-xyz}$$
And also by AM-GM for $\sqrt{xy},\sqrt{yz},\sqrt{zx}$ we have:
$$\sqrt{xy}+\sqrt{yz}+\sqrt{zx} \geq 3 \sqrt[3]{xyz}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability using indicator function There are 10 pairs of different socks in a drawer. You take 5 pairs out of it. What is the expected value of matching socks?
Is it OK to use the indicator function as follows?
Let $X$ be the number of pairs that match. Let $I_i$ be the indicator function that takes $1$ if the the $i$ pair matches, and $0$ if it doesn't.
This allows to express $X$ as $X=I_1 + I_2 + ... + I_5$.
To calculate $I_i$, we use the probability of a random pair being a match, that is $I_i={10\over{20 \choose 2}}$.
Now we can use the linearity of the expectation to compute the result.
Edit: Fix probability
|
I am not familiar with the method of indicator functions, but we may check your solution using basic counting methods. We count cases for each specific number of matching pairs. We will use derangements in our solution, which is notated by $!n.$
The total number of ways to draw the socks is $\frac{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}{5!} = 945.$
Case 1: $0$ matching pairs
Notice that since there are $0$ matching pairs, we can neglect this case in our expected value calculation.
Case 2: $1$ matching pair
$P(1) = \frac{5 \cdot !4}{945} = \frac{45}{945} = \frac{1}{21}.$
This contributes $1 \cdot \frac{1}{21}$ to the expected value.
Case 3: $2$ matching pairs
$P(2) = \frac{\dbinom{5}{2} \cdot !3}{945} = \frac{20}{945} = \frac{4}{189}.$
This contributes $2 \cdot \frac{4}{189}$ to the expected value.
Case 4: $3$ matching pairs
$P(3) = \frac{\dbinom{5}{3} \cdot !2}{945} = \frac{10}{945} = \frac{2}{189}.$
This contributes $3 \cdot \frac{2}{189}$ to the expected value.
Case 5: $5$ matching pairs
$P(5) = \frac{!0}{945} = \frac{1}{189}.$
This contributes $5 \cdot \frac{1}{189}$ to the expected value.
Our expected value is $\boxed{\frac{8}{63}}.$
|
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|
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
|
Let $y=4\csc^2x+9\sin^2 x$
$$y'=-8\csc^2x\cot x+9\sin2 x$$
$$y''=-8\csc^2x(-\csc^2 x)-8\cot x(-2\csc^2 x\cot x)+18\cos2 x$$
$$y''=8\csc^4x+16\csc^2\cot^2 x+18\cos2 x$$
for maxima or minima, $y'=0$ hence, $$-8\csc^2x\cot x+9\sin2 x=0$$
$$2\cos x\left(9\sin x-\frac{4}{\sin^3 x}\right)=0$$
$$\frac{\cos x}{\sin^3 x}\left(9\sin^4 x-4\right)=0$$
$\cos x=0\implies x=\pi/2$
or $9\sin^4x-4=0\iff \sin^2 x=\frac{2}{3}\ \ \ \ \ (\forall \ \ \ \sin x\ne 0)$
One, can easily check that minimum of $y$ occurs for $\sin^2 x=\frac 23$ ($y''>0$),
hence, substituting $\sin^2 x=\frac{2}{3}$ in $y$, the minimum value is
$$y_{\text{min}}=\frac{4}{\sin^2x}+9\sin^2 x=\frac{4}{2/3}+9(2/3)=\color{red}{12}$$
|
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|
How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is $$\frac{\sqrt{2}\sqrt{3}}{9}$$ which isn't equal to $$\frac{2\sqrt{2}-\sqrt{3}}{5}$$
What did I do wrong?
|
\begin{align}
\frac {1}{2\sqrt 2 + \sqrt 3}= \frac {2\sqrt 2-\sqrt 3}{5}&\iff \\\frac {1}{2\sqrt 2 + \sqrt 3}\cdot {2\sqrt 2 + \sqrt 3} = \frac {2\sqrt 2-\sqrt 3}{5}\cdot {(2\sqrt 2 + \sqrt 3)} &\iff\\
1=\frac {(2\sqrt2)^2-(\sqrt 3)^2}{5}&\iff\\
1=\frac {4\cdot 2-3}{5}&\iff 1=\frac 5 5.
\end{align}
|
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|
Prove that if $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle.
Prove that if $a,b,c,$ and $d$ are positive numbers and satisfy $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle.
I find it hard to go from the given inequality to the triangle inequality for $3$ of $a,b,c,d$. Expanding it may help but that would get ugly. I did it and got $d^2 (2 a^2+2 b^2+2 c^2-2 d^2)+b^2 (2 a^2-2 b^2+2 c^2)+a^2 (2 c^2-2 a^2)-2 c^4$. Also doing a ravi substitution here seems almost impossible to do with $4$ variables.
|
We'll prove that $a$, $b$ and $c$ are sides-lengths of a triangle.
Indeed, by C-S
$$3(a^4+b^4+c^4+d^4)=(2+1)(a^4+b^4+c^4+d^4)\geq\left(\sqrt{2(a^4+b^4+c^4)}+d^2\right)^2$$
which gives $a^2+b^2+c^2>\sqrt{2(a^4+b^4+c^4)}$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$
and we are done because $a+b-c<0$ and $a+c-b<0$ impossible.
|
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|
What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$ Here's what I tried:
$$\sum_{n \ge0} {\frac{1}{3} \choose n} x^n= \sum_{n \ge0} = \frac{\frac{1}{3}!}{n!(n-\frac{1}{3})!}x^n=\sum_{n \ge0} \frac{(\frac{1}{3}-1)(\frac{1}{3}-2)\cdot ...\cdot(\frac{1}{3}-(n-1)) }{n!}x^n$$
What to do more, or is this all wrong?
|
Hint: Supplement to the already given answer
If a series
\begin{align*}
A(x)=\sum_{n\geq 0}a_nx^n
\end{align*}
is stated, the coefficient of $x^k$ is $a_k$ without any notion of $x$. A convenient notation for the coefficient of $x^k$ is $[x^k]$.
Since the coefficient of $x^4$ should be obtained, an answer could be stated as
\begin{align*}
[x^4]\sqrt[3]{1+x}&=[x^4]\sum_{n \ge0} {\frac{1}{3} \choose n}x^n\\
&=\binom{\frac{1}{3}}{4}\\
&=\frac{1}{4!}\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)\\
&=-\frac{10}{243}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using contour integrals to evaluate sum - Problem calculating residues
"Compute $$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}}$$ using contour integration"
I have used the function $F(z) = \frac {\pi cot\pi z}{z^2(z+1)^2}$
Which has double poles at $z=0$ and $z=-1$
For the pole at $z=0$, if I calculate the residue by taking the limit of $\frac{dF}{dz}$ as $z \to 0 $, I end up with a $cosec(0)$ term, which is $\infty$
Instead I can try to calculate the residue using the Laurent series, and about $z=0$ again I find;
$\frac {\pi cot(\pi z)}{z^2{(z+1)^2}} = \frac{\pi}{z^2} [(\pi z)^{-1} - \frac {1}{3}(\pi z) - \frac {1}{45}(\pi z)^3 ...][1 - 2z +3z^2 -...]$
And I find the residue to be $3-\frac{1}{3} \pi^2$
To compute the residue at $z=-1$, however, I can't compute the expansion of $\pi cot(\pi z)$ about $z=-1$ using normal Taylor expansion methods, because if:
$g(z) = \pi cot(\pi z)$, then $g(-1) = \frac{1}{0}$
and that's where I'm stuck - computing the residue at $z=-1$.
Once I've found the residue, computing the series is simple:
$$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}} = \frac{1}{2} \sum_{Res}$$
Any help would be greatly appreciated!
|
Consider
$$f(a) = \sum_{n=-\infty}^{\infty} \frac1{(n^2+a^2)[(n+1)^2+a^2]} $$
The sum we want will be $\frac12 \lim_{a \to 0} [f(a) - \frac{2}{a^2 (1+a^2)}]$. (These represent the $n=0$ and $n=-1$ terms in the sum that become singular as $a \to 0$.) We may evaluate this sum using the residue theorem. Recall that, from considering the integral
$$\lim_{N \to \infty} \oint_{C_N} dz \, \cot{\pi z} \, h(z) = 0$$
where $C_N$ is the square with vertices $(\pm (1 \pm i) N/2$. The integral is also equal to $i 2 \pi$ times the sum of the residues, which all must sum to zero. Thus
$$\sum_{n=-\infty}^{\infty} h(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k}[ \cot{\pi z} h(z) ] $$
where the $z_k$ are the non-integer poles of $h$. In this case, we see that
$$f(a) = -\pi \sum_k \operatorname*{Res}_{z=z_k} \frac{\cot{\pi z}}{(z^2+a^2)[(z+1)^2+a^2]}$$
In this case, we have these poles at $\pm i a$ and $-1\pm i a$. Thus,
$$\begin{align}f(a) &= -\pi \left [\frac{-i \coth{\pi a}}{i 2 a (1+i 2 a)} + \frac{i \coth{\pi a}}{-i 2 a (1-i 2 a)} + \frac{\cot{\pi (-1+i a)}}{i 2 a (1-i 2 a)} + \frac{\cot{\pi (-1-i a)}}{-i 2 a (1+i 2 a)} \right ] \\ &= \frac{\pi}{a} \left (\frac{\coth{\pi a}}{1-i 2 a} + \frac{\coth{\pi a}}{1+i 2 a} \right )\\ &= \frac{2 \pi \coth{\pi a}}{a (1+4 a^2)} \end{align}$$
The sum is
$$\begin{align}\sum_{n=1}^{\infty} \frac1{n^2 (n+1)^2} &= \frac12 \lim_{a \to 0} \left [ \frac{2 \pi \coth{\pi a}}{a (1+4 a^2)}- \frac{2}{a^2 (1+a^2)} \right ] \\ &= \frac12 \lim_{a \to 0} \left [\frac{2}{a^2} \left (1+\left (\frac{\pi^2}{3} - 4 \right ) a^2+\cdots \right ) - \frac{2}{a^2} (1-a^2+\cdots) \right ] \end{align} $$
The singular pieces cancel, and we finally have
$$\sum_{n=1}^{\infty} \frac1{n^2 (n+1)^2} = \frac{\pi^2}{3} - 3$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1622229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Looking for "an easy to understand" proof for following Power series I'm looking for proof for the following Power series
$exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
If X is $A_{nxn}$ matrix, then prove the series is converge
Given
$\exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
$\sin(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k}}{2k!}$
$\cos(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k+1}}{(2k+1)!}$
$\cosh(X) = \sum_{k=0}^{n} \frac{X^{2k}}{(2k)!}$
$\sinh(X) = \sum_{k=0}^{n} \frac{X^{2k+1}}{(2k+1)!}$
Here is what I try so far, and I'm looking for "easy to understand" proof
Let $ X =
\begin{bmatrix}
0 & -\beta \\
\beta & 0
\end{bmatrix}
$
$
\exp(X) =
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & -\beta \\
\beta & 0
\end{bmatrix}
+
\frac{1}{2!}
\begin{bmatrix}
-\beta^{2} & 0 \\
0 & \beta^{2}
\end{bmatrix}
+
\frac{1}{3!}
\begin{bmatrix}
0 & \beta^{3} \\
-\beta^{3} & 0
\end{bmatrix}
+
\frac{1}{4!}
\begin{bmatrix}
\beta^{4} & 0 \\
0 & \beta^{4}
\end{bmatrix}
+
\frac{1}{5!}
\begin{bmatrix}
0 & -\beta^{5}\\
\beta^{5} & 0
\end{bmatrix}
$
$
\exp(X) =
\begin{bmatrix}
1 + \frac{1}{2!}(-\beta^{2}) + \frac{1}{4!}\beta^{4} + ... & -\beta + \frac{1}{3!}\beta^{3} + ...\\
\beta + \frac{1}{3!}-\beta^{3} + ... & 1 + \frac{1}{2!}\beta^{2} + \frac{1}{4!}\beta^{4} + ...
\end{bmatrix}
=
\begin{bmatrix}
\sin(\beta) & -\cos(\beta)\\
\cos(\beta) & \sin(\beta)
\end{bmatrix}
$
|
Better to proceed as follows.
Consider for example the series $\sum_{k=0}^\infty\frac1{k!}t^kA^k$, say with $t$ complex or if you prefer take $t\in\mathbb R$. You can verify that the all entries of the series are power series.
So, all that matters for convergence is the radius of convergence. It is easy to show that the radius of convergence of the series of numbers $\sum_{k=0}^\infty\frac1{k!}t^k\|A^k\|$ is infinity, and so the same goes for each entry of the original series.
Conclusion: each entry of $\sum_{k=0}^\infty\frac1{k!}t^kA^k$ converges, for all values of $t$, and you can take say $t=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to compute the integral of $ \frac{\sqrt{(x^2+1)}}{x^2}$? II have spent hours trying to find the right substitution but I have had no results.
I have tried using $x=\tan u$ but it did not give the result shown in my book.
|
$$\int \frac{\sqrt{x^2+1}}{x^2}dx$$
Apply Integration By Parts:$\:\int \:uv'=uv-\int \:u'v$
$u=\sqrt{x^2+1},\:u'=\frac{x}{\sqrt{x^2+1}},\:\:v'=\frac{1}{x^2},\:\:v=-\frac{1}{x}$
Then
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\sqrt{x^2+1}}dx$$
$$=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\sqrt{x^2+1}}dx$$
Then Note:
$\int \:-\frac{1}{\sqrt{x^2+1}}dx=-arcsinh \left(x\right)$
So
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=\color{red}{arcsinh \left(x\right)-\frac{\sqrt{x^2+1}}{x}+C}$$
|
{
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"url": "https://math.stackexchange.com/questions/1623501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find the maximum value of $(12\sin x-9\sin^{2} x)$
The maximum value of $(12\sin x-9\sin^{2} x)$
is equal to
$a.)\ 3 \\
\color{green}{b.)\ 4} \\
c.)\ 5 \\
d.)\ \text{none of these}$
As
$-1\leq \sin x\leq 1 ,\\
12\sin x-9\sin^{2} x \\
=12-9=3 \\
$
But the answer given is $4.$
I am looking for a short and simple way.
I have studied maths up to $12$th grade.
Note : I can't use calculus.
|
Hint: If you complete the square, you will get:
$$12\sin x - 9\sin^2 x$$
$$= 4-4 + 12\sin x - 9\sin^2 x $$
$$= 4-2^2 + 2 \cdot 2 \cdot 3\sin x - (3\sin x)^2 $$
$$= 4-(3\sin x - 2)^2.$$
Now, it is easy to determine the maximum value of this expression.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post linked here.
The second part is where I am confused because there is a 'hence'
so I thought of taking 2 approaches:
either $$ z^6 + \frac 1{z^6} $$
or $$ z^6 $$ and equating real parts.
I will start with my first approach
$$ z^6 + \frac 1{z^6} $$
$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$
$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$
$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$
$$ 2\cos(6x) $$
Which is no where near what I am suppose to prove..
So with my second approach (expanding and equation real parts)
$$ z^6 $$
$$ (\cos(x) + i \sin(x))^6 $$
Using pascals
$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$
Simplifying
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$
Now considering only real
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$
At this point I'm confused , am I on the right approach?
|
$n=1\implies2\cos x=z+\dfrac1z$
$$\cos^6x=\left(\dfrac{z+\dfrac1z}2\right)^6$$
$$64\cos^6x=\left(z+\dfrac1z\right)^6=z^6+\dfrac1{z^6}+\binom61\left(z^4+\dfrac1{z^4}\right)+\binom62\left(z^2+\dfrac1{z^2}\right)+\binom63$$
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$ I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have
$\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right){/}\left((\sqrt{2}+\sqrt{4})(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})\right)=$
$(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})+(\sqrt{2}+\sqrt{4})...(\sqrt{2n}+\sqrt{2n+2})+...+(\sqrt{2}+\sqrt{4})...(\sqrt{2n-2}+\sqrt{2n})$
How to reduce this sum to general form?
|
Hint rationlize you will get a telescoping series for eg after rationalizing first two terms we get $\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$ on rationalizing till $n$ see which terms remain then you will get your steps
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all functions $F(x)$ for which $F (x) + F ((x − 1)/x) = 1 + x$
Let $F (x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $F (x) + F ((x − 1)/x) = 1 + x$. Find $F (x)$.
This functional equation looks like I could do an inverse substitution. Meaning, let $x = \dfrac{x-1}{x}$ then we have $F \left(\dfrac{x-1}{x} \right)+F\left( \dfrac{1}{x-1}\right) = \dfrac{2x-1}{x}$. Thus, $1+x-F(x) = \dfrac{2x-1}{x}-F\left( \dfrac{1}{x-1}\right)$. I am not sure how to proc
|
Solve the system of questions:
$$
\begin{align}
F(x) + F\left(\frac{x-1}{x}\right) &= 1+x \\
F\left(\frac{x-1}{x}\right) + F\left(\frac{1}{1-x}\right) &= \frac{2x-1}{x} \\
F\left(\frac{1}{1-x}\right) + F(x) &= \frac{2-x}{1-x}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1642298",
"timestamp": "2023-03-29T00:00:00",
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|
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$ = $\begin{bmatrix}
a^2 + c^2 & ab + cd \\
ab + cd & b^2 + d^2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle .
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
|
from $A^TA = I$, we konw $(a, c)$ and $(b, d)$ is orthogonal, for some $\theta$, we know $(a, c) = \sqrt{(a^2 + c^2)}(\cos\theta, \sin\theta) = (\cos\theta, \sin\theta)$, the unit vector orthogonal to $(a, c)$ is $(-\sin\theta, \cos\theta)$, so we know, for some scalar $k$, $(b, d) = k(-\sin\theta, \cos\theta)$. $k$ is either $1$ or $-1$. since $\det(A) = 1$,
here we used $\sqrt{a^2 + c^2} = 1$, and $\sqrt{b^2 + d^2} = 1$
Let us calculate
$$\det(A) = \det \begin{bmatrix} \cos\theta & \sin\theta \\ -k\sin\theta & k\cos\theta \end{bmatrix} = k(\cos^2\theta + \sin^2\theta) = k = 1$$
so
$$ A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find parallel line value I've an homework problem that i'm unable to find the right answer.
The problem is:
The line $tx + sy = 2$ goes through point $(2,1)$ and is parallel to line $y = 8 -3x$, find the value of $t^2 + s^2$.
$ A. {32\over49}$ $B.{18\over49}$ $C.{36\over49}$ $D.{25\over49} $ $E.{40\over49} $
I was able to find a parallel line at $ y = -3x + 7 $ but i'm unable to find any of the possible answers to be right.
|
Rewrite $y = -3x + 7$ in the form $tx + sy = 2$: $$\begin{align*}y & = -3x + 7 \\ 3x + y & = 7 \\ \frac{6}{7} x + \frac{2}{7} y & = 2.\end{align*}$$ By comparison of the form of $\frac{6}{7} x + \frac{2}{7} y = 2$ to that of $tx + sy = 2$, we find $t = \frac{6}{7}$ and $s = \frac{2}{7}$. Therefore, $$t^2 + s^2 = {\left(\frac{6}{7}\right)}^2 + {\left(\frac{2}{7}\right)}^2 = \frac{40}{49}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.
|
Taking
$a_k = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}}$
we have
$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}) +\dots = 1$
As per Mario Carneiro's suggestion in the comments, let us instead take
$a_k = \frac{1}{\sqrt{p_{k-1}}} - \frac{1}{\sqrt{p_{k}}}$,
where
$p_0 = 1$ and $p_k$ for $k > 0$ is the k-th prime number ($p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$ etc.), so
$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}) +(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}) +\dots = 1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculation of area in 2 definite integrals given function $y=x^2$ Here is a graph for $y=x^2$
Given that the area in blue is equal to the area in pink, find a in terms of b and solve for a.
My attempt:
From the graph I can see that :$a^2=b$ and $a=\sqrt b$
Since area blue is equal to area in pink:
$$\int^a_1 x^2 dx=\int^b_1 y^\frac{1}{2}dy$$
$$\left[\frac{1}{3}x^3\right]^a_1=\left[\frac{2}{3}y^\frac{3}{2}\right]^b_1$$
$$\frac{1}{3}a^3-\frac{1}{3}=\frac{2}{3}b^\frac{3}{2}-\frac{2}{3}$$
$$a^3-1=2b^\frac{3}{2}-2$$
$$a^3-1=2a^3-2$$
$$a^3=1$$
$$a=1$$
The answer to $a=1$ makes sense because when $a=1$, the area becomes $0$ $units^2$
But my teacher said the answer is $a=1+\sqrt 3$
Is there any idea what has gone wrong in my calculations?
Also please tell me how to get to the answer as well?
|
Note$$\bigg(\int_{1}^{1+\sqrt{3}} x^2 dx\bigg) -\bigg(\int_{1}^{(1+\sqrt{3})^2} \sqrt{y} dy \bigg) = -(3+2\sqrt{3}) \neq 0$$
So your teacher is wrong in this case, and you are right.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
|
\begin{align}
\sin(x)+\cos(x) &= \frac 75 \\
\left(\sin(x)+\cos(x)\right)^2 &= \left( \frac 75 \right)^2 \\
1 + 2 \sin(x) \cos(x) &= \frac{49}{25} \\
\sin(x) \cos(x) &= \frac{12}{25}
\end{align}
\begin{align}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}
&= \frac{\sin(x) + \cos(x)}{\sin(x) \cos(x)}\\
&= \frac{\left(\frac{7}{5}\right)}{\left(\frac{12}{25}\right)}\\
&= \frac{7 \times 25}{5\times 12}\\
&= \frac{35}{12}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integer it has to be $k=101$ cause $gcd(k,k-101)=1\lor101$ and both $k,k-101$ can't be both squares of an integer,so $k=101t$ and $t(t-1)$ is never an square except for $t=1,0$ and $t=0$ is not possible hence $k=101$ is only possible integer solution
EDIT: So if $\gcd(k,k-101)=1$ then $k=h^2,k-101=(h-s)^2$ then $h(2h-s)=101$ which can be $s=1,h=51$ or $s=101,h=51$.$y=51^2+51\cdot50=51\cdot 101,x=51$
And since $x=2k-y$ is integer then $k=\frac{h}{2}$,if $h=2q$ then $k$ is integer otherwise if $h=2q+1$ then $$\frac{2q+1}{2}+\frac{1}{2}\sqrt{4q^2-400q-201}=\frac{2q+1}{2}+\frac{1}{2}\sqrt{(2q-10)^2-301}\\(2q-10)^2-301=r^2\\2q-10=z\\z^2-301=(z-c)^2\\c(2z-c)=301,c=1,z=151,c=7,z=25,c=43,z=25$$
The $z=151$ is impossible cause $2q-10$ is even,and $z=25$ is also impossible because $2q-10$ is even.Hence the only solutions are $(x,y)=(101,101),(51,5151),(5151,51)$ The last one clearly from symmetry
|
$1/x + 1/y = 2/101$
$(x+y)/xy = 2/101$
So $101|xy$. But 101 is prime so 101|x or 101|y or both.
Wolog symmetry assume 101|x.
Let $x = 101x'$.
$(101x' + y)/101x'y = 2/101$
$(101x' + y)/x'y = 2$
So $x'|y$. Let y = x'y'.
$(101 + y')/x'y' = 2$
So $y'|101$.
So $y' = 1$ of $y' = 101$.
So $102/x' = 2$ and $x'=51; y'=1$ or $202/x'101 = 2$ and $x'=1; y' = 101$
So $x=5151;y=51$ or $x=101;y=101$. And removing wolog symmetry; or $x = 51; y= 5151$
|
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"timestamp": "2023-03-29T00:00:00",
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|
To prove $\sum_{n=0}^\infty \binom{r}{x}\binom{N-r}{n-x}=\binom{N}{n}.$ To prove $$\sum_{x=0}^n \binom{r}{x}\cdot \binom{N-r}{n-x}=\binom{N}{n}.$$
I tried comparing the coefficients of
$(1+x)^{(n+k)} = (1+x)^n(1+x)^k$
but couldn't reach the answer.
|
We obtain for $0\leq r\leq N$
\begin{align*}
(1+x)^{N}&=(1+x)^r(1+x)^{N-r}\\
&=\sum_{k=0}^r\binom{r}{k}x^k\sum_{l=0}^{N-r}\binom{N-r}{l}x^l\\
&=\sum_{n=0}^{N}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{r}{k}\binom{N-r}{l}\right)x^n\\
&=\sum_{n=0}^{N}\left(\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}\right)x^n\\
\end{align*}
Since
\begin{align*}
(1+x)^{N}=\sum_{n=0}^{N}\binom{N}{n}x^n
\end{align*}
we compare coefficients and conclude
\begin{align*}
\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}=\binom{N}{n}
\end{align*}
|
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|
Calculus, Finding Integral of a quotient Seems I have forgotten some basic integrating rules, but how do I go about finding the primitive function of $f(R) = \frac {R^2}{B+R^2}$?
|
You have to consider three cases: $B=0$, $B<0$ and $B>0$.
If $B=0$, then for $R\ne 0$, the function becomes $f(R)=1$, and anti-derivative is given by
$$
F(R)=R+C,
$$
where $C$ is a constant.
If $B\ne 0$, you have to write
$$
f(R)=\frac{R^2}{B+R^2}=\frac{B+R^2-B}{B+R^2}=1-\frac{B}{B+R^2}.
$$
If $B<0$, then $B=-A^2$, and
$$
f(R)=1+\frac{A^2}{R^2-A^2}=1-\frac{A^2}{2A}\left(\frac{1}{R-A}-\frac{1}{R+A}\right)=1-\frac{A}{2}\left(\frac{1}{R-A}-\frac{1}{R+A}\right).
$$
An anti-derivative of $f$ is then given by
$$
F(R)=R-\frac{A}{2}\ln\left|\frac{R-A}{R+A}\right|+C=R+\sqrt{-B}\ln\sqrt{\left|\frac{R+\sqrt{-B}}{R-\sqrt{-B}}\right|}+C.
$$
If $B>0$, then an anti-derivative of $f$ is given by
$$
F(R)=R-\sqrt{B}\arctan\left(\frac{R}{\sqrt{B}}\right)+C
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
If $x,y,z$ are positive real number and $x+y+z=1\,$ Then prove that
$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
Let $$f(x,y,z)=xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$
Then $$\frac{f(x,y,z)}{xyz} = \frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}$$
Now Using $\bf{Cauchy-Schwarz\; }$ Inequality, We get$$\frac{f(x,y,z)}{xyz}\geq \frac{4(x+y+z)^2}{x+y+z} = 4\Rightarrow f(x,y,z)\geq 4xyz$$
My Question is How can we solve it Using $\bf{A.M\geq G.M}$ nequality,
plz explain me, Thanks
|
First I edited your proof, you missed the factor $4$ in the answer. To use the AM-GM you can still use what you have there in the proof.
$$\dfrac{f(x,y,z)}{xyz} = \sum_{\text{cyclic}} \dfrac{(1-x)^2}{x}=\sum_{\text{cyclic}} \dfrac{1-2x+x^2}{x} = \sum_{\text{cyclic}} \dfrac{1}{x} - 6 + 1\geq \dfrac{9}{x+y+z} -6+1 = 4$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don't seem to follow. I can understand that there would be some good simplification and the product will hopefully telescope but I'm lacking the right algebra. I also thought of finding a recurrence solution probably from the corresponding DE but that didn't follow as well.
|
Squaring the infinite product we observe that
$$
2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots\\ =
2\cdot\frac{2}{2+\sqrt{2}}\cdot\frac{2+\sqrt{2}}{2+\sqrt{2-\sqrt{2}}}
\cdot\frac{2+\sqrt{2-\sqrt{2}}}{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots \\
=\frac{4}{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}
$$
But
$$
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}= 1.
$$
See: Convergence of $a_{n+1}=\sqrt{2-a_n}$. Thus
$$
2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots = \frac{4}{3}
$$
Finally
$$
\sqrt{2}\sqrt{2-\sqrt{2}}\sqrt{2-\sqrt{2-\sqrt{2}}}\sqrt{(2-\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots = \sqrt{\frac{4}{3}}
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
|
We have
$$
(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab + bc + ca) - 3abc
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
integrate $\int \frac{dx}{1+cos^2x}$
$$\int \frac{dx}{1+\cos^2x}$$
I used $\cos x=\frac{1-v^2}{1+v^2}$ and $dx=\frac{2dv}{1+v^2}$
and got $$2\int \frac{dv}{v^4-v^2+1}=2\int \frac{dv}{(v^2-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4}{\sqrt{3}}arctan(\frac{2v^2-1}{\sqrt{3}})+c$$
How to continue?
|
An alternative method:
Multiplying through by $sec^2(x)$:
$$\int \frac{dx}{1+\cos^2x} = \int \frac{sec^2x}{1+sec^2x}dx$$
Defining $u = tanx$, $du = sec^2x$ $dx$, and using the identity $1+tan^2x = sec^2x$:
$$\int \frac{sec^2x}{1+sec^2x}dx = \int \frac{1}{u^2+2}du$$
The integral is now of the form of a very well-known $\arctan$ integral
|
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|
Random vector with density on triangle and trapezoid I have a random vector $(X,Y)$ with the density $f_{XY}(x,y)=\tfrac{1}{5}$ on the trapezoid $T_1$ with vertex $(0,0),(0,1),(2,1),(3,0)$ and $f_{XY}(x,y)=\tfrac{3}{2}x$ on the triangle $T_2$ with vertex $(0,4),(1,4),(1,3)$.
I need to find:
*
*support and probability function for marginal $X$;
*support and probability function for $(Y|X=x), 0\le{x}\le1$;
*support and distribution for $Z=X+Y$
I started with support for $T_1$ and $T_2$:
$T_1$ ${(x,y):0\le{x}\le{-y+3}, 0\le{y}\le{1}}$
$T_2$ ${(x,y):0\le{x}\le{1}, 3\le{y}\le{-x+4}}$
Now I need to check that
$\iint_{T_1}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y + \iint_{T_2}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y = 1$, so:
$\frac{1}{5}\int_0^{1}(\int_0^{-y+3}{}\mathrm{d}x)\mathrm{d}y + \frac{3}{2}\int_0^1(\int_3^{-x+4}\mathrm{d}y)x\mathrm{d}x = 1$
For $T_1$ I got $\frac{1}{2}$ and for $T_2$ $\frac{1}{4}$, but the sum is not 1. Why?
I need to find the error before proceed with other questions :-)
First edit and first attempt:
*
*support and probability function for marginal $X$:
$f_x(x)\begin{cases}\frac{3}{2}x^2+\frac{1}{5} & 0\leq{x}\leq{1}\\ \frac{1}{5} & 1\leq{x}\leq{2}\\ \frac{-x+3}{5} & 2\leq{x}\leq{3} \end{cases}$
*
*distribution function for marginal $X$:
$F_x(x)\begin{cases}\frac{x^3}{2}+\frac{x}{5} & 0\leq{x}\leq{1}\\ \frac{7}{10} + \frac{x}{5} & 1\leq{x}\leq{2}\\ \frac{7}{10} + \frac{1}{5} -\frac{x^2}{10} + 3x & 2\leq{x}\leq{3} \end{cases}$
|
*
*I think your marginal pdf for $X$ is right.
*I think your cdf for it is wrong, although the question doesn't require it. It should be:
$$ F_X(x) =
\begin{cases}
\int_{0}^{x} (3u^2/2+ 1/5)\;du &= \dfrac{x^3}{2} + \dfrac{x}{5}, & \text{if $0\leq x\leq 1$} \\
\\
F(1) + \int_{1}^{x} (1/5)\;du &= \dfrac{x}{5} + \dfrac{1}{2}, & \text{if $1\leq x\leq 2$} \\
\\
F(2) + \int_{2}^{x} (\frac{-u+3}{5})\;du &= \dfrac{-x^2}{10} + \dfrac{3x}{5} + \dfrac{1}{10}, & \text{if $2\leq x\leq 3.$} \\
\end{cases}$$
$$\\$$
*Conditional pdf of $Y$ given $X$ with $0\leq X\leq 1$:
In this range for $X$ we have two ranges for $Y$ to handle.
For $0\leq y\leq 1$:
\begin{align}
f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{1/5}{1/5 + 3x^2/2} = \dfrac{2}{2 + 15x^2}.
\end{align}
For $4-x\leq y\leq 4$:
\begin{align}
f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{3x/2}{1/5 + 3x^2/2} = \dfrac{15x}{2 + 15x^2}.
\end{align}
$$\\$$
*Support and pdf of $Z=X+Y$:
Assuming you've already drawn the regions above on the $xy-$plane, now imagine a line $x+y=c$ moving across. Values of $c$ are in the support of $Z$ if the line crosses a region of support for $(X,Y)$. So the support for $Z$ is $0\leq Z\leq 3$ and $4\leq Z\leq 5$.
It also shows the cases we should consider:
For $0\leq z\leq 1$:
$$P(Z=z) = \int_{x=0}^{z} P(X=x\cap Y=z-x)\;dx = \int_{0}^{z}\dfrac{1}{5}\;dx = \dfrac{z}{5}.$$
For $1\leq z\leq 3$:
$$P(Z=z) = \int_{x=z-1}^{z} P(X=x\cap Y=z-x)\;dx = \int_{z-1}^{z}\dfrac{1}{5}\;dx = \dfrac{1}{5}.$$
For $4\leq z\leq 5$:
$$P(Z=z) = \int_{x=z-4}^{1} P(X=x\cap Y=z-x)\;dx = \int_{z-4}^{1}\dfrac{3x}{2}\;dx = \dfrac{3}{4}\left(-z^2+8z-15\right).$$
|
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|
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
|
You have made a few mistakes in your calculation.
First of all, since $v=\tan\frac{x}{2}$ with $\tan\frac{\frac{\pi}{4}}{2}=\tan\frac{\pi}{8}=\sqrt 2-1$, you should have
$$\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}=\int_{0}^{\color{red}{\sqrt 2-1}}\frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}}$$
Also, you should have
$$\begin{align}&\frac{1-v^2}{(1+v^2)(-v^2+v+1)}\\&=\color{red}{+}\frac 15\cdot\frac{-2v+4}{v^2+1}+\frac{1}{5}\cdot\frac{-2v+1}{-v^2+v+1}\\&=-\frac 15\cdot\frac{2v}{v^2+1}+\frac 45\cdot\frac{1}{v^2+1}+\frac{1}{5}\cdot\frac{-2v+1}{-v^2+v+1}\end{align}$$
Now these give
$$\begin{align}&\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}\\&=\left[-\frac 15\ln(v^2+1)+\frac 45\arctan v+\frac 15\ln(-v^2+v+1)\right]_{0}^{\sqrt 2-1}\\&=-\frac 15\ln(4-2\sqrt 2)+\frac 45\arctan (\sqrt 2-1)+\frac 15\ln(3\sqrt 2-3)\\&=\frac 15\ln\left(\frac{3\sqrt 2-3}{4-2\sqrt 2}\right)+\frac 45\cdot\frac{\pi}{8}\\&=\frac 15\ln\left(\frac{3\sqrt 2}{4}\right)+\frac{\pi}{10}\approx 0.32594\end{align}$$
|
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|
Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
|
Let $x=\sinh t$, then $dx=\cosh t \, dt$
\begin{align*}
\int \frac{\sqrt{1+x^{2}}}{x} \,dx &=
\int \frac{\cosh t}{\sinh t} \cosh t \, dt \\ &=
\int \frac{1+\sinh^{2} t}{\sinh t} \, dt \\ &=
\int (\sinh t+\operatorname{csch} t) \, dt \\ &=
\cosh t+\int \frac{dt}{2\sinh \frac{t}{2} \cosh \frac{t}{2}} \\ &=
\cosh t+\int \frac{\operatorname{sech}^{2} \frac{t}{2} \, dt}
{2\tanh \frac{t}{2}} \\ &=
\cosh t+\int \frac{d(\tanh \frac{t}{2})}
{\tanh \frac{t}{2}} \\ &=
\cosh t+\ln \left| \tanh \frac{t}{2} \right|+C \\ &=
\cosh t+\ln \left| \frac{\cosh x-1}{\sinh x} \right|+C \\ &=
\sqrt{1+x^{2}}+\ln \left| \frac{\sqrt{1+x^{2}}-1}{x} \right|+C
\end{align*}
|
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|
Show that there exist no $a, b, c \in \mathbb Z^+$ such that $a^3 + 2b^3 = 4c^3$
Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.
Proof: There don't exist any integer solutions for the give equation.
Proof by the Well Ordering Principle.
Let $d$ be the set of all such possible combinations in form $(a, b, c)$.
By the Well Ordering Principle, we that every nonempty set of non-negative integers always has a smallest element. So, there must be a tuple $(a, b, c)$ such that $\max(a, b, c)$ is the least possible value in $d$.
We also know that $a$, $b$ and $c$ must be even. Since $a$ must be even if $4c^3$ has to be even but we know that $4c^3$ is even, so $a$ is the same. Similarly it can be shown that all $a$, $b$ and $c$ are even.
Now, let $a = 2x$, $b = 2y$ and $c = 2z$.
But then,
$$a^3 + 2b^3 = 4c^3$$
$$\implies (2x)^3 + 2*(2y)^3 = 4*(2z)^3$$
$$\implies 8x^3 + 16y^3 = 32z^3$$
$$\implies x^3 + 2y^3 = 4z^3$$
So, $(x, y, z) \in d$. But, $\max(x, y, z) < \max(a, b, c)$. But $(a, b, c)$ was supposed to be the smallest such member. We have a contradiction here. There is no smallest $(a, b, c)$ which satisfy this criterion and hence no $(a, b, c)$ at all and $d$ is empty. Hence, there exist no integers $a$, $b$ and $c$ which satisfy $a^3 + 2b^3 = 4c^3$
I wrote this proof. Is it enough to prove the above question? Am I right? Are there any mistakes in the above proof? If there are any, please help me.
Do you have any other ways of proving it?
|
There is another proof using the well ordering principle.
We use the following fact:
Third powers are can only be congruent to $0 $, $1 $ or $8 \mod 9 $.
Now, we see that the left hand side is congruent to $0$, $1$, $2$, $3$, $6$, $7$ or $8 \mod 9$, while the right hand side is congruent to $0, 4$ or $5 \mod 9$. So the only possibility is that $4c^3 \equiv 0 \mod 9$ and $a^3+2b^3 \equiv 0 \mod 9$. The first modular congruence gives that $3 \mid c$. The second can only hold true if $3 \mid a$ and $3 \mid b$.
If one of those isn't true, then we get either a combination of $0 \mod 9$ and $2$ or $7 \mod 9$, a combination of $0 \mod 9$ and $1$ or $8 \mod 9$, a combination of $1$ or $8 \mod 9$ and $2$ or $7 \mod 9$, none of which add to $0 \mod 9$.
Now suppose that $(a,b,c)$ is the solution with the smallest $a$.
We know that $3 \mid c$, $3 \mid a$ and $3 \mid b$.
Hence write $a=3x, b=3y, c=3z$ and then is $(x,y,z)$ also a solution with smaller $a$.
|
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|
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
|
For the sake of being concise, I have chosen to omit most of the intermediary algebraic simplifications. Letting $u = x+2$, we get
$$\lim_{u \to 4} \frac{{u^{\frac 52} - 4u^{\frac 32} + 4\sqrt{u} - 8}}{4 - u^2 + 4u - 4}$$
By L'hopital, this is equal to
$$\lim_{u \to 4} \frac{\frac {5u^2 - 12u+4}{2 \sqrt{u}}}{4 - 2u} = \lim_{u \to 4} \frac{2 - 5u}{4\sqrt{u}} = -\frac{9}{4}$$
|
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|
Find all possible values of rank(A) as 'a' varies For an assignment I have to solve the following problem:
Find all possible values of rank(A) as 'a' varies:
\begin{bmatrix}a&2&-1\\3&3&-2\\-2&-1&a\end{bmatrix}
I came across several answers on google and stack (Find all possible values of rank(A) as a varies?, and similar question, different matrix: Finding all possible ranks of matrix $A$ as $a$ varies?) but I need to solve the problem using row reduction instead of calculating the determinant. How do I go about?
I'm stuck here:
swapping rows gives:
\begin{bmatrix}-2&-1&a\\3&3&-2\\a&2&-1\end{bmatrix}
reducing rows 2 and 3 gives:
\begin{bmatrix}-2&-1&a\\0&\frac{3}{2}&-2+\frac{3}{2}a\\\\0&2-\frac{1}{2}a&-1+\frac{1}{2}a^2\end{bmatrix}
How to proceed from here?
according to Find all possible values of rank(A) as a varies? the answer should be: rank = 2 for a = 1 or a = 5/3 for all other values rank = 3.
|
Your first step is correct. Now multiply the second row by $\frac{2}{3}(\frac{1}{2}a-2)$ and sum to the third line, so you find an upper triangular matrix:
$$
\begin {bmatrix}
-2&-1&a\\
0&\frac{3}{2}&-2+\frac{3}{2}a\\
0&0&a^2-\frac{8}{3}a+\frac{5}{3}
\end{bmatrix}
$$
If $a^2-\frac{8}{3}a+\frac{5}{3} \ne 0$ the matrix is full rank, and this is done if $a$ is different from 1, and $\frac{5}{3}$ (the roots of $a^2-\frac{8}{3}a+\frac{5}{3} = 0$ ).
Substituting these values you see that the other two rows are linearly independent so the rank of the matrix is $2$ for $a=1$ or $a=\frac{5}{3}$.
|
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|
Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) = \frac{b^{n + 1} - a^{n + 1}}{(b-a)}$
What can I do from here?
|
If you look at $b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n$, you have $n+1$ terms. We have
$$
a^n + a^n +\cdots + a^n < b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n < b^n + b^n + \cdots + b^n
$$
because $a^n < a^kb^l < b^n$ if $k+l = n$ (and $k, l$ are both positive). Use the definition of multiplication to rewrite $a^n + \cdots +a^n = (n+1)a^n$ and $b^n + \cdots +b^n = (n+1)b^n$, and you're done.
|
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|
Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
|
A more rigoures approach.
$$ \int \frac{x^2}{x^3+1} dx $$
$$ u = x^3 +1 $$
$$ \frac{du}{dx} = 3x^2 $$
so:
$$ \frac{1}{3} \int \frac{3x^2}{x^3+1} dx = \frac{1}{3} Ln |x^3 +1| + K$$
|
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|
finding the maxima and minima for $f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ the actual question is to show that the function$f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ for all x belongs to $R$ has a minimum at $x=\left(\frac{\pi}{6}\right)$ and maxima at $x=\left(\frac{\pi}{2}\right)$ and at $x=\left(\frac{-\pi}{6}\right)$.
Actually the $f'(x)$ is coming to be $\sin\left(x-\frac{\pi}{3}\right)\cos\left(x+\frac{\pi}{3}\right) + \cos\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)= \sin(2x)$ $\Rightarrow$x=$\frac{\pi}{2}$.
I am not getting how to find the other 2 points $x=\frac{-\pi}{6}$ and $x=\frac{\pi}{6}$.
there is only- ONE critical point which is $\frac{\pi}{2}$
|
Expanding first the original formula you can see better what is happening:
$$
f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right) =
\frac14-\cos^2 x.
$$
|
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|
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$
$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$
$=6×n(n+1)(2n+1)/6+2×n(n+1)/2$
$=n(n+1)(2n+1+1)$
$=2n^3+2n^2+2n^2+2n$
$=2n(n^2+n+n+1)$
$=2n(n^2+2n+1)$
$a_n=2n(n+1)^2$
for $n=99, a_{99}=2×99×(99+1)^2=198×10^4$
I'm looking for short trick or alternative way, can you explain please?
|
Let $b_{n}=\frac{a_{n}}{2}$.
Then $b_{n}=3n^2+n+b_{n-1}$.
For any $n$th degree polynomial $Q(x)$, one can find a $n+1$th degree polynomial $P(x)$ such that $P(x)-P(x-1)=Q(x)$.
Then, use this to note that $b_{n}-n(n+1)^2=b_{n-1}-(n-1)n^2$, since $n(n+1)^2-(n-1)n^2=3n^2+n$.
Thus $b_{n}-n(n+1)^2$ is a constant. But since $b_{1}=0$, then this implies that $a_{n}=2n(n+1)^2$.
Thus $a_{99}=198 \times 10^4$.
|
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|
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make sense:
Let $n = 1.$ Then $1 + a + a^2 + \ldots +1 < \frac {1 + a + a^2 + \ldots +1 + a}{2} = \frac{2 + 2a + a^2+ \ldots}{2} = 1 + a + \frac{a^2}{2} + \ldots$
Then I did this below, but it's unclear if the difference is positive:
$\frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1} - \frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} = \frac {n + na +na^2+ \ldots + na^{n - 1} + na^n - n - 1 - an - a- na^2 - a^2 - \ldots -na^{n - 1} - a^{n - 1}}{n(n + 1)} = \frac{n(1 + a + a^2+ \ldots + a^{n - 1} + a^n - 1 -a - a^2 -a^{n - 1}) - 1 - a - a^2 - \ldots - a^{n - 1}}{n(n+1)} $
$ = \frac{- 1 + a(n(a^{n - 1}) - 1 - a - \ldots - a^{n - 2})}{n(n+1)}$
What can I try now?
|
Another solution:
Your inequality is equivalent to $\frac{a^n-1}{n}<\frac{a^{n+1}-1}{n+1}$. But this is $\int_1^a x^{n-1}dx< \int_1^a x^n dx$, which is true because $x^{n-1} < x^n, \forall x \in (1, a]$.
|
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|
$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$$
I tried to use
$$\lim_{x \to 2} \frac{(x^2)^n-4^n}{x^2-3x+2}$$ but i can't find anything special
|
Using $a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots + b^{n-1})$, we get
$$
x^{2n}-4^n = x^{2n}-2^{2n}=(x-2)(x^{2n-1}+\cdots + 2^{2n-1}).
$$
Then divide numerator and denominator through $x-2$, then
\begin{align}
\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}&=\lim_{x\to 2}\frac{(x-2)(x^{2n-1}+\cdots + 2^{2n-1})}{x^2-3x+2}\\
&=\lim_{x\to 2}\frac{x^{2n-1}+\cdots+ 2^{2n-1}}{x-1}\\
&=\frac{2^{2n-1}+2\cdot 2^{2n-2}+\cdots+2^{2n-1}}{1}\\
&=n\cdot 2^{2n}.
\end{align}
|
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|
Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right or is there a simple trick?
|
Here is a proof by induction,
*
*setting $n=1$, one should get
$$n^3+11n=1^3+11\cdot 1=12$$
obviously, the above number $12$ is divisible by $6$ hence statement is true for $n=1$.
*Assume that the number $n^3+11n$ is divisible by $6$ for $n=k$ then
$$k^3+11k=6\lambda$$or $$k^3=6\lambda-11k\tag 1$$
*Now, setting $n=k+1$, $$(k+1)^3+11(k+1)$$$$=k^3+3k^2+14k+12$$
$$=6\lambda-11k+3k^2+14k+12$$
$$=6\lambda+3(k^2+k+4)$$
since, $k^2+k+4$ is even for all integer $k$ hence setting $k^2+k+4=2m$,
$$=6\lambda+3(2m)$$
$$=6(\lambda+m)$$
since, $(\lambda +m)$ is an integer hence, the above number $6(\lambda+m)$ is divisible by $6$
Hence, $n^3+11n$ is divisible by $6$ for all integers $n\ge 1$
|
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|
Integrating a function of tan inverse How would I carry out the following integration?
$$\int_0^12\arctan x^2$$
I tried to substitute $x^2 = \tan\theta$
From there I wrote the integral as:
$$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$
Now, is my only option to use integration by parts? Or is there a better method to solve this integral?
|
By using the Taylor series of the arctangent function, $\arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{2n+1}$, we get:
$$ \int_{0}^{1}\arctan(x^2)\,dx = 2\sum_{n\geq 0}\frac{(-1)^n}{(4n+2)(4n+3)}=2\sum_{n\geq 0}(-1)^n\left(\frac{1}{4n+2}-\frac{1}{4n+3}\right),$$
where the series
$$\sum_{n\geq 0}\frac{(-1)^n}{4n+3} = \sum_{n\geq 0}\left(\frac{1}{8n+3}-\frac{1}{8n+7}\right)$$
can be computed through the identity
$$ -\log(1-z) = \sum_{n\geq 1}\frac{z^n}{n} $$
with $z$ being a root of $\frac{z^8-1}{z-1}$, i.e. through the discrete Fourier transform. By collecting the pieces,
$$ \int_{0}^{1}\arctan(x^2)\,dx = \color{red}{\frac{\pi}{4}-\frac{\pi}{2\sqrt{2}}+\frac{1}{\sqrt{2}}\,\log(1+\sqrt{2})}.$$
As an alternative, you may notice that the problem is equivalent, by integration by parts, to computing $\int_{0}^{1}\frac{x^2}{1+x^4}\,dx$, but $$1+x^4 = (1+2x^2+x^4)-2x^2 = (1+x^2)^2-(\sqrt{2} x)^2 $$
gives
$$ \frac{x^2}{1+x^4} = \frac{1}{2\sqrt{2}}\cdot\left(\frac{x}{x^2-\sqrt{2}\,x+1}-\frac{x}{x^2+\sqrt{2}\,x+1}\right)$$
and the primitive of $\frac{x}{x^2\pm\sqrt{2}\,x+1}$ is just:
$$ \frac{1}{2}\,\log(x^2\pm\sqrt{2}\,x+1)\mp\arctan\left(\frac{2x\pm\sqrt{2}}{2}\right).$$
|
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|
Describe the set of all complex numbers $z$ such that $|z-a |+| z-b |=c$
Describe the set of all complex numbers $z$ such that :
$$|z-a |+| z-b |=c$$
where $a,b, c$ are real
At a simple look I immediately recognized that this is some ellipse because it's the same with the definition of the ellipse with two foci, the sum of the distances from a point in the ellipse in this case $z$ remains constant.
I started working out on algebraic manipulations based on the rectangular coordinates of $z$ but the equation gives me only square roots like:
$$\sqrt{(x-a)^{2}+y^2}+\sqrt{(x-b)^2+y^2}=c$$
but this is at least not very convincing, can someone give some hint how to prove this more analytically.
|
Hint:
$$|z-a|+|z-b|=c$$
Square
$$|z-a|^2+2|z-a||z-b|+|z-b|^2=c^2.$$
Move the squares from the LHS to the RHS and square again
$$4|z-a|^2|z-b|^2=(c^2-|z-a|^2-|z-b|^2)^2\\
=c^4+|z-a|^4+|z-b|^4+2|z-a|^2|z-b|^2-2c^2|z-a|^2-2c^2|z-b|^2.$$
Then,
$$0=c^4+|z-a|^4+|z-b|^4\color{red}{-2|z-a|^2|z-b|^2}-2c^2|z-a|^2-2c^2|z-b|^2\\
=c^4+(|z-a|^2-|z-b|^2)^2-2c^2|z-a|^2-2c^2|z-b|^2.$$
Fully developing the last expression, you get the equation of a quadric ($x^2$ and $y^2$ cancel out in the parenthesis). The terms of the second degree are
$$(2(x_a-x_b)x+2(y_a-y_b)y)^2-4c^2(x^2+y^2)\\
=4((x_a-x_b)^2-c^2)x^2+8(x_a-x_b)(y_a-y_b)xy+4((y_a-y_b)^2-c^2).$$
The type of the conic depends on the sign of
$$\Delta=(x_a-x_b)^2(y_a-y_b)^2-((x_a-x_b)^2-c^2)((y_a-y_b)^2-c^2)\\
=c^2((x_a-x_b)^2+(y_a-y_b)^2-c^2)=c^2(|a-b|^2-c^2).$$
|
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|
Methods for efficiently factoring the cubic polynomial $x^3 + 1$ $$x^3 + 1$$
factors as
$$(x^2 - x + 1)(x + 1) .$$
It would have taken me a few minutes to identify this. What are the various approaches to determining rapidly that it is factorable, and factoring it?
|
Here is one approach.
Note that $$x^3+1=x^3+3x^2+3x+1 -3x^2-3x=(x+1)^3-3x(x+1)=(x+1)((x+1)^2-3x)=(x+1)(x^2-x+1)$$
Another approach would be noticing that $x=-1$ is a solution to $x^3+1=0$. Then you can divide $x^3+1$ by $x+1$ with polynomial long division.
The third approach would be to set $x^3+1=(x+a)(x^2+bx+c)$ and solving the system of equations.
A fourth approach would be using the formula for geometric series.
The sum of the geometric series $1$, $-x$, $x^2$ is $\frac{1-(-x)^3}{1-(-x)}=\frac{x^3+1}{x+1}$.
Multiplying $x+1$ on each side gives us that $(1-x+x^2)(1+x)=1+x^3$.
|
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|
Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
|
This is the Fourier series for $f(x) = \sin^7 x$ on the domain $[0,2 \pi]$.
You have $\int_0^{2 \pi} \sin^7 x \sin x dx = d\int_0^{2 \pi} \sin x \sin x dx$.
Here is the answer:
This gives ${35 \over 64 } \pi = d \pi$.
To compute the integral,
\begin{eqnarray}
\int_0^{2 \pi} \sin^8 x &=& { 1\over (2i)^8} \int_0^{2 \pi} (e^{it}-e^{-it})^8 dt \\
&=& { 1\over (2i)^8} \sum_{k=0}^8 \binom{8}{k} \int_0^{2 \pi} e^{it(8-2k)} dt \\
&=& { 1\over (2i)^8} \binom{8}{4} \int_0^{2 \pi} 1 dt \\
&=& {35 \over 64} \pi
\end{eqnarray}
|
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|
triple integral $\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$ I need to solve this
triple integral $$\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$$
where $V$ is all $\Bbb R^3$
I've spent on this task a few hours,
ok firstly
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{-1/2[(x-y)^2+(y-z)^2+(z-x)^2]}$$
I tried to use theorm of the change of variables, but
this substitution doesn't make sense
$$x-y=a$$
$$y-z=b$$
$$z-x=c$$
also
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{(x-1/2y^2-1/2z^2)^2+((\sqrt{3}/2)y-
(\sqrt3/2)z)^2}$$
Maybe here change of variables'll be OK.
Please, help me. It's my homework, so important and I have no idea what I should do next.
|
To pursue the change-of-variables method further, note that with $a$ and $b$ defined as originally done, we have
$$
x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{2} \left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] = \frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right],
$$
since $a + b = x - z$. Rearranging, we then get
$$
\frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right] = \frac{1}{4} \left[ 3 (a + b)^2 + (a - b)^2 \right],
$$
since $a^2 + b^2 = \frac{1}{2}(a + b)^2 + \frac{1}{2}(a - b)^2$. We can therefore use a different set of coordinates:
$$
u = a + b = x - z, \qquad v = a - b = x - 2y + z, \qquad w = z
$$
(Really, $w$ can be pretty much anything—see below.) The integral then becomes, up to an overall constant coming from the Jacobian,
$$
\iiint e^{-\frac{3}{4}u^2 - \frac{1}{4} v^2} \, du \, dv\, dw.
$$
Since there is no $w$-dependence in the integrand, and since its range of integration will run from $-\infty$ to $\infty$, we conclude that the integral does not converge.
The dummy coordinate $w$, by the way, can be chosen to be any coordinate that linearly dependent on $x, y, z$ and that is not degenerate with $u$ and $v$ (i.e., one that doesn't cause the Jacobian matrix to have zero determinant, which basically means something that is not a linear combination of $u$ and $v$.)
|
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evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how do I find the sum?
|
Notice, $$\sum_{n=2}^{\infty}\frac{3}{10^n}$$
$$=3\sum_{n=2}^{\infty}\frac{1}{10^n}$$
$$=3\left(\underbrace{\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+\ldots}_{\text{sum of an infinite G.P.}}\right)$$
$$=3\left(\frac{\frac{1}{10^2}}{1-\frac{1}{10}}\right)$$
$$=3\left(\frac{1}{90}\right)=\color{red}{\frac 1{30}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1680556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How to find bases $\alpha$ and $\beta$ for the following $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$? Suppose a linear transformation $T:P_3(\Bbb{R})\rightarrow P_2(\Bbb{R})$ has the matrix $A=\begin{bmatrix}
1 &2 & 0 & 0 \\
0 & 1 & 2& 1 \\
1 & 1 & 1 & 1 \\
\end{bmatrix} $ relative to the standard bases of $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$. How to to find bases $\alpha$ of $P_3(\Bbb{R})$ and $\beta$ for $P_2(\Bbb{R})$ such
that the matrix $T$ relative to $\alpha$ and $\beta$ is the reduced row-echelon form of $A$?
I assume we have to use change of basis formula. I have difficulty to start this type of question
because it's hard to identity the $[I]$ and $[T]$. Here the question says $A$ is relative to the standard
bases of $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$ so I think $[\alpha]_\gamma$ and $[\beta]_\gamma$ are respectively $[1, x, x^2, x^3]$ and $[1, x, x^2]$. But
then I don't know how to continue because I am not sure how $A$ works and how to write the change
of basis formula.
|
Note that for any matrix $A$ there is an invertible matrix $E$ such that
$\DeclareMathOperator{rref}{rref}EA=\rref A$. One computes $E$ by keeping track of the row reductions. In our case, we have
\begin{align*}
A &=
\left[\begin{array}{rrrr}
1 & 2 & 0 & 0 \\
0 & 1 & 2 & 1 \\
1 & 1 & 1 & 1
\end{array}\right] &
E &=
\left[\begin{array}{rrr}
-\frac{1}{3} & -\frac{2}{3} & \frac{4}{3} \\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
-\frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{array}\right] &
\rref A &=
\left[\begin{array}{rrrr}
1 & 0 & 0 & \frac{2}{3} \\
0 & 1 & 0 & -\frac{1}{3} \\
0 & 0 & 1 & \frac{2}{3}
\end{array}\right]
\end{align*}
Note that $E$ is invertible, being the product of elementary matrices. This implies that $E$ is the change of basis matrix corresponding to the basis
\begin{align*}
p_1 &= -\frac{1}{3}+\frac{2}{3}\,x-\frac{1}{3}\,x^2 &
p_2 &= -\frac{2}{3}+\frac{1}{3}\,x+\frac{1}{3}\,x^2 &
p_3 &= \frac{4}{3}-\frac{2}{3}\,x+\frac{1}{3}\,x^2
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1681834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Calculate the sum $x+y+z$ If positive real numbers $x,y,z$ satisfying the conditions
$$x^2+xy+y^2=25, y^2+yz+z^2=49, z^2+zx+x^2=64$$
then calculate the sum $x+y+z.$
My attempt:
By subtracting equalities are obtained $$(z-x)(x+y+z)=24, (x-y)(x+y+z)=15.$$
And here I stopped. )-:
|
$\angle ADB=\angle CDB= \angle ADC=120^{\circ}$
$D -$ Torricelli Point
$$S_{ADB}+S_{CDB}+S_{ADC}=S_{ABC}$$
$$\frac 12 xy \sin 120^{\circ}+\frac 12 zy \sin 120^{\circ}+\frac 12 xz \sin 120^{\circ}=\sqrt{10 \cdot 3 \cdot 3 \cdot 2}$$
$$\frac {\sqrt 3}{4} xy +\frac {\sqrt 3}{4} zy +\frac {\sqrt 3}{4} xz=6\sqrt{5}$$
$$\frac {\sqrt 3}{4}(xy+yz+zx)=6\sqrt{5}$$
$$xy+yz+zx=\frac{24\sqrt 5}{\sqrt3}$$
$$x^2+xy+y^2+y^2+yz+z^2+z^2+zx+x^2=25+64+49$$
$$2(x^2+y^2+z^2)+4xy+4yz+4zx=138+3 \cdot \frac{24\sqrt 5}{\sqrt3}$$
$$(x+y+z)^2=69+12\sqrt{15}$$
$$$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1682985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If $\sum_{n=-2}^{\infty}\cos^n x=8$, then find $x.$ Let $0<2x<\pi$. If $$\sum_{n=-2}^{\infty}\cos^n x=8,$$ then please find $x.$
I tried $\sum_{n=-2}^{\infty}\cos^n x=\frac{1}{1-\cos x}+\frac{1}{\cos x}+\frac{1}{\cos^2 x}=\frac{1+\cos x}{\cos^2 x \sin^2 x}=8.$ But I cant find $x.$
|
Substituting y to cos(x), we have $\sum_{n=-2}^\infty y^n=-\frac{1}{y^2(y-1)}=8.$
The solutions are $1/2$, $\frac{1-\sqrt{5}}{4}$, $\frac{1+\sqrt{5}}{4}$
Then we have $\cos(x)=1/2$ so $x = \pi/3+2k\pi$ or $x=-\pi/3+2k\pi$ ($k \in \mathbb{Z}$)
the two other solutions give four values for x.
$\pi/5+2k\pi$, $-\pi/5+2k\pi$ and $3\pi/5+2k\pi$, $-3\pi/5+2k\pi$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?
|
You just needed one more step.
$$\begin{align}n(n+1)(n+5) = & (n+1)(n+2)(n+6) \\ = & n^3+9 n^2+20 n+12 \\ = & (n^3+6n^2+5n)+3n^2+15n+12 \\ = & n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$
$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.
Of course you could have done this immediately.
*
*$n(n+1)$ is divisible by $2$.
*If neither $n$ nor $n+1$ are divisible by $3$, then $n+2$ and $n+5$ will be.
*
*exactly one of the three factors, $n$, $n+1$, $n+5$ is divisible by $3$.
*Therefore $n(n+1)(n+5)$ is divisible by $2$ and $3$, and thus by $6$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.
|
You can first remove a few factors
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}\\
=\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt{\tan x}}{x \sqrt{x}}\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}\\
=\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}.$$
Then multiply by the conjugate
$$=\lim_{x \to 0^+}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})},$$
evaluate the finite factor at denominator
$$=\frac12\lim_{x \to 0^+}\frac{1-\cos x(1-\sin^2 x)}{x^2},$$
use trigonometric identitites
$$=\frac12\lim_{x \to 0^+}\frac{2\sin^2\frac x2+\cos x\sin^2 x}{x^2},$$
and conclude
$$=\left(\frac12\right)^2+\frac12.$$
We used
$$\frac{\tan x}x=\frac{\sin x}x\frac1{\cos x}\to 1,$$
$$\frac{\sin ax}x=a\frac{\sin ax}{ax}\to a.$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval
$(A)(-1,0)\cup(0,1)$
$(B)(1,2)$
$(C)(-2,-1)$
$(D)(-\infty,-2)\cup(2,\infty)$
My try:
Let $x-\lfloor x \rfloor = \{x\}= t$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq t<1$. Then
$$\Rightarrow -3t^2+2t+a^2 = 0\Rightarrow 3t^2-2t-a^2 = 0$$
$$\displaystyle \Rightarrow t = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
I do not know how to solve further.
|
EDIT (ELABORATION)
There is probably a typo here; it should say real solutions.
Assume it has solutions.
$$0\leq t<1, t=\frac{1\pm \sqrt{1+3a^2}}{3} \Leftrightarrow 1 \le \sqrt{1+3a^2} < 2 $$
From the fact that $1-\sqrt{1+3a^2}<0$ if $a \neq 0$. From here, we square both sides to get $$0 \le a^2 \le 1$$
This implies $-1 \le a \le 1$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $tan^2 \theta = \frac{x}{y}$ how can we construct the angle $\theta$? If we are given the values of $x$ and $y$ and we know that $\tan^2 \theta = \dfrac{x}{y}$ is it possible for us to construct the angle $\theta$?
|
Set $z=x/y$, for simplicity. Then
$$
z=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-1
$$
so
$$
\cos^2\theta=\frac{1}{z+1}=\frac{y}{x+y}
$$
and
$$
\sin^2\theta=1-\frac{1}{z+1}=\frac{z}{z+1}=\frac{x}{x+y}
$$
Depending on which quadrant $\theta$ lives in, you can choose the signs in
$$
\cos\theta=\pm\sqrt{\frac{y\vphantom{A}}{x+y}},
\quad
\sin\theta=\pm\sqrt{\frac{x\vphantom{A}}{x+y}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1692267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Multiple radicals: $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$ $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
I don't think multiplying these out will work, and I am stuck in the beginning, without a basic concept to get started. Can anyone show me how to do these? I would appreciate more detailed responses as I have a solution to this already but since it does not explain the steps, I cannot learn from it. Thanks
Answer, if you want to check
359
|
HINT:
$$(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
$$=\{(a+b)^2-c^2\}\{c^2-(a-b)^2\}$$
$$=-(a^2+b^2+2ab-c^2)(a^2+b^2-2ab-c^2)$$
$$=-\{(a^2+b^2-c^2)^2-(2ab)^2\}$$
which is symmetric on expansion.
So, WLOG choose $\{a,b,c\}$ from $\{\sqrt{10},\sqrt{11},\sqrt{12}\}$ to find the same result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1693179",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Find the Cartesian equation of the locus described by $\arg \left(\frac{z-2}{z+5} \right)= \frac{\pi}{4}$ My working:
$$ \frac{x + iy - 2}{x + iy + 5} $$
$$ \frac{(x - 2 + iy)(x+5-iy)}{(x + 5 + iy)(x+5-iy)} $$
$$ \frac{x^2+5x-ixy-2x-10+2iy+ixy+5iy+y^2}{x^2+5x-ixy+5x+25-5iy+ixy+5iy+y^2} $$
$$ \frac{x^2+3x-10+y^2+7iy}{x^2+10x+25+y^2}$$
$$ \frac {\Im(z)}{\Re(z)} = \tan \frac{\pi}{4} = 1$$
$$ x^2 + 3x + y^2 - 10 = 7y $$
$$ x^2 + 3x + y^2 - 7y = 10 $$
$$ \left(x+ \frac 32\right)^2 + \left(y- \frac 72\right)^2 = 10 + \frac 94 + \frac {49}{4} $$
$$ \left(x- -\frac 32\right)^2 + \left(y- \frac 72\right)^2 = \left(\frac {7 \sqrt{2}}{2}\right)^2 $$
Is this correct? Is there a better method than what I did here?
|
Let me try this way -
The numerator is the line joining $Z(x,y)$ to $(2,0)$ while the denominator joins $Z(x,y)$ to $(-5,0)$ .
Geometrically, These two points are at a $45^\circ$ angle to each other.
Both the points should be on the locus. (Imagine a very small line at each point making $45^\circ$ angle with the other line)
Now, two points on the locus suspend an angle of $45^\circ$ to any other point on the locus implies - the locus is a circle!! and the center is at $90^\circ$ to these two line.
The center also lies on the perpendicular bisector of these two points, and so, the center is at $(-1.5,3.5)$. The two points are on the circle. So, radius is done too.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ How can I rationalize the following surd
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$
What would be the conjugate of the denominator
|
Use twice the conjugate:
$$\begin{aligned}
\frac{1}{1+\sqrt{2}-\sqrt{3}} &= \frac{1}{(1+\sqrt{2})-\sqrt{3}}\frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})+\sqrt{3}}\\
&=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}}\\
&= \frac{2+\sqrt{2}+\sqrt{6}}{4}
\end{aligned}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
The integral $\int\ln(x)\cos(1+(\ln(x))^2)\,dx$ Help with a integral calculus please!?
The equation is
$$\int\ln(x)\cos(1+(\ln(x))^2)\,dx$$
My teacher told me, i have to use substitution? but i can't still solve it.
I've been solving this last week but still i can't get the answer, please help me guys. Thanks!
|
First, substitute via $u=1+\ln^2(x)$ and use the complex exponential form of cosine to get
\begin{align*}
\int \ln(x) \cos(1+\ln^2(x)) \,dx
&= \int \sqrt{u-1}\, \cos(u)\, \frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du \\
&= \frac{1}{2} \int \cos(u)\,\exp(\sqrt{u-1}) \,du\\
&= \frac{1}{2} \int \frac{1}{2}[\exp(iu)+\exp(-iu)] \,\exp(\sqrt{u-1}) \,du \\
&= \frac{1}{4}
\left[ {\int \exp(iu) \,\exp(\sqrt{u-1}) \,du}
+
\int \exp(-iu) \,\exp(\sqrt{u-1}) \,du
\right]\\
&= \frac{1}{4}
\left[ I_1 + I_2 \right]
\end{align*}
Let's deal with $I_1$ first.
\begin{align*}
I_1
&= {\int \exp(iu) \,\exp(\sqrt{u-1}) \,du}\\
&= \underbrace{-\exp(\sqrt{u-1}) i \exp(iu) }_{\gamma_1}
- \int -i\exp(iu)\frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du\\
&= -\gamma_1 + ie^i\int \exp(a^2i+a) \,da\\
&= -\gamma_1 + ie^i\int \exp\left( i\left[a-\frac{i}{2}\right]^2 +\frac{i}{4} \right) \,da\\
&= -\gamma_1 + i\exp\left(\frac{5i}{4}\right)\int
\exp\left( \left[a\sqrt{i} - \frac{i^{3/2}}{2} \right]^2 \right) \,da\\
&= -\gamma_1 + \sqrt{i}\exp\left(\frac{5i}{4}\right)\int
\exp\left( z^2 \right) \,da\\
&= -\gamma_1 + \frac{\sqrt{i\pi}}{2}\exp\left(\frac{5i}{4}\right)\,\text{erfi}(z)
\end{align*}
where the first step uses integration by parts, then the substitution
$a=\sqrt{u-1}$ is used, followed by the substitution $z=a\sqrt{i} - i^{3/2}/2$.
Recall that $\text{erfi}(z)=-i\,\text{erf}(iz)$.
Now for $I_2$, using a similar strategy.
\begin{align*}
I_2
&= {\int \exp(-iu) \,\exp(\sqrt{u-1}) \,du}\\
&= \underbrace{\exp(\sqrt{u-1}) i \exp(-iu)}_{\gamma_2}
- \int i\exp(-iu)\frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du\\
&= \gamma_2 -
i\int \exp(-ia^2-i+a) \,da\\
&= \gamma_2 -
ie^{-i}\int \exp\left( -i\left[a + \frac{i}{2}\right]^2 -\frac{i}{4} \right) \,da\\
&= \gamma_2 -
i \exp\left( \frac{-5i}{4} \right)
\int \exp\left( -\left[ a\sqrt{i} + \frac{i^{3/2}}{2} \right]^2 \right) \,da\\
&= \gamma_2 -
\sqrt{i} \exp\left( \frac{-5i}{4} \right)
\int \exp\left( -\zeta^2 \right) \,da\\
&= \gamma_2 -
\frac{\sqrt{i\pi}}{2}
\exp\left( \frac{-5i}{4} \right) \,
\text{erf}(\zeta)\\
\end{align*}
using the same subsitution with $a$ and with $\zeta = a\sqrt{i} + i^{3/2}/2$.
Ok, now let's simplify the terms without the error functions:
\begin{align*}
-\gamma_1 + \gamma_2 &=
-\exp(\sqrt{u-1}) i \exp(iu) +
\exp(\sqrt{u-1}) i \exp(-iu) \\
&=
-\underbrace{\exp(\sqrt{u-1})}_{x}i[\underbrace{e^{iu} - e^{-iu}}_{2i\sin(u)}] \\
&=
2x\sin(1+\ln^2(x)) \\
\end{align*}
where we used the complex exponential form of sine.
Next we need to be able to undo the substitutions:
\begin{align*}
z
&= \sqrt{i} a - \frac{i^{3/2}}{2}\\
&= \sqrt{i} \sqrt{u-1} - \frac{i^{3/2}}{2}\\
&= \sqrt{i} \ln(x) - \frac{i^{3/2}}{2}\\
&= \frac{1}{2} \sqrt{i}\left[ 2\ln(x) - i \right]\\
\zeta
&= \sqrt{i} a + \frac{i^{3/2}}{2}\\
&= \sqrt{i} \sqrt{u-1} + \frac{i^{3/2}}{2}\\
&= \sqrt{i} \ln(x) + \frac{i^{3/2}}{2}\\
&= \frac{1}{2} \sqrt{i}\left[ 2\ln(x) + i \right]
\end{align*}
Now we can put it all together:
\begin{align*}
\int & \ln(x) \,\cos(1+\ln^2(x)) \,dx \\
&= \frac{1}{4} \left[ I_1 + I_2 \right]\\
&= \frac{1}{4} \left[
-\gamma_1 + \frac{\sqrt{i\pi}}{2}\exp\left(\frac{5i}{4}\right)\,\text{erfi}(z)
+
\gamma_2 -
\frac{\sqrt{i\pi}}{2}
\exp\left( \frac{-5i}{4} \right)
\text{erf}(\zeta)
\right]\\
&=
\frac{1}{4} \left(
-\gamma_1 + \gamma_2 +
\frac{\sqrt{i\pi}}{2}
\exp\left(\frac{-5i}{4}\right)
\left[
\exp\left( \frac{5i}{2} \right)
\text{erfi}(z)
-
\text{erf}(\zeta)\right]
\right)\\
&=
\frac{1}{4} \left(
2x\sin(1+\ln^2(x)) +
\frac{\sqrt{i\pi}}{2}
\exp\left(\frac{-5i}{4}\right)
\left[
\exp\left( \frac{5i}{2} \right)
\text{erfi}(z)
-
\text{erf}(\zeta)\right]
\right)\\
&=
\frac{1}{8} \left(
4x\sin(1+\ln^2(x)) +
{\sqrt{i\pi}}
e^{\frac{-5i}{4}}
\left[
e^{\frac{5i}{2}}
\text{erfi}\left(\frac{\sqrt{i}\left[ 2\ln(x) - i \right]}{2} \right)
-
\text{erf}\left(\frac{\sqrt{i}\left[ 2\ln(x) + i \right]}{2} \right)\right]
\right)
\end{align*}
A quick check in Mathematica gives:
f[x_] := Log[x] Cos[1 + (Log[x])^2]
FullSimplify[Integrate[f[x], x]]
1/8 ((-1)^(1/4) E^(-((5 I)/4))
Sqrt[\[Pi]] (-Erf[1/2 (-1)^(1/4) (I + 2 Log[x])] +
E^((5 I)/2) Erfi[1/2 (-1)^(1/4) (-I + 2 Log[x])]) +
4 x Sin[1 + Log[x]^2])
Which maybe should have been the whole answer...
See also this question, which solves a similar problem.
|
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|
Real part of $(1+2i)^n$ Is it true that for all $n\in \mathbb{N}$, $n\ge 2$ we have
$$|\textrm{Re}((1+2i)^n)|>1?$$
I do know de Moivre's Theorem.
I do not know how to show that $|\sqrt{5}^n\cos(n\arccos\left ( \frac{1}{5} \right ))|>1$ because the value $\cos(n\arccos\left ( \frac{1}{5} \right ))$ can become (theoretically) arbitrarily small.
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$a_{2n} = Re((1+2i)^{2n}) = Re((-3+4i)^n) = Re((1+4(-1+i))^n)$
Applying the Binomial expansion we get $(1+4(-1+i))^n$ as a series that converges $2$-adically :
$(1+4(-1+i))^n = 1 + 4(-1+i)\binom n1 -32i \binom n2 + \ldots + 4^k(-1+i)^k\binom nk + \ldots$
Since $v_2(4^k/k!) \ge k \to \infty$, this gives a $2$-adic power series
$(1+4(-1+i))^n = 1 + a_1 n + a_2 n^2 + \ldots$ where $a_k \in 2^k \Bbb Z_{(2)}[i]$
Taking the real part we get $a_{2n} = 1 + b_1 n + b_2 n^2 + \dots$
Computing the first few coefficients mod $8$, we get $b_1 = 4 \pmod 8$ and $b_n = 0 \pmod 8$ for $n > 1$, which shows that the map $n \mapsto a_{2n}$, when extended continuously to $\Bbb Z_{(2)}$, is a bijection from $\Bbb Z_{(2)}$ to $1+4\Bbb Z_{(2)}$. So it takes the value $1$ exactly once, at $n=0$.
For $a_{2n+1}$ the same thing works : $(1+2i)^{2n+1} = (1+2i)^{2n}(1+2i)$, so we just multiply each $a_k$ by $(1+2i)$ before taking real parts. Again we end up on $a_{2n+1} = 1 + 4n \pmod 8$, so it is again a bijection from $\Bbb Z_{(2)}$ to $1+4\Bbb Z_{(2)}$, it takes the value $1$ exactly once, at $n=0$.
You can be a bit more precise and show that $v_2(a_{2n} - a_{2m}) = v_2(a_{2n+1} - a_{2m+1}) = v_2(4(n-m)) = 2+v_2(n-m)$
This also implies that $(a_n)$ can only take a particular value at most twice, and so $|a_n| \to \infty$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1697435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Insights needed for the following Lagrange Multipler problem Find the point of the paraboloid $z = \frac{x^2}{4} + \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$.
So this seems like a pretty standard question of Lagrange multiplier, except I ran into some problems but cant figure out where I went wrong.
Attempt:
I maximuse $D^2$ instead of D to remove the square-root
1.) Constraint equation : $\frac{x^2}{4} + \frac{y^2}{25} - z$
2.) $D^2 = (x-3)^2 + y^2 + z^2$
3.) Do the usual Lagrange procedure
4.) $2z = -\lambda \\ 25y = \lambda y \\4(x-3) = \lambda x$
5.) Assume $x \neq 0, y\neq0$, we get $z = -\frac{25}{2}$
6.) This is where I got stuck, how can z be negative where is the sum of 2 positive number?
Which part of my attempt did I commit a mistake, and how do I rectify it?
Any insights and help is deeply appreciated.
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You want to minimize $$F=(x-3)^2+y^2+z^2+\lambda \left(\frac{x^2}4+\frac{y^2}{25}-z\right)$$ So $$F'_x=\frac{\lambda x}{2}+2 (x-3)=0$$ $$F'_y=\frac{2 \lambda y}{25}+2 y=0$$ $$F'_z=2 z-\lambda=0$$ $$F'_\lambda=\frac{x^2}{4}+\frac{y^2}{25}-z=0$$ Eliminating $x,y,z$ from the first derivatives leads to $x=\frac{12}{\lambda +4}$, $y=0$, $z=\frac{\lambda }{2}$.
Plugging these expressions in $F'_\lambda$ leads to $$\frac{36}{(\lambda +4)^2}-\frac{\lambda }{2}=0$$ that is to say $$\lambda ^3+8 \lambda ^2+16 \lambda -72=0$$ $\lambda=2$ is an obvious root by inspection. So,$$\lambda ^3+8 \lambda ^2+16 \lambda -72=(\lambda-2)(\lambda ^2+10 \lambda +36)=0$$ and the quadratic does not show any real solution.
So, $\lambda=2$, $x=2$, $y=0$, $z=1$ which makes $D^2=2$.
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"timestamp": "2023-03-29T00:00:00",
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Quadratic Formula, nature of roots with Trigonometric Functions The original problem:
If $0\le a,b\le 3$ and the equation $$x^2+4+3\cos(ax+b)=2x$$ has at least one real solution, then find the value of $a+b$
$$$$
At first, on rearranging, I got the following expression:
$$x^2-2x+(4+3\cos(ax+b))=0$$ I thought this was a quadratic in $x$, and thus from the quadratic formula(and that at least one real root exists), $D\ge 0$ ie $$4-4(4+\cos(ax+b))\ge 0$$
$$$$
However I'm not really sure about this. I've treated $\cos(ax+b)$ as a constant term even though the argument of the cosine includes $x$: the variable in which the quadratic expression is.
$$$$Under these circumstances, is it correct to use $3\cos(ax+b)$ as a constant? If not, how could I use the quadratic formula to find values of $x$ satisfying $$x^2-2x+(4+3\cos(ax+b))=0$$
Many thanks in anticipation!
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Wih a little bit of manipulation we can rewrite your given equation as $$x^2 - 2x + 4 = -3\cos{\left( ax + b \right)}$$
Let $f(x) = x^2 - 2x + 4$. Differentiating to find the minimum, we get
$$f'(x) = 2x - 2 = 0 \implies x = 1$$
$$f''(x) = 2 > 0 \implies \text{minimum at } x = 1$$
The minimum value of LHS is thus $f(1) = 3$.
The RHS is a cosine whose value oscillates in the range $[-3,3]$. The maximum value of the RHS is thus $3$. So we can see that equality holds if and only if the LHS is minimum and RHS is maximum. We just saw that the LHS, $f(x)$, is minimal only at $x = 1$. Now the value of $x$ is fixed, so the value of the RHS depends only on $a$ and $b$.
Hence, we have that
$$f(1) = 3 = -3\cos(a\times 1 + b) \implies \cos(a+b) = -1$$
I leave it to you to complete the problem from here :)
On a side note: No you can't take $\cos(ax + b)$ as a constant since $x$ is not a constant :)
$a + b = (2n+1)\pi$ but $0 \le a+b \le 6$. So $n = 0$ and $a + b = \pi$
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"timestamp": "2023-03-29T00:00:00",
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