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Does this matrix have a square root, and what is it? let $$M=\Biggl(\matrix{1&0&1\\0&1&0\\1&0&1}\Biggr)$$
does $M$ have a square root?
I believe that $M$ corresponds to a positive linear transformation, because it is a self-transpose, and has all non negative eigenvalues 2, 0, and 1 (according to numpy.linalg.eig). Because of this it should have a square root,
However, scipy.linalg.sqrtm says that the square root is
$$\Biggl(\matrix{\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\0&1&0\\\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}}\Biggr)$$
when I square that matrix in python, it gives:
$$\Biggl(\matrix{\frac{1}{2}&0&\frac{1}{2}\\0&1&0\\\frac{1}{2}&0&\frac{1}{2}}\Biggr)$$
Is there a bug with my brain, or with numpy/scipy?
| Now, I see the problem. In programming languages there are two "squaring" operations on matrices. One squares each individual entry, the other is actual matrix multiplication.
You squared each individual entry of $M^{1/2} := \begin{bmatrix}\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}}\end{bmatrix}$ to get $\begin{bmatrix}\tfrac{1}{2} & 0 & \tfrac{1}{2} \\ 0 & 1 & 0 \\ \tfrac{1}{2} & 0 & \tfrac{1}{2}\end{bmatrix}$.
If you use matrix multiplication, you'll get $M^{1/2}M^{1/2} = \begin{bmatrix}\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}}\end{bmatrix} = \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix} = M$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
Finding Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
what I tried:
let $S = 4x^2+4xy+4y^2+x-5$
$\dfrac{dS}{dx}=8x+4y+1$ and $\dfrac{dS}{dy}=4x+8y$
for center $\dfrac{dS}{dx}=0$ and $\dfrac{dS}{dy}=0$
getting center as $ x=-\dfrac{1}{6}$ and $y=\dfrac{1}{12}$
How do I solve it? Help me please
| We have to turn axes through $\theta$ where
$$ \tan 2 \theta = \frac {h}{b-a}=\frac {2}{4-4}= \infty; \theta = \pi/4$$
in order to remove $xy$ term and have axes parallel to $x-,y-,$ axes
$$ x1=(x-y)/\sqrt2;\, y1=(x+y)/\sqrt2;$$
plug in and taking only quadratic terms as $(x,y)$ translations do not affect eccentricity $e,$
$$ \frac12x_1^2+\frac32y_1^2 + ... = 0 $$
$$ e^2=1-\frac{b^2}{a^2} = 1- \frac{2/3}{2}\rightarrow e=\sqrt{\frac{2}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two distinct answers to same integral? I am wonder if any integral has 2 answers which might not be equivalent.
As far as I know the integral:
∫sec²x tanx dx
has 2 answers using substitution:
When tanx=t
sex²xdx=dt
=∫tdt
=t²/2 + C
=tan²x/2 + C
When secx=t
=secxtanxdx=dt
=∫tdt
=t²/2 + C
=sec²x/2 + C
If both answers above are equivalent then please show it to me how?
| The $C$ constant term in $ \frac{\tan^{2}x}{2} + C $ and $ \frac{\sec^{2}x}{2} + C $ need not be the same number, since these are solutions of an indefinite integral which is geometrically a family of curves - and they may refer to same or different curves depending on the value of their $C$s. It you are wondering how to arrive from one to the other, here is how you can do it:
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\frac{\sin^2x}{\cos^2x} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\frac{\sin^2x}{\cos^2x} +\frac{1}{2} -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{\sin^2x}{\cos^2x} +1\Big) -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{\sin^2x + \cos^2x}{\cos^2x}\Big) -\frac{1}{2} + C $$
and since, $\sin^2{\theta}+\cos^2{\theta}=1$, hence:
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{1}{\cos^2x}\Big) -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{\sec^2x}{2} -\frac{1}{2} + C $$
and $-\frac{1}{2}+C$ can be another constant, say $C_{2}$.
$$ \frac{\tan^2x}{2} + C = \frac{\sec^2x}{2} + C_{2} $$
So answering your original question - are they equivalent, YES they are!
| {
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"url": "https://math.stackexchange.com/questions/3417821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
interpretation on $1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}$ and $1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$ How to interpret the results
$$
1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}
\\
1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}
$$
I want to find a clear argument (combinatorial example,etc.) to prove this, other than induction or merely use the formula.
| Two different combinatorial interpretations of your first identity are given in the answers to this question. My answer (not the accepted one) was based on counting the number of ways you can place two white bishops on an $n\times n$ chessboard so that they guard each other. Calling this number $B_n$, and counting in two different ways, I showed that
$$B_n=2\binom{n+1}3+2\binom n3$$
and
$$B_n=2[1^2+2^2+\cdots+(n-1)^2].$$
For details see my answer to that old question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve binomial coefficient equation My book asks me to solve this equation:
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\begin{pmatrix} 7\\x \end{pmatrix}$$
The solution is $x=3$ and the formula $$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n\\k \end{pmatrix}=\begin{pmatrix} n+1\\k \end{pmatrix}$$ is supposed to reason that solution. What I do not understand however is, if $n=6$ would $n-1$ not equal $5$?
| We have
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{2!4!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
$$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
therefore since $(n)!=(m)!$ implies $n=m$ only when $n,m > 1$ we see that
$$(x-1)! = 2! \implies x-1=2 \implies x=3$$
$$(6-(x-1))!=4! \implies 6 -(x-1)=4 \implies -x=-2-1\implies x=3$$
which gives us our first solution of $x=3$. Next, because
$$\begin{pmatrix} 6\\2 \end{pmatrix}=\begin{pmatrix} 6\\4 \end{pmatrix}$$
we have
$$\begin{pmatrix} 6\\4 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{4!2!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
$$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
hence
$$(x-1)! = 4! \implies x-1=4 \implies x=5$$
$$(6-(x-1))!=2! \implies 6 -(x-1)=2 \implies -x=-4-1\implies x=5$$
which gives us our second solution of $x=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $h(t)=-16t^2+80t$ then simplify $\frac{h(a)-h(1)}{a-1}$ I am to simplify $\frac{h(a)-h(1)}{a-1}$ given $h(t)=-16t^2+80t$. The solution provided is
$\frac{-64+80a-16a^2}{-1+a}$ = $-16a+64$
I cannot see how this was arrived at. Here's as far as I got:
$\frac{h(a)-h(1)}{a-1}$
$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # substitute in the function h(t)
$\frac{-16a^2+80a-64}{a-1}$ # simplify numerator
$\frac{16(-a^2+5a-4)}{1-a}$ # 16 is a common factor in the numerator, attempted to simplify
...
I was not able to factor $-a^2+5a-4$
How can I arrive at the provided solution? More granular baby steps appreciated.
| Sometimes it's helpful (but not really necessary) to have positive leading coefficients on the terms of highest power.
$$\begin{align}
\frac{16(-a^2+5a-4)}{1-a} &= \frac{-16(a^2-5a+4)}{-(a-1)}\\
&= \frac{-1}{-1}\cdot\frac{16(a^2-5a+4)}{a-1}\\
&= \frac{16(a^2-5a+4)}{a-1}\\
\end{align}$$
In order to factor the polynomial in the numerator, we need to find two numbers that when multiplied together give $4$ and when added give $-5$. But, we can also make the guess that the denominator is a factor of the numerator (because the question asks us to "simplify").
$$\begin{align}
\frac{16(a^2-5a+4)}{a-1} &= \frac{16(a^2-a-4a+4)}{a-1}\\
&= \frac{16(a(a-1)-4(a-1))}{a-1}\\
&= \frac{16(a-4)(a-1)}{a-1}\\
&= \frac{16(a-4)}{1}\cdot\frac{a-1}{a-1}\\
&= \frac{16(a-4)}{1}\cdot1\\
&= \frac{16(a-4)}{1}\\
&= 16(a-4)\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that the equation :$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=2$ has no solution in $\mathbb{N^{3}}$ Prove that the following equation :
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=2$
$x,y,z\in\mathbb{N}$
Has no solution in $\mathbb{N^{3}}$
I'm going to use the hint are given :
$\left(\frac{a+b+c}{3}\right)^{3}\geq abc$
But how I use it ??
For example if we take :
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$ then $\prod \frac{x}{y}=1$
this mean :
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3$
| You used the inequality in the special case of $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$. You need to prove the result in general without this assumption, as follows:- $$\left(\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3}\right)^3 \ge \frac{x}{y}\frac{y}{z}\frac{z}{x}=1.$$ So
$$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge 1.$$
Then as per the solution of @John Omielan.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
rx^2-r2x+r+ry^2=1-x^2-y^2 \iff
$$
$$
rx^2+x^2-r2x+r+ry^2+y^2=1 \iff
$$
$$
x^2(r+1)-r2x+r+y^2(r+1)=1 \iff
$$
$$
x^2-x\frac{2r}{r+1}+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 3
$$
Completing the square of $x^2-x\frac{2r}{r+1}$:
$$
x^2-x\frac{2r}{r+1}+\Big (\frac{r}{1+r} \Big )^2-\Big (\frac{r}{1+r} \Big )^2=\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2
$$
Inserting in $(3)$ gives:
$$
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 4
$$
I'm stuck here, what is next?
Second method, starting from $(2)$:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
\Big (\frac{x(r+1)-r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
(x(r+1)-r)^2+y^2(1+r)^2=1 \tag 5
$$
Expand $(x(r+1)-r)^2$:
$$
(x(r+1)-r)^2=x^2(1+r)^2-x2r(1+r)+r^2
$$
Inserting in $(5)$ gives:
$$
x^2(1+r)^2-x2r(1+r)+r^2+y^2(1+r)^2=1 \iff
$$
$$
x^2(1+2r+r^2)-2rx-2xr^2+r^2+y^2(1+2r+r^2)=1\iff
$$
$$
x^2+2rx^2+r^2x^2-2rx-2xr^2+r^2+y^2+2ry^2+r^2y^2=1\iff
$$
Collect $r^2$ and $r$:
$$
r^2(x^2-2x+1+y^2)+r(2x^2-2x+2y^2) = 1-x^2-y^2 \iff
$$
$$
r^2((x-1)^2+y^2)+r(2x(x-1)+2y^2) = 1-x^2-y^2 \tag 6
$$
I'm stuck here, I don't know how to go from $(6)$ to $(1)$.
| Don't forget to get a common denominator first, before you add or subtract fractions.
Starting from (4)
$$
\begin{align}
\Big(x-\frac{r}{1+r}\Big)^2-\Big(\frac{r}{1+r} \Big)^2+\frac{r}{r+1}+y^2&=\frac{1}{r+1}\tag 4\\
\Big(x-\frac{r}{1+r}\Big)^2+y^2&=\frac{1}{r+1} +\Big(\frac{r}{1+r} \Big)^2- \frac{r}{r+1}\\
&=\frac{r+1}{(r+1)^2} +\frac{r^2}{(r+1)^2}- \frac{r(r+1)}{(r+1)^2}\\
&=\frac{r+1+r^2-r^2-r}{(1+r)^2}\\
&=\frac{1}{(1+r)^2} = \Big(\frac{1}{r+1}\Big)^2\\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$
Find $$\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$$
What I did was:
*
*Find the second order Taylor expansion for $xy-\sin(x)\sin(y)$ at $(0,0)$:
$P_2 (0,0)=f(0,0)+df_{(0,0)}+d^2f_{(0,0)}+r_2(0,0)$
*
*$f(0,0)=0$
*$df_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x + \frac{\partial f}{\partial y} (0,0) \Delta y = 0$
*$d^2f_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x^2 + \frac{\partial f}{\partial x \partial y} (0,0) \Delta x \Delta y + \frac{\partial f}{\partial y} (0,0) \Delta y^2 = 0$
$\implies$ $P_2(0,0)=r_2(0,0)$
*
*Now I have $\lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2}$
$\implies \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2} = \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2} \cdot \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{\sqrt{x^2+y^2}} \cdot \frac{\sqrt{x^2+y^2}}{x^2+y^2}= 0$
However, wolframalpha is telling me the limit does not exist...
| We have that
$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\frac{xy}{x^2+y^2}\left(1-\frac{\sin(x)}x\frac{\sin(y)}y\right)\to 0$$
indeed $\frac{xy}{x^2+y^2}$ is bounded and $1-\frac{\sin(x)}x\frac{\sin(y)}y \to 0$
As an alternative by polar coordinates
$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\cos\theta\sin \theta-\frac{\sin(r\cos\theta)}r\frac{\sin(r\sin\theta)}r=$$
$$=\cos\theta\sin \theta-\left(\cos \theta+o(r)\right)\left(\sin \theta+o(r)\right)=o(r) \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
| A slightly faster variant of the same computation using the identity $$\tan^{-1} u \pm \tan^{-1} v = \tan^{-1} \frac{u \pm v}{1 \mp u v}$$ can be performed by observing that in the special case $u = v$ $$2\tan^{-1} u = \tan^{-1} \frac{2u}{1-u^2}.$$ Consequently, we iterate $g(u) = 2u/(1-u^2)$ twice for $u = 1/5$ to obtain $$4 \tan^{-1} \frac{1}{5} = \tan^{-1} g(g(\tfrac{1}{5})) = \tan^{-1} \frac{120}{119}.$$ Now we apply the original formula to obtain $$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} = \tan^{-1} \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{(119)(239)}} = \tan^{-1} 1 = \frac{\pi}{4}.$$ In all, we used three steps instead of four.
It is also worth noting that when $u, v \in \mathbb Q$, we can write $$\tan^{-1} \frac{p}{q} \pm \tan^{-1} \frac{r}{s} = \tan^{-1} \frac{ps \pm qr}{qs \mp pr}.$$ If we think of each rational as being represented by an ordered pair, which in turn is an element of the complex numbers, e.g. $u = p/q$ has the representation $z = q + pi$, and we define the function $$T(z,w) = \tan\left(\tan^{-1} \frac{\Im(z)}{\Re(z)} + \tan^{-1} \frac{\Im(w)}{\Re(w)}\right),$$ then $$T(z,w) = \frac{\Im(zw)}{\Re(zw)}.$$ In fact, the inverse tangent identity is simply a consequence of multiplication in the complex plane: $$\arg(zw) = \arg(z) + \arg(w).$$ I leave the details of this relationship as an exercise for the reader.
From the above, we may then regard Machin's formula as a statement about the existence of a nonzero real number $\rho$ such that $$(5+i)^4 = \rho(1+i)(239+i).$$ What is this number?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Normal functional equation: $ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that for any real numbers $ x $ and $ y $ we have
$$ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $$
What I tried:
$$ P ( 0 , - x ) \implies f \big( f ( 0 ) \big) + f \left( x ^ 2 \right) = f ( 0 ) - x f ( - x ) = f \left( x ^ 2 \right) $$
which implies that $ f $ is odd, because
$$ P ( 0 , x ) \implies f \big( f ( 0 ) \big) + f \left( x ^ 2 \right) = f ( 0 ) + x f ( x ) = f \left( x ^ 2 \right) $$
Also, considering
$$ P ( 1 , 1 ) \implies f \big( 2 f ( 1 ) \big) = f ( 2 ) \implies f ( 1 ) = 1 $$
Let $ a , b \in \mathbb R $ such that $ a ^ 2 = b $.
$$ P \left( a ^ 2 , a \right) \implies f \Big( a ^ 2 f ( a ) + f \left( a ^ 2 \right) \Big) = a f \left( a ^ 2 + a \right) \text . $$
Now with the observation that $ f ( - x ) = - f ( x ) $, we can deduce that $ f ( x ) = x $.
How to proceed?
| Note that for the rest of this proof $a$ and $b$ will be any number in $\mathbb{R}$.
First, consider the following inputs.
\begin{alignat}{3}
P(0,0) &\rightarrow f(f(0)) + f(0) = f(0) + 0 &&\rightarrow f(f(0)) = 0\\
P(0,1) &\rightarrow f(f(0)) + f(1) = f(0) + f(1) &&\rightarrow f(0) = f(f(0)) = 0\\
P(0,a) &\rightarrow f(f(0)) + f(a^2) = f(0) + af(a) &&\rightarrow f(a^2) = af(a)\\
P(0,-a) &\rightarrow f(f(0)) + f(a^2) = f(0) -af(-a) &&\rightarrow af(a) = -af(-a) &\rightarrow f(-a) = -f(a)\\
P(a,0) &\rightarrow f(af(0)+f(a)) + f(0) = f(a) + 0f(a) &&\rightarrow f(f(a)) = f(a)
\end{alignat}
These have yielded some equalities which we will use in the rest of this proof. Next off, we will prove that f(2a) = 2f(a). Consider the inputs (a,a) and (-a,a). These yield the following:
$$P(a,a) \rightarrow f((a+1)f(a)) + af(a) = f(a) + af(2a)$$
$$P(-a,a) \rightarrow f(-(a+1)f(a)) + af(a) = -f(a) + 2f(0) \rightarrow f((a+1)f(a)) = f(a) + af(a)$$
Note that these have the same factor $f((a+1)f(a))$. Setting the other sides equal yields:
$$f(a)+af(2a)-af(a) = f(a) + af(a) \rightarrow af(2a) = 2af(a) \rightarrow f(2a) = 2f(a)$$
Now, consider the inputs (a,b) and (-a,b). These yield the following:
$$P(a,b) \rightarrow f(af(b)+f(a)) + f(b^2) = f(a) + bf(a+b)$$
$$P(-a,b) \rightarrow f(-af(b)-f(a)) + f(b^2) = -f(a) + bf(b-a) \rightarrow \\ f(af(b)+f(a)) = f(a) +bf(b) +bf(a-b)$$
Note that both of these have the same term $f(af(b)+f(a))$. Setting these equal yields:
$$f(a) + bf(a+b) - bf(b) = f(a) + bf(b) + bf(a-b) \rightarrow bf(a+b) = 2bf(b) + bf(a-b) \rightarrow\\ f(a+b) = f(2b) + f(a-b)$$
Taking c = a-b and d=2b, this turns into:
$$f(c+d) = f(d) + f(c)$$
Using this equation on the starting equation yields:
\begin{align*}
&f(xf(y)+f(x)) + yf(y) = f(x) + yf(x+y) \rightarrow\\
&f(xf(y)) + f(f(x)) + yf(y) = f(x) + yf(x) + yf(y) \rightarrow\\
&f(xf(y)) = yf(x)
\end{align*}
Putting $P(1,a)$ into this new equation yields $f(f(a)) = af(1) \rightarrow f(a) = af(1)$. In other words, $f$ is linear. That means that the only functions which could still be $f$, are of the form $f(x) = ax$. Note, however, that we also have $$f(f(x)) = f(x) \rightarrow a(a(x)) = a(x) \rightarrow a^2x = ax \rightarrow a = 1 \lor a = 0$$
In other words, the only functions which statisfy this equation, are $f(x) = 0$ and $f(x) = x$.
| {
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Challenging Sum: compute $\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)$ How to prove
$$\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)=\frac74\zeta(2)\zeta(3)-\frac{279}{16}\zeta(5)+\frac43\ln^3(2)\zeta(2)-7\ln^2(2)\zeta(3)\\+\frac{53}4\ln(2)\zeta(4)-\frac2{15}\ln^5(2)+16\operatorname{Li}_5\left(\frac12\right)$$
where $H_n^{(q)}=\sum_{k=1}^n\frac{1}{k^q}$ is the generalized harmonic number, $\operatorname{Li}_a(x)=\sum_{k=1}^\infty\frac{x^k}{k^a}$ is the polylogarithmic function and $\zeta$ is the Riemann zeta function.
This problem was proposed by Cornel and no solution has been submitted yet. I managed to convert it to a double integral but it seems tough to crack. Here is what I did:
Using the integral representation of the polygamma function:
$$\int_0^1\frac{x^n\ln^a(x)}{1-x}dx=-\psi^{(a)}(n+1)=(-1)^a a!\left(\zeta(a+1)-H_n^{(a+1)}\right)$$
With $a=2$ we have
$$\zeta(3)-H_n^{(3)}=\frac12\int_0^1\frac{x^n\ln^2(x)}{1-x}dx\overset{x=y^2}{=}4\int_0^1\frac{y^{2n+1}\ln^2(y)}{1-y^2}dy$$
multiply both sides by $\frac{H_n}{2n+1}$ then sum up we get
$$\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)=4\int_0^1\frac{\ln^2(y)}{1-y^2}\left(\sum_{n=1}^\infty\frac{y^{2n+1}H_n}{2n+1}\right)dy$$
we have
$$\sum_{n=1}^\infty \frac{y^{2n+1}H_n}{2n+1}=-\int_0^y\frac{\ln(1-x^2)}{1-x^2}dx$$
which follows from integrating
$\sum_{n=1}^\infty x^{2n}H_n=-\frac{\ln(1-x^2)}{1-x^2}$ from $x=0$ to $x=y$.
so
$$\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)=-4\int_0^1\int_0^y\frac{\ln^2(y)\ln(1-x^2)}{(1-y^2)(1-x^2)}dxdy$$
$$=-4\int_0^1\frac{\ln(1-x^2)}{1-x^2}\left(\int_x^1\frac{\ln^2(y)}{1-y^2}dy\right)dx$$
For the inner integral, Mathematica gives
$$\int_x^1\frac{\ln^2(y)}{1-y^2}dy\\=\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)-\ln(x)\operatorname{Li}_2(-x)+\ln(x)\operatorname{Li}_2(x)-\ln^2(x)\tanh^{-1}(x)+\frac74\zeta(3)$$
and the integral turned out very complicated. So any good idea how to approach the harmonic series or the integral?
Thank you.
| A second solution in large steps by Cornel Ioan Valean
Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is
Let $n$ be a non-negative integer number. Then, we have
$$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=\frac{1}{2}H_{2n}^2-2\log(2) H_{2n}+\frac{1}{2}H_{2n}^{(2)}-\frac{1}{4}H_n^2-\frac{1}{4}H_n^{(2)}+\log (2)H_n+\frac{1}{2} \log ^2(2)-\sum_{k=1}^{n-1}\frac{H_k}{2 k+1},$$
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$.
By multiplying both sides of the identity above by $1/n^3$ and considering the summation from $n=1$ to $\infty$, we get
$$\sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{H_{k}}{2 k+1}=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty}\frac{1}{n^3}\frac{H_{k}}{2 k+1}=\underbrace{\sum_{k=1}^{\infty}\frac{H_{k}}{2 k+1}\left(\zeta(3)-H_k^{(3)}\right)}_{\text{The desired series}}$$
$$=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^2}{n^3}-2\log(2) \sum_{n=1}^{\infty}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^{(2)}}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$$
$$+\log (2)\sum_{n=1}^{\infty} \frac{H_n}{n^3}+\frac{1}{2}\log ^2(2)\sum_{n=1}^{\infty}\frac{1}{n^3}-\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x,$$
where we see all the series in the right-hand side are easily reducible to known series which may also be found in the book (Almost) Impossible Integrals, Sums, and Series.
On the other hand, with simple integration by parts, we obtain
$$\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x$$
$$=\frac{1}{2}\log^2(2)\zeta(3)-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(x)}{x}\textrm{d}x-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x)}{x}\textrm{d}x,$$
where the last integrals may be found calculated in the paper The calculation of a harmonic series with a weight $5$ structure, involving the product of harmonic numbers, $H_n H_{2n}^{(2)}$.
A note: The sister of the result above (easy to obtain by recurrence relations and very useful),
$$\int_0^1 x^{2n-1} \frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=2\log(2) H_{2n}-\log(2)H_n+\frac{1}{4}H_n^2+\frac{1}{4}H_n^{(2)}-\frac{1}{2}H_{2n}^2-\frac{1}{2} H_{2n}^{(2)}+\frac{H_{2n}}{2n}-\frac{H_n}{2n}
$$
$$
-\frac{1}{2}\log^2(2)+\sum_{k=1}^{n-1}\frac{H_k}{2 k+1}.
$$
| {
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Find $a$ and $b$ such that the improper integral converges Find $a$ and $b$ so that the integral $$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$ converges.
I know that $\int_1^\infty \frac{1}{x^p}$ converges if $p>1$ and I've tried integration by parts but neither has helped. Can someone give me a hint as to how I should approach this?
| We have that
$$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx+\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$
and
$$\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_0^1 \frac{y+1}{y^a(y^2+2y+2)^b}\, dy$$
which converges for $a<1$ since as $y \to 0^+$
$$ \frac{y+1}{y^a(y^2+2y+2)^b} \sim\frac1{y^{a}}$$
and
$$\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}dx, \quad \frac{x}{(x-1)^a(1+x^2)^b}\sim\frac1{x^{a+2b-1}} \quad x\to \infty$$
converges for $a+2b-1>1 \iff b>1-\frac a 2$.
Therefore the given integral converges for $a<1$ and $b>1-\frac a 2$.
| {
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Limit of $\frac{e^{-xy}}{1+x^2+y^2}$ in $x,y ≥ 0$ I am trying to figure out if $f(x,y)=\frac{e^{-xy}}{1+x^2+y^2}$ goes to zero, in the area x,y ≥ 0, when $x^2+y^2$ goes to infinity.
So I rewrote the function as $\frac{1}{e^{xy}}\frac{1}{1+x^2+y^2}$.
The function $\frac{1}{1+x^2+y^2} $ definitley goes to zero. But what about
$\frac{1}{e^{xy}}$? Since $x,y ≥ 0$ it looks like to me that this function should also go to zero.
So if $x$ or $y$ is equal to zero then we have $\frac{1}{e^{0}}=1$
So in this case I would think you could argue that $f$ goes to $1\cdot0=0$ when $x^2+y^2$ goes to infinity.
When $x,y > 0$ I tried changing to polar coordinates then we have
$$0 <\frac{1}{e^{xy}}=\frac{1}{e^{r^2cos(t)sin(t)}}=\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$$
Since $- \frac{1}{2} ≤ \frac{1}{2}sin(2t) ≤ \frac{1}{2}$
I tried to use this to (maybe) see that $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$ is a function that goes to zero. But with $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}} ≤ \frac{1}{e^{-\frac{r^2}{2}}}$ we cant come to any conclusion since $\frac{1}{e^{-\frac{r^2}{2}}}$ goes to infinity when $r$ goes to infinity.
Could I prove that the limit is zero in some other way? If not, how is it possible that $\frac{1}{e^{xy}}$ $x,y > 0$ does not approach zero when $x^2+y^2$ goes to infinity?
| We have that for $x^2+y^2 \to \infty$ since
$$0< e^{-xy}\le 1$$
$$\frac{e^{-xy}}{1+x^2+y^2}\le \frac{1}{1+x^2+y^2}\to 0$$
| {
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Sum of first n terms is prime.
$a_n$ is a non-decreasing sequence of positive integers. If an positive integer $k$ appears in $a_n$ exactly $k$ times and $S_n$ is the sum of the first n terms, find all $N$ such that $S_N$ is prime.
The sequence is
$$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots$$
If $n$ is a triangular number, then the sum is just the sum of squares. So $S_n=$
$$\frac{n(n+1)(2n+1)}{6}-kn$$
for some non negative integer $k<n$.
$$S_n=\frac{n(2n^2+3n+1-6k)}{6}$$
I have somewhat closed form, but what could I do to find the $n$ such that it's prime?
Thanks!
| Let $n=m(m+1)/2 + j$ with $m$ a positive integer and $0\leq j \leq m$.
Then $S_n=\frac{(2m+1)(m+1)m}{6} + (m+1)j$.
We write this as $\frac{(2m^2+m+6j)(m+1)}{6}$
Suppose $m+1$ is not a divisor of $6$, then the left side must be a divisor of $6$. This is clearly only possible if $m=1$, in which case $j=0$ gives $S_n=1$ and $j=1$ gives $S_n=3$.
In the other cases $m+1$ must be a divisor of $6$.
If $m=2$ then $j=0$ gives $S_n= 5$, $j=1$ gives $8$ and $j=2$ gives $11$.
If $m=5$ then then $j=0$ gives $S_n = 55$, $j=1$ gives,$j=3$ gives $67$ $j=4$ gives $73$, and $j=5$ gives $79$.
Hence the values of $n$ that you want are $2, 3,5,16,17,18,19$
| {
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Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?
| Hint :
\begin{eqnarray*}
31(4^n+5^n+6^n)= 4^{n+2}+5^{n+2}+6^{n+2} +15 \times 4^n+ 6 \times 5^n - 5 \times 6^n.
\end{eqnarray*}
| {
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Evaluating $\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}\right)+\cos^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$
Evaluate
$$\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}\right)+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}\right)+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left[\cot\left(\cot^{-1}\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\right)+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left[\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left[\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left[\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right]$$
$$\sin^{-1}\left[\left(2+\sqrt{3}\right)+\dfrac{5\pi}{12}\right]$$
But domain of $\sin^{-1}$ is $[-1,1]$ and $\left(2+\sqrt{3}+\dfrac{5\pi}{12}\right)$ seems out of the domain.
But actual answer is $0$. What am I missing here?
| I think there are some errors in your expression. If you notice $5\pi /12>1$
Also $$\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}=\dfrac{\pi}{12}$$
So I don't think your expression is correct. But if you put the expression $$\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}$$ inside the $cot$ then it's $0$
So I suppose the correct expression would be
$$\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right)\right]$$
| {
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$
How to consider it?
| We have that
$$\frac{1}{\sqrt{n^2+n-nk}}\le \frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}\le \frac{(e-1)\frac{\log k}k+1}{\sqrt{n^2+n-nk}}$$
and since
$$\frac{1}{\sqrt{n^2+n-nk}}-\frac{1}{\sqrt{n^2-nk}}=$$
$$=\frac{n}{\sqrt{n^2+n-nk}\sqrt{n^2-nk}\left(\sqrt{n^2+n-nk}+\sqrt{n^2-nk}\right)}=$$
$$=\frac{\frac1{n^2}}{\sqrt{1+\frac1n-\frac kn}\sqrt{1-\frac k n}\left(\sqrt{1+\frac1n-\frac kn}+\sqrt{1-\frac kn}\right)}$$
$$\le\frac1{n^2} \frac{1}{\sqrt{1-\frac kn}\sqrt{1-\frac k n}\left(\sqrt{1-\frac kn}+\sqrt{1-\frac kn}\right)}$$
therefore
$$ \lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-nk}}=\lim\limits_{n \to \infty} \frac1n\sum\limits_{k=1}^{n}\frac{1}{\sqrt{1-\frac kn}}=\int_0^1 \frac{1}{\sqrt{1-x}}dx=2$$
then we need to show that
$$ \lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\frac{\log k}k}{\sqrt{n^2+n-nk}}=0$$
| {
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In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$ In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$
My attempt is as follows:-
$$\triangle=\dfrac{1}{2}bc\sin A$$
For having the maximum area,
$$A=90^{\circ}$$
$$\triangle_{max}=\dfrac{1}{2}bc$$
According to the above figure,
$$BE=12$$
$$AC=b$$
$$AE=EC=\dfrac{b}{2}$$
$$CF=9$$
$$BA=c$$
$$BF=FA=\dfrac{c}{2}$$
For $\triangle AEB$,
$$AB^2+AE^2=BE^2$$
$$c^2+\left(\dfrac{b}{2}\right)^2=12^2$$
$$4c^2+b^2=144\cdot4\tag{1}$$
For $\triangle FAC$
$$AC^2+AF^2=CF^2$$
$$b^2+\left(\dfrac{c}{2}\right)^2=9^2$$
$$4b^2+c^2=81\cdot4\tag{2}$$
Solving equations $(1)$ and $(2)$
$$16b^2-b^2=81\cdot4\cdot4-144\cdot4$$
$$15b^2=144(9-4)$$
$$b^2=48$$
$$b=4\sqrt{3}$$
Putting the value of $b$ in $(1)$
$$4\cdot48+c^2=324$$
$$c^2=132$$
$$c=2\sqrt{33}$$
Hence $$\triangle_{max}=\dfrac{1}{2}8\sqrt{99}$$
$$\triangle_{max}=12\sqrt{11}$$
But actual answer is $72$. But this method seems to be correct, why am I not getting the correct answer from this method? Please help me in this.
| Let $m_{1}=12,m_{2}=9$ and $m_{3}$ be the third median.
Let $$s=\frac{m_{1}+m_{2}+m_{3}}{2}=\frac{21+m_{3}}{2}$$
So $$s-m_{1}=\frac{m_{3}-3}{2};s-m_{2}=\frac{m_{3}+3}{2};s-m_{3}=\frac{21-m_{3}}{2}$$
Now, area of triangle the length of whose medians are $m_{1},m_{2},m_{3}$ is given by (http://mathworld.wolfram.com/TriangleMedian.html)
$$\triangle=\frac{4}{3}\sqrt{s(s-m_{1})(s-m_{2})(s-m_{3})}=\frac{4}{3}\sqrt{2916-\frac{(m^2_{3}-225)^2}{16}}\leq 72$$
the maximum area being attained when $m_{3}=15$
| {
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Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?
| It' not a coincidence. Let $f(x)=x + g(x)$ where $g(g(x))=x$, that is, $g$ is an involution. We'll show that $f(x)=f(a-x)$, provided that $f(x)=a$.
Assume $f(x)=a$, that is $a-x=g(x)$. Now
$$f(a-x)=f(g(x))=g(x)+g(g(x))=g(x)+x=f(x)$$.
In our case define
$$g(x)=\frac{3x}{\sqrt{x^2 - 9}}.$$
Indeed, $g$ is an involution as
$$g(g(x))=\dfrac{3\frac{3x}{\sqrt{x^2 - 9}}}{\sqrt{\frac{9x^2}{x^2 - 9} - 9}}=
\dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{x^2}{x^2 - 9} - 1}}=
\dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{9}{x^2 - 9}}}=x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?
| \begin{eqnarray}
\mathcal L &=& \lim_{x\to0}\frac{1-\cos x \sqrt{\cos 2x}}{x\sin x}=\\
&=& \lim_{x\to0}\frac{1-\cos^2 x \cos 2x}{x\sin x (1+\cos x\sqrt{\cos 2x})}=\\
&=& \lim_{x\to0}\frac{1-\cos^2x(2\cos^2x-1)}{2x\sin x}=\\
&=& \lim_{x\to0}\frac{1-2\cos^4x+\cos^2x}{2x\sin x}=\\
&=&\lim_{x\to0}\frac{2\left(\frac12 + \cos^2 x\right)\left(1-\cos^2 x\right)}{2x\sin x}=\\
&=&\lim_{x\to0} \frac32 \frac{1-\cos^2 x}{x\sin x}=\frac32.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Integral $\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$ I want to evaluate the integral $$\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$$
I tried using Feynman’s trick and introduced the parameter $e^{-\sqrt{1+t^2} x}$, then differentiating with respect to $x$ under the integral. However I don’t see a good way to evaluate the resulting integral. Is there a contour method?
This was an AMM problem ($12145$).
| Another solution for an interesting antiderivative.
$$I=\int\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt$$
First write
$$I=\frac12\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt+\frac12\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt$$
For the first integral, use $u=\sqrt{t^2+1}+t$ to get by the end
$$\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt=\int \frac {\sin(u)} u \,du=\text{Si}(u)=\text{Si}(\sqrt{t^2+1}+t)$$ For the second integral, use $u=\sqrt{t^2+1}-t$ to get by the end
$$\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt=-\int \frac {\sin(u)} u \,du=-\text{Si}(u)=-\text{Si}(\sqrt{t^2+1}-t)$$
So
$$I=\frac 12\Big(\text{Si}\left(\sqrt{t^2+1}+t\right)-\text{Si}\left(\sqrt{t^2+1}-t\right)\Big)$$ For the definite integral
$$J=\int_{-p}^p\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt=\text{Si}\left(p-\sqrt{p^2+1}\right)+\text{Si}\left(p+\sqrt{p^2+1}\right)$$
Now, for large values of $p$, composing Taylor series
$$p-\sqrt{p^2+1}=-\frac{1}{2 p}+\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$
$$\text{Si}\left(p-\sqrt{p^2+1}\right)=-\frac{1}{2 p}+\frac{19}{144 p^3}+O\left(\frac{1}{p^5}\right)\qquad \to 0$$
$$p+\sqrt{p^2+1}=2 p+\frac{1}{2 p}-\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$
Now, using for large values of $a$
$$\text{Si}(a)=\frac{\pi}{2} -\frac{\left(a^2-2\right) \cos (a)}{a^3}-\frac{\sin (a)}{a^2}+\cdots$$ then the result.
Using $p=100$, the above truncated expressions lead to $1.5733$ while the numerical integration gives $1.5634$.
| {
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"url": "https://math.stackexchange.com/questions/3450861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Sequence: $u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$
Study the following sequence of numbers:
$$\forall n\in\mathbb{N}, u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$$
I tried to calculate $u_{n+1}-u_n$, but I couldn't simplify the expression.
Plotting the sequence shows arithmetic (or seems to be an) progression.
| We have that
$$\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \ge n\cdot \frac{n}{\sqrt{2n^2}}=\frac{n}{\sqrt 2}$$
form which we conclude that $u_n \to \infty$, we can also obtain that
$$\frac{n}{\sqrt{2}}\le \sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \le \frac{2n}{\sqrt{5}}$$
and by Riemann sum since
$$\lim_{n\to \infty }\frac1n \sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}= \lim_{n\to \infty }\frac1n\sum_{k=n}^{2n}\frac{\frac kn}{\sqrt{1+\frac{k^2}{n^2}}}=\int_1^2 \frac{x}{\sqrt{1+x^2}}dx=\sqrt 5 - \sqrt 2$$
we obtain
$$\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}\sim (\sqrt 5 - \sqrt 2)n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A problem about equilateral triangle In below equilateral triangle $ABC$, $AD=CE$, $DH=GH$. Prove that $BE=BG$
My thoughts: It looksl ike we need to prove that $ABGD$ are on the same circle but I am not sure how to use the condition $DH=GH$
| Let $A,B,C$ be vectors from $O(0,0)$ to $A,B,C$ resp, let $a=A-C,b=B-C$. And let $C$ be $(0,0)$ then.
Let's write all we have:
$D=(1-t)a, E=tb$,
$F$:$F=AE\cap BD$, $AE:ua+(1-u)tb$, $BD:vb+(1-v)(1-t)a$
$$\begin{cases}
u=(1-t)(1-v)\\\
(1-u)t=v
\end{cases}$$
$$u=(1-t)(1-t+ut)$$
$$u-t(1-t)u=(1-t)^2$$
$$u = \frac{(t - 1)^2}{t^2 - t + 1}, v = \frac{t^2}{t^2 - t + 1}$$
$$F=\frac{(t - 1)^2}{t^2 - t + 1}a+\frac{t^2}{t^2 - t + 1}b$$
$H=\frac{w}2F$
$G=xA+(1-x)E=xa+(1-x)tb$, and, finally $DH=HG$:
$2H=D+G=(1-t+x)a+(1-x)tb$
$$\begin{cases}
1-t+x=w\frac{(t - 1)^2}{t^2 - t + 1}\\\
(1-x)t=w\frac{t^2}{t^2 - t + 1}
\end{cases}$$
$$w = 2-t,x=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1},
1-x=-\frac{t (t - 2)}{t^2 - t + 1}$$
$$\hbox{Thus }G=xa+(1-x)tb=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1}a-\frac{t^2 (t - 2)}{t^2 - t + 1}b,$$
$$BE=E-B=(1-t)b, BG=G-B=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1}a-\frac{(t - 1)^2 (t + 1)}{t^2 - t + 1}b$$
$$BG^2=(k_1a+k_2b)^2=k_1^2a^2+2k_1k_2ab+k_2^2b^2,$$
then recall $a^2=b^2$, $ab=|a|\cdot|b|\cos\angle ACB=\frac12a^2,$ thus $BG^2=\left(k_1^2+k_1k_2+k_2^2\right)a^2$ $=(1-t)^2a^2$ $=BE^2$, QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $2x^2-2\lfloor x \rfloor-1=0$ How do I solve $2x^2-2\lfloor x \rfloor-1=0$?
I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfloor x \rfloor\{x\}+2\{x\}^2-2\lfloor x \rfloor-1=0$$ but now I am stuck. How should I proceed?
| Suppose $x = n + \delta$ where $n$ is an integer and $0 \le \delta < 1$
Then
\begin{align}
2x^2-2\lfloor x \rfloor-1&=0 \\
x^2 &= \lfloor x \rfloor + \dfrac 12 \\
x &= \sqrt{\lfloor x \rfloor + \dfrac 12} \\
n+\delta &= \sqrt{n + \dfrac 12} \\
\delta &= \sqrt{n + \dfrac 12} - n \\
\end{align}
\begin{array}{|c|c|c|}
\hline
n & \delta & x \\
\hline
0 & \dfrac{\sqrt 2}{2} & \dfrac{\sqrt 2}{2} \\
1 & \dfrac{\sqrt 6}{2}-1 & \dfrac{\sqrt 6}{2} \\
2 & \color{red}{\dfrac{\sqrt{10}}{2}-2 < 0} \\
\hline
\end{array}
We have to stop at $n=2$ because, for $n \ge 2$, $\sqrt{n + \dfrac 12} - n < 0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $f(x)=x^2+a\cdot x + b$ prove $b\leq -\frac{1}{4}$ Equation $f(f(x))=0$, for $f(x)=x^2+a\cdot x + b$, have four real solutions and sum of two of them is $-1$. Prove that $b\leq-\frac{1}{4}$.
My progress so far:
Let $x_{1}$ and $x_{2}$ be real solutions for $f(x)=0$. We know that discriminant of this equation is positive, so from $a^2-4b\geq 0$, we have $b<\frac{a^2}{4}$.
Now let $y_{1}$ and $y_{2}$ be real solutions for $f(x)=x_{1}$ and $y_{3}$ and $y_{4}$ be real solutions for $f(x)=x_{2}$. From Vieta's formulas we know $y_{1}+y_{2}=-a$ and $y_{3}+y_{4}=-a$. From here I don't know how to continue, had many ideas but none of them worked.
Can anyone help?
| I found out a solution:
Discriminant of equations $f(x)=x_{1}$ and $f(x)=x_{2}$ must be positive so:
$a^2-4b+4x_{1}\ge 0$ and $a^2-4b+4x_{2}\ge 0$, if we sum them we will get:
$2a^2-4a-8b\ge 0$ (by vieta's formula $x_{1}+x_{2}=-a$)
form this we have $b\le \frac{1}{4}(a-1)^2-\frac{1}{4}$ ... (*)
Now discussion by which solutions sum to $-1$:
$1)$ $y_{1}+y_{2}=-1$ or $y_{3}+y_{4}=-1$
In this case by vieta's formula $y_{1}+y_{2}=-a=-1$ or $y_{3}+y_{4}=-a=-1$ and we have $a=1$ and by (*) $b\le -\frac{1}{4}$.
$2)$ $y_{1}+y_{3}=-1$ or any other pair of solutions
we have: $y_{1} = Y$, $y_{2}=-a-Y$, $y_{3}=-1-Y$ and $y_{4}=-a+1+Y$. From $x^2+ax+b-x_{1}$ we got $x_{1}=Y^2+aY+b$ and from $x^2+ax+b-x_{2}$ we got $x_{2}=Y^2-(a-2)Y+b-a+1$.
Now by $x_{1}+x_{2}=-a$ we have $Y^2+Y+b+\frac{1}{2}$. From problem we know that $Y$ is real , so discriminant must be positive or zero. That means $1-4(b+\frac{1}{2})\ge 0$, form that $b\le-\frac{1}{4}$.
Q.E.D.
This solution maybe isn't best or smarter but works!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Find all solutions to equation in $\mathbb{R}$ Find all solutions to equation in $\mathbb{R}$
$$\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}} = \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}$$
Solution: For $1991\leq x<1996$,
$\frac{x-7}{1989}>\frac{x-1989}{7}, \frac{x-6}{1990}>\frac{x-1990}{6}$ and $\frac{x-5}{1991}>\frac{x-1991}{5}$
$\sqrt{\frac{x-7}{1989}}>\sqrt{\frac{x-1989}{7}}, \sqrt{\frac{x-6}{1990}}>\sqrt{\frac{x-1990}{6}}$ and $\sqrt{\frac{x-5}{1991}}>\sqrt{\frac{x-1991}{5}}$
L.H.S. > R.H.S.
For $x>1996$,
$\frac{x-7}{1989}<\frac{x-1989}{7}, \frac{x-6}{1990}<\frac{x-1990}{6}$ and $\frac{x-5}{1991}<\frac{x-1991}{5}$
$\sqrt{\frac{x-7}{1989}}<\sqrt{\frac{x-1989}{7}}, \sqrt{\frac{x-6}{1990}}<\sqrt{\frac{x-1990}{6}}$ and $\sqrt{\frac{x-5}{1991}}<\sqrt{\frac{x-1991}{5}}$
L.H.S. < R.H.S.
$\therefore$ only solution is $x=1996$
Is there another method of solving this? More algebraic
| You have a nice solution.
Anyway, consider that you are looking for the zero of function
$$f(x)=\left(\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}}\right) -\left( \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}\right)$$
Perform a Taylor expansion at $x=1991$ to second order (I shall not write the coefficients). So, you face a quadratic equation in $(x-1991)$ and the solution of it is $1994.58$.
Otherwise, use Newton method and face the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 1991.10 \\
1 & 1992.92 \\
2 & 1995.40 \\
3 & 1995.98 \\
4 & 1996.00
\end{array}
\right)$$
| {
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"url": "https://math.stackexchange.com/questions/3462555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we prove partial sums $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ for all $N$ Can we prove $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ ?
I can figure out how to prove this where $N$ is a power of $2$ by adding the successive inequalities listed below (Cauchy condensation)
$$ \frac{1}{1} > 0 $$
$$ \frac{1}{2} \ge \frac{1}{2}$$
$$ \frac{1}{3} + \frac{1}{4} \ge \frac{1}{4} \cdot 2 = \frac{1}{2}$$
$$ \frac{1}{5} + \cdots + \frac{1}{8} \ge \frac{1}{8} \cdot 4 = \frac{1}{2}$$
$$ \cdots $$
$$ \cdots $$
I am having trouble figuring out how to do this for the partial sums when $N$ is not a power of $2$.
Click here to see a is a graph on desmos showing how the partial sums grow.
If there is a way to prove $\sum_{n=1}^{N} \frac{1}{n} \ge \log_e(N)$, that will also suffice.
Thanks!
| Here is one way.
$$
\sum_{n=1}^N \frac{1}{n} = \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{n}
\geq \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{x}
= \int_{1}^{N+1}\frac{dx}{x} = \Big[\ln x\Big]_1^{N+1}
= \ln (N+1)
$$
(I write $\ln$ for the natural logarithm.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume of ellipsoid in Cartesian co ordinates ( w/o changing to spherical or cylindrical systems) I tried triple integrating over the $$\int_{-a}^{a} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} {d}z {d}y {d}x $$ but getting $4/3 π a^2 b c$. Checking the possibility of deriving it with Cartesian coordinates and not converting to spherical or cylindrical systems. The derivation is as below
$$ z = \pm c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} ; y = \pm b\sqrt{1-\frac{x^2}{a^2}} \{z=0\}; x = \pm \,\, a \{z=0,y=0\}$$
for simplicity let $z = \pm c_1 \, and \, y = \pm b_1$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} z\Big|_{-c_1}^{c_1}\,{d}y{d}x$$
evaluating for z
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,-\,\left[ -c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\right] {d}y{d}x$$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} 2c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,{d}y{d}x$$
which is
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\int_{-b_1}^{b_1} \sqrt{{a^2}{b^2}-{x^2}{b^2}-{a^2}{y^2}}\,{d}y{d}x$$
now we know that
$$\boxed {\int \sqrt{{\alpha^2}-{\chi^2}}\,{dx} = \frac{x}{2}\sqrt{{\alpha^2}-{\chi^2}} +\frac{\alpha^2}{2}sin^{-1}\left(\frac{\chi}{\alpha}\right) + C\,\,}$$
putting $\alpha = \sqrt{{a^2}{b^2}-{x^2}{b^2}}\, and \, \chi = \sqrt{{a^2}{y^2}}$ in above, we get
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\, \frac{ay}{2}\sqrt{(ab)^2-(bx)^2-(ay)^2}+\frac{(ab)^2-(bx)^2}{2}sin^{-1}\left(\frac{ay}{\sqrt{(ab)^2-(bx)^2}}\right)\Big|_{-b_1}^{b_1}\,$$
Substituting the limits;
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\left[\frac{a}{2}\left(b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} + \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}b\sqrt{1-\frac{x^2}{a^2}}\right) - \,\left[\frac{a}{2}\left(-b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} - \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}(-b)\sqrt{1-\frac{x^2}{a^2}}\right) dx$$
Simplifying
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\frac{b}{2}\,\sqrt{a^2-x^2}\,(0) + \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(1)\,-\,\,\frac{-b}{2}\,\sqrt{a^2-x^2}\,(0) - \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(-1) dx$$
we know that $sin^{-1}(1) = \frac{\pi}{2}; sin^{-1}(-1) = \frac{-\pi}{2} $
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,0 + \frac{b^2(a^2-x^2)}{2}\,\frac{\pi}{2}-\,0 - \frac{b^2(a^2-x^2)}{2}\,\left(\frac{-\pi}{2}\right)dx$$
taking the constants out;
$$V_E = \frac{2\pi b^2c}{2ab} \int_{-a}^{a}(a^2-x^2)\,dx$$
Solving for x
$$V_E = \frac{\pi bc}{a}\left( a^2x - \frac{x^3}{3} \Big|_{-a}^{a} \right)$$
$$V_E = \frac{\pi bc}{a}\left(\left[ a^3 - \frac{a^3}{3}\right]-\left[- a^3 - \frac{-a^3}{3}\right]\right)$$
hence
$$V_E = \frac{4\pi a^2bc}{3}$$
Not sure what went wrong
| For starters, I would rescale $x, y, z$ by $X = x/a, \ Y = y/b, \ Z = z/c$. Thus our triple integral becomes
$$ I = abc \int_{-1}^1 \int_{-\sqrt{1 - X^2}}^{\sqrt{1 - X^2}} \int_{-\sqrt{1 - X^2 - Y^2}}^{\sqrt{1 - X^2 - Y^2}} dZ dY dX. $$
Integrating with respect to $Z$
\begin{align*}
I & = 2abc \int_{-1}^1 \int_{-\sqrt{1 - X^2}}^{\sqrt{1 - X^2}} \sqrt{1 - X^2 - Y^2} \ dY dX \\
& = \pi a b c \int_{-1}^1 1- X^2 \ dX \\
& = 2 \pi a b c \left (X - \frac{X^3}{3} \right ) \bigg|_{X=0}^{X=1} \\
& = \frac{4 \pi a b c}{3}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the mistake in my solution to find roots of a parabola? The parabola is,
$\frac{1}{2}g{{t}_{t}}^{2}+u{t}_{t}-2H=0$
By the way, $u=\sqrt{\frac{3gH}{2}}$. I have found the discriminant as below,
$\Delta=b^2-4ac$
$\Delta=u^2-4(\frac{1}{2}g)(-2H)$
$\Delta=u^2+4gH$
$u^2=\frac{3gH}{2} \Rightarrow \Delta=\frac{3gH}{2}+4gH=\frac{11gH}{2}$
Let's find roots,
${{t}_{t}}_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$
${{t}_{t}}_{1,2}=\frac{-u\pm\sqrt{\frac{11gH}{2}}}{g}$
The variable ${{t}_{t}}$ is a variable of time. Time can not be negative, so I use $+$ sign.
${{t}_{t}}=\frac{-u+\sqrt{\frac{11gH}{2}}}{g}$
But, when I put this in the parabola, I can't get $0$ as result. Is there any mistake?
important: $g$ and $H$ are constants. So, $u$ as well a constant.
| Both solutions produce $0$ in $f$.
\begin{align*}
f&\left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) \\
&= \frac{g}{2} \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right)^2 + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H \\
&= \frac{g}{2} \left( \frac{ \left( -u + \sqrt{\frac{11gH}{2}} \right)^2}{g^2} \right) + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H \\
&= \frac{1}{2} \cdot \frac{\left( -u + \sqrt{\frac{11gH}{2}} \right)^2}{g} + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H \cdot \frac{g}{g} \\
&= \frac{1}{2} \cdot \frac{u^2 -2u\sqrt{\frac{11gH}{2}} + \frac{11gH}{2}}{g} + \frac{-u^2 + u \sqrt{\frac{11gH}{2}}}{g} - \frac{2gH}{g} \\
&= \frac{\frac{1}{2} u^2 -u\sqrt{\frac{11gH}{2}} + \frac{11gH}{4}}{g} + \frac{-u^2 + u \sqrt{\frac{11gH}{2}}}{g} - \frac{2gH}{g} \\
&= \frac{\frac{1}{2} u^2 -u\sqrt{\frac{11gH}{2}} + \frac{11gH}{4} -u^2 + u \sqrt{\frac{11gH}{2}} -2gH}{g} \\
&= \frac{\frac{-1}{2} u^2 + \frac{11gH}{4} - 2gH\cdot\frac{4}{4}}{g} \\
&= \frac{\frac{-1}{2} \frac{3gH}{2} + \frac{11gH}{4} - \frac{8gH}{4}}{g} \\
&= \frac{\frac{-3gH}{4} + \frac{11gH}{4} - \frac{8gH}{4}}{g} \\
&= \frac{0}{g} \\
&= 0 \text{.}
\end{align*}
Similarly for the other root.
| {
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"answer_id": 0
} |
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
| Following the logic in your answer
$$
\lim_{n\to\infty}\overbrace{\left(\frac1n+\frac1n+\cdots+\frac1n\right)}^\text{$n$ terms}=0
$$
since each term tends to $0$. This is obviously false since the sum for each $n$ is $1$, so the limit is $1$.
We can say that a finite sum of terms which have a limit equals the finite sum of the limits. However, the same cannot be said for an unlimited sum of terms even though each has a limit. We must have some other theorem that allows us to swap order of the limits.
One proper way to handle the sum in the question is to bound the sum with
$$
\frac{n}{n+1}\le\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right)\le1
$$
and then the Squeeze Theorem says the limit is $1$.
| {
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"answer_count": 7,
"answer_id": 6
} |
Find all primes $p$ such that $x^2\equiv 10\pmod p$ has a solution
Find all primes $p$ such that $x^2\equiv 10$ mod $p$ has a solution
So I want $\left(\frac{10}{p}\right)=\left(\frac{5}{p}\right)\left(\frac{2}{p}\right)=1$
Then $\left(\frac{5}{p}\right)=\left(\frac{2}{p}\right)=1$
or $\left(\frac{5}{p}\right)=\left(\frac{2}{p}\right)=-1$
for the first The quadratic residues of $5$ are $1,4$
And $2$ is a quadratic residue for primes of the form $8k+1,8k+7$
so the primes of the form $5k+1,5k+4$ and either $8k+1,8k+7$ will work
or primes which are the form $5k+3,5k+2$ and $8k+3,8k+5$
I'm not sure if it's possible to narrow this down more?
| Everything you do is correct (modulo noting that $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)$ by Quadratic Reciprocity, which is why you are checking for quadratic residues modulo $5$) and dealing with the primes $p=2$ and $p=5$.
If $p=2$ or $p=5$, then $x^2\equiv 10\pmod{p}$ clearly has solutions.
Otherwise, your development is correct. You need either
*
*$p\equiv 1,-1\pmod{5}$ and $p\equiv 1,-1\pmod{8}$; hence $p\equiv \pm 1\pmod{40}$ or $p\equiv \pm 31\equiv\pm 9\pmod{40}$;
*$p\equiv 2,-2\pmod{5}$ and $p\equiv 3,-3\pmod{8}$; hence $p\equiv \pm 27\equiv \pm13\pmod{40}$ or $p\equiv\pm37\equiv \pm3\pmod{40}$;
*$p=2$;
*$p=5$.
| {
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"answer_id": 0
} |
Proving systems of nonlinear modular equations have no solution I have reason to suspect this system of six nonlinear modular equations has no solution for $2 < x < y < z$ even integers.
$$
\left\{
\begin{aligned}
z(3y+2) \equiv y(3z+2) \equiv 0& \mod x\\
z(3x+2) \equiv x(3z+2) \equiv 0& \mod y\\
x(3y+2) \equiv y(3x+2) \equiv 0& \mod z
\end{aligned}
\right.
$$
Remove any one and numerous solutions are easy to find, so I can't make the system any smaller. Having no solution is also consistent with other empirical results. But naturally I'd like a proof or a counterexample. Since the moduli are not pairwise coprime, I don't see how the CRT can help.
Using the definition of modulus, I could transform these into a system of nonlinear equations with nine variables:
$$\begin{bmatrix}
-K_1 &2 &3y\\
-K_2 & 3z & 2\\
2 & -K_3 &3x\\
3z &-K_4 & 2\\
2 & 3x &-K_5\\
3y & 2 &-K_6
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0
\end{bmatrix}
$$
where the Ki's are positive constants. But this doesn't seem to get me anywhere and may actually make the problem harder.
Any hope of proving something like this? This is research, but I'm a computer scientist by training, not a mathematician. Abstract algebra, discrete math, number theory, linear algebra etc either self-taught or learned back in the Dark Ages by candlelight.
--BF
| Proof outline
*
*We first simplify the system into (almost) coprime modulus using only $3$ of the $6$ equations (ignoring the $\equiv 0$ part).
*This will allow us to derive 2 general classes of solutions.
*We will then use a fourth equation $x(3z+2)\equiv 0 \pmod y$ to show both are not feasible, concluding the proof that there are no solutions.
Since $2<x<y<z$ are even integers, we let $x=2r,y=2s,z=2t$ and part of the system of equations (using the $\equiv 0$ part later) becomes
$$
\begin{align*}
2z-2y&\equiv 0 \pmod x &\implies 2t-2s &\equiv 0 \pmod r\\
2z - 2x &\equiv 0 \pmod y &\implies 2t-2r &\equiv 0 \pmod s\\
2y - 2x &\equiv 0 \pmod z &\implies 2s-2r &\equiv 0 \pmod t
\end{align*}
$$
Let $d = \gcd(s,t)$ and write $s=bd,t=cd$ so that $\gcd(b,c)=1$. Then from
$$
2s - 2r= 0 \pmod t,
$$
we get
$$
2bd - 2r \equiv 0 \pmod{cd}
$$
So that $d$ divides $2r$. Hence we let $2r = ad$. We now simplify the system of equations:
$$
\begin{align*}
2t-2s &\equiv 0 \pmod r &\implies 4t-4s &\equiv 0 \pmod{2r}&\implies 4cd-4bd &\equiv 0 \pmod{ad} &\implies 4c-4b&\equiv 0\pmod a\\
2t-2r &\equiv 0 \pmod s &\implies 2cd-ad &\equiv 0 \pmod{bd} &\implies 2c-a &\equiv 0 \pmod b\\
2s-2r &\equiv 0 \pmod t &\implies 2bd-ad &\equiv 0\pmod{cd} &\implies 2b-a &\equiv 0 \pmod c
\end{align*}
$$
Hence we get a new system
$$
\begin{align*}
4c-4b &\equiv 0 \pmod{a}\\
2c-a &\equiv 0 \pmod{b}\\
2b-a &\equiv 0 \pmod{c}
\end{align*}
$$
Next we prove a lemma that bounds the values of $a,b,c$:
Lemma. Any solution to the system must satisfy
$$
1\leq a,b,c \leq 9
$$
Proof. From the system of equations:
$$
\begin{align*}
2c - a &\equiv 0 \pmod b &\implies 2b + 2c - a &\equiv 0 \pmod b\\
2b - a &\equiv 0 \pmod c &\implies 2b + 2c - a &\equiv 0 \pmod c
\end{align*}
$$
Since $\gcd(b,c)=1$, by CRT we have
$$
2b+2c - a \equiv 0 \pmod{bc}
$$
Since $2b = 2s/d > 2r/d = a$, this means $2b+2c-a > 0$. Therefore we obtain a bound of $bc$:
$$
bc \leq 2b+2c - a < 2b+2c
$$
If $3 \leq b < c$, then
$$
(b-2)c < 2b \implies c < (2b)/(b-2) = 2 + 4/(b-2) \leq 2+4 = 6
$$
giving us a bound of $c \leq 5$. Similarly,
$$
(c-2)b < 2c \implies b < (2c)/(c-2) = 2 + 4/(c-2) \leq 2+4 = 6
$$
Hence we get $b,c \leq 5$. Using $2b > a$ then bounds $a,b,c$ as $a,b,c \leq 9$.
For the remaining case, $b=1$ or $b=2$. If $b=1$ then $2b>a$ forces $a=1$, so the system reduces to
$$
2-1 \equiv 0 \pmod{c}
$$
This forces $c=1$, contradicting $c>b$. Alternatively, if $b=2$ the system becomes
$$
\begin{align*}
4c-8 &\equiv 0 \pmod{a}\\
2c-a &\equiv 0 \pmod{2}\\
4-a &\equiv 0 \pmod{c}
\end{align*}
$$
So we see that $2$ divides $a$. Since $1\leq a < 2b = 4$, this can only be $a=2$. But now
$$
4-2 \equiv 0 \pmod c
$$
forces $c=1,2$, again contradicting $c>b$. This means $b\leq 2$ is not possible, therefore the previous bound
$$
1\leq a,b,c \leq 9
$$
is the only possible one and we are done.
$$
\tag*{$\square$}
$$
Now a brute force search of $1\leq a,b,c \leq 9$, conditioned on $a/2<b<c$ and $\gcd(b,c)=1$ shows that the only solutions are
$$
(a,b,c) = (1,3,5), (2,3,4)
$$
which corresponds to
$$
(x,y) = (2r,2s) = (ad,2bd) = (d,6d), (2d,6d)
$$
($z$ doesn't matter) They must satisfy one of the original equations
$$
x(3z+2) \equiv 0 \pmod y
$$
Hence we must have
$$
\begin{align*}
(d)(3z+2) \equiv 0 \pmod{6d} \implies 3z+2 \equiv 0 \pmod 6\\
(2d)(3z+2) \equiv 0 \pmod{6d} \implies 2(3z+2) \equiv 0\pmod 6
\end{align*}
$$
This is impossible $\pmod 3$, therefore there are no solutions and we are done.
| {
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"answer_id": 0
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Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}=\lim\limits_{x\to 0} \dfrac{\frac{1}{\sqrt{1-x^2}}-1}{5x^4}$ .
Is my approach correct?
| The problem is not too difficult if you compose Taylor series
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\frac{5 x^7}{112}+O\left(x^9\right)$$
$$\sin(x)\sin ^{-1}(x)=x^2+\frac{x^6}{18}+\frac{x^8}{30}+O\left(x^{10}\right)$$ will give not only the limit but also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving a first order ordinary differential equation that is not linear and not separable Problem:
Solve the following differential equations.
$$ ( 3x - y + 1 ) dx - ( 3x - y) dy = 0 $$
Answer:
I am going to use the substitution $z = 3x - y + 1$.
\begin{align*}
\frac{dz}{dx} &=3 - \frac{dy}{dx} \\
dz &=3 \, dx - dy \\
z + (z-1)( 3 - \frac{dz}{dx} ) &= 0 \\
z + 3(z-1) - (z-1)\frac{dz}{dx} &= 0 \\
4z - 1 - (z-1)\frac{dz}{dx} &= 0 \\
4z - 1 &= (z-1)\frac{dz}{dx} \\
\frac{dz}{dx} &= \frac{4z-3}{z-1} \\
dx &= \left( \frac{z-1}{4z-3} \right) \, dz
\end{align*}
Using long division, I find that:
\begin{align*}
\frac{z-1}{4z-3} &= \frac{1}{4} - \frac{ \frac{1}{4} } {4z-3} \\
\int \frac{z-1}{4z-3} \, dz &= \frac{z}{4} - \frac{1}{16} \ln{|4z-3|} + C_1 \\
x &= \frac{z}{4} - \frac{1}{16} \ln{|4z-3|} + C_1 \\
16x &= 4z - \ln{|4z-3|} + C_2 \text{ with } C_2 = 16C_1 \\
16x &= 12x - 4y + 4 - \ln{|12x - 4y + 4 - 3|} + C_2 \\
4x &= -4y - \ln{|12x - 4y + 1|} + C \text{ with } \,\, C = C_2 + 4
\end{align*}
Now, I am going to check the answer by differentiating it.
\begin{align*}
4 &= -4 \frac{dy}{dx}- \frac{12 - 4\frac{dy}{dx}}{12x - 4y + 1} \\
4( 12x - 4y + 1) &= -4( 12x - 4y + 1)\frac{dy}{dx} - 12 + 4 \frac{dy}{dx} \\
48x - 16y + 4 &= ( -48x + 16y - 4)\frac{dy}{dx} - 12 + 4 \frac{dy}{dx} \\
48x - 16y + 16 &= ( -48x + 16y - 4)\frac{dy}{dx} + 4 \frac{dy}{dx} \\
48x - 16y + 16 &= ( -48x + 16y )\frac{dy}{dx} \\
(3x - y + 1 ) \, dx &= ( -3x + y) \, dy \\
(3x - y + 1 ) \, dx - ( -3x + y) \, dy &= 0 \\
\end{align*}
Hence the answers do not check but I am close. It looks like a sign error to me but I cannot seem to find it.
| $$( 3x - y + 1 ) dx - ( 3x - y) dy = 0$$
You made a sign mistake here: $$z{+(z-1)(3-z')}=0$$ it should be $$z\color{blue}{-(z-1)(3-z')}=0$$
$$z-3(z-1)+(z-1)z'=0$$
$$-2z+3+(z-1)z'=0$$
$$(z-1)z'=(2z-3)$$
$$\frac {(z-1)}{(2z-3)}dz=dx$$
Integrate:
$$2z+\ln|z-\frac 32|=4x+C$$
Substitute back $z=3x-y+1$:
$$2x-2y+\ln|3x-y-\frac 12|=C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $n$ is a natural number then how to show that $\frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{\sqrt{5}*2^n} $is also a natural number? This question is from Introduction to Analysis volume 1 by R. Courant, chapter 1, section 1.5, 9th question.
I tried using binomial theorem, root $5$ get an odd power in the numerator and can be cancelled by root $5$ in denominator but I don't have any idea how to get rid of those $n$ choose $k$ terms.
| We derive the formal power series $\sum_{n=0}^\infty F_nz^n$ with
\begin{align*}
F_n=\frac{1}{2^n\sqrt{5}}\left(\left(1+\sqrt{5}\right)^n-\left(1-\sqrt{5}\right)^n\right),\quad n\geq 0\tag{1}
\end{align*}
We do some calculations with its coefficients and conclude that $F_n$ is a natural number for each $n\geq 0$.
We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty F_nz^n}
&=\frac{1}{\sqrt{5}}\sum_{n=0}^\infty\left(\frac{1+\sqrt{5}}{2}\right)^nz^n
-\frac{1}{\sqrt{5}}\sum_{n=0}^\infty\left(\frac{1-\sqrt{5}}{2}\right)^nz^n\\
&=\frac{1}{\sqrt{5}}\,\frac{1}{1-\frac{1+\sqrt{5}}{2}z}-\frac{1}{\sqrt{5}}\,\frac{1}{1-\frac{1-\sqrt{5}}{2}z}\tag{2}\\
&=\frac{1}{\sqrt{5}}\frac{\left(1-\frac{1-\sqrt{5}}{2}z\right)-\left(1-\frac{1+\sqrt{5}}{2}z\right)}
{\left(1-\frac{1-\sqrt{5}}{2}z\right)\left(1-\frac{1+\sqrt{5}}{2}z\right)}\tag{3}\\
&\,\,\color{blue}{=\frac{z}{1-z(1+z)}}\tag{4}\\
\end{align*}
Comment:
*
*In (2) we do a geometric series expansion.
*In (3) we write the expression with common denominator.
*In (4) we do some simplifications.
Next we expand (4) into a geometric series and derive another representation of the coefficients $F_n$. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain from (4) for $n\geq 0$:
\begin{align*}
\color{blue}{F_n}&=[z^n]\frac{z}{1-z(1+z)}\\
&=[z^n]z\sum_{k=0}^\infty z^k(1+z)^k\tag{5}\\
&=\sum_{k=0}^{n-1}[z^{n-1-k}](1+z)^k\tag{6}\\
&=\sum_{k=0}^{n-1}[z^k](1+z)^{n-1-k}\tag{7}\\
&\,\,\color{blue}{=\sum_{k=0}^{n-1}\binom{n-1-k}{k}}\tag{8}
\end{align*}
We have in (8) another (well known) representation of the Fibonacci numbers $F_n$. Since $\binom{n-1-k}{k}$ are non-negative integers we conclude $F_n$ are natural numbers.
Comment:
*
*In (5) we do a geometric series expansion.
*In (6) we apply the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$. We also set the upper index of the sum to $n-1$ since other terms do not contribute.
*In (7) we change the order of summation $k\to n-1-k$.
*In (8) we select the coefficient of $z^k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hard inequality for positive numbers The problem is to prove that for $a,b,c>0$ we have
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}$, and then to use Muirhead's inequalities to prove the remaining inequality - but unsuccessfully, so far...
| I will use the following inequality:
(Vasile Cirtoaje) If $x,y,z$ are positive real numbers, then:
$$(x+y+z)^3 \geq \frac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
Setting $x=\frac{a^2}{b^2}, y = \frac{b^2}{c^2}, z = \frac{c^2}{a^2}$, we find
$$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^3\geq \frac{27}{4}\left(\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2}+1\right)$$
and using Cauchy-Schwarz
$$\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2} \geq \frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}$$
It remains to prove that:
$$\frac{27}{4}\left[\frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}+1\right]\geq \left[\frac{15}{4}-\frac{9abc}{4(a^3+b^3+c^3)}\right]^3$$
If we set
$$t=\frac{a^3+b^3+c^3}{abc}\geq 3$$
this becomes
$$\frac{27}{4}\left(\frac{t^2}{3}+3\right)\geq \left(\frac{15}{4}-\frac{9}{4t}\right)^3$$
Fully expanding this is equivalent with:
$$\frac{9}{64t^3}(t-3)(16t^4+48t^3-183t^2+126t-27)\geq 0$$
which is true as $t \geq 3$.
| {
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"url": "https://math.stackexchange.com/questions/3485380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Solve $\int x^2e^x\sin x$ Solve $\int x^2e^x\sin x$
My attempt is as follows:-
$$I_1=\int e^x\sin x$$
$$I_1=e^x\sin x-\int e^x\cos x$$
$$I_1=e^x\sin x-e^x\cos x-\int\sin (x)e^x$$
$$2I_1=e^x\left(\sin x-\cos x\right)$$
$$I_1=\dfrac{e^x\left(\sin x-\cos x\right)}{2}\tag{1}$$
$$I_2=\int e^x\cos x$$
$$I_2=e^x\cos x+\int e^x\sin x$$
$$I_2=e^x\cos x+e^x\sin x-\int e^x \cos x$$
$$I_2=\dfrac{e^x(\cos x+\sin x)}{2}$$
$$I=\int x^2e^x\sin x$$
$$I=I_1x^2-2\int xI_1$$
$$I=I_1x^2-\int xe^x(\sin x-\cos x)$$
$$I=I_1x^2-\int xe^x\sin x+\int xe^x\cos x$$
$$I=I_1x^2-xI_1+\int I_1+xI_2-\int I_2$$
$$I=I_1x^2-xI_1+xI_2+\int I_1-\int I_2$$
$$I=I_1x^2-xI_1+xI_2+\dfrac{1}{2}\int e^x\left(\sin x-\cos x\right) -\dfrac{1}{2}\int e^x\left(\cos x+\sin x\right)$$
$$I=I_1x^2-xI_1+xI_2+\dfrac{I_1}{2}-\dfrac{I_2}{2} -\dfrac{I_2}{2}-\dfrac{I_1}{2}$$
$$I=I_1x^2-xI_1+xI_2-I_2$$
$$\dfrac{e^x}{2}\left(x^2\sin x-x^2\cos x-x\sin x+x\cos x+x\cos x+x\sin x-\sin x-\cos x \right)+C$$
$$\dfrac{e^x}{2}\left((x^2-1)\sin x-(x-1)^2\cos x \right)+C$$
Is there any better way to solve it which is short and clean. Mine got very long.
| Start with writing $\sin x=\color{red}{\Im}e^{ix}$
$$\int x^2e^x \sin x\ dx=\color{red}{\Im}\int x^2e^{(1+i)x}\ dx$$
$$=\color{red}{\Im}\left(\frac{x^2}{1+i}e^{(1+i)x}-\frac{2x}{(1+i)^2}e^{(1+i)x}+\frac{2}{(1+i)^3}e^{(1+i)x}\right)$$
$$=x^2 \color{red}{\Im} \frac{e^{(1+i)x}}{1+i}-2x \color{red}{\Im}\frac{e^{(1+i)x}}{(1+i)^2}+2\color{red}{\color{red}{\Im}} \frac{e^{(1+i)x}}{(1+i)^3}$$
$$=x^2\left(\frac12e^x\sin x-\frac12e^x\cos x\right)-2x\left(-\frac12e^x\cos x\right)+2\left(-\frac14e^x\sin x-\frac14e^x\cos x\right)$$
$$=\frac12e^x\sin x(x^2-1)-\frac12e^x\cos x(x-1)^2+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Factoring $x^4-2x^3+2x^2+x+4$
I need to show that the polynomial is not irreducible and I am trying to factor the polynomial
$$x^4-2x^3+2x^2+x+4$$
I checked from a calculator that it has a factor but how do I get it by myself?
I tried grouping but It didnt work I got
$x^2(x^2-2x+2)+x+4$ And I dont know how should I proceed. My guts tell me that it should be of the form:
$(x^2-ax+k)(x^2+bx+l)$, should I just try to figure out the constants by trying out?
| $$x^4-2x^3+2x^2+x+4$$
$$=(x^2+1)^2-2x^3+x+3$$
$$=(x^2+1)^2-2x(x^2+1)+3x+3$$
$$=(x^2+1)^2+x(x^2+1)-3x^3+3$$
$$=(x^2+1)^2+x(x^2+1)-3(x-1)(x^2+x+1)$$
$$=(x^2+x+1)(x^2+1-3x+3)$$
and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Problem (1) 1.A of Linear Algebra Done Right by Axler: Multiplicative Inverse My textbook, Linear Algebra Done Right, Third Edition, by Axler, has the following problem:
Suppose $a$ and $b$ are real numbers, not both $0$. Find real numbers $c$ and $d$ such that
$$\dfrac{1}{(a + bi)} = c + di.$$
This is effectively finding the multiplicative inverse $\beta = c + di$:
multiplicative inverse
for every $\alpha \in \mathbb{C}$ with $\alpha \not= 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$.
However, I'm struggling to see how this is done.
I found the following solution online:
Because $(a + bi)(a - bi) = a^2 + b^2$, one has
$$\dfrac{1}{a + bi} = \dfrac{a - bi}{a^2 + b^2}$$
Hence
$$c = \dfrac{a}{a^2 + b^2}, \ \ d = -\dfrac{b}{a^2 + b^2}$$
However, I don't see how $c = \dfrac{a}{a^2 + b^2}, \ \ d = -\dfrac{b}{a^2 + b^2}$ follows from $\dfrac{1}{a + bi} = \dfrac{a - bi}{a^2 + b^2}$?
I would greatly appreciate it if people would please take the time to clarify this.
| You have
$$\begin{equation}\begin{aligned}
\frac{1}{a + bi} & = \frac{a-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \frac{-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \left(\frac{-b}{a^2+b^2}\right)i
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Consider the solution of what $c$ and $d$ are in the original equation of
$$\frac{1}{(a + bi)} = c + di \tag{2}\label{eq2A}$$
Comparing the real & imaginary parts as suggested in the question comment (since $a$ and $b$ are real numbers, then $\frac{a}{a^2 + b^2}$ and $\frac{-b}{a^2 + b^2}$ are also both real values), gives the solution you're asking about, i.e., that
$$c = \frac{a}{a^2 + b^2}, \ \ d = -\frac{b}{a^2 + b^2} \tag{3}\label{eq3A}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\forall n \in N $and $L_2=-1$ is impossible.
if $a_n >0 $ then $a_n+2<3a_n+2 \rightarrow a_{n+1}=\frac{a_n+2}{3a_n+2} <1$ and then all the succession is beween $0$ and $1 $ excluded.
the succession jumps back and forth the value $\frac{2}{3}$ because:
if $a_n<\frac{2}{3}$ then $a_{n+1}>\frac{2}{3}$
if $a_n>\frac{2}{3}$ then $a_{n+1}<\frac{2}{3}$
Once I arrived at this point I don't know how to conclude that the limit of the succesion is if $\frac{2}{3}$
Can someone help me to understand how to procede?
| Observe that:
$$
a_{n+2} = \frac{a_{n+1} + 2}{3a_{n+1} + 2} = \frac{\frac{a_n + 2}{3a_n + 2} + 2}{3\frac{a_n + 2}{3a_n + 2} + 2} = \frac{a_n + 2 + 6a_n + 4}{3a_n + 6 + 6a_n + 4} = \frac{7a_n + 6}{9a_n + 10}
$$
Therefore:
$$
a_{n+2} - a_n = \frac{7a_n + 6}{9a_n + 10} - a_n = \frac{-9a_n^2 - 3a_n + 6}{9a_n + 10} = \frac{-3(a_n + 1)(3a_n - 2)}{9a_n + 10}
$$
If $a_n < \frac{2}{3}$, then $\frac{2}{3} > a_{n+2} > a_n$, and if $a_n > \frac{2}{3}$ then $\frac{2}{3} < a_{n+2} < a_n$. Therefore, both subsequence $(a_{2n})$ and $(a_{2n + 1})$ are bounded and monotone, hence converge. Since the only possible limit for both subsequence is $\frac{2}{3}$, the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Error in different attempts to solve $\int \frac{dx}{(1+x^4)^{1/4}}$
$$
\int \frac{dx}{(1+x^4)^{1/4}}=\frac{1}{2}\bigg[\frac{1}{2}\log\Big|\frac{1+z}{1-z}\Big|-\tan^{-1}z\bigg]+C
$$
Find $z$
Method 1
Set $x^2=\tan t\implies 2x.dx=\sec^2t.dt\implies dx=\dfrac{\sec^2t.dt}{2\sqrt{\tan t}}$
$$
\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{\sec^2t.dt}{2\sqrt{\tan t.\sec t}}=\int\frac{1}{2\cos t\sqrt{\sin t}}dt=\int\frac{\cos t}{2.\cos^2 t\sqrt{\sin t}}dt
$$
Set $y=\sin t\implies dy=\cos t.dt$
$$
I=\int\frac{dy}{2(1-y^2)\sqrt{y}}
$$
Set $z=\sqrt{y}\implies dz=\dfrac{dy}{2\sqrt{y}}$
$$
I=\int\frac{dz}{1-z^4}=\frac{1}{2}\int\frac{dz}{1-y^2}-\frac{1}{2}\int\frac{dz}{1+z^2}=\frac{1}{2}\bigg[\frac{1}{2}\log\Big|\frac{1+z}{1-z}\Big|-\tan^{-1}z\bigg]\\
\sec^2t=1+x^4\implies\cos^2t=\frac{1}{1+x^4}\implies\sin^2t=1-\frac{1}{1+x^4}=\frac{x^4}{1+x^4}\\
\implies\boxed{z=\sqrt{\sin t}=\frac{x}{(1+x^4)^{1/4}}}
$$
Method 2
$$
I=\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{dx}{x(1+\dfrac{1}{x^4})^{1/4}}=\int\frac{x^4.dx}{x^5(1+\dfrac{1}{x^4})^{1/4}}
$$
Set $1+\dfrac{1}{x^4}=z^4\implies 4z^3.dz=\dfrac{-4}{x^5}dx$
$$
I=\int\frac{-z^3.dz}{z(z^4-1)}=\int\frac{-z^2}{z^4-1}dz=\frac{-1}{2}\int\frac{z^2+1+z^2-1}{(z^2-1)(z^2+1)}dz\\
=\frac{-1}{2}\bigg[\int\frac{dz}{z^2-1}+\int\frac{dz}{z^2+1}\bigg]\\
=\frac{-1}{2}.\frac{1}{2}\log|\frac{z-1}{z+1}|-\frac{1}{2}\tan^{-1}z+C\\
=\frac{1}{2}\bigg[\frac{1}{2}\log|\frac{1+z}{1-z}-\tan^{-1}z|\bigg]+C'\\
\implies\boxed{z=\frac{(1+x^4)^{1/4}}{x}}
$$
This type of integrals are attempted to solve at Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$, and Thanx @lab bhattacharjee.
Why do I seem to get different expression for $z$ in methods 1 and 2 ?
Note: My reference gives the solution $z=\dfrac{(1+x^4)^{1/4}}{x}$
| Both solutions only differ by a constant and a mistake in sign because
$$\log\left|\frac{1+\frac{1}{z}}{1-\frac{1}{z}}\right| = \log\left|\frac{1+z}{1-z}\right|$$
and
$$\tan^{-1}\left(\frac{1}{z}\right) = \frac{\pi}{2} - \tan^{-1}(z)$$
which is consistent because there is a sign error in a method, I just can't immediately see where.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Cauchy Schwarz inequality Taken from Titu Andreescu and Bogdan Enescu's Mathematical Olympiad Treasures on page 9 Problem 1.19, to prove,
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geqslant \frac{3}{2}$.
It is easy to see why $LHS$ may yield
$\geqslant\frac{(a+b+c)^2}{2(ab+bc+ca)}$ from Cauchy.
Yet how do one bring in the '$3$' part from the $RHS$ knowing that $a^2+b^2+c^2 \geqslant ab+bc+ca $?
Any help would be appreciated.
| Write $x:=b+c,\,y:=c+a,\,z:=a+b$ so $\frac{a}{b+c}=\frac{y+z-x}{2x}$. The desired result is equivalent to $\sum_\text{cyclic}\frac{y+z}{x}\ge6$, which is trivial because Cauchy-Schwarz implies $q+\frac1q\ge2$ (hint: use $\binom1q,\,\binom{q}{1}$) for $q\in\{\frac{x}{y},\,\frac{y}{z},\,\frac{z}{x}\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$, $a^2-b^2+c^2-d^2=1749$. What are the possible values of $a^2+b^2+c^2+d^2$?
$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$,
$a^2-b^2+c^2-d^2=1749$. What are the possible values of
$a^2+b^2+c^2+d^2$?
So far I've used difference of squares to get: $(a+b)(a-b)+(c+d)(c-d)=1749$ but this doesn't help me very much, and I don't know how to use the inequalities or the fact that they are all primes.
I'm fairly sure guess and check would be a bad idea... but I have to say, I don't have many other ideas...
| From Yahoo Answers:
The inequality can be rewritten as:
$$a ≥ 3b+1, b ≥ 2c+1, c ≥ 2d+1 \tag{i}$$
$$a^2−b^2+c^2−d^2 = 1749 \tag{ii}$$
Using $\text{(i)}$, the left side of $\text{(ii)}$ is always odd and $1749$ is also odd. Therefore $d^2$ must be even which forces $d = 2$.
With $d=2$, $\text{(ii)}$ becomes $1753 = a^2−b^2+c^2 ≥ (3b+1)^2−b^2+c^2 = 8b^2+6b+1+c^2$
Using $\text{(ii)}$ this is $1753 ≥ 8(2c+1)^2+6(2c+1)+1+c^2 \Rightarrow 33c^2+44c−1738 ≤ 0$.
The roots of this quadratic imply the inequality $c≤6.6$ so $c≤5$. From condition $\text{(i)}$, $c ≥ 2(2)+ 1 \Rightarrow c ≥ 5$ so $c=5$.
Repeating this process gives only two values for $b$. Can you continue from here?
| {
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"url": "https://math.stackexchange.com/questions/3497826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Distributing balls into bins randomly Problem:
If $n$ balls are distributed at random into $r$ boxes (where $r \geq 3$), what is the probability that box $1$ at exactly $j$ balls for
$0 \leq j \leq n$ and box $2$ contains exactly $k$ balls for $0 \leq k \leq n$ ?
Answer:
Let $p$ be the probability that we seek. First we consider a special case. If $j + k > n$ then $p = 0$. Let $p_1$ be the probability that
a ball is placed in box $1$. Let $p_2$ be the probability that a ball is placed in box $2$. Let $p_3$ be the probability that a ball is placed in box other than box $1$ and box $2$.
\begin{align*}
p_1 &= \frac{1}{r} \\
p_2 &= \frac{1}{r} \\
p_3 &= \frac{r-2}{r}
\end{align*}
Now we have a multinomial distribution.
\begin{align*}
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^j \left( \frac{1}{r}\right) ^k \left( \frac{r-2}{r} \right)^{n - j - k} \\
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^{j+k} \left( \frac{r-2}{r} \right)^{n - j - k} \\
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \frac{ (r-2)^{n-j-k} } {r^n}
\end{align*}
So for example, if we have: $j = 2$, $k = 2$, $n = 8$ and $r = 8$ then
\begin{align*}
P &= \left( \frac{8!}{2!2!(8-2-2)!} \right) \frac{ (8-2)^{8-2-2} } {8^8} = \left( \frac{8!}{4(8-2-2)!} \right) \frac{ (6)^{4} } {8^8} \\
P &= \left( \frac{8!}{4(4)!} \right) \frac{ (6)^{4} } {8^8} = \left( \frac{8(7)(6)(5)}{4} \right) \frac{ 2^4(3^4) } {8^8} \\
P &= \frac{ 8(7)(6)(5)( 2^4)(3^4) } { 4(8^8) } = \frac{ 2(7)(6)(5)( 2^4)(3^4) } { 4(8^7) } \\
P &= \frac{ 7(6)(5)( 2^4)(3^4) } { 2(8^7) } = \frac{ 7(6)(5)( 2^3 )(3^4) } { (8^7) }\\
P &= \frac{ 7(6)(5)(3^4) } { 8^6 } = \frac{ 17010 } { 262144 } \\
P &= \frac{8505 } { 131072 } \\
P &\doteq 0.064888
\end{align*}
I think I have it right now. Do I?
| You have corrected the mistakes that were your initial solution. Your work is now correct.
| {
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In a ring where $(a-b)^2 = a - b$ for fixed $a,b$, then $(a-b)(a+b) = 1 \iff a^2 - b^2 = 1$. Let $(A,+,\cdot)$ be a ring where $(a-b)^2 = a - b$ for fixed $a,b$. Then $$(a-b)(a+b) = 1 \iff a^2 - b^2 = 1.$$
I was able to prove one implication:
Proof ($"\Rightarrow")$
We have that $(a-b)^2(a+b) = a-b = (a-b)(a+b) = 1$ thus $a-b = 1$ so $a+b=1$. Then $a + b - (a - b) = 1 - 1 = 0 = 2b$;
But since $a-b = 1$ we get $a = b + 1$ so $a^2 = b^2 + 2b + 1 = b^2 + 1$ so
$$a^2 - b^2 = 1$$.
Any tips for the other implication?
| Idempotents are always a joy to work with. For the converse implication denote $a-b=e$ and notice that $e$ is an idempotent by hypothesis; substituting $a=b+e$ in the relation $a^2=b^2+1_A$ leads (after a few simple calculations and cancellations) to $e+be+eb=1_A$ and hence to
$$eb+be=1_A-e \tag{1}$$
Multiplying relation (1) by $e$ on the left as well as on the right yields
$$eb+ebe=ebe+be=0_A \tag{2}$$
and hence to $eb=be=-ebe$.
Since $b$ and $e$ commute, $b$ and $a=b+e$ must also commute and thus one can factor the difference of squares
$$1_A=a^2-b^2=(a-b)(a+b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality Solution Correctness Let $a,b,c \in \Re $ and $ 0 < a < 1 , 0 < b < 1 , 0 < c < 1 $ & $ \sum_{cyc} a = 2$
Prove that
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
My solution
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
or
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$
From $ \sum_{cyc} a = 2$ we can conclude $\prod_{cyc}a \le \frac{8}{27} $ , thus getting a maximum bound on $\prod_{cyc}a$ .
Going back to
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$
We can say $$\prod_{cyc}(1-a) \le \left(\frac{\sum_{cyc}(1-a)}{3}\right )^3=\left(\frac{1}{3}\right )^3=\frac{1}{27}.$$
Which is true from $AM-GM$ inequality .
Is this solution correct ?
| You need to prove that:
$$\prod_{cyc}(1-a)\leq\frac{1}{8}abc.$$
You proved that $$\prod_{cyc}(1-a)\leq\frac{1}{27}.$$
Thus, it's enough to prove that $$\frac{1}{27}\leq\frac{1}{8}abc,$$ which is wrong, which says that your solution is wrong.
Since $$a+b-c=2-2c>0,$$ we see that $a$, $b$ and $c$ are sides-lengths of a triangle.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x+y+z=1$ and $x$, $y$ and $z$ are positives and after homogenization we need to prove that:
$$(x+y)(x+z)(y+z)\geq8xyz,$$ which is true by AM-GM.
| {
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Find solution of system of recurrence relations by using generating function How can I use generating functions to find $\{a_n\}$ and $\{b_n\}$ if
$\left\{ \begin{array}{l}
a_n=a_{n-1}+b_{n-1};\\
b_n=n^2a_{n-1}+b_{n-1};\\
a_0=1, b_0=1.
\end{array} \right.$
Thank you in advance.
| Hint.
Taking
$\left\{ \begin{array}{l}
\sum_{n=1}\left(a_nx^n-a_{n-1}x^n-b_{n-1}x^n\right) = 0\\
\sum_{n=1}\left(b_nx^n-n^2a_{n-1}x^n-b_{n-1}x^n\right)=0\\
\end{array} \right.$
and calling
$$
\sum_{n=0}a_nx^n = A(x)\\
\sum_{n=0}b_nx^n = B(x)
$$
we have
$$
\cases{A(x)-a_0-xA(x)-xB(x) = 0\\
B(x) - b_0 -\sum_{n=1}n^2a_{n-1}x^n-xB(x) = 0
} ~~~~(*)
$$
now regarding
$$
\sum_{n=1}n^2a_{n-1}x^n = x\sum_{n=0}(n^2+2n+1)a_n x^n
$$
and
$$
\sum_{n=0}n a_n x^n = x\frac{d}{dx}\left(\sum_{n=0}a_n x^n\right)=x\frac{d}{dx}A(x)
$$
and consequently
$$
\sum_{n=1}n^2 a_n x^n = x\frac{d}{dx}(x\frac{d}{dx}A(x))= x\frac{d}{dx}A(x)+x^2 \frac{d^2}{dx^2}A(x)
$$
Then,
\begin{align*}
\sum_{n=1}n^2a_{n-1}x^n &= x(x\frac{d}{dx}A(x)+x^2 \frac{d^2}{dx^2}A(x)+2x\frac{d}{dx}A(x)+A(x))\\
&=x^3\frac{d^2}{dx^2}A(x)+3x^2\frac{d}{dx}A(x)
+xA(x)\end{align*}
Finally, we have
$$
(*) \Leftrightarrow \cases{A(x)-xA(x)-xB(x)-1= 0\\
B(x)-x^3\frac{d^2}{dx^2}A(x)-3x^2\frac{d}{dx}A(x)-xA(x)-xB(x)-1 = 0
}~~~~(**)
$$
With those rules in mind, you can build a DE involving $A(x)$ whose solution will give you the closed form to $A(x)$. With $A(x)$ developed in series, the series coefficients will give you the sought $\{ a_k\}$ sequence.
NOTE
$A(x), B(x)$ are called generating functions for the sequences $\{ a_k\},\{b_k\}$ respectively.
The resulting DE for $A(x)$ is
$$
x^4 A''(x)+3 x^3 A'(x)+(2 x-1) A(x)+1=0
$$
This DE can be solved proposing for $A(x) = \sum_{k=0}a_k x^k$ furnishing the following recurrences
$$
\left\{a_0-1=0,a_1-2 a_0=0,a_2-2 a_1=0,-3 a_1-2 a_2+a_3=0,-8 a_2-2 a_3+a_4=0,-15 a_3-2 a_4+a_5=0,-24 a_4-2
a_5+a_6=0,\cdots,\right\}
$$
or $a_0 = 1, a_1 = 2, a_2 = 4$ and for $n\ge 3$
$$
\cases{
a_n = 2a_{n-1}+n(n-2)a_{n-2}\\
b_n = a_{n-1}+n(3n-2) a_{n-2}+n^2(n-1)(n-3)a_{n-3}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\gcd(a,b)=d$ then prove that $\gcd(\frac{a}{c},\frac{b}{c})=\frac{d}{c}$ My attempt:
$ax+by=d$(as gcd(a,b)=d).
Dividing by c
$\frac{a}{c}x+\frac{b}{c}y=\frac{d}{c}$
Let $k$ be the gcd of$ (\frac{a}{c},\frac{b}{c})$ so
$k|\frac{a}{c}$ and $ k|\frac{b}{c}$ then $k|\frac{a}{c} x + \frac{b}{c}y$ and so $k|\frac{d}{c}$.
As $d|a $ implies that $\frac{d}{c}|\frac{a}{c} $ similarly it divides $\frac{b}{c}$ . Hence $\frac{d}{c}|k$(as $k$ is the greatest common factor). Thus $k= \frac{d}{c}$
| Clearly $\frac {d}{c} \mid \frac {a}{c}$ and $\frac {b}{c}$ as $\gcd(a,b)=d$. Let $m$ be a common divisor of $\frac {a}{c}$ and $\frac{b}{c}$. Then $m$ divides the left side which implies that $m \mid \frac{d}{c}$ and $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving recursive function outputs $2^x \cdot {x \choose y}$ Consider following function
$f: \mathbb{N}\times\mathbb{N} \rightarrow \mathbb{N}$:
\begin{align*}
f(x ,y) =
\begin{cases}
0 & \text{if } x < y\\
2^x & \text{if } y = 0\\
2 \cdot (f(x - 1, y - 1) + f(x - 1, y)) & \text{else.}\\
\end{cases}
\end{align*}
I want to prove that $f(x, y) = 2^x \cdot {x \choose y}$, $x, y \in \mathbb{N}$, $x \geq y$.
I do see that $f(x - 1, y - 1)$ resembles ${x - 1 \choose y - 1}$ and $f(x - 1, y)$ resembles ${x - 1 \choose y}$, $\left({x \choose y} = {x - 1 \choose y - 1} + {x - 1 \choose y}\right)$ but I don't know where to go from this.
| short answer: Do induction on the values of $n = x -y$ starting with $n = 0$.
Prove the Base case and the induction step by induction on $y$ starting with $y=0$.
That is the Base case is to prove that the result is true for $x-y=0$ and showing this holds for $f(y,y)$ by doing base case $f(0,0)$ and doint induction on $y$ shoowing true for $f(y,y) \implies $ true for $f(y+1, y+1)$.
Then to the Induction step by assume true for all $x-y=n$ and showing by induction it holds for $f(y+ (n+1),y)$ doing a base case on $f(n+1,0)$ then show true for $f(y+n+1,y) \implies $true for$f((y+1) + (n+1),y+1)$.
Induction within induction.
To wit:
========
Claim 1: For all $n \ge 0$ and all $x = y+n$ then $f(x,y) = 2^x{x\choose y}$.
Well, prove this with induction.
Base case of Claim 1: $n= 0$.
Pf: We'll prove the base case by induction.
Claim: For all $x = y$, $f(x,y) = 2^x{x \choose y}$
Base case: $x = 0$ then $f(0,0) = 2^0 =1$ and ${0\choose 0} =1$.
Induction step: Assume true for $x$ that $f(x,x) = 2^x{x\choose x}$.
Then $f(x+1,x+1) = 2f(x,x) + f(x,x +1) = 2\cdot 2^x{x\choose x} + 0=2^{x+1}{x\choose x}$. But ${x\choose x}= 1={x+1\choose x+1}$
Thus Base case of Claim 1:
Induction step:
Assume it is true that for all $x-y=n$ our result is true:
Claim 2: For all $x = y+ (n+1)$ then $f(x,y) = 2^x{x\choose y}$
We will prove Claim 2 by induction:
Base case of Claim 2: $f(n+1,0) = 2^{n+1}$ and ${n+1\choose 0} =1$.
Induction step of Claim 2: Suppose it is true for $x = y+(n+1)$.
Then $f(x+1, y+1) = 2(f(x,y) + f(x, y+1))$.
$f(x,y) = 2^x{x \choose y}$. But notice $x-(y+1) = n$ and we know it is true for all $x' = y'+n$ so it is true for so $f(x,y+1) = 2^x{x\choose y+1}$.
So $f(x+1,y+1) = 2(2^x{x \choose y} + 2^x{x\choose y+1})=$
$2^{x+1} ({x \choose y} + {x\choose y+1})=$
$2^{x+1} (\frac {x!}{(x-y)!y!} + \frac {x!}{(x-y-1)!(y+1)!})=$
$2^{x+1}(\frac {x!(y+1) + x!(x-y)}{(x-y)!(y+1)!})=$
$2^{x+1}(\frac {x!(x+1)}{((x+1)-(y+1))!(y+1)!})=$
$2^{x+1}(\frac {(x+1)!}{((x+1)-(y+1))!(y+1)!})= 2^{x+1}{x+1\choose y+1}$.
So it is true for all $x= y+(n+1)$.
That proves our induction step of Claim 2.
So Claim 2 is proven.
Proving Claim 2 proves the induction step of Claim 1.
So Claim 1 is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question on Integration solution I am looking through the solution of integrating: $$\int_{}^{} \frac{x^4+1}{x^3+1} dx$$
I understand it but for the following part:
My problem is limited to the first and penultimate column. I would very much appreciate it if someone could explain to me these, especially the penultimate one.
EDIT 0: Clarification
I have a problem with this line:
How do I get those thetas?
My main problem is with this part:
Why 1 = ..., -1 = ... and 0 = ...?
| An introduction to partial fractions.
If we have two fraction and we want to add them together, $\frac {a}{b} + \frac {c}{d} = \frac {ad + bc}{bd}$
We can do the same thing, with rational polynomials.
$ \frac {1}{x} - \frac {1}{x+1} = \frac{x+1}{(x)(x+1)} - \frac {x}{x(x+1)} = \frac {x+1-x}{x(x+1)} = \frac {1}{x(x+1)}$
Supposing you had $\frac {1}{x(x+1)}$ How do you break it into partial fractions?
We already know the solution, but how do you find it if you didn't know?
We know that the denominators are factors of the denominator of the original expression.
So we "guess" that it might look something like this:
$\frac {1}{x(x+1)} = \frac {a}{x} + \frac {b}{x+1}$
Solve for $a,b$
Multiply both sides by $(x)(x+1)$
$\frac {1}{x(x+1)}(x)(x+1) = \frac {a}{x}(x)(x+1) + \frac {b}{x+1}(x)(x+1)\\
1 = a(x+1) + bx\\
1 = ax + a + bx\\
1 = (a+b)x + a$
We have an $x$ term on the right and no $x$ term on the left, we might do something like this to create an x term on the left without changing the value of the expression.
$0x + 1 = (a+b)x + a$
If the LHS equals the RHS, then the coefficient of the x terms must be equal. The constant terms must also be equal.
$0x = (a+b)x\\
1 = a$
There was no reason to carry the $x$ along and we could have cut straight to
$0 = a+b\\
1 =a$
One more increase in difficulty.
Suppose we have.
$\frac {1}{x(x^2+1)}$
When we break it into partial fractions, we should assume that the degree of the numerator in each term is one less than the degree of the denominator. That is:
$\frac {1}{x(x^2+1)} = \frac {a}{x} + \frac {bx + c}{x^2 + 1}$
Then we multiply both sides by $(x)(x^2+1)$ as we did before, and set corresponding coefficients equal to each other.
$\frac {1}{x(x^2+1)}(x)(x^2+1) = \frac {a}{x}(x)(x^2+1) + \frac {bx + c}{x^2 + 1}(x)(x^2+1)\\
1 = a(x^2 + 1) + (bx+c) x\\
1 = ax^2 + a + bx^2+ cx\
0x^2 + 0x + 1 = (a+b) x^2 + cx + a\\
0 = a+b\\
0 = c\\
1 = a$
This is exactly what was done in the problem above
$\frac {1-x}{(x+1)(x^2-x+1)}$
We want the numerators to be of one degree less than each term's denominator.
$\frac {1-x}{(x+1)(x^2-x+1)} = \frac {a}{x-1} + \frac {bx + c}{x^2 - x + 1}$
Multiply through by $(x^3 + 1)$ or $(x+1)(x^2 - x + 1)$ if you prefer.
$\frac {1-x}{(x+1)(x^2-x+1)}(x+1)(x^2 - x + 1) = \frac {a}{x+1}(x+1)(x^2 - x + 1) + \frac {bx + c}{x^2 - x + 1}(x+1)(x^2 - x + 1)\\
1-x = a(x^2 - x + 1)+ (bx + c)(x+1)\\
1-x = ax^2 - ax + a + bx^2 + bx + cx + c\\
0x^2 - x + 1 = (a+b)x^2 + (-a+b+c)x + (a+c)\\
0 = a+b\\
-1 = -a+b+c \\
1 = a+c$
I hope this clears things up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
\le e^{\frac{1}{n^2}} + e^{\frac{2}{n^2}} + ...e^{\frac{n}{n^2}} \le
e^{\frac{n}{n^2}} + e^{\frac{n}{n^2}} + ... e^{\frac{n}{n^2}}$$
$$ n e^{\frac{1}{n^2}}
\le \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
ne^{\frac{1}{n}}$$
$$ -n e^{\frac{1}{n}}
\le - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
- n e^{\frac{1}{n^2}}$$
$$ n - n e^{\frac{1}{n}}
\le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
n - n e^{\frac{1}{n^2}}$$
Here I found that the limit of the left-hand side is equal to $-1$, while the limit of the right-hand side is $0$. So I got that:
$$-1 \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le 0$$
And I cannot draw a conclusion about the exact limit. What should I do?
| Using $e^x=1+x+O(x^2)$ along with $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2=O\left(n^3\right)$ , we assert that
$$\begin{align}
\sum_{k=1}^ne^{k/n^2}&=\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)+O\left(\frac{1}{n}\right)\\\\
&=n+\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)\\\\
&=n+\frac12+O\left(\frac1n\right)
\end{align}$$
Hence, we see that
$$\lim_{n\to\infty}\left(n-\sum_{k=1}^ne^{k/n^2}\right)=-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence $(a_n)$ is convergent.
This is what I did:
We have:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
$$a_n = n\sqrt{n^2+n}-an^2+n\sqrt{n^2-n}$$
$$a_n = n^2\sqrt{1+\dfrac{1}{n}} + n^2 \sqrt{1 - \dfrac{1}{n}} - an^2$$
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$
The only way we could have $a_n$ convergent is if we would have the limit result in an indeterminate form. In this case, we need $\infty \cdot 0$. So we have:
$$\lim_{n \to \infty} \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg ) = 0$$
And from that we can conclude that:
$$a = 2$$
So now that I found $a$, I must find the limit of the sequence. So this limit:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
I tried this:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n^2+n}-2n^2+n\sqrt{n^2-n}$$
$$= \lim_{n \to \infty}( n\sqrt{n^2+n} - n^2) + \lim_{n \to \infty} (n\sqrt{n^2-n} -n^2)$$
And then I multiplied both of those limits with its respective conjugate, but after my calculations, it still results in an indeterminate form, only it's $\infty - \infty$ this time.
So, if my previous calculations aren't wrong, my question is how can I find this limit:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
| $\begin{array}\\
\Delta^2_hf(x)
&= f(x-h)-2f(x)+f(x+h)\\
\text{so}\\
\Delta^2_1f(x)
&= f(x-1)-2f(x)+f(x+1)\\
&\approx (f(x)-f'(x)+f''(x)/2-f'''(x)/6+...)-2f(x)+(f(x)+f'(x)+f''(x)/2+f'''(x)/6+...)\\
&\approx f''(x)/2+O(f''''(x))\\
\end{array}
$
If
$f(x) = x^{1/2},\\
f'(x) = x^{-1/2}/2,\\
f''(x) = -x^{-3/2}/4,\\
f'''(x) = 3x^{-5/2}/8,\\
f''''(x) = -15x^{-7/2}/16,\\
$
so
$\Delta^2_1(\sqrt{n})
=-n^{-3/2}/4+O(n^{-7/2})
$
so
$n^{3/2}\Delta^2_1(\sqrt{n})
=-1/4+O(n^{-2})
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
Then $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left(\frac{1}{2^n}-\frac{1}{4^n}\right)=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{4^n}$$
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}$$
$ii$) We found our series can be written as the difference of two geometric series with $|r|<1$, and the solution is
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}=\frac{\frac{1}{4}}{1-\frac{1}{2}}-\frac{\frac{1}{16}}{1-\frac{1}{4}}=\frac{5}{12}$$
I have given this series to an online series calculator, and it claims the result is not $\frac{5}{12}$ but $\frac{2}{3}$, and yet I can not find the mistake in my procedure. Can anyone point it out to me?
EDIT: Please remember, I'm asking more than "what is the right way to solve this problem"; I'm asking precisely what was the mistake I made. Correlated, but not the same.
| Your error in algebra:
You have
$a^n = a^2 a^{n-1}
$
for $a = \frac12$ and
$\frac14$.
It should be
$a^n = a\cdot a^{n-1}
$.
$a^2 a^{n-1}
=a^{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| Starting from
$$\int \frac{x}{x^4+4}\,dx$$
Make the substitution $u=x^2$. Then, $\dfrac{du}{2}=x\,dx$ and the integral becomes
$$\frac{1}{2}\int \frac{1}{u^2+4}\,du$$
Now let $u=2\tan(\theta)$. Then, $du=2\sec^2(\theta)\,d\theta$ and the integral becomes
$$\frac{1}{4}\int \frac{\sec^2(\theta)}{1+\tan^2{(\theta)}}\,d\theta$$
and since $1+\tan^2{(\theta)}=\sec^2(\theta)$, we may rewrite the integral as
$$\frac{1}{4}\int d\theta=\frac{1}{4}\theta+C$$
where $\theta=\arctan\left(\dfrac{u}{2}\right)$ and $u=x^2$, so we conclude that
$$\frac{1}{4}\theta+C=\frac{1}{4}\arctan\left(\frac{u}{2}\right)+C=\frac{1}{4}\arctan\left(\frac{x^2}{2}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove or disprove inequality between 1-norm, 2-norm and infinity norm. Prove or disprove:
$$
v\in R^n \Rightarrow ||v||_1||v||_{\infty}\leq \frac{1+\sqrt n}{2}||v||^2_2.
$$
Can someone help me with this? I have used several cases to verify it but cannot come up with proof.
| Here is a proof which also shows that the inequality is tight.
Instead of $|x_i|$, consider w.l.o.g. $x_i \ge 0$. Further, let $x_{\rm max} = x_n$.
Denote $\sqrt{\sum_{i=1}^{n-1} x_i^2} = ||\bar v||_2$.
First, by Cauchy-Schwarz
$$||v||_1 = \sum_{i=1}^{n-1} 1 \cdot x_i + x_n\le \sqrt{\sum_{i=1}^{n-1} 1} \sqrt{\sum_{i=1}^{n-1} x_i^2} + x_n= \sqrt{n-1} ||\bar v||_2 + x_n
$$
This gives
$$
\frac{||v||_1||v||_{\infty}}{||v||^2_2} \le \frac{\sqrt{n-1} ||\bar v||_2 x_n+ x_n^2}{||\bar v||_2^2 + x_n^2}
$$
Now the function
$$
f(x_n) = \frac{ax_n + x_n^2}{b + x_n^2}
$$
has a maximum value (see appendix below) of $\frac12 (\sqrt{\frac{a^2+b}{b}} +1)$. Inserting $a = \sqrt{n-1} ||\bar v||_2$ and $b = ||\bar v||_2^2$ gives
$$
\frac{||v||_1||v||_{\infty}}{||v||^2_2} \le
\frac12 (\sqrt{\frac{(n-1) ||\bar v||_2^2 +||\bar v||_2^2}{||\bar v||_2^2}} +1) = \frac12 (\sqrt{n} +1)
$$
This proves the claim. $\qquad \Box$
Note that the bound is tight. From the appendix below, the tight bound is taken
at $$x^*_n = \frac{a}{-1+\sqrt{\frac{b+a^2}{b}}} = ||\bar v||_2\frac{\sqrt{n-1}}{-1+\sqrt{n}} $$ Further, the Cauchy-Schwarz inequality above is tight for $x_1 = x_2 = \cdots = x_{n-1}$. Hence we have that $x^*_n = x_1 (1+\sqrt{n})$ is the selection of values where the claim is obeyed with equality. Indeed, one can check that both sides equal:
\begin{align}
{\rm LHS:}&||v||_1||v||_{\infty} = ((n-1)x_1 + x_n^*)x_n^* = x_1^2 (1+{\sqrt{n}})(n+{\sqrt{n}})\\
{\rm RHS:}&\frac{1+\sqrt n}{2}||v||^2_2 = \frac{1+\sqrt n}{2}((n-1)x_1^2+ (x_n^*)^2) \\
& = x_1^2({1+\sqrt n}) \frac{n-1 + (1+\sqrt{n})^2}{2} = x_1^2 (1+{\sqrt{n}})(n+{\sqrt{n}})
\end{align}
Appendix :
The maximum value of the function
$
f(x) = \frac{ax + x^2}{b + x^2}
$
can be obtained without calculus as follows. For simplicity, let $x = ay$ and $b =a^2 c$. Then we require $
f(y) = \frac{y + y^2}{c + y^2} \le m
$ or alternatively
$$0 \le -(y + y^2) + m(c + y^2) = (m-1)(y - \frac{1}{2(m-1)})^2 + mc - \frac{1}{4(m-1)}$$
To guarantee this inequality, the tight case is that the square term gets zero, and also $mc - \frac{1}{4(m-1)}$ (which is unbounded w.r.t. $m$) gets zero. This gives $m = \frac12 (1 + \sqrt{\frac{c+1}{c}})$. Reinserting $c = b/a^2$ gives the highest value of $f(x)$ which is $\frac12 (\sqrt{\frac{a^2+b}{b}} +1)$, as stated above. The bound is then taken where the square term gets zero, i.e. at $x^* = a y = \frac{a}{2(m-1)} = \frac{a}{-1+\sqrt{\frac{c+1}{c}}} = \frac{a}{-1+\sqrt{\frac{b+a^2}{b}}} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove that both sides are greater than $12$ using AM-GM:
$$LHS \geq \frac{3 \cdot 4ab \cdot 4bc}{4a^2bc} = 12$$
and
$$RHS \geq 7+\frac{5[(a+b)^2+(b+c)^2]}{2(a+b)(b+c)} \geq 7+5 = 12$$
So, substract $12$ from both sides and write the inequality into:
$$3\cdot \frac{(a+b)^2(b+c)^2-16ab^2c}{4ab^2c} \geq 5 \cdot \frac{a^2+b^2+c^2-ab-bc-ca}{(a+b)(b+c)}$$
or
$$3\cdot \frac{(b+c)^2(a-b)^2+4ab(b-c)^2}{4ab^2c} \geq \frac{5}{2}\cdot \frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b)(b+c)}$$
My next idea was to use $(c-a)^2\leq 2[(a-b)^2+(b-c)^2]$ and write it into a sum of square form with only $(a-b)^2$ and $(b-c)^2$. However, I couldn't reach significant progress.
| Another proof:
Since the inequality is symmetric in $a$ and $c$, WLOG, assume that $a \ge c$.
Due to homogeneity, assume that $c = 1$.
Let $a = 1 + s$ for $s \ge 0$.
We split into two cases:
1) $0 < b \le 1$: Let $b = \frac{1}{1+t}$ for $t \ge 0$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(2+t)(st+s+t+2)(1+s)(1+t)^2}f(s,t)
\end{align}
where
\begin{align}
f(s,t) &= 3 s^3 t^6+27 s^3 t^5+9 s^2 t^6+79 s^3 t^4+90 s^2 t^5+9 s t^6+109 s^3 t^3\\
&\quad +281 s^2 t^4+99 s t^5+3 t^6+78 s^3 t^2+412 s^2 t^3+314 s t^4+36 t^5\\
&\quad +28 s^3 t+324 s^2 t^2+452 s t^3+112 t^4+4 s^3+140 s^2 t+304 s t^2\\
&\quad +152 t^3+28 s^2+76 s t+76 t^2.
\end{align}
Clearly, $f(s,t) \ge 0$. True.
2) $b > 1$: Let $b = 1+r$ for $r > 0$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(1+s)(1+r)^2(2+s+r)(2+r)}g(s,r)
\end{align}
where
\begin{align}
g(s, r) &= 3 r^6+9 r^5 s+9 r^4 s^2+36 r^5+22 r^4 s+44 r^3 s^2\\
&\quad +112 r^4+4 r^3 s+44 r^2 s^2+76 r^3+9 r s^2+9 s^2+\tfrac{15}{16} s^3\\
&\quad + 19 (2 r-s)^2+19 r (2 r-s)^2+3 (r-1)^2 r s^3+4 s^3 (r-\tfrac78)^2.
\end{align}
Clearly, $g(s,r) > 0$. True.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^{x^2} - 1 \sim x^2 \,\,\,\,\text{when}\,\,\,\,x\rightarrow 0,$$
and then
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2xe^{x^2}-2x+\frac{2}{3}x^3}=\lim_{x\to 0}\frac{x^4}{4(e^{x^2}-1)+\frac{4}{3}x^2}\\
&=\lim_{x\to 0}\frac{x^4}{4x^2+\frac{4}{3}x^2}=\lim_{x\to 0}\frac{3}{16}x^2=0.
\end{align*}
Later, inspecting this other approach combining infinitesimals and the Taylor expansions
\begin{align*}
e^{x^2} &= 1+x^2+\frac{x^4}{2}+o(x^4),\\
\sin 2x &= 2x - \frac{4x^3}{3} + \frac{4x^5}{15} + o(x^5),\\
e^{x^2}\sin 2x &= 2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5),
\end{align*}
I get this other result
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5)-2x+\frac{2}{3}x^3}\\
&=\lim_{x\to 0}\frac{x^5/2}{-x^5/15 + o(x^5)}=-\frac{15}{2}.
\end{align*}
Can someone help me to identify what am I getting wrong?
Thanks.
| Both computations are incorrect. The first method amounts to writing
\begin{equation}
\begin{array}{l}
1 - \cos(x^2) = \frac{x^4}{2} + o(x^4)\cr
\arctan(x) = x + o(x)\cr
\sin(2x)= 2x + o(x)\cr
e^{x^2}= 1 + x^2 + o(x^2)
\end{array}
\end{equation}
but substituting these values in the fraction cannot tell us the limit because it gives
\begin{equation}
\frac{\frac{x^5}{5}+ o(x^5)}{o(x)}
\end{equation}
which limit is not known.
In the second method, as @JoséCarlosSantos pointed out, there is a wrong sign. It is most probable that the sign before $\frac{2}{3}x^3$ is already wrong in the initial fraction. Apart from this point, the second method should work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $1−a\cos x−b\sin x−A\cos2x−B\sin2x\geq0$ for all real $x$, then $a^2+b^2\leq2$ and $A^2+B^2\leq1$
Four real constants $a$, $b$, $A$, $B$ are given, and
$$
f(x) = 1 − a\cos(x) − b\sin(x) − A\cos(2x) − B\sin(2x)
$$
Prove that if $f(x)\geq 0$ for all real $x$, then
$$
a^2+b^2\leq 2 \quad \mbox{and} \quad A^2 + B^2 \leq 1
$$
Edit: I know that
$$-\sqrt{a^2+b^2}\leq a\sin(x)+b\cos(x) \leq \sqrt{a^2+b^2}$$
But I don't know how to apply this one to prove the question.
| Observe
\begin{align}
a \cos()+b\sin()
&= \sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos(x)+\frac{b}{\sqrt{a^2+b^2}}\sin(x) \right)\\
&= \sqrt{a^2+b^2} \sin(x-\theta_0)
\end{align}
for some $\theta_0$ and likewise
\begin{align}
A \cos(2)+B\sin(2) = \sqrt{A^2+B^2} \sin(2x-\phi_0).
\end{align}
Then we see that
\begin{align}
f(x)= 1-\sqrt{a^2+b^2} \sin(x-\theta_0)-\sqrt{A^2+B^2} \sin(2x-\phi_0)\geq 0
\end{align}
for all $x$.
Set $x=\theta_0+\frac{3\pi}{4}$ then, we see that
\begin{align}
1-\frac{1}{\sqrt{2}}\sqrt{a^2+b^2}+\sqrt{A^2+B^2}\cos(2\theta_0-\phi_0)\geq 0.
\end{align}
If we also set $x=\theta_0+\frac{\pi}{4}$ then we get
\begin{align}
1-\frac{1}{\sqrt{2}}\sqrt{a^2+b^2}-\sqrt{A^2+B^2}\cos(2\theta_0-\phi_0)\geq 0.
\end{align}
Finally, we see that
\begin{align}
2-\frac{2}{\sqrt{2}}\sqrt{a^2+b^2} \geq 0
\end{align}
which is the desired inequality.
The other inequality is derived in a similar manner, i.e. set $x=\frac{\phi_0}{2}+\frac{\pi}{4}$ and $ \frac{\phi_0}{2}+\frac{5\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unifying the sum of two error functions. Let's propose you have the sum of two error functions:
$f(x) = \text{erf}(ax)+\text{erf(bx)}$
If you wanted to solve for x, you'd first unify them into a third function
$f(x) = \text{erf}(cx)$
Or something similar. How could you approach this problem?
| As already said in comments and answer, you will need some numerical method for solving for $x$ the nonlinear equation
$$y=\text{erf}(a x)+\text{erf}(b x)$$
I shall consider the case where $a$ and $b$ are positive (the other situations being more complex).
The first point is to notice that $y$ is bounded by $2\text{erf}(a x)$ and $2\text{erf}(b x)$.
Now, to have bounds, we could use the simple approximation
$$\text{erf}(t)\approx \sqrt{1-\exp\Big(-\frac 4 {\pi}t^2 \Big)}$$ which makes the solution to be between
$$\frac{\sqrt{\pi }}{2} \sqrt{-\frac{\log \left(1-\frac{y^2}{4}\right)}{a^2}}\qquad \text{and} \qquad \frac{\sqrt{\pi }}{2} \sqrt{-\frac{\log \left(1-\frac{y^2}{4}\right)}{b^2}}$$
Start at the midpoint and use Newton method. Using for example $a=0.5$, $b=1.5$ and $y=1.2345$, the iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.81835949 \\
1 & 0.67741900 \\
2 & 0.69292373 \\
3 & 0.69314665 \\
4 & 0.69314670
\end{array}
\right)$$
What we could also do is to expand as Taylor series
$$y=\frac{2 x (a+b)}{\sqrt{\pi }}-\frac{2 x^3 \left(a^3+b^3\right)}{3 \sqrt{\pi
}}+\frac{x^5 \left(a^5+b^5\right)}{5 \sqrt{\pi }}-\frac{x^7
\left(a^7+b^7\right)}{21 \sqrt{\pi }}+O\left(x^9\right)$$ and use series reversion to get
$$x=t+\alpha t^3+\beta t^5+\gamma t^7+O\left(t^9\right)\qquad \text{and} \qquad t=\frac{\sqrt{\pi } }{2 (a+b)}y$$
$$\alpha=\frac{1}{3} \left(a^2-a b+b^2\right)$$
$$\beta=\frac{1}{30} \left(7 a^4-17 a^3 b+27 a^2 b^2-17 a b^3+7 b^4\right)$$
$$\gamma=\frac{1}{630} \left(127 a^6-519 a^5 b+1191 a^4 b^2-1471 a^3 b^3+1191 a^2 b^4-519 a b^5+127 b^6\right)$$
For the worked example, this would give, as an estimate to start with, $x=0.685654$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find:$\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$
Evalute: $$\lim_{x\to
0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$
My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},\;\;\lim_{x\to 0}\frac{\tan x}{x}=1,\;\;\lim_{x\to 0}\frac{e^x-1}{x}=1$$
$$$$
$L=\displaystyle\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}=$
$$$$
$\displaystyle\lim_{x\to 0}\left[\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{e^{1-\cos^3x}-e^{1-\cos^4x}}\cdot\left(\frac{e^{1-\cos^3x}-1}{1-\cos^3x}\cdot\frac{1-\cos^3x}{x^2}-\frac{e^{1-\cos^4x}-1}{1-\cos^4x}\cdot\frac{1-\cos^4x}{x^2}\right)\cdot\frac{x}{\arctan x}\right]$
Substitution: $$[t=\arctan x\implies x=\tan t\;\;\&\;\; x\to 0\implies t\to 0]$$
$$\lim_{x\to 0}\frac{x}{\arctan x}\iff\lim_{t\to 0}\frac{\tan t}{t}=1$$
The next step:$$1-\cos^3x=(1-\cos x)(1+\cos x+\cos^2x)$$
$$1-\cos^4x=(1-\cos x)(1+\cos x)(1+\cos^2x)$$
Now I obtained:
$$L=1\cdot\left(1\cdot\frac{3}{2}-1\cdot 2\right)\cdot 1=-\frac{1}{2}$$
Is this correct?
| Here is another way after your first step (removing the $\sin() $). Note that if $A, B$ tend to $1$ then $$A-B=B\cdot\frac{\exp(\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and the first two factors above tend to $1$. Therefore $A-B$ can be replaced by $\log A - \log B$.
Thus the numerator in your case gets replaced by $$(1-\cos^3x)-(1-\cos^4x)=-\cos^3x(1-\cos x) $$ and therefore we can safely replace it by $-(1-\cos x) $. The denominator $x\arctan x$ can be replaced by $x^2$ because of limit $\lim_{x\to 0}\dfrac {\arctan x} {x} =1$. It should be now clear that the desired limit is $-1/2$.
See the usage of this technique here, here and here.
| {
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"url": "https://math.stackexchange.com/questions/3523158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find cube roots of 1 modulo power of two (if they exist)? It would be useful for efficiently implementing an algorithm if I could find a $c > 1$ where $c^3 \equiv 1 \pmod {2^{64}}$. It's plausible that such a $c$ exists because $2^{64} \equiv 1 \pmod 3$, so all non-zero values could be partitioned into groups of three (each $x$ along with $cx \pmod {2^{64}}$ and $c^2 x \pmod {2^{64}}$).
Is there a known way to find these more efficiently than brute force (probably infeasible, no solutions for $2^{32}$)? Or is it known to have no solution? Or are there known solutions, perhaps for other powers of two between $2^{32}$ and $2^{64}$?
| You can do this by induction. We begin with $1^3 \equiv 1 \pmod {2^k}$ where the start is $k=1.$ In paerticular, there are no others. Do we get any additional roots as $k$ increases? Just two choices.
This works: $1^3 \equiv 1 \pmod {2^{k+1}}$
Maybe:
$$ (1+2^k)^3 = 1 + 3 \cdot 2^k + 3 \cdot 2^{2k} + 2^{3k} \equiv 1 + 3 \cdot 2^k \pmod {2^{k+1}} \; , \; $$
so that
$$ (1+2^k)^3 \equiv 1 + 2^k \pmod {2^{k+1}} \; , \; $$
because $2 \cdot 2^k \equiv 0 \pmod {2^{k+1}}$ This second choice fails, as the exponent of $2$ increases, the only cube root of $1$ remains $1$
The alternatives that fail to give cube roots of one resemble this:
$$ 3^3 = 27 \equiv 3 \pmod 4 $$
$$ 5^3 = 125 \equiv 5 \pmod 8 $$
$$ 9^3 = 729 \equiv 9 \pmod {16} $$
$$ 17^3 = 4913 \equiv 17 \pmod {32} $$
$$ 33^3 = 35937 \equiv 33 \pmod {64} $$
$$ 65^3 = 274625 \equiv 65 \pmod {128} $$
$$ 129^3 = 2146689 \equiv 129 \pmod {256} $$
$$and \; \; so \; \; on...$$
This procedure comes under the name of Hensel Lifting. And this gives a complete proof that there is just one cube root of one $\pmod {2^{64}}$
| {
"language": "en",
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"source": "stackexchange",
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Generating function of binary string with minimum lengths for 0s and 1s a) Show that the generating function by length for binary strings where every block of 0s has length at least 2, each block of ones has length at least 3 is:
$$\frac{(1-x+x^3)(1-x+x^2)}{1-2x+x^2-x^5}$$
b) Give a recurrence relation and enough initial conditions to determine coefficients of power series.
So for a), I came up with the block decomposition $((0^*00)^*(1^*111)^*)^*$ and found the generating function, using the fact that $0\leadsto x$, $1\leadsto x$, and $a^*\leadsto\frac{1}{1-a}$ where a is some binary string:
$$\frac{(1-x)^2}{(1-x-x^2)(1-x-x^3)}$$
which, clearly, does not equal the expected result. Could someone clear up for me where I went wrong?
Also, for b), how would you find a recurrence relation, since the degree of the numerator and denominator are the same, so there would be no general $a_n$ term.
Thanks in advance for any help!
| robjohn has given a generating function. Here is a recurrence based approach.
Suppose $b_n$ is the number of strings ending with $0$ and $c_n$ the number ending with $1$, and $a_n=b_n+c_n$ being what you want. It seems obvious that
*
*$a_n=b_n+c_n$ by definition
*$b_n=b_{n-1}+c_{n-2}$ by adding $0$ to an existing $0$ or $00$ to an existing $1$
*$c_n=b_{n-1}+c_{n-1}$ by adding $1$ to an existing $0$ or $1$
so
*$c_n=a_{n-1}$ from (3) and (1) and so $c_{n-2}=a_{n-3}$
*$b_{n-1}=c_n-c_{n-1}=a_{n-1}-a_{n-2}$ from (3) and (4) and so $b_{n}=a_{n}-a_{n-1}$
*$a_{n}-a_{n-1}=a_{n-1}-a_{n-2}+a_{n-3}$ from (2) and (5) and (4) giving $$a_{n}=2a_{n-1}-a_{n-2}+a_{n-3}$$
That will lead to a denominator in the generating function of $1-2x+x^2-x^3$
Since $b_1=0, b_2=1, c_1=1, c_2=1$, we can find $a_0=1, a_1=1, a_2=2, a_3=4, a_4=7$ etc. But $\frac{1}{1-2x+x^2-x^3}=1+2x+3x^2+5x^3+9x^4+\cdots$ so we need to subtract $\frac{x}{1-2x+x^2-x^3} = x+2x^2+3x^3+5x^4+\cdots$ and then add $\frac{x^2}{1-2x+x^2-x^3}= x^2+2x^3+3x^4+\cdots$ to match the coefficients, with a result of $$\frac{1-x+x^2}{1-2x+x^2-x^3}$$ as robjohn found more directly
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the trigonometric equation $483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$
I'm trying the value of $\alpha \in [0,\pi]$ that is solution to this trigonometric equation: $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
I've tried to write down $\sin\left(\alpha+\frac\pi{3}\right)$ and $\sin\left(2\alpha+\frac{\pi}{3}\right)$with the formula $\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha$, but after that I'am stuck and I don't have completely any idea of how to proceed.
| Starting from @the_candyman's answer
$$\begin{cases}
\frac{483}{2}Y + \frac{483\sqrt{3}}{2}X + 16\sqrt{3}XY + 48X^2 - 4 = 0 \\
X^2 + Y^2 = 1
\end{cases}$$
eliminate $Y$ from the first equation
$$Y=\frac{-96 X^2-483 \sqrt{3} X+8}{32 \sqrt{3} X+483}$$ Plug it in the second to end with
$$12288 X^4+123648 \sqrt{3} X^3+928548 X^2-38640 \sqrt{3} X-233225=0$$ which can be solved exactely using radicals.
The formulae are messy; the usual tests for quartic equations shows only two reals roots which numerically are
$$X_1=-0.492379182948765 \qquad \text{and} \qquad X_2=+0.506762587904079$$
$$Y_1=+0.870380801832569 \qquad \text{and} \qquad Y_2=-0.862085656707476$$
The solution for $\alpha \in [0,\pi]$ then corresponds to $(X_i,Y_1)$.
In fact, the solution is so close to $\frac 23 \pi$ that we could make one single Newton iteration for the original equation
$$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
$$\alpha\sim\frac 23 \pi+\frac{4}{16 \sqrt{3}-483}$$
| {
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value of expression having variables $p,q,r,x,y,z$
If $p,q,r,x,y,z$ are non zero real number such that
$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$
Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is
what try
$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$
$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2$
$2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2$
How do i solve it Help me please
| By C-S $$0=px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}\geq px+qy+rz+|px+qy+rz|$$ and since
$$px+qy+rz+|px+qy+rz|\geq0,$$ we obtain $$px+qy+rz+|px+qy+rz|=0,$$ which gives
$$px+qy+rz\leq0.$$
Also, the equality occurs for $$(x,y,z)||(p,q,r),$$ which says that there is $k<0$, for which
$$(p,q,r)=k(x,y,z).$$
Thus, $$\sum_{cyc}\frac{py}{qx}=\sum_{cyc}\frac{kxy}{kyx}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $x^2 + ax + b > 0$ and $x^2 + (a + np)x + (b + nq) > 0$, prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
Given that $$\large \left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R \ (p, q \in \mathbb R \setminus \{0\}, n \in \mathbb N, n > 1)$$
Prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
We have that $$\left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R$$
$$ \iff \left\{ \begin{align} a^2 - 4b < 0 \ (1)\\ (a + np)^2 - 4(b + nq) < 0 \ (2) \end{align} \right.$$
For $(2)$, we obtain that $p^2n^2 + 2(ap - 2q)n + (a^2 - 4b) < 0$
$$\implies (ap - 2q)^2 - p^2(a^2 - 4b) > 0 \iff q^2 - apq + bp^2 > 0 \implies \left\{ \begin{align} (ap)^2 - 4bp^2 < 0\\ (aq)^2 - 4q^2 < 0 \end{align} \right.$$
$\implies \left\{ \begin{align} a^2 - 4b < 0\\ a^2 - 4 < 0 \end{align} \right.$. And that's a dead-end.
| U could always check out D.See if leading coefficient >0 and D is less than 0 the quadratic is >0 for all real x.
Given : $ a^2 - 4b <0$
Given : $ (a+np)^2 - 4(b+nq)= a^2-4b + n(np^2 +2(ap-2q))<0$
(as n decreases $np^2 +2(ap-2q)$ also decreases) $\Rightarrow$ the expression is lesser for lesser 'n' and hence less than 0.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and replace $y$ in the quadratic equation with $2x-t$ to get a function but then, I get confused.
| By C-S:
$$3=\sqrt{2x^2-4xy+6y^2}=\frac{2}{3}\sqrt{\left(2(x-y)^2+4y^2\right)\left(2+\frac{1}{4}\right)}\geq$$
$$\geq\frac{2}{3}\sqrt{(2(x-y)+y)^2}=\frac{2}{3}|2x-y|,$$ which gives
$$-\frac{9}{2}\leq2x-y\leq\frac{9}{2}.$$
The equality occurs for $$\left(\sqrt2(x-y),y\right)||\left(\sqrt2,\frac{1}{2}\right),$$ which with the condition gives
$$(x,y)=\left(\frac{5}{2},\frac{1}{2}\right)$$ for the right inequality and
$$(x,y)=\left(-\frac{5}{2},-\frac{1}{2}\right)$$ for the left inequality,
which says that we got a maximal and the minimal value.
| {
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If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p {\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q. If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p{\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q.
So I take the second term $4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2}) (r)$ , where r is the common ratio.
$4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2})( 2{\sqrt3} + {\sqrt2 })$
And found that the common ratio, r = $2{\sqrt3} + {\sqrt2 }$
To find the third term, I multiplied the second term with the common ratio.
$(4-{\sqrt 6})( 2{\sqrt3} + {\sqrt2 })= p{\sqrt 3} - q {\sqrt 2}$
$8{\sqrt 3} + 4{\sqrt2} - 6 {\sqrt 2} - 2{\sqrt 6} = p{\sqrt 3} - q {\sqrt 2}$
I am unable to proceed beyond this step.
| Note
$$(4- {\sqrt 6})^2=({\sqrt 3} - {\sqrt 2})( p {\sqrt 3} - q {\sqrt 2})$$
or,
$$22-8\sqrt6 = 3p +2q -(p+q)\sqrt6$$
Therefore,
$$22= 3p +2q,\>\>\>\>\>p+q=8$$
Solve to obtain $p=6$ and $q=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \sin{x}\ +\ \frac{1}{2}\sin{2x}\ +\ \frac{1}{3}\sin{3x}\ +\ \frac{1}{4}\sin{4x}\ =\ \frac{2}{3}\left(\cos{x}+1\right)\left(\sin^5x\ +\ 4\right)$ SORRY IF MY TITTLE IS UNCLEAR WITH ONLY MATH FUNCTIONS. IT'S MORE THAN 150 CHARACTERS
*
*This is my math problem
$$ \sin {x}\ +\ \frac{1}{2}\sin {2x}\ +\ \frac{1}{3}\sin {3x}\ +\ \frac{1}{4}\sin {4x}\ =\ \frac{2}{3}\left(\cos {x}+1\right)\left(\sin ^5x\ +\ 4\right) \left(*\right)
$$
*
*This is my effort
$$ \left(*\right) <=> 12\sin {x\left(1+\cos x\right)\ +\ 12\sin {x}\ -\ 16\sin ^3x\ +\ 12\sin {\left(x\right)\cos {\left(x\right)}\cos {\left(2x\right)}}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)
$$
$$ <=> 12\sin {x\left(1+\cos x\right)\ + 4\sin {x\left(2\cos {\left(2x\right)\left(1+\cos {x}\right)\\ +\ \left(1+\cos {x}\right)\left(2\cos ^2x\ -\ 2\cos {x\ +\ 1}\right)}\right)}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)
$$
$$ <=> 4\sin {x\left(1+\cos {x}\right)\left(6\cos ^2x\ -\ 2\cos {x\ +\ 2}\right)\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)}
$$
Now
$$\cos {x} = -1
$$
or
$$ 6\sin {x}\cos ^2x\ -\ 2\cos {x}\sin {x}\ +\ 2\sin {x}\ -\ 2\sin ^5x\ -\ 8\ =\ 0
$$
To here I tried many ways like converting all to \sin , group somes together but it didn't work. Please give me some HINTS
| An algorithmic approach for many of these problems can be:
You can call $t=\tan(\frac{x}{2})$. Then $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t}{1+t^2}$. Making this substitution your equation
$$6\sin(x)cos^2(x) − 2\cos(x)\sin(x) + 2\sin(x) − 2\sin^5(x) − 8 = 0$$
becomes a polynomial equation.
$$6(2t)(1-t)^2(1+t^2)^2− 2(1-t)(2t)(1+t^2)^3 + 2(2t)(1+t^2)^4−2(2t)^5 − 8(1+t^2)^5=0$$
With Sturm's theorem we check that this polynomial doesn't have real solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Splitting field of $\alpha=(2+\sqrt{2})^{1/3}$ Given $\alpha=(2+\sqrt{2})^{1/3}$ The problem asks to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and find it's splitting field. I have the solution and I understood until the point where it shows that the minimal polynomial is $$x^6-4x^3+2=0$$
Now, since we got this polynomial developing the equation $\alpha=(2+\sqrt{2})^{1/3}$ it is clear that $(2+\sqrt{2})^{1/3}$ is a root of the minimal polynomial.
I don't understand the following: Why are $\left( 2-\sqrt{2} \right) ^{1/3},\left( 2-\sqrt{2} \right) ^{1/3}w,\left( 2-\sqrt{2} \right) ^{1/3}w^2,\left( 2+\sqrt{2} \right) ^{1/3}w,\left( 2+\sqrt{2} \right) ^{1/3}w^2
$ the other roots of the polynomial? $w$ is the cubic root of unity ($w^3=1$).
| First, remember that if $\alpha$ is a (real or complex) root of $x^n-a=0$, $a\neq 0$, then the other roots are all of the form $\lambda^i\alpha$, where $\lambda$ is a primitive $n$th root of unity. That is, a complex number such that $\lambda^n=1$ but $\lambda^k\neq 1$ for $1\leq k\leq n-1$. This follows simply by noting that all of them are roots of $x^n-a$, they are all distinct, since $\lambda^k\alpha = \lambda^j\alpha$ implies $\lambda^k=\lambda^j$, and therefore that $\lambda^u=1$ where $u=|k-j|$. That means $n|k-j$, and so you get equality if both $k$ and $j$ are between $0$ and $n-1$.
Now, your polynomial is
$$x^6 - 4x^3 + 2.$$
This is a quadratic polynomial in $x^3$. So you can easily find the two values of $x^3$ that are roots. Call them $a_1$ and $a_2$. So then you want to solve $x^3=a_i$, or equivalently find a root of $x^3-a_i$, $i=1,2$.
Since $Y^2 - 4Y + 2= 0$ has roots $2+\sqrt{2}$ and $2-\sqrt{2}$, then you want to find the solutions to
$$x^3 - (2+\sqrt{2})\qquad\text{and}\qquad x^3-(2-\sqrt{2}).$$
One root of the first is $\sqrt[3]{2+\sqrt{2}}$; so the three roots are $\sqrt[3]{2+\sqrt{2}}$, $\omega\sqrt[3]{2+\sqrt{2}}$, and $\omega^2\sqrt[3]{2+\sqrt{2}}$.
Similarly, since $\sqrt[3]{2-\sqrt{2}}$ is a root of the second polynomial, the other two roots are $\omega\sqrt[3]{2-\sqrt{2}}$ and $\omega^2\sqrt[3]{2-\sqrt{2}}$.
| {
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Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$ I have to solve the following integral:
$$\int\frac{1}{(x^4+1)^2}dx$$
I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex roots and got stuck. Is there a trick for this kind of integral or some kind of helpful substitution? Thanks.
EDIT:
I did the following:
Let $x^2=\tan\theta$, then $x = \sqrt{\tan\theta}$ and $dx=\frac{\sec^2\theta}{2x}d\theta$
Then:
$$I=\int\frac{1}{(x^4+1)^2}dx = \int\frac{1}{(\tan^2\theta+1)^2} \frac{\sec^2\theta}{2x}d\theta=\int\frac{1}{\sec^4\theta} \frac{\sec^2\theta}{2x}d\theta$$
$$I=\frac{1}{2}\int{\frac{1}{\sec^2\theta \sqrt{\tan\theta}}}d\theta$$.
After this I don't know how to proceed.
| Use $\left(\frac x{x^4+1}\right)' = -\frac3{x^4+1} + \frac 4{(x^4+1)^2} $ to rewrite the integral as
$$I = \int \frac 1{(x^4+1)^2}dx=\frac x{4(x^4+1)}+\frac34\int\frac1{x^4+1} dx$$
where
\begin{align}\int\frac2{x^4+1} dx = &\int\frac{1+x^2}{x^4+1} dx + \int\frac{1-x^2}{x^4+1} dx\\
= &
\int\frac{d(x-\frac1{x})}{(x-\frac1{x})^2+2} - \int\frac{d(x+\frac1{x})}{(x+\frac1{x})^2-2}\\
=&\ \frac1{\sqrt2} \tan^{-1}\frac{x^2-1}{\sqrt2x} + \frac1{\sqrt2} \coth^{-1}\frac{x^2+1}{\sqrt2x}
\end{align}
Thus
$$I = \frac x{4(x^4+1)}+\frac3{8\sqrt2} \tan^{-1}\frac{x^2-1}{\sqrt2x} + \frac3{8\sqrt2} \coth^{-1}\frac{x^2+1}{\sqrt2x} + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which arose from the deduction that $\frac{1}{4}\sqrt{\frac{256\sin^{4}40^{\circ}-80\sin^{2}40^{\circ}+12-\ 8\sqrt{3}\sin40^{\circ}}{\left(16\sin^{4}40^{\circ}-4\sin^{2}40^{\circ}+1\right)}}=\cos50^{\circ}$. Despite the apparent simplicity of the relationship, it seems quite tricky to prove. I managed to prove it by solving the equation as a quadratic in $(\sin\frac{\pi}{9})$ and then using the identity $\sqrt{\sec^2 x-1}=|\tan x|$, the double angle formulae and finally that $\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x$ can be written in the form $\sin\left(x+\frac{2\pi}{3}\right)$.
But it seems like quite a neat problem. So, does anyone have a better way of proving it?
| Evaluate
$$
\begin{aligned}
4\cos ^2\frac{π}{9}&\left( LHS-RHS \right) \\
= \ &16\sin ^2\frac{π}{9}\cos ^2\frac{π}{9}-8\sqrt{3}\sin \frac{π}{9}\cos ^2\frac{π}{9}+4\cos ^2\frac{π}{9}-1\\
=\ &4\sin ^2\frac{2π}{9}-4\sqrt{3}\sin \frac{2π}{9}\cos \frac{π}{9}+2\left( 1+\cos \frac{2π}{9} \right) -1 \\
= \ &2\left( 1-\cos \frac{4π}{9} \right) -2\sqrt{3}\left( \frac{\sqrt3}2+\sin \frac{π}{9} \right) +2\cos \frac{2π}{9}+1\\
= \ &2\cos \frac{2π}{9}-2\cos \frac{4π}{9}-2\sqrt{3}\sin \frac{π}{9}\\
= \ &4\sin \frac{π}{3}\sin \frac{π}{9}-2\sqrt{3}\sin \frac{π}{9}=0
\end{aligned}
$$
where $\sin\fracπ3 = \frac{\sqrt3}2$ is used.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{(n-2)}]$$$$\implies\lim_
{n\to\infty}(-1)^n\cdot\frac{1}{3}[1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+...] $$
From the options it was evident that the sum had a relation with $\log_{e}2$, however I couldn't manage to solve the problem. What was I missing?
| Like Steven Sadnicki said, you factored out the $(-1)^n$ term even though it's inside the sum, so the summands are of alternating sign. Here's a solution:
$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 + n - 2 } = \sum_{n=2}^{\infty} \frac{(-1)^n}{(n-1)(n+2)} = \sum_{n=2}^{\infty} (-1)^n [ \frac{1}{3} \cdot ( \frac{1}{n-1} - \frac{1}{n+2})] = \frac{1}{3} \cdot \lim_{n \to \infty} (1 - \frac{1}{4} - \frac{1}{2} + \frac{1}{5} + \frac{1}{3} - \frac{1}{6} - \frac{1}{4} + \frac{1}{7} + \frac{1}{5} - \dots) = \frac{1}{3} \lim_{n \to \infty}(1 - \frac{1}{2} + \frac{1}{3} + 2\sum_{n=4}^{\infty}\frac{(-1)^{n-1}}{n}) = \dots = -\frac{5}{18} - \frac{2}{3}\ln{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
| Method 1. For any numbers $a,b,c$ let $F(c)=c^3-3abc-(a^3+b^3).$ Observe that $F(a+b)=0$ so $c-(a+b)$ is a factor of the polynomial $F(c).$ Using synthetic division we obtain $$F(c)=[c-(a+b)]\cdot [c^2+c(a+b)+(a^2-ab+b^2)].$$ Now if $a,b,c$ are positive and $c<a+b$ then $$c-(a+b)<0$$ while $$c^2+c(a+b)+(a^2-ab+b^2)>(a^2-ab+b^2)=(a-b)^2+ab>0,$$ so $F(c)<0.$
Method 2. The inequality is obvious if $c\le \max (a,b).$ With $F(c)$ as in Method 1., with positive $a,b:$ We have $F'(c)=3(c^2-ab)>0$ whenever $c>\max(a,b),$ so $F(c)$ is strictly increasing for $c>\max (a,b).$ So if $a+b>c>\max(a,b)$ then $F(c)<F(a+b)=0.$
| {
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Find the values of $a$ and $b$ if the limits exist. Could someone solve this
$$f(x) =
\begin{cases}
x^3+a, & x<0 \\
a\sin\frac{\pi{x}}{3}+b, & 0\leq {x}<2 \\
3, & x=2 \\
\log_2x^{b+1}, & x>2\\
\end{cases}$$
If $\lim_{x\to0}$ and $\lim_{x\to2}$ both exist, find the value of a and b.
| We have $$\lim_{x\rightarrow 0^-}f(x)=a; \lim_{x\rightarrow 0^+}f(x)=b;$$ $$\lim_{x\rightarrow 2^-}f(x)=b+a\frac{\sqrt{3}}{2}; \lim_{x\rightarrow 2^+}f(x)=b+1$$
Because $\exists \lim_{x\rightarrow 0}f(x)$ and $\exists \lim_{x\rightarrow 2}f(x)$, then
$a=b$ and $b+1=b+a\frac{\sqrt{3}}{2}$ so $a=b=\frac{2\sqrt{3}}{3}$
| {
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Find a $3\times 3$ matrix $P$ such that $g_{j} = \sum_{i=1}^{3}P_{ij}f_{i}$ Let $V$ be the vector space over the complex numbers of all functions from $\textbf{R}$ into $\textbf{C}$, i.e., the space of all complex-valued functions on the real line. Let $f_{1}(x) = 1$, $f_{2}(x) = e^{ix}$ and $f_{3}(x) = e^{-ix}$.
(a) Prove that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.
(b) Let $g_{1}(x) = 1$, $g_{2}(x) = \cos(x)$ and $g_{3}(x) = \sin(x)$. Find a $3\times 3$ matrix $P$ such that
\begin{align*}
g_{j} = \sum_{i=1}^{3}P_{ij}f_{i}
\end{align*}
MY ATTEMPT
(a) The given functions are indeed linear independent. In fact, we have
\begin{align*}
& af_{1} + bf_{2} + cf_{3} = a + b\cos(x) + c\cos(x) + i(b\sin(x) - c\sin(x)) = 0
\end{align*}
If $x = 0$, then $a + b + c = 0$. If $x = \pi/2$, then $b = c$ and $a = 0$. Consequently, $a = b = c = 0$, from whence we conclude that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.
(b) Such sought matrix $P$ is the change of basis matrix, that is to say
\begin{align*}
P = \begin{bmatrix}
1 & 0 & 0 \\
0 & \frac{1}{2} & \frac{1}{2i} \\
0 & \frac{1}{2} & -\frac{1}{2i}
\end{bmatrix}
\end{align*}
Could someone double-check if the matrix $P$ is expressed correctly?
| (a) Your reasoning is not sound. Scalars $a,b,c$ are complex, so $a = -1 - i$, $b = 1$, $c = i$ satisfy your two equations also. A hint is that you are on the right track—just go one step further.
(b) I got the same $P$.
| {
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In how many ways can $6$ prizes be distributed among $4$ persons such that each one gets at least one prize? My understanding:
First select $4$ prizes and distribute it among the $4$ people in $^6C_4\times4!$ ways and then distribute the remaining $2$ prizes in two cases: when $2$ people have $2$ prizes: $\frac{4!}{2!}$ways or when one person has $3$ prizes: $4$ ways. So totally: $^6C_4\times4!\times( \frac{4!}{2!}+4)$ ways $= 5760$ ways.
However the answer is $1560$ ways.
How?
| We can also count this using a generating function approach.
Since the prizes are not identical, we compute the exponential generating function. To remove the possibility of anyone getting no prizes we use $e^x-1$ in each factor. The number of ways to distribute $6$ prizes among $4$ people is
$$
\begin{align}
6!\left[x^6\right]\left(e^x-1\right)^4
&=6!\left[x^6\right]\left(\binom44e^{4x}-\binom43e^{3x}+\binom42e^{2x}-\binom41e^{x}+\binom{4}{0}\right)\\
&=6!\left(\binom44\frac{4^6}{6!}-\binom43\frac{3^6}{6!}+\binom42\frac{2^6}{6!}-\binom41\frac{1^6}{6!}+\binom40\frac{0^6}{6!}\right)\\
&=\binom444^6-\binom433^6+\binom422^6-\binom411^6+\binom400^6\\[6pt]
&=1560
\end{align}
$$
| {
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Evaluate $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$ Denote $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$, then how can we prove that
*
*$f(n,1)=\cdots=f(n,n-1)=0,$
*$f(n,n)=\frac{1}{2} (2 n)!,\ \ f(n,n+1)=\frac{1}{12} n (n+1) (2 n+1) (2 n)!, \cdots$
Moreover is there a general closed-form for $f(n,j)$ when $j>n+1$? Any help is appreciated.
| I would like to fill in the details for @MHZ. We seek to evaluate
$$F_{n,j} = \sum_{k=0}^n k^{2j} (-1)^{n-k} {2n\choose n-k}.$$
where $j\ge 1.$ With this in mind we introduce the function
$$F_n(z) = \frac{(2n)!}{2} z^{j-1} \prod_{q=1}^n \frac{1}{z-q^2}.$$
This has the property that the residue at $z=k^2$ where $1\le k \le n$ is
the desired sum term. We find
$$\mathrm{Res}_{z=k^2} F_n(z) = \frac{(2n)!}{2} k^{2j-2}
\prod_{q=1}^{k-1} \frac{1}{k-q} \frac{1}{k+q}
\prod_{q=k+1}^{n} \frac{1}{k-q} \frac{1}{k+q}
\\ = \frac{(2n)!}{2} k^{2j-2} \frac{1}{(k-1)!} \frac{k!}{(2k-1)!}
\frac{(-1)^{n-k}}{(n-k)!} \frac{(2k)!}{(n+k)!}
\\ = \frac{(2n)!}{2} k^{2j-1} \frac{1}{(2k-1)!}
\frac{(-1)^{n-k}}{(n-k)!} \frac{2k (2k-1)!}{(n+k)!}
\\ = (2n)! k^{2j} (-1)^{n-k} \frac{1}{(n-k)!} \frac{1}{(n+k)!}
\\ = \sum_{k=0}^n k^{2j} (-1)^{n-k} {2n\choose n-k}.$$
We now use the fact that residues sum to zero, so our sum becomes
$$- \mathrm{Res}_{z=\infty} F_n(z).$$
Note however that when $n-(j-1) \ge 2$ or $n\ge j+1$ this residue is
zero. Therefore we continue with an Iverson bracket $[[j+1\gt n]]$
(smallest $j$ is $n$)
$$- \mathrm{Res}_{z=\infty} F_n(z)
= \mathrm{Res}_{z=0} \frac{1}{z^2} F_n(1/z)
\\ = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{j+1}} \prod_{q=1}^n \frac{1}{1/z-q^2}
\\ = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{j+1-n}}
\prod_{q=1}^n \frac{1}{1 - q^2 z}.$$
We thus have immediately by inspection
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n} = \frac{1}{2} (2n)!.}$$
We also get
$$F_{n,n+1} = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{2}}
\prod_{q=1}^{n} \frac{1}{1 - q^2 z}
= \frac{(2n)!}{2} \sum_{q=1}^{n} q^2.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n+1} = (2n)!
\times \frac{1}{12} n (n+1) (2n+1).}$$
Do one more to get
$$F_{n,n+2} = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{3}}
\prod_{q=1}^{n} \frac{1}{1 - q^2 z}
\\ = \frac{1}{2} (2n)!
\left(\sum_{p=1}^n \sum_{q=p+1}^n p^2 q^2
+ \sum_{q=1}^n q^4\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n+2} = (2n)!
\times \frac{1}{720}
n (n+1) (n+2) (2n+1) (2n+3) (5n-1).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| $$x^2(x^2+6)=(x^2+3)^2-3^2$$
WLOG $x^2+3=3\csc t,t\to0^+$
$$\lim_{t\to0^+}3(\csc t-1)(3\csc t-3\cot t)$$
$$=\lim\dfrac{9(1-\sin t)(1-\cos t)}{\sin^2t}$$
$$=9\lim\dfrac{1-\sin t}{1+\cos t}=?$$ as $1-\cos t\ne0$ as $t\ne0$ as $\to0$
| {
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} |
Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
| Here's an alternative answer if you absolutely have to use L'Hopital's rule:
First rewrite the expression inside the limit as follows:
$$x^3\Big(\sin(\frac{1}{x+2})-2\sin(\frac{1}{x+1})+\sin(\frac{1}{x})\Big)=x^3\Big[(\sin(\frac{1}{x+2})-\frac{1}{x+2})-2(\sin(\frac{1}{x+1})-\frac{1}{x+1})+\sin(\frac{1}{x})\Big]+x^3\Big(\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}\Big)$$
We've written the expression in this way, suggestively, so that for each individual term in parentheses the limits exist. Then we compute the limits as follows:
$$\lim_{x\to\infty}x^3(\sin(\frac{1}{x})-\frac{1}{x})=\lim_{u\to 0}\frac{\sin(u)-u}{u^3}=-\frac{1}{6}$$
by applying L'Hopital's rule twice.
Also
$$\lim_{x\to\infty}x^3(\sin(\frac{1}{x+1})-\frac{1}{x+1})=\Big[\lim_{x\to\infty}(\frac{x}{x+1})^3\Big]\Big[\lim_{x\to\infty}(x+1)^3(\sin(\frac{1}{x+1})-\frac{1}{x+1})\Big]=-\frac{1}{6}$$
and similarly
$$\lim_{x\to\infty}x^3(\sin(\frac{1}{x+2})-\frac{1}{x+2})=-\frac{1}{6}$$
Finally
$$\lim_{x\to\infty}x^3\Big(\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}\Big)=\lim_{x\to\infty}\frac{2x^3}{x(x+1)(x+2)}=2$$
and hence we find by adding all those limits together that
$$\lim_{x\to\infty}x^3\Big(\sin(\frac{1}{x+2})-2\sin(\frac{1}{x+1})+\sin(\frac{1}{x})\Big)=-\frac{1}{6}+2\frac{1}{6}-\frac{1}{6}+2=2$$
The takeaway from this manipulation is that applying L'Hopital's rule is not straightforward, but there is a way avoid lengthy calculations , by which one has to add and subtract terms that amount to known or easily derived limits. However, in my personal opinion, expanding in a Taylor series is the only foolproof prescription for taking limits of that sort.
| {
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Closed form of integral $\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx$ Originally this integral was interesting for me as it represents the specific harmonic number $H_{-\frac{1}{5}}$ and also because Mathematica returned the wrong value $0$ for the integral. I posted the probem here https://mathematica.stackexchange.com/questions/215089/possible-bug-integrate1-x-1-5-1-x-x-0-1-0
Later I found a way to get a reasonable result from Mathematica which, finally I could simplify to this form.
$$\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx=\frac{\pi}{2} \sqrt{1+\frac{2 \sqrt{5}}{5}}-\frac{5}{4} \log (5)-\frac{\sqrt{5}}{2} \log \left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)$$
So a presumed bug by has turned finally into a nice statement. Can you prove it?
| My working is quite long but I got the following in the end:
$$\begin{align}F(x) = \int\dfrac{1 - \frac1{\sqrt[5]x}}{1 - x}\,\mathrm dx =&\dfrac{-\left(5 + \sqrt 5\right)\ln\left(2\sqrt[5]{x^2} + \left(1 + \sqrt{5}\right)\sqrt[5]x + 2\right) - \left(5 - \sqrt5\right)\ln\left(2\sqrt[5]{x^2} + \left(1 - \sqrt{5}\right)\sqrt[5]x + 2\right)}{4} \\ &+ \dfrac{\left(5 - \sqrt5\right)\arctan\left(\frac{4\sqrt[5]x + \sqrt5 + 1}{\sqrt2\sqrt{5 - \sqrt5}}\right)}{\sqrt2\sqrt{5 - \sqrt5}} + \dfrac{\sqrt{5 + \sqrt5}\arctan\left(\frac{4\sqrt[5]x - \sqrt5 + 1}{\sqrt2\sqrt{5 + \sqrt5}}\right)}{\sqrt2} + C\end{align}$$
And with the help of Matlab, $$F(1) - F(0) = -0.387792901804606$$
Matlab code (in case you want to verify the computation):
a = @(x)((-1*(5+sqrt(5))*log(2*x^(2/5) + (1 + sqrt(5)) * x^(1/5) + 2) - (5 - sqrt(5))*log(2*x^(2/5) + (1 - sqrt(5)) * x^(1/5) + 2))/4);
b = @(x)((5 - sqrt(5))*atan((4*x^(1/5) + sqrt(5) + 1)/(sqrt(2)*sqrt(5-sqrt(5)))) / (sqrt(2)*sqrt(5-sqrt(5))));
c = @(x)(sqrt(5 + sqrt(5))*atan((4*x^(1/5) - sqrt(5) + 1)/(sqrt(2)*sqrt(5+sqrt(5)))) / sqrt(2));
format long;
ans = (a(1) + b(1) + c(1)) - (a(0) + b(0) + c(0))
| {
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Beginner Proof by Induction: Is this correct? I am asked to prove $\frac{1}{1.4} + \frac{1}{4.7} + ... + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}, n\geq 1$ by induction.
Can someone verify that I did this correctly. I am unsure about my Inductive Step.
Proof by induction
Inductive Hypothesis Let $P(k) = \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} = \frac{k}{3k+1}$
Base case $n = 1$
\begin{align*}
P(1) &= \sum\limits_{i=1}^{1} \frac{1}{(3(1)-2)(3(1)+1)} = \frac{1}{3(1)+1} \\
&= \frac{1}{(3-2)(3+1)} = \frac{1}{3+1} \\
&= \frac{1}{(1)(4)} = \frac{1}{4} \\
&= \frac{1}{4} = \frac{1}{4} \checkmark
\end{align*}
Inductive Step
For $k \geq 1$, show that $P(k) \to P(k+1)$ is true.
Assume $P(k)$ is true, assume $\sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} = \frac{k}{3k+1}$ is true.
Prove $P(k+1)$ is true, examine $\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} = \frac{k+1}{3(k+1)+1}$.
\begin{align*}
\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \frac{k+1}{3(k+1)+1} \\
\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \frac{k}{3(k+1)} \\
\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)}
\end{align*}
Can someone check that this is the correct way to prove the theorem by induction?
| I don't understand what you're doing in the last couple of lines for $P(k + 1)$, for example between the third & second last, you change the RHS value but leave the LHS the same.
Instead, I would normally go from one side of the equation, expand or otherwise manipulate it as need be so can use the assumption that $P(k)$ is true, and then do various simplifications to show how it's the same as the other side. In general, I've found it's easiest to start with the summation. As such, here's how I would handle the inductive step.
$$\begin{equation}\begin{aligned}
\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} & = \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} + \frac{1}{(3(k+1)-2)(3(k+1)+1)} \\
& = \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)} \\
& = \frac{k(3k+4) + 1}{(3k+1)(3k+4)} \\
& = \frac{3k^2 + 4k + 1}{(3k+1)(3k+4)} \\
& = \frac{(3k + 1)(k + 1}{(3k+1)(3k+4)} \\
& = \frac{k + 1}{3(k+1) + 1} \\
& = P(k+1)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
| {
"language": "en",
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Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ Question: Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ by Lagrange's method of Solution.
Progress: Lagrange's auxiliary equations are.
$$\dfrac{dx}{3x+y-z}=\dfrac{dy}{x+y-z}=\dfrac{dz}{2(z-y)}$$ The three ratios are equal to $\dfrac{-dx+3dy+dz}{0}$. So we have $-dx+3dy+dz=0\implies x-3y-z=C_1$. How can I find the other one. It is given that other solution is $\dfrac{x-y+z}{\sqrt{x+y-z}}=C_2$.
How can I find this?
| We can find that,
$$\text{Each ratio} = \frac{dx-dy+dz}{2x-2y+2z} = \frac{dx+dy-dz}{4x+4y-4z}$$
or
$$\frac{d(x-y+z)}{x-y+z} = \frac{d(x+y-z)}{2(x+y-z)}$$
On integrating,
$$\ln(x-y+z) = \frac{1}{2}\ln(x+y-z) +\ln C_2 \implies C_2 = \frac{x-y+z}{\sqrt{x+y-z}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$ I have to find this integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$$
My attempt was to split the integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{x\arctan x+1}{x\sqrt{x^2+1}}+\int_{\frac{1}{4}}^4\frac{\arctan x-1}{x\sqrt{x^2+1}}$$
The first integral is:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{1}{x}(x\arctan x)'dx=\arctan x|_{\frac{1}{4}}^4+\int_{\frac{1}{4}}^4\frac{1}{x}\arctan x dx$$
but I am stuck here.
| The bounds suggest the subtitution $u=\frac{1}{x}\Rightarrow dx=-\frac{1}{u^2}du$
$$\int_{1/4}^4 \frac{(x+1)\arctan x}{x\sqrt{x^2+1}} \, dx=\int_{1/4}^4 \frac{(1+u)\arctan \frac{1}{u}}{u\sqrt{u^2+1}} \, du$$
Therefore, using $\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}$ for $x>0$, we get:
$$
\begin{aligned}
\int_{1/4}^4 \frac{(x+1)\arctan x}{x\sqrt{x^2+1}} &= \frac{1}{2}\int_{1/4}^4 \frac{(x+1)(\arctan x+\arctan\frac{1}{x})}{x\sqrt{x^2+1}} \, dx\\
&= \frac{\pi}{4}\int_{1/4}^4 \frac{x+1}{x\sqrt{x^2+1}} \, dx
\end{aligned}
$$
Can you end it now?
| {
"language": "en",
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"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\lfloor k^4\rfloor$ where $k^3 - 5k^2 + 2 = 0$. Problem :
Find a value of $\lfloor k^4 \rfloor$ where $k$ is the biggest real root of equation $k^3 - 5k^2 + 2 = 0$.
Can't use calculator.
Let $f(x) = x^3 - 5x^2 + 2$. Then $f(4)f(5)<0$. so $k\in(4,5)$.
And I bounded more like : $$\frac{9}{2}<k<5.$$
But this is not enough to Find $\lfloor k^4 \rfloor$ because I have $$410<\left(\frac{9}{2}\right)^4 < k^4<5^4=625$$
and this is too rough to find $\lfloor k^4 \rfloor$.
I tried Newton's method with $x_0 = \frac{9}{2}$ but this is so messy, I got $x_1=\frac{612}{126}.$
So, I think I need some tight inequality for $k$ or some tricky technique, But I don't know how should I do. Thanks for any help.
| If $a<b<c$ are roots of the equation, so by the Viete's theorem $a^4+b^4+c^4=585$ and since for another roots $x_i$ we have $|x_i|<0.7,$ we obtain the answer: $[c^4]=584.$
Indeed, $a+b+c=5,$ $ab+ac+bc=0$ and $abc=-2$.
Thus, $$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=$$
$$=(a+b+c)^4+4(a+b+c)abc=625-40=585.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Transformation of homogeneous coordinates to Euclidean coordinates? I understand that two vectors
$$
v_{1} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}
\quad
\text{,}
\quad
v_{2} = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}
$$
in homogeneous coordinates are equivalent if
$$
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=
\lambda
\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}
\quad,\quad
\lambda \neq 0.
$$
I visualize these equivalence classes as "rays" from the coordinate system origin (with the origin itself excluded).
To convert a vector from homogeneous to Euclidean coordinates, one divides all components by a number such that $z\rightarrow 1$. Example:
$$
\begin{pmatrix} 0 \\ 8 \\ 4 \end{pmatrix}_\text{hom}
\quad
\rightarrow
\quad
\begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix}_\text{eucl}
$$
Question: What is the correct transformation of $v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_\text{hom}$ to Euclidean coordinates?
UPDATE - I'm adding the source for this question. This is the slide in question:
Take a look at how $P _{4} \,$ and $P _{5} \,$ are transformed:
$$
P_{4, \,hom} = \begin{pmatrix} 1 \\ 0 \\ 0.0001 \end{pmatrix}
\quad
\rightarrow
\quad
P_{4, \,eucl} = \begin{pmatrix} 10000 \\ 0 \\ 1 \end{pmatrix}
$$
$$
P_{5, \,hom} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
\quad
\rightarrow
\quad
\text{"at infinity} \,\textbf{on x-axis}"
$$
| $v$ is a point at infinity. There is no equivalent point in euclidean space, but you can think it as the point of intersection of the lines of a parallel bundle.
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
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Find minimum value of $\frac{cb}{1-c} +\frac{ac}{1-a} + \frac{ba}{1-b}$ How do I minimize the expression
$$f(a,b,c)=\frac{cb}{1-c} +\frac{ac}{1-a} + \frac{ba}{1-b}$$
subject to the constraint $abc=(1-a)(1-b)(1-c)$ with $a$, $b$, $c \in (0,1)$. Conceptually, the Lagrange multiplier procedure can be utilized. But, the resorting algebra seems messy and hard to figure. It is not hard to guest that 3/2 is the minimum given the symmetry. So, I suspect the inequality of certain AM-GM sort could be the answer, yet don't know how to apply. Would appreciate some ideas.
| Let $\frac{a}{1-a}=\frac{y}{x},$ $\frac{b}{1-b}=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $\frac{c}{1-c}=\frac{x}{z}$, $a=\frac{y}{x+y}$ and by Nesbitt we obtain:
$$\sum_{cyc}\frac{ab}{1-b}=\sum_{cyc}\left(\frac{y}{x+y}\cdot\frac{z}{y}\right)=\sum_{cyc}\frac{z}{x+y}\geq\frac{3}{2}.$$
The equality occurs for $a=b=c=\frac{1}{2},$ which says that we got a minimal value.
| {
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"url": "https://math.stackexchange.com/questions/3568387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $(4-x^2)y''+2y=0$ by series I know that there are other ways to solve this, but I want to solve the ODE $(4-x^2)y''+2y=0$ by series close to $0$, that is, I'm looking for solutions of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$
Substituing in the ODE, I got:
$$(8a_{2}+2a_{0})+(24a_{3}+2a_{1})x+\sum_{n=2}^{\infty}[4(n+2)(n+1)a_{n+2}+2a_{n}-n(n-1)a_{n}]x^{n}=0\implies \\
a_2=-\frac{a_0}{4};\quad a_3=-\frac{a_{1}}{12}; \quad a_{n+2}=\frac{(n-2)a_{n}}{4(n+2)}. $$
But this recorrence relation give me a problem: $a_4=0\implies a_{2n}=0$ for all $n>2 $.
My question is: Did I made a mistake? Or this solution considers, for even $n$, only the terms $a_2$ and $a_0$, that is, $y_{1}(x)=a_{0}\left(1-\frac{1}{4}x^{2}\right)$?
| We have a differential equation that doesn’t have a constant coefficient for the second derivative.
$p\left( x \right) = 4-{x^2} \hspace{0.25in}p\left( 0 \right) = 4 \ne 0$
So, assume that $x_0$ is an ordinary point for this differential equation. We first need the solution and its derivatives,
$y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \hspace{0.25in}y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}}$
Plug these into the differential equation.
$\left( 4-{{x^2}} \right)\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} +2\sum\limits_{n = 0}^\infty {{a_n}{x^{n}}}= 0$
Now, break up the first term into two so we can multiply the coefficient into the series and multiply the coefficients of the second and third series in as well.
$4\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n-2}}} - x^{2}\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} + 2\sum\limits_{n = 0}^\infty {{a_n}{x^{n}}}=0$
$4\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n-2}}} - \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^n}} + 2\sum\limits_{n = 1}^\infty {{a_n}{x^{n}}}=0$
We will only need to shift the first and third series down by two and one respectively to get all the exponents the same in all the series.
$4\sum\limits_{n = 0}^\infty {(n+2)\left( {n + 1} \right){a_{n+2}}{x^n}} - \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^n}} + 2\sum\limits_{n = 0}^\infty {{a_{n+1}}{x^{n}}}=0$
At this point we could strip out some terms to get all the series starting at $n=2$, but that’s actually more work than is needed. Let’s instead note that we could start the second series at $n=0$ if we wanted to because that term is just zero. Likewise, the terms in the second series are zero for both $n=1$ and $n=0$ and so we could start that series at $n=0$. If we do this all the series will now start at $n=0$ and we can add them up without stripping terms out of any series.
$\begin{align*}\sum\limits_{n = 0}^\infty {\left[ {4{(n+2)}\left( {n + 1} \right){a_{n+2}} - n\left( {n - 1} \right){a_{n}} + 2{a_{n}}} \right]{x^n}} & = 0\\
\sum\limits_{n = 0}^\infty {\left[ {4{(n+2)}\left( {n + 1} \right){a_{n+2}} + \left(n+2 \right)\left(n-1\right){a_{n}}} \right]{x^n}} & = 0\\ \end{align*}$
Now set coefficients equal to zero.
${4{(n+2)}\left( {n + 1} \right){a_{n+2}} + \left(n+2 \right)\left(n-1\right){a_{n}}}= 0,\hspace{0.25in}n = 0,1,2, \ldots$
Solving this gives,
${a_{n + 2}} = - \frac{{\left( {n + 2} \right)\left( {n - 1} \right){a_n}}}{4{\left( {n + 2} \right)\left( {n + 1} \right)}},\hspace{0.25in}n = 0,1,2, \ldots$
Now, we plug in values of $n$,
$n = 0:\hspace{0.25in}{a_2} = \frac{a_0}{4}\\
n = 1:\hspace{0.25in}{a_3} = 0{a_1} = 0\\
n = 2:\hspace{0.25in}{a_4} = - \frac{1}{12}{a_2}\neq0\\
n = 3:\hspace{0.25in}{a_5} = - \frac{1}{8}{a_3} = 0\\
n = 4:\hspace{0.25in}{a_5} = - \frac{3}{20}{a_4}\neq0$
Now, from this point we can see that $a_{2n}=0$,for all $n>2$.
Thus,for solution considers, for even $n$, only the terms $a_2$ and $a_0$, that is,
$y(x)=a_{0}\left(1+\frac{1}{4}x^{2}\right)$.
| {
"language": "en",
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Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the residual theorem:
I started with $\gamma: [0,\pi] \to \mathbb{C}, t \to e^{2it}$. Since $\cos(t) = \frac{1}{2}\left(e^{it}+e^{-it}\right)$ and $\gamma'(t) = 2ie^{2it}$ we find that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(e^{it}+e^{-it}\right)} \cdot \frac{2ie^{2it}}{2ie^{2it}}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t
\end{align}
I did this with the aim to use
\begin{align}
\int_{\gamma} f(z) \mathrm{d}z = \int_{a}^{b} (f\circ \gamma)(t)\gamma'(t) \mathrm{d}t,
\end{align}
so we find
\begin{align}
\int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t = \int_{\gamma} \frac{-i}{6z+2z\sqrt{z}+2\sqrt{z}} \mathrm{d}z
\end{align}
At this point I don't know how to continue. Can anyone help?
| Use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx.~~~(1)$$
$$I=\int_{0}^{\pi} \frac{dt}{3+2 \cos t}~~~~(2)$$
Using (1), we get
$$I=\int_{0}^{\pi} \frac{dt}{3-2 \cos t}~~~~(3)$$
Add (2) and (3), then
$$2I=\int_{0}^{\pi} \frac{6\,dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{\sec^2 t}{9\tan^2 t+5}=\frac{4}{3} \int_{0}^{\infty}\frac{du}{u^2+(\sqrt{5}/3)^2}$$
$$2I=\frac{4}{3} \frac{3}{\sqrt{5}}\tan^{-1}(3u/\sqrt{5})|_{0}^{\infty}=\frac{2\pi}{\sqrt{5}}\implies I=\frac{\pi}{\sqrt5} $$ In above we have used $\tan t =u$.
| {
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Why after depressing a cubic does it have different roots? The cubic $(x-1)(x-2)(x-2)=0$ will have the roots 1 and 2. Expanding will give $x^3-5x^2+8x-4=0$ which is in the form $ax^3+bx^2+cx+d=0$. Depressing it by substituting $x = t - \frac{b}{3a} = t+\frac{5}{3}$ will give $t^3-\frac{1}{3}t+\frac{2}{27}=0$. This is in the form $t^3+pt+q=0$.
Using Cardano's formula $x=\sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}}$ for the above, the following root is gotten.
\begin{align}
x &= \sqrt[3]{-\frac{\frac{2}{27}}{2} + \sqrt{\left(\frac{\frac{2}{27}}{2}\right)^2+\left(\frac{-\frac{1}{3}}{3}\right)^3}} + \sqrt[3]{-\frac{\frac{2}{27}}{2} - \sqrt{\left(\frac{\frac{2}{27}}{2}\right)^2+\left(\frac{-\frac{1}{3}}{3}\right)^3}}\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{\left(\frac{1}{27}\right)^2+\left(-\frac{1}{9}\right)^3}} + \dots\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{\frac{1}{729}+\left(-\frac{1}{729}\right)}} + \dots\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{0}} + \sqrt[3]{-\frac{1}{27} - \sqrt{0}}\\
&= \sqrt[3]{-\frac{1}{27}} + \sqrt[3]{-\frac{1}{27}}\\
&= -\frac{1}{3} + -\frac{1}{3}\\
&= -\frac{2}{3}\\
\end{align}
However, the initial cubic didn't have $-\frac{1}{6}$ as one of its roots. If the depressed cubic $t^3-\frac{1}{3}t+\frac{2}{27}=0$ is plotted, it's completely different to the original, the other root being $\frac{1}{3}$.
I thought that depressing a cubic was simply rewriting it and not changing it. And if it does change what the cubic is, then why is it said that Cardano's formula works for all cubics, shouldn't it be said that it only works for depressed cubics?
| As you said in the first paragraph, you get the depressed cubic by substituting
$x = t - \frac{b}{3a} = t+\frac53$.
Conversely, that means $t = x - \frac53$ in your example.
So if $x = 1$ is a root of the original cubic, it had better be true that
$t = x - \frac53 = -\frac23$ is a root of the depressed cubic.
And if $t = -\frac23$ is a root of the depressed cubic, then $x = t+\frac53 = 1$
is a root of the original cubic.
And as it turns out that's exactly what happened.
| {
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Prove that $1-3^x+5^x-7^x = 0 \Leftrightarrow x=0$ Problem :
Prove that equation $$1-3^x+5^x-7^x = 0$$
has unique root $x=0$.
Since $\lim_{x\to-\infty}(1-3^x+5^x-7^x) = 1$ and $\lim_{x\to\infty}(1-3^x+5^x-7^x) = -\infty$,
I tried to show $x \to 1-3^x+5^x-7^x$ is decreasing function, but it wasn't easy to determine its derivative is negative.
Thanks for any help.
| Note that $$1-3^x+5^x-7^x = 0 \iff 1+5^x = 3^x+7^x$$
Since the exponential functions involved are increasing functions we have the following inequalities.
$$x>0\implies 3^x+7^x > 1+5^x$$
and $$x<0\implies 3^x+7^x < 1+5^x$$
Thus the equality happens if and only if $x=0$
| {
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"source": "stackexchange",
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Solve indefinite integral $ \int \frac{x^4}{\sqrt{x^2+4x+5}}dx $ I need to solve the next problem:
$$
\int \frac{x^4}{\sqrt{x^2+4x+5}}dx
$$
I know the correct answer is
$$
(\frac{x^3}{4}-\frac{7x^2}{6}+\frac{95x}{24}-\frac{145}{12})*\sqrt{x^2+4x+5}\space+\space\frac{35}{8}\ln{(x+2+\sqrt{x^2+4x+5})}+C
$$
still, I cannot find the solution.
I've already tried substituting $u=x+2$ and that has't given any satisfying result
| Hint:
$$\int\frac{x^4\mathrm dx}{\sqrt{x^2+4x+5}}=\int\frac{x^4\mathrm dx}{\sqrt{(x+2)^2+1}}=\int(\tan\theta -2)^{4}\sec\theta\mathrm d\theta$$
Set $\tan\theta=x+2$, $\sec^2\theta\mathrm d\theta=\mathrm dx$. Can you proceed?
| {
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Greatest common divisor of $(x+1)^{4n+3} + x^{2n}$ and $x^3-1$. I have to find the greatest common divisor of
$$(x+1)^{4n+3} + x^{2n}$$
and
$$x^3-1$$
I know I can express the second polynomial as:
$$x^3-1 = (x-1)(x^2+x+1)$$
So I would have to check if the first polynomial is divisible by $(x^3-1)$, $(x^2+x+1)$ or $(x-1)$ and if it is not divisible by any of those, then the two polynomials do not have a common divisor except for $1$. But I don't know how I can divide the polynomial
$$(x+1)^{4n+3} + x^{2n}$$
by those $3$ other polynomials and therefore can't check the greatest common divisor.
| Only $x^2+x+1$ can be an answer because $x-1$ is not valid.
Modulo $x^2+x+1$ we obtain:
$$(x+1)^{4n+3}+x^{2n}\equiv x^{2n}-x^{8n+6}=x^{2n}(1-(x^3)^{2n+2})\equiv0.$$
| {
"language": "en",
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Find the general solution to $\csc \theta + \sec \theta = 1$
Find the general solution to $$\csc\theta + \sec\theta =1$$
This is how I solved.
We have,
\begin{align}
\csc\theta + \sec\theta &=1\\
\frac1{\sin\theta} + \frac1{\cos\theta}& =1\\
\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} &=1\\
(\sin\theta + \cos\theta)^2 &= (\sin\theta\cos\theta)^2 \\
1 + 2\sin\theta\cos\theta &= \frac{4\sin^2\theta\cos^2\theta}4\\
1 + \sin2\theta &= \frac{(2\sin\theta\cos\theta)^2 }4\\
4 + 4\sin2\theta &= \sin^2 2\theta\\
\sin^2 2\theta - 4\sin2\theta - 4 &= 0\\
\sin2\theta &= 2 - 2\sqrt2\end{align}
Now here I am stuck. Can someone please help me proceed further?
| Hint
Again avoid squaring as it immediately introduces
When do we get extraneous roots?
$\sin x\cos x=\sin x+\cos x=y,$
$y=\sqrt2\cos(x-45^\circ)\implies-\sqrt2\le y\le?$(say)
$$y^2=1+2\sin x\cos x$$
Let $y^2-1=2y\iff y^2-2y-1=0$
$y=1\pm\sqrt2$
$\implies y=1-\sqrt2$
$\implies\cos(x-45^\circ)=\dfrac1{\sqrt2}-1$
| {
"language": "en",
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Properties of three terms of a geometric series I’m [still!] working on the equation in this question, namely
$$(b^2+2)^2=(a^2+2c^2)(bc-a). \tag{$\star$}$$
where $a,b,c$ are integers. Evidently, $(\star)$ implies
$$\frac{b^2+2}{bc-a} = \frac{a^2+2c^2}{b^2+2}, \tag{1}$$
which is to say that $\{bc-a,b^2+2,a^2+2c^2\}$ are three consecutive terms of a geometric series.
QUESTION: Does that fact provide any information that would help in solving $(\star)$? i.e., are there properties of geometric series that can be brought to bear on the problem?
Each fraction in $(1)$ is actually an integer, in case that provides more leverage/structure.
EDIT: The reason I know this is that I derived this equation from the equation $x^3=y^2+2$, where $x=(b^2+2)/(bc-a)$ is a positive integer by assumption.
| If I have understood the additional constraints to your original problem correctly, the problem is now to find all integral solutions to
$$(b^2+2)^2=(a^2+2c^2)(bc-a), \tag{$\star$}$$
such that
$$\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2},$$
and moreover $x:=\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2}$
is an integral solution to $x^3=y^2+2$ for some integer $y$.
A standard argument in $\Bbb{Z}[\sqrt{-2}]$, which is a UFD, shows that then $x=3$. Hence we can express the problem as a system of two simultaneous diophantine equations:
\begin{eqnarray}
a^2+2c^2&=&3(b^2+2)\tag{1.1}\\
b^2+2&=&3(bc-a).\tag{1.2}
\end{eqnarray}
Equation $(1.2)$ shows that $3a=3bc-b^2-2$, and hence from $(1.1)$ we find that
$$27(b^2+2)=9(a^2+2c^2)=(3a)^2+18c^2=(3bc-b^2-2)^2+18c^2.$$
Expanding and collecting like terms shows that this is equivalent to
$$b^4-6b^3c+9b^2c^2-23b^2-12bc+18c^2-50=0,$$
which in turn we can rewrite as
$$(b^2+2)(b-3c)^2=25(b^2+2).$$
Of course $b^2+2\neq0$ and so it follows that $(b-3c)^2=25$, or equivalently
$$b=3c\pm5\qquad\text{ and hence }\qquad a=\frac{3bc-b^2-2}{3}=\mp5c-9.$$
This shows that every solution to your system of diophantine equations is of the form
$$(a,b,c)=(\mp5c-9,3c\pm5,c).$$
Note that these are precisely the solutions you already found in your original question, i.e. the parametric family of solutions
$$(5d+1,3d+1,d+2)\qquad\text{ with }\qquad d\in\Bbb{Z},$$
along with their involutions given by $(a,b,c)\ \longmapsto\ (a,-b,-c)$.
| {
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"question_score": "4",
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What is the area of the smaller right triangle?
The two diagonal lines are parallel, and the area of the region between them is $42$
What is $A,$ the area of the smaller right triangle?
So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$
Solving the equation we get:
$3x + 4y = 72$
I don't know what's next here so I just assumed that $x = 12, y = 9$ which satisfies the equation and resulting to $A = 54$. My question is, am I missing something so that I can solve the equation?
|
So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$
Continuing your equation, we get:
$\dfrac{xy + 3x + 4y + 12 - xy}{2} = 42$
$xy + 3x + 4y + 12 - xy = 2 \cdot 42$
$3x + 4y + 12 = 84$
$\boxed{3x + 4y = 72}$
My question is, am I missing something so that I can solve the equation?
The lines are parallel, thus their slopes will be equal
$\implies \dfrac{x}{y} = \dfrac{x + 4}{y + 3}$
$(y + 3)(x) = (x + 4)(y)$
$xy + 3x = xy + 4y$
$3x = 4y$
$\color {red}{3x + 4y = 72},3x + (3x) = 72$
$x = 12, y = 9$
$\boxed{Area = \dfrac{xy}{2} = \dfrac{12 \cdot 9}{2} = \dfrac{108}{2} = 54}$
| {
"language": "en",
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"question_score": "5",
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how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| $$x^2-x+e^x(x-2)=0 \implies e^{-x}=\frac{x-2}{x(1-x)}$$ and the solution cen write using the generalized Lambert function (have a look at equation $(4)$).
By inspection or graphing, the positive root is close to $x=2$. To have a better value, expand the lhs as a Taylor series
$$y=x^2-x+e^x(x-2)=2+\left(3+e^2\right) (x-2)+\left(1+e^2\right) (x-2)^2+\frac{1}{2} e^2
(x-2)^3+O\left((x-2)^4\right)$$
Use series reversion to get
$$x=2+\frac{y-2}{3+e^2}-\frac{\left(1+e^2\right)
(y-2)^2}{\left(3+e^2\right)^3}+\frac{\left(4+5 e^2+3 e^4\right) (y-2)^3}{2
\left(3+e^2\right)^5}+O\left((y-2)^4\right)$$ Make $y=0$ to get
$$x=\frac{2 \left(136+257 e^2+196 e^4+76 e^6+14
e^8+e^{10}\right)}{\left(3+e^2\right)^5}\approx 1.771$$ while the exact solution is $1.768$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators. My question is: Do we have an splitting equation where we can produce fractions with odd denominators?
To split an Egyptian fraction to Egyptian fractions, we can use the splitting equation below:
$\frac{1}{n}= \frac{1}{n+1}+\frac{1}{n(n+1)}$
The key limitation of the above equation is the following:
If $n$ is even, then $n+1$ is odd and $n(n+1)$ is even, otherwise $n+1$ is even and $n(n+1)$ is even.
Either way, the splitting equation produces with at least one even Egyptian fraction.
An example of a splitting to Odd Egyptian fraction is given below:
$\frac{1}{3}= \frac{1}{5}+\frac{1}{9}+\frac{1}{45}$
$\frac{1}{5}= \frac{1}{9}+\frac{1}{15}+\frac{1}{45}$
$\frac{1}{7}= \frac{1}{15}+\frac{1}{21}+\frac{1}{35}$
$\frac{1}{7}= \frac{1}{9}+\frac{1}{45}+\frac{1}{105}$
$\frac{1}{9}= \frac{1}{15}+\frac{1}{35}+\frac{1}{63}$
$\frac{1}{11}= \frac{1}{21}+\frac{1}{33}+\frac{1}{77}$
The link below is useful for further details about egyptian fraction: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#section9.5
| Just a thought:
If you limit the problem to just the examples you gave; that is, can any given odd Egyptian fraction be split into three different odd Egyptian fractions?
The examples you gave have two main algebraic forms:
$$\frac{1}{n}=\frac{1}{b}+\frac{1}{an}+\frac{1}{abn}\tag{1}$$
and
$$\frac{1}{n}=\frac{1}{ab}+\frac{1}{an}+\frac{1}{bn}\tag{2}$$
where $a$, $b$ and $n$ are all odd positive integers.
Then one way of proceeding is to study congruence patterns for the two basic forms (1) and (2), for example for $n=4m+1$ and $n=4m+3$ in the first instance.
| {
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$a+ b+ c+ d+ e = 12$, and $a^4b^3c^3de=4\times 6^6$, $a,b,c,d,e$ are positive. Find value of $4a + 3b + 3c + d + e$ Given that $$a+ b+ c+ d+ e = 12\quad \&\quad a^4b^3c^3de=4\times 6^6$$ where $a,b,c,d,e$ are positive integers, find the value of $4a + 3b + 3c + d + e$
I tried by trial and error by putting values but it takes to long.
I also tried by taking logarithm on both sides(in equation 2nd) to obtain a similar expression as asked in question but couldn't go any further.
| Since nobody wants to type an answer:
as suggested in the comment, use AM-GM inequality on the $12$ numbers $a/4, a/4, a/4, a/4, b/3, b/3, b/3, c/3, c/3, c/3, d, e$.
This gives:
$$\frac{a + b + c + d + e}{12} \geq \sqrt[12]{\frac{a^4b^3c^3de}{4\times 6^6}},$$ where equality holds if and only if all the $12$ numbers are equal.
But the conditions say that both sides are equal to $1$, hence equality does hold, and all the $12$ numbers are equal.
This means that $a/4 = b/3 = c/3 = d = e$. Then from the fact that $a + b + c + d + e = 12$, it follows that $a = 4, b = 3, c = 3, d = 1, e = 1$.
| {
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The parametric ratio $\frac{x}{y}$ with known $x+y$ and $x\cdot y$ $x$ and $y$ are in fact $\lambda_1$ and $\lambda_2$, the bigger and smaller eigenvalues of a parametric matrix $A'A$, and $t$ is a very small constant. I have that
$$
\begin{split}
x+y &= 1+ \frac{t^2}{2} \\
xy &= \frac{t^2}{4}
\end{split}
$$
How would I prove that $\frac{x}{y}\ge \frac{1}{t}$? The solutions of the system are here, but it gets pretty messy. Is there an elegant way to go about this?
| So $x=\frac{1}{4}(t^2+2+\sqrt{t^4+4})$ and $y=\frac{1}{4}(t^2+2-\sqrt{t^4+4})$, $\dfrac{x}{y}=\dfrac{(t^2+2+\sqrt{t^4+4})^2}{4t^2}$ say $t>0$.
To prove that $$\dfrac{(t^2+2+\sqrt{t^4+4})^2}{4t}\ge 1\Leftrightarrow t^4+4+4t^2+t^4+4+2(t^2+2)(\sqrt{t^4+4})-4t\ge 0$$ but $t^2-t+1\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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