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Evaluate $\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$
Evaluate
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$$
Does anyone have any smart ideas how to evaluate such a sum? I know one solution with complex numbers and complex analysis but I'm looking for some more smart or sophisticated methods.
| I would not say that it is elegant, but:
The form $n^4+1$ in the denominator suggests that one should be able to get this series by expanding a combination of a hyperbolic and trigonometric function in a Fourier series.
Indeed, after some trial and error, the following function seems to work:
$$
\begin{gathered}
\left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)-\sin
\left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)\right)\cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right) \\
+ \left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)+\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh
\left(\frac{\pi }{\sqrt{2}}\right)\right)\sin \left(\frac{x}{\sqrt{2}}\right) \sinh
\left(\frac{x}{\sqrt{2}}\right)
\end{gathered}
$$
It is even, and its cosine coefficients are
$$
\frac{\sqrt{2}\bigl(\cos(\sqrt{2}\pi)-\cosh(\sqrt{2}\pi)\bigr)(-1)^{n+1} n^2}{\pi(1+n^4)},\quad n\geq 1.
$$
(The zero:th coefficient is also zero). Evaluating at $x=0$ (the series converges pointwise there) gives
$$
\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n^2}{1+n^4}=
\frac{\pi\left(\sin
\left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)-\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)\right)}{\sqrt{2}\bigl(\cosh(\sqrt{2}\pi)-\cos(\sqrt{2}\pi)\bigr)}\approx 0.336.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| From the fact that $x^2-2x$ is rational, with $x$ irrational, we conclude that $x$ has degree $2$ over $\Bbb Q$, so we can write it in the form $a+b\sqrt{D}$ for some rational $a$ and $b$ and squarefree $D\in\Bbb{Z}$.
Now we can write: $$x^2-2x = (a^2-2a+Db^2)+2b(a-1)\sqrt{D}\in\Bbb Q$$
Since $b\ne 0$, we can conclude that $a=1$, so $x=1+b\sqrt{D}$. Another calculation: $$x^3-5x = (3b^2D-4) + b(b^2D-2)\sqrt{D}\in\Bbb Q$$
We conclude that $b^2D=2$, so: $$x^3-5x=3b^2D-4=6-4=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the determinant of this $5 \times 5$ matrix
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
& \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0
& 0 & 1 & a^2
& b^2 \\ 0 & 0 & 0 & a
& b \\ 0 & 0 & 0 & -b
& a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$
I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.
I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /
As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.
So then I have the matrix
$$\begin{pmatrix}
\sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\
0 & 0 & 1 & a^2 & b^2\\
0 & 0 & 0 & a & b\\
0 & 0 & 0 & 0 & a^2+b^2
\end{pmatrix}$$
$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$
But this is wrong and I don't see how to get the correct determinant...?
| same as determinant of $$W=\begin{pmatrix} \sin\alpha
& \cos\alpha & 0 & ? & ?? \\
-\cos\alpha & \sin\alpha & 0 & ??? & ???? \\ 0
& 0 & 1 & 0
& 0 \\ 0 & 0 & 0 & a
& b \\ 0 & 0 & 0 & -b
& a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Solving a fourth degree equation Well, I want to find the roots of the equation:
$$x^4+6x^3-9x^2-162x-243=0\tag1$$
And Wolfram Alpha tells me that I can factor this and get:
$$(x^2-3x-9)(x^2+9x+27)=0\tag2$$
Then it will be easy to solve... But my question is, how can I go from $(1)$ to $(2)$?
| In case anyone ever looks at this question again...it is best to try things that might give a factorization first. What I notice is the powers of 3 in $x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \;$ This suggests taking $x=3t,$ then dividing through by the resulting constant factor $81,$ resulting in
$$ t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
The two things to try next are
$$ (t^2 + At + 1) (t^2 + B t - 3) $$
and
$$ (t^2 + Ct - 1) (t^2 + D t + 3) $$
The first one fails, you get $A = 8/5$ and $B = 2/5,$ the coefficient of $t^2$ is wrong.
The second one works, $C+D=2,$ then $3C-D= -6.$ Add these to get $4C = -4,$ then $C = -1,$ then $D = 3.$ The $t^2$ term also works out correctly.
$$ (t^2 -t - 1) (t^2 + 3 t + 3) = t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
To get back to $x$ I suggest multiplying by $9 \cdot 9 = 81$ for
$$ (9t^2 -9t - 9) (9t^2 + 27 t + 27) = 81t^4 + 162 t^3 - 81 t^2 - 486 t - 243 \; . \; $$
Finally $t = x/3$ becomes
$$ (x^2 -3x - 9) (x^2 + 9 x + 27) = x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \; $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| $$\int{\frac 12 \cos(2u)+\frac12 \space du}=\frac12\int{\cos(2u)+1\space du}$$
$$=\frac 12 \bigg(\frac 12 \sin(2u)+u\bigg)=\frac 14\sin(2u)+\frac 12$$
You just missed that when we differentiate or integrate trigonometric functions, their arguments don't change.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate $\int_0^{\pi/2}\ln^2(\sin\theta)\,\mathrm d\theta$ How can I evaluate
$$\int_0^{\pi/2}\ln^2(\sin\theta)\,\mathrm d\theta$$
With Wolfram Alpha (see) I get to the next result
$$ \int_0^{\pi/2}\ln^2(\sin\theta)\,\mathrm d\theta = \frac{1}{24} \left(\pi^3 + 3\pi \ln^2(4)\right)$$
How to prove these ?
| Using the Fourier expansion, we have that
\begin{align}
\log(\sin x) = -\log 2-\sum^\infty_{k=1}\frac{\cos(2kx)}{k}
\end{align}
for $0 \leq x \leq \pi/2$. Then we see that
\begin{align}
\log^2(\sin x) = \left(\log 2 +\sum^\infty_{k=1}\frac{\cos(2kx)}{k} \right)^2
\end{align}
which means
\begin{align}
&\int^{\pi/2}_0\left( \log 2 +\sum^\infty_{k=1}\frac{\cos(2kx)}{k} \right)^2\ dx \\
&= \int^{\pi/2}_0\log^2 2+2\log 2\sum^\infty_{k=1}\frac{\cos(2kx)}{k}+\sum^\infty_{k=1}\sum^\infty_{n=1} \frac{\cos(2kx)\cos(2nx)}{kn}\ dx\\
&= \frac{\pi}{2} \log^2 2+2\log 2 \sum^\infty_{k=1}\int^{\pi/2}_0 \frac{\cos(2kx)}{k}\ dx+\sum^\infty_{k=1}\sum^\infty_{n=1}\int^{\pi/2}_0 \frac{\cos(2kx)\cos(2n x)}{kn}\ dx\\
&= \frac{\pi}{2}\log^2 2 + \sum^\infty_{n=1} \int^{\pi/2}_0 \frac{\cos^2(2nx)}{n^2}\ dx = \frac{\pi}{2}\log^2 2 +\frac{\pi}{4}\sum^\infty_{n=1}\frac{1}{n^2}\\
&= \frac{\pi}{2}\log^2 2 +\frac{\pi^3}{24} =\frac{1}{24}\left(\pi^3+3\pi\log^2 4 \right)
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing the surface integral of the octant of a sphere with polar coordinate substitution Let me first describe where I start:
$$\iint_Sz^2\,dS$$
We want to compute the surface integral of the octant of a sphere $S$.
The radius = 1.
The sphere is centered at the origin.
$$S=x^2+y^2+z^2=1.$$
$$z=f(x,y)=\sqrt{1-x^2-y^2}$$
$R$ is the projection of $S$ on the $xy$-plane.
Now we compute the normalization factor used to project the integral on the $xy$-plane. First we compute the derivative $\frac{\partial f}{\partial x}$:
If
$$m=1-x^2-y^2$$
and
$$n=\sqrt{m}\,,$$
then the derivative of $n$ is
$$n'=\frac12m^{-\frac12},$$
and the derivative of $m$ (with regard to $x$) equal to:
$$m'=-2x.$$
Now we can compute $\frac{\partial f}{\partial x}$ using the chain rule:
$$\frac{\partial f}{\partial x}=n'\cdot m'=\frac12m^{-\frac12}\cdot m'=\frac12(1-x^2-y^2)^{-\frac12}\cdot-2x=-{\frac{x}{\sqrt{(1-x^2-y^2)}}}.$$
Since $\sqrt{(1-x^2-y^2)}=z$:
$$\frac{\partial f}{\partial x}=-{\frac xz}$$
And in the same manner (using the derivative of $m$ with regard to $y$) we can calculate $\frac{\partial f}{\partial y}$:
$$\frac{\partial f}{\partial y}=-{\frac yz}$$
Now we can calculate the normalization factor:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\sqrt{1+\frac{x^2}{z^2}+\frac{y^2}{z^2}}=\frac1z\sqrt{x^2+y^2+z^2}.$$
Since $x^2+y^2+z^2=1$:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\frac1z$$
Now we can calculate the projection $R$ of the sphere $S$ on the $xy$-plane:
$$\iint_Sz^2\,dS=\iint_R z^2\frac1z \,dx\,dy=\iint_R z\;dx\,dy.$$
Substituting for $z$ finishes the conversion of the surface integral (remember $z=f(x,y)=\sqrt{1-x^2-y^2}$):
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy.$$
This is where my problem starts:
The book I'm reading says if we convert this to polar coordinates, the integration should be trivial.
So we convert to polar coordinates:
$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$z=f(x,y)=f(r\cos\theta,r\sin\theta)$$
Calculate the Jacobian determinant:
$$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\\ \end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\,(\cos^2\theta + \sin^2\theta)=r$$
Substitute:
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy=\iint_T\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta}\cdot r \cdot dr\,d\theta$$
$$\iint_T\sqrt{1-r^2(\cos^2\theta+\sin^2\theta)}\cdot r \cdot dr\,d\theta = \iint_T\sqrt{1-r^2}\cdot r \cdot dr\,d\theta=\iint_T\sqrt{r^2-r^4}\cdot dr\,d\theta$$
Since the radius $r=1$ it is easy to see that:
$$\int^1_0\sqrt{r^2-r^4}\cdot dr=\int^1_0(r^2-r^4)^{\frac12}\cdot dr = \int^1_0 \frac{(r^2-r^4)^{\frac32}}{\frac32}=0$$
According to the book the result of the calculation of the surface of the sphere in the first octant should be $\pi/6$.
That won't happen if $\int^1_0\sqrt{r^2-r^4}\cdot dr=0$.
The domain of $\theta$ is:
$$0\le\theta\le\frac12\pi$$
So where am I going wrong?
| The integration you are doind at last step is wrong. You can apply power rule for linear polynomial in $x$. For example you can do this $\int (4x + 2)$ or $\int (1-x)
$
But if there is quadratic polynomial in $x$, then there are different rules for integration. Usually you substitute. In you case there was quadratic polynomial in $x$ present with linear polynomial in $x$. So you know $d(x^2)=2x$. So put $x^2=t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Limit $\lim_{x \to \infty}\left (1- \frac1x\right)^x = ?$ $$\lim_{x \to \infty} \left(1+ \frac1x\right)^x = e$$
, then$$\lim_{x \to \infty} \left(1- \frac1x\right)^x =\; ?$$
I tried $\lim_{x \to \infty} \left(1- \frac1x\right)^x = \lim_{x \to \infty}\left(\frac{x-1}{x}\right)^x$, but this is not helpful.
| Try this:
$$
1-\frac{1}{x}=\frac{1}{\frac{x}{x-1}}=\frac{1}{1+\frac{1}{x-1}}
$$
so
$$
\lim_{x\rightarrow \infty} \left(1-\frac{1}{x}\right)^x=\lim_{x\rightarrow \infty} \left(\frac{1}{1+\frac{1}{x-1}}\right)^x=\lim_{x\rightarrow \infty} \frac{1}{\left(1+\frac{1}{x-1}\right)^{x-1}} \left(\frac{1}{1+\frac{1}{x-1}} \right)=\frac{1}{e}\cdot 1=\frac{1}{e}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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What is the coefficient of $x^n$ in the expansion of $(1+x+x^2+x^3+\dots)^4$? I have just come across a question (in a past exam paper for a module that I will be taking soon) which asks what the coefficient of $x^n$ is in the expansion of
$$
(1+x+x^2+x^3+x^4+\dots)^4
$$
Can anyone give me an idea as to how this might be done? Using Wolfram Alpha to expand the brackets and then by inspecting the coefficient of each power manually, I was able to deduce that the coefficient of $n^\text{th}$ term is given by $C(n+2, 3)$. How might I have been able to work this out for myself?
| Combinatoric Approach
Suppose that we pick $x^a$ from the first factor of $\left(1+x+x^2+x^3+\cdots\right)$, $x^b$ from the second, $x^c$ from the third, and $x^d$ from the fourth. Therefore, we need to count how many ways we can choose $a,b,c,d\ge0$ so that $a+b+c+d=n$ to count how many terms we get whose product is $x^n$.
To do this, we can use Stars and Bars ($3$ bars and $n$ stars).
Binomial Theorem
$$
\begin{align}
\left(1+x+x^2+x^3+\cdots\right)^4
&=(1-x)^{-4}\\
&=\sum_{n=0}^\infty\binom{-4}{n}(-x)^n\\
&=\sum_{n=0}^\infty\binom{n+3}{3}x^n\\
\end{align}
$$
since $\binom{-4}{n}=(-1)^n\binom{n+3}{n}=(-1)^n\binom{n+3}{3}$ as shown in this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int\frac{2t^2}{t^4+1}dt$
Integrate $\int\frac{2t^2}{t^4+1}dt$
While evaluating the integral $\int\sqrt{\tan x}dx$ in Evaluating the indefinite integral $\int\sqrt{\tan x}dx$. using the substitution $t^2=\tan x\implies2tdt=\sec^2x.dx$, thus
$$
\int\sqrt{\tan x}dx=\int\frac{2t^2}{t^4+1}dt
$$
This is solved using partial fractions, Check answers of @Bhaskara-III, @Harish Chandra Rajpoot. But, what if I try the following
My Attempt
$$
\int\frac{2t^2}{t^4+1}dt=\int\frac{2t^2}{(t^2+i)(t^2-i)}dt\\
\frac{2t^2}{(t^2+i)(t^2-i)}=\frac{A}{t^2+i}+\frac{B}{t^2-i}\\
2t^2=A(t^2-i)+B(t^2+i)\implies A=1, B=1\\
\color{red}{\frac{2t^2}{(t^2+i)(t^2-i)}=\frac{1}{t^2+i}+\frac{1}{t^2-i}}\\
$$
$$
\int\frac{2t^2}{t^4+1}dt=\int\frac{1}{t^2+i}dt+\int\frac{1}{t^2-i}dt=\int\frac{1}{t^2+(\sqrt{i})^2}dt+\int\frac{1}{t^2-(\sqrt{i})^2}dt
$$
Is it possible to somehow finish the integration with my substitution ?
Pls check: integrating square root of $\tan x$, answer by @Mhenni Benghorbal, seems to be a similar substitution as in my attempt.
| Hint
You have $$\int \frac {2t^2dt}{t^4+1}$$
Which can be written as $$\int \frac {(t^2+1+t^2-1)dt}{t^4+1}$$
$$\int \frac {(t^2+1)dt}{t^4+1}+\int \frac {(t^2-1)dt}{t^4+1}$$
$$\int \frac {(1+t^{-2})dt}{t^{-2}+t^2}+\int \frac {(1+t^{-2})dt}{t^2+t^{-2}}$$
For the first one substitute $$u=x-\frac 1x$$ and for second one substitute $$p=x+\frac 1x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Basis Transformation Matrix Let $A=
\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix}$ be a transformation $T:\mathbb{R}^3\to \mathbb{R}^3$ by the elementary basis $B$ and let $V=\{(1 ,0 ,0),(1 ,1 ,0),(0 ,0 ,1)\}$ basis such that the $A$ is diagonal.
I am trying to show that $[I]^{B}_{V}[T]_{B}=[T]_V$
So we have to solve
$(1,0,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
$(0,1,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
$(0,01,1)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
which is $\begin{pmatrix}
1 & -1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ but $\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix}\cdot \begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix}\neq \text{diagonal}$
| Let
$$S=
\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$$
then by the change of basis $y=Sv$ and $x=Su$, we have that
$$y=Ax \implies Sv=ASu \implies v=S^{-1}ASu$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| Following @LoveInvariants's hint, we get the following. $$\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2=a+2\implies x+\frac1x=\sqrt{a+2}$$ since $x>0$.
Next, we have $$\left(x^2+\frac1{x^2}\right)\left(x+\frac1x\right)=x^3+\frac1{x^3}+x+\frac1x=a\sqrt{a+2}$$ so $$x^3+\frac1{x^3}=(a-1)\sqrt{a+2}$$
Now do $$\left(x^3+\frac1{x^3}\right)\left(x^2+\frac1{x^2}\right)=x^5+\frac1{x^5}+x+\frac1x=a(a-1)\sqrt{a+2}$$ and therefore we have $$x^5+\frac1{x^5}=[a(a-1)-1]\sqrt{a+2}$$
| {
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"answer_id": 1
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Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, Hence:
$A=-\frac{1}{4}$ and $B=\frac{1}{4}$
But is this correct?
| Denote $F(x)$ this rational function. A priori, the decomposition into partial fractions over $\mathbf R$ has the form:
$$F(s)= \frac{As+B}{s^2+2s+2}+ \frac{Cs+D}{s^2-2s+2}.\tag{1}$$
We can determine the coefficients a little faster than with the basic method, if we observe that $F(s)$ is an odd function, so that
$$ F(-s)= \frac{-As+B}{s^2-2s+2} + \frac{-Cs+D}{s^2+2s+2}=-F(s)=-\frac{As+B}{s^2+2s+2}- \frac{Cs+D}{s^2-2s+2}$$
By identification, we obtain that $\;-As+B=-Cs-D$, whence
$$A=C,\quad B=-D.$$
Now, multiply both sides of $(1)$ by $s$ and let $s\to\infty$, we obtain the limits
$$0= A+C,\enspace\text{whence }\; A=C=0.$$
Next, set $x=1$:$$F(1)=\frac1{5\cdot 1}=\frac B5 +\frac D1=\frac{4D}5,\enspace\text{whence }\enspace D=\frac14,\;B=-\frac 14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solving $2^{x-z} + 2^{y-z} = 1$ Suppose we have an equation $2^{x-z} + 2^{y-z} = 1$,where $x, y, z$ are integers from $0$ to $9$. $x, y, z$ can have same values. I guess $x$ and $y$ should have $9$ possible values like $x=y=0$ and $z=1$ and so on . Or should it have more values?
| WLOG, let's assume $x\leq y$ and since $2^{x-z} + 2^{y-z} = 1 \iff 2^x+ 2^y=2^z \implies z \gt x, z\gt y$.
Since, $y \geq x \implies y-x \geq 0 \implies 2^{y-x} \geq 1$ and $2^{y-x} \in \mathbb{Z}$ and
similarily, $z\gt x \implies 2^{z-x}>1$ and $2^{z-x} \in \mathbb{Z}$
Now, $2^x+2^y=2^z \iff 2^{z-x} - 2^{y-x} =1$
Since, the difference of two even numbers can't be odd, so, $2^{y-x}$ has to be $1$ which $\implies y-x=0\implies x=y$ and
thus, $2^{z-x} = 1+1 = 2 \implies z-x = 1 \implies z = x+1$.
Thus solutions for $(x,y,z)$ are $(0,0,1),(1,1,2),(2,2,3),(3,3,4),(4,4,5),(5,5,6),(6,6,7),(7,7,8),(8,8,9)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to determine the fourth coefficient in the power series expansion of $(1 + \ln(1-x))^{-1}$?
$\dfrac{1}{1+\ln(1-x)}=\sum\limits_{n=0}^{+\infty}a_n x^n$, then $a_4=$?
My approach:
\begin{align*}\dfrac{1}{1+\ln(1-x)}&=\sum\limits_{n=0}^{+\infty}(-1)^n(\ln(1-x))^n\\
&=\sum\limits_{n=0}^{+\infty}(-1)^n\left(\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\end{align*}
Then $$a_4=(-1)^1\cdot\dfrac{1}{4}+(-1)^2\cdot\left(\dfrac{1}{2}\cdot \dfrac{1}{2}+\binom{2}{1}\cdot 1\cdot\dfrac{1}{3}\right)+ (-1)^3 \cdot\binom{3}{1}\cdot 1\cdot1\cdot\dfrac{1}{2}+ (-1)^4\cdot 1\cdot1\cdot 1\cdot 1=\dfrac{1}{6}$$
But the answer is $\dfrac{11}{3}$, ignoring $(-1)^n$. What’s the problem?
| Not smart at all.
In this problem, you are not asked for $a_n$ or even for $a_{123}$ but just for $a_4$.
Being lazy, what I would have done is to use
$$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac{1}{1+\ln(1-x)}=\frac 1 {1-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}+O\left(x^6\right)}$$ and then long division to get
$$\frac{1}{1+\ln(1-x)}=1+x+\frac{3 x^2}{2}+\frac{7 x^3}{3}+\frac{11 x^4}{3}+\frac{347
x^5}{60}+O\left(x^6\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim\limits_{n→∞}\left(\frac{\sqrt{n^2+n}-1}n\right)^{2\sqrt{n^2+n}-1}$ I have the limit
$$\lim_{n\to\infty}\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}.$$
I simplified the limit by using $(f(x)-1)\cdot g(x)$, where $f(x)= \sqrt{n^2+n}-1$ and $g(x) = 2\sqrt{n^2+n}- 1$ since it is in the form of $1^\infty$. I ended up getting $$\frac{(3+2n)(n-(\sqrt{n^2-n})+1}{n},$$ which tends to $\infty$.
Please help how to proceed further or a different way to solve it.
| We have
$$\frac{\sqrt{n^2+n}-1}{n}=\sqrt{1+1/n}-1/n=1+\frac1{2n}+o\left(\frac1{n}\right)-\frac1n=1-\frac1{2n}+o\left(\frac1{n}\right)$$
then
$$\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}=\left[\left(1-\frac1{2n}+o\left(\frac1{n}\right)\right)^{\frac1{-\frac1{2n}+o\left(\frac1{n}\right) }}\right]^{(2\sqrt{n^2+n}-1)\left(-\frac1{2n}+o\left(\frac1{n}\right)\right)}\to \frac1e$$
indeed
$$(2\sqrt{n^2+n}-1)\left(-\frac1{2n} +o\left(\frac1{n}\right)\right)=-\frac{\sqrt{n^2+n}}{n}+o(1)\to -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
formula for $\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}$ Is there some sort of formula for the infinte sum:
\begin{equation}
\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}
\end{equation}
(one can assume $a,b>1$)
What I've got so far:
$$\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}\le\sum_{n = 1}^{\infty}{\frac{1}{a^n}}=\frac{1}{a-1}<\infty$$
Therefore, the series converges.
if:
\begin{equation}
f(a,b)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}
\end{equation}
then:
*
*$f(a,b)=f(b,a)$
*$f(a,0)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+0^n}}=\sum_{n = 1}^{\infty} {\frac{1}{a^n}}=\frac{1}{a-1}$
*$f(a,a)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+a^n}}=\sum_{n = 1}^{\infty}{\frac{1}{2a^n}}=\frac{1}{2}\sum_{n = 1}^{\infty}{\frac{1}{a^n}}=\frac{1}{2}\cdot\frac{1}{a-1}$
*For every constant $b$, $\lim_{a\to\infty}{f(a,b)}=0$
Also:
\begin{equation}
\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}=\sum_{n = 1}^{\infty}{\prod_{m=0}^{n-1}{\frac{1}{a-b\cdot e^{\frac{1+2m}{n}\pi i}}}}
\end{equation}
Maybe that would help...
Does anyone know the answer?
EDIT:
I calculated the first few values:
*
*$f(1, 0)=\infty$
*$f(2, 0)=1.0$
*$f(2, 1)=0.7644997803484442$
*$f(3, 0)=0.5$
*$f(3, 1)=0.40406326728086184$
*$f(3, 2)=0.32135438719750625$
*$f(4, 0)=0.3333333333333333$
*$f(4, 1)=0.27940026240596016$
*$f(4, 2)=0.2355002196515558$
*$f(4, 3)=0.1978825074467063$
| definitions
$g(a,b) = \sum_{n=1}^{\infty}{\frac{1}{a^n+b^n}}$
$f(x,r) = \sum_{n=1}^{\infty}{\frac{x^n}{1+r^n}}$
lemma 1
$ g(a,b) = f(\frac{1}{a},\frac{b}{a})$
Proof: simple subsitution
lemma 2
$ f(x,r) + f(x \cdot r, r) = \frac{x}{1-x} $
Proof
$ = f(x,r) + f(x \cdot r, r) $
$ = \sum_{n=1}^{\infty}{\frac{x^n}{1+r^n}} + \sum_{n=1}^{\infty}{\frac{(x \cdot r)^n}{1+r^n}} $
$ = \sum_{n=1}^{\infty}{(\frac{x^n}{1+r^n} + \frac{(x \cdot r)^n}{1+r^n})} $
$ = \sum_{n=1}^{\infty}{\frac{x^n+(x \cdot r)^n}{1+r^n}} $
$ = \sum_{n=1}^{\infty}{\frac{x^n(1 + r^n)}{1+r^n}} $
$ = -x^0 + \sum_{n=0}^{\infty}{x^n} $
$ = \frac{1}{1-x} - 1 $
$ = \frac{x}{1-x} $
lemma 3
$f(x,r) = (-1)^{(2^N)}f(x \cdot r^{(2^N)},r) + \sum_{n=0}^{2^N-1}{(-1)^n\frac{r^nx}{1-r^nx}} $
Proof
$ f(x,r) + f(x \cdot r, r) = \frac{x}{1-x} $
$ f(x,r) = \frac{x}{1-x} - f(x \cdot r, r) $
$ f(x,r) = \frac{x}{1-x} - \frac{rx}{1-rx} + f(x \cdot r^2 ,r) $
Each additional subsitution yields double the terms, after $N$ subsitutions we achieve the result
for $r<1$ this equation simplifies to $f(x,r) = \sum_{n=0}^{\infty}{(-1)^n\frac{r^nx}{1-r^nx}}$
lemma 4
$ f(x,r) = \frac{1}{log(r^2)}(\psi_{r^2}(log_{r^2}(x)) - \psi_{r^2}(log_{r^2}(x \cdot r)))$
Proof
$ = \frac{1}{log(r^2)} \left(
-log(1 - r^2) +
log(r^2)\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x))}}{1-(r^2)^{(n+log_{r^2}(x))}}} - \left(
-log(1 - r^2) +
log(r^2)\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x\cdot r))}}{1-(r^2)^{(n+log_{r^2}(x\cdot r))}}}
\right)
\right)$
$ = \left(
\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x))}}{1-(r^2)^{(n+log_{r^2}(x))}}} -
\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x\cdot r))}}{1-(r^2)^{(n+log_{r^2}(x\cdot r))}}}
\right)$
$ = \left(
\sum_{n=0}^{\infty}{\frac{(r^2)^n \cdot (x) }{1-(r^2)^n \cdot (x) }} -
\sum_{n=0}^{\infty}{\frac{(r^2)^n \cdot (x\cdot r) }{1- (r^2)^n \cdot (x \cdot r) }}
\right)$
$ = \left(
\sum_{n=0}^{\infty}{\frac{r^{2n}x }{1-r^{2n}x }} -
\sum_{n=0}^{\infty}{\frac{r^{2n+1}x }{1- r^{2n+1}x }}
\right)$
$ = \sum_{n=0}^{\infty}{(-1)^n\frac{r^nx}{1-r^nx}}$
Solution
$\sum_{n=1}^{\infty}{\frac{1}{a^n+b^n}} = \frac{
\psi_{\frac{a}{b}^2}(-log_{\frac{a}{b}^2}(a)) -
\psi_{\frac{a}{b}^2}(log_{\frac{a}{b}^2}(b/a^2)
}{2(log(b)-log(a))}$
$ = \frac{
\psi_{r^2}(-log_{r^2}(a)) -
\psi_{r^2}(log_{r^2}(r/a))
}{2(log(r))}, r=b/a$
Assumptions
in most it is assumed that $x$ and $r$ are positive and smaller than one however if $a>b>1$ this is always the case.
Final notes
the final result could probably be further simplified and made symmetrical, I unfortunately lack the skills to do so.
References:
[q-PolyGamma function] http://mathworld.wolfram.com/q-PolygammaFunction.html
| {
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"url": "https://math.stackexchange.com/questions/2811384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
And I am stuck...
| From here
$$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
divide by $\cos^2 \frac{\alpha}{2} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.
| Following up on my comment from above, I'm setting out to prove that
$$
\frac{(n-1)^4}4\leq 1+\cdots +(n-1)^3\leq\frac{n^4}4
$$
for suitable $n$ (say $n\geq 1$). (I feel that this ought to be easier as I only have one long sum of cubes to contend with rather than two.)
The base case is easily shown: $0\leq 0\leq \frac14$.
As for the induction step, assume $k\geq 1$ and that
$$
\frac{(k-1)^4}4\leq 1+\cdots +(k-1)^3\leq\frac{k^4}4
$$
Adding $k^3$ to all sides, we get
$$
\frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4
$$
Now in the numerator on the right-hand side we have
$$
k^4 + 4k^3\leq k^4 + 4k^3+6k^2+4k+1 = (k+1)^4
$$
and on the left-hand side we have
$$
(k-1)^4 + 4k^3 = k^4 + 6k^2-4k+1 = k^4 + 1 + 2k(3k-2)\geq k^4
$$
(where that last inequality is true because both $1$ and $2k(3k-2)$ are positive).
Gathering it all up, this yields
$$
\frac{k^4}4\leq \frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4\leq \frac{(k+1)^4}4
$$
and this finishes the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove By Induction $U_n=2^n+1.$ Given that $U_1=3,U_2=5,$ and $U_{n+2}-3U_{n+1}+2U_n=0.$
Show that $U_n=2^n+1.$
I'm stuck at showing that if $P(n+1)$ is true if $P(n)$ is true.
| It's hard using induction. The characteristic equation is $$z^2-3z+2=0$$which has two roots $$z_1=1\\z_2=2$$then the equation of $U_n$ is $$U_n=a\cdot2^n+b\cdot1^n$$which by substituting $U_1$ and $U_2$ yields the same result.
alternative way (induction)
Consider $w_n=u_{n+1}-u_{n}$ therefore$$w_{n+1}=2w_n$$with $w_1=2$ which leads to $$w_n=2^n$$also $$u_{n+1}=u_n+2^n$$ or $$u_{n+1}=u_1+2+2^2+2^3+2^4+\dots +2^n=2+1+2+2^2+2^3+2^4+\dots +2^n=2^{n+1}-1+2=2^{n+1}+1$$or alternatively$$u_n=2^n+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the value of $\cot(16)\cot(44)+\cot(44)\cot(76)-\cot(76)\cot(16)$ Find the value of $$S=\cot(16)\cot(44)+\cot(44)\cot(76)-\cot(76)\cot(16)$$
Note:All angles are in degrees
My method:
I used the identity
$$\tan(x)\tan(60+x)\tan(60-x)=\tan(3x)$$
So choosing $x=16$ we get
$$\tan76 \tan44 \tan 16=\tan48$$
hence
$$S=\frac{\tan(76)+\tan(16)-\tan(44)}{\tan(48)}$$
$$S=\frac{\frac{\sin(76)}{\cos(76)}-\frac{\sin(44)}{\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{\sin(32)}{\cos(76)\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{2\sin(32)}{2\cos(76)\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{2\sin(32)}{\frac{-1}{2}+\cos(32)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
can i need a clue from here?
| Let us see how the problem came into being in my understanding.
From Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
if $\cot3x=\cot3y$ for non-zero finite $\tan x,$
$$\tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$
$$\iff\cot^3x\tan3y-3\cot^2x-3\tan3y\cot x+1=0$$
whose roots are $\cot\left(x+60^\circ n\right), n=0,1,2$
$$\implies\cot x\cot\left(x+60^\circ\right)+\cot\left(x+60^\circ\right)\cot\left(x+120^\circ\right)+\cot\left(x+120^\circ\right)\cot x=\dfrac{-3\tan3y}{\tan3y}$$
$=-3$ for for non-zero finite $\tan3y$
Set $x=16^\circ$ and $\cot\left(16^\circ+120^\circ\right)=\cot(180^\circ-44^\circ)=-\cot44^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Using the Lagrange Multipliers Method to prove $\ |h(x,y)|\leq 1$
Using the Lagrange multipliers method, I have found that the maximum and minimum values of the function
$$f(x,y)=xy$$
on the curve
$$x^2-yx+y^2=1$$ are
$$1 \ \text{at} \ (\pm1,\pm1) \ \ \ \text{and} \ \ -\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big) \ \ \text{respectively.}$$
Using this, prove that $$\Big|\frac{xy}{x^2-yx+y^2}\Big|\leq1 \ \ \ \forall(x,y)\neq0$$
I don't really know where to start. I thought of multiplying across as the inequality will be preserved, which yields
$$|xy|\leq|x^2-yx+y^2|$$
then
$$|xy|-|x^2-yx+y^2|\leq 0$$
At this point, I'm relying on my algebraic manipulation skills to yield something true. I'm not using the information above obtained by the Lagrange multipliers method.
| Equivalently, you prove: $(xy)^2 - (x^2-xy+y^2)^2 \le 0\iff (xy-x^2+xy-y^2)(xy+x^2-xy+y^2) \le 0\iff(2xy-x^2-y^2)(x^2+y^2) \le 0\iff -(x-y)^2(x^2+y^2) \le 0$ which is true.
We can show the max/min result without bothering Lagrange ! We have:
$(x-y)^2 \ge 0 \implies x^2-2xy+y^2 \ge 0 \implies x^2 -xy+y^2 \ge xy \implies 1 \ge xy\implies \text{max} = 1$
And observe $xy \ge -\dfrac{x^2-xy+y^2}{3}= -\dfrac{1}{3}\implies \text{min} = -\dfrac{1}{3}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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Proving trigonometric problem from given trigonometric equations If $p$ & $q$ are the solutions of $$a \cos x + b \sin x = c$$
Then how do I prove that, $$\cos (p + q) = \dfrac{a^2-b^2}{a^2 + b^2} $$
I tried all the adjustments I could think of, like dividing by $ \cos x $ and extracting $a$ and $b$ from the 2 equations. Also tried adding/subtracting and all the basics I know to no avail.
Any help is appreciated, thank you :)
| Y'all no worries! I figured it out – (by a fluke?), I just want someone to confirm this is right :)
Converting to it to original question in my textbook. Let $p= \alpha$ and $q = \beta$
$\alpha$ and $\beta$ are solutions of $a\cos\theta+b\sin\theta=c$
$$a\cos\alpha+b\sin\alpha=c=a\cos\beta+b\sin\beta $$
$$b(\sin\alpha-\sin\beta)=-a(\cos\alpha-\cos\beta)$$
$$ 2b\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)=2a\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$
$$\frac{b}{a}=\frac{\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}=\tan\left(\frac{\alpha+\beta}{2}\right) $$
$$ \tan^2\left(\frac{\alpha+\beta}{2}\right)=\frac{b^2}{a^2}$$
$$\cos(\alpha+\beta)=\frac{\cos^2\left(\frac{\alpha+\beta}{2}\right)-\sin^2\left(\frac{\alpha+\beta}{2}\right)}{\cos^2\left(\frac{\alpha+\beta}{2}\right)+\sin^2\left(\frac{\alpha+\beta}{2}\right)}$$
Dividing the numerator and denominator by $\cos^2\left(\frac{\alpha+\beta}{2}\right)$ , we get,
$$\cos(\alpha+\beta)=\frac{1-\tan^2\left(\frac{\alpha+\beta}{2}\right)}{1+\tan^2\left(\frac{\alpha+\beta}{2}\right)}=\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$$
$$\Rightarrow\qquad \cos(\alpha+\beta)=\frac{a^2-b^2}{a^2+b^2}$$
Thanks to everyone for their efforts :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Gaussian elimination with an unknown constant How do I do Gaussian elimination on this system of equations?
$$
\begin{cases}
x_1+x_2+x_3=0\\
x_1+2x_2+ax_3=1\\
x_1+ax_2+2x_3=-1
\end{cases}
$$
When I try to do this I end up with a coefficient with an $a^2$ term which I'm assuming is not supposed to be there.
| You equation is
$$ \left[
\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
1 & 2 & a & 1 \\
1 & a & 2 & -1 \\
\end{array}
\right]$$
By doing $R_2 - R_1 \to R_2$, $R_3 - R_1 \to R_3$ you get
$$ \left[
\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & a-1 & 1 \\
0 & a-1 & 1 & -1 \\
\end{array}
\right]$$
Now, $R_3 - (a-1)R_2\to R_3$
$$ \left[
\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & a-1 & 1 \\
0 & 0 & 1 - (a-1)^2 & -1 -(a-1) \\
\end{array}
\right]$$
Simplify
$$ \left[
\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & a-1 & 1 \\
0 & 0 & -a^2 + 2a & -a \\
\end{array}
\right]$$
Now can divide the last row by $-a$ (assuming $a\ne0$), that is, $-\frac1a R_3\to R_3$
$$\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & a-1 & 1 \\
0 & 0 & a - 2 & 1 \\
\end{array}
\right]$$
I assume you can finish from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
| Here is a way to calculate the integral without Feynman's trick or trigonometric substitutions.
By using the substitution previously used by Shashi we have:
$$\underbrace{\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx}_{x=\frac{1}{\sqrt{1+t^2}}}=\underbrace{\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt}_{t=\frac{1}{t}}$$
$$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt-\int _0^{\infty \:}\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt$$
$$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\frac{1}{2}\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$
$$=\frac{1}{2}\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt+\frac{1}{2}\underbrace{\int _1^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt}_{t=\frac{1}{t}}=\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$
$$=\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(1+t^2\right)}{t}\:dt}_{t=t^2}$$
$$=\frac{3}{4}\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt=\frac{3}{4}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}=\frac{3}{8}\zeta \left(2\right)$$
Thus:
$$\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx=\frac{3}{8}\zeta \left(2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
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Angle between two vectors given magnitudes only I have a physics problem about vectors:
Two vectors $\boldsymbol{A}$ and $\boldsymbol{B}$ have precisely equal magnitudes. For the magnitude of $\boldsymbol{A} + \boldsymbol{B}$ to be $100$ times larger than the magnitude of $\boldsymbol{A} - \boldsymbol{B}$, what must be the angle between them?
I have tried to draw a diagram and calculate the problem with geometrical methods with two simultaneous equations of the form $c^2 = a^2 + b^2 - 2ab \cos θ$:$$
|\boldsymbol{A} + \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² - 2|\boldsymbol{A}||\boldsymbol{B}|\cos θ\\
|\boldsymbol{A} - \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² - 2|\boldsymbol{A}||\boldsymbol{B}|\cos(π - θ)
$$
Equating these two equations in terms of $θ$ gives$$
\cos θ = -\frac{9999|\boldsymbol{A} + \boldsymbol{B}|²}{|4|\boldsymbol{A}|²|}.
$$
This is as far as i could get, any help solving the problem will be greatly appreciated
| You haven't used the fact that $|A| = |B|$
$|A+B| = 100|A-B|\\
|A|^2 + |B|^2 + 2|A||B|\cos \theta = 100^2(|A|^2 + |B|^2 - 2|A||B|\cos \theta)$
let $a = |A| = |B|$
$a^2 + a^2 + 2a^2\cos\theta = 100^2(a^2 + a^2 - 2a^2\cos\theta)$
Isolate $\cos \theta$ and simplify
$2(100^2+ 1)a^2 \cos \theta = 2a^2(100^2-1)\\
\theta = \arccos \frac {100^2 - 1}{100^2 + 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Generalization of Alternating Harmonic Series Consider the following sum for positive integers $a$ and $b$
$$\sum_{n = 0}^\infty \frac{(-1)^n}{an + b}. $$
What is the closed form of this sum? Solutions, partial progress, and suggestions appreciated.
Note: The original problem asks for $a = 8, b = 3$, though we haven't solved this case yet.
Things my friends and I tried: For the case $a = 8, b = 3$, it is equivalent to $\int_0^1 \frac{x^2}{1 + x^8} \, dx$, not much progress here. Next, we tried writing $f(x) = \sum \frac{x^n}{an+b}$ or $f(x) = \sum \frac{\cos(nx)}{an + b}$ and writing differential equations.
| $$\sum_{n = 0}^\infty \frac{(-1)^n}{a\,n + b}=\frac 1 a\,\Phi \left(-1,1,\frac{b}{a}\right)$$ where appears the Lerch transcendent function.
For $a=8$ and $b=3$, the result expresses in terms of the digamma function; it is
$$\frac{1}{16} \left(\psi ^{(0)}\left(\frac{11}{16}\right)-\psi
^{(0)}\left(\frac{3}{16}\right)\right)\approx 0.273898$$ For
$$I=\int_0^1 \frac{x^2}{1 + x^8} \, dx$$ $$I==\frac{1}{8} \left(\cos \left(\frac{\pi }{8}\right) \left(\pi -2 \tanh
^{-1}\left(\sin \left(\frac{\pi }{8}\right)\right)\right)-\sin \left(\frac{\pi
}{8}\right) \left(\pi -2 \tanh ^{-1}\left(\cos \left(\frac{\pi
}{8}\right)\right)\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the value of $x^2 + y^2 + z^2$. If the real numbers $x, y, z$ are such that $x^2 +4y^2 +16z^2=48$ and $xy+4yz+2zx=24$, What is the value 0f $x^2+y^2+z^2 ?$.
The value of $x+2y+4z = \left\lvert 12 \right\rvert$. I don't know how to proceed after that.
| $$x^2+4y^2+16z^2=48$$
$$xy+4yz+2zx=24$$
Now,
$$2(x^2+4y^2+16z^2)-4(xy+4yz+2zx)=0$$
$$(x-2y)^2+(2y-4z)^2+(x-4z)^2=0$$
Let say $x=2y=4z=k$
$$x=k,y=\frac k2,z=\frac k4$$
Since $k^2+\frac{4k^2}{4}+\frac{16k^2}{16}=48$
$$k=4$$
$$x=4,y=2,z=1$$
$$x^2+y^2+z^2=16+4+1=21$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that the equation $ xyz=xy+xz+yz+x+y+z$ has finite numbers of natural solutions. Prove that the equation
$$\begin{equation}
xyz=xy+xz+yz+x+y+z\tag{*}
\end{equation}$$
has finitely many natural solutions.
My attempt. Suppose that $x \leq y \leq z$ then
$$
xyz\leq xz+xz+yz+3z \implies xy \leq 2x+y+3 \implies x \leq \frac{y+3}{y-2}
$$
Since $$x \leq \frac{y+3}{y-2}=\frac{y-2+5}{y-2}=1+\frac{5}{y-2} \implies (x-1)(y-2) \leq 5,$$
we can conclude that the last inequality has finite number of natural solutions, so there are only finite numbers for $x,y.$
For any fixed $x,y$ the equation (*) reduces to linear equation for $z$ with finite numbers of solutions.
Is it ok?
Is there any other solutions?
| Add $xyz+1$ to both sides to obtain $$2xyz+1=xyz+xy+yz+zx+x+y+z+1=(x+1)(y+1)(z+1)$$
The left-hand side is odd, so the right-hand side must also be odd, whence $x,y,z$ are all even.
Now divide by $xyz$ to obtain $$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{xyz}\gt 2$$
Now note that $\left(\frac 54\right)^3=\frac {125}{64}\lt 2$, so that one of $x,y,z$ at least must be equal to $2$ - say $x$.
We then have $$\frac 32\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{2yz}$$ which gives $$\left(1+\frac 1y\right)\left(1+\frac 1z\right)\gt \frac 43$$ from which it follows that one of $x$ and $y$ must be less than $8$. Note that it can't be $6$, because that would not work in the top equation of this answer taken mod $3$. Once $y$ is fixed, there is at most one possibility for $z$ (the equation is linear in $z$) and any answer must be a permutation of one with $x\le y\le z$.
Using this method, you should be able to find the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\cos(\frac{2\pi}{5})$ in terms of $\sin(\frac{\pi}{5})$ I intend to calculate $\cos\left(\frac{2\pi}{5}\right)$ via the formula
$$\sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$$
This could be expressed as:
$\sqrt{\frac{5-\sqrt{5}}{8}}=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$. Squaring both sides and manipulate a bit, this will give $\cos\left(\frac{2\pi}{5}\right)=-\frac{1-\sqrt{5}}{4}$. However, the true result is $\frac{\sqrt{5}-1}{4}$. What is wrong with my derivation?
| Note that
$$-\frac{1-\sqrt{5}}{4}=(-1)\cdot\frac{1-\sqrt{5}}{4}=\frac{(-1)\cdot1-(-1)\cdot(\sqrt{5})}{4}=\frac{\sqrt{5}-1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of $\sum_{k=0}^{49} (-1)^{k} \binom{99}{2k}$ I have a finite sum to evaluate. I do compute the sum using Euler's formula. I would appreciate a combinatorial argument, or an argument using a binomial identity.
Problem
Among all the sums obtained from adding the coordinates to the integral solutions of
\begin{equation*}
\sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} = x^{y} ,
\end{equation*}
which is the biggest?
Solution
According to Euler's formula,
\begin{align*}
&2^{99/2} \left(\frac{-1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) \\
&\qquad = (2^{99/2}) e^{i \frac{3\pi}{4}} \\
&\qquad = 2^{99/2} e^{i \frac{99}{4} \pi} \\
&\qquad = (1 + i)^{99} \\
&\qquad = \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} + i \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k+1} .
\end{align*}
So,
\begin{equation*}
-2^{49} = -2^{99/2} \left(\frac{1}{\sqrt{2}}\right) = \sum_{k=0}^{49} (-1)^{k} \binom{99}{2k} .
\end{equation*}
Since $-2$ is a prime number, the only integral solutions to the given equation are $(-2, \, 49)$, $(-2^{7}, \, 7)$, and $(-2^{49}, \, 1)$. The biggest sum obtained from adding the coordinates to the integral solutions is $-2 + 49 = 47$.
| The way is right.
Really,
$$\sum\limits_{k=0}^{49}(-1)^k\binom{99}{2k} = \Re (1+i)^{99} = \Re (2i)^{49}(1+i)= -2^{49}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$A^2+B^2 +AB=36. B^2+C^2+BC=49. C^2+A^2+AC=64.$ Find $(A+B+C)^2$
$$A^2+B^2 +AB=36.\\ B^2+C^2+BC=49.\\ C^2+A^2+AC=64.$$ Find $(A+B+C)^2$.
I have tried it by using geometry I.e constructing a triangle and marking a point inside it which is making 120 ° and then using cosine rule
But have difficulty in solving further
Please use geometry
| Here is a purely algebraic way. Subtract the equations in pairs to get$$(C-B)(A+B+C)=28,$$$$(C-A)(A+B+C)=13,$$$$(A-B)(A+B+C)=15.$$Let us write $x=(A+B+C)^2$ and $y=BC+CA+AB$. Then squaring and adding the above three equations yields$$x(x-3y)=589.$$Summing the original equations gives$$2x-3y=149.$$Eliminating $3y$ between these equations results in the quadratic equation$$x^2-149x+589=0.$$Solving this, we get$$x=\tfrac12(149\pm63\surd5),$$where the larger root corresponds to the case when $A$, $B$, and $C$ are all positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Polynomial division and geometric series By long division $$\dfrac{5x^2+2x-4}{x-2} =5x+12+\color{red}{\dfrac{20}{x-2}}$$
If we continue dividing we get
$$\dfrac{5x^2+2x-4}{x-2} =5x+12+\color{red}{ 20x^{-1}+40x^{-2}+ \dfrac{80x^{-2}}{x-2}}$$
I'm trying to understand two items here
1) What does the expansion with negative exponents represent ?
2) If we expand the same using geometric series formula I get a different result (with out negative exponents) :
$$\color{Red}{\dfrac{20}{x-2} = -10\dfrac{1}{1-(x/2)}=-10\left(1+(x/2) + (x/2)^2 + \cdots\right)} $$
Why did the expansion using long division differ from geometric series ?
3) Since geometric series only makes sense when $|x/2|\lt 1$, I'm not so sure if the above expansion is valid..
| This expansion is the Laurent series of the function $\dfrac{5x^2+2x-4}{x-2}$ around $x = 0$ on the annulus $|x| < 2$.
Actually, you are getting the same result both ways:
\begin{align}
\frac{20}x + \frac{40}{x^2} + \frac{80}{x^2(x-2)} &= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2}\frac1{1-\frac{x}2}\\
&= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2}\sum_{n=0}^\infty \frac{x^n}{2^n}\\
&= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2} - \frac{20}x - 10\sum_{n=2}^\infty \frac{x^{n-2}}{2^{n-2}}\\
&= - 10\sum_{n=0}^\infty \frac{x^{n}}{2^{n}}\\
\end{align}
The Laurent series on $|x| > 2$ is
\begin{align}
\dfrac{5x^2+2x-4}{x-2} &=5x+12+\color{red}{\dfrac{20}{x-2}}\\
&= 5x+12+\color{red}{\frac{20}{x}\frac{1}{1-\frac2x}}\\
&= 5x+12+\color{red}{\frac{20}{x}\sum_{n=0}^\infty \frac{2^n}{x^n}}\\
&= 5x+12+\color{red}{10\sum_{n=1}^\infty \frac{2^n}{x^n}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843241",
"timestamp": "2023-03-29T00:00:00",
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The set of integers $n$ expressible as $n=x^2+xy+y^2$
Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$
Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?
I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$
I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$.
The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity.
Any ideas ?
| Denote $a^2+ab+b^2$ by $L(a, b)$. Numbers of the form $L$ are known as Loeschian numbers. What you are trying to prove is that the set $S$ of Loeschian numbers is multiplicative.
Here's a direct proof that the product of two Loeschian numbers is Loeschian.
\begin{align*}
L(a, b)L(c, d) &=(a^2+ab+b^2)(c^2+cd+d^2)\\
&= a^2c^2+a^2cd+a^2d^2+abc^2+abcd+abd^2+b^2c^2+b^2cd+b^2d^2\\
&=[a^2c^2+2abc^2+b^2c^2+2abcd+2b^2cd+b^2d^2]+{}\\
&\qquad +[a^2cd-abc^2+abcd-b^2c^2+abd^2-b^2cd]+{}\\
&\qquad +[a^2d^2-2abcd+b^2c^2]\\
&=(ac+bc+bd)^2+(ac+bc+bd)(ad-bc)+(ad-bc)^2\\
&=L(ac+bc+bd, ad-bc)
\end{align*}
In addition,
\begin{align*}
L(a, b)L(c, d)
&= (a^2+ab+b^2)(c^2+cd+d^2)\\
&= a^2c^2+a^2cd+a^2d^2+abc^2+abcd+abd^2+b^2c^2+b^2cd+b^2d^2\\
&=[a^2c^2-2abcd+b^2d^2]+{}\\
&\qquad +[a^2cd+abc^2+abcd-abd^2-b^2cd-b^2d^2]+{}\\
&\qquad +[a^2d^2+2abcd+b^2c^2+2abd^2+2b^2cd+b^2d^2]\\
&=(ac-bd)^2+(ac-bd)(ad+bc+bd)+(ad+bc+bd)^2\\
&=L(ac-bd, ad+bc+bd)
\end{align*}
It can also be proved that $n$ has the form $a^2+ab+b^2$ iff $n$ has the form $u^2+3v^2$. (Sketch of a proof: Where $a+b+c=0$, $L(a, b)=L(b, c)=L(c, a)$. Moreover, at least one of $a, b$ and $c$ is even. $L(a, 2v)=a^2+2av+4v^2=u^2+3v^2$ where $u=a+v$. Conversely, $u^2+3v^2=L(a, b)$ where $b=2v$ and $a=u-v$. Thus $L$ and $u^2+3v^2$ represent the same set of integers.)
Thus the set of numbers of the form $u^2+3v^2$ is multiplicative. Again there is a direct proof, and it is slightly simpler than the proof above.
| {
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Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| Note that the integrand has a period of $\pi/2$ since $$f\left(x+\frac{\pi}{4}\right) = \sin^4\left(x+\frac{\pi}{2}\right) + \cos^4\left(x+\frac{\pi}{2}\right) = (-\cos x)^4 + (\sin x)^4 = f(x) $$
So it's sufficient to compute
$$ \int_0^{2\pi} f(x)\ dx = 4\int_0^{\pi/2} f(x)\ dx $$
The comments have noted that
$$ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - \frac{1}{2}\sin^2(2x) $$
with that in mind
$$ \frac{1}{\sin^4 x + \cos^4 x} = \frac{2}{2-\sin^2 (2x)} = \frac{2\sec^2 (2x)}{2\sec^2 (2x) - \tan^2(2x)} $$
The substitution $u = \tan (2x)$, then $du = 2\sec^2 (2x)\ dx$ and $\sec^2(2x) = u^2+1$, therefore
$$ \frac{2\sec^2 (2x)\ dx}{2\sec^2 (2x) - \tan^2(2x)} = \frac{du}{u^2 + 2} $$
A potential problem with this substitution is $\tan(2x)$ has a singularity at $x=\pi/4$. This is fine as the integrand is symmetric about this point, i.e.
$$ f\left(\frac{\pi}{2}- x\right) = \sin^4\left(\frac{\pi}{2}- x\right) + \cos^4 \left(\frac{\pi}{2}- x\right) = \cos^4 x + \sin^4 x = f(x) $$
So we can write
\begin{align} \int_0^{\pi/2} f(x)\ dx &= \int_0^{\pi/4} f(x)\ dx + \int_{\pi/4}^{\pi/2} f(x) \ dx \\
&= \int_0^{\pi/4} f(x)\ dx - \int_{\pi/4}^0 f\left(\frac{\pi}{2}-x\right)\ dx \\
&= 2\int_0^{\pi/4} f(x)\ dx
\end{align}
Continuing the integration, we find
$$ \int_{0}^{2\pi} f(x)\ dx = 8\int_0^{\pi/4} f(x)\ dx = 8\int_{0}^{\infty} \frac{1}{u^2+2} du = \frac{8}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right)\Bigg|_0^\infty = \frac{8\pi}{2\sqrt{2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
Solving Quadratics Using Continued Fractions/Nonsimple to Simple Continued Fractions Let's say we want to find the continued fraction that solves the equation $x^2 - 2x - 1 = 0$.
Solution:
$$
x = 2 + \frac1x = 2 + \dfrac1{2 + \dfrac1x} = 2 + \dfrac1{2 + \dfrac1{2 + \dfrac1x}} = [2;\overline2]
$$
However, what happens if the quadratic is a bit more complicated, say, $2x^2 - 5x + 1 = 0$. If we use non-simple continued fractions to solve this, we get
$$
x = \frac52 -\dfrac{\frac12}{\frac52-\dfrac{\frac12}{\frac52-\ddots}} \stackrel{how?}= [ 2; \overline{3, 1, 1} ]
$$
My question is: How to convert non-simple continued fractions to simple continued fractions and in general how to solve quadratics using simple continued fractions?
| I believe the methods above overcomplicate things.
If $x = a + \frac{b}{x}$, then $x^2 - ax - b = 0$ which is a quadratic. Dividing both sides of $2x^2-5x+1 =0$ by $2$ gives $x^2 - \frac{5}{2}x + \frac{1}{2}$, and so comparing coefficients, $a = \frac{5}{2}, b = - \frac{1}{2}$.
Thus $x = \frac{5}{2} - \frac{1/2}{x}$ gives both solutions to $x$. We can replace the $x$ in the denominator by itself to get:
$$x = \frac{5}{2} - \frac{1/2}{x} = \frac{5}{2} - \frac{1/2}{\frac{5}{2} - \frac{1/2}{x}} = \frac{5}{2} - \frac{1/2}{\frac{5}{2} - \frac{1/2}{\frac{5}{2} + \frac{1/2}{x}}}$$
$$= \frac52 -\dfrac{\frac12}{\frac52-\dfrac{\frac12}{\frac52-\ddots}}$$
Assuming that this infinite, periodic continued fraction converges to a limit $L$, then we have $L = \frac{5}{2} - \frac{1/2}{L}$ and we can solve the quadratic.
However, this only converges to one of the roots of the given quadratic. This is the larger root in this case as if $L < \frac{5}{4}$, we have that $\frac{5}{2} - \frac{1/2}{L} < \frac{5}{4} \implies L < \frac{2}{5}$ and the maximum value of L keeps getting smaller each time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit with unknown parameter What possible values can take $a,b\in\mathbb{R}$ such that
$$
\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}-\log
_{5}\left( 3^{n}+4^{n}+5^{n}\right) -\sqrt[4]{n^{4}+1}\right) =1~?
$$
Denote $\left( a_{n}\right) $ the expresion inside limit. My idea is to
manipulate the expresions one-by-one inside the limit and I get that
\begin{align*}
\lim\limits_{n\rightarrow\infty}\left( a_{n}-1\right) & =\lim
\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}-1-\log_{5}%
5^{n}\left( \left( \dfrac{3}{5}\right) ^{n}+\left( \dfrac{4}{5}\right)
^{n}+1\right) -\sqrt[4]{n^{4}+1}\right) \\
& =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}%
-1-n+\log_{5}\left( \left( \dfrac{3}{5}\right) ^{n}+\left( \dfrac{4}%
{5}\right) ^{n}+1\right) -\sqrt[4]{n^{4}+1}\right) \\
& =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}%
+1}-1-n-\sqrt[4]{n^{4}+1}\right) \\
& =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}%
+1}-1-2n-\left( \sqrt[4]{n^{4}+1}-n\right) \right) \\
& =\lim\limits_{n\rightarrow\infty}\left( \dfrac{an^{3}+bn^{2}+1-1}
{\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n-\lim
\limits_{n\rightarrow\infty}\dfrac{1}{n^{\alpha}+...}\right) \\
& =\lim\limits_{n\rightarrow\infty}\left( \dfrac{an^{3}+bn^{2}}
{\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n\right) .
\end{align*}
where I have used $\log_{5}\left( \left( \dfrac{3}{5}\right) ^{n}+\left(
\dfrac{4}{5}\right) ^{n}+1\right) \rightarrow\log_{5}1=0.$
The limit $\lim\limits_{n\rightarrow\infty}\left( a_{n}-1\right) $ should be
equal with $0$, but I do not think that my last relation would imply this
thing. Am I wrong something?
| Write:
$$\left\{\begin{array}{lcl}
\sqrt[3]{an^{3}+bn^{2}+1}&=& n\, \sqrt[3]{a+\frac{b}n+\frac1{n^2}}\\
\log_{5}\left( 3^{n}+4^{n}+5^{n}\right) &=&
%\log_{5}\left( 5^n\left({\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)\right)=
n + \log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)\\
\sqrt[4]{n^{4}+1} &=& n\,\sqrt[4]{1+\frac1{n^4}}
\end{array}\right.
.$$
Then, we have
\begin{align}
a_n
&=
n\, \sqrt[3]{a+\frac{b}n+\frac1{n^2}}
-n -\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right)
-n\,\sqrt[4]{1+\frac1{n^4}}
\\&=
n\left(\sqrt[3]{a+\frac{b}n+\frac1{n^2}} - 1 - \sqrt[4]{1+\frac1{n^4}}\right)
-\frac1n\,\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right).
\end{align}
Now, as $n\to\infty$, we have that
$$\frac1n\,\log_{5}\left( {\left(\frac35\right)}^{n}+{\left(\frac45\right)}^{n}+1\right) \to 0.$$
Moreover, as $n\to\infty$ we have
$$\sqrt[3]{a+\frac{b}n+\frac1{n^2}} - 1 - \sqrt[4]{1+\frac1{n^4}}
\to \sqrt[3]a - 2.$$
Hence, if $a > 8$, then $a_n\to+\infty$ and if $a < 8$, then $a_n\to-\infty$.
If $a=8$, then we have an indeterminate.
Consider
$$f(x) =
\frac{\displaystyle
\sqrt[3]{8+\frac{b}x+\frac1{x^2}} - 1 - \sqrt[4]{1+\frac1{x^4}}
}{1/x}.$$
As $x\to\infty$, $f(x)$ is an indeterminate expression of the type $0/0$.
We apply L'Hôpital, obtaining that
\begin{align}
\lim_{x\to\infty}f(x)
&=
\lim_{x\to\infty}
\frac{\displaystyle
-\frac{b}{3x^2{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
-\frac{2}{3x^3{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
+\frac{1}{x^5{\left(1+\frac1{x^4}\right)}^{3/4}}
}{-1/x^2}
\\&=\lim_{x\to\infty}
\frac{b}{3{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
+\frac{2}{3x{\left(8+\frac{b}x+\frac1{x^2}\right)}^{2/3}}
-\frac{1}{x^3{\left(1+\frac1{x^4}\right)}^{3/4}}
\\&=
\frac{b}{3\cdot 8^{2/3}} = \frac b{12}.
\end{align}
It follows that
$$\lim_{n\to\infty}a_n
=\left\{\begin{array}{lccr}
-\infty&&&a<8\\
b/12&&&a=8\\
+\infty&&&a>8
\end{array}
\right.,$$
so $\lim_{n\to\infty}a_n = 1$ if and only if $(a,b) = (8, 12)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck on Generalized Wilson's Theorem I have to prove the following,
$$\prod_{\substack{n=1\\ \gcd(n,m)=1}}^{m}n\equiv \begin{cases}-1\pmod{m}\quad\text{if }m\text{ has a primitive root.}\\ +1\pmod{m}\quad\text{otherwise.}\end{cases}$$
I can prove the first part but I'm having trouble with the second part, any hints ? I tried proving the contrapositive, namely that $\Pi \equiv -1\pmod{m}$ implies $m$ has a primtive root but I didn't get far.
| The congruence
$$x^2\equiv 1\pmod{m}\Rightarrow (x-1)(x+1)\equiv 0\pmod{m}\tag{1}$$
is possible for each modulus $m$. Dispensing with the trivial cases $m=1$ or $2$, we see the $r$ roots must fall into two pairs of size $r/2$, since for any root $\rho$ we must have $-\rho$ a root too. Now by $(1)$, $\gcd(\rho,m)=\gcd(-\rho,m)=1$, and so they are incongruent modulo $m$:
$$\rho\not\equiv -\rho\pmod{m}\Rightarrow 2\rho\not\equiv 0\pmod{m}\tag{2}$$
We also have $(-\rho)(\rho)= -\rho^2\equiv-1\pmod{m}$, so it follows the product of all roots $r$ is $+1$ or $-1$ dependent on whether $4\mid r$ or not:
$$\prod_{\substack{i=1\\ 4\mid r}}^{r}\rho_i\equiv1\pmod{m}\quad\text{else}\quad\prod_{\substack{i=1\\ 4\nmid r}}^{r}\rho_i\equiv-1\pmod{m}\tag{3}$$
Now consider the number of roots of congruence $(1)$: When $m=1$ or $2$ the number of roots, $r$, is $1$;
$$(1-1)(1+1)\equiv0\pmod{m}$$
When $m$ is the power of an odd prime, $p^a$ say, or double such, $2p^a$, or $4$, then the number of roots $r$ is $2$.
\begin{align*}
(1-1)(1+1)\equiv &((2^2-1)-1)((2^2-1)+1)\equiv0\pmod{2^2}\\
(1-1)(1+1)\equiv &((p^a-1)-1)((p^a-1)+1)\equiv0\pmod{p^a}\\
(1-1)(1+1)\equiv &((2p^a-1)-1)((2p^a-1)+1)\equiv0\pmod{2p^a}\\
\end{align*}
We now need to show that otherwise $4\mid r$. This shows $r=2$ in the right-handside of $(3)$. To see this consider $m=2^b$, where $b>2$. Then we wish to find the roots
$$x^2-1=(x-1)(x+1)\equiv0\pmod{2^b}$$
Now both $(x-1)$ and $(x+1)$ are even, so we may write
$$\frac{x-1}{2}\cdot\frac{x+1}{2}\equiv0\pmod{2^{b-2}}$$
Since the difference of $\frac{x-1}{2}$ and $\frac{x+1}{2}$ is $x$ which is odd, this means one of the factors is even and the other odd. Therefore one has to be divisible by $2^{b-2}$, and we get
$$x\equiv1\pmod{2^{b-1}}\quad\text{or}\quad x\equiv-1\pmod{2^{b-1}}$$
Modulo $2^b$ we get the four incongruent roots $1$, $1+2^{b-1}$ and $-1$, $-1-2^{b-1}$. Now for a general $m=2^bp_1^{a_1}\cdots p_j^{a_j}$ we use The Chinese Remainder Theorem to find the number of roots of $x^2-1$ in $\mathbb{Z}/(m)$ by finding the number of roots in each $\mathbb{Z}/(p_i^{a_i})$, of which there will be two for each odd prime power, and the cases for $\mathbb{Z}/(2^{b})$ have been covered: namely $1$ for $\mathbb{Z}/(2^{1})$, $2$ for $\mathbb{Z}/(2^{2})$, and finally $4$ for $\mathbb{Z}/(2^{b})$ where $b>2$. So in all cases we get $4\mid r$.
Now of $\varphi(m)$ numbers $j$, $1\le j\le m-1$, that are relatively prime to $m$, we have the $r$ roots of $(1)$, and any other non-roots, $u$, $v$ say, will pair up as multiplicative inverses as they are relatively prime to $m$:
$$uv\equiv1\pmod{m}$$
Note we won't get $u^2\equiv1\pmod{m}$ as then $u$ satisfies $(1)$ and so is a root, this also means $u\not\equiv v\pmod{m}$, and it follows the product of these $\varphi(m)-r$ non-roots is $1$.
Now if we multiply all $\varphi(m)$ amount of numbers together we get the result:
$$
\prod_{\substack{n=1\\ \gcd(n,m)=1}}^{m}n\equiv
\begin{cases}
0\pmod{m}\quad\text{if $m=1$,}\\
-1\pmod{m}\quad\text{if $m=4$, $p^a$ or $2p^a$,}\\
+1\pmod{m}\quad\text{otherwise.}
\end{cases}
$$
By a primitive root modulo $m$ we mean a number $g$, such that for every $a$ relatively prime to $m$, there exists some power of $g$ having $a\equiv g^k\pmod{m}$, where we call $k$ the index or discrete logarithm of $a$ to the base $g$ modulo $n$.
The multiplicative group of integers modulo $m$, $(\mathbb{Z} /(m))^{\times}$, is cyclic if and only if $m$ is $1$, $2$, $4$, $p^a$ or $2p^{a}$, where $p$ is an odd prime and $a>0$. Now a group is cyclic if and only if it has a generator. Hence the primitive root $g$ is a generator in such cases, with $g^0$, $g^1$, $g^2,\dotsc,g^{\varphi{(m)}-1}$ giving the full set of possible residues.
This condition is equivalent to that derived before, and so
$$
\prod_{\substack{n=1\\ \gcd(n,m)=1}}^{m}n\equiv
\begin{cases}
0\pmod{m}\quad\text{if $m=1$,}\\
-1\pmod{m}\quad\text{if $m$ has a primitive root, $m\neq2$,}\\
+1\pmod{m}\quad\text{otherwise.}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Cyclic Algebra. Suppose that $a,b,c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$ and $a^3 + b^3 + c^3 = 1$.
Find all possible value(s) of $a+b+c$.
My solution :
$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)$
$\Rightarrow a+b+c=1+c(1-c^2)+a(1-a^2)+b(1-b^2)$
$\Rightarrow a+b+c=1+c-c^3+a-a^3+b-b^3$
$\Rightarrow a+b+c=1+c+a+b-1$
$\Rightarrow a+b+c=c+a+b$
This led me to the same expression $a+b+c$.
Is there a smarter way to solve this problem?
| For a collection $a_1,...,a_n$ of real numbers, let $m=\max|a_i|$. We notice that $m \leq \big(\sum_i |a_i|^2\big)^{1/2}$ with equality iff there exists an index $i$ such that $|a_i| = m$ and $a_j=0$ for $j \neq i$. Moreover, since $|a_i|^3 \leq m|a_i|^2$ we notice that $$\bigg( \sum_i |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot m^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot \bigg(\sum_i |a_i|^2 \bigg)^{ \frac12 \cdot \frac13} = \bigg( \sum_i |a_i|^2 \bigg)^{1/2}.$$ Now if we know that $\sum a_i^2 = \sum a_i^3 =1$, then it holds that $$1 = \bigg(\sum a_i^3 \bigg)^{1/3} \leq \bigg( \sum |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum a_i^2 \bigg)^{1/2} = 1,$$ and therefore equality must hold throughout both of the previous expressions. This forces one to have $m= \big(\sum a_i^2 \big)^{1/2}$. Hence some $|a_i|=m$ and all other $a_j$ are zero. This forces $a_i=1$, hence the sum must be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $ How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $
I know that $\sin \theta \leq \theta, \theta$ very small
But if take $ f(x) = \sin (\frac{1}{x}) - \frac{1}{2x}, f'(x) = \frac{-1}{x^2} \cos (\frac{1}{x}) + \frac{1}{2x^2} $
But if i take $x=\frac 4\pi, x \in (0, \pi/2), $ i am getting $f'(x) < 0$ which should be other way around. Am I missing something? I got this doubt while reading Does $\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right) $ absolutely converge?
If i say since $\sin (x)$ converges to $x$, i will have $\sin x$ values slightly less than $x$ and slightly more than $x$. But it depends on whether $f(x)$ is increasing/decreasing (local maxima or local minima)
Pls clarify
| We know that for $x=\frac1n>0$
$$x-\frac16x^3\le \sin x \le x$$
[ refer to Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ ]
and
$$\frac12x\le x-\frac16x^3 \iff \frac12x-\frac16 x^3\ge 0 \iff x^2\le 3 \iff0<x<\sqrt 3$$
which is true since $0<x=\frac1n \le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluating Limit of $\frac{xyz}{x^2 + y^2 + z^2}$ using $\epsilon$ - $\delta$ method I've been working on this for a little while.
I understand the way to do this is using spherical coordinates and the limit = 0, but I'm also wondering if it's possible to solve it using the epsilon-delta method.
Here is some of what I have:
Since I know that
$$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2 + y^2 + z^2} = 0$$
I start with
$$\left|\frac{xyz}{x^2 + y^2 + z^2}\right|<\epsilon$$
and
$$\sqrt{x^2 + y^2 + z^2} < \delta$$
Which seems to indicate
$$x < \delta \quad ; \quad y < \delta \quad ; \quad z < \delta$$
And thus
$$ \frac{\delta ^3}{x^2 + y^2 + z^2} < \epsilon$$
Since
$$ x^2 + y^2 + z^2 < \delta^2$$
This implies
$$ \frac{\delta^3}{\delta^2} < \frac{\delta ^3}{x^2 + y^2 + z^2} < \epsilon$$
And finally
$$ \delta < \epsilon$$
Which is fine.. but if this is the "game" that some people use to explain the method, I've only claimed that the $\delta$ I provide will be smaller than the $\epsilon$ provided. Am I interpreting this correctly, or does this mean: "any number larger than $\sqrt{x^2 + y^2 + z^2}$ will be smaller than $\epsilon$"? Which I suppose means any finite number > 0 will work.
From the examples that I've seen in my book and online, they usually set $\delta = g(\epsilon)$ by solving for an inequality like $\sqrt{x^2+y^2+z^2} < g(\epsilon)$ using the inequality that starts like $|f(x) - L| < \epsilon$. I can't feel but think I made some mistake.
I appreciate any help!
| The idea is: given $\varepsilon>0$, to find a $\delta>0$ such that $\sqrt{x^2+y^2+z^2}<\delta\implies\left|\frac{xyz}{x^2+y^2+z^2}\right|<\varepsilon$. So, fix some $\varepsilon>0$. Now, let $r=\sqrt{x^2+y^2+z^2}$. Then $|x|,|y|,|z|\leqslant r$ and therefore$$\left|\frac{xyz}{x^2+y^2+z^2}\right|\leqslant\frac{r^3}{r^2}=r.$$So, take $\delta=\varepsilon$. This will work, because if $\sqrt{x^2+y^2+z^2}=r<\delta=\varepsilon$, then $\left|\frac{xyz}{x^2+y^2+z^2}\right|\leqslant r<\varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Range of $(\sin x)^6+ (\cos x)^6$ I need to find the range of the function $y = (\sin x)^6+ (\cos x)^6$
I did find the answer but working in a crude way rather than a methodical step by step approach. I give below the steps I used , please help with a methodical approach to such problems.
1) To find the max value of the function I noticed that in the range where x is $[0,2\pi]$ , when $\cos x$ hits $+1$ then $\sin x$ is $0$ , when $\cos x$ is $0$
then $\sin x$ is $+1$ ..etc so the max value at any of these points could be either $1^6+0^6$ or $(-1)^6+0^6$ so the max value is 1.
to find the minimum I differentiated the function
$f' (x) = 6(\sin x)^5\cos x- 6(\cos x^5)\sin x =0$ , equating this to zero
we have $(\sin x)^4 = (\cos x)^4 => x = \pi/4 =>$ min value of function is $(\sin(\pi/4))^6 + (\cos(\pi/4))^6 = 1/4$ . so the range is $(\frac{1}{4},1)$.
Please can someone help with how can this type of problems be methodically approached ? - Thanks.
| With $$a=\sin(x)$$ $$b=\cos(x)$$ we have $$a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)=(a^2+b^2)((a^2+b^2)^2-3a^2b^2)$$
Because of $a^2+b^2=1$ we have $$\sin^6(x)+\cos^6(x)=1-3\sin^2(x)\cos^2(x)=1-\frac{3}{4}\sin^2(2x)$$
Obviously the range of $\sin^2(2x)$ is $[0,1]$, so we get range $[\frac{1}{4},1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int_{0}^{\infty} \frac{dx}{1 + x^n}$ using the Residue Theorem I'm trying to compute the following integral:
\begin{align}
\int_{0}^{\infty} \frac{dx}{1 + x^n}, \quad n \geq 2.
\end{align}
Consider the function $f(z) = \frac{1}{1 + z^n}$. Consider the following contour, $\Gamma_n$:
We then have,
\begin{align}
\int_{\Gamma_n} = \int_{0}^{R} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{1}} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{2}} \frac{dz}{1 + z^n},
\end{align}
where $\Gamma_n^{1}$ denotes an arc of the circle, $R e^{i \theta}, \theta \in [0, 2\pi] $. I have shown that the integral along this arc goes to zero as $R$ goes to $\infty$. I'm having trouble with the integral along the arc $\Gamma_n^{2}$. Here's what I have done:
Parametrize $\Gamma_n^{2}$ as,
\begin{align}
z(t) = \Big (1 - \frac{t}{R} \Big ) R e^{\frac{2 \pi i}{n}}, \quad t \in [0, R].
\end{align}
We then have,
\begin{align}
dz(t) & = - e^{\frac{2 \pi i}{n}} dt, \\
1 + z(t)^n & = 1 + R^n \Big (1 - \frac{t}{R} \Big )^n = 1 + R^n \sum_{k = 0}^{n} (-1)^k \Big ( \frac{t}{R} \Big )^k \\
& = 1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n.
\end{align}
Hence, we have that,
\begin{align}
\int_{\Gamma_n^{2}} \frac{dz}{1 + z^n} = - e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dt}{1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n}.
\end{align}
I'm not sure how to proceed from here. The answer given states that this integral converges to,
\begin{align}
- e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dx}{1 + x^n}.
\end{align}
| Note that $\Gamma_n^1$ is an arc of a circle, not a semicircle (in general).
Besides,\begin{align}\int_{\Gamma_n^2}\frac{\mathrm dz}{1+z^n}&=\int_0^R\frac{-e^{\frac{2\pi i}n}}{1+\left(1-\frac tR\right)^nR^n}\,\mathrm dt\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+(R-t)^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+t^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dz}{1+z^n}.\end{align}Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the least value of $\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$? What is the least value of this expression?
$$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$$
Will putting $\theta=45^{\circ}$ give right answer?
| $$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta=3+2(\tan^2\theta+\cot^2\theta)$$
Now $\tan^2\theta+\cot^2\theta=(\cot\theta-\tan\theta)^2+2\ge?$
the equality occurs if $\cot\theta-\tan\theta=0\iff\tan^2\theta=1\iff\cos2\theta=0\iff2\theta=(2m+1)\dfrac\pi2$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $A^n , n \ge 1$ Let $u = (1, 2, 3)$ and $v = (1, \frac{1}{2},\frac{1}{3})$ · Let $A= u^tv.$ Find $A^n , n \ge 1$
My attempts :$ A = u^tv$ = $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}[ 1 \frac{1}{2} \frac{1}{3} ]$.
$A = \begin{bmatrix} 1 &1& \frac{2}{3} \\ 2& 1& \frac{2}{3}\\3& \frac{3}{2} & 1 \end{bmatrix}$.
i don't know How to find $A^n $?
| $A^2 = u^tvu^tv= u^t (vu^t)v\\
A^n = u^tvu^tv= u^t (vu^t)^{n-1}v\\
(vu^t) = 3\\
A^n = 3^{n-1} A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
multivariable calculus, Cartesian coordinate and polar coordinate It's a question about multivariable calculus. My solution differs from the one suggested by the book, and I can understand the solution from the book, yet I cannot figure out where I am wrong...
The question:
$$\iint_D\left|x^2+y^2-2y\right|d\sigma,\ D=\left\{(x,y)\mid x^2 + y^2 \le 4\right\}$$
And my solution:
$$\begin{align}
\iint_D\left|x^2+y^2-2y\right|d\sigma&=2\iint_{D_1}\left|x^2+y^2-2y\right|d\sigma,\, D_1=\left\{(x,y)\mid x^2 + y^2 \le 4,\ x\le0\right\} \\
&=2\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}d\theta\int_0^2\left|\rho^2-2\rho\sin\theta\right|\rho\,d\rho
\end{align}$$
(switch to polar coordinates)
Then I calculated $\int_0^2\left|\rho^2-2\rho\sin\theta\right|\rho\,d\rho$
$$\begin{align}
\int_0^2\left|\rho^2-2\rho\sin\theta\right|\rho\,d\rho&=\int_0^{2\sin\theta}(2\sin\theta-\rho)\,\rho^2\,d\rho+\int_{2\sin\theta}^2(\rho-2\sin\theta)\,\rho^2\,d\rho \\
&=\left.\left(\frac{2\rho^3\sin\theta}3-\frac{\rho^4}4\right)\right|_0^{2\sin\theta}+\left.\left(\frac{\rho^4}4-\frac{2\rho^3\sin\theta}3\right)\right|_{2\sin\theta}^2 \\
&=4-\frac{16\sin\theta}3+\frac{8\sin^4\theta}3
\end{align}$$
Inserting the result back, we have:
$$\iint_D\left|x^2+y^2-2y\right|\,d\sigma
=2\int_{\frac{\pi}2}^{\frac{3\pi}2}\left(4-\frac{16\sin\theta}3+\frac{8\sin^4\theta}3\right)d\theta
=10\pi$$
Here is the solution suggested by the book:
$$\begin{align}
\iint_D\left|x^2+y^2-2y\right|d\sigma
=&\iint_{D_1}-(x^2+y^2-2y)\,d\sigma+\iint_{D_2}(x^2+y^2-2y)\,d\sigma \\
D_1=\left\{(x,y)\mid x^2+(y-1)^2\le1\right\}&,\ D_2=\left\{(x,y)\mid x^2+y^2\le4,\ x^2+(y-1)^2>1\right\} \\
=&\iint_D(x^2+y^2-2y)\,d\sigma-2\iint_{D_1}(x^2+y^2-2y)\,d\sigma \\
=&\int_0^{2\pi}d\theta\int_0^2(\rho^2-2\rho\sin\theta)\,\rho\,d\rho-2\int_0^{\pi}d\theta\int_0^{2\sin\theta}(\rho^2-2\rho\sin\theta)\,\rho\,d\rho \\
=&\int_0^{2\pi}\left(4-\frac{16\sin\theta}3\right)d\theta-2\int_0^{\pi}\left(-\frac{4\sin^4\theta}3\right)d\theta \\
=&\ 9\pi
\end{align}$$
Thanks in advance!
| We have that
$$x^2+y^2-2y=x^2+(y-1)^2-1=0$$
represents a circle $D_1$ centered at $(0,1)$, therefore the function $f(x,y)=x^2+y^2-2y$ is $\le 0$ inside the circle and $\ge 0$ outside.
therefore the correct set up is that given from the book or as an alternative
$$\iint_D (x^2+y^2-2y)\, dS+2\iint_{D_1} (-x^2-y^2+2y)\, dS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve by a series of powers around the given ordinary point Solve by a series of powers around the given ordinary point, $$ x^{2}y'' + xy'-y = 0 \hspace{5mm}x_{0} = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=\sum_{n=0}^{\infty}c_{n}(x-2)^{n}\\y'=\sum_{n=1}^{\infty}c_{n}n(x-2)^{n-1}\\y''=\sum_{n=2}^{\infty}c_{n}n(n-1)(x-2)^{n-2}$$
| Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
\begin{align}
y &= \sum_{n=0}^\infty c_nt^n \\ \\
(t+2)y' &= (t+2)\sum_{n=1}^\infty c_nnt^{n-1} \\
&= \sum_{n=1}^\infty nc_nt^n + \sum_{n=1}^\infty 2nc_nt^{n-1} \\
&= \sum_{n=1}^\infty nc_nt^n + \sum_{n=0}^\infty 2(n+1)c_{n+1}t^n \\ \\
(t+2)^2y'' &= (t^2+4t+4)\sum_{n=2}^\infty n(n-1)c_nt^{n-2} \\
&= \sum_{n=2}^\infty n(n-1)c_nt^n + \sum_{n=2}^\infty 4n(n-1)c_nt^{n-1} + \sum_{n=2}^\infty 4n(n-1)c_nt^{n-2} \\
&= \sum_{n=2}^\infty n(n-1)c_nt^n + \sum_{n=1}^\infty 4(n+1)nc_{n+1}t^n + \sum_{n=0}^\infty 4(n+2)(n+1)c_{n+2}t^n
\end{align}
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + \sum_{n=2}^\infty \big[4(n+2)(n+1)c_{n+2} + 2(n+1)(2n+1)c_{n+1} + (n^2-1)c_nt^n \big] $$
Then we have the recurring relations
\begin{align}
c_2 &= \frac18(c_0-2c_1) \\
c_3 &= -\frac12c_2 \\
c_{n+2} &= \frac{2n+1}{2(n+2)}c_{n+1} + \frac{n-1}{4(n+2)}c_n
\end{align}
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + \frac{b}{x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I solve this algebra problem The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 \tag{1}$
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 \tag{2}$
I tried simplifying the first equation to,
$$ x^5 + 5\left[ \left(xy+1 \right) \left( y - x^2 \right) \right] = 16 $$
and second equation to,
$$ y^5 + 5 \left[ \left(xy+1 \right) \left( y^2 + x \right) \right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?
Thanks
| COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\\y^5+5(xy+1)(y^2+x)=-57$$ From which
$$\frac{y^2+x}{y-x^2}=\frac{y^5+57}{x^5-16}\qquad(1)$$
A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$\frac{y^2+x}{y-x^2}=a\qquad(2)$$ $$\frac{y^5+57}{x^5-16}=a\qquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$
Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.
I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.
| Hint: Use mathematical induction. Also, for $n > 2$, $$\left(\frac{1+i}{2}\right)^{2^{n}} = \frac{1}{2^{2^{n-1}}},$$
and
$$\left(1-\frac{1}{2^{2^{n-1}}}\right)\left(1+\frac{1}{2^{2^{n-1}}}\right)=1-\frac{1}{2^{2^{n}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?
| Let $y=f(x)$, then
\begin{align}
0 &= x^4+(5-y)x^2+(7-3y) \\
x^2 &= \frac{y-5 \pm \sqrt{(5-y)^2+4(\color{red}{3y-7})}}{2} \\
&= \frac{y-5 \pm \sqrt{y^2+2y-3}}{2} \\
&= \frac{y-5 \pm \sqrt{(y+3)(y-1)}}{2}
\end{align}
Now,
$$\Delta=(y+3)(y-1) \ge 0$$
Together with $f(x)>0$,
$$y \ge 1$$
in which $\Delta$ is increasing with $y$.
*
*When $y \ge 5$,
\begin{align}
3y-7 & \ge 3(5)-7 \\
& = 8 \\
& > 0 \\
\Delta & > \sqrt{(5-y)^2} \\
& = y-5 \\
\end{align}
*
*When $1 \le y \le 5$,
\begin{align}
y-5+\sqrt{(y+3)(y-1)} & \ge 0 \tag{$x^2 \ge 0$} \\
(y+3)(y-1) & \ge (5-y)^2 \\
4(\color{red}{3y-7}) & \ge 0 \\
y & \ge \frac{7}{3} \\
\end{align}
The non-negativity of $x^2$ imposes $$x^2=\frac{y-5+\sqrt{(y+3)(y-1)}}{2}$$
which is increasing with $y$.
The required minimum is $\frac{7}{3}$ which is achieved when
$$x^2=\frac{\frac{7}{3}-5+5-\frac{7}{3}}{2}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\int\sqrt{\frac{2-x}{x-3}} dx$
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
Need help in spotting my mistake:
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
$x-2 = t^2$
$\implies dx = 2t dt$
$$2 \int \sqrt{\dfrac{1}{1-t^2}} t^2dt$$
$t= \sin \theta $
$dt = \cos\theta d\theta$
$$\int (1- \cos 2\theta )d\theta $$
$=\theta - \dfrac{\sin 2\theta}{2}$
$$= \arcsin(t)- t\sqrt{1-t^2}$$
$= \arcsin (\sqrt{x-2})- \sqrt{(x-2)(3-x)}+C$
But answer given is: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$
| My HINT: Substituting the square root we get
$$t=\sqrt{\frac{2-x}{x-3}}$$
then we get $$x=\frac{2+t2}{t^2+1}$$ and $$dx=-\frac{2t}{(t^2+1)^2}dt$$
Try this!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
| This Problem is well known.
Note that $$\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)^2-4ab}{2(a+b)}$$
after squaring two times you will get
$$\left(\frac{(a+b)^2}{2}-\frac{(a-b)^4}{4(a+b)^2}\right)^2\geq 4ab\left(\frac{a^2+b^2}{2}\right)$$
Doing this Algebra you will get
$$\frac{1}{16}\frac{(a-b)^8}{(a+b)^4}\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
| If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^{-1}$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^{-1}=1\,.\tag{*}$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^{-1}=1\,.$$
Thus,
$$a^{-1}\big(1-b(1-a)\big)=b^{-1}=1-b-a(1-a)\,.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a)\,.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1)\,.\tag{#}$$
Thus, $a\notin\{0,1,4\}$ for the equation above to be possible (noting that $a\neq 0$ and $b\neq 0$). You are left with four choices of $a$, namely, $a\in\{2,3,5,6\}$.
If $a=2$, then, using (#), we get $b=3^{-1}\cdot 5=5\cdot 5=4$, so $b^{-1}=2$, but then $$a(1-a)+b+b^{-1}=2\cdot(-1)+4+2=4\neq 1\,.$$
If $a=3$, then we have by (#) that $b=5^{-1}\cdot20=-3=4$, whence
$$a(1-a)+b+b^{-1}=3\cdot(-2)+4+2=0\neq 1\,.$$
If $a=5$, then $b=3^{-1}\cdot (-1)=-5=2$, so $b^{-1}=4$, but then
$$a(1-a)+b+b^{-1}=5\cdot(-4)+2+4=0\neq 1\,.$$
Finally, if $a=6$, then $b=4^{-1}\cdot3=2\cdot 3=-1$, whence $b^{-1}=-1$, and
$$a(1-a)+b+b^{-1}=6\cdot(-5)+(-1)+(-1)=-4\neq 1\,.$$
Thus, (*) cannot be satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Show that, for all $m\geq 2$, there exists $C_m>0$ such that, for all $k\geq 1$, $\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$ Define $\phi:\mathbb{N}\to \mathbb{N}$ by $\phi(k)=k^m$, where $m\geq 2$ is some fixed number. I want to investigate, if there exists $C_m>0$ such that $$\frac{\phi(k+2)-\phi(k+1)}{\phi(k+1)-\phi(k)}=\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$$ for all $k\geq 1$. Since
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=\frac{\left ( \frac{1+2/k}{1+1/k} \right )^m-1}{1-\left ( \frac{1}{1+1/k} \right )^m},
$$
we define $f:(0,1]\to\mathbb{R}$ by
$$
f(x)=\frac{\left ( \frac{1+2x}{1+x} \right )^m-1}{1-\left ( \frac{1}{1+x} \right )^m}
$$
I suspect that $f(x)$ is increasing, and so $f(1/x)$ must be decreasing (through some calculator graphs), which implies that
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq \frac{(1+2)^m-(1+1)^m}{(1+1)^m-1^m}=\frac{3^m-2^m}{2^m-1}=:C_m
$$
for all $k\geq 1$. The question is, how do I show the monotonicity of $f(x)$ or $f(1/x)$ in a simplest way? Determining $f'$ and then checking if it's greater than $0$ on $(0,1]$ takes a lot time.
| Playing naively...
$\begin{array}\\
r_m(k)
&=\dfrac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\\
&=\dfrac{(1+2/k)^m-(1+1/k)^m}{(1+1/k)^m-1}\\
&=\dfrac{1+2m/k+O(1/k^2)-(1+m/k+O(1/k^2)}{(1+m/k+O(1/k^2))-1}\\
&=\dfrac{m/k+O(1/k^2)}{m/k+O(1/k^2)}\\
&=1+O(1/k)\\
\end{array}
$
So $1$ is a candidate.
If
$f(x) = x^m$,
then
$f'(x) = mx^{m-1}$
and
$f''(x) = m(m-1)x^{m-2}$
so
$f''(x) \ge 0$
is $m \ge 2$.
Therefore
$\frac12(f(x)+f(x+2))
\ge f(x+1)
$
so
$f(x)+f(x+2)
\ge 2f(x+1)
$
or
$f(x+2)-f(x+1)
\ge f(x+1)-f(x)
$.
Therefore
$r_m(k) \ge 1$.
$r_m(1)
=\dfrac{3^m-2^m}{2^m-1}
=\dfrac{(3/2)^m-1}{1-2^{-m}}
\gt (3/2)^m-1
$.
If we can show that
$r_m(k)$
is a decreasing function of $k$,
we are essentially done.
$\begin{array}\\
r_m(1/x)
&=\dfrac{(1+2x)^m-(1+x)^m}{(1+x)^m-1}\\
&=\dfrac{(1+2mx+m(m-1)4x^2)-(1+mx+m(m-1)x^2/2)+O(x^3)}{(1+mx+m(m-1)x^2/2)-1+O(x^3)}\\
&=\dfrac{mx+m(m-1)4x^2/2)-m(m-1)x^2/2)+O(x^3)}{mx+m(m-1)x^2/2+O(x^3)}\\
&=\dfrac{mx+3m(m-1)x^2+O(x^3)}{mx+m(m-1)x^2/2+O(x^3)}\\
&=\dfrac{1+3(m-1)x+O(x^2)}{1+(m-1)x/2+O(x^2)}\\
&=(1+3(m-1)x+O(x^2))(1-(m-1)x/2+O(x^2))\\
&=1+2(m-1)x+O(x^2)\\
\end{array}
$
Therefore,
for small enough $x$,
$f(1/x)$ is increasing
so $f(k)$
is decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Volume enclosed by paraboloid and plane The volume, $V$, enclosed by paraboloid $z=x^2 + y^2$ and the plane $z=3-2y$ can be expressed as a triple integral. Determine the limits describing the enclosed volume. By evaluating the integral, show the volume is $8\pi$.$\\$
There is then a hint that $\cos^4\theta = \frac{1}{8}\cos4\theta + \frac{1}{2}\cos2\theta + \frac{3}{8}$.
I worked out the limits in Cartesian as
$V=\int_{y=-3}^{y=1}{\int_{x=-\sqrt{4-(y+1)^2}}^{x=\sqrt{4-(y+1)^2}}{\int_{z=x^2+y^2}^{z=3-2y}dz}dx}dy$
This looks pretty hard to compute though, so I then made the polar limits as
$V=\int_{R=0}^{R=2}{\int_{\theta=0}^{\theta=2\pi}{\int_{z=R^2}^{z=3-2Rsin\theta}R dz}d\theta}dR$,
but when I compute this I don't get $8\pi$. I'm wondering if this is me making a calculation error, or if the limits are incorrect. I also never used the hint, so was wondering if anyone could shine a light on when that comes in handy.
Thanks!
| If u integrate what u got in cartesian coordinates, u get the answer
Given integral = $\int_{y=-3}^1(1-\frac{2}{3}y-\frac{y^2}{3})4 \sqrt {4-(y+1)^2} dy $
Now put $y+1 = 2 \sin t$, we get
given integral = $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{4}{3}-\frac{8}{9} \sin t - \frac{4}{9} \sin^2 t)16 \cos^2 t dt \\ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{64}{3} \cos^4 t dt = 8 \pi$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
| HINT.-If you want to have another way then you can do as follows:
$$y^2+xy-30=0\iff 2y=-x\pm\sqrt{x^2+120}$$ the only possibilities are $x=\pm1\text{ or } \pm7$ and $x=\pm1$ is not compatible. This proves that $(7,3)$ and $(-7,-3)$ are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the values for $a,b,c,d$
Given $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$ for any
real number $x$ and $y$ , find the value of $a,b,c,d$
| $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$
cancel out $x^3$ and $y^2$
our equation becomes
$$4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy=0$$
write it as following
$$4x^2y-bx^cy+axy^2+7xy^2+dxy+3xy=0$$
take common terms
$$y(4x^2-bx^c)+xy^2(a+7)+xy(d+3)=0$$
this equation has to be true for any real x and y thus
you can clearly see these terms has to cancel out each other
$$4x^2-bx^c=0$$
$$a+7=0$$
$$d+3=0$$
we obtain
$$a=-7$$
$$ b=4$$ $$c=2$$ $$d=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it. Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it.
Using the ratio test, I could show that the series is convergent. What is an easy way to evaluate it? I was thinking about estimating it from below and above by series with equal value but coulnd't find any familar series that I could use for this.
| Working it out
in as elementary a way
as possible.
$\begin{array}\\
\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}
&=\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}\\
&=\sum_{n=1}^\infty \frac{n^2}{n!2^n} +\sum_{n=1}^\infty \frac{1}{n!2^n}\\
&=\sum_{n=1}^\infty \frac{n}{(n-1)!2^n} +\sum_{n=0}^\infty \frac{(1/2)^n}{n!}-1\\
&=\sum_{n=0}^\infty \frac{n+1}{n!2^{n+1}} +e^{1/2}-1\\
&=\sum_{n=0}^\infty \frac{n}{n!2^{n+1}}+\sum_{n=0}^\infty \frac{1}{n!2^{n+1}} +e^{1/2}-1\\
&=\sum_{n=1}^\infty \frac{n}{n!2^{n+1}}+\frac12\sum_{n=0}^\infty \frac{1}{n!2^{n}} +e^{1/2}-1\\
&=\sum_{n=1}^\infty \frac{1}{(n-1)!2^{n+1}}+\frac12e^{1/2} +e^{1/2}-1\\
&=\sum_{n=0}^\infty \frac{1}{n!2^{n+2}}+\frac32e^{1/2} -1\\
&=\frac14\sum_{n=0}^\infty \frac{1}{n!2^{n}}+\frac32e^{1/2} -1\\
&=\frac14 e^{1/2}+\frac32e^{1/2} -1\\
&=\frac74 e^{1/2} -1\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/2884192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The integral problem: $\int_{0}^{16}\frac{dx}{\sqrt{x^2+9}-\sqrt{x}}$ I have met this kind of problem today.in fact, I spent hours trying to solve this problem on its own. My method was just to calculate the indefinite integral. I looked at Wolfram Alpha after I failed. Wolfram couldn't evaluate this integral. I don't know the spesific reason.
The integral:
$$\int_{0}^{16}\frac{dx}{\sqrt{x^2+9}-\sqrt{x}}$$
| According to Maple 18, the integral is equal to
\begin{align*}
&\frac15\arctan \left( \frac{\sqrt {5}}{5} \right) \sqrt {5}+\frac{1}{14}\sqrt {7} \ln \left( \sqrt {7}+1 \right) -\frac{1}{14}\sqrt {7}\ln \left( \sqrt {7}-1 \right) \\
&-\frac{1}{14}\sqrt {7}\ln \left( \sqrt {5}\sqrt {7}\sqrt {53}+19 \right) +\frac{1}{14}\sqrt {7}\ln \left( \sqrt {5}\sqrt {7}\sqrt {53}-19 \right) \\
&+\frac{1}{5}\sqrt {5}\arctan \left( {\frac {13\,\sqrt {53}}{265}} \right)
+\frac{1}{5}\sqrt {5}\arctan \left( \frac85\sqrt {5}-\frac{1}{5}\sqrt {7}\sqrt {5} \right) \\
&+\frac{1}{5}\sqrt {5}\arctan \left( \frac85\sqrt {5}+\frac{1}{5} \sqrt {7}\sqrt {5} \right)+\frac{1}{14}\sqrt {7}\ln \left( 19-4\,\sqrt {7} \right) \\
&-\frac{1}{14}\sqrt {7}\ln \left( 19+4\,\sqrt {7} \right) +\ln \left( 3 \right) -\ln \left( -16+\sqrt {5}\sqrt {53} \right).
\end{align*}
Enjoy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is the graph of $y=(x^2+x-6)/(x-2)$ the same as the graph of $y=x+3$, and a continuous function?
Why is the graph of $y=(x^2+x-6)/(x-2)$ the same as the graph of $y=x+3$, and a continuous function?
I understand that this is because
$$
\frac{x^2+x-6}{x-2} = \frac{(x+3)(x-2)}{x-2} = x+3.
$$
But my problem with this is that in $(x-2)$, when $x=2$ the denominator will be $0$. So shouldn't there be a break here, because if $y=(x^2+x-6)/(x-2)$, then $y$ should then be undefined? So why can $y=(x^2+x-6)/(x-2)$ still be considered continuous and equal to $y=x+3$?
Please try to explain this simply. I'm still just beginning Khan Academy calculus. Thank you.
| Their graphs are not the same, nor is $f(x) = (x^2+x-6)/(x-2)$ continuous at $x=2$.
Indeed, as you say $f(x) = (x^2+x-6)/(x-2)$ is undefined at $x=2$ and therefore has a sort of a "puncture" at $x = 2$.
In particular, $f$ cannot be continuous at $x=2$, since the definition of continuity at a point requires the function to be defined at that point.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$. prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$.
I have tried squaring both sides and obtianed a biquadratic equation. According to the question it should have 2 solutions. From the given equation it is evident that the solutions will be positive. After this can not figure out how to proceed.
$\sqrt{a+\sqrt{x+a}} = x$
$\sqrt{a+x} = x^2 -a$
$x+a = (x^2-a)^2$
need help
| We have $$\sqrt{a+x}= x^2-a\implies a^2-a(2x^2+1)+x^4-x=0$$
Solving this equation on $a$ we get, discriminat = $(2x+1)^2$ so $$ a= {2x^2+1\pm (2x+1)\over 2}$$
*
*case: If $a= x^2+x+1$ we get $$ x= {-1\pm \sqrt{4a-3}\over 2}$$
which does not count since $a< 0$.
*case: If $a= x^2-x$ we get $$ x= {1\pm \sqrt{4a+1}\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve $(2x-3)(2^x-4)+(x^2-3x+2)(2^x\cdot\log2)=0$ I want to find when this function equals to $0$:
$$f'(x)=(2x-3)(2^x-4)+(x^2-3x+2)(2^x\cdot\log2)=0$$
This is the derivative of $(x^2-3x+2)(2^x-4)$ and I set it equal to $0$ so I can get the "critical points" of the derivative.
However it's not entirely clear to me how I would have to get the solutions. I developed it this way:
$$2x\cdot2^x-8x-3\cdot2^x+12+(x^2\cdot2^x\cdot\log2-3x\cdot2^x\cdot\log2+2^{x+1}\cdot\log2)=0$$
I understand this should be an exponential equation because there's $2^x$ in there although I'm not exactly sure what I need to group (maybe $2^x$). Any hints?
| Amazingly, we could find rather good approximations of the second solution approximating $f'(x)$ by a $[2,n]$ Padé approximant built at $x=2$ and get the appoximate solution at the price of a linear equation.
The second degree in numerator is required since $x=2$ is a known solution. So, we shall have, as an approximation
$$f'_{(n)}(x)=(x-2)\frac {a^{(n)}_0+a^{(n)}_1(x-2) } {1+\sum_{k=1}^n b_k (x-2)^k}$$ in which all coefficients are expressed in terms of higher derivatives. Moreover, for any $n$, we shall have $a^{(n)}_0=8 \log(2)$ and, then
$$x_{(n)}=2-\frac{a^{(n)}_0 } {a^{(n)}_1 }=2-\frac{8 \log(2) } {a^{(n)}_1 }$$ Moreover $a^{(n)}_1$ is itself a polynomial in $\log(2)$ with no constant term making the expression
$$x_{(n)}=2-\frac{8 } {\frac{a^{(n)}_1}{\log(2)} }=2-\frac 8 {c_{n}}$$
The first terms would be
$$\left(
\begin{array}{cc}
n & c_n \\
0 & {6 \left(2 +\log (2)\right)} \\
1 & \frac{2 \left(108 +60 \log (2)+11 \log ^2(2)\right)}{9
(2+\log (2))} \\
2 & \frac{2 \left(648 +396 \log (2)+86 \log ^2(2)+5 \log
^3(2)\right)}{ 108+60 \log (2)+11 \log ^2(2)} \\
3 & \frac{2 \left(58320 +38880 \log (2)+9720 \log ^2(2)+840 \log
^3(2)-43 \log ^4(2)\right)}{15 \left(648+396 \log (2)+86 \log ^2(2)+5
\log ^3(2)\right)}
\end{array}
\right)$$ and the decimal representation of the successive $x_{(n)}$ would then be
$$\left(
\begin{array}{cc}
n & x_{(n)} \\
0 & 1.50492 \\
1 & 1.37399 \\
2 & 1.35835 \\
3 & 1.35785
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Given that $z=\cos\theta+i\sin\theta$, show that $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
Given that $z=\cos\theta+i\sin\theta$, show that
$Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
For this question I had to show that the real part of $\frac{z-1}{z+1}=0$
To find that I first substituted $z$ with $\cos\theta+i\sin\theta$ to get
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}$$
I then multiplied by the conjugate of the denominator
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\cdot\frac{\cos\theta-i\sin\theta+1}{\cos\theta-i\sin\theta+1}$$
Which, when expanded, gives me
$$\frac{\cos^2\theta+\sin^2\theta+2i\sin\theta-1}{\cos^2\theta+\sin^2\theta+2i\cos\theta+1}$$
After that,
$$\frac{2i\sin\theta}{2\cos\theta+2}=\frac{i\sin\theta}{\cos\theta+1}$$
How do I proceed?
Edit: Just realized (after being told by a commenter) that $\frac{i\sin\theta}{\cos\theta+1}$ is imaginary. I won't delete the question. Ínstead I'll leave it here, as a testament to my stupidity, for everyone to see. May God have mercy on my soul during my exam.
| Note that we can resort to trigonometric formulae.
$$\frac{z-1}{z+1}=\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{1-2\sin^2\frac{\theta}{2}+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}-1}{2\cos^2\frac{\theta}{2}-1+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}+1}=\frac{2i\sin\frac{\theta}{2}\cdot e^{i\theta/2}}{2\cos\frac{\theta}{2}\cdot e^{i\theta/2}}=i\tan\frac{\theta}{2}\implies\Re\left(\frac{z-1}{z+1}\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ Prove that $f(x)=(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ for all $x>0.$
We have $f(x)=f(\frac{1}{x}), f'(x)=-\frac{1}{x^2}f'(\frac{1}{x}),$ so we only need to prove $f'(x)＞0$ for $0 < x < 1.$
| My ugly proof based on the following statements.
By your work it's enough to prove that
$$(1+x)^{\frac{1}{x}}\left(\frac{1}{x+x^2}-\frac{\ln(1+x)}{x^2}\right)+\left(1+\frac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)\geq0$$ for all $0<x\leq1$ or
$$(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x\left(\frac{1}{x+x^2}-\frac{\ln(1+x)}{x^2}\right)+\left(1+\frac{1}{x}\right)^{2x}\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)\geq0.$$
Now, easy to prove the following statements. For all $0<x\leq1$ we have:
*
*$$(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x\leq4,$$
*$$\frac{1}{x+x^2}-\frac{\ln(1+x)}{x^2}\leq0,$$
*$$\left(1+\frac{1}{x}\right)^{2x}\geq1+4x-x^2$$ and
*$$\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}>0.$$
Id est, it's enough to prove that:
$$4\left(\frac{1}{x+x^2}-\frac{\ln(1+x)}{x^2}\right)+(1+4x-x^2)\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)\geq0,$$
which is smooth.
The proof of $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x\leq4$ see here:
Monotonicity of the function $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A point $(a, b)$ on a circle with an equation $x^2+y^2=1$ defines another point $(x_1,y_1)$. What is the equation of the curve and is it a circle? The other point $(x_1, y_1)$ is defined as a point that lies on the tangent of a circle to the point $(a, b)$ and is $2$ units away from the point $(a, b)$. The question is, what is the equation of a curve the collection of all possible points $(x_1, y_1)$ trace? It seems to be a circle, but I don't understand how.
You can imagine this curve by getting two sticks that are connected at a 90 degree angle to one another, one lenght 1 and the other length 2, and spinning the lenght 1 stick in a circle around the end that isn't connected to the 2 stick, and then seeing what kind of curve the 2 stick's end that isn't connected to the 1 stick traces.
The equation of a tangent line is $y=\frac{a^2-ax+b^2}{b}$, and the distance of 2 units is given by the circle with radius 2 from the point $(a, b)$ or $(x-a)^2+(y-b)^2=4$. There are 2 intersections at points $$x=\frac{a^3\pm2\sqrt{b^2(a^2+b^2)}+ab^2}{a^2+b^2}$$
$$y=\frac{a^2b^2\pm2a\sqrt{b^2(a^2+b^2)}+b^4}{b(a^2+b^2)}$$
Knowing that $a^2+b^2=1$, substituting this into our equations we get the folowing parametric equation for this curve:
$$x=a\pm2\sqrt{1-a^2}$$
$$y=\frac{\pm(a^2\pm2\sqrt{1-a^2}a+1)}{\sqrt{1-a^2}}$$
$$-1<=a<=1$$
How do I prove that this is a circle, if it even is one? Also, is there a faster and better solution to understanding why it is a circle, if it is one?
|
The equation of a tangent line is $y=\frac{a^2-ax+b^2}{b}$, and the distance of 2 units is given by the circle with radius 2 from the point $(a, b)$ or $(x-a)^2+(y-b)^2=4$.
Expand the squares, rearrange the equations, and use that $\,a^2+b^2=1\,$:
$$
\begin{align}
\begin{cases}
ax + by &= a^2 + b^2 \\
x^2+y^2 -2(ax+by) + a^2+b^2 &= 4
\end{cases}
\\[5px] \iff\quad
\begin{cases}
ax + by &= 1 \\
\color{blue}{x^2+y^2} &= 2(ax+by) + 3 = 2 \cdot 1 + 3 \color{blue}{=5}
\end{cases}
\end{align}
$$
| {
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Determine the Jordan normal form of a complex matrix I want to determine the Jordan normal form of a complex matrix $A$ with characteristic polynomial $\chi_A(x)=(x+1)^4(x+2)^2$, minimal polynomial $m_A(x)=(x+1)^2(x+2)^2$ and that has the property that the matrix $A+I_6$ has order $3$.
$$$$
I have done the following:
The eigenvalues of $A$ are the roots of $\chi_A$. So the eigenvalues of $A$ are $-1$ and $-2$, where the multiplicity of the eigenvalue $-1$ is $4$ and the multiplicity of the eigenvalue $-2$ is $2$.
From the minimal polynomial we get the information that there is a Jordan block for the eigenvalue $-1$ that is $2\times 2$ (since we have the power $2$). We also have a Jordan block for the eigenvalue $-2$ that is $2\times 2$ (since we have the power $2$).
Is everything correct so far?
What information do we get from the fact that the matrix $A+I_6$ has order $3$ ?
| The information you have so far is correct. This narrows it down to two forms:
$$
\begin{bmatrix}
-1 & 1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & -1 & 0 & 0\\
0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & -2\\
\end{bmatrix} \quad\text{ and } \quad \begin{bmatrix}
-1 & 1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 1 & 0 & 0\\
0 & 0 & 0 & -1 & 0 & 0\\
0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & -2\\
\end{bmatrix}$$
Now the first matrix has $\operatorname{rank} (A + I) = 3$ and the second one has $\operatorname{rank} (A + I) = 4$.
Therefore the first one is the one you need.
| {
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How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$? How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$ ? I know that if it is a square root I will multiply by the conjugate. but if it is a cubic root what shall I do?
Thanks
| Formally substitute $n=1/x^{3/2}$ to get
$$
f(x)=\sqrt[3]{\frac{1}{x^3}+1}-\sqrt[3]{\frac{1}{x^3}+\frac{1}{x^{3/2}}}=
\frac{\sqrt[3]{1+x^3}-\sqrt[3]{1+x^{3/2}}}{x}=
\frac{1-1-x^{3/2}/3+o(x^{3/2})}{x}=-\frac{x^{1/2}}{3}+o(x^{1/2})
$$
Then $\lim_{x\to0^+}f(x)=0$. Since
$$
\sqrt[3]{n^2+1}-\sqrt[3]{n^2+n}=f\left(\frac{1}{n^{2/3}}\right)
$$
you have
$$
\lim_{n\to\infty}f\left(\frac{1}{n^{2/3}}\right)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ $ I want to know how to integrate this function I have tried many things many substitutions but none works.
I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles
| $$I=\int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx=\cdots=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx$$
As $\dfrac{d(\sin x+\cos x)}{dx}=\cos x-\sin x$
if we choose $\sin x+\cos x=u, 2\sin x\cos x=u^2-1$
$$I=2\int\dfrac{1+2(u^2-1)}{u(2-(u^2-1))}du$$
Now use Partial Fraction Decomposition, $$\dfrac{2u^2-1}{u(3-u^2)}=\dfrac Au+\dfrac{Bu+C}{3-u^2}$$
$\implies3A=-1,B-A=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Inequality proof concerning two variables $k,q$ Let $q,k$ positive integers satisfying the following inequalities
$$k\geq3$$
$$q\geq2$$
$$k\geq q$$
I have checked that the following inequality holds for a few values of $k,q$ I have tried but I have not been able to prove it.
$$A\triangleq kq{{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq kq\cdot q^{2k-4}(q-1)-q^{k-1}(q-1)\frac{(q-1)^3+1}{q}k\triangleq B$$
I tried to simplify $B$ as
\begin{eqnarray*}
B&=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})\frac{q^3-3q^2+3q}{q}k\\
&=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})(q^2-3q+3)k\\
&=&kq\cdot q^{2k-4}(q-1)-(q^{k+2}-3q^{k+1}+3q^k-q^{k+1}+3q^k-3q^{k-1})k\\
&=&kq\cdot q^{2k-4}(q-1)-kq^{k+2}+4kq^{k+1}-6kq^k+3kq^{k-1}\\
&=&kq\cdot(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})\\
\end{eqnarray*}
Now, I need to show that
$${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})$$
and I am stuck at this point.
| The inequality does hold for $k\ge 3$, $q\ge 2$ and $k\ge q$.
We can directly see that the inequality holds for $(k,q)=(3,2),(3,3),(4,2),(4,3),(4,4),(5,2),(6,2)$ respectively.
Here, we use two lemmas.
Lemma 1 : ${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\ge q^{q^{k-3}}(q-1)$.
Proof for lemma 1 :
$$\begin{align}&{{q^{k-2}}\choose{q^{k-3}+1}}(k-1)
\\\\&\ge {{q^{k-2}}\choose{q^{k-3}+1}}(q-1)
\\\\&=\frac{q^{k-2}\times (q^{k-2}-1)\times \cdots\times (q^{k-2}-q^{k-3})}{(q^{k-3}+1)\times q^{k-3}\times \cdots \times 1}\times(q-1)
\\\\&=\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times \frac{q^{k-2}-2}{q^{k-3}-1}\times\frac{q^{k-2}-3}{q^{k-3}-2}\times \cdots\times\frac{q^{k-2}-q^{k-3}}{1}\times (q-1)
\\\\&\ge\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times q\times q\times\cdots\times q\times q\times (q-1)
\\\\&=\frac{q^{k-2}}{q^{k-3}}\times\frac{q^{k-2}-1}{q^{k-3}+1}\times q^{q^{k-3}-1}\times (q-1)
\\\\&\ge q\times 1\times q^{q^{k-3}-1}\times (q-1)
\\\\&=q^{q^{k-3}}(q-1)\quad\square
\end{align}$$
Lemma 2 : $8\cdot q^{2k-3}\geq q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}$.
Proof for lemma 2 :
$$\begin{align}8\cdot q^{2k-3}&=q^{2k-3}+4q^{2k-3}+3q^{2k-3}
\\\\&\ge q^{2k-3}+4q^{k}+3q^{k-2}
\\\\&\ge q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}\quad\square
\end{align}$$
Now, from the two lemmas, in order to prove the inequality, it is sufficient to prove
$$q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\tag1$$
We can directly see that $(1)$ holds for $(k,q)=(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(6,6)$ respectively.
Lemma 3 : $(k-3)\log 2-\log(2k)\ge 0$ for $k\ge 7$.
Proof for lemma 3 : Let $f(k)=(k-3)\log 2-\log(2k)$. Then, $f'(k)=\log 2-\frac 1k$. So, $f(k)$ is increasing for $k\ge 7$ with $f(7)=\log\frac 87\gt 0$. $\ \ \square$
So, for $k\ge 7$, $q\ge 2$ and $k\ge q$, we can get, from the lemma 3,
$$\begin{align}\log q\ge \log 2\ge\frac{\log(2k)}{k-3}
&\implies(k-3)\log q\ge\log(2k)
\\\\&\implies q^{k-3}\ge 2k
\\\\&\implies q^{k-3}-2k+3\ge 3
\\\\&\implies q^{q^{k-3}-2k+3}\ge q^3\ge 8
\\\\&\implies q^{q^{k-3}-2k+3}(q-1)\ge 8(q-1)\ge 8
\\\\&\implies q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\end{align}$$
which is $(1)$.
Therefore, the inequality holds for $k\ge 3$, $q\ge 2$ and $k\ge q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show Sylvester sequence is the smallest solution with n terms to sum of unit fractions equalling 1 I want to show that a prefix of Sylvester's sequence gives the "smallest" solution to the equation where the sum of n unit fractions equals 1.
$$\sum_{i=1}^{n-1}{\frac{1}{x_i}} + \frac{1}{x_n - 1} = 1$$
Where the terms can be defined in two equivalent ways
$$x_n = \prod_{i=1}^{n-1}{x_i} + 1 = x_{n-1}(x_{n-1} - 1) + 1$$
and our base case is $x_1 = 2$.
For example, with 4 terms
$$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} = 1$$
I define the smallest solution to be the same ordering you may give a vector or tuple, the smaller of two solutions is the one with the first smaller denominator. As examples of this, the below table of fractions (which are not solutions) is sorted from smallest to largest.
$$\begin{array}{|c|c|}
\hline
\frac{1}{2} + \frac{1}{3} + \frac{1}{7} & (2, 3, 7) \\ \hline
\frac{1}{2} + \frac{1}{3} + \frac{1}{8} & (2, 3, 8) \\ \hline
\frac{1}{3} + \frac{1}{7} + \frac{1}{20} & (3, 7, 20) \\ \hline
\frac{1}{3} + \frac{1}{8} + \frac{1}{10} & (3, 8, 10) \\ \hline
\end{array}$$
I can show via induction that the first n terms of Slyvester's sequence always yields a valid solution, but I'm having trouble showing it is the smallest out of all possible solutions with n terms.
What I tried
I began by trying to prove that if you take the smallest $n-1$ term solution and extended it by a term through splitting the $n-1^{th}$ term in two, using the greedy method, you would get the same solution as is given by the first $n$ terms of Slyvester's sequence. While this works out, I can't justify that this method gives the smallest solution with $n$ terms.
Edit
Looking at the above wiki pages, it mentions the first n terms of the slyvester sequence gives the closest approximation to 1 possible with n unit fractions. Is it possible to build a proof based on this?
| If I read your question correctly, then it is a different question than what Mathlove answered. You are asking for the following:
Suppose that $ a_1 \leq a_2 \leq \ldots \leq a_n $ are integer solutions to$ \sum \frac{1}{a_i} = 1 $. Order them in lexicographic order. Then, the smallest solution is $\{s_i \} = ( x_1, x_2, \ldots x_{n-1}, x_n-1) $.
If so, then you essentially have the proof, but just need to verbalize it out.
Claim 1: $ \sum_{i<n} \frac{1}{x_i} + \frac{1}{x_n-1} = 1 $.
Proof by induction. Observe that $\frac{1}{x_n -1 } = \frac{1}{ x_n} + \frac{1}{x_n(x_n -1 )} = \frac{1}{x_n} + \frac{1}{x_{n+1} - 1 }$.
Claim 2: $s_j$ is the smallest integer strictly larger than $ \frac{1}{ 1 - \sum_{i< j} \frac{1}{s_i}}$.
Proof: This is what the greedy algorithm means. It follows directly from Claim 1 since $ 1 - \sum_{i<j} \frac{1}{s_i} = \frac{1}{s_j - 1 }$, so the smallest integer strictly larger than the inverse of that is clearly $s_j$.
Corollary: Suppose that $\sum \frac{1}{ a_i} = 1 $, and that $\forall i$ satisfying $ 1 \leq i \leq j < n-2$, we have $a_i = s_i$. Then we can conclude that $a_{j+1} \geq s_{j+1}$.
Note: This is what the greedy algorithm means.
Proof: For $j<n-2$, we have $\frac{1}{a_{j+1}} = 1 - \sum_{i<j+1} \frac{1}{s_i} - \sum_{i>j+1} \frac{1}{a_i} < 1 - \sum_{i<j+1} \frac{1}{s_i}$.
Hence, by claim 2, $s_{j+1} \leq a_{j+1}$.
Corollary: The smallest lexicographic solution is $\{s_i\}$.
Proof: By claim 1, $\{s_i\}$ works.
By the previous corollary, the first $n-1$ terms have to be equal to $s_i$. Hence, the last term is equal to $s_n$.
Note: Mathlove is trying to answer:
Suppose that $a_1 \leq a_2 \leq \ldots \leq a_n$ are integers satisfying $ \sum \frac{1}{a_i} < 1 $. Then $\sum\frac{1}{a_i} \leq 1 - \frac{1}{a_{n+1} -1 }$.
I believe this is a correct claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.
| It seems like a better start if you eliminate a variable (or two...)
Thus $z= 10-x-y$, so $x^2+y^2+100-2(x+y)+(x+y)^2=36$, that is, $x^2+y^2+xy-x-y+64=0$.
You are looking for the maximum of $xy(10-x-y)$.
If you do not want to look for anything creative, Lagrange multiplier surely works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Computing a Geometrical ratio $\frac{a}{b}$. $XYZ$ is an equilateral triangle as shown on the image below.
The aim is to find the ratio $\frac{a}{b}$.
So far from the picture, it is easy to see that $b= \frac{YZ}{2}$.
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.
|
Another way to find the ratio
is to solve
the $\triangle ROS$,
for which we know two sides and an angle:
\begin{align}
\triangle ROS:\quad
|OR|&=r=\tfrac13\cdot2b\cdot\tfrac{\sqrt3}2
=\tfrac{\sqrt3}3\,b
,\\
|OS|&=2r=\tfrac{2\sqrt3}3\,b
,\\
\angle SRO&=\angle ZRO+\angle SRZ=90^\circ+60^\circ
=150^\circ
,
\end{align}
so we can apply the cosine rule to get
\begin{align}
|OS|^2&=|OR|^2+|RS|^2-2\cdot|OR|\cdot|RS|\cos\angle SRO
,\\
\tfrac43\,b^2&=
\tfrac13\,b^2+a^2-2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos150^\circ
,\\
\tfrac43\,b^2&=
\tfrac13\,b^2+a^2+2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos30^\circ
,
\end{align}
\begin{align}
b^2
-b\cdot a
-a^2
&=0
,\\
(\tfrac{b}a)^2
-
\tfrac{b}a
-1
&=0
,
\end{align}
which gives the answer (positive root) as
\begin{align}
\frac{b}a
&=\frac{1+\sqrt{5}}{2}
,
\end{align}
which is indeed the Golden ratio.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equations with factorials Solve the following equation :
$$4 (x+1)!= x! (2 x-6)!$$
My turn :$$24(x+1)!=6 x! (2x-6)!$$
$$4!(x+1)!=3!x!(2x-6)!$$
$$\frac{(x+1)!}{3!x!}=\frac{(2x-6)!}{4!}$$
I tried to get a formula of a permutation in both sides but i could not do it
| Bringing $x!$ to the left hand side gives $\frac{4(x+1)!}{x!} = (2x - 6)! \implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 \geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 \implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $\frac{4x+8}{4x+4} = 1+\frac{4}{4x+4}$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 \geq 0$ we must have $x \geq 3$ and one sees that $x = 3,4$ do not work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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Evaluate $\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}$ I want to evaluate the following limit
$$\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}$$
I know the result is $3$. It's very tempting for me to do $\frac{t^2-2t-1}{-t-2}\sim-t$ but I know this is not allowed because of the final $+t$ there.
I've tried multiplying the brackets like this:
$$\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}=\lim_{t\rightarrow+\infty}{\frac{t^2-2t-1}{-t-2}}+\frac{t^2-2t-1}{(2+t)(-t-2)}+t=\dots
$$
Any hints on how to find the limit?
| There are a few ways you can tackle this. Using the working you've done,
\begin{align*}
&\frac{t^2-2t-1}{-t-2} + t + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{t^2-2t-1}{-t-2} + \frac{t(-t - 2)}{-t - 2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{t^2-2t-1}{-t-2} + \frac{-t^2 - 2t}{-t - 2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{-4t-1}{-t-2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{-4-\frac{1}{t}}{-1-\frac{2}{t}} + \frac{1-\frac{2}{t}-\frac{1}{t^2}}{(\frac{2}{t}+1)(-1-\frac{2}{t})} \\
\to \; & 4 - 1 = 3
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating the integral $\int_0^{\infty}\frac{x^3}{x^2+a^2}\,\mathrm{d}x$
If $\displaystyle\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x=\large\displaystyle\dfrac1{ka^6}$, then find the value of $\displaystyle\dfrac{k}{8}$.
I tried a lot but finally stuck at an intermediate form :
$$\begin{align}
&\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x, \text{with}\, {x^2=t},{2x~\mathrm{d}x=\mathrm{d}t}\\
&=\frac12\int_0^{\infty}\dfrac{(x^2)(2x)}{x^2+a^2}\,\mathrm{d}x=\frac12\int_0^{\infty}\dfrac{t}{t+a^2}\,\mathrm{d}t=\frac12\int_0^{\infty}\dfrac{t+a^2-a^2}{t+a^2}\,\mathrm{d}t\\
&=\frac12\left[\int_0^{\infty}\mathrm{d}t-\int_0^{\infty}\dfrac{a^2}{t+a^2}\,\mathrm{d}t\right]=\frac12\left[t|_0^{\infty}-a^2\ln(a^2+t)|_0^{\infty}\right]
\end{align}$$
| $$I=\int_0^\infty\frac{x^3}{x^2+a^2}dx$$
firstly, let $u=x^2+a^2$ so $\frac{du}{dx}=2x \therefore dx=\frac{du}{2x}$
For $x=0$, $u=a^2$. For $x\to\infty$, $\lim_{x\to \infty}\left(x^2+a^2\right)\to\infty$
so our integral becomes:
$$I=\frac{1}{2}\int_{a^2}^\infty\frac{x^3}{x^2+a^2}.\frac{du}{x}=\frac{1}{2}\int_{a^2}^\infty\frac{x^2}{x^2+a^2}dx=\frac{1}{2}\int_{a^2}^\infty\frac{u-a^2}{u}du=\frac{1}{2}\int_{a^2}^\infty du-\frac{a^2}{2}\int_{a^2}^\infty\frac{1}{u}du$$
Both parts to this integral are divergent and so the integral cannot be calculated
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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About the proof that every real number in the unit interval is the limit of a sequence of dyadic numbers
Given $x \in (0,1)$, show there exists a sequence $(x_n) \subset \{0,1\}$ such that $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$.
After running into difficult in trying to solve this problem, I found this MSE post, which, to my chagrin, gave me more difficulties. I'm trying to follow John Ma's reasoning. Here's how I understand it:
If $x < \frac{1}{2}$, the choose $x_1 =0$ and therefore $|x-\frac{x_1}{2}| = |x| = x < \frac{1}{2}$, and in fact $x-\frac{x_1}{2} \ge 0$, so in this case we can find $x_1$ for which $0 \le x - \frac{x_1}{2} < \frac{1}{2}$. If, however, $x \ge \frac{1}{2}$, then choose $x_2 =1$, and $|x-\frac{x_1}{2}| < \frac{1}{2}$ holds if and only if $- \frac{1}{2} < x - \frac{1}{2} < \frac{1}{2}$ if and only if $0 < x < 1$. Since $0 < x < 1$ is in fact true, so must $|x-\frac{x_1}{2}| < \frac{1}{2}$. Also, we have $x - \frac{x_1}{2} \ge 0$ since we assumed $x \ge \frac{1}{2}$. Hence, in either case we found $x_1 \in \{0,1\}$ for which $0 \le x-\frac{x_1}{2} < \frac{1}{2}$ holds. Now assume that $x_1,...x_n \in \{0,1\}$ have been chosen such that $0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$. From this we want to show that it is possible to choose $x_{n+1} \in \{0,1\}$. If $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$, then choose $x_{n+1} = 0$ and we obtain $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$. If, however, $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$, choose $x_{n+1} = 1$. Then $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} = (x - \sum_{k=1}^n \frac{x_k}{2^k}) - \frac{1}{2^{n+1}}$
But here's the problem: why is $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k}$ nonnegative? If that can't be shown, then I cannot appeal to Squeeze lemma to show $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$. I could really use some help. Thanks in advanced!
| All that is necessary is to highlight/expand on what the OP has already written out.
Let $P(n)$ be the inductive hypothesis and we write as true
$\tag 1 P(n): \quad 0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}, \; \text{ with } x_1,...x_n \in \{0,1\}$
We need to set $x_{n+1} = 0$ or $x_{n+1} = 1$ such that $P(n+1)$ is true, i.e.
$\tag 2 0 \le x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$
Case 1: $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$
In this case $\text{(2)}$ will be true by simply setting $x_{n+1} = 0$.
Case 2: $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$
Using $\text{(1)}$ we can write
$\tag 3 \frac{1}{2^{n+1}} \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$
Subtracting $\frac{1}{2^{n+1}}$ we can write
$\tag 4 \frac{1}{2^{n+1}} - \frac{1}{2^{n+1}} \le [x - \sum_{k=1}^n \frac{x_k}{2^k}] -\frac{1}{2^{n+1}} < \frac{1}{2^n} - \frac{1}{2^{n+1}}$
or
$\tag 5 0 \le [x - \sum_{k=1}^n \frac{x_k}{2^k}] -\frac{1}{2^{n+1}} < \frac{1}{2^{n+1}}$
So if we set $x_{n+1} = 1$ statement $\text{(2)}$ will be true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\int\sqrt{\sin x+\cos x}\,dx $ I am trying to solve the question
$$
\int\sqrt{\sin x+\cos x}\,dx
$$
Is their any substitution by which I can get the answer. I tried different substitution like i multiplied both numerator and denominator by $\sqrt{\sin x+\cos x} $ and uses sinx -cos x=t but getting complicated
| Elliptic Integral of the Second Kind:
$$E(\varphi,k) = \int_0^\varphi \sqrt {1 - k^2(\sin\vartheta)^2}\, \mathrm d\vartheta$$
$$ I = \int \sqrt{\sin(x) + \cos(x)}\,dx$$
$$\sqrt[4]{2}\int \sqrt{\frac{1}{\sqrt{2}}\sin(x) + \frac{1}{\sqrt{2}}\cos(x)}\,dx$$
$$\sqrt[4]{2}\int \sqrt{\sin(x)\cos\left(\frac{\pi}{4}\right) + \cos(x)\sin\left(\frac{\pi}{4}\right)}\,dx$$
Set $u = \left(x + \frac{\pi}{4}\right) \to du = dx$
$$\sqrt[4]{2}\int \sqrt{\sin(u)}\,du$$
So the elliptic integral can also be written as
$$\int \sqrt{\sin(u)}\,du = \int \sqrt{1 - 2\sin^2\left(\frac{\pi}{4}-\frac{\pi}{2}\right)}\,du$$
$$-2\int\sqrt{1-2\sin^2(u)}$$
$$-2\,E(u\,|\,2) + C$$
$$-2\,E\left(\frac{\pi}{4} - \frac{u}{2}\,|\,2\right) + C$$
Back substitution $u = \left(x + \frac{\pi}{4}\right)$
$$-2\,E\left(\frac{\pi}{4} - \frac{\left(x + \frac{\pi}{4}\right)}{2}\,|\,2\right) + C$$
$$-2\,E\left(\frac{\pi}{8} - \frac{x}{2}\,|\,2\right) + C$$
$$-2\,E\left(\frac{1}{8}\left(\pi - 4x\right)\,|\,2\right) + C$$
Therefore
$$I = -2\sqrt[4]{2}\,E\left(\frac{1}{8}(\pi - 4x)\,|\,2\right) + C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability with pins in a box The question is as follows:
There are 7 red pins, 7 black pins, and 7 green pins in a box. Pick 5 at random without replacement. What is the probability that at least one color was not picked?
My thinking is that this could happen in two ways. Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors. In each case, at least one color is excluded.
Clearly, there are $\binom{21}{5}$ ways of choosing the 5 pins from the box.
In the first case, we choose the pins so that they all have the same color. This equates to $$\binom{7}{5} + \binom{7}{5} + \binom{7}{5}= 3\binom{7}{5}.$$
In the second case, the 5 pins are chosen from two colors, excluding one. So, we have $14$ pins to choose from and we choose $5.$ We can either pick from the red and black, the black and green, or the red and green. This then equates to
$$\binom{14}{5} + \binom{14}{5} + \binom{14}{5} = 3\binom{14}{5}.$$
The final answer would then be
$$\frac{3\binom{7}{5}+3\binom{14}{5}}{\binom{21}{5}} \approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!
| There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $\binom{14}5$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $\binom{7}{5}\times3$.
Therefore the probability is $\frac{3\times(\binom{14}{5}-\binom75)}{\binom{21}5}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx$? How would one evaluate the integral $$\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx$$?
I was told it had a nice closed form and could have been solved with differentiation under the integral sign; however, I tried to set $$I(\alpha) = \int_0^\infty \frac{\arctan(\alpha x)}{x(1+x^2)}~dx$$ and got nowhere (the resulting integral was very messy). Is there a much more clever substitution that can be used to tackle the integral?
| Note
$$\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx=\frac12\int_0^\infty \frac1x d\arctan^2(x)=\frac12\int_0^\infty \frac{\arctan^2(x)}{x^2} \,dx.$$
Now define
$$ I(a,b)=\int_0^\infty \frac{\arctan(ax)\arctan(bx)}{x^2} \,dx $$
and then
\begin{eqnarray}
\frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_0^\infty \frac{1}{(1+a^2x^2)(1+b^2x^2)} \,dx\\
&=&\frac{1}{a^2-b^2}\int_0^\infty\bigg(\frac{a^2}{1+a^2x^2}-\frac{b^2}{1+b^2x^2}\bigg)\;dx\\
&=&\frac{\pi}{2(a+b)}.
\end{eqnarray}
So
$$ \int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx=\frac12\int_0^\infty \frac{\arctan^2(x)}{x^2} \,dx=\frac\pi4\int_0^1\int_0^1\frac{1}{a+b}=\frac{\pi}{2}\ln2. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How fast is the area increasing for an equilateral triangle under the given conditions? Question:.
The sides of an equilateral triangle are increasing at the rate of $\sqrt{3}$ cm/sec. How fast is the area increasing when its side is $6$ cm?
I have done this problem using calculus. Is there any alternative way to solve the above problem?
Given the sides the equilateral triangle = $a$ cm.
Area of the equilateral triangle is
$A=\dfrac{\sqrt{3}}{4}a^2$
differentiating w.r.t $t$ we get,
$\dfrac{dA}{dt}=\dfrac{\sqrt{3}}{4}\times 2a\times \dfrac{da}{dt}$
But it is given by
$\dfrac{da}{dt}=\sqrt{3}$ cm/sec
when $a=6$ cm.
$\therefore \left[\dfrac{dA}{dt}\right]_{a=6}=\dfrac{\sqrt{3}}{4}\times 2\times 6\times \sqrt{3}=9\; cm^2/sec$.
| This still uses calculus as its underlying mechanic, but we can pretend we don't know any of it. We transform the problem into a physic problem.
An object is accelerating at constant $a = \sqrt{3}\ \text{cm/sec}^2$. Using the formula for constant acceleration we find that for initial velocity $0$ that $x = \frac{1}{2}at^2$ where $x$ is the distance traveled. We know $x = 6\ \text{cm}$ and $a = \sqrt{3}\ \text{cm/sec}^2$ so we solve for $t$:
$$6 = \frac{1}{2} \cdot \sqrt 3 \cdot t^2$$
$$t = \sqrt{\frac{2 \cdot 6}{\sqrt 3}} = 2\sqrt[4]{3}$$
Now we can calculate the velocity at $t = 2\sqrt[4]{3}$ using $v = at$ to find $v = 2 \cdot 3^{3/4}$.
So we know that the side of the equilateral triangle when it reaches $6\ \text{cm}$ is increasing by $2 \cdot 3^{3/4}\ \text{cm/sec}$, and thus the area of the equilateral triangle is increasing by $9\ \text{cm}^2/\text{sec}$
| {
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"timestamp": "2023-03-29T00:00:00",
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P.T. $\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$
If $abc=1$ where $a,b,c$ are positive real. Prove that ,$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$.
I tried to multiply the LHS by $abc$ to make the relation homogeneous but......
There is a $3$ in the RHS ,So I applied AM-GM and found that it is enough to show that
$(a+b)(b+c)(c+a) \le 8$ where $abc=1$.
I know it is a very well-known inequality but I couldn't answer it . Please help me.
Source: It is derived from a question came in any Olympiad question (I cannot remember, now).
It is a good problem for the application of AM-GM or, Cauchy-Schartz.Jensen's inequality can also help.
| Let
$$a=\frac1x, \, b=\frac1y, \, c=\frac1z \implies xyz=1$$
$$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32 \iff \frac{x^2}{y+z} +\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac 32$$
by Jensen's Inequality with $f(x)=x^{-1}$ convex we have that
$$\frac{\frac{x^2}{y+z} +\frac{y^2}{x+z}+\frac{z^2}{x+y}}{x+y+z}=\frac{xf\left(\frac{y+z}x\right) +yf\left(\frac{x+z}y\right) +zf\left(\frac{x+y}z\right)}{x+y+z}\ge f\left(\frac{(y+z)+(x+z)+(x+y)}{x+y+z}\right)=\frac12 $$
and by AM-GM $x+y+z\ge 3\sqrt[3]{xyz}=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Values of this integeration The values of $$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+x^2}dx$$
My Failed Attempt
I tried applying F(x) =F(a+b-x) in hopes the denominator will cancel out but it's not. I tried to further simplify the expression but with no luck
$$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+({1-x})^2}dx$$
Adding both integral isn't helping me much
| $$\int_{0}^1 x^4(1-x)^4\dfrac{1}{1+x^2}$$
First compute the integral without boundaries $$\int x^4(1-x)^4\dfrac{1}{1+x^2}$$
Apply long division on $\dfrac{x^4(1-x)^4}{x^2+1}$ and we get in the form of $x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}$
Now you can easily compute the integral $$\int x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}\ dx$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving/Proving Recurrence Relations Given $a_{k-1}=7*5^{k-1}-3*8^{k-1}$ and
$a_k=7*5^k-3*8^k$
for the recurrence $a_n=13a_{n-1}-40a_{n-2}$
prove that the next term is $a_{k+1}=7*5^{k+1}-3*8^{k+1}$
I am kind of stuck right now. This is what I have so far. If anyone could give me advice on where I should go from here. Thanks
$a_{k+1}=13(7*5^k-3*8^k)-40(7*5^{k-1}-3*8^{k-1})$
$=13*7*5^k-13*3*8^k-40*7*5^{k-1}+40*3*8^{k-1}$
$=7(13*5^k-40*5^{k-1})+3(-13*8^k+40*8^{k-1})$
$=7(13*5-40)*5^{k-1}+3(-13*8+40)*8^{k-1}$
| Express $a_{k+1}$ as a linear combination of $5^{k-1}$ and $3^{k-1}$, rework the coefficients and extract the powers of $5$ and $3$.
$$13\,(7\cdot5^k-3\cdot8^k)-40\,(7\cdot5^{k-1}-3\cdot8^{k-1})
\\=(13\cdot7\cdot5-40\cdot7)\,5^{k-1}-(13\cdot3\cdot8-40\cdot3)\,3^{k-1}
\\=175\cdot5^{k-1}-192\cdot 3^{k-1}
\\=7\cdot5^2\cdot5^{k-1}-3\cdot3^2\cdot8^{k-1}
\\=7\cdot5^{k+1}-3\cdot8^{k+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let a, b and c be positive real numbers satisfying Let a, b and c be positive real numbers satisfying
$\frac{1}{a+2019}$ + $\frac{1}{b+2019}$ +$\frac{1}{c+2019}$ = $\frac{1}{2019}$
Show that abc ≥$4038^3$.
My first impression is to use arithematic mean ≥ geometric mean.
| With the help of the Lagrange multipliers the question can be stated as:
determine the stationary points for
$$
L(a,b,c,\lambda) = a b c - (2N)^3 + \lambda\left(\frac{1}{a+N}+\frac{1}{b+N}+\frac{1}{c+N}-\frac{1}{N}\right)
$$
giving the stationary conditions
$$
\left\{
\begin{array}{rcl}
b c-\frac{\lambda }{(a+N)^2}=0 \\
a c-\frac{\lambda }{(b+N)^2}=0 \\
a b-\frac{\lambda }{(c+N)^2}=0 \\
\frac{1}{a+N}+\frac{1}{b+N}+\frac{1}{c+N}-\frac{1}{N}=0 \\
\end{array}
\right.
$$
easily solved given the feasible values
$$
a = b = c = 2N, \lambda = 36N^4
$$
then
$$
abc \ge (2N)^3
$$
because the solution point is a minimum point.
NOTE
Substituting $c = \frac{N^2(a+b+2N)}{N^2-ab}$ into $f(a,b,c) = a b c - (2N)^3$ we get
$$
g(a,b) = N^2 \left(\frac{a b (a+b+2 N)}{a b-N^2}-8 N\right)
$$
and the hessian gives
$$
H_g = \left(
\begin{array}{cc}
-\frac{2 b N^4 (b+N)^2}{\left(N^2-a b\right)^3} & -\frac{2 N^5 (a+N) (b+N)}{\left(N^2-a b\right)^3} \\
-\frac{2 N^5 (a+N) (b+N)}{\left(N^2-a b\right)^3} & -\frac{2 a N^4 (a+N)^2}{\left(N^2-a b\right)^3} \\
\end{array}
\right)
$$
which at the solution point has the eigenvalues $\{2N, \frac{2N}{3}\}$ hence $H_g$ is definite positive, characterizing the minimum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n. Q: Prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n.
In fact, I already have a method, but it seems too long. So I would first (1)is there any problem with my logic? (2) Is there any better/faster method to do these kinds of questions?
My solution:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$$
$$=\frac{1}{15}(3n^5+5n^3+7n)$$
Then my next time is to use M.I to prove that $(3n^5+5n^3+7n)$ is divisible by $15$ for positive integers. And for negative integer, I switch it into -$(3n^5+5n^3+7n)$ and it also satisfies the divisibility of $15$.
But the above seems too clumsy. This course is "Intro to number theory". Is there any beter method to do it?
| Use induction for $n\in \mathbb Z^+$.
Base case:
$$P(1)=\frac{1}{15}(3\cdot 1^5+5\cdot 1^3+7\cdot 1)=1.$$
Inductive hypothesis:
$$P(n)=\frac{1}{15}(3n^5+5n^3+7n) \in \mathbb Z^+.$$
Inductive step:
$$P(n+1)=\frac{1}{15}(3(n+1)^5+5(n+1)^3+7(n+1))=\\
=\underbrace{\frac{1}{15}(3n^5+5n^3+7n)}_{P(n)\in \mathbb Z^+}+\\
\frac{1}{15}(15(n^4+2n^3+2n^2+n)+15(n^2+n)+15) \in \mathbb Z^+.$$
Since the powers of $n$ in $P(n)$ are odd, it will be true for negative integers as well.
| {
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"url": "https://math.stackexchange.com/questions/2917871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Decomposition of polynomials like $1+x^4$ What is the "trick" to get from $1+x^4$ to $$(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)?$$ Of course I can calculate it's true, but I don't understand what steps to take to get from one to the other.
Next to this specific question, I am also looking for the general rules for such decompositions?
Cheers!
| Use the binomial formula $x^4+1=(x^2+i)(x^2-i)$ and compute the roots of the factors
$$x_{1,2} = \pm \tfrac{1}{2}\sqrt{2}(1+i)$$
$$x_{3,4} = \pm \tfrac{1}{2}\sqrt{2}(1-i)$$
Now multiply
$$(x-x_1)(x-x_3) = x^2 - \sqrt{2}x +1$$
$$(x-x_2)(x-x_4) = x^2 + \sqrt{2}x +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Polynomial equation, cannot solve for $x$
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$\begin{align}
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\
a &= 3 \\[0.2ex]
b &= -6 \\
c &= -6
\end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$\begin{align}
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{(-6)^2-4(3)(-6)}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{36-72}}{2a}
\end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}\,$ per my textbook's given solutions?
| $$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=\frac{2\pm\sqrt{4+8}}{2}=\frac{2\pm 2\sqrt 3}{2}=1\pm\sqrt 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$. Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$.
I know that $(1+2\sqrt{3}) = 1^3 + 3(1)^2(\sqrt{2})^2 + 3(1)(\sqrt{2})^2 + (\sqrt{2})^3 = 7+5\sqrt{2}$
so $a_n = 7, b_n = 5$ by binomial expansion.
So my strategy is to prove this via mathematical induction. The base case is easy. What I need for the inductive step is to get the relation between $(a_n+1, bn+1)$ to $(a_n, b_n)$. So we get:
$$a_{n+1}, b_{n+1}\sqrt{2} = (1+\sqrt{2})(1+\sqrt{2})^n = (1+\sqrt{2})(a_n +b_n\sqrt{2}) = (a_n + 2b_n)+(a_n+b_n)\sqrt{2}$$
So we have $a_{n+1} = a_n + 2b_n, b_{n+1} = a_n + b_n$
So now I'm trying to use the euclidean algorithm to show that $\gcd(a_{n+1}, b_{n+1}) = 1$. So we get:
$\gcd(a_{n+1}, b_{n+1}) = (a_n + 2b_n,a_n + b_n)=$
$a_n+2b_n= 1 *(a_n + b_n) + b_n =$
$a_n + b_n = 1 * (b_n) + a_n$
$b_n = 0 * (a_n) + bn$
But, where do I go from here?
Any help would be appreciated!
-IdleMathGuy
| worth emphasizing that multiplying a (column) vector of integers by an integer (square) matrix with integer inverse (so determinant is $\pm 1$) preserves the gcd of the vector... Your (row) vector $(a,b)$ is mapped to $(a+2b, a+b);$ as columns, the matrix is
$$
\left(
\begin{array}{cc}
1 & 2 \\
1 & 1
\end{array}
\right)
$$
with determinant $-1$
If I demand
$$
\left(
\begin{array}{cc}
1 & 2 \\
1 & 1
\end{array}
\right)
\left(
\begin{array}{c}
u \\
v
\end{array}
\right) =
\left(
\begin{array}{c}
x \\
y
\end{array}
\right)
$$
it is obvious that $\gcd(u,v)$ divides both $x,y$ so that $\gcd(u,v) | \gcd(x,y).$ If I then point out that
$$
\left(
\begin{array}{cc}
-1 & 2 \\
1 & -1
\end{array}
\right)
\left(
\begin{array}{c}
x \\
y
\end{array}
\right) =
\left(
\begin{array}{c}
u \\
v
\end{array}
\right)
$$
we see that $\gcd(x,y) | \gcd(u,v),$ therefore
$$\gcd(x,y) = \gcd(u,v) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove by induction that ${2n \choose n} > \frac{2^{2n-1}}{n}$ for every natural number n≥2 When I plug n+1 and elaborate on that I end up with half a sheet of paper full of numbers but I can't get to anything conclusive on the right side. Any help would be appreciated.
| Write
$$
\binom{2(n+1)}{n+1} = \frac{(2n)! \cdot (2n+1)(2n+2)}{(n+1)!(n+1)!}
\\= \frac{(2n)!(2n+1)(2n+2)}{(n!)^2(n+1)^2} >
\frac{2^{2n-1}}{n} \frac{(2n+1)(2n+2)}{(n+1)^2} =
\frac{2^{2n-1}}{n+1} \frac{2(n+1/2)}{n} \frac{2(n+1)}{n+1}
$$
where we identified $\frac{(2n)!}{(n!)^2} = \binom{2n}{n}$ to make use of inductive hypothesis, and in the last equality we have
$$
n + 1/2 > n, \Rightarrow \frac{2(n + 1/2)}{n} > 2, \;
\frac{2(n+1)}{n+1} = 2
$$
so the final product is equal to $\frac{2^{2n-1}}{n+1} \cdot 2^{2} = \frac{2^{2n+1}}{n+1} = \frac{2^{2(n+1)-1}}{n+1}$, as required.
In the proof above, we've made use of the fact that $(n+1)! = n! (n+1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$ to $\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$ I'm trying to bring this expression:
$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$$
To this one:
$$\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$$
Where $\log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
| $$\frac{\color{blue}{5}}{6}\log \frac{5}{4} - \frac{1}{6}\log 2 = \frac{\color{blue}{6-1}}{6}\log\frac{5}{4} - \frac{1}{6}\log2 = \log \frac{5}{4} - \frac{1}{6}\left(\color{green}{\log\frac{5}{4} + \log 2 }\right)$$$$ = \log \frac{5}{4} - \frac{1}{6}\color{green}{ \log \left( \frac{5}{4} \cdot 2 \right)} = \log \frac{5}{4} - \frac{1}{6} \log \frac{5}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $|z^2|+\text{Re}(az)+b = 0.$
Fix $a\in\mathbb{C}$ and $b\in\mathbb{R}.$ Show that the equation $|z^2|+\text{Re}(az)+b = 0$ has a solution iff $|a^2|\geq 4b.$ When solutions exist, show the solution set is a circle.
A seemingly easy problem. Letting $z=x+yi$ and $a = c+di$ the equation can be rewritten as $$x^2+y^2+cx-dy+b = 0.$$
I don't know how to proceed.
| It is a matter of completing squares:
\begin{align*}
& x^{2} + y^{2} + cx - dy + b = 0\Longleftrightarrow \left(x^{2} + cx + \frac{c^{2}}{4}\right) + \left(y^{2} - dy + \frac{d^{2}}{4}\right) + b - \frac{c^{2} + d^{2}}{4} = 0 \\
&\therefore \left(x+\frac{c}{2}\right)^{2} + \left(y - \frac{d}{2}\right)^{2} = \frac{c^{2}+d^{2}}{4} - b \geq 0 \Longleftrightarrow c^{2} + d^{2} \geq 4b \Longleftrightarrow |a|^{2} \geq 4b
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$ I want to find the limit $$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$$
My efforts:
$$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}=\lim_{x\rightarrow 0}\frac{5^x((3/5)^x-1)}{3^x((2/3)^x-1)}$$
Multiplying and dividing numerator and denominator by $x$ we get, $$\lim_{x\rightarrow 0}\frac{5^x\frac{((3/5)^x-1)}{x-0}}{3^x\frac{((2/3)^x-1)}{x-0}}\tag{1}$$
Let,
$$f(x)=(3/5)^x-1, g(x)=(2/3)^x-1$$
Now $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=f'(x)$$
and similarly for $g$, we can rewrite $(1)$ as $$\lim_{x\rightarrow 0} \frac{5^x}{3^x}\times \frac{f'(0)}{g'(0)}\tag{2}$$
We know if $h(x)=a^x,$ then $h'(x)=a^x \log(a)$
Computing and putting everything in piece, we write $(2)$ as
$$\lim_{x\rightarrow 0}\frac{\log(3/5)}{\log(2/3)}$$
So we get limit equal to $\frac{\log(3/5)}{\log(2/3)}$.
Is my computation correct?
| Yes, it is correct, but it is not correct to say that$$\lim_{x\to0}\frac{5^x\frac{((3/5)^x-1)}{x-0}}{3^x\frac{((2/3)^x-1)}{x-0}}=\lim_{x\to0}\frac{5^xf'(0)}{3^xg'(0)}.$$That is, you can't replace some $x$'s by $0$, while keeping others as $x$.
And you could have used L'Hopital's rule$$\lim_{x\to0}\frac{5^xf(x)}{3^xg(x)}=\lim_{x\to0}\frac{5^x}{3^x}\times\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}=\frac{\log\left(\frac35\right)}{\log\left(\frac23\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding limit $\ \lim_{x \to 2} \frac{\cos \frac{\pi}{x}}{x-2}$ without l'hopital's Find limit of
$$\ \lim_{x \to 2} \frac{\cos \frac{\pi}{x}}{x-2} $$
by using $\ t = \frac{\pi}{2} - \frac{\pi}{x} $
$$\ \lim_{x \to 2} \frac{\cos \frac{\pi}{x}}{x-2} = \lim_{t \to 0} \frac{\cos( \frac{\pi}{2}-t)}{-\frac{\pi}{t}} = \lim_{t \to 0} -\frac{t \cdot \cos(\frac{\pi}{2} - t)}{\pi} = \frac{0 \cdot \cos \frac{\pi}{2}}{\pi} = \frac{0}{\pi}$$
Not sure what am I missing here as the correct answer is $\ \frac{\pi}{4} $ according to wolframalpha.This is exactly the same question, yet even using the way suggested there I didn't get to the answer.
Please if you can show solution using the $\ t$ value I gave here and trig identities. I guess there are other ways using l'hopital's rule or something but that's not my intent.
| We have that by $x=y+2\,$ with $\,y \to 0$
$$\lim_{x \to 2} \frac{\cos \frac{\pi}{x}}{x-2}=\lim_{y \to 0} \frac{\cos \frac{\pi}{y+2}}{y}=\lim_{y \to 0} \frac{\sin\left( \frac{\pi}{2}-\frac{\pi}{y+2}\right)}{y}=\lim_{y \to 0} \frac{\sin\left( \frac{\pi y }{2(y+2)}\right)}{y}=$$$$=\lim_{y \to 0} \frac{\sin\left( \frac{\pi y }{2(y+2)}\right)}{ \frac{\pi y }{2(y+2)}}\frac{\pi }{2(y+2)}=1\cdot \frac{\pi}4=\frac{\pi}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $a^2 + 2b^2 - 3c^2 - 6d^2 = 1$ and $ab=3cd$ over integers I'd like to solve this set of equations over $\mathbb{Z}$. Consider this pair of conic sections:
\begin{eqnarray*}
a^2 + 2b^2 - 3c^2 - 6d^2 &=& 1 \\
ab - 3 cd&=& 0
\end{eqnarray*}
If these two equations were over $\mathbb{R}$ , this should define the intersection of a pair of quadrics in four-dimensional space.
One possible starting point is to try that $\frac{a}{c} \times \frac{b}{d} = 3 $ so these numbers are proportional.
| From $ab=3cd$ it follows that $3 \mid ab$. If $3\mid a$ then we get that $1=a^2+2b^2-3c^2-6d^2 \equiv 2b^2 \pmod 3$. Multiplying by $2$ we get $2 \equiv 4b^2\equiv b^2 \pmod 3$ which is a contradiction because there are no squares of integers congruent to $2$ modulo $3$.
Therefore $3 \mid b$. Write $b=3b'$. Then $ab'=cd$. This means that $a=kl, b'=mn, c=km, d=ln$ for some integers $k,l,m,n$. Substituting this to the first equation we get
$$(k^2-6n^2)(l^2-3m^2)=1.$$
It follows that $k^2-6n^2=1=l^2-3m^2$ or $k^2-6n^2=-1=l^2-3m^2$. The second case is impossible, as then $l^2 \equiv 2 \pmod 3$. Therefore we have $k^2-6n^2=1=l^2-3m^2$.
These equations are known as Pell equations and all their solutions are known. See e.g. here.
So, all solutions to the OP's question are quadruples $(a,b,c,d)=(kl,3mn,km,ln)$, where $(k,n)$ is a solution to the Pell equation $k^2-6n^2=1$, and $(l,m)$ is a solution to the Pell equation $l^2-3m^2=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find an equation of the plane that passes through the points $(1, 2, 5), (5, 4, 8) \textrm{ and } (2, 4, 8)$ Find an equation of the plane that passes through the points $(1, 2, 5), (5, 4, 8) $, and $(2, 4, 8)$.
$v_1 = [1,2,5] - [5,4, 8] = [-4, -2, -3]$
$v_2 = [2,4,8] - [5,4,8] = [-3,0,0]$
$v_1 \times v_2 = [0,9,-6]$
thus,
$ax + by + cz = d, a = 0, b = 9, c = -6$
to get d we plug $0(5) + 9(4) - 6(8) = -12$
Therefore the equation is $9y - 6z = -12$
Right?
| Yes, your answer is correct.
Note that if $P(x,y,z)$ is on the plane passing through your given points then then the volume generated by vectors $$ (x-1,y- 2,z- 5), (x-5,y- 4,z- 8), (x-2,y- 4,z- 8)$$ is zero.
Thus the equation of the plain is:
$$ \det \begin{bmatrix} x-1&y-2 &z-5\\x-5&y-4&z-8\\x-2&y-4&z-8\end{bmatrix} =0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite sum of $nk^n$ How do you evaluate a sum of the form $\sum\limits_{n=0}^\infty nk^n$, for $|k|<1$?
If we take $k=\frac{1}{2}$, the sum is $\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \dots$, which isn't like any series I know. Wolfram Alpha evaluates this to $\frac{k}{(k-1)^2}$, but I don't know how it finds that.
| If you don't want to use derivative of power series:
\begin{align}
\frac{z-z^{n+1}}{1 - z}& =& &z+&z^2+&z^3+...+&z^n \\
\frac{z^2-z^{n+1}}{1 - z}& =& &&z^2+&z^3+...+&z^n \\
\frac{z^3-z^{n+1}}{1 - z}& =& &&&z^3+...+&z^n \\
\frac{z^n-z^{n+1}}{1 - z}& =& &&&&z^n \\
\end{align}
Adding them together you get
$$\sum_{k=1}^n kz^k = \frac{z+z^2+..+z^n-nz^{n+1}}{1-z}= \frac{z\frac{1-z^{n}}{1-z}-nz^{n+1}}{1-z}=\frac{z-(n+1)z^{n+1}+nz^{n+2}}{(1-z)^2}$$
Now, use the fact tha for $|z|<1$ we have
$$\lim_n (n+1)z^{n+1}=\lim_n nz^{n+2}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find real $x$ that satisfies $\frac{1}{1+x}>\frac{x}{x-1}$ $\dfrac{1}{1+x}>\dfrac{x}{x-1}$
Multiply both sides with $(x+1)(x-1)$.
$x-1>x(x+1)$
Subtract $(x-1)$ from both sides.
$0>x^2+1$
This seems to have no real answers, although I have been told there should be. What am I doing wrong?
| $$\frac{1}{x+1}-\frac{x}{x-1}=\frac{-x^2-1}{(x-1)(x+1)}.$$
Thus, $(x-1)(x+1)<0,$ which gives $-1<x<1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Check the convergence of $\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$ and if it converges find the sum Check whether the series converges or not. If yes, find the sum of the series $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
My Efforts
Observe $n^3<n^4+n^2+1$
$\therefore$
$\frac{1}{n^4+n^2+n+1}<\frac{1}{n^3}$ which further implies that $\frac{n}{n^4+n^2+n+1}<\frac{n}{n^3}=\frac{1}{n^2}$
Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we conclude by comparison test that the series converges.
I am not able to find the sum of this series? Any hints or directions will be appreciated.
| As an alternative since
$$\frac{n}{n^4+n^2+1}\sim \frac1{n^3}$$
we can conclude that the series converges by limit comparison test with $\sum \frac1{n^3}$.
For the evaluation let consider
$$\frac{n}{n^{4}+n^{2}+1} = \frac{1}{2(n^{2}-n+1)}-\frac{1}{2(n^{2}+n+1)}$$
and use telescoping, indeed for $n=k+1$ we have $$\frac{1}{n^{2}-n+1}=\frac{1}{k^{2}+k+1}$$
and therefore
$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}=\frac12\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=$$
$$=\frac12\left(1+\sum_{n=2}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12\left(1+\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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summation of $\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}$ So I need to show that this sum converges to 4.5. But when i did this is got the sum converges to 2.5.
$$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}$$
My workings:
$$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=1}^\infty \frac {2^n}{6^n}+\sum_{n=1}^\infty \frac {4^n}{6^n}$$
The two summations are then two converging geometric series, whihc the first is 0.5 and the second is 2 so the overall sum converges to 2.5 - but the sum converges to 4.5. Any help would be great.
| Your result seems correct indeed recall that for $|r|<1$
$$\sum_{k=0}^\infty r^n = \frac1{1-r} \implies \sum_{k=1}^\infty r^n = \frac1{1-r} -1$$
therefore
$$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=1}^\infty \left(\frac13\right)^n+\sum_{n=1}^\infty \left(\frac23\right)^n=\frac12+2=\frac52$$
while
$$\sum_{n=0}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=0}^\infty \left(\frac13\right)^n+\sum_{n=0}^\infty \left(\frac23\right)^n=\frac32+3=\frac92$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$ How do I find the value of the following infinite series?
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} $$
My attempt at a solution:
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} = \sum_{n=0}^\infty \frac{1}{((2n-1)(2n+1))^2} = \sum_{n=0}^\infty \left(\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)\right)^2 = \frac{1}{4}\sum_{n=0}^\infty \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)^2 $$
I then tried to compute the partial sums of this series, but with no luck. Does anyone else know how to do it?
| Let's continue with the approach you took. We have
$$\begin{align}\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}
&=\frac{1}{4}\sum_{n=0}^\infty \bigg(\frac{1}{2n-1}-\frac{1}{2n+1}\bigg)^2\\
&=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}-\frac{2}{(2n-1)(2n+1)}
\end{align}$$
Now, using the fact that
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}$$
we have
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+2}{8}-\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}$$
Now we may use telescoping to evaluate the latter series as
$$\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}=-\frac{1}{4}$$
so that we have
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+4}{8}=\frac{1}{2}+\frac{\pi^2}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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