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Maximization problem with partial derivative Problem: Find the volume of the largest rectangular box (with faces parallel to the coordinate planes) that can be inscribed inside the ellipsoid \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \end{align*}
Attempt at a solution: Since the box is inside the ellipsoid, the volume of the box is given by $V= 2x \cdot 2y \cdot 2z = 8xyz$. The box and the ellipsoid have several coordinates in common (at the respective vertices), so choose $(x,y,z)$ in the first octant. We can solve the equation for the ellipsoid for $z$ so that we are left with only two variables instead of three. \begin{align*} 1 &= \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \\ z^2 &= c^2 (1- \frac{y^2}{b^2} - \frac{x^2}{a^2}) \\ z &= c \sqrt{1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}} \end{align*} We put this in our equation for the volume and get: \begin{align*} V &= 8xyc \sqrt{1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}} \\ V^2 &= 64x^2 y^2 c^2 \Big(1- \frac{y^2}{b^2}- \frac{x^2}{a^2} \Big) \\ &= 64c^2 \Big(x^2 y^2 - \frac{x^2 y^4}{b^2} - \frac{x^4 y^2}{a^2} \Big) \end{align*} I guess now we need to take partials with respect to $x$ and $y$. \begin{align*} \frac{\partial V^2}{\partial x} &= 64c^2 \Big(x^2 y^2 - \frac{x^2 y^4}{b^2} - \frac{x^4 y^2}{a^2} \Big)' \\ &= 64c^2 \Big(2xy^2 - \frac{2xy^4}{b^2} - \frac{4x^3 y^2}{a^2}\Big) \\ &= 128c^2 xy^2 \Big(1- \frac{y^2}{b^2} - \frac{2x^2}{a^2} \Big) \\ \frac{\partial V^2}{\partial y} &= 128c^2x^2 y \Big(1- \frac{2y^2}{b^2} - \frac{x^2}{a^2} \Big) \end{align*} How can I proceed now? I know I have to equate the partials with zero and solve, but there's just too many variables?
| Since $x,y>0$, you can conclude that $\displaystyle\frac{2x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\displaystyle\frac{x^2}{a^2}+\frac{2y^2}{b^2}=1$.
Subtracting these two equations gives $\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}$, and then substituting back into either equation gives
$\;\;\;\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{1}{3}$, so $\displaystyle\frac{z^2}{c^2}=\frac{1}{3}$ and $\displaystyle x=\frac{a}{\sqrt{3}}, y=\frac{b}{\sqrt{3}}, z=\frac{c}{\sqrt{3}}.$
Therefore the largest volume is given by $\displaystyle V=8xyz=\frac{8abc}{3\sqrt{3}}$.
Here is an alternate approach:
We want to maximize $V=8xyz$ subject to $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, so using Lagrange multipliers gives
$\displaystyle\;\;\;8yz=\lambda\cdot\frac{2x}{a^2}, \;\;8xz=\lambda\cdot\frac{2y}{b^2}, \;\;8xy=\lambda\cdot\frac{2z}{c^2}$.
Multiplying by x in the first equation, y in the second, and z in the third gives
$\;\;\displaystyle\lambda\left(\frac{2x^2}{a^2}\right)=\lambda\left(\frac{2y^2}{b^2}\right)=\lambda\left(\frac{2z^2}{c^2}\right)$, so $\lambda\ne0\implies\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}$.
Since $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,\;\;\; \displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{1}{3}\implies x=\frac{a}{\sqrt{3}}, y=\frac{b}{\sqrt{3}}, z=\frac{c}{\sqrt{3}}$.
| {
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Prove by induction that $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$ Prove by induction $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$
Step 1: Show true for n = 0:
LHS: $\frac{0}{2^{0}}$ = 0
RHS = $2 - \frac{0+2}{2^{0}}$ = 0
Step 2: Show that it is true for $n = p$, it is true for $n = p + 1$
Starting with the LHS of the $n = p + 1$ expression, breaking out the largest term and substituting in the $n = p$ equality gives:
$\sum_{k=0}^{p+1} \frac{k}{2^{k}} = \frac{p+1}{2^{p+1}} + \sum_{k=0}^{p} \frac{k}{2^{k}} = \frac{p+1}{2^{p+1}} + 2 - \frac{p + 2}{2^{p}}$
Making everything besides the 2 have the same denominator $2^{p+1}$
$\frac{p+1}{2^{p+1}} + 2 - \frac{p + 2}{2^{p}} = 2 + \frac{p + 1 - 2(p+2)}{2^{p+1}} = 2 - \frac{p - 3}{2^{p+1}} $
...I get lost here.
Question: Where have I gone wrong in the above attempt?
| You've just made a sign mistake:
$$p+1-2(p+2)=-p-3=-(p+3)$$
and not $-(p-3)$.
And then you're done, since $p+3=(p+1)+2=n+2$.
Good job otherwise.
| {
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Minimum of $f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|$
For $z\in\mathbb{C}$, calculate the minimum value of
$$
f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|
$$
My Attempt:
Let $z= x+iy$. Then
$$
\begin{align}
z^2+z+1 &= (x+iy)^2+(x+iy)+1\\
&= (x^2-y^2+x+1)+i(2xy+y)
\end{align}
$$
and
$$
\begin{align}
z^2-z+1 &= (x+iy)^2-(x+iy)+1\\
&= (x^2-y^2-x+1)+i(2xy-y)
\end{align}
$$
Define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by
$$
f(x,y) = \sqrt{\big(x^2-y^2+x+1\big)^2+(2xy+y)^2}+\sqrt{\big(x^2-y^2-x+1\big)^2+(y-2xy)^2}
$$
Using the Triangle Inequality, we know that
$$
f(x,y) \geq 2\sqrt{\left(x^2-y^2+1\right)^2+y^2}
$$
How can the problem be solved from this point?
| Your analysis starts the right way. Substituting $z=x+iy$, you get as you write:
$$f(x,y)=\sqrt{(x^2-y^2+x+1)^2+(2xy+y)^2}+\sqrt{(x^2-y^2-x+1)^2+(y-2xy)^2}$$
The above now is a real multivariable function, so standard calculus techniques apply. In particular, the critical points are determined by looking first at the system:
$$\frac{df}{dx}=0$$
$$\frac{df}{dy}=0$$
This system has the solutions:
$$(x,y)=(0,0)$$
and
$$(x,y)=\left(0,\pm \frac{\sqrt{2}}{2}\right)$$
The null solution is rejected because it is not a global extremum and you are left with the second solution which after switching back to complex variables corresponds to:
$$\left(x=0,y=\pm\frac{i}{\sqrt{2}}\right)$$
as Cristian points out.
Here's the corresponding graph in two variables:
| {
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Find the volume of the region of a sphere bounded by two planes Calculate the volume of a sphere $x^2+y^2+z^2=R^2$ which is bounded by $z=a$ and $z=b$, where $0\leq a<b<R$ using double integral. I can imagine the picture but I don't know how to set it up.
| The projection of the solid in the xy-plane is the region bounded by the circle $x^2+y^2=R^2-a^2$,
and the height of the solid is constant inside the circle $x^2+y^2=R^2-b^2$.
Using a double integral in polar coordinates, we get
$V=\displaystyle\int_0^{2\pi}\int_0^{\sqrt{R^2-b^2}}(b-a)r\;dr d\theta+\int_0^{2\pi}\int_{\sqrt{R^2-b^2}}^{\sqrt{R^2-a^2}}\left(\sqrt{R^2-r^2}-a\right)r\;dr d\theta$
$\;\;\;\;\displaystyle=2\pi(b-a)\left(\frac{R^2-b^2}{2}\right)+2\pi\left[\frac{1}{2}\cdot\frac{2}{3}(b^3-a^3)-\frac{a}{2}(b^2-a^2)\right]$
$\;\;\;=\displaystyle\pi(b-a)(R^2-b^2)+\frac{\pi}{3}\left[2(b^3-a^3)-3a(b^2-a^2)\right]$
$\;\;\;=\displaystyle\frac{\pi}{3}(b-a)\left[3R^2-b^2-ab-a^2\right]$.
As pointed out in the comments, it is easier to calculate this volume as a solid of revolution, which gives
$\displaystyle V=\int_a^b \pi(R^2-z^2)\;dz=\pi\left[R^{2}z-\frac{z^3}{3}\right]_a^b=\pi\left(R^2 b-\frac{b^3}{3}-R^2 a+\frac{a^3}{3}\right)$.
| {
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Prove the inequality $\sum\limits_{\text{cyc}} \frac{x}{y+z+\sqrt{x}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}$ How do I show the following?
$$
\frac{x}{y+z+\sqrt{x}}+\frac{y}{z+x+\sqrt{y}}+\frac{z}{x+y+\sqrt{z}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}
$$
Any hint would do; I don't require a complete solution.
| It is just the Cauchy-Schwarz inequality in the form:
$$ \frac{a^2}{A}+\frac{b^2}{B}+\frac{c^2}{C}\geq\frac{(a+b+c)^2}{A+B+C}$$
for $a=x,b=y,c=z$ and $A=xy+xz+x\sqrt{x},B=xy+zy+y\sqrt{y},C=xz+yz+z\sqrt{z}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determine the radical of an ideal Determine the radical of the ideal $(x^3-y^6,xy-y^3)$ in $C[x,y]$.
I used Nullstellensatz theorem $\sqrt{I}=I(V(I))$.
Factorization gives:
$$x^3-y^6=(x-y^2)(x+(\frac{1}{2}+\frac{\sqrt{3}}{2}i) y^2)(x+(\frac{1}{2}-\frac{\sqrt{3}}{2}i) y^2)$$
$$xy-y^3=y(x-y^2)$$
so $V(I)={(x,y)|x-y^2=0}$
Hence $\sqrt{I}=(x-y^2)$. Is this right?
| The result is right.
It is a good idea to factorise your polynomials. The correct factorisation is the following:
$x^3-y^6=(x-y^2)(y^4+xy^2+x^2)$ and $xy-y^3=y(x-y^2)$
Then, a point $(x,y)$ lies in $V(I)$ if and only if one of the following hold:
$x=y^2$ or
$y=0$ and $y^4+xy^2+x^2=0$
As this latter only gives $(0,0)$, already in $V(x-y^2)$, you have indeed $\sqrt{I}=I(V(I))=(x-y^2)$.
| {
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Evaluate $\lim_{x \to 4}\frac{(2x)^{1/3} - 2}{\sqrt{x} - 2}$ without L'Hospital rule I know how to solve this limit with the L'Hospital rule:
$$\lim_{x \rightarrow 4} \frac{(2x)^{1/3} - 2}{\sqrt{x} -2}$$
The answer is $\frac{2}{3}$. I am trying to solve it without the L'Hospital rule.
| You already get many answers, so I shall try something different hoping that you know Taylor series (if this is not the case, just discard my answer).
Built at $x=b$, you have by Taylor $$x^a=b^a+a b^{a-1} (x-b)+\frac{1}{2} (a-1) a b^{a-2} (x-b)^2+O\left((x-b)^3\right)$$ So, the expansion of the numerator is $$\frac{x-4}{6}-\frac{1}{72} (x-4)^2+O\left((x-4)^3\right)$$ and for the denominator $$\frac{x-4}{4}-\frac{1}{64} (x-4)^2+O\left((x-4)^3\right)$$ Performing the long division, you will the get that, close to $x=4$$$\frac{(2x)^{1/3} - 2}{\sqrt{x} -2}\approx \frac{2}{3}-\frac{x-4}{72}$$ which shows both the limit and how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate via substitution and derivation rule I have to solve
this integral
$$\int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x$$
with substitution and then the trick that $\dfrac 1 {\sqrt{1-x^2}} = \dfrac{\mathsf d\;\arcsin(x)}{\mathsf d\;x\qquad\quad\;\,} $
can someone give me a tip on what I should substitute in order to continue?
| First use $x=r \sin(u)$ so, $\mathrm d x = r \cos u\, \mathrm d u$
$$\begin{align}
& \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}}\; \operatorname d y \operatorname d x
\\
& = \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{\frac{r^2}{r^2-x^2-y^2}} \;\operatorname d y \operatorname d x
\\
& = \int_{-\pi/2}^{+\pi/2}r|r| \cos u \int_{-r \cos u}^{+r\cos u} \sqrt{\frac{1}{r^2\cos^2 u - y^2}} \;\operatorname d y \operatorname d u
\end{align}$$
Now can you see how to use the identity?
| {
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An $n \times n$ matrix with rational entries such that $A^{n+1}=I$ I'm working on finding $A \in M_n(\mathbb{Q})$ such that $A^{n+1}=I$. If $n$ is odd, $A=-I$ satisfies the condition. When $n$ is even, clearly it should have eigenvalues $e^{2 \pi ik/(n+1)}(k=1,\cdots ,n)$ as a complex matrix. Now I proved this matrix should be similar to $\begin{pmatrix} R_\theta & 0 & \cdots & 0\\0&R_{2\theta} & \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & R_{n\theta/2} \end{pmatrix} $, where $R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ and $\theta = \frac{2\pi}{n+1}$. Unfortunately, however, $R_\theta$ is not rational if $n \ge 4$. Is there any other method?
| In general, we can take the companion matrix to the polynomial
$$
p(x) = x^n + x^{n-1} + \cdots + 1 = \frac{x^{n+1} - 1}{x - 1}
$$
That is, the matrix $A \in M_n(\Bbb Q)$ given by
$$
A =
\pmatrix{
0&\cdots& &-1\\
1&0&\cdots&-1\\
&1&\ddots&\vdots \\
&&\ddots&-1
}
$$
will satisfy $A^{n+1} = I$
So, for example, you can verify that
$$
\pmatrix{
0&0&0&-1\\
1&0&0&-1\\
0&1&0&-1\\
0&0&1&-1
}^5 =
\pmatrix{
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
}
$$
| {
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Evaluate $\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$ I would like this integral:
$$\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$$
I used substitution $x+ \frac{a}{2k} = \frac{\sqrt{4kb-a^{2}}}{2k} \tan \theta$.
This will reduce the denominator to
$$
\left(\left(\frac{\sqrt{4kb-a^{2}}}{2k}\tan\theta \right)^{2} + \frac{4kb-a^{2}}{4k^{2}}\right)^{2} = \left(\frac{4kb-a^{2}}{4k^{2}}\right)^{2} \sec^{4}\theta ,$$
and integral is reduced to
$$\frac{1}{4k^{2}}\frac{16k^{4}}{(4kb-a^{2})^{\frac{3}{2}}2k} \int \cos^{2} \theta d\theta = \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}}\left(\frac{\theta}{2}+\frac{1}{4}\sin(2\theta)+c\right)$$
$$= \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}} \left(\frac{\arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{2}+\frac{\sin(2 \arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{4}\right).$$
Is there a simpler solution?
| I don't know the demonstration but I have this formula :
$$
\int \frac{dx}{(ax^2+bx+c)^2}=\frac{2ax+b}{(4ac-b^2)(ax^2+bx+c)}+\frac{2a}{4ac-b^2}\int \frac{dx}{ax^2+bx+c}
$$
And after you can use one of this for the second term (it depends on your discriminant)
$$
\int \frac{dx}{ax^2+bx+c}=\frac{1}{\sqrt{b^2-4ac}}ln\left(\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right)
$$
Or
$$
\int \frac{dx}{ax^2+bx+c}=\frac{2}{\sqrt{4ac-b^2}}*tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)
$$
Shadock
| {
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Prove $ \frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right) $ $ \frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right) $
This is what I have so far:
I know that $A + B + C = 180^\circ$, so $C = 180^\circ - (A+B)$. Plugging this in, I get that $\sin C = \sin(A+B)$ and $\cos C = -\cos(A+B)$. When I plug this back into the equation, I get:
$\frac{a\sin A + b\sin B + c \sin(A+B)}{a \cos A + b \cos B - c \cos (A+B)}$.
Expanding out $c \sin(A+B)$ and $c \cos(A+B)$ gives me:
$\frac{a\sin A + b\sin B + c \cos A \cos B - c \sin A \sin B}{a \cos A + b \cos B - c \cos A \cos B + c \sin A \sin B}$.
If I use the Extended Law of Sines, then I get: $ a = 2R * \sin A$, $b = 2R * \sin B$, and $c = 2R * \sin C$, and plugging that in gives me:
$\frac{2R*\sin^{2} A + 2R * \sin^{2} B + 2R \sin C \cos A \cos B - 2R \sin C \sin A \sin B}{2R \sin A \cos A + 2R \sin B \cos B - 2R \sin C \cos A \cos B + 2R \sin C \sin A \sin B}$.
I can factor out all of the $2R's$ to get:
$\frac{\sin^{2}A + \sin^{2} B + \sin C \cos A \cos B - \sin C \sin A \sin B)}{\sin A \cos A + \sin B \cos B - \sin C \cos A \cos B + \sin C \sin A \sin B}$.
Now I'm stuck. What do I do next, to get the end result of $R (\frac{a^2+b^2+c^2}{abc})$?
| Your use of Extended Law of Sines is correct.
Get rid of all the trigo ratios in the numerator.
For the denominator, try to prove that $$a\cos A+b\cos B+c\cos C=R(\sin(2A)+\sin(2B)+\sin(2C))=4\sin A \sin B \sin C$$
Then, use the Extended Law of Sines again! :D
| {
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How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
| $$\sqrt{1 + x + x^2} = 1 + \dfrac{1}{2}(x+x^2) + \cdots = 1 + \dfrac{1}{2}x + \cdots$$ so the
$$\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}} =
\lim \limits_{x \to 0}\dfrac{1 + 1/2 x + \cdots - 1}{x} = \dfrac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate rational function with multiple complex roots I want to integrate
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx.
$$
By WolframAlpha I know the solution is
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx = \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}.
$$
Do derive it on my own, I did a partial fraction decomposition and found
$$
\frac{x^2}{(1+n^2x^2)^2} = \frac{\frac{1}{2n^2}}{1+n^2x^2} + \frac{\frac{1}{4n^2}}{(x-i/n)^2} + \frac{\frac{1}{4n^2}}{(x+i/n)^2}.
$$
I checked this multiple times, but it must be wrong, because when I proceed with it I get
\begin{align*}
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx
& = \frac{1}{n^2} \left( \frac{1}{2} \int_{-1}^1 \frac{1}{1+n^2x^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x-i/n)^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x+i/n)^2} d x \right) \\
& = \frac{1}{n^2}\left( \frac{1}{2} \frac{\arctan(xn)}{n}\bigg\vert_{-1}^1 + \frac{1}{4} \left( \frac{-1}{x-i/n}\bigg\vert_{-1}^1 + \frac{-1}{x+i/n}\bigg\vert_{-1}^1 \right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} + \frac{1}{4} \left( \frac{-2}{1+1/n^2} + \frac{-2}{1+1/n^2}\right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} -\frac{1}{1+1/n^2} \right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} - \frac{n^2}{n^2+1} \right) \\
& = \frac{\arctan(n)}{n^3} - \frac{1}{n^2+1}
\end{align*}
This result is consistent which what I found with WolframAlpha, but surely
$$
\frac{\arctan(n)}{n^3} - \frac{1}{n^2+1} \ne \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}
$$
so what went wrong, I am sitting here since hours and do not see any fault...
| You have a strange form for your partial fraction decomposition. I would be inclined to stay in the field of real numbers and write:
$$
\frac{x^2}{(1+n^2x^2)^2} = \frac{Ax+B}{1+n^2x^2} + \frac{Cx+D}{(1+n^2x^2)^2}
$$
It turns out $B=\frac{1}{n^2}$ and $D = -\frac{1}{n^2}$, while $A=C=0$.
To integrate $\int\frac{dx}{(1+n^2x^2)^2}$, use the substitution $nx=\tan\theta$. Then $n\,dx = \sec^2\theta\,d\theta$ and $(1+n^2x^2)^{\color{green}{2}} = \sec^4\theta$. You get
\begin{align*}
\int\frac{dx}{(1+n^2x^2)^2} &= \frac{1}{n}\int\frac{\sec^2\theta\,d\theta}{\sec^4\theta} \\
&= \frac{1}{n} \int\cos^2\theta\,d\theta \\
&= \frac{1}{2n} \int(1+ \cos(2\theta))\,d\theta\\
&= \frac{\theta}{2n} + \frac{1}{4n}\sin(2\theta) + C \\
&= \frac{\arctan(nx)}{2n}+\frac{1}{2n}\sin\theta \cos\theta + C \\
&= \frac{\arctan(nx)}{2n}+\frac{1}{2n}\frac{nx}{\sqrt{1+n^2x^2}}\frac{1}{\sqrt{1+n^2x^2}} + C \\
&= \frac{\arctan(nx)}{2n}+\frac{1}{2}\frac{x}{1+n^2x^2} + C \\
\end{align*}
[Green exponent of $2$ added thanks to Stefan's careful eye.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Fourier transform of $f(x) = \chi\cos^n(\pi x)$ I ran across an abandon post from 2013 where the OP has no work shown but just a problem statement. The OP was last seen May 2013 so I doubt they will be returning to edit their post with relevant work. However, the problem seems interesting so I am posting it with my own work. The original post can be found here
. On the post, @DilipSarwate comments saying write cosine in exponential form and use the Binomial theorem.
We are trying to find the Fourier transform of
$$
f(x) = \chi\cos^n(\pi x)
$$
where $n\in\mathbb{N}$, $x\in\mathbb{R}$, and
$$
\chi = \begin{cases} 1, & -1/2 <x< 1/2\\
0, & x\not\in (-1/2, 1/2)
\end{cases}
$$
Using the following defintion, $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx$, we have
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\cos^n(\pi x)e^{i\omega x}dx &=
\frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\Bigl[\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}e^{i\pi x(n-k)}e^{-i\pi xk}\Bigr]e^{i\omega x}dx\\
&= \frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx
\end{align}
Can we swap the sum and integral here? If so, we have
$$
\frac{1}{\sqrt{2\pi}}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\int_{-1/2}^{1/2}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx =
\frac{1}{\sqrt{2\pi}}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\frac{2\sin\bigl[1/2(\pi n - 2\pi k + \omega)\bigr]}{\pi n - 2\pi k + \omega}
$$
From Mathematica, I know that sum evaluates to
$$
\sum = -\frac{ n! \sin \bigl(\frac{1}{2} (\pi n+\omega )\bigr) \Gamma \bigl(-\frac{\pi n+\omega }{2 \pi }\bigr)}{2^n\pi \Gamma \Bigl[(\frac{1}{2} \bigl(n-\frac{\omega }{\pi }+2\bigr)\Bigr]}
$$
but I don't know how to evaluate it analytically. This whole part was under the assumption the sum and integral can be interchanged.
| Since $n$ is finite, there is just a finite sum. Therefore it is allowed to interchange the summation and integration.
To calculate the summation, we start with this, without having interchanged summation and integration:
\begin{align}
\sum&=\int_{-1/2}^{1/2}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx\\
&=\int_{-1/2}^{1/2}\frac{1}{2^n}\left[\sum_{k=0}^n\binom{n}{k}e^{-2i\pi kx}\right]e^{i(n\pi+ \omega)x}dx\\
&=\frac{1}{2^n}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^ne^{i(n\pi+ \omega)x}dx\equiv \frac{1}{2^n}I(n,n\pi+\omega)\\
&=\frac{1}{2^n}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^n\left[\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]'dx\\
&=\frac{1}{2^n}\left[(1+e^{-2i\pi x})^n\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]_{-1/2}^{1/2}+\frac{1}{2^n}\int_{-1/2}^{1/2}2in\pi e^{-2i\pi x}(1+e^{-2i\pi x})^{n-1}\left[\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]dx\\
&=\frac{1}{2^n}\frac{2\pi n}{n\pi+\omega}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^{n-1}e^{i(n\pi+\omega-2\pi)x}dx\\
&=\frac{1}{2^n}\frac{2\pi n}{n\pi+\omega}I(n-1,(n-2)\pi+\omega)
\end{align}
Repeating this $n$ times, it follows:
\begin{align}
I(n,n\pi+\omega)&=n!\prod_{k=0}^{n-1}\frac{1}{\frac{\omega}{2\pi}+\frac{n}{2}-k}I(0,\omega-n\pi)\\
&=n!\prod_{k=0}^{n-1}\frac{-1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\int_{-1/2}^{1/2}e^{i(\omega-n\pi)x}dx\\
&=n!(-1)^{n}\prod_{k=0}^{n-1}\frac{1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\left[\frac{e^{i(\omega-n\pi)x}}{2\pi i(\frac{\omega}{2\pi}-\frac{n}{2})}\right]_{-1/2}^{1/2}\\
&=n!(-1)^{n+1}\prod_{k=0}^{n}\frac{1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\left[\frac{2i\sin\left(\frac{1}{2}(\omega-n\pi)\right)}{2\pi i}\right]\\
&=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(n+1-\frac{\omega}{2\pi}-\frac{n}{2})}\sin\left(\frac{1}{2}(\omega-n\pi)\right)\\
&=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}\sin\left(\frac{1}{2}(\omega-n\pi)\right)\\
&=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}(-1)^n\sin\left(\frac{1}{2}(\omega+n\pi)\right)\\
\end{align}
Conclusion:
$$\sum=-\frac{n!\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}\sin\left((\frac{1}{2}(\omega+n\pi)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume that $n$ is a positive number $> 1$ and that $P(n)$ holds. Hence,
$$
\frac{1}{2}+\ldots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I now add ${\displaystyle \frac{n + 1}{2\cdot 2^n}}$ to both sides to get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2\cdot 2^n}=2-\frac{2+n}{2^n}+\frac{n+1}{2\cdot 2^n}.
$$
But after manipulating the right side I get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{3n + 5}{2^{n+1}}.
$$
This is definitely not right since I should be getting
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n + 3}{2^{n+1}}=2-\frac{2+(n+1)}{2^{n+1}}
$$
to prove that $P(n)$ implies $P(n+1)$.
What am I doing wrong?
| For $n\geq 1$, let $P(n)$ denote the statement
$$
P(n) : \frac{1}{2}+\frac{2}{2^2}+\cdots+\frac{n}{2^n} = 2-\frac{2+n}{2^n}.
$$
Then, we want to show that $P(n)\to P(n+1)$; that is, we want to show that
$$
P(n+1) : \frac{1}{2}+\frac{2}{2^2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}} = 2-\frac{2+(n+1)}{2^{n+1}}.
$$
The statement $P(1)$ is clearly true: $\frac{1}{2}=2-\frac{2+1}{2^1}$. For the inductive step, fix some $k\geq 1$ and suppose that
$$
P(k) : \frac{1}{2}+\frac{2}{2^2}+\cdots+\frac{k}{2^k} = 2-\frac{2+k}{2^k}
$$
holds. Needed to be shown is that
$$
P(k+1) : \frac{1}{2}+\frac{2}{2^2}+\cdots+\frac{k}{2^k}+\frac{k+1}{2^{k+1}} = 2-\frac{2+(k+1)}{2^{k+1}}
$$
follows. Starting with the left-hand side of $P(k+1)$,
\begin{align*}
\frac{1}{2}+\frac{2}{2^2}+\cdots+\frac{k}{2^k}+\frac{k+1}{2^{k+1}}&= \left(2-\frac{2+k}{2^k}\right)+\frac{k+1}{2^{k+1}}\tag{ind. hyp}\\[1em]&=2-\frac{2(2+k)}{2^{k+1}}+\frac{k+1}{2^{k+1}}\tag{manipulate}\\[1em]&=2-\frac{4+2k-k-1}{2^{k+1}}\tag{simplify}\\[1em]&=2-\frac{3+k}{2^{k+1}}\tag{simplify}\\[1em]&=2-\frac{2+(k+1)}{2^{k+1}},
\end{align*}
which equals the right-hand side of $P(k+1)$. This completes the inductive step. By mathematical induction, for every $n\geq 1, P(n)$ is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$ Let $a,b,c\in{\mathbb{R^+}}$. Prove that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$
I tried to expand both, but I did not get anything useful.
| On expanding, this is the same as showing
$$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3$$
which follows using a series of AM-GMs. Specifically:
$$\begin{align}
LHS &\ge \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +3 \\
&= \sum_{cyc}\left(\frac{a^2}{b^2}+1 \right) \ge \sum_{cyc}2\frac{a}b \\
&\ge \sum_{cyc} \frac{a}b+3 = RHS
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Chinese Remainder Theorem with with non-pairwise coprime moduli I have a system of congruences:
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 5\pmod 6 \\
x\equiv 0\pmod 7
$$
2, 3, 4, 5, 6, 7 are non-pairwise coprime, so i can't use Chinese Reminder Theorem in this case.
$x\equiv 3\pmod 4$ yields $x\equiv 1\pmod 2$ so i can just drop 3-rd equation from the system.
Using this principle i got new system:
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 0\pmod 7 \\
$$
Solving this system o got the answer: $497 + 42k$, which is actually wrong.
Right answer i can get here: http://maciejkus.com/chinese_remainder/ , which is: $119 + 420k$
How can i solve this system right?
| $$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 5\pmod 6 \\
x\equiv 0\pmod 7
$$
$x\equiv 5\pmod 6$ if and only if $x\equiv 1\pmod 2$ and $x\equiv 2\pmod 3$. and both conditions are already on the list. This leaves us with
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 0\pmod 7
$$
$x\equiv 3\pmod 4$ implies that $x\equiv 1\pmod 2$ but not vice-versa. So we keep the first and remove the second.
$$
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 0\pmod 7
$$
The solution is $x = 119$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^4-3x^2+1=0$ in terms of cosine. I put the equation in the form of a quadratic:
$(x^2)^2-3x^2+1=0$
Then using the quadratic formula,
$x^2=\frac{3\pm\sqrt{9-4}}{2}$
$x^2=\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$
$x=\pm\frac{1+\sqrt{5}}{2}$ and $\pm\frac{1-\sqrt{5}}{2}$
So there are four roots as expected given the equation is a quartic. But I really don't know how to put the answers in terms of cosine. Any hints?
| Noting that a diagonal of a pentagon is $\frac{1+\sqrt5}{2}$, we can draw a diagonal from 2 non-adjacent verticies of a unit pentagon and get a triangle with degree measures $36,36,108$ and side lengths $1,1,\frac{1+\sqrt5}{2}$. Now if we take a cosine of a 36 degree angle, we get that $\cos(36^\circ)=\frac{\frac{1+\sqrt5}{2}}{1}=\frac{1+\sqrt5}{2}$. You can find the other solution by taking $180^\circ-36^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_{-\pi}^{\pi}\sin \sin x \,dx=0$ without using the fact that $\sin(x)$ is odd. Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.
Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?
| Another silly answer, using complex analytic methods (similar to sos440's answer, but avoiding use of Bessel functions):
Rewrite the integrand using Euler's formulas and put $z = e^{ix}$, thus mapping $[-\pi,\pi]$ to the unit circle (some algebra omitted):
$$
\int_{-\pi}^\pi \sin \sin x \, dx =
-\frac1{2} \int_{|z|=1} \frac{\exp\Big( \frac12 ( z - \frac1z ) \Big) - \exp\Big( \frac12 ( -z + \frac1z ) \Big)}{z}\,dz.
$$
The integrand has an essential singularity at $z=0$, but we can still compute the relevant residue. Thanks to the $z$ in the denominator, we only have to compute the $0$:th terms of the Laurent series for the numerator.
We have
\begin{align}
\exp\Big( \frac12 ( z - \frac1z ) \Big) &=
e^{z/2} \cdot e^{-1/(2z)} \\
&= \Big( 1 + \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\Big( 1 - \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\end{align}
Hence, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Similarly
\begin{align}
\exp\Big( \frac12 (-z + \frac1z ) \Big) &=
e^{-z/2} \cdot e^{1/(2z)} \\
&= \Big( 1 - \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\Big( 1 + \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\end{align}
And again, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Summing up, the $0$:th term in the Laurent series for the numerator vanishes, and by the residue theorem, the integral, unsurprisingly is $0$.
Of course, this approach also uses, albeit implicitly, since I bothered to write out a number of unnecessary calculations, that the integrand is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
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Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction
My proof so far:
Let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$
Base Case
$P(1):$
LHS = $1^3 = 1$
RHS = $(1)^2 = 1$
Since LHS = RHS, therefore base case holds
Induction Hypothesis
Let $n \in \mathbb{N}$ be arbitrary
Assume $P(n)$ holds
Induction Step
Prove $P(n+1)$ holds:
$$
\begin{align}
& 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1)
\end{align}
$$
This is where I get stuck. I don't know how to prove that my last step is equivalent to:
$$(1 + 2 + \cdots + \;n + (n+1))^2$$
| So basically you want that last expression to turn out to be $\big(1+2+\ldots+n+(n+1)\big)^2$, so you want $(n+1)^3$ to be equal to the difference
$$\big(1+2+\ldots+n+(n+1)\big)^2-(1+2+\ldots+n)^2\;.$$
That’s a difference of two squares, so you can factor it as
$$(n+1)\Big(2(1+2+\ldots+n)+(n+1)\Big)\;.\tag{1}$$
To show that $(1)$ is just a fancy way of writing $(n+1)^3$, you need to show that
$$2(1+2+\ldots+n)+(n+1)=(n+1)^2\;.$$
Do you know a simpler expression for $1+2+\ldots+n$?
Work forward from this...
Okay so going forward.
Basically, Induction works like this, I'll use your question as an example:
Consider the case when $n = 1$. If so, then we will have $1^3 = 1^2$.
Now suppose that $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$.
Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$.
Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$.
Hence, the statement holds for the $n + 1$ case. Thus by the mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
$\mathbb Q.\mathbb E.\mathbb D$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
How is $x^2+1=(1/{x^2})[1-{1}/{x^2}+{1}/{x^4}-{1}/{x^6}+\cdots]$? The author of my book writes:
$$x^2+1=x^2\left(1+\frac{1}{x^2}\right)$$
$$=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]$$
I do not understand the last step. How did the author write the last step. Please help.
| You may write $$x^2 + 1 = x^2(1 + \frac{1}{x^2})$$
as $$\frac{1}{(1 + \frac{1}{x^2})} = \frac{x^2}{1 + x^2}$$
and
$$\frac{1}{1 + \frac{1}{x^2}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{x^{2n}}$$
Edit: To avoid any more downvotes, the author has made a mistake, clearly the equation does not hold, this was just an attemptive to reinterpret whatever the author's intention was.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Generating Functions for Two Variables Find the generating function for the number of words, from the standard 26-letter
alphabet, that have $k$ letter with exactly 1 A and at least 2 Bs. ($k$ will vary)
Workings:
For the time being I'm going to ignore $k$.
Exactly 1 A:
$x$
At least 2 Bs:
$(x^2 + x^3 + x^4 + ... x^{26})$
Letters C to Z:
$(1 + x + x^2 + ... x^{26})^{24}$
So
$g(x) = x(x^2)(1 + x + x^2 + ... + x^{24})(1 + x + x^2 + ... x^{26})^{24}$
Now I'm not sure how to deal with the varying $k$. Any help will be appreciated.
| We need $k\geq3$.
The number of words of length $k$ with exactly one $A$ are $\binom{k}{1}25^{k-1}$ (choose the position for the $A$ and count all possibilities for the letters in the other positions).
the generating function of this sequence is
$$g_1(x):=\sum_{k=3}^{\infty}k25^{k-1}x^k=\frac{x}{25}\left(\frac{1}{1-25x}\right)'=\frac{x}{(1-25x)^2}-x+50x^2$$
The number of words with exactly one $A$ and exactly one $B$ is $2\binom{k}{2}24^{k-2}$ (choose the positions for the $A$ and the $B$ and count all possibilities for the letters in the other positions).
Its generating function is
$$g_2(x)=\sum_{k=3}^{\infty}k(k-1)24^{k-2}x^k=\frac{x^2}{24^2}\left(\frac{1}{1-24x}\right)''=\frac{2x^2}{(1-24x)^3}-2x^2$$
The number of words with exactly one $A$ and no $B$ is $\binom{k}{1}24^{k-1}$.
Its generating function is
$$g_3(x)=\sum_{k=3}^{\infty}k24^{k-1}x^k=\frac{x}{24}\left(\frac{1}{1-24x}\right)'=\frac{x}{(1-24x)^2}-x+48x^2$$
The number of words with exactly one $A$ and at least two $B$ is
$$a_k:=\binom{k}{1}25^{k-1}-2\binom{k}{2}24^{k-2}-\binom{k}{1}24^{k-1}=k25^{k-1}-k(k-1)24^{k-2}-k24^{k-1}$$
Now we sum $$\begin{align}f(x)&:=g_1(x)-g_2(x)-g_3(x)\\&=\frac{x}{(1-25x)^2}-x+50x^2-\frac{2x^2}{(1-24x)^3}+2x^2-\frac{x}{(1-24x)^2}+x-48x^2\\&=\frac{x}{(1-25x)^2}-\frac{2x^2}{(1-24x)^3}-\frac{x}{(1-24x)^2}+4x^2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $2^x\equiv 5\pmod{13}$ I know the solution is $x\equiv 9 \pmod{12}$.
I worked it by doing the donkey work of taking powers of $2$ : $2^2, 2^2, 2^3, \ldots, 2^{12}$ and picking the one that reduces to $5$. Just wondering if there is any other efficient method to solve the congruence when the variable is in exponent ? Thanks!
| $$2^x\equiv5\equiv-8\equiv-2^3\pmod{13}$$
$$\iff2^{x-3}\equiv-1$$
Now $2^{12}\equiv1\pmod{13}\iff13|(2^6+1)(2^6-1)$ and we actually have $2^6\equiv-1\pmod{13}$
Clearly,$6$ is the smallest positive exponent
$\implies x-3\equiv6\pmod{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric functions - finding solutions Question:
Find the general solution for the equation:
$$\sin x + 2\sin2x - \sin3x = 3$$
Approach:
Well using the identity of $\sin a - \sin b $, I merged together $\sin x - \sin3x$
And as $\sin 2x = 2\sin x\cos x$, I got up till here:
$$2\sin x(2\cos x - \cos 2x) = 3$$
Edit:
Considering $\sin x - \sin 3x = -2\cos 2x\sin x$
The equation becomes:
$$-2\cos 2x\sin x+2\sin 2x = 3$$
$$(-2\sin x)\cos 2x +2\sin 2x = 3$$
The maximum value of the equation $asinx + bcosx$ is $\sqrt{a^2 + b^2}$
Using this in here: (considering $b = -2\sin x$ and $a = 2$)
$$3 \leq \sqrt{4\sin^2 x + 4}$$
$$3/2 \leq \sqrt{1 + sin^2x}$$
Obviously cannot happen as maximum value of $\sqrt{1 + sin^2x}$ is $\sqrt2$
| By using the trigonometric identities:
$$\sin(2x) = 2\sin(x)\cos(x)$$
$$\sin(3x) = 3\sin(x)-4\sin^{3}(x),$$ we have
$$\sin(x) +4\sin(x)\cos(x) -3\sin(x) + 4\sin^3(x) = 3 \implies \cos(x) = \frac{3 + 2\sin(x) -4\sin^3(x)}{4\sin(x)}.$$
By squaring the equation, the solutions are kept. Hence,
$$1-\sin^2(x) = \frac{16\sin^6(x)-16\sin^4(x)-24\sin^3(x)+4\sin^2(x)+12\sin(x) +9}{16\sin^2(x)}.$$
Take $y = \sin(x)$. Thus,
$$(1-y^2) = \frac{16y^6-16y^4-24y^3+4y^2 +12y+9}{16y^2}$$ $$\implies 16y^6-24y^3-12y^2 +12y+9 =0,$$ which has no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that $a^2+b^2+c^2+d^2+e^2 > a(b+c+d+e)$ Prove that $a^2+b^2+c^2+d^2+e^2 > a(b+c+d+e)$
Seems to be easy but, cannot see the method right now.
Tried adding known things like $a^2+b^2>=2ab$ and so on with other letters.Maybe I didn't found the right combination.
Please give me only a little hint.
| First, notice that
$$\left(\dfrac{a}{2}-b\right)^2\ge 0\Longleftrightarrow \dfrac{a^2}{4}+b^2\ge ab$$
Apply the same property to the other variables:
$$\dfrac{a^2}{4}+c^2\ge ac$$
$$\dfrac{a^2}{4}+d^2\ge ad$$
$$\dfrac{a^2}{4}+e^2\ge ae$$
Summing up everything:
$$a^2+b^2+c^2+d^2+e^2\ge a(b+c+d+e)\quad\Box$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $((x^2-1)(x+1))^{-2/3}$ using the substitution $u=(x-1)/(x+1)$? I was asked to find the indefinite integral $$\int \frac{1}{((x^2-1)(x+1))^{2/3}} dx$$
using the substitution of $u=(x-1)/(x+1)$.
How do I make this substitution?
I attempted to solve this problem as follows: First I factored the denominator and distributed the exponent. Then I multiplied by one in the form of (x-1)/(x-1) to yield $$\int {(x-1)^{-4/3}((x-1)/(x+1))^{2/3}(x+1)^{-2/3}} dx$$ At this point I made the u substitution and replace all x with $((-u-1)/(u-1))$ I did some rearranging and got stuck with (1/2) $$\int u^{2/3}(-2/(u-1))^{4/3}(-2u/(u-1))^{-4/3} du$$
| $$
du = \frac{2}{(x+1)^2}dx
$$
Also realise that
$$
\left(x^2-1\right)(x+1) = (x-1)(x+1)(x+1) = (x -1)(x+1)^2
$$
$$
\int\frac{1}{\left( (x -1)(x+1)^2\right)^{2/3}}dx = \\
\int\frac{1}{(x+1)^{4/3}(x-1)^{2/3}}dx =\\
\int \frac{(x+1)^{2/3}}{(x+1)^{6/3}}\frac{1}{(x-1)^{2/3}}dx=\\
\int \frac{1}{(x+1)^2}\left(\frac{x+1}{x-1}\right)^{2/3}dx
$$
Using the sub now
$$
\frac{1}{2}\int \left(\frac{1}{u}\right)^{2/3}du
$$
Which you can solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is ... A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.
| Note that
$$\frac{2^{k+1}}{3^{2^{k+1}}-1} - \frac{2^{k}}{3^{2^{k}}-1} = \frac{2^{k}}{3^{2^{k}}-1} \left ( \frac{2}{3^{2^{k}}+1} - 1\right ) = -\frac{2^k}{3^{2^{k}}+1}$$
Thus the finite sum is a telescoping sum and depends only on the end terms of the series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine a matrix knowing its eigenvalues and eigenvectors I read through similar questions, but I couldn't find an answer to this:
How do you determine the symmetric matrix A if you know:
$\lambda_1 = 1, \ eigenvector_1 = \pmatrix{1& 0&-1}^T;$
$\lambda_2 = -2, \ eigenvector_2 = \pmatrix{1& 1& 1}^T;$
$\lambda_3 = 2, \ eigenvector_3 = \pmatrix{-1& 2& -1}^T;$
I tried to solve it as an equation system for each line, but it didn't work somehow.
I tried to find the inverse of the eigenvectors, but it brought a wrong matrix.
Do you know how to solve it?
Thanks!
| Writing the matrix down in the basis defined by the eigenvalues is trivial. It's just
$$
M=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 2
\end{array}
\right).
$$
Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely:
$$
S = \left(
\begin{array}{ccc}
1 & 1 & -1 \\
0 & 1 & 2 \\
-1 & 1 & -1 \\
\end{array}
\right).
$$
This is just the matrix whose columns are the eigenvectors. We can change to the standard coordinate bases by computing $SMS^{-1}$. We get
$$
SMS^{-1} = \frac{1}{6}\left(
\begin{array}{ccc}
1 & -8 & -5 \\
-8 & 4 & -8 \\
-5 & -8 & 1 \\
\end{array}
\right).
$$
You can check that this matrix has the desired eigensystem. For example,
$$
\frac{1}{6}\left(
\begin{array}{ccc}
1 & -8 & -5 \\
-8 & 4 & -8 \\
-5 & -8 & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
-1 \\ 2 \\ -1
\end{array}
\right)
=
\left(
\begin{array}{c}
-2 \\ 4 \\ -2
\end{array}
\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 0
} |
Clever use of Pell's equation Find infinitely many triples $(a,b,c)$ of positive integers such that $a,b,c$ are in arithmetic progression and such that $ab+1$, $bc+1$, and $ca+1$are perfect squares.
The solution is:
Consider the Pell's equation $x^{2}-3y^{2}=1$. If $(r,s)$ is a solution, then the triple
$(a,b,c)=(2s-r,2s,2s+r)$ is in arithmetic progression and $(2s-r)(2s)+1 = (r-s)^{2}$,
$(2s-r)(2s+r)+1 = s^{2}$, and $(2s)(2s+r)+1 = (r+s)^{2}$
Can someone explain to me how the author came up with this solution? How did he know to use $x^{2}-3y^{2}=1$?
| (Old answer revised.)
I. Method.
Here is a sketch of how one can find that Pell equation (and others) from first principles. Let,
$$\begin{aligned}
ab+1\;& = (p_1x+p_2y)^2\\
ac+1\;& = (p_3x+p_4y)^2\\
bc+1\;& = (p_5x+p_6y)^2
\end{aligned}\tag1$$
Since we wish $a,b,c$ to be in arithmetic progression, assume,
$$a,\,b,\,c = -q_1x+q_2y,\;q_3y,\;q_1x+q_2y$$
for unknown integers $p_i,\,q_i$.Expand $(1)$ and collect powers of $x,y$,
$$\begin{aligned}
r_1x^2+r_4xy+r_7y^2 \;& =1\\
r_2x^2+r_5xy+r_8y^2 \;& =1\\
r_3x^2+r_6xy+r_9y^2 \;& =1\\
\end{aligned}$$
where the $r_i$ are in terms of the $p_i,\,q_i$. The above is a clue that a Pell equation may be involved. Then solve the system,
$$\begin{aligned}
& r_1 = r_2 = r_3=1\\
& r_4 = r_5 = r_6=0\\
& r_7 = r_8 = r_9\\
\end{aligned}$$
which is quite easy to do.
II. Solution.
We find,
$$\begin{aligned}
ab+1\;& = \big(x-(m+n)y\big)^2\\
ac+1\;& = (ny)^2\\
bc+1\;& = \big(x+(m+n)y\big)^2
\end{aligned}$$
where $a,\,b,\,c = -x+my,\;2(m+n)y,\;x+my,\,$ and $x,y$ solve the more general Pell equation,
$$x^2-(m^2-n^2)y^2=1\tag2$$
Since $a,b,c$ are to be in arithmetic progression, let $m = 2(m+n)$, so $m,n = 2,-1$ and $x^2-3y^2=1$ pops out.
III. Others.
Using the same method,
$$\begin{aligned}
ab-1\;& = (my)^2\\
ac-1\;& = \big(x+(m+n)y\big)^2\\
bc-1\;& = \big(x+(m-n)y\big)^2
\end{aligned}$$
where $a,\,b,\,c = x+ny,\;x-ny,\;2(x+my),\,$ and $x,y$ solve the similar Pell equation,
$$x^2-(m^2+n^2)y^2=1\tag3$$
IV. Higher Powers
Also by solving a system of equations, given,
$$x^2-17y^2 =\pm1\tag2$$
then,
$$(13x^2 + 12x y - 17y^2)^4 + (13x^2 - 12x y - 17y^2)^4 = (239x^4 - 14x^2 y^2 - 289y^4)^2+1$$
As you may notice, Pell equations are useful with other Diophantine equations where one term is set equal to 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (assuming I'm not the right way)
EDIT
Following the anwer:
$$\int \sum_{n=0}^\infty x^{2n-1} dx = \int \frac{\sum_{n=0}^\infty (x^2)^n}{x} dx = \int \frac{\frac{1}{1-x^2}}{x} dx= \int \frac{1}{x(1-x^2)}dx = \int \frac{1-x^2+x^2}{x(1-x^2)} = \int \frac{1}{x} dx + \int \frac{x}{1-x^2}dx = \ln(x) + \int \frac{1}{2}\frac{1}{1-t}dt + C = \ln(x) - \frac{\ln(x^2)}{2} + C$$
*
*Is that right?
*How to evaluate $C$?
| Consider,
$$\sum_{n\ge 1} x^{n-1} = \frac{1}{1-x}$$
Let $x \rightarrow x^2$ and then integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Let $p^3+q^3=4$ and $pq=2/3$ . Find $p+q$.
Let $p^3+q^3=4$ and $pq=\frac{2}{3}$ . Find $p+q$.
A graphing calculator can find values of $p$ and $q$ numerically. As one can see from the graph below, the two solutions are approximately $(0.4, 1.6)$ and $(1.6, 0.4)$:
However, I am interested in a symbolic solution. Is there a method to solve this problem quickly without having to use a graphing calculator?
| we have $p^3 + q^3 = 4$ and $pq = 2/3$ so $p^3 q^3 = 8/27.$ therefore
$$0= (x - p^3)(x-q^3) = x^2 - (p^3 + q^3)x + p^3q^3 = x^2 - 4x + 8/27$$ solving this quadratic equation we find
$x = \dfrac{4 \pm \sqrt{16 - 4*8/27}}{2} =\dfrac{6\sqrt 3 \pm 10}{ 3\sqrt 3}$
so $$p+q = \left( \dfrac{6\sqrt 3 + 10}{ 3\sqrt 3}\right)^{1/3} + \left( \dfrac{6\sqrt 3 - 10}{ 3\sqrt 3}\right)^{1/3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify this arithmetic expression I'm trying to simplify:
$\left[(\frac{3}{4}\right)^{7}\cdot$ $\left(\frac{3}{4}\right)^{-4}]^{2}$ $\cdot4^5$
The only advance that I have done is
$\left[(\frac{3}{4}\right)^{14}\cdot$ $\left(\frac{3}{4}\right)^{-8}]$ $\cdot4^5$
and then $\left[(\frac{3}{4}\right)^{6}]$ $\cdot4^5$
the answer is$$\frac{3^6}{4}=\frac{729}{4}$$
I do not know what to do next, can someone please guide me in how to solve this exercise.
| $[(3/4)^{7-4}]^2 \times 4^5 = [(3/4)^3]^2 \times 4^5 = (3/4)^6 \times 4^5 = (3^6)/(4^6) \times 4^5 = (3^6)/4 = 729/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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how to compute this limit
compute $I=\lim\limits_{n\to+\infty}\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}$
attempt: I tried to evaluate the integral
$$\begin{align}
\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx&=\int\limits_0^1x^{n+1}\left(1-x-x^2-x^3+x^5+\cdots\right)dx\\
&=\int\limits_0^1x^{n+1}-x^{n+2}-x^{n+3}-x^{n+4}+x^{n+6}+\cdots dx\\
&=\left.\frac{x^{n+2}}{n+2}-\frac{x^{n+3}}{n+3}-\frac{x^{n+4}}{n+4}-\frac{x^{n+5}}{n+5}+\frac{x^{n+6}}{n+6}+\cdots\right|_0^1\\
&=\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}-\frac{1}{n+5}+\frac{1}{n+6}+\cdots
\end{align}$$
attempt 2:
I think I can use mean value theorem for integrals
if $f$ are continuous into $[a,b]$ then exists $\xi\in[0,1]$ such that $\int\limits_a^b f(x)dx=f(\xi)(b-a)$
then using it for $[a,b]=[0,1]$ since the function inside integral are continuous, then exists $\xi\in[0,1]$ such that
$$\begin{align}
\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-0)}\\
&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}
\end{align}$$
then making for $n\in\mathbb{N}^*$
$$\begin{align}
a_n&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\\
a_{n+1}&=\sqrt[n+1]{\xi^{n+2}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\left[\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\right]^n(1-\xi^{n+1})}\\
&=\sqrt[n+1]{a_n^n\xi(1-\xi^{n+1})}\\
a_{n+1}^{n+1}&=a_n^n\xi(1-\xi^{n+1})
\end{align}$$
and $\xi\in(0,1)\Rightarrow\xi^n\in(0,1)\Rightarrow 1-\xi^n\in(0,1)\Rightarrow a_n>0$ and $a_n\in(0,1)$
then
$$\begin{align}
a_{1}&=\xi^2(1-\xi)\\
a_{2}&=\sqrt{\xi^3(1-\xi)(1-\xi^2)}\\
&=\sqrt{\xi^3(1-\xi)^2(1+\xi)}\\
&=\xi(1-\xi)\sqrt{\xi(1+\xi)}
\end{align}$$
for $\xi\in(0,1)\Rightarrow \xi(1-\xi)>0$
$$\begin{align}
0<\xi&<1+\xi\\
0<\xi^2&<\xi(1+\xi)=\xi+\xi^2\\
\xi&<\sqrt{\xi(1+\xi)}\\
\xi^2(1-\xi)&<\xi(1-\xi)\sqrt{\xi(1+\xi)}\\
a_1&<a_2
\end{align}$$
then i think if i can proof that $a_{n+1}>a_n,n\in\mathbb{N}^*$ and $\xi\in(0,1)$ then the limit would be $1$
| We have:
$$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$
while on the other hand, if we set:
$$ A(n,m)\triangleq\int_{0}^{1} x^n \prod_{k=1}^{m}(1-x^k)\,dx,\qquad B(n,m)=\int_{0}^{1} x^n (1-x^m)^m\,dx \tag{2}$$
we have:
$$\begin{eqnarray*} I_n = A(n+1,n) &=& A(n+1,n-1)-A(2n+1,n-1)\\&=&A(n+1,n-2)- A(2n,n-2)-A(2n+1,n-1)\end{eqnarray*} $$
and since $A(2n+2-i,n-i)\leq B(2n+2-i,n-i)\leq\frac{1}{(n+2)^3}$ it follows that:
$$ I_n = A(n+1,1)-\sum_{k=1}^{n-1}A(2n+2-k,n-k) \geq \frac{1}{(n+2)(n+3)}-\frac{n-1}{(n+2)^3}\tag{3}$$
so $I_n\geq\frac{2}{(n+2)^3}$ and this inequality, together with $(1)$, implies:
$$ I_n^{\frac{1}{n}}\xrightarrow[n\to +\infty]{}\color{red}{1}. \tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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How to evaluate the limit $\lim_\limits{x\to 0+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right )$? $$\lim_{x\to 0^+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right ) =\ ?$$
I rearranged it as
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x\sqrt{x}\sin x} = \lim_{x\to0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x}$$
Which gives an indetermination of the form $0/0$. Then, I tried L'Hospital:
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x} = \lim_{x\to 0^+ } \frac{1-\cos x}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}} \cos x}$$
Should I continue to apply L'Hospital or is there a simpler way to solve it?
| With MacLaurin formulas it's straightforward:
$$\lim_{x\to 0^+}\frac1{\sqrt{x}}\left(\frac1{\sin x} - \frac1{x}\right) = \lim_{x\to 0^+}\frac{x - \sin x}{x\sqrt x\sin x} = \lim_{x \to 0^+}\frac{\frac16x^3 + o(x^3)}{x^{5/2} + o(x^{5/2})} = \lim_{x \to 0^+}\frac{\sqrt x}6 = 0$$
We used the fact that
$$\sin x = x - \frac{x^3}{3!} + o(x^3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $f: \mathbb{R^3} \to \mathbb{R^3}$ with characteristic polynomial $-\lambda(\lambda-3)^2$ and $f(1,0,0)=(1,0,-1)$ and $f(0,1,0)=(2,3,1)$ Let $f: \mathbb{R^3} \to \mathbb{R^3}$ be a diagonalizable endomorphism with characteristic polynomial $-\lambda(\lambda-3)^2$ such that
$f(1,0,0)=(1,0,-1)$ and $f(0,1,0)=(2,3,1)$. Given these data, how can I find the explicit expession on $f(x,y,z)$ and a basis of $\mathbb{R^3}$ of eigenvectors of $f$?
| You know that the matrix of the endomorphism is
$$
A=\begin{bmatrix}
1 & 2 & a \\
0 & 3 & b \\
-1 & 1 & c
\end{bmatrix}
$$
for some $a,b,c$. Elimination gives
$$
\begin{bmatrix}
1 & 2 & a \\
0 & 3 & b \\
0 & 3 & c+a
\end{bmatrix}
$$
and, since we know that the rank must be $2$, we can conclude that $b=c+a$. We also know that $3$ is an eigenvalue of $f$ with multiplicity $2$, so the rank of $A-3I$ must be $1$ (because $A$ is supposed to be diagonalizable):
$$
A-3I=
\begin{bmatrix}
-2 & 2 & a \\
0 & 0 & a+c \\
-1 & 1 & c-3
\end{bmatrix}
$$
Let's do elimination again:
$$
\begin{bmatrix}
-2 & 2 & a \\
0 & 0 & a+c \\
-1 & 1 & c-3
\end{bmatrix}
\to
\begin{bmatrix}
1 & -1 & -a/2 \\
0 & 0 & a+c \\
-1 & 1 & c-3
\end{bmatrix}
\to
\begin{bmatrix}
1 & -1 & -a/2 \\
0 & 0 & a+c \\
0 & 0 & c-3-a/2
\end{bmatrix}
$$
Can you say what the conclusion is?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that $ 17/4 \leq (a+b+c)\leq 1+ \sqrt{32}. $ If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that
$$
\frac{17}{4} \leq (a+b+c)\leq 1+ \sqrt{32}.
$$
My attempt
Tried using Vieta but it didn't work. Also I used some standard inequalities but got $ \geq\sqrt{15} $.
| Let $p=a+b+c,\,q=ab+bc+ca=5,\,r=abc=1$ we have
$$\begin{aligned}0 \leqslant (a-b)^2(b-c)^2(c-a)^2 & = -27r^2-2p(2p^2-9q)r+q^2(p^2-4q) \\&= (17-4p)(p^2-2p-31).\end{aligned}$$
Now, solve inequalities
$$(17-4p)(p^2-2p-31) \geqslant 0,$$
we get
$$\frac{17}{4} \leqslant p \leqslant 1 + 4 \sqrt 2.$$
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Partial fractions and trig functions A long time ago I wrote down a silly problem. It starts with
Attempt to write $$\frac{1}{\sin(x)\cos(x)}$$ using partial fractions.
and then goes on to prove a trig identity.
I was wondering if there is actually a way to do this? Is there a way to write a "trig rational function" as a partial fraction? I would assume that the form (in general) is simply as follows, as if $\sin(x)=:y$ and $\cos(x)=:z$ and following your nose?
$$\frac{1}{\sin(x)\cos(x)}=\frac{A\sin(x)+B\cos(x)+C}{\sin(x)}+\frac{D\sin(x)+E\cos(x)+F}{\cos(x)}$$
| OK, let's try the tangent half-angle substitution:
\begin{align}
\tan\frac\theta 2 & = t \\[8pt]
\theta & = 2\arctan t \\[8pt]
\sin\theta & = \sin(2\arctan t) = 2\sin(\arctan t)\cos(\arctan t) \\
& = 2\frac{t}{\sqrt{t^2+1}} \cdot \frac{1}{\sqrt{t^2+1}} \\[6pt]
& = \frac{2t}{t^2+1} \\[8pt]
\cos\theta & = \cos(2\arctan t) = \cos^2\arctan t - \sin^2\arctan t \\
& = \left(\frac{1}{\sqrt{t^2+1}}\right)^2 - \left(\frac{t}{\sqrt{t^2+1}}\right)^2 \\[6pt]
& = \frac{1-t^2}{1+t^2}
\end{align}
Then:
$$
\frac 1 {\sin\theta\cos\theta} = \frac{(t^2+1)^2}{2t(1-t^2)} = \frac{t^4+2t^2+1}{2t(1-t)(1+t)}
$$
Long division of polynomials gives us a first-degree polynomial in $t$ plus $\dfrac{\cdots}{2t(1-t)(1+t)}$, where the numerator is at most a second-degree polynomial, and the fraction becomes $\dfrac A t+ \dfrac B{1-t} + \dfrac C{1+t}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to caluclate the integral of $\int \frac{1}{\sqrt{4x^{2}+1}}dx$ using a trig substitution? I am trying to determine the following integral: $\int \frac{1}{\sqrt{4x^{2}+1}} dx$ using a suitable substitution.
My progress:
let $x = \frac{1}{2} \tan \theta$
$dx = \frac{1}{2}\sec^{2} \theta d\theta$
$$\int \frac{1}{1+\tan^{2}\theta}\times \frac{1}{2}\times \sec^{2} \theta d\theta$$
$$= \frac{1}{2} \times \int 1 d\theta$$
$$ = \frac{1}{2} \theta + c$$
I know I have to get my answer back in terms of $x$, but I am a little stuck. Can someone please help?
| You forgot the square root
$$x=\frac{1}{2} \tan \theta \implies dx= \frac{1}{2} \sec^2 \theta d\theta $$
The integral becomes:
$$\int \frac{1}{\sqrt{1+tan^2 \theta}} \cdot \frac{1}{2} \sec^2 \theta d\theta$$
$$\Large{\text{Not:}}$$
$$\bbox[8pt, border: crimson 4pt solid]{\int \frac{1}{1+tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta d\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
System of equations with 2 parameters I have no idea how even to start!
\begin{align*}
(u^2+v^2)(u+v)&=15uv \\
(u^4+v^4)(u^2+v^2)&=85u^2v^2
\end{align*}
| The system
\begin{align}
(u^2+v^2)(u+v)&=15 u v, \\
(u^4+v^4)(u^2+v^2)&=85u^2 v^2.
\end{align}
can be rewritten,
using Newton's identities:
\begin{align}
p_1 & = e_1 \\
p_2 &= e_1 p_1 - 2 e_2 \\
p_3 &= e_1 p_2 - e_2 p_1 + 3 e_3 = e_1 p_2 - e_2 p_1 \\
p_4 &=e_1 p_3-e_2 p_2 + e_3 p_1 -4 e_4 = e_1 p_3-e_2 p_2
\end{align}
as
\begin{align}
p_2 e_1&=15 e_2, \\
p_4 p_2&=85 e_2^2,
\end{align}
where
\begin{align}
p_i&=u^i+v^i, \\
e_1&=u+v,\ e_2=u v, \ e_i=0,\quad \forall i>2.
\end{align}
Now we can express $p_2,p_3$ and $p_4$ in terms of $e_1$ and $e_2$
and find out that
\begin{align}
e_2 = \frac{e_1^3}{2 e_1+15}.
\end{align}
Then it boils down to equation in terms of $e_1$:
\begin{align}
e_1^4 (3e_1+35)(e_1-6)=0,
\end{align}
hence
\begin{align}
e_1&=0,6,-\frac{35}{3}; \\
e_2&=0,8,\frac{1715}{9}.
\end{align}
Thus solutions to the original system are:
\begin{align}
(u,v)&=(0,0),(2,4),\left(\frac{-35+7i\sqrt{115}}{6},\frac{-35-7i\sqrt{115}}{6}\right),
\end{align}
and since the system is symmetric, we also need to include pairs $(v,u)$.
The illustration shows points of real solutions:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sequence proof if it exists For every integer r>=3 there exists a sequence $a_{1,\space }a_{2,}.....,a_r$ of nonzero integers with the property that
$a_1^2+a_2^2+....+a_{r-1}^2=a_r^2$
I tried to prove this with proof by induction (as to whether it was false or not) I'm not sure if that was the correct method. any help with solving this would be greatly appreciated
| Let us start with the good old $3^2+4^2=5^2$, that is, $a_1=3$, $a_2=4$, and $a_3=5$, which gives a sequence of length $3$, and produce a sequence of length $4$.
The idea is to multiply the first two terms of our previous sequence through by $3$, and add terms $4\cdot 5$ and $5\cdot 5$. So for $r=4$ our sequence is $3\cdot 3$, $3\cdot 4$, $4\cdot 5$, and $5\cdot 5$.
For a sequence of length $5$, multiply the first three terms of the previous sequence by $3$, and add $4\cdot 5\cdot 5$ and $5\cdot 5\cdot 5$.
Continue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
find eigen values of given matrix
If $w$ is a non-real cube root of unity, then eigenvalues of the matrix:
$$ \begin{pmatrix} 1&1&1 \\ 1&w&w^2 \\ 1&w^2&w^4 \end{pmatrix}
\begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&0 \end{pmatrix}
\begin{pmatrix} 1&1&1 \\ 1&1/w&1/w^2 \\ 1&1/w^2&1/w^4 \end{pmatrix} $$
are which of these:
(a) $1,-1,0$
(b) $\frac{1}{3},-\frac{1}{3},0$
(c) $1,w,w^2$
(d) $3,-3,0$
Please suggest some trick to find eigen value of this matrix given as product of 3 matrices, because to actually compute product of these matrix would not be a feasible idea in a exam.
| Hint: note that the matrix
$$
\frac 1{\sqrt{3}}\pmatrix{
1&1&1\\
1&\omega & \omega^2\\
1 & \omega^2 & \omega^4
}
$$
is unitary (since its columns are orthonormal). From there, it's easy to see that
$$
\left[
\frac 1{\sqrt{3}}\pmatrix{
1&1&1\\
1&\omega & \omega^2\\
1 & \omega^2 & \omega^4
}
\right]^{-1}=
\frac 1{\sqrt{3}}\pmatrix{
1&1&1\\
1&\omega^{-1} & \omega^{-2}\\
1 & \omega^{-2} & \omega^{-4}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Definite Integrals problem The question is to find the value of : $$\frac{\displaystyle29\int_0^1 (1-x^4)^7\,dx}{\displaystyle4\int_0^1 (1-x^4)^6\,dx}$$ without expanding. According to the book, the answer is 7.
I tried taking $I=\displaystyle\int_0^1(1-x⁴)^7\,dx$ and integrating it by parts taking 1 as the function to be integrated, but I could not manage the $-4x^3$ terms that I was getting by the chain rule.
I cannot figure out how to get the answer.
| For the integration by parts, you apply a trick. You don't take the constant $1$ as the factor, but you calculate
\begin{align}
\int_0^1 (1-x^4)^7\,dx &= \int_0^1 (1-x^4)\cdot(1-x^4)^6\,dx\\
&= \int_0^1 (1-x^4)^6\,dx - \int_0^1 x^4\cdot (1-x^4)^6\,dx\\
&= \int_0^1 (1-x^4)^6\,dx + \frac{1}{28} \int_0^1 x\bigl(7(1-x^4)^6\cdot (-4x^3)\bigr)\,dx\\
&= \int_0^1 (1-x^4)^6\,dx + \frac{1}{28}\biggl(\Bigl[x(1-x^4)^7\Bigr]_0^1 - \int_0^1 (1-x^4)^7\,dx\biggr)\\
&= \int_0^1 (1-x^4)^6\,dx - \frac{1}{28}\int_0^1 (1-x^4)^7\,dx.
\end{align}
Moving the last term to the left and dividing by the factor, we obtain
$$\int_0^1 (1-x^4)^7\,dx = \frac{28}{29} \int_0^1 (1-x^4)^6\,dx,$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Generalized Vandermonde's identity Can you please provide a reference to the following generalization of Vandermonde's identity?
Given a positive integer $k$ and nonnegative integers $n_1, n_2, \ldots, n_k$ and $m$, it holds that $$\sum_{i_1+i_2+\cdots+i_k=m} \binom{n_1}{i_1} \binom{n_2}{i_2} \cdots \binom{n_k}{i_k} = \binom{n_1+n_2+\cdots+n_k}{m}.$$ The proof is well-known and based on the idea of counting in two different ways the coefficient of $x^m$ in the polynomial $(1+x)^{n_1+n_2+\cdots+n_k}$.
| \begin{array}{l}
\sum\limits_{i_1 + i_2 + \cdots + i_k = m} {\left( \begin{array}{c}
n_1 \\
i_1 \\
\end{array} \right)\left( \begin{array}{c}
n_2 \\
i_2 \\
\end{array} \right)\left( \begin{array}{c}
n_3 \\
i_3 \\
\end{array} \right) \cdots \left( \begin{array}{c}
n_k \\
i_k \\
\end{array} \right)} = \\
= \sum\limits_{i_1 + i_2 = m - \left( {i_3 \cdots + i_k } \right)} {\left( {\left( \begin{array}{c}
n_1 \\
i_1 \\
\end{array} \right)\left( \begin{array}{c}
n_2 \\
i_2 \\
\end{array} \right)} \right)\left( \begin{array}{c}
n_3 \\
i_3 \\
\end{array} \right) \cdots \left( \begin{array}{c}
n_k \\
i_k \\
\end{array} \right)} = \\
= \sum\limits_{i_3 = m - \left( {i_4 \cdots + i_k } \right)} {\left( {\left( \begin{array}{c}
n_1 + n_2 \\
m - \left( {i_3 \cdots + i_k } \right) \\
\end{array} \right)} \right)\left( \begin{array}{c}
n_3 \\
i_3 \\
\end{array} \right) \cdots \left( \begin{array}{c}
n_k \\
i_k \\
\end{array} \right)} = \\
= \cdots \\
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Proving the inequality $\frac12\frac34....\frac{2n-1}{2n}<\frac1{\sqrt{2n+1}}$ How to show this inequality:
$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2n-1}{2n}<\dfrac{1}{\sqrt{2n+1}}$
Using induction the inequality is verified for $n=1$
now assume that that the inequality holds for $n$,to show it for $n+1$
Then
$\dfrac{1}{2}.\dfrac{3}{4}....\dfrac{2n-1}{2n}\dfrac{2n+1}{2(n+1)}<\dfrac{1}{\sqrt{2n+1}}.\dfrac{2n+1}{2(n+1)}=\dfrac{\sqrt{2n+1}}{2(n+1)}<\dfrac{\sqrt{2(n+1)}}{2(n+1)}=\dfrac{1}{\sqrt{2(n+1)}}$
but I have to make it less than $\dfrac{1}{\sqrt{2n+3}}$ which is not coming.Any help
| Asuume this is true for $n=k$ i.e. $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k-1}{2k}<\dfrac{1}{\sqrt{2k+1}}$$ then you have to show that $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k+1}{2k+2}<\dfrac{1}{\sqrt{2k+3}}$$. But notice that $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k+1}{2k+2}<\dfrac{1}{\sqrt{2k+1}}\cdot \dfrac{2k+1}{2k+2}$$ but $$\dfrac{1}{\sqrt{2k+1}}\cdot \dfrac{2k+1}{2k+2}=\frac{\sqrt{2k+1}}{2k+2}<\frac{1}{\sqrt{2k+3}}$$ The last inequality you can check by cross multiplying and simplifying terms. Thus result is true by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Evaluate this limit $\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}$ Evaluate the limit
$$\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}.$$
I've tried and it always comes out $\frac{0}{0}$, and l'Hopital doesn't seem to help me much here, what would you do?
| $$\lim_{x\to-\infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}=\lim_{x\to\infty}\frac{4^{3-x}-3^{2-x}-2^{1-x}}{4^{1-x}+3^{2-x}+2^{3-x}}=\lim_{x\to\infty}\frac{4^3\cdot6^x-3^2\cdot 8^x-2\cdot12^x}{4\cdot6^x+3^2\cdot8^x+2^3\cdot12^x}$$
$$\lim_{x\to\infty}\frac{4^3\cdot(\frac{6}{12})^x-3^2\cdot(\frac{8}{12})^x-2}{4\cdot(\frac{6}{12})^x+3^2\cdot(\frac{8}{12})^x+2^3}=\frac{4^3\cdot0-3^2\cdot0-2}{4\cdot0+3^2\cdot0+2^3}=-\frac{2}{8}=-\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
For positive $x,y$ such that $x+y \leq x^2+y^2$ prove inequality for every real $t \ge 1$ $x^t+y^t \leq x^{t+1}+y^{t+1}$.
For positive $x,y$ such that $x+y \leq x^2+y^2$ prove inequality for every real $t \ge 1$ $x^t+y^t \leq x^{t+1}+y^{t+1}$.
I see that $\frac{(x+y)}{2} \leq \frac {x^{t+1}+y^{t+1}}{(x^t+y^t)}$
| It is enough to show that for $x, y > 0$:
$$(x^t+y^t)(x^2+y^2) \le (x^{t+1}+y^{t+1})(x+y)$$
$$\iff x^t+y^t\ge x^{t-1}y+y^{t-1}x$$
which follows from Rearrangement inequality as $x \ge y \iff x^{t-1} \ge y^{t-1}$ for $t \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Diophantine: $(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$ For which positive integer values of $n$ does the following diophantine equation have no nontrivial solutions? :
$(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$
Note: I discovered this problem while trying to solve another one. I have attempted this and I know it is true for e.g. $n=3$, but it is hard to find all such $n$'s.
| The original equation is equivalent to:
$$(15n-5)(x^2+y^2+z^2)=(9n+1)(x+y+z)^2,$$
so, if $n\not\equiv 1,2\pmod{5}$, both $x+y+z$ and $x^2+y^2+z^2$ must be multiples of $5$. Since the quadratic residues $\!\!\pmod{5}$ are $\{-1,0,1\}$, we can assume without loss of generality that $x\equiv 0\pmod{5}$. However, this implies $y\equiv -z\pmod{5}$ and $y^2\equiv -z^2\pmod{5}$, so $x\equiv y\equiv z\equiv 0\pmod{5}$ leads to an infinite descent.
This proves that for $n\not\equiv 1,2\pmod{5}$ there are no non-trivial solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction proof that $1^3+2^3+...+n^3=\frac{n^2(n+1)^2}4$ Prove that:
$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$
So I am thinking that I need to do a proof by mathematical induction. Here's my attempt:
Let S(n) be the statement $1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$. When n=1,
$$1^3=\frac{1^2(1+1)^2}{4}$$
$$1=\frac{(2)^2}{4}$$
$$1=\frac{4}{4}$$
So P(1) is true.
Suppose that k is any integer with $k \ge 1$ such that $p(k)1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$ for $k \in N$. Then adding $(k+1)^3$ to both sides,
$$1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2+4(k+1)^3}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4(k+1)}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4k+4)}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+1+1)^2}{4}=P(k+1)$$
Thus, since we have proved that P(k+1) is true, we can conclude that P(k) is true for all $k \in N$.
How does it look, any edits I should make?
| Ok, It is good.
I would do it as this. In the first step for $(k+1)$,
Left side:
$$P_{k+1} - P_k = (k+1)^3$$
Right side:
$$P_{k+1}-P_k={{(k+1)^2}{(k+2)^2} \over 4}-{{(k)^2}{(k+1)^2} \over 4}$$
$$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({(k+2)^2-(k)^2})$$
$$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({4k+4})$$
$$P_{k+1}-P_k={(k+1)^3}$$
Therefor:
$L=R$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\frac{a-bc}{a+bc}+\frac{b-ac}{b+ac}+\frac{c-ab}{c+ab}\le \frac32$ Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\le \dfrac32$.
My attempt:
First I thought if each term could be less than $\dfrac12$. That did not help. So, a little manipulation yielded
$\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\ge \dfrac34$.
Again I thought that each term is greater than $\dfrac14$.
Proceeding that way yielded $3(ab+bc+ca)\ge 1$
Now, $a^2+b^2+c^2+2ab+2bc+2ca\ge 3(ab+bc+ca)\ge 1$.
Here, we get $1\ge 1$, here, I am confused whether the solution is done or not. I don't think it is. So, please help. Thank you.
| Hint: $a +bc = 1-b-c+bc = (1-b)(1-c)$,
$\displaystyle \sum_{\text{cyclic}} (a-bc)(1-a) = \sum a - \sum a^2 - \sum bc + 3abc$
You might need to use Schur inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$? When integrating $\int \frac{1-x}{x^2-1} dx$ Maple rewrote it as $-\int\frac{1}{x+1}dx$
How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$?
| hint:
$$x^2-1=(x+1)(x-1)\\
\frac{1-x}{x^2-1}=\frac{-(x-1)}{(x+1)(x-1)}=?$$
| {
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Prove that the sum is always greater than $1$ Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+ \frac{1}{n^2} \ge 1$, for all natural numbers $n$.
I tried to use mathematical inducion but failed and I tried to figure out a short formula for the sum but I couldnt find any.
| Using some calculus you can see that $$\sum_{k=n}^{n^2} \frac{1}{k} \ge \int_{n}^{n^2} \frac{1}{x} dx = \ln(n^2) - \ln(n) = \ln(n).$$
Certainly this is larger than 1 for $n\ge 3$. For $n=1$ the claim follows automatically. For $n=2$ we check $$\frac12 + \frac13 + \frac14 = \frac{6+4+3}{12} = \frac{13}{12}.$$
Since we know this holds for $n=2$, another approach is to show that $f(n)=\sum_{k=n}^{n^2}1/k$ is an increasing function. Since between $f(n)$ and $f(n+1)$ we lose $1/n$ and gain $$\frac{1}{n^2+1} + \frac{1}{n^2+2} + \cdots + \frac{1}{(n+1)^2}$$ we need to show that the sum of the new terms is larger than $1/n$.
We can get a lower bound on this tail: $$\frac{1}{n^2+1} + \frac{1}{n^2+2} + \cdots + \frac{1}{(n+1)^2} > \frac{2n+1}{(n+1)^2}.$$ If this left term is larger than $1/n$ we win. Using some precalculus type reasoning we can see that $$\frac{2n+1}{(n+1)^2} \ge \frac1n$$ is equivalent to $$n^2 - n - 1 \ge 0.$$
This is satisfied for $n \ge 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$
If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$
I have tried the following,in decreasing order of success:
1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$
2)Substituting $1=abc$ yields nothing
3)Substituting $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ yields something weird
4)Rearrangement inequality on the sequences $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}$
5)Titu's lemma doesn't help
A little nudge in the right direction would help.
| (I) By rearrangement inequality:
$\displaystyle \begin{align} &\sum_{cyc} \frac{1}{b(a+b)} \ge \sum\limits_{cyc} \frac{1}{b(a+c)} \tag{1} \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{b(a+c)}\right) = \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{c(a+b)}\right) \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \frac{1}{2}\sum_{cyc} \frac{b+c}{bc(a+b)} \end{align}$
By Am-Gm Inequality :
$$\sum_{cyc} \frac{b+c}{bc(a+b)} \ge 3\sqrt[3]{\prod\limits_{cyc} \frac{b+c}{bc(a+b)}} = 3$$
Thus establishing desired inequality.
Note: $(1)$ can be viewed as a consequence of CS as well.
$$\sum_{cyc} \left(\frac{1}{b(a+b)} - \frac{1}{b(a+c)}\right) \ge 0 \iff \sum_{cyc} \frac{c-b}{b(a+b)(a+c)} \ge 0 \\ \iff \sum_{cyc} \frac{c^2-b^2}{b} \ge 0 \iff \sum_{cyc} \frac{c^2}{b} \ge \sum_{cyc} b$$
(II) Substituting $\displaystyle a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$:
The inequality required to prove becomes:
$$\sum\limits_{cyc} \frac{x^2}{z^2+xy} \ge \frac{3}{2}$$
We may rewrite LHS as $\displaystyle \sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y}$ and apply Cauchy-Schwarz Inequality:
$$\sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y} \ge \frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc} x^2z^2 + \sum\limits_{cyc} x^3y}$$
So it suffices to prove that: $$2(x^2+y^2+z^2)^2 \ge 3\sum\limits_{cyc} x^2z^2 + 3\sum\limits_{cyc} x^3y \\ \iff (x^4+y^4+z^4) + \sum\limits_{cyc} (x^4 + x^2y^2) \ge 3\sum\limits_{cyc} x^3y$$
This is the consequence of adding the following:
(i) The Rearrangement Inequality: $x^4+y^4+z^4 \ge x^3y+y^3z+z^3x$
(ii) The Am-Gm Inequality: $\displaystyle \sum\limits_{cyc} (x^4 + x^2y^2) \ge 2\sum\limits_{cyc} \sqrt{x^6y^2} = 2\sum\limits_{cyc} x^3y$
| {
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"url": "https://math.stackexchange.com/questions/1141774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do four natural numbers exist which satisfy these constraints? Do four natural numbers $a,b,c$ and $d$ exist such that the following three conditions are true?
$$a^2+b^2+2d^2=c^2$$
$$\sqrt{a^2+d^2}\in\mathbb{N}$$
$$\sqrt{b^2+d^2}\in \mathbb{N}$$
| For any Pythagorean triple $(x,y,z)$ (such that $x^2+y^2=z^2$)
denote
$$
a=x^2,\\
b=y^2,\\
c=z^2,\\
d=xy.
$$
Then
$$
a^2+b^2+2d^2 = a^2+b^2+2ab = (a+b)^2=z^4=c^2,
$$
$$
a^2+d^2=x^4+x^2y^2=x^2(x^2+y^2)=(xz)^2,
$$
$$
b^2+d^2=y^4+x^2y^2=y^2(y^2+x^2)=(yz)^2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation
$$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$
has a (real) root, prove the inequality
$$ a^2 + b^2 \geq 8. $$
I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out.
| The proof from user109899 is wonderful. Here is another one and the idea is the same.
Note
\begin{eqnarray}
&&x^4 + ax^3 + 2x^2 + bx + 1\\
&=&x^2(x^2+ax)+2x^2+bx+1\\
&=&x^2(x+\frac a2)^2-\frac{a^2}{4}x^2+2x^2+bx+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}x^2+bx+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x^2+\frac{4b}{8-a^2}x\right)+1\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+1-\frac{b^2}{8-a^2}\\
&=&x^2(x+\frac a2)^2+\frac{8-a^2}{4}\left(x+\frac{2b}{8-a^2}\right)^2+\frac{8-a^2-b^2}{8-a^2}
\end{eqnarray}
Hence if $a^2+b^2<8$, then $8-a^2>0, 8-a^2-b^2>0$ and thus
$$ x^4 + ax^3 + 2x^2 + bx + 1>0. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show by induction on $n$ that: $$1^4 + 2^4 +\cdots+n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
I proved true for case $n=1$ , assumed true for $n=k$ , but cannot get things to work out. I tried putting the right side over a common denominator and going from there but I'm sure how much it helps.
| $$\sum_{k=1}^{n+1}k^{4}=\sum_{k=1}^{n}k^{4}+\left(n+1\right)^{4}=\frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}+n^{4}+4n^{3}+6n^{2}+4n+1
$$$$=\frac{6n^{5}+15n^{4}+10n^{3}-n}{30}+\frac{30n^{4}+120n^{3}+180n^{2}+120n+30}{30}=\frac{6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30}{30}$$
and see that$$6\left(n+1\right)^{5}+15\left(n+1\right)^{4}+10\left(n+1\right)^{3}-\left(n+1\right)$$$$=\left(6n^{5}+30n^{4}+60n^{3}+60n^{2}+30n+6\right)
+\left(15n^{4}+60n^{3}+90n^{2}+60n+15\right)+\left(10n^{3}+30n^{2}+30n+10\right)-n-1=6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deriving an expression for $\cos^4 x + \sin^4 x$
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$
I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.
| $\cos^4x+\sin^4 x\\=(\cos^2x+\sin^2x)^2-2(\sin x\cos x)^2\\=1-2(\frac{1}{2}\sin(2x))^2\\=1-\frac{1}{2}\sin^2(2x)\\=1-\frac{1}{2}\frac{1-\cos(4x)}{2}\\=\frac{1}{4}\cos(4x)+\frac{3}{4}$
| {
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"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
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Proving an inequality involving reduction sine I'm trying to prove that $I_{2n + 2} \leqslant I_{2n + 1} \leqslant I_{2n}$ where
$$
I_k = \int_0^{\pi/2} \sin^k x\, dx = \frac {k - 1} {k}\int_0^{\pi/2} \sin^{k - 2} x \, dx
$$
From this I have worked out that
$$
I_{2n + 2} = \frac{2n + 1} {2n + 2} \bullet \frac{2n - 1} {2n} \bullet \frac{2n - 3} {2n - 2} \bullet \dotsi \bullet \frac{5} {6} \bullet \frac{3} {4} \bullet \frac{1} {2} \bullet \frac{\pi} {2}
$$
$$
I_{2n + 1} = \frac{2n} {2n + 1} \bullet \frac{2n - 2} {2n - 1} \bullet \frac{2n - 4} {2n - 3} \bullet \dotsi \bullet \frac{6} {7} \bullet \frac{4} {5} \bullet \frac{2} {3}
$$
$$
I_{2n} = \frac{2n - 1} {2n} \bullet \frac{2n - 3} {2n - 2} \bullet \frac{2n - 5} {2n - 4} \bullet \dotsi \bullet \frac{5} {6} \bullet \frac{3} {4} \bullet \frac{1} {2} \bullet \frac{\pi} {2}
$$
Then I found that
$$
\frac{2n + 1} {2n + 2} I_{2n} \leqslant \frac{2n} {2n + 1} \bullet \frac{2n - 2} {2n - 1} \bullet \dotsi \bullet \frac{6} {7} \bullet \frac{4} {5} \bullet \frac{2} {3} \leqslant I_{2n}
$$
And from there I haven't worked out anything conclusive besides dividing everything by $I_{2n}$ which doesn't help me that much. So could anyone tell me what to do next or maybe point out a mistake?
| The reduction formula is not needed. On our interval $0\le \sin^{k+1} x\le \sin^k x$. The result follows.
| {
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A system of quadratic Diophantine equations with four variables Is the following system has any positive integer solution $(x,y,u,v)$?
$$\begin{cases} x^2+y^2=u^2\\ x^2-y^2=v^2 \end{cases}$$
I can prove that any pair of these integers can be relatively prime, but I couldn't find any solution. Any hint?
Thanks in advance!
| From solutions of the Pythagorean equation we have $x = 2ab, y = a^2 - b^2$, so we must have $$(a^2 - b^2 - 2ab)(a^2 - b^2 + 2ab) = v^2$$ we have $(a^2 - b^2 - 2ab, a^2 - b^2 + 2ab) = 1$, so $$a^2 - b^2 - 2ab = (a - b)^2 - 2b^2 = x^2$$ $$a^2 - b^2 + 2ab = (a + b)^2 - 2b^2 = y^2$$ your equation has solution if and only if the Bell equations, $x^2 - 2b^2 = m^2$, $y^2 + 2b^2 = n^2$ have solutions for $(m, n)$ such that $2|m+n$, you can easily find all solutions of this equations and I think you can find some solutions this way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Tangents in Harmonic Progression Here is a problem I am working on:
In triangle ABC, $\tan(A)$, $\tan(B)$, and $\tan(C)$ form a harmonic progression. Also, $BC=189$ and $AB=459$. Find AC.
My progress so far:
Because tanA, tanB, and tanC are in harmonic progression, $\tan(B)=\frac{(2\tan(A)\cdot\tan(C))}{(\tan(A)+\tan(C))}$. Also, we can compute AC using law of Cosines if we can find $\cos(B)$.
However, I am stuck from here. Any tips?
| We know that $\cot C,\cot B,\cot A$ are in arithmetic progression, so $\tan A,\tan B,\tan C$ are $\frac{1}{x+y},\frac{1}{x},\frac{1}{x-y}$ and since $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ must hold, we must have
$$ 3x^2-y^2 = 1,$$
so:
$$\cot A = x+\sqrt{3x^2-1},\qquad\cot B = x,\qquad \cot C=x-\sqrt{3x^2-1}$$
and since $\frac{1}{\sin^2 z}=\cot^2 z+1$ we have:
$$\frac{1}{\sin^2 A}=4x^2+2x\sqrt{3x^2-1},\qquad \frac{1}{\sin^2 B}=x^2+1,\qquad \frac{1}{\sin^2 C}=4x^2-2x\sqrt{3x^2-1},$$
but since we know the ratio between $BC$ and $AB$ we also know the ratio between $\sin^2 A$ and $\sin^2 C$. From that, we can compute $x$, then $\sin B$, then $AC$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
| By setting $x=\arctan t$, so that $dx=\frac{dt}{1+t^2}$, we get:
$$\int \frac{dx}{3\sin^2 x+5\cos^2 x} = \int \frac{dx}{3+2\cos^2 x} = \int \frac{dt}{(1+t^2)\left(3+\frac{2}{1+t^2}\right)}=\int\frac{dt}{5+3t^2}$$
so:
$$\int \frac{dx}{3\sin^2 x+5\cos^2 x} = C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} t\right)=C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} \tan x\right).$$
| {
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Let $d$ be any positive integer not equal to $2, 5,$ or $ 13$ , then $\exists a, b \in \{2, 5, 13, d\}$ such that $ab − 1$ is not a perfect square? Let $d$ be any positive integer not equal to $2, 5,$ or $ 13$. Then can we always find
distinct $a, b \in \{2, 5, 13, d\}$ such that $ab − 1$ is not a perfect square ?
| Note that $2\cdot 5-1=9$, $2\cdot13-1=25$, and $5\cdot15-1=64$. These are all perfect squares. Thus, there exists $a$ and $b$ such that $ab-1$ is a perfect square if and only if at least one of
$$2d-1, 5d-1, \text{ or } 13d-1$$
is a perfect square.
We will continue by contradiction. Assume temporarily that all of these are perfect squares. We then have
$$2d-1=x^2,$$
$$5d-1=y^2,$$
and
$$13d-1=z^2$$
for positive integers $x$, $y$, and $z$.
We will now split this into two cases: $d$ is even and $d$ is odd.
If $d$ is even, then $x^2\equiv 3\pmod{4}$. This is not possible, so $d$ must be odd. If $d\equiv 3\pmod{4}$, then $y^2\equiv 2\pmod{4}$. This is not possible either so, then $d\equiv 1\pmod{4}$.
Let $d=4n+1$, where $n$ is an integer. Substituting this into our three equations, we get
$$8n+1=x^2,$$
$$20n+4=y^2,$$
and
$$52n+12=z^2.$$
Since $4$ is a perfect square, an integer divisible by $4$ is a perfect square iff you get a perfect square when you divide it by $4$. So we can factor out a $4$ from the second two equations. So we have
$$5n+1=y'^2$$
and
$$13n+3=z'^2$$
where $y'$, and $z'$ are integers. Using the last two equations and taking mod $4$, we find that $n+1$ and $n+3$ are both equivalent to either $0$ or $1 \pmod{4}$. This is not possible, so we have a contradiction. $\blacksquare$
| {
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Minimum value of a big function For a positive integer $n≥2$, let
$$f(n)=\frac{(log_32)(log_33)(log_34)⋯(log_3n)}{9^n}$$
Let $m$ denote the minimum value of $f(n)$, and let $N$ be the sum of all values of $n$ for which $f(n)=m$. Find $N$. I need help simplifying the function.
| Note that for $n>2$
$$f(n)=f(n-1)\cdot\dfrac{\log_3 n}{9}.$$
For $n<3^9$ $\quad f(n)<f(n-1)$.
For $n=3^9$ $\quad f(n)=f(n-1)\cdot \dfrac{\log_3 3^9}{9}=f(n-1)$.
For $n>3^9$ $\quad f(n)>f(n-1)$.
So there are two minimal values: $f(3^9-1)$ and $f(3^9)$.
$N = (3^9-1)+3^9$.
| {
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"url": "https://math.stackexchange.com/questions/1161125",
"timestamp": "2023-03-29T00:00:00",
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If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.
| Let $b=M-d$ and $c=M+d$ then from $ a \le {M-d + M+d \over 2 } $ we get the reformulation $$ \tag 1 a \le M $$
Now take the squares
$$\tag{lhs} a \to a^2 $$
$$ \tag{rhs} {(M-d)^2 + (M+d)^2 \over 2 } = M^2+ d^2 $$
So if $ a \le M $ as assumed in $(1)$ then surely $$ \tag 2 a \le M = {b+c \over 2}\to a^2 \le M^2+d^2 = { b^2+c^2\over 2}$$
which answers the question in generality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.
My attempt:
Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$
$$1=\frac{6}{6}.$$
So $P(1)$ is true.
Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following:
$$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
$$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$
$$=\frac{2k^3+9k^2+13k+6}{6}$$
$$=\frac{(k^2+3k+2)(2k+3)}{6}$$
$$=\frac{(k+1)(k+2)(2k+3)}{6}$$
$$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
$$=P(k+1).$$
Thus $P(k)$ is true for $k \in \mathbb{N}$.
Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$.
| I am going to provide what I think is a nice way of writing up a proof, both in terms of accuracy and in terms of communication. You be the judge(s).
Claim: For $n\geq 1$, let $S(n)$ be the statement
$$
S(n) : 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.
$$
Base step $(n=1)$: The statement $S(1)$ says $1^2=1(2)(3)/6$ which is true.
Inductive step $(S(k)\to S(k+1))$: Fix some $k\geq 1$ and suppose that
$$
S(k) : 1^2+2^2+3^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}
$$
holds. To be shown is that
$$
S(k+1) : 1^2+2^2+3^2+\cdots+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &= 1^2+2^2+3^2+\cdots+k^2+(k+1)^2\tag{definition}\\[1em]
&= \frac{k(k+1)(2k+1)}{6}+(k+1)^2\tag{by $S(k)$}\\[1em]
&= (k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]\\[1em]
&= (k+1)\frac{k(2k+1)+6(k+1)}{6}\\[1em]
&= (k+1)\frac{2k^2+k+6k+6}{6}\\[1em]
&= (k+1)\frac{2k^2+7k+6}{6}\\[1em]
&= (k+1)\frac{(k+2)(2k+3)}{6}\\[1em]
&= \frac{(k+1)(k+2)(2(k+1)+1)}{6}\\[1em]
&= \text{RHS},
\end{align}
the right-hand side of $S(k+1)$ follows. This completes the inductive step.
Thus, by mathematical induction, for every $n\geq 1, S(n)$ is true. $\Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum and Divisibility Puzzle I have $5$ positive integers: $a,b,c,d,e$.
$a,b,c,d,e$ are all different, and $a\mid b\mid c\mid d\mid e$, in other words the ratios
$$\frac ba, \frac cb, \frac dc, \frac ed $$ are all integers.
$$a+b+c+d+e = 47.$$
I need to find out what $a,b,c,d,e$ are. Apparently there is only one solution to this.
I did some trial and error and came to an answer of $1,2,4,8,32$. But I really have no idea how to come to this conclusion more formally.
The problem comes from a section of my book that talks about prime factorization.
I can figure that the direction that I need to go in is looking at the fact that 47 is prime and that adding the prime factorizations of a,b,c,d,e will give me 47 in some elegant way.
Can someone provide some direction?
| First, if $a \geq 2$, then you'd have $b \geq (2)(2) = 4$, and $c\geq (2)(4) = 8$, and $d\geq (2)(8) = 16$ and $e\geq (2)(16) = 32$, since each of the "adjacent ratios" must be at least $2$. But then $a + b + c + d + e \geq 62$, which doesn't work. Hence $a = 1$.
Next, if $b = 3$, then $b, c, d, e$ are all divisible by $3$, which means $a + b + c + d + e$ is congruent to $1$ modulo $3$, which doesn't work since $47$ is congruent to $2$ rather than $1$. Also, if $b \geq 4$, then $a+b+c+d+e > 47$ as above. Therefore we must have $b = 2$, since $b > a = 1$.
Now $c = bk = 2k$ for some $k>1$. If $c = (2)(3) = 6$, then $a+b+c+d+e = 0$ mod $3$, which again doesn't work, while if $c \geq (2)(4) = 8$, then $a+b+c+d+e > 47$ again. Hence $c = (2)(2) = 4$.
At this point we have $(a, b, c) = (1, 2, 4)$, so $a+b+c=7$ and therefore $d+e = 40$. Also, $d = 4m$ and $e = dn = 4mn$ for some integers $m, n > 1$.
Thus $40 = 4m + 4mn = 4(m + mn)$, so $10 = m + mn = m(1 +n)$. The only way to write $10$ as the product of two integers both greater than $1$ is $10 = (2)(5)$, so either $m = 2$ and $1+n = 5$, or $m=5$ and $1+n = 2$. The latter possibility is eliminated as it would mean $n=1$, so we conclude $m=2$ and $1+n=5$, so $n=4$. That is, $d=8$ and $e=32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
| (Here I'll assume that $-1^k$ here ought to be $(-1)^k$.) Hint:
$$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4} = -\sum_{n = 1}^{\infty} \frac{1}{n^4} + 2 \sum_{n > 1 \text{ even}}^{\infty} \frac{1}{n^4} = - \sum_{n = 1}^{\infty} \frac{1}{n^4} + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n)^4}.$$
This is $$- \sum_{n = 1}^{\infty} \frac{1}{n^4} + \frac{1}{8} \sum_{n = 1}^{\infty} \frac{1}{n^4} = - \frac{7}{8} \sum_{n = 1}^{\infty} \frac{1}{n^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| Since $x=y$ is clearly a root, one may divide $x^3-y^3$ by $x-y$ directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\lim\limits_{n\to\infty} 1 + 1/2 + 1/3+\cdots+1/2n- \ln n$ I know the proof of $$\lim_{n\to\infty} 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n = \gamma$$ but how do I use that proof to find the value of the limit above.
| \begin{align}
& 1 + \frac 12 + \frac 13 + \cdots + \frac 1 {2n} -\ln n \\[10pt]
= {} & \Big(\ \underbrace{1 + \frac 12 + \frac 13 + \cdots + \frac 1 {2n} -\ln(2n)}_{\text{This part approaches $\gamma$.}}\ \Big) + \ln 2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove by induction that $3^{3n+1} + 2^{n+1}$ is divisible by 5 How do I do this? I've tried using logarithms, factoring, but nothing seems to work.
| HINT:
If $f(n)=3^{3n+1}+2^{n+1}$
$$f(m+1)-a\cdot f(m)=3^{3m+1}(3^3-a)+2^{m+1}(2-a)$$
If we set
$3^3-a=0\iff a=3^3\implies f(m+1)-3^3f(m)=\cdots\equiv0\pmod5$
or
$2-a=0\iff a=2\implies f(m+1)-2f(m)=\cdots\equiv0\pmod5$
$\implies5|f(m+1)\iff5|f(m)$
Non-inductive methods:
$(1)$Binomial Expansion:
$3^3=25+2\implies3^{3n+1}+2^{n+1}=2^{n+1}+3(2+25)^n$
$=2^{n+1}+3[2^n+\sum_{r=1}^n\binom nr2^{n-1}25^r]=2^n(2+3)+($ a multiple of $25)$
$(2)$Modular Arithmetic
$3^3\equiv2\pmod5\implies3^{3n+1}=3(3^3)^n\equiv3\cdot2^n\pmod5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating the tail exponent of a Pareto Distribution (difficult integrals). Looking for two difficult integrals, whichever is solved first:
With $\alpha>0,L>0, \mu>0,\sigma>0, a>0, b>0, x>L$ ,
$$\phi_l (x;\mu,\sigma)=\frac{1}{\sqrt{2 \pi } \sigma }\int_0^{\infty } \alpha L^{\alpha } x^{-\alpha -1} e^{-\frac{(\alpha -\mu )^2}{2 \sigma ^2}} \, \mathrm{d}\alpha$$
and
$$\phi_g (x;a,b)=\frac{b^{-\frac{a}{b}} }{\Gamma \left(\frac{a}{b}\right)}\int_0^{\infty } e^{-\frac{\alpha }{b}} L^{\alpha } x^{-\alpha -1} \alpha ^{a/b}\, \mathrm{d}\alpha$$
Background: these correspond to the density of a Pareto distribution $\alpha L^{\alpha } x^{-\alpha -1} $ with its tail exponent $\alpha$: 1) Lognormally distributed in the first case, and 2) following a Gamma distribution in the second case.
With gratitude.
$\textbf{Later Comment}$: It turned out that the first integral was not the Lognormally distributed exponent (I copied the wrong equation), but I leave here for its calculus/integration interest, rather than distributional importance. –
| Let's start with $\phi_l$:
\begin{align}
\phi_l(x;\mu,\sigma) &= \frac{L}{\sqrt{2\pi \sigma^2}}\int_0^\infty \alpha \left(\frac{L}{x}\right)^{\alpha - 1} e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty y^\alpha e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2} + \alpha \ln y}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}}\frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty e^{-\tfrac{(\alpha - \mu - \sigma^2\ln y)^2 + \mu^2 - (\mu + \sigma^2 \ln y)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} e^{-\tfrac{\mu^2 - (\mu + \sigma^2 \ln y)^2}{2\sigma^2}} \int_0^\infty e^{-\tfrac{(\alpha - \mu - \sigma^2\ln y)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} e^{\mu\ln y\, + \frac{\sigma^2}{2}\ln^2 y} \sqrt{2\pi \sigma^2}\operatorname{erfc}\left(-\frac{\mu + \sigma^2\ln y}{\sqrt{2\sigma^2}}\right)\\
&= x\exp\left\{-\mu\ln \frac{x}{L} + \frac{\sigma^2}{2}\ln^2 \frac{x}{L}\right\}\left(\mu - \sigma^2 \ln \frac{x}{L}\right) \operatorname{erfc}\left(-\frac{\mu + \sigma^2 \ln y}{\sqrt{2\sigma^2}}\right)\\
&\quad + \frac{\sigma x}{\sqrt{2}}\exp\left\{-\mu\ln \frac{x}{L} + \frac{\sigma^2}{2}\ln^2 \frac{x}{L}-\frac{(\mu - \sigma^2 \ln \frac{x}{L})^2}{2\sigma^2}\right\}.
\end{align}
Now with $\phi_g$:
\begin{align}
\phi_g(x;a,b) &= \frac{b^{-\tfrac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \int_0^\infty \alpha^{\tfrac{a}{b}} e^{-\left(\frac{1}{b} + \ln \frac{x}{L}\right)\alpha}\, d\alpha\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}\int_0^\infty u^{\frac{a}{b}} e^{-u}\, du, \quad u = \left(\frac{1}{b} + \ln \frac{x}{L}\right)\alpha\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1} \Gamma\left(\frac{a}{b} + 1\right)\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1} \frac{a}{b}\Gamma\left(\frac{a}{b}\right)\\
&= \frac{a}{x}b^{-\frac{a}{b} - 1} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}\\
&= \frac{a}{x}\left(1 + b\ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lagrange Multipliers with two constraints The problem is to find the maximum value of $ \ f(x,y,z) \ = \ x+y+z \ $ subject to the two constraints $ \ g(x,y,z) \ = \ x^2+y^2+z^2 \ = \ 9 \ $ and $ \ h(x,y,z) \ = \ \frac{1}{4}x^2+\frac{1}{4}y^2+4z^2 \ = \ 9 \ $.
I got these equations:
$$1 \ = \ 2x \ λ \ + \ \frac{1}{2} x \ μ \ \ , \ \ 1 \ = \ 2y \ λ + \frac{1}{2}y \ μ \ \ , \ \
1 \ = \ 2z \ λ + 8z \ μ \ \ . $$
And from here, I'm not sure what I can solve for or simplify.
| Applying the indicated "Lagrange equations", we can subtract the second from the first to produce
$$ ( \ 2 \lambda \ + \ \frac{1}{2} \ \mu \ ) \ ( \ x \ - \ y \ ) \ = \ 0 \ \ . $$
This makes it a bit more evident that one's intuition about the symmetry of the functions is helpful. As Ted Shifrin already notes, one result from this is that $ \ x \ = \ y \ $ . Inserting this into the constraint equations, we obtain
$$ 2 \ x^2 \ + \ z^2 \ = \ 9 \ \ , \ \ \frac{1}{2} \ x^2 \ + \ 4 \ z^2 \ = \ 9 \ \ \Rightarrow \ \ \frac{3}{2} \ x^2 \ = \ 3 \ z^2 $$
$$ x^2 \ = \ 2 \ z^2 \ \ \Rightarrow \ \ x \ = \ y \ = \ \pm \sqrt{2} \ z $$
$$ 5 \ z^2 \ = \ 9 \ \ \Rightarrow \ \ z \ = \ \pm \frac{3 \sqrt{5}}{5} \ \ , \ \ x \ = \ y \ = \ \pm \frac{3 \sqrt{10}}{5} \ \ . $$
Hence, the maximum value of the function $ \ f(x, \ y, \ z) \ = \ x + y + z \ $ is found in the first octant as $ \ \frac{3 \sqrt{5}}{5} \ + \ \frac{3 \sqrt{5}}{5} \ + \ \frac{3 \sqrt{10}}{5} \ = \ \frac{3 \sqrt{5}}{5} \ ( \ 2 \sqrt{2} \ + \ 1 \ ) \ $ . Because the constraint surfaces are symmetric about the origin, and the function to be extremized is anti-symmetric about the origin, we would expect that the minimum value for the function is $ \ -\frac{3 \sqrt{5}}{5} \ ( \ 2 \sqrt{2} \ + \ 1 \ ) \ $ , found in the opposing octant at the point on the constraint surfaces intersected by a line through the origin from the point we just found.
The other "result" from the first equation is $ \ \lambda \ = \ -4 \ \mu \ $ . Inserting this into the third Lagrange equation gives us $ \ 2 \ z \ \lambda \ = \ 8 \ ( \ -4 \ \lambda \ ) z \ = \ 1 $ $ \ \Rightarrow \ \lambda \ = \ -\frac{1}{30 \ z} \ \ , \ \ \mu \ = \ \frac{2}{15 \ z} \ $ . This, however, doesn't appear to take us any further.
$$ \ \ $$
the constraint sphere and constraint oblate spheroid, with one intersection circle marked in yellow (the other is located symmetrically "below the equator"); the minimum of the function to be extremized lies on the far side of the surfaces depicted, on the "lower" circle [the "beveling" near the equator of the oblate spheroid is an artifact of the grapher]
If we look at the geometric situation, the two constraint surfaces are a sphere and a (highly) oblate spheroid, which intersect at two circles in planes parallel to the $ \ xy-$ plane and disposed symmetrically about it (producing the rather saturnian -- or perhaps flying-saucerish -- configuration in the diagram above). We can locate those planes by solving for the intersections of the spheroids:
$$ z^2 \ = \ 9 \ - \ x^2 \ - \ y^2 \ = \ \frac{1}{4} \ (9 \ - \ \frac{1}{4} \ x^2 \ - \ \frac{1}{4} \ y^2) \ $$
$$ 144 \ - \ 16 \ x^2 \ - \ 16 \ y^2 \ = \ 36 \ - \ x^2 \ - \ y^2 \ \ \Rightarrow \ \ 15 \ (x^2 \ + \ y^2) \ = \ 108 $$
$$ \ \ \Rightarrow \ \ x^2 \ + \ y^2 \ = \ \frac{108}{15} \ = \ \frac{36}{5} \ \ \Rightarrow \ \ z^2 \ = \ 9 \ - \ \frac{36}{5} \ = \ \frac{9}{5} \ \ . $$
This yields the values we found above. We can "reduce" the function to that of a single variable, $ \ f(x) \ = \ x \ \pm \ \left( \ \frac{36}{5} \ - \ x^2 \ \right)^{1/2} \ \pm \frac{3 \sqrt{5}}{5} \ $ to locate extrema thus:
$$ 1 \ \pm \ \frac{1}{2} \ \left( \ \frac{36}{5} \ - \ x^2 \ \right)^{-1/2} \cdot \ ( -2x ) \ = \ 0 \ \ \Rightarrow \ \ \mp \frac{x}{\sqrt{\frac{36}{5} \ - \ x^2}} \ = \ 1 $$
$$ \frac{36}{5} \ - \ x^2 \ = \ x^2 \ \ \Rightarrow \ \ x^2 \ = \ \frac{18}{5} \ \ , $$
from which follow the results described above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the series of $\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}$ converge? Does the following sum converge?
$$
\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}
$$
Have tried ratio test and root test, inconclusive. alternating series test not useful.
| If $\sum_{n=2}^{\infty}(-1)^n \frac{n+1}{n-1}$ converge, then there exist a $N$ such that $\forall n \geq N$
$$L-\frac{1}{3} < \sum_{n=2}^{2N}(-1)^n \frac{n+1}{n-1} < L+\frac{1}{3}$$
But then
$$L-\frac{1}{3}-\frac{2N+2}{2N} < \sum_{n=2}^{2N+1}(-1)^n \frac{n+1}{n-1} < L+\frac{1}{3} - \frac{2N+2}{2N}$$
But $\frac{2N+2}{2N}> \frac{2}{3}$, then
$$L-1 < \sum_{n=2}^{2N+1}(-1)^n \frac{n+1}{n-1} < L-\frac{1}{3} $$
contradiction
This is the idea of the proof of why the general term must converge to 0
| {
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"timestamp": "2023-03-29T00:00:00",
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The value of the logarithmic expression can never be $\ldots$ The value of the logarithmic expression $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x},\text{where}\quad x\geq y>1\quad$ can never be
$\bf\text{options}$ a.) $-1\quad$ b.)$\quad0.5\quad$ c.) $\quad0\quad$ d.) $\quad1$
$$\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\\
&=\frac{2\log_x y}{\log_x y}-\frac{(\log_x y)^2}{\log_x y} -\frac{1}{\log_x y}\\
&=-\frac{ (\log _{x}y-1)^{2} }{ \log _{x}y}\end{align}$$
here if i input $x=y=2$ then option $0$ is removed , not sure how to deduce the rest.
Wolfram does gives the answer as option $1$
| \begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\end{align}
Let $y^p = x$, therefore $\log_y x =p$ and $ \log_x y = \frac1p $ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$ So, $1-\log_x y +1-\log_y x = 2 - (p + \frac1p)$
For the RHS of above expression to equate to -1, 0.5, 0, 1, we must have p + $\frac1p$ = 3, 1.5, 2, 1 respectively.
So, we have to solve for $p$ the corresponding quadratic equations which work out to be $\quad$ $p^2-3p+1=0$, $\quad$ $2p^2-3p+2=0$, $\quad$ $p^2-2p+1=0$, $\quad$ $p^2-p+1=0$. $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ On solution we obtain $p=(3+\sqrt{5})/2 $ & $p=(3-\sqrt{5})/2$ for the first, $\quad$$\quad$ $p=1$ & $p=\frac12$ for the second, $\quad$$\quad$ $p=1$ for the third, $\quad$$\quad$ and $p=(1+\sqrt3i)/2$ & $p=(1-\sqrt3i)/2$ for the fourth.
Since $y^p = x$, imaginary $y$ would be out of domain of the problem and also make $x$ outside the domain of the problem (domain is x,y > 1, i.e. x, y real) and hence solution $p=1$ which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=1$ is not possible. So, answer is option d.) $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ Also notice that for second equation, $p=1$ is a false root, and $p=\frac12$ would make $y \gt x$ which is also outside the domain of the problem. So, this equation which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=\frac12$ is also not possible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$ How do I evaluate
$$\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)\ ?$$
I'm stuck. I tried using the sandwich theorem but I was getting nowhere.
Since $n \lt n+k \lt 2n$, I think it should be $ \infty$ .
| Since
$\begin{array}\\
\frac1{k}
&> \int_k^{k+1} \frac{dx}{x}
> \frac1{k+1}\\
\sum_{k=n+1}^{2n}\frac1{k}
& > \sum_{k=n+1}^{2n}\int_k^{k+1} \frac{dx}{x}
> \sum_{k=n+1}^{2n}\frac1{k+1}\\
or\\
\sum_{k=n+1}^{2n}\frac1{k}
& > \int_{n+1}^{2n+1} \frac{dx}{x}
> \sum_{k=n+1}^{2n}\frac1{k}-\frac1{n+1}+\frac1{2n+1}\\
\end{array}
$
Since
$\int_{n+1}^{2n+1} \frac{dx}{x}
=\ln(2n+1)-\ln(n+1)
=\ln(2-\frac1{n+1})
=\ln(2)-\ln(1-\frac1{2n+2})
$,
if
$d(n)
=\ln(2)-\sum_{k=n+1}^{2n}\frac1{k}
$,
$d(n)
< \ln(1-\frac1{2n+2})
< 0
$
and
$d(n)
>\ln(1-\frac1{2n+2})-\frac1{n+1}+\frac1{2n+1}
$.
Setting
$k=2n+1$
in
$\frac1{k}
> \int_k^{k+1} \frac{dx}{x}
> \frac1{k+1}
$
we get
$\frac1{2n+1}
> -\ln(1-\frac1{2n+1})
> \frac1{2n+2}
$.
Therefore
$d(n)
>-\frac1{2n+1}-\frac1{n+1}+\frac1{2n+1}
=-\frac1{n+1}
$.
so
$$-\frac1{n+1}
< \ln(2)-\sum_{k=n+1}^{2n}\frac1{k}
< 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How to solve $\sec^2x+3\csc^2x=8$ I have tried substitution involving the Pythagorean Identities, as well as converting to just sin and cos. I cannot seem to manipulate the equation into a form which has a clear solution process. Please provide help.
$$\sec^2x+3\csc^2x=8$$
| it is some time easier to the unit circle. we will take $$x = \cos t, y = \sin t $$ the important relation to remember is $$x^2 + y^2 = 1. \tag 1$$ we can convert $$\sec^2 t + 3\csc^2 t = 8 $$ into $$\frac 1{x^2} + \frac{3}{y^2} = 8$$ so you have $$3x^2 + y^2 = 8x^2 y^2 \tag 2 $$
you can solve $(1)$ and $(2)$ by eliminating ,say, $y$ to get $$3x^2 + 1 - x^2 =8x^2(1-x^2)$$ this is a quadratic equation in $x^2.$ that is $$8x^4+6x^2 - 1=0$$ has roots $$x^2 = \frac{-6 \pm\sqrt{68}}{16}, \, x = \pm\sqrt{\frac{-6 +\sqrt{68}}{16}}, t = 81.92^\circ, 98.07^\circ, 261.92^\circ, 278.07^\circ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let $f$ be a function such that $ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $ Let $f$ be a function such that $$ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $$ for $x > 1$. In that domain, $f(x)$ has the form $\frac{a+\sqrt{cx+d}}{b},$ where $a,b,c,d$ are integers and $a,b$ are relatively prime. Find $a+b+c+d.$
| Let's write $f$ for $f(x)$.
\begin{align}
\sqrt{x-\sqrt{x+f}}&=f\\
x-\sqrt{x+f}&=f^2\\
\sqrt{x+f}&=-f^2+x\\
x+f&=f^4-2xf^2+x^2\\
0&=f^4-2xf^2-f+x^2-x
\end{align}
This looks like a very ugly equation to solve. We want to factor that somehow (maybe there's some difference of squares in there? I can't see it...) Since we have nothing else to do, let's complete the square on the right. (Why? Well, I'm hoping for a difference of two squares somewhere, so getting a square in there seems like a good start.)
\begin{align}
0&=f^4-2xf^2-f+x^2-x\\
0&=f^4-2xf^2-f+\Big(x-\frac12\Big)^2-\frac14
\end{align}
Let's ignore the $-\frac14$ at the end for just one second, and look at the rest. What do we have here? Well, $f^4-\textit{something}+(x-\frac12)^2$. This is almost a square! (Remember that $(a-b)^2=a^2-2ab+b^2$; we have the $a^2$ and $b^2$ in here.)
So, let's try seeing what $(f^2-(x-\frac12))^2$ looks like. (Maybe we'll get lucky and find out that our thing actually is a square.)
\begin{align}
\Big(f^2-\left(x-\tfrac12 \right)\Big)^2&=f^4-2f^2 \left(x-\frac12 \right)+\Big(x-\frac12\Big)^2\\
&=f^4-2f^2x+f^2+\left(x-\frac12\right)^2
\end{align}
Drats — that's almost what we have. What we have is that thing, but with a $-f$ instead of a $+f^2$. In other words, we have $\textit{that thing}-f^2-f$. Well, let's put it into our equation above anyway:
\begin{align}
0&=f^4-2xf^2-f+\Big(x-\frac12\Big)^2-\frac14\\
&=\left(\left(f^2-\left(x-\tfrac12\right)\right)^2-f^2-f \right)-\frac14\\
&=\Big(f^2-\left(x-\tfrac12 \right)\Big)^2-f^2-f-\frac14
\end{align}
Woah! We can factor that...
\begin{align}
&=\Big(f^2-\left(x-\tfrac12 \right)\Big)^2-\left(f+\tfrac12 \right)^2
\end{align}
And suddenly it's the difference of two squares:
\begin{align}
&=\Big(f^2-\left(x-\tfrac12 \right)+\left(f+\tfrac12 \right)\Big)\Big(f^2-\left(x-\tfrac12 \right)-\left(f+\tfrac12 \right)\Big)\\
&=(f^2+f-x+1)(f^2-f-x)
\end{align}
Woohoo! We factored that beast. The end is in sight.
We now split this into two equations: $f^2+f-x+1=0$ and $f^2-f-x=0$. Each give us two possible solutions. After using the quadratic formula, we end up with
$$\frac{-1\pm\sqrt{4x-3}}2,\frac{1\pm\sqrt{4x+1}}2$$
Now, we need to test each of these to see which one works. Actually, this isn't too bad — we can just choose a good choice for $x>1$ for each of these (preferably a choice of $x$ that gives us lots of integers) and see if they work. (Why $x>1$? 'Cause that's what the problem says.)
So, let's pick good values of $x$ and plug them into the original equation: $\sqrt{x-\sqrt{x+f}}=f$.
Working with $x=3$, the first two solutions give me $1=1$ and $\sqrt2=-2$ respectively; working with $x=2$, the last two solutions give me $0=2$ and $1=-1$ respectively. Clearly, only the first one works; thus, the solution is:
$$f=\frac{-1+\sqrt{4x-3}}2$$
EDIT: Technically, they want you to find $a+b+c+d$. So, the answer is $(-1)+(2)+(4)+(-3)=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proof of no primes such that $x^2 + y^2 = z^2$ I'm in a pretty simple "CS Math" course for year 1 Comp Sci, and I came across this:
Disprove, $x^2 + y^2 = z^2$, such that $x, y, z$ are primes
I thought of this as, if n is a prime, then prime factorization of n must be:
$n = z*1$
$n^2 = (z*1)(z*1) = (z)(z)$
So I tried to derive a contradiction:
Assume $x, y, z$ are primes such that $x^2 + y^2 = z^2$
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$ #Contradiction, prime factorization of $x^2$ must be uniquely represented.
But then I noticed that this statement is true if $z-y = 1$ and $z+y = x^2$, So:
Then: $y = -1+z = y = z-1$
Because: $x^2 + y^2 = z^2 $
Then: $(z-1)(z-1) + z+y = z^2$
Then: $z^2 - 2z + 1 + z + y = z^2$
Then: $z^2 - z + y + 1 = z^2$ # subtract $z^2$ from both sides
Then: $-z + y + 1 = 0$
Then: $y = -1 + z$ #Loop??
I remember reducing this to $x + y = z$, which I think is false?
EDIT: Thought of this:
z - y = 1
Then: z = y + 1
Then z must be 3, y must be 2
Because: z is greater than y by 1, since z is a prime and y is a prime, z and y must be 3 and 2 respectively since no other prime number is 1 apart because one of them would be even if greater than 2, and any even number has at least 3 divisors (2, the even number, and 1) thus not a prime.
Then: $x^2 + 2^2 = 3^2$ ,Because $2^2 = 4, 3^2 = 9$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction: x is a prime, but sqrt(5) is not a prime!
Is this a solid evidence proof in that can I say that only prime numbers 1 apart are 2 and 3? I'm pretty sure there is a more simpler and intuitive way to prove this, are there any other ways to prove this via contradiction?
| If $x,y,z\in\mathbb{N}_{\geq 1}$ and $x^2+y^2=z^2$, then at least one number between $x$ and $y$ is even and at least one number between $x$ and $y$ is a multiple of three, since the only possibilities for a square $\!\!\pmod{3}$ or $\!\!\pmod{4}$ are to be $0$ or $1$. So, in order to prove the statement it is sufficient to check that $2^2+3^2$ is not a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$ I've first transformed the integral to
$$\int\frac{5x^2+3x+2}{x(x^2+2x+1)}dx$$
Which gave me
$$\frac{A}{x}+\frac{Bx+C}{x^2+2x+1}$$
$$=\frac{A(x^2+2x+1)+Bx^2+Cx}{x(x^2+2x+1)}$$
$$\frac{5x^2+3x+2}{x(x^2+2x+1)}=\frac{(A+B)x^2+(2A+C)x+A)}{x(x^2+2x+1)}$$
So I've found the corresponding variables
$$A = 2$$
$$A+B = 5, B = 3$$
$$2A+C=3, C=-1$$
So the final integral is
$$2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}$$
$$=2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1)$$
However, the expected answer is
$$2ln(x)+3ln(x+1)+\frac{4}{x+1}$$
What is my error ?
| You may use this:
$$\frac{5 x^2+3 x+2}{x (x+1)^2}=\frac{2}{x}+\frac{3}{x+1}-\frac{4}{(x+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluate the sum $\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$ $$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$
I am having difficulty finding the function that represents this series.
I have only found radius of convergence which is $(-\infty,+\infty)$ from the ratio test.
| Let $f(x)$ be the function represented by your series. Then
$$f(x) = \frac{1}{2}\sum_{n = 0}^\infty \frac{1 - (-1)^n}{n!}x^n$$
since $1 - (-1)^n$ is $0$ when $n$ is even and $2$ when $n$ is odd. Thus
$$f(x) = \frac{1}{2}\sum_{n = 0}^\infty \frac{x^n}{n!} - \frac{1}{2}\sum_{n = 0}^\infty \frac{(-x)^n}{n!}, $$
which simplifies to
$$f(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Expressing $\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}$ with rational Denominator could you please help me express this with a rational denominator
$\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}$
Thank you
| $$\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}=\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3}-\sqrt{5}}=$$
$$=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt6}=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt6}\frac{\sqrt6}{\sqrt6}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $ Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $
Can someone please explain to me how to go about doing this?
| you can write $$\sqrt 3 \sin t - \cos t = 2\left(\frac{\sqrt 3}2 \sin t - \frac 12 \cos t\right) = 2\left(\sin t\sin \left(\frac{\pi}3 \right) - \cos\left(\frac {\pi} 3\right) \cos t\right)= -2\cos \left(t + \frac{\pi}3\right) = 2 \cos \left(t - \frac{2\pi}3\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Show that the sum of the $x$-coordinates of three points on the graph of $y = x^2$ whose normal lines intersect at a common point is $0$. Suppose that three points on the graph of $y = x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I've done a bit of work in trying to explain this, but I'm not so certain I can visualise it to begin with.
The the line tangent to the curve $y = x^2$ at a point $T(a, a^2)$ is $$y - a^2 = 2a(x - a) \implies y = a^2 + 2ax - 2a^2 \implies y = 2ax - a^2$$
In order for the lines tangent to two points to intersect at a common point $P(x, y)$, which will be along the $y$-axis, their $y$-intercepts must be equivalent, and $x = 0$.
The two points must therefore be equidistant from the origin, such $|c| = |a|$ in order for the lines tangent to their coordinates $T_2(c, c^2)$ and $T_1(a, a^2)$ will intersect at point $(x, y)$, as their $y$-intercepts are expressed by $-c^2$ and $-a^2$, respectively.
I feel a bit as though I'm rambling, to a certain extent. I'm having trouble thinking of a way to describe this scenario as it would seem that the normal line to each point would have to pass through the origin in order for the lines to meet at one point, as a linear function $f(x)$ may only meet another linear function $g(x)$ at one point assuming that $f(x) \neq g(x)$ and $m_f \neq m_g$ (which would mean infinitely many points of intersection, and no points of intersection, respectively).
Edit:
Let there be three points $A(a,a^2)$, $B(b,b^2)$, $C(c,c^2)$ whose normal lines intersect at a point $P(x_p, y_p)$
The tangent to the curve at $A$ is $y - a^2 = 2a(x - a)$
The normal to the curve at $A$ is $y - a^2 = -\frac{1}{2a}(x - a) \implies x+2ay=2a^3+a$
$$
\begin{align*}
\\ &\left[y = \dfrac{2a^3 + a - x}{2a} \right] \text{ } \left[y = \dfrac{2b^3 + b - x}{2b} \right] \text{ } \left[y = \dfrac{2c^3 + c - x}{2c} \right]
\\
\\ &\left[x = 2a^3 + a - 2ay \right] \left[ x = 2b^3 + b - 2by \right] \left[ x = 2c^3 + c - 2cy \right]
\end{align*}
$$
$I_{A \cdot B} = $ the intersection of $line_{normal_A}$ and $line_{normal_B} \implies$ $I_{A \cdot B}([-2ab(a+b)], [a^2+ab+b^2+\frac{1}{2}])$
$I_{A \cdot C} = $ the intersection of $line_{normal_A}$ and $line_{normal_C} \implies$ $I_{A \cdot C}([-2ac(a+c)], [a^2+ac+c^2+\frac{1}{2}])$
From here, I'm not certain how to proceed.
| Suppose that three points on the graph of $y = x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
Let the points be $A(a,a^2)$, $B(b,b^2)$, $C(c,c^2)$
The tangent to the curve at $A$ is $y - a^2 = 2a(x - a)$
The normal to the curve at $A$ is $y - a^2 = -\frac{1}{2a}(x - a) \implies x+2ay=2a^3+a$
\begin{align*}
\\ &\left[y = \dfrac{2a^3 + a - x}{2a} \right] \text{ } \left[y = \dfrac{2b^3 + b - x}{2b} \right] \text{ } \left[y = \dfrac{2c^3 + c - x}{2c} \right]
\\ &\left[x = 2a^3 + a - 2ay \right] \left[ x = 2b^3 + b - 2by \right] \left[ x = 2c^3 + c - 2cy \right]
\end{align*}
$I_{A \cdot B} = $ intersection of the normal lines of $A$ and $B$ $\implies$ $I_{A \cdot B}([-2ab(a+b)], [a^2+ab+b^2+\frac{1}{2}])$
$I_{A \cdot C} = $ intersection of the normal lines of $A$ and $C$ $\implies$ $I_{A \cdot C}([-2ac(a+c)], [a^2+ac+c^2+\frac{1}{2}])$
$$
\\ \begin{align}
\\ I_{A \cdot B} = I_{A \cdot C} \implies x_{I_{A \cdot B}} = x_{I_{A \cdot C}} \implies \frac{x_{I_{A \cdot B}}}{x_{I_{A \cdot C}}} = 1 \implies \dfrac{-2ab(a+b)}{-2ac(a+c)} &= 1
\\
\\ b(a + b) &= c(a + c)
\\ ba+b^2 &= ca + c^2
\\ a(b - c) &= c^2 - b^2
\\ a(b - c) &= (c + b)(c - b)
\\ a(b-c) &= (c+b)(b-c)(-1)
\\ \frac{a(b-c)}{(b-c)} &= -c - b
\\ a + b + c &= 0
\\
\\\end{align}
$$
This shows the sum of $a+b+c$ is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
| \begin{align}
A &= \int_{0}^{\pi}\left(\:\frac{1\:-\:\sin (x)}{\sin (x)\:+\:1}\:\right)dx \\
F(x) &= \int\left(\:\frac{1\:-\:\sin (x)}{\sin(x)\:+\:1 }\cdot\frac{\sin(x)\:-\:1}{\sin (x)\:-\:1 }\:\right)dx \\
&= \int\left(\:\frac{ -\sin^2(x) + 2\cdot sin(x)-1}{\sin^2(x)\:-\:1 }\:\right)dx \\
&= \int\left(\:\frac{ -(\sin^2(x) - 2\cdot sin(x)+1)}{-(1-\sin^2(x)) }\:\right)dx \\
&= \int\left(\:\frac{ (\sin^2(x) - 2\cdot sin(x)+1)}{(1-\sin^2(x)) }\:\right)dx \\
&= \int\left(\:\frac{\sin^2(x) - 2\cdot sin(x)+1}{\cos^2(x) }\:\right)dx \\
&= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}-\frac{ 2\cdot sin(x)}{\cos^2(x)}+\frac{1}{\cos^2(x)}\:\right)dx \\
&= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}\right)dx-\int\left(\frac{2\cdot sin(x)}{\cos^2(x)}\right)dx+\int\left(\frac{1}{\cos^2(x)}\:\right)dx \\
F(x) &= \tan(x) - x - 2\cdot\sec(x) + \tan(x)\\
A &= F(\pi) - F(0)\\
F(\pi) - F(0) &= (\tan(\pi) - \pi - 2\cdot\sec(\pi) + \tan(\pi))-(\tan(0) - 0 - 2\cdot\sec(0) + \tan(0))\\
&= (0 - \pi - 2\cdot-1 + 0)-(0 - 0 - 2\cdot1 + 0)\\
&= (-\pi + 2)-(-2)\\
&= -\pi + 2+2\\
&= 4-\pi\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 2
} |
Prove an inequality on natural number Show that if $ a,b\in N$ and $a < b$,
then $$\frac{a^a}{(a+1)^{a+1}} > \frac{b^b}{(b+1)^{b+1}}.$$
| Using Bernoulli's Inequality, we get
$$
\begin{align}
\frac{(a+1)^{a+1}}{a^a}\frac{(a-1)^{a-1}}{a^a}
&=\frac{a+1}{a-1}\left(1-\frac1{a^2}\right)^a\\
&\ge\frac{a+1}{a-1}\left(1-\frac1a\right)\\
&=\frac{a+1}a\\[8pt]
&\gt1
\end{align}
$$
Therefore,
$$
\frac{a^a}{(a+1)^{a+1}}\lt\frac{(a-1)^{a-1}}{a^a}
$$
which says that $\dfrac{a^a}{(a+1)^{a+1}}$ is a strictly decreasing function of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to prove that the equation $3(a^2+b^2)=7(c^2+d^2)$ has no solution
Question 1: Let $a,b,c,d$ be positive integers, show that
$$3(a^2+b^2)=7(c^2+d^2)$$ has no solution.
question 1 is from Mathlove, Curious How prove it?
My question 2: Find the least positive integer $k$ such that
$$a^2+b^2=k(c^2+d^2)$$ has no solution.
| $$\frac{a^2+b^2}{c^2+d^2}=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}$$
Use Brahmagupta–Fibonacci identity for the numerator to get $$\frac{a^2+b^2}{c^2+d^2}=\left(\frac{ac\pm bd}{c^2+d^2}\right)^2+\left(\frac{ad\mp bc}{c^2+d^2}\right)^2$$
But $3,7\equiv-1\pmod4$ can not be expressed as the sum of two squares
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. What am I doing wrong?
| Since $f(1) = 0$ and $f(x) = f(x-1)$, we have $f(2) = f(1) = 0$. Hence if $\deg f \leq 2$,
$$
f(x) = a(x-2)(x-1),
$$
and so
$$
a(x-2)(x-1) = a(x-3)(x-2) \\
a(x-1) = a(x-3)
$$
for all $x$. Hence, $a=0$, so $f = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ I am trying to find this limit
$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$
My so far method is this
*
*$f(x)>0.$
*$f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$
*For every $0<y<1$ the equation $f(x)=y$ has a unique solution(found in maple):
$$x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}$$
The previous facts implies that the limit is $0$.
I am wondering if there is any way easier than this to find the limite.
thanks in advance.
| Yet another method: let $a = (x+1)^{1/3}$, $b = x^{1/3}$, so $$f(x) = (x+1)^{1/3} - x^{1/3} = a - b = \frac{(a-b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = \frac{a^3 - b^3}{a^2 + ab + b^2}.$$ But $a^3 - b^3 = (x+1) - x = 1$, hence $$f(x) = \frac{1}{a^2 + ab + b^2}.$$ Now since $a \to \infty$ and $b \to \infty$ as $x \to \infty$, it is obvious that $$\lim_{x \to \infty} f(x) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Prove that the algebraic expressions are equivalent.
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$
with steps make left hand side = right hand side by modifying one or both expressions
Thanks for your help guys, I solved it like this:
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+1}-1}{2} + \frac{2*3^{k+1}}{2} = \frac{3^{k+2}-1}{2} $$
| hint: $3^{k+1} = \dfrac{2\cdot 3^{k+1}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding a basis for the null space of a matrix The problem is to find the null space of the matrix
$$
A = \begin{bmatrix}1&2&1&-3\\2&4&4&-1\\3&6&7&1\end{bmatrix}.
$$
Does this have something to do with linear independence or dependence? A hint or solution would be appreciated!
| First of all, if someone asks you to find a vector space satisfying some conditions, then you have to provide the basis of that space only. Because using that basis he or she can generate the whole space.
Now, the row-echelon form of $A$ is $\begin{pmatrix}1 & 2 & 1 & -3\\ 0 & 0 & 2 & 5\\ 0 & 0 & 0 & 0\end{pmatrix}$. So $rank(A) = 2$.
As we know, $rank(A) + dim(nullspace(A)) = \text{No. of columns of A}$, it gives $dim(nullspace(A)) = 4 - 2 = 2$, which means a particular basis of $nullspace(A)$ contains two linearly independent vectors.
Again, $nullspace(A) = \left\{\underline{x} : A\underline{x} = \underline{0}\right\}$. Take any arbitrary $\underline{y} = (y_1, y_2, y_3, y_4)'$, where $\underline{y}$ is not necessarily a null vector. This $\underline{y}$ will belong to the $nullspace(A)$ iff it satisfies $A\underline{y} = \underline{0}$. Since rank of $A$ is 2, we will be able to have solution of only two variables freely, rest of the variables will be expressed in terms of other two variables.
Let's find the $y_i$'s. Here it is enough to solve the system,
$$
\begin{pmatrix}1 & 2 & 1 & -3\\
0 & 0 & 2 & 5\\
0 & 0 & 0 & 0\end{pmatrix}\underline{y} = \underline{0}
$$
which gives, $y_1 + 2y_2 + y_3 - 3y_4 = 0$ and $2y_3 + 5y_4 = 0$, which I have obtained as,
$$
y_3 = -\frac{1}{2}y_1 - y_2
$$ and
$$
y_4 = \frac{1}{5}y_1 + \frac{2}{5}y_2
$$
Hence the general solution is,
$$
\underline{y} = \begin{pmatrix}y_1\\ y_2\\ -\frac{1}{2}y_1 - y_2\\ \frac{1}{5}y_1 + \frac{2}{5}y_2\end{pmatrix} = y_1\begin{pmatrix}1\\ 0\\ -\frac{1}{2}\\ \frac{1}{5}\end{pmatrix} + y_2\begin{pmatrix}0\\ 1\\ -1\\ \frac{2}{5}\end{pmatrix}\qquad(*)
$$, where $y_1$ and $y_2$ are arbitrary. Now it is clear that every vector belonging to nullspace will be of the form $(*)$, which is nothing but the linear combination of $(1, 0, -\frac{1}{2}, \frac{1}{5})'$ and $(0, 1, -1, \frac{2}{5})'$. Hence these two vectors form the basis of $nullspace(A)$.
Please check that these two vectors are linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving an expression is perfect square I have this expression I got in one larger exercise:
$$\frac{(2+\sqrt3)^{2n+1}+(2-\sqrt3)^{2n+1}-4}{6}\frac{(2+\sqrt3)^{2(n+1)+1}+(2-\sqrt3)^{2(n+1)+1}-4}{6}+1$$
and i need to prove it is perfect square.
I tried many different approaches but couldn't find way to show it is square. Interesting fact is $(2+\sqrt3)(2-\sqrt3)=1$ so I tried replacing $(2+\sqrt3)=x$ and $(2-\sqrt3)=1/x$ to see if I would get an idea.
Alternative form I got after some steps and using equality giving $1$ I got:
$$\frac{(2+\sqrt3)^{4n+4}(1-16((2- \sqrt3)^{2n+2})+66((2- \sqrt3)^{2n+2})^2-16((2- \sqrt3)^{2n+2})^3+((2- \sqrt3)^{2n+2})^4)}{36}$$
which is interesting as I have "rising exponent" but this coefficients doesn't make sense to me.
Any ideas?
EDIT:
I actually need to prove $$(y_{n+1}^2-1)(y_{n+2}^2-1)+1$$ is perfect square where $y_0=1$, $y_1=3$ and $y_{n}=4y_{n-1}-y_{n-2}$. Expression from above I got solving this recursion and using exact expression for $y_n$. I also tried solving directly using induction and this recursion but didn't get result.
| Let $y_n=4y_{n-1}-y_{n-2}$ with $y_0=1$ and $y_1=3$.
First we show by induction that
$$
y_n^2+y_{n+1}^2-4y_ny_{n+1}+2=0 \; \forall n \in \mathbb{N}. \label{*}\tag{*}
$$
Clearly it holds for $n=0$ and
\begin{align*}
& y_{n+1}^2+y_{n+2}^2-4y_{n+1}y_{n+2}+2 =\\
& (y_{n+2}-y_{n+1})^2-2y_{n+1}y_{n+2}+2 =\\
& (3y_{n+1}-y_n)^2-2y_{n+1}y_{n+2}+2 =\\
& y_n^2+y_{n+1}^2-4y_ny_{n+1}+2+8y_{n+1}^2-2y_ny_{n+1}-2y_{n+1}y_{n+2} =\\
& 8y_{n+1}^2-2y_{n+1}(y_n+y_{n+2}) =\\
& 8y_{n+1}^2-2y_{n+1}4y_{n+1} = 0.
\end{align*}
Finally $\eqref{*}$ implies the desired result:
\begin{align*}
-y_n^2-y_{n+1}^2 & = -4y_ny_{n+1} + 2 \\
y_n^2y_{n+1}^2-y_n^2-y_{n+1}^2+2 &= y_n^2y_{n+1}^2-4y_ny_{n+1} + 4 \\
(y_n^2-1)(y_{n+1}^2-1)+1 & = (y_ny_{n+1}-2)^2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\beta$ is a zero of $f(x)=x^2+x+2$ over $\mathbb{Z}_3$, find the other zero If $\beta$ is a zero of $f(x)=x^2+x+2$ over $\mathbb{Z}_3$, find the other zero.
What I tried:
Suppose $B$ is a zero of $f(x)$, so $f(\beta)=\beta^2+\beta+2=0$. I know that $f(x)$ is irreducible in $\mathbb{Z}_3$. Any idea how to proceed?
| Call the other root $\gamma$. We have:
$x^2 + x + 2 = (x - \beta)(x - \gamma) = x^2 - (\beta + \gamma)x + \beta\gamma$.
Thus we have:
$\gamma = \dfrac{2}{\beta} = 2 - \beta$
Note that $\dfrac{2}{\beta} = 2 - \beta$ tells us that:
$\beta^2 + 2 = 2\beta$, adding $\beta$ to both sides gives us:
$\beta^2 + \beta + 2 = 0$, as we already knew.
We can check that both forms of $\gamma$ give an actual root:
$\left(\dfrac{2}{\beta}\right)^2 + \dfrac{2}{\beta} + 2 = \dfrac{1}{\beta^2} + \dfrac{2\beta}{\beta^2} + \dfrac{2\beta^2}{\beta^2} = \dfrac{2(1 + 2\beta + 2\beta^2)}{2\beta^2}$
$= \dfrac{2 + \beta + \beta^2}{2\beta^2} = \dfrac{0}{2\beta^2} = 0$, and:
$(2 - \beta)^2 + (2 - \beta) + 2 = 1 - \beta + \beta^2 + 2 - \beta + 2 = \beta^2 - 2\beta + 2 = \beta^2 + 3\beta - 2\beta + 2$
$ = \beta^2 + \beta + 2 = 0$.
Finally, it should be clear that:
$(x - \beta)(x - (2 - \beta)) = x^2 - (2 - \beta + \beta)x + (\beta)(2 - \beta)$
$= x^2 - 2x + 2\beta - \beta^2 = x^2 + x -(\beta + \beta^2) = x^2 + x + 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Conics - required to show $SR \times S'R' = b^2$
Consider the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$.
$R$ and $R'$ are the feet of the perpendiculars from the foci $S$ and $S'$ on to the tangent at $P(a\cos\theta, b\sin\theta)$.
Show that $SR\times S'R'=b^2$.
(You may assume the tangent to ellipse at the point $P(a\cos\theta,b\sin\theta)$ has equation $bx\cos\theta+ay\sin\theta-ab=0$.)
so basically in this question, i know you have to do the perpendicular distances of SR and S'R' but then after that, I'm not sure how to go about simplifying it to get the answer. thanks for the help :)
| We have $bx\cos\theta+ay\sin\theta-ab=0$. Distances from $(\pm ae,0)$ would be:
$$\frac{|b(\pm ae)\cos\theta+a(0)\sin\theta-ab|}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}}=\frac{ab(1\mp e\cos\theta)}{\sqrt{a^2\sin^2\theta+a^2(1-e^2)\cos^2\theta}}=\frac{ab(1\mp e\cos\theta)}{a\sqrt{1-e^2\cos^2\theta}}$$
Product of which is:
$$\frac{a^2b^2(1-e^2\cos^2\theta)}{a^2(1-e^2\cos^2\theta)}=b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove $U = \{u(x) \in P_4 | u(0) + u(1) = 0\}$ is a subspace of $P_4$ I am trying to prove $U = \{u(x) \in P_4 | u(0) + u(1) = 0\}$ is a subspace of $P_4$
For U is nonempty I have:
Let $u(x) = 0x^4 + 0x^3 + 0x^2 + 0x + 0$
For U is closed under $+$ I have:
Let $x, y \in U$, then
$x(0) + x(1) = 0$ and $y(0) + y(1) = 0$
$x + t = (x(0) + x(1)) + (y(0) + y(1)) = 0 + 0 = 0$
$\implies x + y \in U$
For $U$ is closed under $\cdot$ I have:
Let $x \in U \implies x(0) + x(1) = 0$
$\alpha x = \alpha ((x(0) + x(1)) = \alpha (0) = 0$
$\implies \alpha x \in U$
Is this sufficient?
| $$p(x)=ax^4+bx^3+cx^2+dx+e$$
$$p(0)+p(1)=0$$
$$e+a+b+c+d+e=0$$
$$d=-2e-a-b-c$$
$$ax^4+bx^3+cx^2+dx+e$$
$$ax^4+bx^3+cx^2+x(-2e-a-b-c)+e$$
$$ax^4+bx^3+cx^2-2xe-ax-bx-cx+e$$
$$a(x^4-x)+b(x^3-x)+c(x^2-x)+e(-2x+1)$$
Base for your subspace is:$$(x^4-x,x^3-x,x^2-x,-2x+1)$$ Dimension of your subspace is 4.
So dimension of complementary subspace of your subspace is 1.
Becase $dim P_4=5$. Sum of dimension of subspace and their complementary subspace must be equal to total dimension, and their intersection must be $(0)$. The base for $P_4$ is $(x^4,x^3,x^2,x,1)$ so the base of complementary subspace can be $({1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the arc length of the curve. How can I find the length of the loop in polar coordinates
$$r=\frac{a}{\cos^{4}( \frac{\varphi}{4})} $$
I do not even know where to start. Help please.
| Parameterising
\begin{align}
\varphi(t) &= t \\
r(t) &= \frac{a}{\cos^{4}( \frac{t}{4})}
\end{align}
and transforming to Cartesian coordinates
$$
x = r \cos \phi = \frac{a}{\cos^4(\frac{t}{4})} \cos t \quad\quad
y = r \sin \phi = \frac{a}{\cos^4(\frac{t}{4})} \sin t
$$
The image shows a plot for $a = 2$ and $t \in [-5,5]$.
$t=0$ is the point $(a, 0)$ we need to parameterise to the left crossing at $(x(t), 0)$ for $t = \pm \pi$.
The arc lenght is
$$
ds^2 = dx^2 + dy^2 \quad \Rightarrow \quad
ds =
\sqrt{
\left(\frac{dx}{d\varphi}\right)^2 +
\left(\frac{dy}{d\varphi}\right)^2} d\varphi
$$
and the derivatives are
$$
\frac{dx}{d\varphi} = \frac{dr}{d\varphi} \cos \varphi - r \sin \varphi
\quad\quad
\frac{dy}{d\varphi} = \frac{dr}{d\varphi} \sin \varphi + r \cos \varphi
$$
so we get the differential arc lenght in polar coordinates
$$
\left(\frac{ds}{d\varphi}\right)^2 =
\left(\frac{dr}{d\varphi}\right)^2 + r^2
$$
and using the parameterisation for $r$ in terms of $\varphi$:
$$
r'(\varphi)
= -4 \frac{a}{\cos^5(\frac{t}{4})}
\left(-\sin\left(\frac{t}{4}\right) \frac{1}{4}\right)
= a \frac{\sin(\frac{t}{4})}{\cos^5(\frac{t}{4})}
$$
Returning to the differential arc length element $ds$ we get
\begin{align}
s'(\varphi)^2 &=
r'(\varphi)^2 + r^2 \\
&=
\left(\frac{a\sin(\frac{\varphi}{4})}{\cos^5(\frac{\varphi}{4})}\right)^2 +
\left(\frac{a}{\cos^{4}( \frac{\varphi}{4})}\right)^2 \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\left(\tan^2\left({\frac{\varphi}{4}}\right)+ 1\right) \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\frac{1}{\cos^2\left({\frac{\varphi}{4}}\right)} \\
&=
\frac{a^2}{\cos^{10}(\frac{\varphi}{4})}
\end{align}
which turned out simpler than feared. So we can now give the arc length:
\begin{align}
s
&= \int\limits_{-\pi}^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2\int\limits_0^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2a \int\limits_0^\pi \frac{d\varphi}{\cos^{5}(\frac{\varphi}{4})} \\
&= 8a \int\limits_0^{\pi/4} \frac{du}{\cos^{5}(u)} \\
&= 8a (F(\pi/4) - F(0))
\end{align}
This has a longish anti-derivative which Maxima computes to
$$
F(\varphi) =
\frac{3\,\ln\left( \sin\left( u\right) +1\right) }{16}-\frac{3\,\ln\left( \sin\left( u\right) -1\right) }{16}-\frac{3\,{\sin\left( u\right) }^{3}-5\,\sin\left( u\right) }{8\,{\sin\left( u\right) }^{4}-16\,{\sin\left( u\right) }^{2}+8}
+ C
$$
and
\begin{align}
s
&=
a \left(
\frac{3\,\ln\left( \frac{1}{\sqrt{2}}+1\right)}{2}-
\frac{3\,\ln\left( 1-\frac{1}{\sqrt{2}}\right)}{2}+
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left((\sqrt{2}+1)^2 \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
3 \ln\left( \sqrt{2}+1 \right) +
7\,\sqrt{2}
\right) \\
&= 12.54361569767029 \, a
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that minimum of two numbers is less than 4 Suppose I have to choose two numbers from set $$S=\{1,2,3,4,5,6 \}$$ without a replacement , then what is the probability that minimum of two is less than $4$?
I made two groups for this problem $A= \{1,2,3 \}$ and $B=\{4,5,6 \}$
.There are two possibilities , either both are from $A$ or one from $A$ and one from $B$ to satisfy our requirement. Hence
$$P(E)=\frac{3}{6}.\frac{2}{5} + \frac{3}{6}.\frac{3}{5}$$
but answer is incorrect. Help?
| We have $\binom{6}{2}$ possibilities, but $\binom{3}{2}$ of them are not what we want, more precisely, those events that 2 elements are chosen from $\{4,5,6\}$, hence, $P(E)=\frac{\binom{6}{2}-\binom{3}{2}}{\binom{6}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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How can we show that the functions are differentiable? Show that the following functions $$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}} \\ f(x, y)=\frac{x^2y}{x^4+y^2}$$ are differentiable at each point of the domain. Determine which of them is $C^1$.
$$$$
The domain of the functions is $\mathbb{R}^2 \setminus \{(0,0)\}$.
How can we show that the functions are differentiable at each point of the domain??
Do we have to find the partial derivatives??
$$$$
EDIT:
$$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}}: \\ \frac{\partial{f}}{\partial{x}}=\frac{y\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2x}{x^2+y^2}=\frac{y(x^2+y^2)-x^2y}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{y^3}{\sqrt{x^2+y^2}(x^2+y^2)} \\ \frac{\partial{f}}{\partial{y}}=\frac{x\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2y}{x^2+y^2}=\frac{x(x^2+y^2)-xy^2}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{x^3}{\sqrt{x^2+y^2}(x^2+y^2)}$$
Since the partial derivatives are continuous at $\mathbb{R}^2 \setminus \{ (0,0)\}$, $f$ is differentiable.
Is this correct??
Could I improve something at the fomrmulation??
| Since $(0,0)$ isn't in the domain, then we don't need to worry about that point. The necessary
and sufficient conditions for multi-variable differentiation are the existence of all the partials, and their continuity at a given point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Range of $f(x) =\frac {x -1}{x^2 -2x + 3} $? Is my solution for finding the range of
$$f(x) = \frac{x-1}{x^2 -2x + 3} $$
correct?
Since its Domain is $ \mathbb{R} $, so transforming this equation into $x$ in terms of $y$ , we get $$ yx^2 - (2y +1)x + (3y+1) = 0 $$
Now, since $ x \in \mathbb {R} $ , so its discriminant $\mathbb{D}$ must be $ \ge 0 $, so we have, $$ (2y + 1)^2 - 4y(3y + 1) \ge 0 $$
Solving it, we get
$$ y \in \left[-{1 \over {2 \sqrt2}}, {1 \over {2 \sqrt2}}\right] $$
Kindly solve my problem.
| We can more easily find out maxima/minima to find its range:
To find where extrema occur by Quotient Rule:
$$ \frac{x-1}{x^2-2 x +3} = \frac{1}{2 x -2 } $$
Solving,
$$ x = 1 \pm \sqrt 2 $$
Extreme values are:
$$ \pm \dfrac{\sqrt 2}{4} $$
which is the range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that matrices $\tiny\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix},\begin{pmatrix} 2& 0 \\ 1&2 \\ \end{pmatrix} $ are similar. Error in my method? Show that the matrix
$$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} $$ is similar to a triangular matrix of the form $$ \begin{pmatrix} \lambda& 0 \\ 1&\lambda \\ \end{pmatrix} $$ where $\lambda$ is an eigen value.
Attempt:
The eigenvalues of the matrix $\begin{pmatrix} 2&-1 \\ 0&2 \\
\end{pmatrix} $ are clearly $2,2$.
Hence, we need to show that the matrix $\begin{pmatrix} 2&-1 \\0&2 \\
\end{pmatrix} $ is similar to $\begin{pmatrix}2&0 \\1&2 \\\end{pmatrix} $
Now, we know that two matrices are similar if and only if they represent the same linear transformation.
But: for any two dimensional vector $(x ~~y)^T : $
$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} \begin{pmatrix}x \\y \\
\end{pmatrix} = \begin{pmatrix} 2x-y \\ 2y \\ \end{pmatrix} $
And
$\begin{pmatrix} 2&0 \\1&2 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} = \begin{pmatrix} 2x \\x+2y \\\end{pmatrix} $
Clearly, $\begin{pmatrix} 2x \\x+2y \\\end{pmatrix} $ and $\begin{pmatrix}
2x-y \\2y \\ \end{pmatrix} $ don't represent the same linear transformation.
So, how can these matrices be similar?
What could be the fault in my reasoning?
Thank you very much for your help in this regard.
| we can show that $A=\pmatrix{0&-1\\0&0}$ and $B= \pmatrix{0&0\\1&0}$ similar by explicitly displaying $$\pmatrix{1&0\\0&0}=AU = \pmatrix{0&-1\\0&0} \pmatrix{0&1\\-1&0} = \pmatrix{0&1\\-1&0} \pmatrix{0&0\\1&0} = UB= \pmatrix{1&0\\0&0}.$$
now, $$AU=UB \implies (A+2I)U=U(B+2I) .$$ therefore $A+2I$ and $B+2I$ are similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Volume of a solid between sphere and cylinder Find the volume of a solid inside a sphere $x^{2} + y^{2} + z^{2} = 4$ and the cylinder $x^{2} + y^{2} + 2y = 0$
Any help is appreciated
| The simplest way is evaluate the integral in cylindrical polar coordinate.
$$(r,\theta,z)\quad\mapsto\quad (x,y,z) = (r\cos\theta,r\sin\theta,z)$$
$$\begin{align}\verb/Volume/ &= \int_{-\pi}^0 \left[\int_0^{-2\sin\theta} 2\sqrt{4-r^2}\, r dr \right] d\theta = \int_{-\pi}^0 \left[ -\frac{2}{3}\sqrt{4-r^2}^3 \right]_0^{-2\sin\theta} d\theta\\ &= \frac{16}{3}\int_0^\pi (1 - |\cos\theta|^3)d\theta = \frac{16}{3}\left(\pi - \frac{4}{3}\right)\\ \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove this limit using epsilon delta method? i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.
$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$
My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon\implies\left|\frac{x^2-c^2}{1+c^2+x^2+x^2c^2}\right|<\epsilon\implies\left|\frac{(x-c)(x+c)}{1+c^2+x^2+x^2c^2}\right|<\epsilon$$
But i dont how to proceed after this?
| $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|=\left|\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right|\le \left|\frac{1+|x||c|}{(1+x^2)(1+c^2)} \right||x-c|$$
Suppose $c\not=0$, then we do some analysis first.
If we require $|x-c|<\frac{|c|}{2}$, then $|1+x^2|\ge 1+\frac{c^2}{4},|x|<\frac{3|c|}{2}$, that is $\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|\le M(c)|x-c|$, where $M(c)>0$ is a fixed constant. Hence it suffices to further choose $|x-c|<\frac{\epsilon}{M(c)}$.
Combine above two parts together, let $\delta=\min(\frac{|c|}{2},\frac{\epsilon}{M(c)})$, and then $|x-c|<\delta\implies \left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon$
If $c=0$, you are proving $\lim_{x\to 0}\frac{x}{1+x^2}=0$, which is easier by squeezing theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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