Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$
which can be simplified
$$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$
i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator
getting
$$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$
and combining them both we get
$$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$
ps
i never used arcsin before and am compeletely unfamilliar with it
| Here a substitution can help out with calculations. Let $ x = \sin t$, i.e. $t = \arcsin x$. First note that $x\in \langle -1,1 \rangle$, so $t\in \langle -\pi/2,\pi/2 \rangle$ and $\cos t>0$. This is important because we can then simplify $\sqrt{\cos^2t} = |\cos t| = \cos t$, which is used couple of times from now on.
Write $$y =\frac{t\sin t}{\sqrt{1-\sin^2t}}=t\tan t. $$ We can now use the chain rule \begin{align}\frac{dy}{dx} &= \frac{dy}{dt} \cdot
\frac{dt}{dx}\\ &= (\tan t+\frac{t}{\cos^2t})\frac{dt}{dx} = \left(\frac{x}{\sqrt{1-x^2}}+\frac{\arcsin x}{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}=\frac{x}{1-x^2}+\frac{\arcsin x}{(1-x^2)^{3/2}}\end{align}
where we used $\cos t = \sqrt{1-x^2}$ which is derived directly from $\sin^2t+\cos^2t = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $0 < a \le b$. Then $a^2 \le \left (\frac{2ab}{a+b} \right) ^2 \le ab \le \left( \frac{a+b}{2} \right) ^2 \le b^2$ In a introductory text about real analysis, in a chapter about the axioms of order (I hope this is the correct translation), I was faced with the following
Problem: Let $0 < a \le b$. Show that $a^2 \le \left (\frac{2ab}{a+b} \right) ^2 \le ab \le \left( \frac{a+b}{2} \right) ^2 \le b^2$. If, at any point of this chain of inequalities, we have equality, then $a = b$.
There was no solution given. I came to the following
Solution: The first inequality follows from
$$ \begin{align} a^2 & \le ab\\
a^2 + ab & \le 2ab \\
a(a+b) & \le 2ab \\
a & \le \frac{2ab}{a+b} \\
a^2 & \le \left( \frac{2ab}{a+b} \right) ^2 \end{align}$$
By the same reasoning, if equality holds, $a^2=ab$ implies that $a=b$.
We get the second inequality from
$$ \begin{align} 0 & \le (a-b)^2\\
0 & \le a^2 - 2ab + b^2 \\
4ab & \le a^2 + 2ab + b^2 \\
\frac{4a^2b^2}{ab} & \le (a+b)^2 \\
\frac{4a^2b^2}{(a+b)^2} & \le ab \\
\left( \frac{2ab}{a+b} \right) ^2 & \le ab \end{align}$$
an similarly for the third inequality with
$$ \begin{align} 0 & \le (a-b)^2\\
0 & \le a^2 - 2ab + b^2 \\
4ab & \le a^2 + 2ab + b^2 \\
ab & \le \frac{(a+b)^2}{4} \\
ab & \le \left( \frac{(a+b)}{2} \right) \\ \end{align}$$
In both cases, $0 = (a-b)^2$ implies that $a=b$.
Finally we have
$$ \begin{align} a & \le b \\
a + b & \le 2b \\
\frac{a+b}{2} & \le b \\
\left( \frac{a+b}{2} \right) ^2 & \le b^2 \end{align}$$
and equality directly leads to $a=b$. $\blacksquare$
I have three questions to this:
*
*Is my solution correct?
*Do I have to take additional steps for the second part of the problem, that equality implies $a=b$?
*Are there shorter and more elegant solutions to this?
Thanks in advance!
| It look quite correct for me, and the equality part is good since $a,b$ are both positive.
Just for showing a shorter (maybe) version for this by using two important but simple inequalies:
(1). $(a+b)^2 \geq 4ab$
And its variation in case of $a>0,b>0$ :
(2). $ 1 \geq \frac{4ab}{(a+b)^2}$
We have:
$$
b^2 \geq (\frac{a+b}{2})^2 \ \ (\text{because } b \geq a > 0) \\
= \frac{(a+b)^2}{4} \geq \frac{4ab}{4}=ab \ \ \ (\text{because (1) }) \\
= ab \times 1 \geq ab \times \frac{4ab}{(a+b)^2} = (\frac{2ab}{a+b})^2 \ \ \ (\text{because (2) }) \\
\geq (\frac{2ab}{b+b})^2 = a^2 \ \ \ (\text{because } b \geq a >0 )
$$
Actually, the first and the last inequalites use $a\geq b>0$, the second and third ones use HM-GM-AM-QM inequalities mentioned by @mrtechtroid.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to solve $|z|^2 + z|z| + 2 = 2i Im(z)$? this is the process I follow:
*
*$z|z| = 2iImz -|z|^2- 2 $
*$z \sqrt{Re(z)^2 + Im(z)^2} = 2iy - (Re(z)^2 + Im(z)^2) - 2$
*$z\sqrt{x^2 + y^2} =2iy -(x^2+y^2)-2$
*$z=(2iy -(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and $Re(z)=(-(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
what should i do next?
| You may simplify the notation from the beginning. Call $z=a+bi$ and your equation becomes:
$$a^2+b^2+(a+bi)\sqrt{a^2+b^2}+2=2bi$$ or
$$[a^2+b^2+a\sqrt{a^2+b^2}+2]+[bi\sqrt{a^2+b^2}]=2bi$$
So, you can split it into the real part and the imaginary part.
$$a^2+b^2+a\sqrt{a^2+b^2}+2=0\quad (1)$$
$$b\sqrt{a^2+b^2}=2b \quad (2).$$
From $(2)$ you get $b=0$ or $\sqrt{a^2+b^2}=2$.
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I can prove it.
$$ab^2aab^3a=ab^5a$$ so I'm thinking something here. we get $b^5=ab^5a$.
| If $a^2 = e$ and $a^{-1} b^2 a = b^3$, we can substitute $a^2 = e$ into the second equation to get $$a^{-1} b^2 a = b^3 = (a^2)^{-1} b^2 a = e^{-1} b^2 a = b^2 a$$
Then, multiplying both sides of this equation by $b^{-2}$, we get $$b^{-2} b^2 a = b^{-2} (b^2 a) = a$$
Now, we can substitute this result, $a = b^{-2} (b^2 a)$ into the equation $a^{-1} b^2 a = b^3$ to get $$b^3 = (b^{-2} (b^2 a))^{-1} b^2 (b^{-2} (b^2 a)) = a^{-1} b^2 a = b^3$$
We can also substitute $a = b^{-2} (b^2 a)$ into the equation $a^2 = e$ to get $(b^{-2} (b^2 a))^2 = e$, which simplifies to $b^{-4} b^4 a^2 = e$.
Finally, we can substitute $a^2 = e$ into this equation to get $b^{-4} b^4 e = e$, which simplifies to $b^4 = e^{-1} = e$.
Therefore, if $a^2 = e$ and $a^{-1} b^2 a = b^3$, we can conclude that $b^5 = e$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$ Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written as $$\dfrac32\cdot2^{2x}=\dfrac{4}{\sqrt{3}}\cdot3^x,$$ or $$\dfrac{2^{2x}}{3^x}=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}\iff\left(\dfrac43\right)^x=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}$$ How do I find $x$ from here, as it is obviously a difficulty for me? What's the approach supposed to be?
| Alternative approach:
*
*Let $r = \log_2(3).$
*Note that $2^a = 2^b \iff a = b.$
Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The LHS can be re-written as
$$2^{2x} \times \left[1 + \frac{1}{2}\right] = 2^{2x} \times \frac{3}{2}.$$
The RHS can be re-written as
$$3^{x + 0.5} \times \left[1 + \frac{1}{3}\right] = 3^{x + 0.5} \times \frac{4}{3}$$
$$ = \left(2^{r}\right)^{x+0.5} \times \frac{4}{3} = \left[2^{r \times (x+0.5)}\right] \times \frac{4}{3}.$$
Since the LHS and RHS are equivalent, you know that
$$2^{2x-3} = \frac{2^{2x}}{8} = \frac{\left[2^{r \times (x+0.5)}\right]}{9} = 2^{r \times (x + 0.5 -2)}.$$
Therefore,
$$2^{2x-3} = 2^{r \times \left[x - \frac{3}{2}\right]} \implies $$
$$2x - 3 = r \times \left[x - \frac{3}{2}\right] \implies $$
$$x(2 - r) = 3 - \left[r \times \frac{3}{2}\right] \implies $$
$$x = \frac{3 - \left[r \times \frac{3}{2}\right]}{2 - r} = \frac{3}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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polynomial approximation of a circle First of all, I am not a mathematician and my mathematics are fairly rusty. I would appreciate some help
I believe there is not polynomial equation of a circle (is that right?) but take a look at this picture
In it we can see two equations. One a circle $30x^2+30y^2=300000$ and the other one a parabola $y=120+-0.003x^2$
Obviously they are different but the parabola in the vicinity of (0,100) seems fairly parallel to the circle
This coefficients are randomly chosen but my question is if I have a circle of radius r , how can I find the coefficients of a parabola that is a bit parallel in:
*
*a point in the axis
*any point of the circle?
| Expanding upon my comment about osculating circles ...
Suppose we have the second-degree polynomial function
$$y=ax^2+bx+c \tag1$$
that we want to be "parallel" to an origin-centered circle of radius $r$ near a point $R$ on that circle; let's say that the distance between the curves at $R$ should be $s$, and write $S$ for the corresponding point on the curve. We want to find $a$, $b$, $c$ such that the origin-centered, $(r+s)$-radius circle is the polynomial's osculating circle at $S$.
The Wikipedia entry tells us that the center of the osculating circle is given by
$$\left(\;x-y'\frac{1+y'^2}{y''}, \; y+\frac{1+y'^2}{y''}\;\right) \tag2$$ where $y'=2ax+b$ and $y''= 2a$ are the first and second derivatives of $y$. We want the center to be the origin, which gives us two conditions:
$$\begin{align}
2ax-(2ax+b)(1+(2ax+b)) &= 0 \tag3\\
1 + b^2 + 2 a c + 6 a b x + 6 a^2 x^2 &= 0 \tag4
\end{align}$$
Together with $(1)$, we can solve for $a$, $b$, $c$ to get
$$
a=-\frac{x^2 + y^2}{2 y^3} \qquad
b= \frac{x^3}{y^3} \qquad
c= -\frac{(x^2 + y^2) (x^2 - 2 y^2)}{2 y^3} \tag{5}$$
Note that point $Q$ has coordinates $((r+s)\cos\theta,(r+s)\sin\theta)$ for some $\theta$. Substituting those into $(5)$ gives
$$a = -\frac{\csc^3\theta}{2 (r + s)} \qquad b=\cot^3\theta \qquad c= \frac12 (r + s) (2 - 3 \cos^2\theta) \csc^3\theta \tag{6}$$
so that the general form of the quadratic curve becomes
$$y = -\frac{\csc^3\theta}{2(r+s)} \left( x^2 -2x(r+s)\cos^3\theta - (r + s)^2 (2 - 3 \cos^2\theta) \right) \tag7$$
Here are some examples:
and a nifty animation:
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$
Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.
Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.
| By Cauchy and AM-GM inequality,
$$\sum\frac{a^3}{b+c+d}\ge\frac{\left(\sum a^2\right)^2}{2\sum ab}\ge\frac{\left(\sum a^2\right)^2}{3\sum a^2}=\frac{\sum a^2}3\ge\frac13.$$
The last "$\ge$" is equivalent to $\sum a^2\ge1$. By rearrangement inequality
$$a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1.$$We're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the volume of the following region $E= (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) $
Find the volume of the following region $E= \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) \} $
I figured that the region E is formed by the points that belong in the gap between the two paraboloids
$z=\sqrt{2}(x^2 + y^2)$ $\hspace{20pt}$ and $\hspace{20pt}$ $ z= \sqrt{6}(x^2 + y^2)$
that are as well inside of the sphere $x^2 + y^2 + z^2 = 1$.
$E$ can be obtained by making a whole rotation around the $z$-axis in the plane-$xy$ then using Guldino's theorem can be appropriate.
Considering now cylindrical coordinates we've that $E$ becomes the following:
$F= \{ (\rho, \theta, z): \rho^2 + z^2 ≤1, \sqrt{2}\rho^2 ≤z≤ \sqrt{6}\rho^2, 0≤ \theta≤ 2\pi \}$
Then we've that:
$vol \hspace{2pt} E = \int _{F} \rho \hspace{2pt} d\rho d\theta dz = 2\pi \int\int _{D} \rho d\rho dz $
I'm struggling to find the region $D$. Any hint?
| To expand on my comment, we first observe that the upper paraboloid $z=\sqrt6\,(x^2+y^2)$ meets the sphere $x^2+y^2+z^2=1$ in a cylinder with radius $\frac1{\sqrt3}$ :
$$\begin{cases}z = \sqrt6\,(x^2+y^2) \\ x^2 + y^2 + z^2 = 1 \end{cases} \implies \frac{z}{\sqrt6} + z^2 = 1 \implies z = \sqrt{\frac23} \\
\implies x^2 + y^2 + \frac23 = 1 \implies x^2 + y^2 = \left(\frac1{\sqrt3}\right)^2$$
Now, if $(x,y,z)$ belongs to the cylinder's interior $x^2+y^2\le\frac13$, then the paraboloids determine the "height" of the solid and we have $\sqrt2\,(x^2+y^2)\le z \le \sqrt6\,(x^2+y^2)$.
Otherwise, if $(x,y,z)$ is outside this cylinder, then $\sqrt2\,(x^2+y^2)\le z\le\sqrt{1-x^2-y^2}$.
Where the lower paraboloid and sphere meet will help set up the limits for $x$ and $y$. We have
$$\begin{cases} z = \sqrt2\,(x^2+y^2) \\ x^2+y^2+z^2 = 1 \end{cases} \implies \frac z{\sqrt2} + z^2 = 1 \implies z = \frac1{\sqrt2} \\
\implies x^2+y^2 + \frac12 = 1 \implies x^2+y^2 = \left(\frac1{\sqrt2}\right)^2$$
which is to say, the whole solid lies inside a cylinder of radius $\frac1{\sqrt2}$, so we can choose e.g. $-\frac1{\sqrt2}\le x\le\frac1{\sqrt2}$ and $|y| \le \sqrt{\frac12-x^2}$.
The volume is then given by the triple integral,
$$\iiint_E dV = \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \int_{-\sqrt{\frac12-x^2}}^{\sqrt{\frac12-x^2}} \int_{\sqrt2\,(x^2+y^2)}^{\min\left\{\sqrt6\,(x^2+y^2),\sqrt{1-x^2}\right\}} \, dz \, dy \, dx$$
By symmetry, it's the same as
$$\iiint_E dV = 4 \int_0^{\frac1{\sqrt2}} \int_0^{\sqrt{\frac12-x^2}} \int_{\sqrt2\,(x^2+y^2)}^{\min\left\{\sqrt6\,(x^2+y^2),\sqrt{1-x^2}\right\}} \, dz \, dy \, dx$$
Converting to cylindrical coordinates from here should be straightforward. Expanding this any further in Cartesian is overkill.
| {
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"timestamp": "2023-03-29T00:00:00",
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How fast does the sequence $a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx$ decays as $n \to \infty$? The question is as in the title.
The sequence $\{a_n \}$ define by
\begin{equation}
a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx
\end{equation}
decays to $0$ as $n \to \infty$, which is a direct consequence of the Dominated Convergence Theorem.
But, when I run the Wolfram Alpha to evluate each $a_n$, I obtain the hypergeometric functions:
\begin{equation}
a_n=2^{-n-7/2}\Bigl(4 {_2}{F}_1\bigl(\frac{1}{2}, -n -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{2}\bigr)-{_2}{F}_1\bigl(\frac{1}{2}, -n, -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{32}\bigr) \Bigr)
\end{equation}
But I cannot find information on the asymptotics of the above hypergeometric functions as $n \to \infty$.
Could anyone please help me?
| With the substitution $t = \log \left( {\frac{3}{4}} \right) - \log \left( {x^2 + \frac{1}{2}} \right)$, your integral becomes
$$
a_n = \left( {\frac{3}{4}} \right)^{n + 3/2} \int_0^{\log (16/11)} {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {3 - 2{\rm e}^t } }}{\rm d}t} .
$$
Now
$$
\frac{{{\rm e}^{ - t} }}{{\sqrt {3 - 2{\rm e}^t } }}=
1 + \frac{3}{2}t^2 + \frac{5}{2}t^3 + \frac{{23}}{4}t^4 + \ldots
$$
when $|t|<\log(3/2)$. Thus, by Watson's lemma,
$$
a_n \sim \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n}\left(
1 + \frac{3}{{n^2 }} + \frac{{15}}{{n^3 }} + \frac{{138}}{{n^4 }} + \frac{{1545}}{{n^5 }} + \ldots \right)
$$
as $n\to +\infty$.
Alternatively,
$$
\frac{1}{{\sqrt {3 - 2{\rm e}^t } }} = \sum\limits_{k = 0}^\infty {c_k \frac{{t^k }}{{k!}}}
$$
for $|t|<\log (3/2)$, where
$$
c_k = \sum\limits_{j = 0}^k {(2j - 1)!!S(k,j)}.
$$
Here $j!!$ is the double factorial and $S(k,j)$ denotes the Stirling numbers of the second kind (cf. $\mathrm{A}305404$). Thus, by Watson's lemma,
\begin{align*}
a_n & \sim \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n+1}\sum\limits_{k = 0}^\infty {\frac{{c_k }}{{(n + 1)^k }}} \\ & = \left( {\frac{3}{4}} \right)^{n + 3/2} \frac{1}{n+1}\left( {1 + \frac{1}{{n + 1}} + \frac{4}{{(n + 1)^2 }} + \frac{{25}}{{(n + 1)^3 }} + \ldots } \right)
\end{align*}
as $n\to +\infty$.
| {
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Finding Taylor Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ in $1<|z|<2$. I have to find Taylor series of
$f(z)=\frac{z^2-2z+5}{(z-2)(z^2+1)}$, for $1<|z|<2$.
I have started with
\begin{eqnarray*}
f(z)&=&\textstyle\frac{z^2 -2z +5}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{1}{z-2}-\frac{2}{z^2+1}\\[4pt]
&=&\textstyle\frac{1}{-2\left(1-\frac{z}{2}\right)}-\frac{2}{z^2\left(1+\frac{1}{z^2}\right)}\\[4pt]
&=&-\sum\limits_{n=0}^{\infty}\frac{z^n}{2^{n+1}}-2\sum\limits_{n=0}^{\infty}(-1)^nz^{-2n-2}.
\end{eqnarray*}
But, this looks like Laurent series. Am I right about that?
| First decompose your expression into partial fractions. $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{A}{z-2}+\frac{Bz+C}{z^2+1}.$$ You will find that $A=1,B=0,C=-2.$ Then $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{1}{(-2)(1-\frac{z}{2})}+\frac{-2}{1+z^2}$$ $$=(-1/2)(1-\frac{z}{2}+\frac{z^2}{4}-...)-2(1-z^2+z^4-...)$$ $$=\sum_{n=0}^{\infty}a_nz^n$$ where $a_n=\frac{-1}{2^{n+1}}$ if $n$ is odd and $$a_n=\frac{-1}{2^{n+1}}+2(-1)^{1+n/2}\text { if $n$ is even.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
| By the Buffalo way, WLOG $a=\min\{a,b,c\}$, then set $a=1+k$, $b=1+k+u$ and $c=1+k+v$. Along with the condition $a\ge1$, we get $k$, $u$, $v\ge0$. Plug in and we get
\begin{align}
&(a+b)(b+c)(c+a)(\text{left side}-\text{right side})\\[.6em]
={}&24k^5+40k^4u+40k^4v+120k^4+30k^3u^2+50k^3uv
\\{}+{}&160k^3u+30k^3v^2+160k^3v+168k^3+12k^2u^3\\
{}+{}&18k^2u^2v+90k^2u^2+36k^2uv^2+150k^2uv+168k^2u
\\{}+{}&12k^2v^3+90k^2v^2+168k^2v+72k^2+2ku^4+2ku^3v\\
{}+{}&24ku^3+8ku^2v^2+36ku^2v+72ku^2+14kuv^3+72kuv^2\\
{}+{}&96kuv+48ku+2kv^4+24kv^3+72kv^2+48kv\\
{}+{}&2u^4+2u^3v+12u^3+2u^2v^3+8u^2v^2+9u^2v+16u^2\\
{}+{}&2uv^4+14uv^3+27uv^2+8uv+2v^4+12v^3+16v^2\ge0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$
I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.
| As this Art of Problem Solving thread suggests, write this using a form of partial fractions:
$$(Ar + B)(r^2 - r + 1) + (Cr + D)(r^2 + r + 1) = r^2 - 1$$
Now comparing the coefficients of $r^3$, $r^2$, $r$, and the constant term, we end up with $A + C = 0, B - A + D + C = 1, -B + A+D+C=0$, and $B + D = -1$.
Adding the middle two equations gives $D + C = 1/2$, and so $B - A = 1/2$. Now $B + (C) = 1/2$ and so we obtain $2(B + C + D) = 0$, which leads to $B = -1/2$, $C = 1$, $D = -1/2$, and $A = -1$.
Hence the summand can be written as:
$$\frac{-r - 1/2}{r^2 + r + 1} + \frac{r-1/2}{r^2 - r + 1}$$
which is telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Which solution is correct: $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$? I was solving this question $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$.
Now, this is solved using $\lim_{x \to a} [1+f(x)]^\frac{1}{g(x)}=e^{\lim_{x \to a} \frac{f(x)}{g(x)}}$.
However, there seem to be two methods of approaching this problem and both yield different answers. In the first approach we multiply by $\frac{1}{2}$ outside the limit and by $2$ inside the limit and add and subtract $1$ to get in the form of the formula. After solving we get the value of the limit as $e$. In the second approach we add and subtract $1$ inside the limit thus getting it in this form and then evaluating the limit we get $e^{-\frac{1}{2}}$.
Now, the book prefers the first approach and so I am assuming my second approach is wrong, but I can't find for the life of me where it is wrong.
| Here is how your second approach should work:
$$\frac{x^2+2x+3}{2x^2+x+5} = 1 + \left(\frac{x^2+2x+3}{2x^2+x+5}-1\right)$$$$ = 1 + \underbrace{\frac{-x^2+x-2}{2x^2+x+5}}_{\stackrel{x\to\infty}{\longrightarrow}-\frac 12}$$
So, you could not use "your rule with $e$", because it would require this term to tend to zero.
So, you actually get
$$\left( 1 +\frac{-x^2+x-2}{2x^2+x+5}\right)^{\frac{3x-2}{3x+2}}\stackrel{x\to\infty}{\longrightarrow}\left(1-\frac 12\right)^1 = \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
sum of 100 terms of logarithmic expression
Calculate value of
$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$
My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\displaystyle =\sum^{100}_{k=1}\ln\bigg[\frac{(1+2(2k+1)+2(2k+1)^2)(1-2(2k+1)+2(2k+1)^2)}{(1+2(2k)+2(2k)^2)(1-2(2k)+2(2k)^2)}\bigg]$
How can I decompose that complex expression into partial fractions?
| $$g(k)=\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{(4k+2)^4+4\cdot1^4}{(4k)^4+4\cdot1^4}$$
Using your formula only,
$\displaystyle x^4+4y^4=(x^2+2y^2)^2-(2xy)^2=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
$x=4k+2, y=1$
$\displaystyle\implies x^2+2xy+2y^2=(4k+2)^2+2(4k+2)+2=16k^2+24k+10\ \ \ \ (1)$
$x^2-2xy+2y^2=(4k+2)^2-2(4k+2)+2=16k^2+8k+2\ \ \ \ (2)$
$x=4k, y=1$
$\displaystyle\implies x^2+2xy+2y^2=(4k)^2+2(4k)+2=?\ \ \ \ (3)$ which cancels $(2)$
$\displaystyle\implies x^2-2xy+2y^2=(4k)^2-2(4k)+2=16k^2-8k+2=(4k-2)^2+2(4k-2)+2\ \ \ \ (4)$
$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\prod_{k=1}^n\dfrac{(4k+2)^2+2(4k+2)+2}{(4(k-1)+2)^2+2(4(k-1)+2)+2}=\prod_{k=1}^n\dfrac{f(k)}{f(k-1)}$$ where $f(m)=(4m+2)^2+2(4m+2)+2$
$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{f(n)}{f(0)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How do I solve this, do I have to continue calculating with rational fractions? $$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$
I tried to use different substitutions such as $t=\cos(x)$, $t=\sin(x)$, $t=\tan(x)$, and after expressing $\sin$ and $\cos$ through $\tan(\frac{x}{2})$, I've got $ \int -4 \frac{t^2dt}{(1+t^2)^2(t^2-t-1)}.$
Rational fractions didn’t work.
| It is possible to solve it without rational fractions. The denominator can be written as
$$\sin(x)+2\cos(x) = \sqrt 5\left(\frac{1}{\sqrt 5}\sin x + \frac{2}{\sqrt 5}\cos x\right) = \sqrt 5 \sin\left(x+\theta\right)$$
where $\sin\theta = \frac{2}{\sqrt 5}$. Now by the change of variable $x+\theta\rightarrow y$ we get
$$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)} = \frac{1}{\sqrt 5} \int \frac{\sin^2(x)dy}{\sin y} = \frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2x\right) dy}{\sin y} = \\\frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2y-2\theta\right) dy}{\sin y} = \frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2y\right)\cos\left(2\theta\right) - \sin\left(2y\right)\sin\left(2\theta\right) dy}{\sin y} =\\
\frac{1}{2\sqrt 5} \left(\int \frac{dy}{\sin y} - \cos\left(2\theta\right)\int \frac{dy}{\sin y} + 2\cos\left(2\theta\right)\int \sin y dy - 2\sin\left(2\theta\right) \int \cos ydy \right)$$
and I believe you can do the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find $f(\frac{2\pi k}{2^n \pm1})$ given $f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$. We are required to find $f(\frac{2\pi k}{2^n \pm1})$ given
$f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$
Let's start by removing the product notation
$\Rightarrow f(x)=(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)$
Now multiplying and dividing by $2\cos x+1$,
$\Rightarrow f(x)=\frac{(2\cos x+1)(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
We know, $a^2-b^2=(a+b)(a-b)$
$\Rightarrow f(x)=\frac{(4\cos^2x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
Now, $\cos2x=2\cos^2x-1$
$\Rightarrow f(x)=\frac{(2\cos2x+1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
$\Rightarrow f(x)=\frac{(2\cos 2^2x+1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
We can now repeatedly apply the difference of squares formula and our function reduces to
$\Rightarrow f(x)=\frac{4\cos^2 2^{n-1}x-1}{2\cos x+1}$
$\Rightarrow f(x)=\frac{2\cos 2^nx+1}{2\cos x+1}$
How do I proceed from here on out?
Edit- We just need to know that $\cos(2k)=\cos(2^nk)$. Thus the denominator and the numerator cancel out and we get
$f(\frac{2\pi k}{2^n \pm1})=1$ and that's the answer.
| In fact you have solved roughly all of it .
You just need to know that:
*
*$\frac{2^n2\pi k}{2^n +1} +\frac{2\pi k}{2^n+1}=2\pi k$
*$\frac{2^n2\pi k}{2^n -1} -\frac{2\pi k}{2^n-1}=2\pi k$
So in the end you get the answer 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Generating Functions - Counting how many solutions for an exact $x^n$ Having trouble with this problem:
$$ x_1+x_2+x_3+x_4+x_5+x_6 = 13 $$
$$ x\ne 3 $$
Thanks in advance for any further help.
Being able to get to this point:
$$\begin{split}
f(x) &= (x^0+x^1+x^2+x^4+x^5+\cdots)^6\\
&= [(-x^3)+{x^0+x^1+x^2+(x^3)+x^4+x^5+\cdots}]^6\\
&= \left[{(-x^3)} + \frac{1}{(1-x)}\right]^6\\
&=\left[ \frac{-x^3(1-x)+1}{(1-x)}\right]^6\\
&=\left[\frac {x^4-x^3+1}{1}\cdot\frac {1}{1-x}\right]^6\\
&=(x^4-x^3+1)^6 \frac {1}{(1-x)^6}\\
\end{split}$$
Now if I long divide $(x^4-x^3+1)$ with $(1-x)$ I get $[(-x^3 + \frac {1}{1-x}) (1-x)]^6 $ thus:
$$\begin{split}
f(x)&=\left(\frac {1}{1-x} -x^3\right)^6 \frac {(1-x)^6}{(1-x)^6}\\
&=\left(\frac {1}{1-x} -x^3\right)^6\\
\end{split}$$
That's where I get stuck, honestly.
| The P.I.E approach needs to prepare some "ingredients".
There are solutions containing $4$ "bad" $x$'s, $3, 2, 1$ or none "bad" $x$'s.
The numbers involved are the following :
$$ {6\choose 4}{2\choose 1} = 30$$
$$ {6\choose 3}{6\choose 2} = 300$$
$$ {6\choose 2}{10\choose 3} = 1800$$
$$ {6\choose 1}{14\choose 4} = 6006$$
$$ {6\choose 0}{18\choose 5} = 8568$$
For example in $ {6\choose 2}{10\choose 3} $, $ {6\choose 2} $comes by requiring to have at least two bad $x$'s, which can be any two of the six $x$'s and ${10\choose 3} $ comes from stars and bars, while placing $13-3-3$ items in $4$ slots = $ {7+4-1\choose 4-1} $
This solution was not required, but the "coincidence" is worth noting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solving $4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$ for real $x$ I was given the following equation:
$$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6; x\in \mathbb{R}$$
Here is how I tried to solve it:
$$2^{2x+2\sqrt{x^2-2}+1}-5\cdot 2^{\sqrt{x^2-2}}=12$$
$$2^{\sqrt{x^2-2}}(2^{2x+1+\sqrt{x^2-2}}-5)=2^2\cdot 3$$
Where $$2^{\sqrt{x^2-2}}=2^2 \land 2^{2x+1+\sqrt{x^2-2}}-5=3$$
$$\Updownarrow$$
$$\sqrt{x^2-2}=2 \land 2x+1+\sqrt{x^2-2}=3.$$
As well as we must note that $$x\in \left( -\infty ; -\sqrt2 \right]\cup \left[ \sqrt{2};\infty\right)$$
However
$$2x+1+\sqrt{x^2-2}=3 \Leftrightarrow -3x^2+8x-6=0$$
$$x_{1,2}=\frac{8\pm\sqrt{-8}}{6}$$
Thus, it's not satisfying given domain.
So I tried to use substitution
Let $\sqrt{x^2-2}=t, t\geq 0,$ then $x^2=t^2+2 \Leftrightarrow x=\pm \sqrt{t^2+2}$
If $$x=\sqrt{t^2+2}$$ then $$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$$
$$\Updownarrow$$ $$2^{2\sqrt{t^2+2}+2t}-5\cdot 2^{t-1}=6$$
$$\Updownarrow$$
$$2^{2\sqrt{t^2+2}+2\sqrt{t^2}}-5\cdot 2^{\sqrt{t^2}-1}=6$$
From here if I continued using substitution to make it into a quadratic equation, I would reach a result, however, I doubt this is the right way as I only had 10 minutes to solve this in class and I'm only up to high school. Can someone provide me with any solution?
| Comment: As you modified we have:
$2^{2x+1+2\sqrt{x^2-2}}-5\cdot 2^{\sqrt {x^2-2}}-12=0$
$2^{2x+1+2\sqrt{x^2-2}}=2^{2x+1}\cdot 2^{2\sqrt {x^2-2}}=2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2$
$2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2-5\cdot 2^{\sqrt{x^2-2}}-12=0$
Let $2^{\sqrt{x^2-2}}=y$ , we can write:
$(2^{2x+1})\cdot y^2-5y-12=0$
$y=\frac {5\pm\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$
Now you have to search whether following system of equations is consistent or not:
$\begin {cases}\Delta=25+48\cdot 2^{2x+1}=k^2\\y=2^{\sqrt{x^2-2}}\end {cases}$
For solutions in $ R$ we take only positive solutions for y:
$y=\frac {5+\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$
Update: we can see that:
$2^{2x+1}=11\rightarrow 2x+1\approx 3.5\rightarrow x=1.25$
$y=\frac{5+17}{2\times 11}=1=2^{\sqrt{x^2-2}}\Rightarrow X^2-2=0$
which gives $x=\sqrt 2\approx 1.4...$
1- $\sqrt 2$ gives $LHS=5.6$
2-$x=2$, gives $LHS=39.3$
3- In fact we must have $\sqrt 2<x< 2$, $x=1.6$ works:
$LHS\approx 12$
No need a complicated analytic solution; these three steps can be done using a calculator in less that ten minutes. What is more important is to find the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)}
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) $$
Proof:
For $a,b>0$,
Using $\ln \left(a^2+b^2\right)=2 Re(\ln (a+b i))$, we can reduce the power $4$ to $2$.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x
& = 2\int_{0}^{\infty} \frac {Re\left[\ln \left(x^2+a ^2i\right)\right]}{b^2+x^2} d x \\
& =2 Re\left(\int_0^{\infty} \frac{\ln \left(x^2+\left[\left(\frac{1+i}{\sqrt{2}}\right) a\right]^2\right)}{b^2+x^2} d x\right)
\end{aligned}
$$
Using (*), we have $$
\begin{aligned}\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x&=2 Re\left[\frac{\pi}{b} \ln \left(\frac{1+i}{\sqrt{2}} a+b\right)\right] \\&=\frac{2 \pi}{b} R e\left[\ln \left(\frac{a}{\sqrt{2}}+b+\frac{a}{\sqrt{2}}i\right)\right] \\&= \boxed{\frac{\pi}{b} \ln \left(a^2+b^2+a b \sqrt{2}\right)}\end{aligned}
$$
In general, for any $a, b \in \mathbb{R} \backslash\{0\}$, replacing $a$ and $b$ by $|a|$ and $|b|$ yields
$$\boxed{\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) }$$
For example, $$
\int_0^{\infty} \frac{\ln \left(x^4+16\right)}{9+x^2} d x= \frac{\pi}{3} \ln (13+6 \sqrt{2})
$$
Comments and alternative methods are highly appreciated.
| Note that $x^4+a^4= (x^2+\sqrt2 ax+a^2) (x^2-\sqrt2 ax+a^2)$
\begin{align}
&\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{x^2+b^2} d x \\
=&\int_{-\infty}^{\infty} \frac{\ln (x^2+\sqrt2 ax+a^2)}{x^2+b^2} \ d x\\
=& \int_{-\infty}^{\infty}\bigg( \int_0^{\pi/4}\frac{2ax\cos t}{x^2+2ax \sin t+a^2} dt+ {\ln (x^2+a^2)}\bigg) \frac{dx}{x^2+b^2}\\
=& \int_0^{\pi/4}\frac{-2\pi a\sin t}{a^2+2ab\cos t+b^2} dt\ + \frac{2\pi}b \ln(a+b)\\
= &\ \overset{}{\frac{\pi}{ b} }\ln \left(a^2+b^2+\sqrt{2} a b\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int \frac1{(x^2+a^2)^{m}}dx$ The integral $$\int\frac{1}{(x^2+a^2)^m}dx$$ can be expressed by a recursive formula of $$\frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{dx}{(x^2+a^2)^{m-1}}$$ I do not understand how integration by part leads to this result. Specifically, since $$\int u dv = uv - \int v du$$ I want to know what is $u$, $du$, $v$, $dv$.
| We have
$$J = \int\dfrac{dx}{(x^2+a^2)^m} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^m}dx$$
and hence
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{m-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^m}dx$$
which gives
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{m-1}} + \dfrac{1}{2a^2(m-1)}\int x\,d\, \dfrac1{(x^2+a^2)^{m-1}}$$
Integrating by parts, we get
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{m-1}} + \dfrac{1}{2a^2(m-1)} \dfrac {x}{(x^2+a^2)^{m-1}} - \dfrac{1}{2a^2(m-1)}\int \dfrac{dx}{(x^2+a^2)^{m-1}}$$
and hence
$$J = \dfrac{1}{2a^2(m-1)} \dfrac {x}{(x^2+a^2)^{m-1}} + \dfrac{2m-3}{2a^2(m-1)}\int \dfrac{dx}{(x^2+a^2)^{m-1}}$$
| {
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Proving $\sum_{cyc}\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}} \leq 3$ for $x$, $y$, $z$ the sides of a triangle
Let $x,y$ and $z$ be the length of a triangle. Prove that
$$\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}}
+ \frac{\sqrt{z + x − y}}{\sqrt{z} + \sqrt{x} − \sqrt{y}}
+ \frac{\sqrt{x + y − z}}{\sqrt{x} + \sqrt{y} − \sqrt{z}} \leq 3$$
I tried to prove that one of the expressions $\leq1$
$x\leq y\leq z$
$∵x-y\leq 0, x-z\leq 0$
$∴(x-y)(x-z) \geq 0$
$x^2-(y+z)x+yz \geq 0$
$x^2+yz \geq x(y+z)$
$x^2+yz \geq xy+xz$
$x^2+yz+2x\sqrt{yz} \geq xy+xz+2x\sqrt{yz}$
$(x+\sqrt{yz})^2 \geq (\sqrt{xy}+\sqrt{xz})^2$
$x+\sqrt{yz} \geq \sqrt{xy}+\sqrt{xz}$
$x \geq \sqrt{xy}+\sqrt{xz}-\sqrt{yz}$
$-2x \leq -2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})$
$-x \leq -2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})+x$
$∵y+z > 0$
$y+z-x \leq x+y+z-2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})$
$y+z-x \leq (\sqrt{y}+\sqrt{z}-\sqrt{x})^2$
$∴ \sqrt{y+z-x} \leq \sqrt{y}+\sqrt{z}-\sqrt{x}$
$\frac{\sqrt{y+z-x}}{\sqrt{y}+\sqrt{z}-\sqrt{x}} \leq 1$
Similarly, the other two expressions can also be proven the same way.
Therefore
$$\frac{\sqrt{ + − }}{\sqrt{} + \sqrt{} − \sqrt{}} +
\frac{\sqrt{ + − }}{\sqrt{} + \sqrt{} − \sqrt{}} +
\frac{\sqrt{ + − }}{\sqrt{} + \sqrt{} − \sqrt{}} \leq 3$$
Is my proof correct？
$$\frac{\sqrt{ + − }}{\sqrt{} + \sqrt{} − \sqrt{}} \geq 1$$
and
$$\frac{\sqrt{ + − }}{\sqrt{} + \sqrt{} − \sqrt{}} \leq 1$$
My proof is incomplete because the 2nd term does not sastify my proof. May I know what is the next steps?
| For the second term, when we'll do the proving we'll shake hands with the expression $$z+x-y \leq (\sqrt{z}+\sqrt{x}-\sqrt{y})^2$$ This is not true for the triplet $(n-1,n,n+1)$
I'll leave it for the OP to answer the reason for this.
| {
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
| Here is one proof, using how ${ \log (1+x) }$ is "sandwiched" between its Taylor polynomials on ${ [0,1] }.$
Consider ${ \log(1+x) }$ on ${ (-1, \infty) }.$ Its Taylor polynomials about $0$ are ${ S _n (x) = x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n}. }$
Notice errors ${ \varepsilon _n (x) := \log(1+x) - \left( x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n} \right) }$ satisfy ${ \varepsilon _n (0) = 0, }$ and ${ \varepsilon _n ' (x) }$ ${ = \frac{1}{1+x} -(1-x + \ldots + (-1) ^{n+1} x ^{n-1}) }$ ${ = \frac{1 - (1+ (-1) ^{n+1} x ^n)}{1+x} }$ ${ = \frac{ (-1) ^n x ^n}{1+x} }$ is $\geq 0$ for ${ \lbrace x \in [0, \infty); n \text{ even} \rbrace }$ and $\leq 0$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ \varepsilon _{n} (x) }$ is ${ \geq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ even} \rbrace }$ and ${ \leq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }$ for ${ x \in [0, \infty), n \in \mathbb{Z} _{\gt 0} }.$
Further ${ S _{2(n+1) -1} (x) - S _{2n-1} (x) }$ ${ = - \frac{x ^{2n}}{2n} + \frac{x ^{2n+1}}{2n+1} }$ ${ = x ^{2n} ( \frac{x}{2n+1} - \frac{1}{2n}) \leq 0}$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n-1} (x)) _{n \geq 1} }$ is decreasing if ${ x \in [0,1] }.$
Similarly ${ S _{2(n+1)} (x) - S _{2n} (x) }$ ${ = \frac{x ^{2n+1}}{2n+1} - \frac{x ^{2n+2}}{2n+2} }$ ${ = x ^{2n+1} (\frac{1}{2n+1} - \frac{x}{2n+2}) \geq 0 }$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing if ${ x \in [0,1] }.$
Fix ${ x \in [0,1] }.$
Now sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing and ${ (S _{2n-1} (x) ) _{n \geq 1} }$ decreasing, with ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }.$
So say ${ \lim _{n \to \infty} S _{2n} (x) = \ell _e }$ and ${ \lim _{n \to \infty} S _{2n-1} (x) = \ell _o }.$ But ${ S _{2n-1} (x) - S _{2n} (x) = \frac{x ^{2n}}{2n} \leq \frac{1}{2n} \to 0 }$ as ${ n \to \infty }.$ So ${ \ell _o - \ell _e = 0 },$ ie ${ \ell _o = \ell _e }.$
This gives ${ \ell _o = \ell _e = \log(1+x), }$ making ${ \log(1+x) = \lim _{j \to \infty} S _j (x). }$
Especially ${ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots }$ converges, and to ${ \log(2) }.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:
$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$
For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations, one for each root.
Is there a general formula for solving equations of the following form [Quartic Equations]?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?
| What has not been mentioned thus far is that one can in fact use any number of "auxiliary cubics" in the solution of the quartic equation. Don Herbison-Evans, in this page (Wayback link, as the original page has gone kaput), which was adapted from his technical report, mention five such possible auxiliary cubics.
Given the quartic equation
$$x^4 + ax^3 + bx^2 + cx + d = 0$$
the five possible auxiliary cubics are referred to in the document as
Christianson-Brown:
$$y^3 +\frac{4a^2b - 4b^2 - 4ac + 16d - \frac34a^4}{a^3 - 4ab + 8c}y^2 + \left(\frac3{16}a^2 - \frac{b}{2}\right)y - \frac{1}{64}(a^3 + 4a b - 8c) = 0$$
Descartes-Euler-Cardano:
$$y^3 + \left(2b - \frac34 a^2\right)y^2 + \left(\frac3{16}a^4 - a^2b + ac + b^2 - 4d\right)y - \frac{1}{64}(a^3 + 4a b - 8c)^2 = 0$$
Ferrari-Lagrange
$$y^3 + by^2 + (ac - 4d)y + a^2d + c^2 - 4bd = 0$$
Neumark
$$y^3 - 2by^2 + (ac + b^2 - 4d)y + a^2d - abc + c^2 = 0$$
Yacoub-Fraidenraich-Brown
$$(a^3 - 4ab + 8c)y^3 + (a^2b - 4b^2 + 2ac + 16d)y^2 + (a^2c - 4bc + 8ad)y + a^2d - c^2 = 0 $$
See the page for how to obtain the quadratics that will yield the solutions to the original quartic equation from a root of the auxiliary cubic.
Let me also mention this old ACM algorithm in Algol. Netlib has a C implementation of that algorithm.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to build a linear equation system? How can one build a $3\times 3$ linear equation system ($3$ equations with $3$ variables) if the variables are known, for example, $a=1, b=2$ and $c=3$?
Thank you in advance!
| Another possibility is to start with a non-singular upper-left triangular matrix $T$ and the desired solution $\vec{x}=(x_1,x_2,x_3)^t$. Now one adds multiples of upper rows to lower ones to obtain a full matrix $A$, e.q.
\begin{gather*}
T=\begin{pmatrix}1& 2 & 3\cr 0& 4 & 5\cr 0 & 0 & 6\end{pmatrix},\;
\vec{x}=\begin{pmatrix}1\cr 2\cr3\end{pmatrix}\\
T_1=\mbox{3rd row}+2\cdot\mbox{2nd row}
=\begin{pmatrix}1& 2 & 3\cr 0& 4 & 5\cr 0 & 8 & 16\end{pmatrix}\\
T_2=\mbox{2nd row}+3\cdot\mbox{1st row}
=\cdots\\
T_3=\mbox{3nd row}+5\cdot\mbox{1st row}
=\begin{pmatrix}1& 2 & 3\cr 3& 10 & 14\cr 5 & 18 & 31\end{pmatrix}
\end{gather*}
If the desired solution ist $\vec{x}=(1,2,3)^t$ then $T\cdot \vec{x}$ yields the right hand side $\vec{b}=(14,\,23,\,18)^t$. Transforming $\vec{b}$ according to $T$ gives the linear system
\begin{gather*}
\begin{pmatrix}1& 2 & 3\cr 3& 10 & 14\cr 5 & 18 & 31\end{pmatrix}
\begin{pmatrix}x\cr y\cr z\end{pmatrix}
=\begin{pmatrix}14\cr 65\cr 134\end{pmatrix}
\end{gather*}
Advantage of this method is that there exists a solution for the students with 'easy' numbers.
| {
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For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$
| Note that $$\frac{1}{2} > \sqrt{n+\frac{1}{2}}-\sqrt{n} > \sqrt{n+1}- \sqrt{n+\frac{1}{2}} > 0,$$
since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$
So we can write
$$ 1> \left( \sqrt{n+\frac{1}{2}}-\sqrt{n} \right) - \left( \sqrt{n+1}- \sqrt{n+\frac{1}{2}} \right) > 0.$$
Hence $$1 > 2\sqrt{n+\frac{1}{2}} - \left( \sqrt{n+1}+\sqrt{n} \right) > 0.$$
From which the result follows.
For clarity we've shown that we can write
$$\sqrt{n+1}+\sqrt{n} + r = 2\sqrt{n+\frac{1}{2}} \quad \textrm{ for } 0 < r < 1,$$
where we note that we do not straddle an integer since $\sqrt{4n+1} < \sqrt{n} + \sqrt{n+1},$ and there are no integers between $\sqrt{4n+1}$ and $\sqrt{4n+2}.$
| {
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mystery regarding power series of $\frac{1}{\sqrt{1+x^{x}}}$ In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$
which is
\begin{align*}
\frac{1}{\sqrt{1+x^{x}}} =& \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}} -\frac{x^{2}\log^{2}(x)}{32\sqrt{2}}+ \frac{5x^{3}\log^{3}(x)}{384\sqrt{2}}\\\
\\\
&+ \frac{17x^{4}\log^{4}(x)}{6144\sqrt{2}} - \frac{121x^{5}\log^{5}(x)}{122880\sqrt{2}} - \frac{721x^{6}\log^{6}(x)}{2949120\sqrt{2}} \ldots
\end{align*}
Before I realized that I couldn't really use this to help me with the sum, I found that the denominators (ignoring the $\sqrt{2}$, because all of them have it in common) correspond to $4^{n}n!$, what is baffling is that the numerators appear to correspond to the coefficients in the exponential generating function for $f(x)=e^{\tanh^{-1}(\tan(x))}$ (I believe that the 7th entry should be 1369 and not 6845), and I'm curious what the explanation is, because $f(x)=e^{\tanh^{-1}(\tan(x))}$ is a mighty weird looking function.
| Well,
$$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$
where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$.
Then
$$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$
where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients
are rationals over $\sqrt2$.
Now consider $f(x)=\exp(\tanh^{-1}(\tan x))$. Now
$$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$
so that
$$e^z=\sqrt{\frac{1+\tanh z}{1-\tanh z}}.$$
Hence,
$$\exp(\tanh^{-1} t)=\sqrt{\frac{1+t}{1-t}}.$$
Putting $t=\tan x$ gives
$$f(x)=\sqrt{\frac{1+\tan x}{1-\tan x}}=\sum_{n=0}^\infty a_n x^n.$$
Then
\begin{eqnarray*}
&&(1+i)f(ix)+(1-i)f(-ix)\\
&=&
2\sum_{m=0}^\infty
(a_{4m} x^{4m}-a_{4m+1} x^{4m+1}-a_{4m+2} x^{4m+2}+a_{4m+3} x^{4m+3}).
\end{eqnarray*}
But
$$f(ix)=\sqrt{\frac{1+i\tanh x}{1-i\tanh x}}
=\frac{1+i\tanh x}{\sqrt{1+\tanh^2x}}$$
and so
\begin{eqnarray*}
&&(1+i)f(ix)+(1-i)f(-ix)\\
&=&
\frac{2-2\tanh x}{\sqrt{1+\tanh^2 x}}
=\frac{2\cosh x-2\sinh x}{\sqrt{\cosh^2x+\sinh^2x}}\\
&=&\frac{2e^{-x}}{\sqrt{(e^{2x}+e^{-2x})/2}}=\frac{2\sqrt2}{\sqrt{1+e^{4x}}}
\end{eqnarray*}
which explains why the coefficients in the two series are the same
up to signs and powers of 4.
| {
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If $X$ and $Y$ are not divisible by 3, then $X^2 - Y^2$ is divisible by 3 Prove: If $X$ is not divisible by 3 and $Y$ is not divisible by 3, then $X^2 - Y^2$ is divisible by 3.
I cannot find a good way to prove this.. Tried contrapositive and contradiction but cannot see the answer. Anyone have any ideas?
| If $x$ and $y$ are not divisible by 3 then they must be of the form $x = 3k \pm 1$ and $y = 3t \pm 1$. Therefore $x^2 - y^2 = 9k^2 \pm 6k + 1 -9t^2 \mp 6t -1 = 3(3k^2 - 3t^2 \pm 2k \mp 2t).$ So $x^2 - y^2$ is divisible by 3.
| {
"language": "en",
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"source": "stackexchange",
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$2^{k} \mid {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}$ I have to prove this question:
$$2^{k} \,\left|\, {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}\right.$$
What i did was to get a closed form for this summation, and i haven't been able to get one. Any ideas, or is there any other method for proving this.
| Let's talk about some general methods. This is a special case of the following problem: given that you understand a sequence $a_n$, how can you understand the sequence $b_n = \sum_{k=0}^n {n \choose k} a_k$? Generating function techniques will come to your rescue. If you let $A(x) = \sum_{n \ge 0} a_n x^n$, then
$$B(x) = \sum_{n \ge 0} b_n x^n = \sum_{n \ge 0} \sum_{k=0}^{n} {n \choose k} a_k x^n.$$
At this point we employ one of the simplest but most effective tools in generatingfunctionology, namely we exchange the order of summation. This gives
$$B(x) = \sum_{k \ge 0} \sum_{n \ge k} {n \choose k} a_k x^n.$$
Since we want an answer in terms of $A$, it seems natural to factor $a_k x^k$ out of the inner sum. This gives
$$B(x) = \sum_{k \ge 0} a_k x^k \sum_{n \ge k} {n \choose k} x^{n-k}.$$
Now a generatingfunctionologist will immediately recognize the inner sum as $\sum_{m \ge 0} {m+k \choose k} x^m = \frac{1}{(1 - x)^{k+1}}$. This gives
$$B(x) = \sum_{k \ge 0} a^k \frac{x^k}{(1 - x)^{k+1}} = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$
Beautiful. Now, in this problem we are dealing with the sequence $a_{2k} = 3^k, a_{2k+1} = 0$. This has generating function $A(x) = \frac{1}{1 - 3x^2}$, hence
$$B(x) = \frac{1}{1 - x} \left( \frac{1}{1 - \frac{3x^2}{(1 - x)^2} } \right) = \frac{1 - x}{(1 - x)^2 - 3x^2} = \frac{1 - x}{1 - 2x - 2x^2}.$$
Now, what we want to prove is that the even coefficients of $B \left( \frac{x}{\sqrt{2}} \right)$ are integral. To extract the even coefficients we appeal to another general tool in generatingfunctionology: if $B(x) = \sum_{n \ge 0} b_n x^n$ is a generating function you want to extract the even coefficients from, then these are given by
$$\sum_{n \ge 0} b_{2n} x^{2n} = \frac{B(x) + B(-x)}{2}.$$
This is a special case of the discrete Fourier transform. Once you know this, it is completely mechanical to verify that
$$\frac{B \left( \frac{x}{\sqrt{2}} \right) + B \left( \frac{-x}{\sqrt{2}} \right)}{2} = \frac{1 - x^2}{1 - 4x^2 + x^4}$$
which has integer coefficients (since the constant term of the denominator is $1$). This is a bit tedious, but my point here is that once one has picked up a few generating function techniques most of the work in solving problems like these is automatic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$
$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$
$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$
$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$
$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$
Remark:
This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$
| {
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Integral of $\frac1{\cos x}$ using t substitution Okay, so I'm trying to find $ \int \frac1{\cos x}\mathrm{d}x$ using the substitution $t = \tan\left(\frac{x}{2}\right)$.
I sub in the trig identity for $\sec$ as $\frac{1+t^2}{1-t^2}$ and then rearrange and substitute $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac12 \left(1+ \tan^2\left(\frac{x}{2}\right)\right)$ so I am left with
$\frac2{1-t^2}$
I then used partial fractions to find
$\frac2{1-t^2} = \frac1{1+t} + \frac1{1-t}$
and therefore integrating I get
$\int \frac1{\cos x}dx = \ln(t+1) - \ln(t-1)$
But subbing in $t = \tan\left(\frac{x}{2}\right)$ doesn't seem to get me anywhere close to the solution that I want to find, which is:
$ \int \frac1{\cos x}\mathrm{d}x = \ln(\sec x + \tan x) + C$
Any help on this would be greatly appreciated.
Thanks!
| We know that $$\cos x=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2} }=\frac{1-\tan^{2}\frac{x}{2}}{\sec^2 \frac{x}{2} }$$ Given that $$\int \frac{1}{\cos x}dx$$$$=\int \frac{1}{\frac{1-\tan^{2}\frac{x}{2}}{\sec^2 \frac{x}{2} }}dx$$ $$=\int \frac{\sec^2 \frac{x}{2}}{1- \tan^{2}\frac{x}{2}}dx$$ Let $$\tan \frac{x}{2}=t \implies \frac{1}{2}\sec^2\frac{x}{2}dx=dt \implies \sec^2\frac{x}{2}dx=2dt $$ Now , we have $$\int \frac{\sec^2 \frac{x}{2}}{1- \tan^{2}\frac{x}{2}}dx=\int \frac{2dt}{1- t^2}$$$$=\frac{2}{2}\ln\left|\frac{1+t}{1-t}\right|$$
$$=\ln\left|\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right|$$
$$=\ln\left|\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right|$$
$$=\ln\left|\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}\right|$$
$$=\ln\left|\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}\right|$$
$$=\ln\left|\frac{1+\sin x}{\cos x}\right|$$
$$\color{blue}{=\ln\left|\sec x+\tan x\right|+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/8564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
| $$\int \frac{1}{\sin x\cos x} dx = \int( \tan x+\cot x)\, dx= \int\frac{\sin x}{\cos x}\, dx+\int\frac{\cos x}{\sin x} \,dx$$ from where it should be fairly simple....(choose correctly the variables though)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/9075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 9
} |
What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| OK, first rewrite it as
$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$
Multiply both sides by $2 \cos(A/2)$
$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$
Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:
$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$
Using Simpson's rule on the second hand of the equation to sum the sines gives:
$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$
Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus
$$ \cos(A/2) = \cos((B-C)/2) $$
Or $A+C=B$ (If $B>C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/10545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this
| The quadratic formula:
$$ x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1} $$
4 + 96 in the square root.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Combinatorial proof of a binomial coefficient summation: $\sum_{k=1}^n \binom nk \binom n{k-1} = \frac12\binom{2n+2}{n+1} - \binom{2n}n$ Let $n$ and $k$ be integers with $1 \leq k \leq n$. Show that:
$$\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac12{2n+2 \choose n+1} - {2n \choose n}$$
I was told this is supposed to use a combinatorial proof and while I'm not that comfortable with that, many similar proofs use a mathematical proof that shows equality. Any guidance would be much appreciated.
| Using standard binomial identities, this can be proven as follows:
$$
\begin{align}
\sum_{k=1}^n \binom{n}{k}\binom{n}{k-1}&=\sum_{k=1}^n \binom{n}{n-k}\binom{n}{k-1}\\
&=\binom{2n}{n-1}\\
&=\binom{2n+1}{n}-\binom{2n}{n}\\
&=\frac{n+1}{2n+2}\binom{2n+2}{n+1}-\binom{2n}{n}\\
&=\frac{1}{2}\binom{2n+2}{n+1}-\binom{2n}{n}
\end{align}
$$
Each identity can be given a simple combinatorial justification.
1.
$$
\binom{n}{k}=\binom{n}{n-k}
$$
The number of ways to choose $k$ items from $n$ is the number of ways to choose the complement of a set of $k$ items from $n$.
2.
$$
\sum_{k=1}^n \binom{n}{n-k}\binom{n}{k-1}=\binom{2n}{n-1}
$$
Given a set of green marbles, numbered $1...n$ and red marbles numbered $1...n$, count the choices of $n-1$ marbles from the aggregate $2n$ by counting the choices of $n-k$ green marbles and $k-1$ red marbles.
3.
$$
\binom{2n}{n-1}+\binom{2n}{n}=\binom{2n+1}{n}
$$
Given a set of green marbles numbered $1...2n$ and $1$ red marble, a choice $n$ marbles from the aggregate $2n+1$ either has the red marble, and thus $n-1$ of the $2n$ green marbles, or doesn't have the red marble, and thus $n$ of the $2n$ green marbles.
4.
$$
(2n+2)\binom{2n+1}{n}=(n+1)\binom{2n+2}{n+1}
$$
Putting $n+1$ marbles from a bag of $2n+2$ into a box and then choosing $1$ from the box is the same as choosing $1$ from the bag of $2n+2$ marbles and then putting $n$ of the remaining $2n+1$ marbles into the box.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Solving systems of linear equations over a finite ring I want to solve equations like this (mod $2^n$):
$$\begin{array}{rcrcrcr} 3x&+&4y&+&13z&=&3&\pmod{16} \\ x&+&5y&+&3z&=&5&\pmod{16} \\ 4x&+&7y&+&11z&=&12&\pmod{16}\end{array}$$
Since we are working over a ring and not a field, Gaussian elimination doesn't work. So how can I still solve these types of equations?
| You can still use Gaussian elimination as long as you don't "divide" by things that are not relatively prime to the modulus. In this case, you can "divide" by any odd number, and perform all the usual computations. In this case, you can perform Gaussian pretty well:
\begin{align*}
\left(\begin{array}{ccc|c}
3 & 4 & 13 & 3\\
1 & 5 & 3 & 5\\
4 & 7 & 11 & 12
\end{array}\right) &\rightarrow
\left(\begin{array}{ccc|c}
1 & 5 & 3 & 5\\
3 & 4 & 13 & 3\\
4 & 8 & 11 & 12
\end{array}\right) && \rightarrow
\left(\begin{array}{ccr|c}
1 & 5 & 3 & 5\\
0 & 5 & 4 & 4\\
0 & 4 & -1 & 8
\end{array}\right)\\
&\rightarrow
\left(\begin{array}{ccr|r}
1 & 5 & 3 & 5\\
0 & 1 & 5 & -4\\
0 & 4 & -1 & 8
\end{array}\right)
&&\rightarrow
\left(\begin{array}{ccr|r}
1 & 5 & 3 & 5\\
0 & 1 & 5 & -4\\
0 & 0 & 11 & 8
\end{array}\right).
\end{align*}
So here you get that $11z\equiv 8 \pmod{16}$. Since $11^{-1} \equiv 3\pmod{16}$, this means $z \equiv 24 \equiv 8\pmod{16}$. Then you can backsubstitute and solve. (Assuming I didn't make any mistakes with my modular arithmetic, anyway...)
If you are unlucky enough to get a congruence in which all the coefficients are even, then you can divide through by $2$ and get a congruence modulo $8$ (instead of $16$); that will lead to two solutions modulo $16$ (if you get $x\equiv 4 \pmod{8}$, that means $x\equiv 4 \pmod{16}$ or $x\equiv 8+4=12\pmod{16}$, for instance).
Basically, so long as you are careful, you can certainly do Gaussian elimination. You can even do it over more general rings, through in that case you have to other restrictions on what you can or cannot conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
Partial fraction integration So I want to find all antiderivaties of
$\frac{x}{x^3-1}$
Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.
I've started by doing:
$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$
Hence,
$x = A(x-1)+B(x^2+x+1)$
Then setting $x = 1$ to get $3B = 1 => B = \frac{1}{3}$
So I have $\int({\frac{A}{x^2+x+1}}+{\frac{1}{3(x-1)}})$
However, I'm not sure how to continue from here.
| You can do partial fractions in this way: $$\frac{x}{x^{3}-1} = \frac{Ax+B}{x^{2}+x+1} + \frac{C}{x-1}$$
Now by equating the coefficients you get:
*
*$A+C=0$
*$B+C-A=1$ and;
*$C-B=0$.
Solving you get the values of $A= -\frac{1}{3}$, $B=\frac{1}{3}$ and $C=\frac{1}{3}$.
So, the required integral breaks down as $$ \int\frac{x}{x^{3}-1} \ dx = -\frac{1}{3}\int \frac{x -1}{x^{2}+x+1} \ dx + \frac{1}{3} \int \frac{1}{x-1} \ dx$$
Now $$\int\frac{x-1}{x^{2}+x+1} \ dx = \frac{1}{2} \int \frac{2x-2}{x^{2}+x+1} \ dx =\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1} -\frac{1}{2} \int \frac{3}{x^{2}+x+1} \ dx$$
The first part of the integral is easy. For the second part use $$x^{2}+x+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \Bigl(\frac{\sqrt{3}}{2}\Bigr)^{2}$$ and then put $t=x+\frac{1}{2}$ and then the trignometric substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
| We have
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}.$$
So you want to know how much is $\sin x\cos x$.
Squaring the equality you know, we have:
\begin{align*}
(\sin x + \cos x)^2 &= \left(\frac{\sqrt{3}+1}{2}\right)^2\\
\sin^2 x + 2\sin x\cos x + \cos^2 x & = \frac{4 + 2\sqrt{3}}{4}\\
1 + 2\sin x \cos x &= 1 + \frac{\sqrt{3}}{2}\\
2\sin x \cos x &= \frac{\sqrt{3}}{2}\\
\sin x \cos x &= \frac{\sqrt{3}}{4}\\
\tan x + \cot x = \frac{1}{\sin x\cos x} &= \frac{4\sqrt{3}}{3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/19131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$ I need some help understanding why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$.
I would be grateful if someone could explain. Be explicit.
Thank you!
| $$\frac{x}{x+1} = \frac{x \cdot 1}{x \cdot (1+\frac{1}{x})}= \frac{x}{x} \cdot \frac{1}{1+\frac{1}{x}} = 1 \cdot \frac{1}{1+\frac{1}{x}} = \frac{1}{1+x^{-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/22685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$n^3 \equiv n^5 \pmod{12} $? I am proving that
$$5n^3 + 7n^5 \equiv 0 \pmod{12}$$
It would suffice to show
$$n^3 \equiv n^5 \pmod{12}$$
How would I go about doing that?
I suppose I could just go through each $n \equiv r \pmod{12}$ with $r$ from $1$ to $11$ and show that $n^3 \equiv n^5 \pmod{12}$ for each, but that would be tedious. Surely there's a better way.
| $$\begin{align}n^5-n^3&=(n-2)(n-1)n(n+1)(n+2)-4n^3+4n\\\\
&=(n-2)(n-1)n(n+1)(n+2)-4(n-1)n(n+1)\end{align}$$
The first expression is a product of 5 consecutive numbers, hence is divisible by $5!$ i.e., by 120. Similarly, the second expression is divisible by $4\cdot (3!)$ i.e., by 24.
Also, $n^5-n^3 = n^2(n-1)n(n+1)$. Clearly, a product of 3 consecutive numbers $(n-1)n(n+1)$ is divisible by 3.
Again, $n^5-n^3 = n^3(n^2-1)$.
If $n$ is even, the R.H.S. is divisible by 8.
If $n$ is odd (equals to $2m+1$, say), then $$n^2-1 = (2m+1)^2 - 1 = 8m(m+1)/2$$ i.e, is divisible by 8 (See If $n$ is an odd natural number, then $8$ divides $n^{2}-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/24100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluating $\int_{0}^{2}\frac{1}{(x+2)(x-1)}dx$ $\int_{0}^{2}\frac{1}{(x+2)(x-1)}$
Will I be able to use partial fractions on the indefinite integral and then evaluate my answer as f(2)-f(0)?
Any help is appreciated as always. :)
EDIT: I looked this up and apparently the integral doesn't converge so now I need help finding where I went wrong.
I had solved for 1 = $\frac{A}{x+2} + \frac{B}{x-1}$
sub in x = 1 and $B=\frac{1}{3}$
sub in x=-2 and $A=\frac{-1}{3}$
Giving:
$\frac{-1}{3}\int\frac{dx}{x+2} + \frac{1}{3}\int\frac{dx}{x-1}$
Simple u-substitution gives:
$\frac{-1}{3}ln|x+2|$ + $\frac{1}{3}ln|x-1| + C $
Which I was going to then evaluate f(2)-f(0) but somewhere I went wrong.
| Hint 1:
$$\frac{1}{(x+2)(x-1)}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$$
Hint 2: Does this converge or diverge? Think about $$\int_{-1}^{1}\frac{1}{x}dx.$$ Does the value of this integral even make sense?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?
Thanks,
| Any integer is going to be congruent to one of $0,1,2,3,4$ modulo $5$. As the testing shows, only those integers which are congruent to $1$ modulo $5$ are solutions, and any integer congruent to $1$ mod $5$ is a solution.
Those tests are using an arbitrary $x$ each time, so you shouldn't think of them as a system of congruences of a single variable $x$ but instead they imply the result described above.
Here's another thing to add which may help. Don't just check $0,1,2,3,4$. Take $x$ to be any arbitrary integer, not necessarily one of those just listed. So like I said before, this arbitrary $x$ is congruent to one of $0,1,2,3,4$ since they form a complete residue system. If $x\equiv 0\pmod{5}$, then
$$
2x^3+7x-4\equiv 2(0)^3+7(0)-4\equiv -4\equiv 1\pmod{5}
$$
so $x$ is not a solution. Likewise if $x\equiv 2,3,4\pmod{5}$. So the only choice which works is when $x\equiv 1\pmod{5}$. Hopefully that's clearer?
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the first derivative of $y= \sin^2(\sqrt{x})$? I'm trying to find out what the graph of the function $y= \sin^2(\sqrt{x})$ looks like. Can someone please help me?
| The graph of $ y = \sin^2(\sqrt{ x }) $ on $x \in [0,15]$ from MATLAB is:
The first derivative of $\displaystyle y = \sin^2(\sqrt{ x })$ is $\displaystyle y^{\prime} = 2 \sin (\sqrt{ x }) \cos( \sqrt{x} ) \frac{1}{2}x^{ - \frac{1}{2} } $.
Simplifying gives $\displaystyle y^{\prime} = \frac{ \sin{ \sqrt{x} } \cos{ \sqrt{ x } } }{ \sqrt{x} }$.
Which means the critical points are (when the numerator or denominator is 0):
$ \sin{ \sqrt{x} } = 0 $
Due to the graph of $\sin(x)$:
you know that's whenever the argument (value given) to $\sin$ is an integer multiple of $\pi$:
$\displaystyle \sqrt{x} = k \pi $
$\displaystyle x = k^2 \pi^2,~~ k \in {I} $
Also when $ \cos{ \sqrt{ x } } = 0 $, which is when
$ \sqrt{x} = k \dfrac{\pi}{2} $
$ x = \dfrac{ k^2 \pi^2 }{ 4 },~~ k \in {I} $
So you can see the first critical point is when k=1, and then we have:
From the $\cos$ equation:
$ x = \dfrac{ k^2 \pi^2 }{ 4 } $
$ x = \dfrac{ \pi^2 }{ 4 } \approx 2.47 $
Which you can see is the first peak in the graph,
From the $\sin$ equation:
$ x = k^2 \pi^2 $
$ x = \pi^2 \approx 9.87 $
Which is the second peak in the graph, at about $x=10$.
Here's a larger graph for $ k \in [ 0, 100 ] $:
| {
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Infinitely many integer solutions for the equations $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$ How do you show that the equation $x^3+y^3+z^3=1$ has infinitely many solutions in integers? How about $x^3+y^3+z^3=2$?
| For $x^3+y^3+z^3=1$ it is trivial - an infinite family of solutions is $(1,n,-n)$, and permutations of that.
For $x^3+y^3+z^3=2$ I'm not so sure there are infinitely many solutions. Are you just hypothesizing this, or do you know it to be true?
| {
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Integral of floor function: $\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$ How would you go about solving integral of a floor? The particular problem I have is:
$$\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$$
| The function:
$$\left\lfloor\frac{1}{x}\right\rfloor$$
is equal to $n$ on the interval $\left(\frac{1}{n+1},\frac{1}{n}\right)$,so if we try to determine the integral from $t>0$ to $1$, we can let $n=\left\lfloor\frac{1}{t}\right\rfloor$ and we have constant value $1$ on range $(\frac{1}{2},1)$, constant value $2$ on range $(\frac{1}{3},\frac{1}{2})$, etc. So since $t<\frac{1}{n}$, we get terms for each interval $(\frac{1}{k+1},\frac{1}{k})$ when $k<n.$ The length of the $k$th interval is $\frac{1}{k(k+1)}$ and the value of the function is $k$ on this interval, so the integral on this interval is $\frac{1}{k+1}$. So the integral from $\frac{1}{n}$ to $1$ is $1/2 + 1/3 + 1/4 + ... + 1/n$. Then then integral from $t$ to $\frac{1}{n}$ is the length of the interval times $n$, which is $n(\frac{1}{n} - t) = 1-nt$. So the total is:
$$\int_t^1 \,\left\lfloor\frac{1}{x}\right\rfloor\, dx = 1 - t{\left\lfloor\frac{1}{t}\right\rfloor} +
\sum_{i=2}^{\left\lfloor\frac{1}{t}\right\rfloor}\frac{1}{i}$$
The indefinite integral, then, is the opposite of this:
$$x\left\lfloor\frac{1}{x}\right\rfloor - \sum_{i=2}^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{i} + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$ I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way.
I am interested if there is a way only using elementary methods to prove it.
Using the software Mathematica confirmed this inequality is correct.
| Since $abc = 1$, any product term such as $ab$ or $\frac{1}{bc}$ can be rewritten as a singleton term (i.e., $ab = \frac{1}{c}$ and $\frac{1}{bc} =a$. Thus the inequality is equivalent to $$\left(a-1+\frac{1}{c}\right)\left(b-1+\frac{1}{a}\right)\left(c-1+\frac{1}{b}\right) \leq 1.$$
This inequality was Problem 2 on the 2000 International Mathematical Olympiad. Knowing that, solutions should be easy to find.
For example, here's a nice one due to Robin Chapman that can be found here.
We have $$\left(b - 1 + \frac{1}{a}\right) = b\left(1 - \frac{1}{b} + \frac{1}{ab}\right) = b\left(1 + c - \frac{1}{b}\right).$$ Hence, $$\left(c - 1 + \frac{1}{b}\right)\left(b - 1 + \frac{1}{a}\right) = b\left(c^2 - \left(1 - \frac{1}{b}\right)^2\right) \leq b c^2.$$
Thus $$\left(a-1+\frac{1}{c}\right)^2\left(b-1+\frac{1}{a}\right)^2\left(c-1+\frac{1}{b}\right)^2 \leq b c^2 a b^2 c a^2 = 1,$$
proving the inequality in the case that every factor on the left side of the inequality is positive.
Now, suppose one of the factors on the left-hand side is negative; i.e., $a - 1 + \frac{1}{c} < 0$. Then $a < 1$ and $c > 1$. Thus $b-1+\frac{1}{a} > 0$ and $c-1+\frac{1}{b} > 0$. So the left-hand side of the inequality is negative, and the inequality still holds.
| {
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
| Since $(n-1)^3 + n^3 + (n+1)^3 = 3n(n^2+2)$ we know that it is divisible by $3$. If we show that $n(n^2+2)$ is also divisible by $3$ then $(n-1)^3 + n^3 + (n+1)^3$ is divisible by $3^2 = 9$. There are three cases to check:
*
*$0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$
*$1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$
*$2(2^2+2) \equiv 2 \cdot 0 \equiv 0 \pmod 3$
| {
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Finding $\alpha$ such that $\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^{\circ})-2}$ I am preparing for my college entry test and I ran into this problem in my book:
$$\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^{\circ})-2}$$
I should find the angle $\alpha$. Can anyone help me please? I tried solving it, but I just can't get any solution. By the way, the correct answer is $\alpha = 42^{\circ}$
Thanks!
| $$\begin{align*}
V&=\frac{\tan a\tan b+\tan a+\tan b-1}{\tan a\tan b-\tan a-\tan b-1}\\
&=\frac{\sin a \cos b + \cos a \sin b + \sin a\sin b - \cos a\cos b}{\sin a\sin b-\cos a\cos b-\sin a\cos b-\cos a\sin b}\\
&=\frac{\sin (a+b)-\cos(a+b)}{-\cos(a+b)-\sin(a+b)}
\end{align*}$$
Now I will use:
$$\sin x-\cos y = \sin x-\sin(90^\circ-y)=2\sin\frac{x+y-90^\circ}2\cos\frac{90^\circ+x-y}2,$$ which yields for $y=x$
$$\sin x-\cos x = -2\cos45^\circ\sin(45^\circ-x)$$
and
$$\cos x +\sin y = \sin(90^\circ-x)+\sin y = 2\cos\frac{90^\circ-x-y}2\sin\frac{90^\circ-x+y}2$$
which yields for $y=x$
$$\cos x+\sin y = 2\sin45^\circ\cos(45^\circ-x)$$
Plugging this into the above formula (for $x=a+b$) I get
$$V=\frac{-2\cos45^\circ\sin(45^\circ-a-b)}{-2\sin45^\circ\cos(45^\circ-a-b)}=\tan(45^\circ-a-b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the next section. (Simmons' Calculus)
The hint given with the problem is to integrate another series. This series must be $$1-\frac{ 1}{2} x^2+\frac{ 1\cdot 3}{2\cdot 4} x^4-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^6+\cdots$$
But where does it come from? [How would I know to integrate that series, if it weren't for the hint?]
The derivative of $\ln(x+\sqrt{1+x^2})$ is $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$,
but I can't see how to get from $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$ to that series.
The derivative can be written as $\displaystyle \frac{ 1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, but that looks even worse.
| You can use the fact, as you write, that the derivative of such function is $$y' = \frac{1}{\sqrt{1+x^2}}$$
But from the binomial theorem you have that
$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {-\frac{1}{2} \choose k} x^{2k}$$
You can check by yourself that
$${-\frac{1}{2} \choose k} ={\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$$
using induction over the following expression
$$\frac{1}{{k!}}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)\left( { - \frac{5}{2}} \right) \cdots \left( {\frac{1}{2} - k} \right)$$
Where I simply replaced the value of $n$ in $$\frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{k!}}$$
As a consequence you get
$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} x^{2k}$$
Upon integration you have
$$ \log\left({x+\sqrt{1+x^2}}\right) = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} \frac{x^{2k+1}}{2k+1} = x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$
Remember that by definition $0!! = 1$ and $(-1)!! = 1$
| {
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Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as:
$$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$
But how is this possible?
| André's approach is pretty simple. In case you really must follow the hint, here's how to get it: note, for example, that in the first fraction,
$$\frac{ab+ca}{a^2+ab+ca}$$
the only thing that is missing from having the numerator exactly equal to the denominator is an $a^2$; so adding it and taking it away again we get
$$\frac{ab+ca}{a^2+ba+ca} = \frac{ab+ca+a^2-a^2}{a^2+ba+ca} = \frac{a^2+ab+ca}{a^2+ab+ca} - \frac{a^2}{a^2+ba+ca} = 1-\frac{a^2}{a^2+ba+ca}.$$
The same thing can be done with
$$\frac{ab+cb}{b^2+ab+bc}$$
by adding and subtracting $b^2$ to the numerator. And also with
$$\frac{ac+cb}{c^2+ac+bc}$$
by adding and subtracting $c^2$ to the numerator. Doing all three, we have:
$$\begin{align*}
\frac{ab+ca}{a^2+ba+ca} &+ \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc} \\&= \left(1 - \frac{a^2}{a^2+ba+ca}a\right)\\
&\qquad\quad\mathop{+}\left(1 - \frac{b^2}{b^2+ab+bc}\right)\\
&\qquad\quad\mathop{+}\left(1 - \frac{c^2}{c^2+ac+bc}\right)\\
&= 3 - \frac{a^2}{a^2+ba+ca}a - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc},
\end{align*}$$
which is exactly the formula your hint gives.
As to how the hint gives the answer any better than the straightforward method suggested by André (which would be my own approach, to simplify first), I couldn't tell you, but that is how the equation can be rewritten in that form.
| {
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How to prove that $a_n = \frac{2n+1}{n+1}$ is bounded? How do I prove that $\displaystyle a_n = \frac{2n+1}{n+1}$ is bounded?
According to the key it works as follows. However, I don't really understand the meaning.
Upper bound:
$\displaystyle a_n = \frac{2n+1}{n+1} < \frac{2n+2}{n+1} = \frac{2(n+1)}{n+1} = 2$
Lower bound:
$\displaystyle a_n = \frac{2n+1}{n+1} > 0$
| For the upper bound part you could also reason as follows:
$$\frac{2n+1}{n+1}=\frac{n+n+1}{n+1}=\frac{n}{n+1}+\frac{n+1}{n+1}=\frac{n}{n+1}+1$$
Since $\dfrac n {n+1}$ comes as monotonically increasing on the natural numbers, and the limit of $\dfrac n {n+1}$ as $n$ increases without bound equals 1, it follows that the limit of $\dfrac{2n+1}{n+1}$ as $n$ increases without bound equals $2$. Since this limit happens for a monotonically increasing function, it qualifies as a least upper bound. So, we have an upper bound.
One could also precede as follows:
*
*$0<1$
*$(2n+1)+0<(2n+1)+1$ by monotonicity of $+$
*$2n+1<(2n+1)+1$ by $0$ as the neutral for $+$
*$2n+1<2n+(1+1)$ by association of $+$
*$2n+1<2n+2$ by $1+1=2$
*$2n+1<2(n+1)$ by distribution of multiplication over addition, and $2*1=2$
*$\dfrac{2n+1}{n+1}<2$ several steps here actually, and it only works since $n+1>0$ by definition.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
| Edit: The main idea is to use the expansion
$$\sqrt[6]{1+x^{-1}}=1+\frac{1}{6}x^{-1}+O\left( x^{-2}\right) $$
which can be generated by changing variables $x=1/z$, expanding about $z=0$, $$\sqrt[6]{1+z}=1+\frac{1}{6}z+O\left( z^{2}\right) $$ and restoring $x$ via $%
z\rightarrow 1/x$.
Let $f(x)=\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$. If $\lim_{x\rightarrow +\infty }(x)=L$, then $\lim_{x\rightarrow -\infty }f(x)=-L$ because $f(-x)=-f(x)$. We have
$$f(x) =\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}=\left\vert x\right\vert \left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right). $$
Since
$$\sqrt[6]{1+x^{-1}} =1+\frac{1}{6}x^{-1}+O\left( x^{-2}\right)\quad\text{ and }\quad\sqrt[6]{1-x^{-1}} =1-\frac{1}{6}x^{-1}+O\left( x^{-2}\right),$$
we get
$$\sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}=\frac{1}{3}x^{-1}+O\left(
x^{-2}\right)$$
and for $x>0$
$$x\left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right) =\frac{1}{3}+O\left(
x^{-1}\right).$$
Therefore
$$L=\lim_{x\rightarrow +\infty }f(x) =\lim_{x\rightarrow +\infty }\left\vert x\right\vert \left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right)=\lim_{x\rightarrow +\infty }\frac{1}{3}+O\left( x^{-1}\right) =\frac{1}{3}.$$
Edit 2: graph of $f(x)$
| {
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about $\operatorname{SO}(2)$ group When
$$
A =\begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad a,b,c,d \in \mathbb{R}, \quad A^2 -2aA + I = 0
$$
Is $ A \in \operatorname{SO}(2)\; $ if $A^n \in SO(2)\;$ for some $n \in \Bbb N$?
$\operatorname{SO}(2)$ is special orthogonal group.
| The answer is yes I believe. Your requirement that $A^2-2aA+I=0$ forces $\det(A)=1$, so we know that if $A^n \in SO(2)$, then $\det(A^n)=\det(A)=1$, so we've eliminated the problem that Qiaochu mentioned. Now, it can be shown using a little bit of algebra that any matrix $B$ in $SO(2)$ is of the form $$\begin{pmatrix} \sin x & -\cos x \\ \cos x & \sin x \end{pmatrix}\ $$ Looking at a few powers of this matrix, we see that $$ B^2= \begin{pmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{pmatrix}\ \text{and } B^3= \begin{pmatrix} -\sin 3x & \cos 3x \\ -\cos 3x & -\sin 3x \end{pmatrix}\ $$ Thus, if we have $A^n \in SO(2)$, then if we finagle the signs and possibly switch $\sin $ and $\cos $ in a very natural way, it seems like $A$ will also be of the proper form with angle $\frac{x}{n}$, implying $A \in SO(2)$.
This isn't a rigorous proof, but I believe it at least gives the method of it.
Edit: As Shiyu pointed out, if $B$ is instead written in the form $$ \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}\ $$ then $$ B^k = \begin{pmatrix} \cos kx & -\sin kx \\ \sin kx & \cos kx \end{pmatrix}\ $$ so if $A^n \in SO(2)$ then $$ A^n = \begin{pmatrix} \cos y & -\sin y \\ \sin y & \cos y \end{pmatrix}\ \implies A= \begin{pmatrix} \cos \frac{y}{n} & -\sin \frac{y}{n} \\ \sin \frac{y}{n} & \cos \frac{y}{n} \end{pmatrix}\ $$ so $A \in SO(2)$
| {
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$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$ Prove :
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$
For $a,b,c,d>0$
| Since for $x>0$ we have $$\left(\frac{x^2+1}{x+1}\right)^3-\frac{x^3+1}{2}=\frac{(x-1)^4(x^2+x+1)}{2(x+1)^3}\geq0$$
and by P-M $$\left(\frac{x+1}{2}\right)^3\leq\frac{x^3+1}{2},$$
we obtain:
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\frac{a^3+d^3}{2}+\frac{b^3+c^3}{2}=$$
$$=\frac{a^3+b^3}{2}+\frac{c^3+d^3}{2}\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$
and we are done!
| {
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Supremum of a Set containing terms of a sequence I am trying to prove that the sequence defined by $x_n = a_n - b_n$ converges to zero where $a_n$ and $b_n$ are sequences such that
$a_n = \sqrt{a_{n-1}b_{n-1}}$ and $ b_n = \frac{a_{n-1} + b_{n-1}}{2}$.
By definition $a_1 = \sqrt{ab}$ and $b_1 = \frac{a + b}{2}$ where $a,b$ are real numbers with $a > b > 0$.
Now I have proven that $x_n$ is strictly increasing, so I just need to show that the supremum of the set of all $x_n's$ is zero. How can I do this?
I tried using contradiction namely to show that $\forall \epsilon > 0$, $\exists n$ such that $-\epsilon < x_n < 0$, but to even produce such an $n$ is tough.
If we complete the square we find that $x_n = -\frac{1}{2}(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})^2 \leq 0.$
That's all I've got at the moment. How can I find the supremum of this?
| To show that $x_n\le0$,
$$
\begin{align}
x_n&=a_n-b_n\\
&=\frac{a_n^2-b_n^2}{a_n+b_n}\\
&=\frac{a_{n-1}b_{n-1}-\frac{1}{4}(a_{n-1}+b_{n-1})^2}{a_n+b_n}\\
&=\frac{-\frac{1}{4}(a_{n-1}-b_{n-1})^2}{a_n+b_n}\\
&\le0
\end{align}
$$
But instead of showing that $a_n-b_n\to0$, it is simpler to show that $b_n^2-a_n^2\to0$:
$$
\begin{align}
|b_n^2-a_n^2|&=\left|\frac{1}{4}(a_{n-1}+b_{n-1})^2-a_{n-1}b_{n-1}\right|\\
&=\frac{1}{4}|b_{n-1}-a_{n-1}|^2\tag{1}
\end{align}
$$
Divide $(1)$ by $|b_{n-1}^2-a_{n-1}^2|$:
$$
\begin{align}
\left|\frac{b_n^2-a_n^2}{b_{n-1}^2-a_{n-1}^2}\right|
&=\frac{1}{4}\left|\frac{(b_{n-1}-a_{n-1})^2}{b_{n-1}^2-a_{n-1}^2}\right|\\
&=\frac{1}{4}\left|\frac{b_{n-1}-a_{n-1}}{b_{n-1}+a_{n-1}}\right|\\
&\le\frac{1}{4}\tag{2}
\end{align}
$$
because $a_n,b_n\ge0$.
Therefore, $|b_n^2-a_n^2|\le4^{-n}|b_0^2-a_0^2|$.
| {
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Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
ADDED I am adding the given answer options:
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
| For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.
Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly above the main ‘diagonal’. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \le k \le p$ let $$\begin{align*}C_k &= \{\langle m,n \rangle \in S:m=k \lor n=k\}\\ &= \{\langle k,n \rangle:k+1 \le n \le q+1\} \cup \{\langle m,k \rangle:1 \le m \le k-1\};\end{align*}$$ clearly $\vert C_k \vert = q+1-k+k-1=q$, and if $1 \le i < k \le p$, $C_i \cap C_k = \{\langle i,k \rangle\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\vert S \vert$.
Columns $2$ through $p$ of the grid contain $\sum_{k=1}^{p-1} k = \binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so $$\begin{align*}\vert S \vert &= \binom{p}{2} + p(q+1-p)\\
&= \frac{p^2 - p}{2} + pq + p - p^2\\
&= pq - \frac{p^2-p}{2}\\&= pq - \binom{p}{2}.
\end{align*}$$
When $p > q+1$ the same approach works, but it’s no longer possible to get every pair of committees to overlap. First form $\left\lfloor \frac{p}{q+1}\right\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\binom{q+1}{2}=\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2}$ students. The remaining $r = p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor$ committees will then require another $rq - \binom{r}{2}$ students, for a grand total of $$\begin{align*}
\vert S \vert &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + rq - \binom{r}{2}\\ &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + \left(p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor\right)q - \binom{r}{2}\\
&= pq - \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} - \binom{r}{2}
\end{align*},$$ which does not appear to simplify greatly.
I see that this is essentially Alex’s solution, but expressed a little more concretely.
| {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| $$\ S =\sum_{k=1}^n k^3 $$
\begin{align}
k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\
k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D
\end{align}
Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and
\begin{align}
S =& \sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\
S =&\sum_{k=1}^n k(k+1)(k+2)-3\sum_{k=1}^nk(k+1)+\sum_{k=1}^nk \\
S =&\sum_{k=1}^n6\binom{k+2}{3}-3\sum_{k=1}^n2\binom{k+1}{2}+\sum_{k=1}^n\binom{k}{1}\\
S=&6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}\\
S=&\left\lgroup\frac{n(n+1)}{2}\right\rgroup^2.
\end{align}
| {
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"source": "stackexchange",
"question_score": "67",
"answer_count": 16,
"answer_id": 13
} |
Proof of the equality $\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6$ Show that for $k$ running over positive integers
$$ \sum_{k=1}^\infty \frac{k^2}{2^k}=6 .$$ We can use finite calculus.
| Solution
For convenience, we denote $$a_k=\frac{k^2+2k+3}{2^{k-1}},~~~k=1,2,\cdots$$
Then $$\frac{k^2}{2^k}=a_k-a_{k+1}.$$
Thus
\begin{align*}\sum_{k=1}^n \frac{k^2}{2^k}&=(a_1-a_2)+(a_2-a_3)+\cdots+(a_n-a_{n+1})=a_1-a_{n+1}\\&=6-\frac{(n+1)^2+2(n+1)+3}{2^n}\\&=6-\frac{n^2+4n+6}{2^n}.\end{align*}
It follows that \begin{align*}\sum_{k=1}^\infty\frac{k^2}{2^k}&=\lim_{n \to \infty}\sum_{k=1}^n \frac{k^2}{2^k}\\&=\lim_{n \to \infty}\left(6-\frac{n^2+4n+6}{2^n}\right)\\&=6-0\\&=6.\end{align*}
| {
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The square of an integer has form $3n$ or $3n+1$ Prove that if $n\in\mathbb Z$, then $n^2$ is of the form $3q$ or $3q+1$ for some $q\in\mathbb Z$
I would like to show that 3q+2 is = 3q+1 thus $n^2$ can be of the form of 3q or 3q+1.
Case one
$(3k)^2=(3k)(3k)=9k^2=3(3k^2)$ and is still of the form $3q$
When $q=3k^2$
$3q$
Case two
$(3k+1)^2= (3k+1)(3k+1)= 9k^2+6k+1$
$(9k^2+6k)+1 =3(3k^2+2k)+1$
this is of the form $3q+1$ when $q=3k^2+2k$
$3q+1$
Case three
$(3k+2)^2= (3k+2)(3k+2) =9k^2+12k+4$
$(9k^2+12k+4) = 3(3k^2+4k+1)+1$
This is of the form $3q+1$ when $q=3k^2+4k+1$
$3q+1$
using a direct proof with cases we see that when $n$ is of the form $3k$ it's in the form $3q$ after squaring. Also when n is in the form $3k+1$, $n$ squared is still in the form $3q+1$ after squaring. Lastly we saw that when $n$ was in the form $3k+2$ we could simplify to the form $3q+1$.
| Let me try a slightly different tack... Let us use the terminology of modular arithmetic, which is the same thing you wrote, but denoted differently. Let $n \equiv 0\ \mathrm{mod}\ 3$. Then $n^2 \equiv 0\ \mathrm{mod}\ 3$. Next, for $n \equiv 1\ \mathrm{mod}\ 3$, we have $n^2 \equiv 1\ \mathrm{mod}\ 3$. Lastly, for $n \equiv 2\ \mathrm{mod}\ 3$, we have $n^2 \equiv 4 \equiv 1 \ \mathrm{mod}\ 3$. Therfore, the possibility that $n^2 \equiv 2\ \mathrm{mod}\ 3$ is impossible; i.e., it cannot be of the form $3q+2$.
| {
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Properties of Fermat primes Fermat primes 17 and 257 appear a lot in the prime composition of numbers of the form $a^{2^n}+1$. For example, $11^8+1$ is divisible by 17 and $11^{32}+1$ is divisible by 257.
I have verified the following statement for 3, 5, 17 and 257 but can't prove it. I would be grateful if someone could provide a hint.
Let $F = 2^{2^k} + 1$ be a Fermat prime.
Then for $n=0, 1, ..., 2^k-1, x^{2^n} + 1 \equiv 0 \mod F$ has exactly $2^n$ solutions.
For example, when k=2 we have:
$(n=0) \hspace{10pt} x^1 + 1 \equiv 0 \mod 17$ has one solution (x=16).
$(n=1) \hspace{10pt} x^2 + 1 \equiv 0 \mod 17$ has two solutions (x=4 and 13).
$(n=2) \hspace{10pt} x^4 + 1 \equiv 0 \mod 17$ has four solutions (x=2, 8, 9 and 15).
$(n=3) \hspace{10pt} x^8 + 1 \equiv 0 \mod 17$ has eight solutions (x=3, 5, 6, 7, 10, 11, 12 and 14).
| Obviously $x^1+1\equiv 0 \pmod p$ has exactly one solution.
For each $n$ it must hold that $x^{2^{n+1}}+1\equiv 0 \pmod p$ has at most twice as many solutions as $x^{2^n}+1\equiv 0\pmod p$ -- otherwise one of the $2^n$th roots must have more than two square roots in $\mathbb F_p$, which is is impossible.
Fermat's little theorem ensures that $x^{2^{2^k}}\equiv 1\pmod p$ for all nonzero $x$. But this means that $x^{2^{(2^k-1)}}$ must always be a square root of $1$ modulo $p$, that is, either $1$ or $-1$. An analogous argument to the one above shows that there cannot be more than $2^{(2^k-1)}$ instances of either $1$ or $-1$, so that must be the number of solutions to $x^{2^{(2^k-1)}}+1\equiv 0\pmod p$. But that determines the number of $2^n$th roots of $-1$ uniquely for $1\le n\le 2^k-1$.
In general, this reasoning means that $x^{2^n}\equiv a \pmod p$ with $a\not\equiv 0$ always has either exactly $2^n$ solutions or none at all, depending on how many times one can square $a$ modulo $p$ without reaching $1$.
| {
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Is there a more efficient method of trig mastery than rote memorization? I would like to get alot better at trig than I am. What is the best/most efficient method?
Thanks much in advance
Joe
| I would emphasize how to derive trigonometric identities from a few ones. Learn:
*
*how to derive the relations between the direct functions of the same angle from the definition of the trigonometric functions and the Pythagorean formula;
*maxima, minima, zeroes and period of each function;
*if a function is odd or even;
*the trigonometric functions of 0º, 30º, 45º, 60º and 90º;
*the relations between functions of symmetric, complementary and supplementary angles;
*the relations between functions of angles whose difference is 180º;
*the relations between functions of angles whose sum is 360º;
*the inverse trigonometric functions;
*the addition formulas of sin and cos;
*how to derive the subtraction formulas of sin and cos;
*how to derive the addition and subtraction formulas of tan and cot;
*how to derive the double and half angle formulas;
*how to derive the sum to product formulas;
*how to solve some elementary trigonometric equations;
*(triangle) sin and cos laws;
*Heron's formula;
*derivatives of direct and inverse trigonometric functions.
Added. Examples. From
$$\sin (\alpha +\beta )=\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin
\beta ,\tag{A}$$
if we set $\alpha =\beta =a$, we get
$$\sin 2a=2\sin a\cdot \cos a.\tag{1}$$
And from
$$\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin
\beta \tag{B}$$
for $\alpha =\beta =a$, we have
$$\cos 2a=\cos ^{2}a-\sin ^{2}a.\tag{2}$$
Using the Pythagorean identity
$$\cos ^{2}a+\sin ^{2}a=1,\tag{C}$$
if $\cos a\neq 0$, then
$$\begin{eqnarray*}
\sin 2a &=&2\sin a\cdot \cos a=2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a+\sin
^{2}a} \\
&=&\dfrac{2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin ^{2}a
}{\cos ^{2}a}}=\dfrac{2\dfrac{\sin a}{\cos a}}{1+\dfrac{\sin ^{2}a}{\cos ^{2}a}}
\\
&=&\dfrac{2\tan a}{1+\tan ^{2}a}.
\end{eqnarray*}\tag{3}$$
Similarly
$$\begin{eqnarray*}
\cos 2a &=&\cos ^{2}a-\sin ^{2}a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos
^{2}a+\sin ^{2}a} \\
&=&\dfrac{\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin
^{2}a}{\cos ^{2}a}}=\dfrac{1-\dfrac{\sin ^{2}a}{\cos ^{2}a}}{1+\dfrac{\sin ^{2}a
}{\cos ^{2}a}} \\
&=&\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.
\end{eqnarray*}\tag{4}$$
Then
$$\tan 2a=\dfrac{\sin 2a}{\cos 2a}=\dfrac{\dfrac{2\tan a}{1+\tan ^{2}a}}{\dfrac{
1-\tan ^{2}a}{1+\tan ^{2}a}}=\dfrac{2\tan a}{1-\tan ^{2}a}.\tag{5}$$
Added 2. The linear equation in $\sin x$ and $\cos x$
$$
A\sin x+B\cos x=C\tag{6}
$$
can be solved by a resolvent quadratic equation in $\tan \frac{x}{2}$, by
writting the $\sin x$ and the $\cos x$ functions in terms of $\tan \frac{x}{2
}$ (set $x=2a$ in $(3)$ and $(4)$):
$$
\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}},\tag{7}
$$
$$
\cos x=\dfrac{1-\tan ^{2}\dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}.\tag{9}
$$
The equation $(6)$ is equivalent to
$$\begin{eqnarray*}
A\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}+B\frac{1-\tan ^{2}\dfrac{x}{
2}}{1+\tan ^{2}\dfrac{x}{2}} &=&C, \\
2A\tan \dfrac{x}{2}+B-B\tan ^{2}\dfrac{x}{2} &=&C+C\tan ^{2}\dfrac{x}{2}, \\
\left( B+C\right) \tan ^{2}\dfrac{x}{2}-2A\tan \dfrac{x}{2}+C-B &=&0.
\end{eqnarray*}\tag{10}$$
| {
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How can I differentiate this? I was trying to differentiate this: $\frac{1}{2} \sin^{-1} \frac{2x}{1+x^2}$ but I am really stuck.I obtained an answer that does not match with the one given in the book.Your help is appreciated.[Edit: I used the formula $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ to get $\frac{1+x^2}{2(1-x^2)}$]
Thank you.
| $y=\sin(x)\Rightarrow\frac{\mathrm{d}y}{\mathrm{d}x}=\cos(x)=\sqrt{1-y^2}\therefore\frac{1}{\sqrt{1-y^2}}=\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\mathrm{d}}{\mathrm{d}y}\sin^{-1}(y)$
The rest is the chain-rule:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2}\sin^{-1}\left(\frac{2x}{1+x^2}\right)
&=\frac{1}{2}\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\frac{\mathrm{d}}{\mathrm{d}x}\frac{2x}{1+x^2}\\
&=\frac{1}{2}\frac{1+x^2}{|1-x^2|}\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\\
&=\frac{1}{2}\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}\\
&=\frac{\operatorname{sgn}(1-x^2)}{1+x^2}\tag{1}
\end{align}
$$
Let $x=\tan(\theta)$, then $\frac{2x}{1+x^2}=\sin(2\theta)$. When $|\theta|<\frac{\pi}{4}$, $|x|<1$, so $\operatorname{sgn}(1-x^2)=1$. Then, equation $(1)$ is the reciprocal of
$$
\frac{\mathrm{d}}{\mathrm{d}\theta}\tan(\theta)=\sec^2(\theta)
$$
When $|\theta|\in(\frac{\pi}{4},\frac{\pi}{2})$, $|x|>1$, so $\operatorname{sgn}(1-x^2)=-1$, but $\sin^{-1}(\sin(2\theta))=\pi\operatorname{sgn}(\theta)-2\theta$. Then, equation $(1)$ is the reciprocal of
$$
\frac{\mathrm{d}}{\mathrm{d}\theta}(-\tan(\theta))=-\sec^2(\theta)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble with elementary number theory I'm trying to solve the following two problems:
Show $n^{12}\equiv 1\pmod{72}$ when $(72,n)=1$
and
Compute ${7^8}^9\pmod{100}$
For the first, I saw that $n^{24}=n^{\varphi(72)}\equiv1\pmod{72}$ by Euler's theorem, but this leaves us with $n^{12}n^{12}\equiv 1\pmod{72}$. This could mean $n^{12}\equiv 1\pmod{72}$ or $n^{12}\equiv 71\pmod{72}$. I'm not sure how to argue correctly to conclude it's only $n^{12}\equiv 1\pmod{72}$.
For the second one, I practically did brute force. ${7^8}^9\pmod{100}$ can be solved by considering $8^{9}\pmod{\varphi(100)=40}$. Since $8$ is cyclic with order $4: 8^5\equiv 8\pmod{40}$, we know $8^9=8^5\cdot8^4\equiv 8^5\equiv 8$. So look at $7^8\pmod{100}$. 7 is also cyclic with $7^4\equiv 1\pmod{100}$. So $7^8=(7^4)^2\equiv 1\pmod{100}$. I knew these "cyclic" things by manually computing and reducing, though.
Thank you for any help... (note this is not homework)
| For problem 2, you can do better than manually checking by using the Chinese Remainder Theorem.
Indeed, you first want to compute $x^9$ modulo $\phi(100)=\phi(2^2\times 5^2) = 2\times 4\times 5$. But working modulo $40$ is equivalent, by the Chinese Remainder Theorem, to working modulo $8$ and modulo $5$. The advantage is that $8^9\equiv 0 \pmod{8}$, and $8^9 \equiv 8\pmod{5}$ (since $\varphi(5)=4$, so $8^4\equiv 1\pmod{5}$). Since $8^9\equiv 8\pmod{8}$ and $8^9\equiv 8\pmod{5}$, then it follows that $8^9\equiv 8\pmod{40}$.
Then you can do the same thing with $7^8\pmod{100}$. Working modulo $4$ we have $7^8\equiv (-1)^8 \equiv 1\pmod{4}$, and working modulo $25$ we have $7^8\equiv (49)^4\equiv (-1)^4\equiv 1\pmod{25}$. So $7^8\equiv 1\pmod{100}$, and we are done.
The first problem is likewise simpler to do by working modulo $8$ and modulo $9$ separately, and then applying the Chinese Remainder Theorem.
| {
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Compute $1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
Sol:
$P(n):\ 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
For $n=n_1 = 1$
$$P(1) = \frac{1}{3}{3} = (1)^2.$$
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$
$$P(k): 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 = \frac{1}{3}k(2k-1)(2k+1).$$
We have to prove that it is true for $P(k+1)$.
$$P(k+1) = 1^1+3^2+5^2+\cdots+(2k+1)^2 = \frac{1}{3}(k+1)(2k+1)(2k+3)\tag{A}.$$
Taking LHS:
$$\begin{align*}
1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\
&= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\
&= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\
&=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\
&=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\
&=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\
&=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\
&=\frac{1}{3}(2k+1)(k+1)\left(k+\frac{3}{2}\right) \tag{B}
\end{align*}$$
EDIT:
Solving EQ (A):
$=(1/3)(2k^2+5K+3) (2K+1) \tag{C}$
Comparing EQ(B) and EQ(C)
Hence proved that it is true for $n = k+1.$
Thus the proposition is true for all $n >= 1$.
Thanks.
| Everything is OK except for the very last line. You somehow lost a factor of two. The penultimate line is already the result you want, since $2k^2+5k+3=(k+1)(2k+3)$.
| {
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Find critical points of a function: $y = x \sqrt{4 - x^2}$ I am having trouble finding the critical points of this function, I was wondering if someone could help me out.
So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points.
$$\begin{align*}
y &= x \sqrt{4 - x^2}\\
y' &= x \frac{d}{dx}\sqrt{4 - x^2} + \sqrt{4 - x^2}&& \text{(product rule)}\\
&= x\left(\frac{1}{2}(4 - x^2)^{-1/2}(-2x)\right) + \sqrt{4 - x^2} &&\text{(chain rule on square root)}\\
&= \frac{\frac{1}{2}x(-2x)}{(4-x^2)^{1/2}} + \sqrt{4-x^2}\\
&\qquad\qquad\text{(moved to denominator changed the exponent sign)}
\end{align*}$$
This is where I am stuck, I cant seem to figure out how to proceed further, any help would be appreciated
Thanks
| First, notice that the domain of your function is $[-2,2]$. We will only be working there.
Now, when you take the derivative, there are other critical points besides the stationary points: you must also exclude any point in the domain where the derivative is not defined. In this case, notice that if $x=-2$ or $x=2$, then the derivative is not defined, so $x=-2$ and $x=2$ are critical points.
To move further, make the expression into a single fraction:
$$\begin{align*}
\frac{\frac{1}{2}x(-2x)}{\sqrt{4-x^2}} + \sqrt{4-x^2} &= \frac{-x^2}{\sqrt{4-x^2}} + \frac{\sqrt{4-x^2}\sqrt{4-x^2}}{\sqrt{4-x^2}}\\
&= \frac{-x^2 + (4-x^2)}{\sqrt{4-x^2}}\\
&= \frac{4-2x^2}{\sqrt{4-x^2}}.
\end{align*}$$
Now, determine the points in the domain where the numerator is $0$ (we already excluded the points where the denominator is $0$).
| {
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Left and Right Cosets Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. I need to give a counterexample or proof of the following statement:
If $aH = bH$, then $Ha = Hb$
Proof:
For every $h \in H, ah = bh$
$
\begin{align}
ah &= bh \newline
ahh^{-1} &= bhh^{-1} \newline
a &= b \newline
ha &= hb
\end{align}
$
Could someone critic my proof?
Thanks in advance.
Edit
Look at the answer below.
| Counterexample: We want to use a non-Abelian group such as $S_{3}$
$\mu_{1} = \pmatrix{1&2&3\\1 &3 &2}$, $\mu_{2} = \pmatrix{1&2&3\\3 &2 &1}$,
$\mu_{3} = \pmatrix{1&2&3\\2 &1 &3}$
$\rho_{0} = \pmatrix{1&2&3\\1 &2 &3}$, $\rho_{1} = \pmatrix{1&2&3\\2 &3 &1}$,
$\rho_{2} = \pmatrix{1&2&3\\3 &1 &2}$
Let $H = \{\rho_{0}, \mu_{2}\}$, $a = \mu_{1}$ and $b = \rho_{1}$
$aH = \mu_{1}\{\rho_{0}, \mu_{2}\} = \{\mu_{1}, \rho_{1}\}$
$bH = \rho_{1} \{\rho_{0}, \mu_{2}\} = \{\rho_{1}, \mu_{1}\}$
But
$Ha = \{\rho_{0}, \mu_{2}\}\mu_{1} = \{\mu_{1}, \rho_{2}\}$
$Hb = \{\rho_{0}, \mu_{2}\}\rho_{1} = \{\rho_{1}, \mu_{3}\}$
So $Ha \neq Hb$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
quotient rule difficulties I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have
$$\begin{align*}
\frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac {d}{dr} \sqrt{r^2+1}} {r^2+1} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac{1}{2}(r^2+1)^{-1/2}2r} {(r^2+1)}\\\\\\\\
&= \frac {\sqrt{r^2+1} - r^2 (r^2+1)^{-1/2}} {(r^2+1)}\\\\\\\\
&= (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}
\end{align*}$$
However, the correct answer is just $$(r^2+1)^{-3/2} $$
I've been over it a number of times now, but I can't see the error. I'm pretty new to the quotient rule.
| You are all right, just only one step away from the finish line.
$$ (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2} $$
$$ = \dfrac{1}{\sqrt {r^2+1}} - \dfrac {r^2}{ (r^2+1)^{3/2}}= \dfrac{(r^2+1)}{ (r^2+1)^{3/2}} - \dfrac {r^2}{ (r^2+1)^{3/2}}=\dfrac {1}{ (r^2+1)^{3/2}}. $$
| {
"language": "en",
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"source": "stackexchange",
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Proving that $f(n)$ is an integer using mathematical induction I want to prove that
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$
is an integer for every integer $n \geq 1$.
I define P(n) to be:
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$ is an integer.
For my basis step, P(1) is true because
$$\frac{1^3}{3}+\frac{1^5}{5}+\frac{7}{15}=1$$
which is an integer.
The inductive step is what's tripping me up...
Let k be an arbitrary positive integer. Assume that P(k) is true, that is,
$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}$$
is an integer.
So based on that assumption, I need to now show that P(k+1) is true, i.e., that
$$\frac{(k+1)^3}{3} +\frac{(k+1)^5}{5} +\frac{7 (k+1)}{15}$$
is an integer.
At this point, I am stuck as to where to go next...
I have tried rewriting the assumption:
$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}=15 m$$
for some integer m. Then I solve for m:
$$\frac{1}{15} \left(\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}\right)=m$$
But this looks like a dead-end, seems there's nothing I can do with this to the "to prove" equation.
I have also tried re-writing the "to show" equation as this, but I get a dead end there and am not sure where to go next:
$$\frac{1}{15} \left(5 (k+1)^3+3 (k+1)^5+7 (k+1)\right)$$
| Here is a proof not by induction, but which may be instructive.
Using repeated differences on the first few values of $P(n)$ we get
$$
\begin{array}{llll}
0 & 1 & 10 & 59 & 228 & 669 & 1630 & \\
1 & 9 & 49 & 169 & 441 & 961 & \\
8 & 40 & 120 & 272 & 520 & \\
32 & 80 & 152 & 248 & \\
48 & 72 & 96 & \\
24 & 24 & \\
0 & \\
\end{array}
$$
Newton's interpolation formula then gives us
$$
P(n)=
0 \binom{n}{0} + 1 \binom{n}{1} + 8 \binom{n}{2} + 32 \binom{n}{3} + 48 \binom{n}{4} + 24 \binom{n}{5}
$$
This is clearly an integer for all $n\ge 0$.
In general, if a polynomial of degree $d$ and with rational coefficients takes integer values for $d+1$ consecutive integers, then it takes integers values for all integer arguments because all repeated differences are integers and so are the coefficients in Newton's interpolation formula.
| {
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"source": "stackexchange",
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Laurent series around certain points I want to compute the laurent series of
i) $(z-3)\sin(\frac{1}{z+1}); (z\ne -1)$ at $z_{0}=-1 $.
ii) $\displaystyle{\frac{z}{(z-1)(z-2)}}; ( z\ne 1,2) $ at $z_{0}=1$
and classify their singularities.
A Laurent series has the form: $\displaystyle{f(z)= \sum_{n=1}^{\infty}c_{n}\frac{1}{(z-z_{0})^{n}}}+\sum_{n=0}^{\infty} b_{n}(z-z_{0})^{n}$
my attempts :
i) $\displaystyle{\sin(z):= \sum_{n=0}^{\infty} \frac{(-1)^{n}(x+1)^{2n+1}}{(2n+1)!}}$ at $z_{0}=-1 $ so for $x=\frac{1}{z+1}$ it follows :
$\displaystyle{(z-3)(\sin(x)) = (z-3) \sum_{n=0}^{\infty} \frac{(-1)^{n}(\frac{1}{1+z}+1)^{2n+1}}{(2n+1)!}}$ then I get stuck .
ii) partial fraction decomposition leads to: $\displaystyle{\frac{2}{z-2}-\frac{1}{z-1}}$ I can write these as taylor series . It is : $\displaystyle{\frac{1}{1-x} = \sum_{n=0}^{\infty}(x-1)^{n}}$
So it follows: $\frac{2}{z-2}-\frac{1}{z-1} = 2\sum_{n=0}^{\infty}(z-4)^{n} + \sum_{n=0}^{\infty}(z-2)^{n}$
then I get stuck
What is the right path?
| You're series expansion of $\sin(z)$ isn't quite right. It should be $\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}$. Keep in mind that the expansion of $z-3$ about $z=-1$ is $z-3=-4+(z+1)$ Therefore,
$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \left(-4+(z+1)\right)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n+1}}$$
This is
$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \sum_{n=0}^\infty \frac{4(-1)^{n+1} }{(2n+1)! (z+1)^{2n+1}} + \sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n}}$$
For the second problem, you only want to expand $2/(z-2)$ not both.
$$\frac{z}{(z-1)(z-2)} = \frac{2}{z-2}-\frac{1}{z-1} = \frac{2}{(z-1)-1} - \frac{1}{z-1}$$
$$\frac{z}{(z-1)(z-2)} = -2\sum_{n=0}^\infty (z-1)^n - \frac{1}{z-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Could someone please help me write $z= \frac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form? Could someone please help me write $z = \dfrac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form?
$|z|=\sqrt{2}$
But how do I proceed to get the argument?
$\mathrm{arg}(z)=\mathrm{arg}(2+2i)^3 - \mathrm{arg}(1+i\sqrt{3})^4$
Thanks /David
| Hints: Use $\arctan 1=\frac{1}{4}\pi $ and $\arctan \sqrt{3}=\frac{1}{3}\pi $ in
$$\arg \left( \left( 2+2i\right) ^{3}\right) =3\arg \left( 2+2i\right)
=3 \arctan \frac{2}{2}=3\times \frac{1}{4}\pi $$
and
$$\arg \left( (1+i\sqrt{3})^{4}\right) =4\arg (1+i\sqrt{3})=4\arctan \frac{\sqrt{3}}{1}=4\times \frac{1}{3}\pi .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hard algebra problem Given
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
Now, it is necessary to find
$$\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=?$$
Is this possible and how? a,b,c are given constants.
I think, ? is probably a complicated function $F=F(a,b,c)$
| Without loss of generality we can assume $a\ge b\ge c>0$.
Then
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}\le\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}.$$
Noting that
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}
=\frac1{a^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)$$
$$\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}=\frac1{c^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right),$$
we have
$$\frac1{a^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{c^2},$$
where equality holds, for example, for $(x,y,z)=(a,0,0)$ and $(x,y,z)=(0,0,c)$. (There are other sets of values $(x,y,z)$ for which the equality holds).
More generally (i.e. if we do not assume $a\ge b\ge c>0$),
$$\frac1{\max(|a|,|b|,|c|)^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{\min(|a|,|b|,|c|)^2}.$$
Edit. We can further show that, when $a\ge b\ge c>0$, $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ can take any value $k\in \left[\frac1{a^2}, \frac1{c^2}\right]$. To show this we can consider $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ (where $(x,y,z)=(at,\ 0,\ c(1-t))$) as a continuous function of $t\in [0,1]$ and use the intermediate value theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many ways a composite number can be resolved into two factors which are prime to each other? Let N denote the number, and suppose $N=a^p \times b^q \times c^r \cdots$, where $a,b,c,\cdots$, are different prime numbers and $p,q,r,\cdots$ are positive integers.Then it is clear that each term of the product
$$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots$$
is a divisor of the given number,and that no other number is a divisor; then how could we use this to show/find that the number of ways in which this $N$ can be resolved into two factors which are prime to each other is $2^{n-1}$,where $n$ is the number of different prime factors in $N$?
| We obtain all ordered pairs $(x,y)$ such that $xy=N$ and $\gcd(x,y)=1$ by choosing a subset of the set of prime divisors of $N$, and giving all these primes to $x$, and giving the rest of the primes to $y$. A set of $n$ elements has $2^n$ subsets, so there are $2^n$ ways to produce a suitable $x$, and hence $2^n$ ordered pairs $(x,y)$.
We now have a count of the ordered pairs. But we want to count the number of factorizations as a product of two relatively prime numbers, and for example $(20)(3)$ is considered to be the same factorization as $(3)(20)$. For $n\ge 1$, to count the unordered pairs, divide the number of ordered pairs by $2$. We get $2^{n-1}$. For a more familiar example, there are $(52)(51)$ ordered pairs of $2$ cards, but there are only $(52)(51)/2$ "hands" of two cards.
The only case where we do not divide by $2$ is the case $N=1$, that is, $n=0$, where the only factorization is $(1)(1)$. So the exact result is that the number of factorizations of the desired type is $2^{n-1}$ if $n\ge 1$ and $1$ if $n=0$.
Comment: Consider the product
$$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots \qquad(\ast)$$
Look at a typical full term $1+p+p^2+\cdots +p^k$. We produce the suitable divisors $x$ of $N$ by deciding whether we will use the $1$ or the $p^k$. No other choice is possible, because if we choose $p^i$ where $0<i<k$, then $N/x$ (in our notation, $y$) will be divisible by $p$, so $x$ and $y$ will not be relatively prime.
Thus at every term in the product $(\ast)$, we have $2$ choices, "None" or "All." There are $n$ terms in the product, so there are $2^n$ ways to produce $x$. Now argue as before that if $n \ge 1$, every suitable factorization is produced twice by this process.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
| It is false.
Take $(c,d)=(1,1)$. Then $a^2 + b^2 = a^3 + b^3 = 2$ has real solutions with $(a+b) \neq 2$.
| {
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"source": "stackexchange",
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Equivalence class of an ideal Let $K$ denote the number field $\mathbb{Q}(\sqrt{15}).$ According to standard lore, we have that $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{15}]$. Moreover,
$2\mathcal{O}_{K} = \langle 2, 1+\sqrt{15}\rangle^{2}$
and
$3\mathcal{O}_{K} = \langle 3, \sqrt{15}\rangle^{2}.$
Resorting to the law of quadratic reciprocity, I have proven that the ideal $\langle 3, \sqrt{15}\rangle$ is non-principal. Since $h(\mathbb{Q}(\sqrt{15}))=2$, it must be the case that
$\langle 2, 1+\sqrt{15}\rangle \langle \alpha \rangle = \langle 3, \sqrt{15}\rangle \langle \beta \rangle$
for certain $\alpha, \beta \in \mathcal{O}_{K}\setminus \{0\}$.
Is there a quick way to determine such a pair $(\alpha, \beta)$?
Thanks!
| Starting from $(2, 1+\sqrt{15})(\alpha) = (3, \sqrt{15})(\beta)$, we can multiply both sides by $(3,\sqrt{15})$ to get:
$$(2, 1+\sqrt{15})(3,\sqrt{15}) = (3\beta/\alpha)$$
Multiplying the ideals on the left together
$$(6, 2\sqrt{15}, 3 + 3\sqrt{15}, 15 + \sqrt{15}) = (3\beta/\alpha)$$
So the ideal on the left side is a principal ideal, and we just need to find a generator. We repeatedly use the property $(x,y) = (x, y+zx)$ for ideals (adding a multiple of one generator to another doesn't change the ideal) to simplify the list of generators.
$$\begin{eqnarray}
& (6, 2 \sqrt{15}, 3 + 3 \sqrt{15}, 15 + \sqrt{15}) \\
= & (6, 2 \sqrt{15}, -42, 15 + \sqrt{15}) & \mbox{subtract 3 times 4th generator from third} \\
= & (6, 2 \sqrt{15}, 15 + \sqrt{15}) & \mbox { 42 is a multiple of 6 }\\
= & (6, -30, 15 + \sqrt{15}) & \mbox { subtract 2 times 3rd generator from 2nd } \\
= & (6, 15 + \sqrt{15}) & \mbox { 30 is a multiple of 6 } \\
= & (6, 3 + \sqrt{15}) & \mbox { subtract 2 times 6 from 2nd generator } \\
= & (3 + \sqrt{15})\\
\end{eqnarray}
$$
For the last equality, note that $(3 + \sqrt{15})(3 - \sqrt{15}) = -6$.
So we can take $\alpha = 3$, $\beta = 3 + \sqrt{15}$.
| {
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Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion? Is there a way to evaluate this limit:
$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?
Thank you,
| You should know the following limits:
$$\begin{split}\lim_{y\to 0} \frac{\sin y}{y} &= 1\\ \lim_{y\to 0} \frac{e^y-1}{y} &= 1\\ \lim_{y\to 0} \frac{\tan y}{ y} &= 1\\ \lim_{y\to 0} \frac{(1+y)^\theta -1}{y} &= \theta \qquad \text{(}\forall \theta \in \mathbb{R}\text{)}\\
\lim_{y\to 0} \frac{1-\cos y}{y^2} &= \frac{1}{2}\end{split}$$
which can be proved using only elementary Calculus tools (i.e. without any Differential Calculus technique).
These five limits are usually written as asymptotic relations in the following manner:
$$\tag{1} \sin y \approx y$$
$$\tag{2} e^y-1 \approx y$$
$$\tag{3} \tan y \approx y$$
$$\tag{4} (1+y)^\theta -1 \approx \theta\ y$$
$$\tag{5} 1-\cos y \approx \frac{1}{2}\ y^2$$
as $y\to 0$. Using asymptotics (1) - (5) you find:
$$\begin{split} \sin(e^{\tan^ 2 x} - 1) &\approx e^{\tan^2 x}-1 &\quad \text{by (1)}\\
&\approx \tan^2 x &\quad \text{by (2)}\\
&\approx x^2 &\quad \text{by (3)}\end{split}$$
$$\begin{split}\cos^{3/5}(x) - \cos(x) &= \Big(\big(1+(\cos x-1)\big)^{3/5} -1\Big) + \Big(1-\cos x\Big)\\ &\approx \frac{3}{5}\ (\cos x-1) + (1-\cos x) &\text{by (4) with } \theta =3/5\\
&= \frac{2}{5}\ (1-\cos x)\\
&\approx \frac{1}{5}\ x^2 &\text{by (5)}\end{split}$$
hence:
$$\lim_{x \to 0} \frac{\sin(e^{\tan ^ 2 {x}} - 1)}{\cos^{\frac{3}{5}}(x) - \cos(x)} = \lim_{x\to 0} \frac{x^2}{\frac{1}{5}\ x^2}=5\; .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lfloor (2+\sqrt{3})^n \rfloor $ is odd
Let $n$ be a nonnegative integer. Show that $\lfloor (2+\sqrt{3})^n \rfloor $ is odd and that $2^{n+1}$ divides $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $.
My attempt:
$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$
$$ 0\leq (2-\sqrt{3})^n \leq1$$
$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$
$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$
$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$
| For the first claim:
Here is a more general discussion.
As is often the case,
none of this is original,
and I did work it out
on my own for fun.
To make one root of
$f(x)
=x^2-2ax+b
=0$
with
$a, b > 0$
in $(0, 1)$.
The roots are
$x_{\pm}
=\dfrac{2a\pm\sqrt{4a^2-4b}}{2}
=a\pm\sqrt{a^2-b}
$.
$\begin{array}\\
x_{-}
&=a-\sqrt{a^2-b}\\
&=(a-\sqrt{a^2-b})\dfrac{a+\sqrt{a^2-b}}{a+\sqrt{a^2-b}}\\
&=\dfrac{a^2-(a^2-b)}{a+\sqrt{a^2-b}}\\
&=\dfrac{b}{a+\sqrt{a^2-b}}\\
\end{array}
$
So we want
$0 < b < a+\sqrt{a^2-b}
$
so that
$ b < a^2$
and
$(b-a)^2 < a^2-b
$
or
$b^2-2ab+a^2 < a^2-b
$
or
$b^2+b < 2ab
$
or
$b< 2a-1$.
If $a=2$ then
$b < 3$
so, if $b$ is an integer,
$b \in \{1, 2\}
$
so
$a^2-b
\in \{3, 2\}
$
so
$\{x_-, x_+\}
=2\pm\sqrt{3},
2\pm\sqrt{2}
$.
Let $c = a^2-b$.
$\begin{array}\\
x_+^n+x_-^n
&=(a+\sqrt{c})^n+(a-\sqrt{c})^n\\
&=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}+\sum_{k=0}^n \binom{n}{k}a^k(-1)^{n-k}c^{(n-k)/2}\\
&=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}(1+(-1)^{n-k})\\
&=\sum_{k=0}^n \binom{n}{k}a^{n-k}c^{k/2}(1+(-1)^{k})\\
&=\sum_{k=0}^{\lfloor n/2 \rfloor}2\binom{n}{2k}a^{n-2k}c^{k}\\
&=2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}\\
\end{array}
$
Since
$0 < x_-^n < 1$.
$\lfloor x_+^n-1 \rfloor
$
is odd and
$\lfloor x_+^n\rfloor
=2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}-1
$
so
$\begin{array}\\
x_+^n-\lfloor x_+^n \rfloor
&=x_+^n-2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}+1\\
&=1-x_-^n\\
&\to_- 1\\
\end{array}
$
If $a=2, b=1$
then
$x_+^n-\lfloor x_+^n \rfloor
=1-(2-\sqrt{3})^n
=1-\dfrac{1}{(2+\sqrt{3})^n}
$.
| {
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"source": "stackexchange",
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approximation formula for the integral Get an approximation formula for the following integral:
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
| $$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
We have that
$$B(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1} \theta \cos^{2y-1} \theta d\theta$$
So we can write your expression as
$$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right)$$
But we also know that
$$B\left( {x,y} \right) = \frac{{\Gamma \left( x \right)\Gamma \left( y \right)}}{{\Gamma \left( {x + y} \right)}}$$
so that
$$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{1}{2}\frac{{\Gamma \left( {k - \frac{1}{2}} \right)\Gamma \left( {n - k + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}$$
And we have closed formulas for two of the three $\Gamma$ functions there, namely:
$$\eqalign{
& \Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n - 1} \right)!!}}{{{2^n}}}\sqrt \pi \cr
& \Gamma \left( {k - \frac{1}{2}} \right) = \Gamma \left( {k - 1 + \frac{1}{2}} \right) = \frac{{\left( {2k - 3} \right)!!}}{{{2^{k - 1}}}}\sqrt \pi \cr} $$
For the last one, we simply put
$$\Gamma \left( {n - k + 1} \right) = \left( {n - k} \right)!$$
So we get
$$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$$
And then
$$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$$
or
$$\frac{{{2^n}}}{{\left( {2n - 1} \right)!!}}\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^k}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.
I want to know the detail about the proof. Thanks in advance.
Actually I know any two of them and three of them are linearly independent.
| I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in $\mathbb Q$ are $x=y=0$.)
This is equivalent to showing that if
$$a+b\sqrt6=c\sqrt3+d\sqrt2$$
then $a=b=c=d=0$.
By squaring both sides of the above equation we get
$$
\begin{align*}
a^2+6b^2+2ab\sqrt6&=3c^2+2d^2+2cd\sqrt6\\
2(ab-cd)\sqrt6=3c^2+2d^2-a^2-6b^2
\end{align*}
$$
Since $a,b,c,d\in\mathbb Q$, this implies
$$
\begin{align*}
ab-cd&=0\\
3c^2+2d^2-a^2-6b^2&=0
\end{align*}
$$
which is the same as
$$
\begin{align*}
ab&=cd\\
3c^2+2d^2&=a^2+6b^2
\end{align*}
$$
Suppose that $b\ne0$, $c\ne0$. (I'll leave the solution of these cases to the reader.) Then we can rewrite the first equation as $\frac ac = \frac db = x$, where $x\in\mathbb Q$. Now the second equation becomes
$$
\begin{align*}
3c^2+2x^2b^2&=x^2c^2+6b^2\\
x^2(2b^2-c^2)&=3(2b^2-c^2)\\
(x^2-3)(2b^2-c^2)&=0
\end{align*}$$
This implies that $x^2=3$ or $2b^2=c^2$. None of them has non-zero solutions in rational numbers.
Alternatively, we could solve the case $b=0$ first. Once we're looking only at $b\ne0$, we can assume w.l.o.g. that $b=1$. (After dividing both sides with $b$.)
So we're looking at $a+\sqrt6=c\sqrt2+d\sqrt2$. And squaring, as above, gives us now $a=cd$. Thus we get
\begin{align*}
cd-c\sqrt2-d\sqrt3+\sqrt6&=0\\
(c-\sqrt3)(d-\sqrt2)&=0
\end{align*}
which means that $c=\sqrt3$ or $d=\sqrt2$. In either case, it is the contradiction with the assumption $c,d\in\mathbb Q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
A version of Hardy's inequality involving reciprocals. How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have
$$\sum_{n=1}^\infty \frac{n}{a_1+a_2+a_3+\cdots+a_n}\le 2\sum_{n=1}^\infty \frac{1}{a_n}$$
Added later:
Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. (See the wiki for what $p$ is).
The case $p=2$ appears here: Proving $A: l_2 \to l_2$ is a bounded operator
| Here is a proof.
We try using induction, but as usual, a direct approach seems to fail and we have to try and prove a stronger statement.
So we try and pick a positive function $f(n)$ such that
$$ \frac{f(n)}{a_1 + a_2 + \dots + a_n} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
Let $S = a_1 + a_2 + \dots + a_n$ and let $x = a_{n+1}$.
In order to prove that $n$ implies $n+1$ it would be sufficient to prove
$$\frac{2}{x} + \frac{f(n)}{S} \ge \frac{f(n+1) + n+1}{S+x}$$
This can be rearranged to
$$2S^2 + f(n) x^2 + (f(n) + 2) Sx \ge (f(n+1) + n+1) Sx$$
Since $$ 2S^2 + f(n) x^2 \ge 2 \sqrt{2 f(n)} Sx$$
it is sufficient to prove that $f(n)$ satisfies
$$ f(n) + 2 + 2\sqrt{2 f(n)} \ge f(n+1) + n + 1$$
Choosing $f(n) = \dfrac{n^2}{2}$ does the trick.
We can easily verify the base case for this choice of $f(n)$.
Thus we have:
$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/98483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
integral resulting in Bessel function Prove that
$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)$$ where $J_{1}$ is the Bessel function of the first kind of order 1.
Some calculations I have done
$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx= \int_{0}^{\infty} \sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k+1}}{2k+1!} \cdot \sum_{l=0}^{\infty }(-1)^{l}\frac{a^{2l+1}x^{-2l-1}}{2l+1!} \ dx$$
$$= \int_{0}^{\infty} \sum_{l=0}^{\infty } \sum_{k=0}^{\infty }(-1)^{k+l}\frac{x^{2(k-l)}}{(2k+1)!(2l+1)!} a^{2l+1} \ dx$$
$$\frac{\pi \sqrt{a}}{2}J_{1}(2sqrt{a})=\frac{\pi \sqrt{a}}{2} \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{2l+1}l!(1+l)!} 2^{l+\frac{1}{2}}a^{l+\frac{1}{2}}$$
$$=\pi \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{l+\frac{3}{2}}l!(1+l)!} a^{l+1}$$
| Let $f(a) = \int_0^\infty \sin(x) \sin\left(\frac{a}{x}\right) \mathrm{d} x$, where $a\in \mathbb{R}$. Without loss of generality we can assume $a > 0$.
Observe that
$$ \begin{eqnarray}
a f^{\prime\prime}(a) = \int_0^\infty \sin(x) \left(-\sin\left(\frac{a}{x} \right) \right) \frac{a}{x} \frac{\mathrm{d} x}{x} \stackrel{x \to a/y}{=} \int_0^\infty \sin\left( \frac{a}{y} \right) (-\sin(y)) \mathrm{d} y = -f(a)
\end{eqnarray}
$$
The differential equation so obtained, $a f^{\prime\prime}(a) + f(a) = 0$, reduces to Bessel differential equation, with general solution
$$
f(a) = c_1 \cdot \sqrt{a} J_1(2\sqrt{a}) + c_2 \cdot \sqrt{a} Y_1(2\sqrt{a})
$$
where $J_1$ and $Y_1$ are Bessel functions of the first and the second kind. Since $f(0)=0$ and $\lim_{a \downarrow 0^+} \sqrt{a} Y_1(2 \sqrt{a} ) = -\frac{1}{\pi}$, we get $c_2 = 0$.
By splitting the integration range into $(0,1)$ and $(1,\infty)$ and changing variables in
the first one to $x \to 1/x$ we get
$$
f(a) = \int_{1}^{\infty} \sin(1/x) \sin(a x) \frac{\mathrm{d} x}{x^2} + \int_1^\infty \sin(x) \sin(a/x) \mathrm{d} x
$$
From here we see that small $a$ expansion first term is
$$
f(a) = a \left( \int_{1}^\infty \frac{1}{x} \sin(x) \mathrm{d} x + \int_{1}^\infty \sin(1/x) \frac{\mathrm{d} x }{x} \right) + \mathcal{o}(a) = \frac{\pi}{2} a + \mathcal{o}(a)
$$
This fixes $c_1 = \frac{\pi}{2}$, proving the requested equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/99919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
I have reduced this problem to $$ 2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$
but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$
and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?
| Let $u=\sqrt{\tan(x)}$. Then $u^2 = \tan(x)$ and $2 u \mathrm{d} u = (1+ \tan^2(x)) \mathrm{d} x$. Thus
$$
\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u
$$
Since $1+u^4 = (1 + \sqrt{2} u + u^2)( 1- \sqrt{2} u + u^2)$, partial fraction decomposition applies:
$$
\frac{2u^2}{1+u^4} = \frac{1}{\sqrt{2}} \left( \frac{u}{u^2-\sqrt{2} u+1}-\frac{u}{u^2+\sqrt{2} u+1} \right)
$$
Hence
$$ \begin{eqnarray}
\int \frac{2u^2}{1+u^4} \mathrm{d} u &=& \frac{1}{2 \sqrt{2}} \log \left(\frac{u^2-\sqrt{2} u+1}{u^2+\sqrt{2} u+1}\right) + \\
&\phantom{=}& \frac{\tan
^{-1}\left(\sqrt{2} u+1\right) -\tan ^{-1}\left(1-\sqrt{2} u\right) }{\sqrt{2}}
\end{eqnarray}
$$
Applying the fundamental theorem of calculus:
$$
\int_0^{\pi/2} \sqrt{\tan(x)} \mathrm{d} x = \frac{\pi}{\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 8,
"answer_id": 5
} |
Help calculating the probability of dealing a custom deck of cards Imagine a deck that consists of two types of cards. There are 2 type A cards and 9 Type B cards, totaling 11 cards. Assuming that 1 card will be dealt to 5 players. What are the probabilities that Type A will be dealt zero, one, and two times?
I hope I've asked that clearly. Any reply would be appreciated.
| Note that three events $A_0$, $A_1$, and $A_2$ of players receiving 0, 1 or 2 cards of type A are exclusive and partition the entire event space (there are only 2 cards of type A). This means that
$$
\mathbb{P}(A_0) + \mathbb{P}(A_1) + \mathbb{P}(A_2) = 1
$$
There total $\binom{11}{5}$ ways to deal 1 card to each of 5 players. There are $\binom{11-2}{5}$ ways to deal 5 cards with no type A card in it (just throw out 2 type A card out of the deck), that makes
$$
\mathbb{P}(A_0) = \frac{\binom{11-2}{5}}{\binom{11}{5}} = \frac{\frac{9!}{4! 5!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \frac{6!}{4!} = \frac{6 \cdot 5}{11 \cdot 10} = \frac{3}{11}
$$
To compute $\mathbb{P}(A_2)$, deal type A cards first. What remains is to deal 9 type B cards to 3 remaining players, thus
$$
\mathbb{P}(A_2) = \frac{\binom{9}{3}}{\binom{11}{5}} = \frac{\frac{9!}{6! 3!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \cdot \frac{5!}{3!} = \frac{5 \cdot 4}{11 \cdot 10} = \frac{2}{11}
$$
Thus
$$
\mathbb{P}(A_1) = 1- \mathbb{P}(A_0) -\mathbb{P}(A_2) = \frac{6}{11}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ Could someone please explain how to solve this:
$$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$
not the answer only, but a step-by-step solution.
I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.
Thank you so much in advance!
| The idea is to represent the polynomial $p(x) = x^4 - 10x^3 + 21x^2 + 40x - 100$ as a product of simple factors. By Vietta formula the free terms equals to the product of roots. Thus, if we confine ourselves to integer roots, they should divide $100$. Start with $x=2$, and verify that $p(2) = 16 - 10 \times 8 + 21 \times 4 + 40 \times 2 - 100 = 16 -80 + 84 - 80 + 100 = 0$.
Now applying long division we compute
$$
p(x) = (x-2)\left( x^3 - 8 x^2 + 5 x + 50 \right)
$$
Now try $x=5$. $5^2(5-8) + 5 ( 5+ 10) = 5( -15 + 15) = 0$. Apply long division again:
$$
p(x) = (x-2)(x-5) \left(x^2 - 3 x - 10\right) = (x-2)(x+2)(x-5)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/103988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Diagonalize a matrix We have a matrix $A =
\left(
\begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{array}
\right)$.
How do you find a diagonal matrix $D$ and an orthogonal matrix $P$ so that:
$D=P^tAP$?
| $A =
\left(
\begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{array}
\right)$.
You know
$D =
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{array}
\right)$.
Now,
$Ax = \lambda x$
x = 2 (Case 1)
$$Ax = 2x$$
$$Ax - 2x = 0$$
$$(A - 2I)x = 0$$
$$A -2I=
\left(
\begin{array}{ccc}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2 \\
\end{array}
\right)$$
$$\left(
\begin{array}{ccc}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2 \\
\end{array}
\right)x=0$$
Take it to echelon form you get
$$\left(
\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
\end{array}
\right)x=0$$
Thus, if x = (a b c)'
a = b = c
Similarly, do for the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/105643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
value of $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$? Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?
| Yes, in fact, everything is known about it. Step 1: get a formula for the denominators (I assume the 6th one is supposed to have a 6 in it, not stop at 5 like the 5th one). Step 2: apply partial fractions to get a telescoping series. Enjoy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
| from user21436 you have
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Number of 5 letter words over a 3 letter alphabet using each letter at least once This is similar to my previous question, Number of 5 letter words over a 4 letter group using each letter at least once.
The only difference is that there are 3 letters to choose from instead of 4.
However, I've run into a problem.
Using inclusion exclusion I get:
$3^5 - 3 \cdot 2^5 + 3 \cdot 1^5 = 150$
Which is the correct answer.
But trying to solve it directly by counting the permutations of the strings $abccc$, $abbbc$ and $aaabc$ gives:
$3 \cdot \frac{5!}{3!}=60$
What am I missing in the combinatorial approach?
Also, could I solve this using generating functions?
So this is really two questions - bonus points for answering both :-)
| You forgot that you can have strings with two copies of one letter, two of another, and one of the third. There are $3$ ways to choose which of the letters is used only once, and $5$ ways to place it in the string. There are then $\binom42=6$ ways to choose which two of the remaining four places get the two copies of the first of the other two letters in alphabetical order, for a total of $3\cdot 5\cdot 6=90$ strings, exactly the number that you’re missing.
Added: To do it with generating functions, let $a_n$ be the number of words of length $n$ over a $3$-letter alphabet that use each letter at least once. The first step is to find a recurrence for $a_n$. For brevity call such words good. Each good word of length $n-1$ can be extended to a good word of length $n$ in $3$ different ways, since each of the three letters can be added at the end of the word. In addition, each $(n-1)$-letter word that contains exactly two of the three letters can be extended to a good word in one way, by adding the missing letter. How many of these $(n-1)$-letter words are there? There are $3$ ways to choose which of the three letters is missing, and each of the $n-1$ positions in the word can contain either of the two remaining letters. However, we have to exclude the two possibilities that contain only one of the two letters, so there are $3(2^{n-1}-2)$ such words. This leads us to the recurrence
$$a_n=3a_{n-1}+3(2^{n-1}-2)\tag{1}\;.$$
Clearly $a_0=a_1=a_2=0$ and $a_3=3!=6$. However, if we assume that $a_n=0$ for $n<0$, $(1)$ requires a small adjustment to get $a_0$ and $a_1$ right: as it stands, it yields the values $-9/2$ and $-3$. To correct this, we add a couple of Iverson brackets to get
$$a_n=3a_{n-1}+3(2^{n-1}-2)+\frac92[n=0]+3[n=1]\tag{2}\;.$$
Now multiply through by $x^n$ and sum over $n$ to get the generating function $f(x)$:
$$\begin{align*}f(x)=\sum_na_nx^n&=3\sum_n\left(a_{n-1}+2^{n-1}-2+\frac32[n=0]+[n=1]\right)x^n\\
&=3\left(\sum_na_{n-1}x^n+\sum_n2^{n-1}x^n-2\sum_nx^n+\frac32+x\right)\\
&=3\left(x\sum_na_{n-1}x^{n-1}+\frac12\sum_n(2x)^n-\frac2{1-x}+\frac32+x\right)\\
&=3\left(x\sum_na_nx_n+\frac{1/2}{1-2x}-\frac2{1-x}+\frac32+x\right)\\
&=3xf(x)+\frac32\left(\frac{7x-3}{(1-x)(1-2x)}+2x+3\right)\\
&=3xf(x)+\frac{6x^3}{(1-x)(1-2x)}\;,
\end{align*}$$
so $$(1-3x)f(x)=\frac{6x^3}{(1-x)(1-2x)}\;,$$ and
$$\begin{align*}
f(x)&=\frac{6x^3}{(1-x)(1-2x)(1-3x)}\\
&=x^3\left(\frac3{1-x}-\frac{24}{1-2x}+\frac{27}{1-3x}\right)\\
&=3x^3\left(\sum_nx^n-8\sum_n2^nx^n+9\sum_n3^nx^n\right)\\
&=\sum_n\left(3-3\cdot 2^{n+3}+3^{n+3}\right)x^{n+3}\\
&=\sum_n\left(3-3\cdot 2^n+3^n\right)x^n\;.
\end{align*}$$
Equating coefficients, we finally have $a_n=3^n-3\cdot 2^n+3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$? Polynomials of the form :
$ T_n(x) =2^{-1} \cdot ((x+\sqrt {x^2-1})^n+ (x-\sqrt {x^2-1})^n)$
are known as Chebyshev polynomials of the first kind .
Consider the polynomials of the form :
$P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$
Have these polynomials some special name ?
First few polynomials of this form are :
$P_0(x) = 2$
$P_1(x) = x$
$P_2(x) =x^2- 2$
$P_3(x) =x^3-3x$
$P_4(x) = x^4-4x^2+2$
$\vdots$
| They are called Pedja polynomials of the first kind. They satisfy $P_{n+1}(x) = x P_n(x) - P_{n-1}(x).$ The Pedja polynomials of the second kind are $$R_{-1}= 0, \; R_0(x) = 1, \; R_1(x)=x, \; R_2(x) = x^2 - 1$$ and solve $R_{n+1}(x) = x R_n(x) - R_{n-1}(x).$ Together they have
$$ P_n^2(x) - (x^2 - 4) R_{n-1}^2(x) = 4.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $1+\frac{1}{1^k+2^k}+\frac{1}{1^k+2^k+3^k}+\frac{1}{1^k+2^k+3^k+4^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k}\cdots$ telescope? after the post $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$, I had to ask
Can we telescope the series $$1+\frac{1}{1^k+2^k}+\frac{1}{1^k+2^k+3^k}+\frac{1}{1^k+2^k+3^k+4^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k+6^k}+\cdots$$ ?
| This is not a perfect answer, but a hint. In sum of powers I've a (a bit old) treatize on the Eulerian numbers, which allow, in connection with the binomial-coefficients, to express the sums-of-like-powers as polynomials. Inserting these in the denominators of your sum should allow to do partial sums like in your linked other question. Unfortunately, sum of powers is in german, but the formulae are so simple, that it should be evident, how to proceed on your own. Maybe the same can be found using the wikipedia-article
| {
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Solve $\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$ I need to solve the limit
$$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$$
when $a,b,c \gt 0$.
I'm looking for ways to avoid $\frac{1}{x}$ power.
| HINT:
$$
\left( a^x + b^x - c^x \right)^\frac{1}{x} = a \left( 1 + \left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x \right)^\frac{1}{x} = a \left( 1 + x \cdot \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x} \right)^\frac{1}{x}
$$
Compute $\lim_{x \to 0} \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x}$, then use the golden limit $\lim_{x\to 0} \left( 1 + \alpha x\right)^\frac{1}{x}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
| I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$.
So given that $x=y^3$, we have that $\frac{dx}{dy}=3y^2$ (either using the power rule or a simpler limit computation). That makes $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.
| {
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how can I write T in matrix form Let V be $ C[x]/(x^3+5x^2+6x+2) $ and let $ T:V→V $ be defined by
$T(p(x))=(x+1)p(x) $. I need to find the bases for the kernel (null space) and image (range) of T.
I have to write T in matrix form right? After that we can easily find the bases. How can we write it in matrix form? Also one more puzzling thing is that do we take $p(x)=x^3+5x^2+6x+2$.
| Here's a more naive approach, for when you don't see how the kernel factors: Using $\{1,x,x^2\}$ as an ordered basis, $1$ maps to $1+x$, $x$ maps to $x+x^2$, and $x^2$ maps to $x^2+x^3=x^2+x^3-x^3-5x^2-6x-2=-4x^2-6x-2$, so the matrix is: $$\begin{bmatrix} 1 & 0 & -2 \\ 1 & 1 & -6 \\ 0 & 1 & -4 \end{bmatrix}$$
The image of a matrix is always the span of the columns, in this case those polynomials $a+bx+cx^2$ with $b=c+a$ and the kernel is found by solving $$\begin{bmatrix} 1 & 0 & -2 \\ 1 & 1 & -6 \\ 0 & 1 & -4 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
So $a+bx+cx^2$ maps to $0$ if and only if $a=2c$ and $b=4c$
| {
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Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$
Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something?
| You don't have to use the hint. Let $x = {\pi \over 5}$. Then $\cos(3x) = -\cos(2x)$ since $3x + 2x = \pi$. Since $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ and $\cos(2x) = 2\cos^2(x) - 1$, you get
$$4\cos^3(x) - 3\cos(x) = - ( 2\cos^2(x) - 1)$$
This simplifies to
$$4\cos^3(x) + 2\cos^2(x) - 3\cos(x) - 1 = 0$$
This factors into
$$(4\cos^2(x) - 2\cos(x) - 1)(\cos(x) + 1) = 0$$
Since $\cos(x)$ is not $-1$, you get the desired equation $4\cos^2(x) - 2\cos(x) - 1 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem about parametric integral How to solve the following integral.
$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|<1$
| I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1}{a \cos x+1} d x \\&=\frac{\pi}{a}-\frac{1}{a} \int_{0}^{\pi} \frac{d x}{a \cos x+1} \\&\stackrel{t=\tan \frac{x}{2}}{=} \frac{\pi}{a}-\frac{1}{a} \int_{0}^{\infty} \frac{1}{1+\frac{a\left(1-t^{2}\right)}{1+t^{2}}} \cdot \frac{2 d t}{1+t^{2}} \\&=\frac{\pi}{a}-\frac{2}{a} \int_{0}^{\infty} \frac{d t}{(1-a) t^{2}+(1+a)} \\&=\frac{\pi}{a}-\frac{2}{a \sqrt{1-a^{2}}} \tan^{-1}\left[\frac{\sqrt{1-a} t}{\sqrt{1+a}}\right]_{0}^{\infty} \\&=\frac{\pi}{a}-\frac{\pi}{a \sqrt{1-a^{2}}}\end{aligned}\tag*{} $
Integrating both sides w.r.t. $a$ yields
\begin{aligned}\int I^{\prime}(a) d a &=\pi\int\left(\frac{1}{a}-\frac{1}{a \sqrt{1-a^{2}}}\right) da \\& \stackrel{a=\sin \theta}{=} \pi\int\left(\frac{1}{\sin \theta}-\frac{1}{\sin \theta \cos \theta}\right) \cos \theta d \theta \\&=\pi\int \frac{\cos \theta-1}{\sin \theta} d \theta\\&I(a) =\pi \int \frac{-\sin ^{2} \theta}{\sin \theta(\cos \theta+1)} d \theta\\&=\pi \ln (1+\cos \theta) +C\end{aligned}
Putting $a=0$ gives $C=-\pi\ln 2$ and hence
$$
\boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =\pi \ln \left[1+\cos \left(\sin ^{-1} a\right)\right]= \pi \ln \left(\frac{1+\sqrt{1-a^{2}}}{2} \right)}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Writing $3\sin(x)+4\cos(x)$ in the form of $r \sin(x-a)$ I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.
Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$
Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.
Getting $r$:
$$\begin{align*}
3^2 + 4^2 &= r^2 \cos(a)^2 + r^2 \sin(a)^2\\
25 &= r^2 (\cos(a)^2 + \sin(a)^2)\\
25 &= r^2\\
r &= -5, 5
\end{align*}$$
Getting $a$:
$$\begin{align*}
\frac{-r\sin(a)}{ r\cos(a)} &= \frac{3}{4}\\
\tan(a) &= -\frac{3}{4}\\
a &= \arctan(-3/4)\\
a &= -36.87, 143.13\\
\end{align*}$$
The questions:-
There are two values for both $r$ and $a$, how should I choose the values to be in the final form?
| $$\begin{align*}
f(x) &= 3 \sin(x) + 4 \cos(x) \\
f(x) &= 5\left(\frac{3}{5} \sin(x) + \frac{4}{5} \cos(x) \right)
\end{align*}$$
Let $\sin(y) = 4/5$ and $\cos(y) =3/5$, for some real number $y$.
$$\begin{align*}
f(x) &= 5\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\
f(x) &= 5\sin(x+y)&\text{(where }y = \arcsin(4/5)\text{)}
\end{align*}$$
Using trigonometric tables, $y = 36.87$.
$$f(x) = 5\sin(x + 36.87^{\circ})$$
which can also be written as
$$f(x) = 5 \sin (180 - x - 36.87) = -5\sin(x - 144.13).$$
This justifies the 2 values of $a$ and $r$ in $r\sin(x−a)$.
| {
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"url": "https://math.stackexchange.com/questions/116209",
"timestamp": "2023-03-29T00:00:00",
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Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?
Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$
Side work:
$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = -3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\sqrt{9\csc^2{\theta} - 9}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\tan{\theta} \ d\theta$
$-\int \frac{3 \cdot \sqrt{\csc^2{\theta} - 1}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\cot{\theta}\csc{\theta}\cot{\theta}}{3 \csc^3{\theta}} \ d\theta$
$-\frac{1}{3} \int \cos^2{\theta} \ d\theta$
$-\frac{1}{3} \int \frac{1}{2} \left(1 + \cos{\left(2\theta\right)}\right) \ d\theta$
$-\frac{1}{6} \int 1 + \cos{\left(2\theta\right)} \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{6} \int \cos{\left(2\theta\right)} \ d\theta$
Sidework:
$u = 2\theta$
$du = 2 \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{3} \int \cos{u} \ du$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{u} + C$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{2\theta} + C$
Sidework:
$\sin{\left(2\theta\right)} = 2 \sin{\theta} \cos{\theta}$
$\sin{\theta} = \frac{3}{x}$
$\cos{\theta} = \frac{\sqrt{x^2 - 9}}{x}$
$\theta = \sin^{-1}{\left(\frac{3}{x}\right)}$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2}{3} \cdot \frac{3}{x} \cdot \frac{\sqrt{x^2 - 9}}{x} + C$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2\sqrt{x^2 - 9}}{x^2} + C$
Thank you for your time.
| Added: It has been pointed out that there may be some ambiguity about the integral the OP is trying to find. The introduction has an $x^2$ at the bottom, but in the solution the OP seems to be using $x^3$. That is much easier. The substitution $x=3\sec\theta$ transforms the integrand into a constant times $\frac{\tan^2\theta}{\sec^2\theta}$, which is just $\sin^2\theta$.
Original post: If we are going to use a trigonometric substitution, I would let $x=3\sec\theta$. (Secant is a bit more familiar than cosecant. You are therefore less likely to differentiate $\sec\theta$ incorrectly; incorrect differentiation of $\csc\theta$ spoiled our integration.)
Then $dx=3\sec\theta\tan\theta\,d\theta$. Plug in. From now on, I will forget about constants. So everything is a little wrong, but only by a constant factor.
On top we get $(\tan\theta)(\sec\theta\tan\theta)$. On the bottom we get $\sec^2\theta$.
So we are interested in
$$\frac{\tan^2\theta\sec\theta}{\sec^2\theta}.$$
Put everything in terms of sines and cosines. We get $\dfrac{\sin^2\theta}{\cos\theta}$.
There is a trick that one can use when there is an odd power of sine or cosine. In this case, multiply top and bottom by $\cos\theta$, and replace the $\cos^2\theta$ now at the bottom by $1-\sin^2\theta$.
We have arrived at
$$\int \frac{\cos\theta \,\sin^2\theta}{1-\sin^2\theta}\,d\theta.$$
Make the substitution $u=\sin\theta$.
We arrive at
$$\int \frac{u^2}{1-u^2}\,du.$$
The integrand is a rational function. It can be rewritten as
$$\frac{1}{1-u^2}-1.$$
But (partial fractions) we find that $\dfrac{1}{1-u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)$.
Remark: As usual, there are many other approaches. For example, we can change $\frac{\tan^2\theta}{\sec\theta}$ to $\frac{\sec^2\theta-1}{\sec\theta}$, and then to $\sec\theta -\cos\theta$. If we happen to know the integral of $\sec\theta$, we are essentially finished. More strikingly, if we guess somehow that $F(x)$ is an antiderivative, then the magic substitution $u=F(x)$ solves the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Absolute maximum value of $\sin^2(x)-\sin(x)$ in $[0,\frac{3\pi}{2}]$ I thought it does not have absolute maximum, but wanted to just check and see why
| We are interested in $w^2-w$, with the proviso that $-1\le w\le 1$.
Completing the square, we find that
$4w^2-4w=(2w-1)^2+3$. So the minimum value of $4w^2-4w$ is $3$, reached when $w=1/2$. The maximum value in our interval is reached when $w$ is as negative as possible, namely at $w=-1$.
So the minimum value of $w^2-w$ is $\frac{3}{4}$, reached when $\sin x=\frac{1}{2}$, that is, at $x=\frac{\pi}{6}$ and also at $x=\pi-\frac{\pi}{6}$.
The maximum value of $w^2-w$ is $\frac{12}{4}$, reached when $\sin x=-1$, that is, at $x=\frac{3\pi}{2}$.
Remark: You came close to being right on no absolute maximum. If we replace the interval $[0,\frac{3\pi}{2}]$ by $(0,\frac{3\pi}{2})$, there is no longer an absolute maximum.
| {
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"url": "https://math.stackexchange.com/questions/119496",
"timestamp": "2023-03-29T00:00:00",
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How does $(x_n - \sqrt{2})^2/(2x_n ) \geq 0$ mean $x_n \geq \sqrt{2}$ when $x_1=1$? I don't understand one step of proving of the Babylonian Method for $a=2,x_1=1$, and
$$x_{n+1} =\frac{1}{2}( x_n + \frac{a}{x_n}). \quad (n=1,2,\ldots)$$
$$x_{n+1} - \sqrt{2} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - \sqrt{2} = \frac{(x_n^2-2\sqrt{2} x_n+2)}{2x_n} = \frac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$$
For all values of $x_n>0$. Conclude $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.
After using $x_n \geq \sqrt{2}$ we find
$$x_{n+1} - x_n = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - x_n = \frac{x_n^2+2-2x_n^2}{2x_n} =\frac{2-x_n^2}{2x_n} \leq 0,$$
which means the sequence $(x_n)$ is monotonically decreasing. Now we know $(x_n)$ is bounded below by $\sqrt{2}$ and $(x_n)$ is monotonically decreasing so $(x_n)$ has a limit.
My problem is I don't understand how could we say $x_n \ge \sqrt{2}$ by using $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$?
| you have concluded that for $x_n > 0$,$$x_{n+1} - \sqrt{2} = {{((x_n−\sqrt{2})^2)}\over{(2x_n)}}≥0. $$ This means $$x_{n+1} - \sqrt{2} ≥ 0$$ Now since n is arbitrary, any value of n will satisfy this equation , put $n = n -1$ in this equation, you will get
$$x_{n} - \sqrt{2} ≥0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to change degree of elements in a field I'm working through an assignment and need to write several elements as polyomials of degree <= 2.
($x^2+1$)($x+1$)
within the field
$\mathbb Z$$_3$[x] / ($x^3 + 2x + 1$)
And am unsure how exactly to go about doing this - could anyone give me some pointers as to what the question is asking me to do?
| Divide the polynomial $(x^2+1)(x+1)$ (multiplied out) by the polynomial $x^3+2x+1$, remembering to work with the right field of coefficients. The remainder (or more properly, its equivalence class) is what you want.
How to do the division? Basically like ordinary polynomial division. $x^3+2x+1$ goes into $x^3+x^2+x+1$ (our product) once. Subtract. We get $(x^3+x^2+x+1)-(x^3+2x+1)$, which is $x^2-x$, or equivalently $x^2+2x$.
Another way: In this case, we can take a shortcut, well, maybe not a shortcut, just another way of thinking. The polynomial $x^3+2x+1$ is equivalent to $0$, so $x^3$ is equivalent to $-2x-1$. So $x^3+x^2+x+1$ is equivalent to $(-2x-1)+(x^2+x+1$, which simplifies to $x^2-x$, or, equivalently, $x^2+2x$.
| {
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Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show me how to do this step by step.
The answer from the book is:
$$ A_2=\sqrt\frac{A_1^2q^2}{2A_1^2gh+q^2} $$
The closest I can get is the following:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
$$ \frac{q^2}{A_1^2} = \frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Invert:
$$ \frac{A_1^2}{q^2} = \frac{(\frac{A_1}{A_2})^2-1}{2gh} $$
Multiply both sides by $2gh$:
$$ 2gh\frac{A_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
$$ \frac{2ghA_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
Add 1 to both sides and re-arrange:
$$ \frac{A_1^2}{A_2^2} = \frac{2ghA_1^2}{q^2} +1 $$
Invert again:
$$ \frac{A_2^2}{A_1^2} = \frac{q^2}{2ghA_1^2} +1 $$
Multiply by $A_1^2$:
$$ A_2^2 = \frac{A_1^2q^2}{2ghA_1^2} +1 $$
Get the square root:
$$ A_2 = \sqrt{\frac{A_1^2q^2}{2ghA_1^2}+1} $$
I cannot see where the $q^2$ on the bottom of the textbook answer comes from.
| \begin{align}
q &= A_1\sqrt\frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\
\frac{q^2}{A_1^2} &= \frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\
\left(\frac{A_1}{A_2}\right)^2-1 &=\frac{2ghA_1^2}{q^2} \\
\frac{A_1}{A_2} &= \sqrt\frac{2ghA_1^2 + q^2}{q^2} \\
A_2 &= A_1\sqrt\frac{q^2}{2ghA_1^2 + q^2} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Show $f(x) = x^3 - \sin^2{x} \tan{x} < 0$ on $(0, \frac{\pi}{2})$ This is the last of a homework problem set from Principles of Mathematical Analysis (Ch. 8 #18(a)) that I've been working/stuck on for a few days:
Define $f(x) = x^3 - \sin^2{x}\tan{x}.$
Find out whether it is positive or negative for all $x \in (0, \frac{\pi}{2})$, or whether it changes sign. Prove your answer.
I've thought of a couple possible ways to solve it but have gotten stuck each time.
*
*Power series: This would be super easy, since the polynomial is gone and all of the other terms are negative. The problem: I'd have to calculate the $\tan$ power series and show various properties of the Bernoulli numbers. Big hassle.
*Straight up approximation: I've spent most of the time on this method, but I always overestimate, causing me to lose the strict inequality. This was using various trig identities and just basic stuff like $\sin{x} < x < \tan{x}$ on $(0, \frac{\pi}{2})$.
*Integrating and more approximating. Running into the same difficulty as above; overestimation.
I'm kind of just running around in circles at this point, probably missing something simple. Can I get a hint?
| Power series approach works, though it requires some numerical computation. Since you seemed to be interested in asymptotics and approximations, you might find this interesting.
It is enough to show that
$$ \sin ^3 x \gt x^3 \cos x,\quad 0 \lt x \lt \frac{\pi}{2} $$
Now the power series for $\displaystyle \sin^3 x$ is
$$ \sin^3 x = x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024} + \frac{671x^{11}}{604800}- \frac{73x^{13}}{1140480} + \dots $$
Note that $\displaystyle \sin^3 x = -\frac{1}{4} (\sin 3x - 3 \sin x)$ and so the $\displaystyle (2k+1)^{th}$ derivative at $\displaystyle 0$ is
$$ c_{2k+1} = \frac{(-1)^{k+1}}{4}(3^{2k+1} - 3)$$
Now for $\displaystyle k \ge 4$, it we can show that
$$ \left|\frac{c_{2k+1}}{(2k+1)!}\right| \gt \left|\frac{\pi^2}{4} \frac{c_{2k+3}}{(2k+3)!}\right|$$
(It is enough to verify for $\displaystyle k=4$, as $\displaystyle 4|c_{2k+1}| (2k+2)(2k+3) - \pi^2|c_{2k+3}|$ in an increasing function of $\displaystyle k$).
Notice that $\displaystyle \frac{c_{2k+1}}{(2k+1)!}$ is nothing but the coefficient in the Taylor series.
Thus for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$, we have that (alternating series with decreasing absolute value of the terms)
$$ \sin ^3 x \gt x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024} $$
Similarly for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that
$$ x^3 \cos x \lt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$
Now it is easy to see that for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that
$$x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024} \gt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
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