Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Maximum value of function $f(x)=\frac{x^4-x^2}{x^6+2x^3-1}$ when $x >1$ What is the maximum value of the $$f(x)=\frac{x^4-x^2}{x^6+2x^3-1}$$ where $x > 1$ .
My try
Unable to solve further.
| Let $x-\frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain:
$$\frac{x^4-x^2}{x^6+2x^3-1}=\frac{x-\frac{1}{x}}{x^3-\frac{1}{x^3}+2}=\frac{t}{t^3+3t+2}=$$
$$=\frac{1}{t^2+\frac{2}{t}+3}\leq\frac{1}{3\sqrt[3]{t^2\left(\frac{1}{t}\right)^2}+3}=\frac{1}{6}.$$
The equality occurs for $t=1$, which says that we got a maximal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Easier way to parametrize an ellipse. I want to calculate the parametrization of the curve $$E=\{(x,y,z):z=x^2+y^2,x+y+z=1\}$$
To do so I did a change of variables taking as basis the vectors $\frac{1}{\sqrt{2}}(-1,1,0),\frac{1}{\sqrt{6}}(-1,-1,2),\frac{1}{\sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=\sqrt{3}/3$, and $$\begin{cases}x=-x'/\sqrt{2}-y'/\sqrt{6}+z'/\sqrt{3}\\y=x'/\sqrt{2}-y'/\sqrt{6}+z'/\sqrt{3} \\z=2y'/\sqrt{6}+z'/\sqrt{3}\end{cases}$$
Now using this change of variables into $E$ I have that $$\frac{(x')^2}{\frac{1}{(3^{(1/4)})^2}}+\frac{(y'-\sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}\cos\theta,\;y'=\sqrt{2/3}+3^{1/4}\sin\theta,\; z'=\sqrt{3}/3,\; 0\leq \theta \leq2\pi$$
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
| You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 \iff z = 1-x-y \tag{$\star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 \implies 1-x-y=x^2+y^2 \iff \left(x+\tfrac{1}{2}\right)^2+\left(y+\tfrac{1}{2}\right)^2=\tfrac{3}{2}$$
This is easily parametrized as $x+\tfrac{1}{2}=\sqrt{\tfrac{3}{2}}\cos t$ and $y+\tfrac{1}{2}=\sqrt{\tfrac{3}{2}}\sin t$ and $z$ follows from $(\star)$:
$$z = 1-x-y=1-\left(\sqrt{\tfrac{3}{2}}\cos t-\tfrac{1}{2}\right)-\left(\sqrt{\tfrac{3}{2}}\sin t-\tfrac{1}{2}\right)=2-\sqrt{\tfrac{3}{2}}\cos t-\sqrt{\tfrac{3}{2}}\sin t$$
to get:
$$\gamma : [0,2\pi]\to\mathbb{R}^3: t \mapsto
\left( \sqrt{\tfrac{3}{2}}\cos t-\tfrac{1}{2} ,
\sqrt{\tfrac{3}{2}}\sin t-\tfrac{1}{2} ,
2-\sqrt{\tfrac{3}{2}}\cos t-\sqrt{\tfrac{3}{2}}\sin t \right)$$
| {
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"url": "https://math.stackexchange.com/questions/3075737",
"timestamp": "2023-03-29T00:00:00",
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If $x$ and $y$ are acute, and $\sin y = 3 \cos (x+y) \sin x$, then find the maximum value of $\tan y$
Given $x,y$ are acute angles such that
$$\sin y = 3 \cos(x+y)\sin x$$
Find the maximum value of $\tan y$.
Attempt:
We have
$$\begin{aligned} 3(\cos x \cos y - \sin x \sin y) \sin x & = \sin y \\ 3 \cos x \sin x - 3 \sin^2 x \tan y & = \tan y \\ 3 \cos x \sin x & = \tan y(1 +3 \sin^2 x) \\ \tan y & = \dfrac{3 \sin x \cos x} {1+3 \sin^2 x} \end{aligned}$$
Now, how about the next step? Or maybe I did some mistakes?
| By AM-GM $$\tan{y}=\frac{3\sin{x}\cos{x}}{1+3\sin^2x}=\frac{3\sin{x}\cos{x}}{\cos^2x+4\sin^2x}\leq\frac{3\sin{x}\cos{x}}{2\sqrt{\cos^2x\cdot4\sin^2x}}=\frac{3}{4}.$$
The equality occurs for $\cos{x}=2\sin{x},$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^2 + y^2 + z^2 = 27$ and $xy + xz + yz = 11$, then what is $x + y + z$ (all values positive)?
Given that $x$, $y$, and $z$ are positive numbers and given that:
$$x^2 + y^2 + z^2 = 27 \quad\text{and}\quad xy + xz + yz = 11$$
What is the value of $x + y + z$?
I'm aware that one way to solve this problem is to square $x + y + z$ since $$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz)$$ However, I'm wondering how you would solve this without the aforementioned shortcut?
| One of the suboptimal ways is to make a variable replacement. For example, $u=x+y+z$, $v=(x-y)\sqrt{3/2}$, $w=-(x+y-2z)/\sqrt{2}$. After the replacement:
$$
x^2+y^2+z^2=\frac{u^2+v^2+w^2}3=27,\qquad xy+yz+zx=\frac{2u^2-v^2-w^2}6=11.
$$
And we need to find the value of $u$.
Then we make replacement $s=u^2, t=v^2+w^2$:
$$
s+t = 3\cdot27,\qquad2s-t=6\cdot 11,
$$
and we need to find the value of $\sqrt{s}$. Equations above are linear. So we can find $s$, and then calculate $\sqrt{s}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the convergence of $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ I want to find what the series $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ converges to exactly or show that it diverges.
By taking the partial sum of the series $S_N$ = $ \sum_{n=1}^{N} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ then $S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... + (N+1)^\frac{1}{3} - N^\frac{1}{3}$
And at the end I'm left with $S_N = -1 + (N+1)^\frac{1}{3}$ and $\lim_{N \to \infty} S_N = -1 + \infty= \infty $
So $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3} = \infty$
Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?
Thank you in prior.
| we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
$$ \small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} \cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
Just a visual thing, the pairs that cancel are next to each other this way
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
| Set
$ p(x) = x^2 - 3x + 1$
and observe that
$ p(x) + x = (x - 1)^2.$
Then for $r \ge 0$,
$$r = \sqrt{r^2} \\
= \sqrt{p(r + 1) + r + 1} \\
= \sqrt{p(r + 1) + \sqrt{(r + 1)^2}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + r + 2}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{(r + 2)^2}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + r + 3}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \sqrt{p(r + 4) + r + 4}}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \sqrt{p(r + 4) + \sqrt{p(r + 5) + r + 5}}}}} \\
= \ldots
$$
Now define for $q \ge 0$:
$$ F(r,q,x_0) = \begin{cases}
\sqrt{p(r + 1) + F(r + 1, q, x_0)}, & \mbox{if } r + 1 \le q\\
x_0, & \mbox{if } r + 1 \gt q.
\end{cases}
$$
So in particular:
$$\begin{array}{llll}
& F(r, r + 1, 0) &=& \sqrt{p(r + 1) + 0} \\
<& F(r, r + 2, 0) &=& \sqrt{p(r + 1) + \sqrt{p(r + 2) + 0}} \\
<& F(r, r + 3, 0) &=& \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + 0}}} \\
<& \lim_{n \rightarrow \infty} F(r, r + n, 0) &=_{\mbox{def}}& \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \ldots}}}
\end{array}$$
and also
$$\begin{array}{llll}
F(r, r + 1, 0) &< F(r, r + 1, r + 1) &= \sqrt{p(r + 1) + r + 1} &= r\\
F(r, r + 2, 0) &< F(r, r + 2, r + 2) &= \sqrt{p(r + 1) + \sqrt{p(r + 2) + r + 2}} &= r \\
F(r, r + 3, 0) &< F(r, r + 3, r + 3) &= \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + r + 3}}} &= r \\
F(r, r + n, 0) &< F(r, r + n, r + n) &= \ldots &= r.
\end{array}$$
So by the monotone convergence theorem, the limit
$$F(r, \infty, 0) =_{\mbox{def}} \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \ldots}}}$$
actually exists (and it's at most $r$). As it exists, it must satisfy the equation
$$F(r, \infty, 0) = \sqrt{p(r + 1) + F(r + 1, \infty, 0)}.$$
Now this is clearly satisfied for $F(r, \infty, 0) = r$, but that doesn't mean (yet) that $r$ is the answer. For $r = 0$, of course, we know
$$0 \le F(0,\infty,0) \le 0.$$
We also know
$$F(r + 1, \infty, 0) = F^2(r, \infty, 0) - p(r + 1).$$
If we use $F(r, \infty, 0) = r$ as an induction hypothesis, we obtain
$$\begin{array}{ll}
F(r + 1, \infty, 0) &= r^2 - p(r + 1)\\
&= r^2 - ((r + 1 - 1)^2 - (r + 1))\\
&= r + 1\\
\end{array}$$
for $r\in\mathbb N$ #.
Note on $0 \le r \lt \frac{1 + \sqrt 5}{2}$. When $p(r + 1) < 0$, we have to resort to the complex square root to make everything work. In that case, the monotonous convergence theorem has to be applied separately to the real and imaginary part of $F$. In particular (after a few terms), the real part increases towards $r$ (as expected), while the imaginary part decreases towards $0$.
$$\begin{array}{rcl}
F(0,1,0) &=& i\\
F(0,2,0) &=& e^{3\pi i/8}\\
F(0,3,0) &=& i\\
F(0,4,0) &\approx& 0.3259i\\
F(0,5,0) &\approx& 0.1062i\\
F(0,6,0) &\approx& 0.04115i\\
F(1,2,0) &=& i\\
F(1,3,0) &=& 0\\
F(1,4,0) &\approx& 0.8938\\
F(1,5,0) &\approx& 0.9852\\
F(1,6,0) &\approx& 0.9983
\end{array}$$
A key observation here is that $-1 + \sqrt{1 + \sqrt 5} > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximizing $x^4+y^4+z^4 + p(xy+xz+yz)$ on a sphere What values of $x,y,z$ maximize $f(x,y,z,p) = x^4+y^4+z^4 + p(xy+xz+yz)$ with constant $p \geq 0$ with the constraint $x^2+y^2+z^2=1$?
Some preliminary studies in Mathematica showed that the behavior of the solution changes around $p=1$. For $p>1$, the problem seems solved by $x=y=z=\frac{1}{\sqrt 3}$. For $0 \leq p \leq 1$ it appears the solution (up to permutations of x,y,z) looks like x > y = z. These solutions appear to favor one or the other of the two terms in $f$, and I find the $p>1$ behavior striking and would appreciate a proof that $x=y=z=\frac{1}{\sqrt 3}$ indeed maximizes.
UPDATE
For at least $p>4/3$, one can show that $x=y=z=\frac{1}{\sqrt 3}$ maximizes f.
The proof for $1<p<4/3$ evades me currently (the left bound, as mentioned in comments, may be slightly higher than 1).
I'll assume without loss of generality that $x,y,z$ are all positive. Note that $f(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}, p) = p+\frac{1}{3} $, so if we show that for some $p$ we enjoy $f(x,y,z,p) \leq p+\frac{1}{3}$, we've shown $x=y=z=\frac{1}{\sqrt 3}$ maximizes $f$.
Now consider that via the constraint, $$f(x,y,z,p) = 1-2(x^2y^2+x^2z^2+y^2z^2) + p(xy+xz+yz)$$
The root mean square compared to arithmetic mean inequality tells us that for nonnegative $a,b,c$ we enjoy $\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3},$ so $-(a^2+b^2+c^2) \leq -\frac{(a+b+c)^2}{3}$. Then $$f(x,y,z,p) \leq 1 + p(xy+xz+yz) -\frac{2}{3}(xy+xz+yz)^2.$$
Now, let's call $(xy+xz+yz)=q$. Note that it is easy to prove with our constraints that $0\leq q \leq 1$.
Then we can bound the maximum of $f(x,y,z,p)$ by finding the maximum of $g(q) = 1+pq-\frac{2}{3}q^2$ on $0\leq q \leq 1$.
The derivative $g'(q) = p-\frac{4}{3}q$, so $g(q)$ increases up to $q=\frac{3}{4}p$. Then on $0\leq q \leq 1$, we have $q_{max} = min(\frac{3}{4}p, 1)$.
Thus for $p \geq \frac{4}{3}$ we have $q_{max}=1$, so $$ f(x,y,z,p) \leq 1+p*1-\frac{2}{3}*1^2 = p + 1/3$$
Thus since $f(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}, p) = p+\frac{1}{3}$, we have for $p \geq \frac{4}{3}$ that $x=y=z=\frac{1}{\sqrt 3}$ maximizes $f$.
I have attempted the method of Lagrange multipliers, which led me to three cubic equations:
I set out to maximize
$4(x^3+y^3+z^3)+p(xy+xz+yz)+\lambda(x^2+y^2+z^2-1)$.
This yielded
$$4x^3+2x\lambda + p(y+z)=0$$
$$4y^3+2y\lambda + p(x+z)=0$$
$$4z^3+2z\lambda + p(y+x)=0$$
But now I feel a little at a loss on how to solve this system of equations.
One can verify that $x=y=z=\frac{1}{\sqrt{3}}$ solves these equations at any p with the right choice of lamda. I'm unsure about pairing Hessian techniques to verify the local nature of the critical point when using lagrange multipliers.
I have also tried incorporating the constraint by writing $z=\cos(\theta)$, $x=\sin(\theta)\cos(\phi)$ and $y=\sin(\theta)\sin(\phi)$.
This yields
$$f(\theta, \phi, p) = \cos(\theta)^4+(\sin(\theta)\cos(\phi))^4+(\sin(\theta)\sin(\phi))^4+p(\cos(\theta)\sin(\theta)\sin(\phi)+\cos(\theta)\sin(\theta)\cos(\phi)+(\sin(\theta))^2\sin(\phi)\cos(\phi) ) $$
which I am unsure how to maximize. However, with motivation from Michael Behrend, I've found that studying $\phi=\frac{\pi}{4}$ shows interesting critical point behavior slightly above p=1 where three critical points undergo some sort of bifurcation leaving just a single critical point at the eternal $x=y=z=\frac{1}{3}$ .
| Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+\lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,\;L'_y=0,\;L'_z=0,\;L'_x=0,\;L'_\lambda=0$$
we get that if $p\ge1$ global maximum is
$$f_{max}=p+\frac13$$
Maximum points is:
if $p>1$ then $[x=\frac{1}{\sqrt{3}},y=\frac{1}{\sqrt{3}},z=\frac{1}{\sqrt{3}}]$ or
$[x=-\frac{1}{\sqrt{3}},y=-\frac{1}{\sqrt{3}},z=-\frac{1}{\sqrt{3}}]$
if $p=1$ then $[x=\frac{\sqrt{6}}{3},y=\frac{1}{\sqrt{6}},z=\frac{1}{\sqrt{6}}],[x=-\frac{\sqrt{6}}{3},y=-\frac{1}{\sqrt{6}},z=-\frac{1}{\sqrt{6}}],[x=\frac{1}{\sqrt{6}},y=\frac{\sqrt{6}}{3},z=\frac{1}{\sqrt{6}}],[x=\frac{1}{\sqrt{6}},y=\frac{1}{\sqrt{6}},z=\frac{\sqrt{6}}{3}],[x=-\frac{1}{\sqrt{6}},y=-\frac{\sqrt{6}}{3},z=-\frac{1}{\sqrt{6}}],[x=-\frac{1}{\sqrt{6}},y=-\frac{1}{\sqrt{6}},z=-\frac{\sqrt{6}}{3}],[x=\frac{1}{\sqrt{3}},y=\frac{1}{\sqrt{3}},z=\frac{1}{\sqrt{3}}],[x=-\frac{1}{\sqrt{3}},y=-\frac{1}{\sqrt{3}},z=-\frac{1}{\sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=\pm1,y=0,z=0],[x=0,y=\pm1,z=0],[x=0,y=0,z=\pm1]$.
If $0<p<1$ we can't exact solve system.
| {
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Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:
Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$
For example, the case for $n=3$ can be proven like this:
Let $a,b,c \in \Bbb{R}^+$. Then we have:
$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $
$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $
By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:
$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $
However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.
Here is my attempt:
Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$
Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.
Inductive hypothesis: I assume $P(n)$ is true.
Inductive step:
$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$
$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$
And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.
Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?
| Also, you can use AM-GM:
$$\sum_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}\geq n\sqrt[n]{\prod_{k=1}^na_k}\cdot \frac{n}{\sqrt[n]{\prod\limits_{k=1}^na_k}}=n^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$. The following question is from Abstract Algebra by Dummit and Foote.
Determine the degree of the extension $\mathbb{Q}(\sqrt{3 +
2\sqrt{2}})$ over $\Bbb Q $.
This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake.
My attempt : I started out by constructing a polynomial $f (x)\in \Bbb{Q}[x]$ such that $\sqrt{3 +
2\sqrt{2}}$ is a root of $f(x)$.
$$x=\sqrt{3 +2\sqrt{2}}\;\implies x^2=3 +2\sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $\sqrt{3 +2\sqrt{2}}$ is a root of $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $\pm1$ and neither of these satisfy $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $\sqrt{3 +2\sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $\sqrt{3 +2\sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[\mathbb{Q}(\sqrt{3 +2\sqrt{2}}):\Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake.
Thank you.
| Because $$3+2\sqrt2=(1+\sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing monotonicity for ratio of binomial pmf and tail cdf I'm interested in showing for $X\sim\text{Bin}(n,p)$, $p\in(0,1)$ that when $x\geq np$,
$$
\frac{P(X=x)}{P(X\geq x)}\leq \frac{P(X=x+1)}{P(X\geq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out.
Below I plot the ratio $P(X=x)/P(X\geq x)$ for values of $x\geq \lfloor np\rfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.
Numerical Simulations
n=500
p=0.3
x=seq(floor(n*p),n)
ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)
diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)
plot(x,ratio)
plot(x,diff)
| We know that
$$
\frac{P(X=x+1)}{P(X=x)}=\frac{n-x}{x+1}\frac{p}{1-p}
$$
Therefore, we will show that
$$
\frac{P(X\geq x+1)}{P(X\geq x)}\leq \frac{n-x}{x+1}\frac{p}{1-p}
$$
Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
\begin{align}
\frac{1-p}{p}\frac{P(X\geq x+1)}{P(X\geq x)}&=\frac{1-p}{p}\frac{\sum_{k\geq x+1}\dbinom{n}{k}p^k(1-p)^{n-k}}{\sum_{k\geq x}\dbinom{n}{k}p^k(1-p)^{n-k}}\\
&=\frac{\binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+\cdots}{\binom{n}{x}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+\cdots}\tag{1}
\end{align}
Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that
$$
\frac{\binom{n}{x+1}}{\binom{n}{x}}=\frac{n-x}{x+1}
$$
and for $k\geq 1$
$$
\frac{\binom{n}{x+1+k}}{\binom{n}{x+k}}=\frac{n-x-k}{x+1+k}\leq \frac{n-x}{x+1}
$$
Combining the last two results, we may write for $k\geq 0$
$$
\binom{n}{x+1+k}\leq \frac{n-x}{x+1}\binom{n}{x+k}
$$
Substituting this inequality for each binomial coefficient in the numerator of (1), we have
\begin{align}
\frac{1-p}{p}\frac{P(X\geq x+1)}{P(X\geq x)}&=\frac{\binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+\cdots}{\binom{n}{x}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+\cdots}\\
&\leq \frac{n-x}{x+1}\frac{\sum_{k=x+1}^n \binom{n}{k-1}p^k(1-p)^{n-k} }{\left[\sum_{k=x+1}^n \binom{n}{k-1}p^k(1-p)^{n-k}\right] +p^{n+1}(1-p)^{-1} }\\
&\leq \frac{n-x}{x+1}
\end{align}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find convergence interval of $f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $ $$f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $$
so looks like what i have to do is use the formal
$$\lim_{n\to∞}\left|\frac{a_{n+1}}{a_n}\right|=a$$ with $a_n=\sin\frac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
| You may recall the fact that $\sin' = \cos$, which implies the standard limit
$$
\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin x - \sin 0}{x- 0} = \sin' 0 = \cos 0 = 1 \tag{1}
$$
from which we have, as $1/n^2 \to 0$ as $n\to\infty$,
$$
\lim_{n\to\infty} \frac{\sin \frac{2}{n^2}}{\frac{2}{n^2}} = 1 \tag{2}
$$
Therefore, we get, for $a_n \stackrel{\rm def}{=} \sin \frac{2}{n^2}$ (for all $n\geq 1$),
$$\begin{align}
\lim_{n\to\infty} \frac{a_{n+1}}{a_n} &=
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\sin \frac{2}{n^2}}
=
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\frac{2}{(n+1)^2}}\cdot\frac{\frac{2}{(n+1)^2}}{\frac{2}{n^2}}\cdot \frac{\frac{2}{n^2}}{\sin \frac{2}{n^2}}
\\&=
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\frac{2}{(n+1)^2}}\cdot\lim_{n\to\infty} \frac{n^2}{(n+1)^2}\cdot \lim_{n\to\infty} \frac{\frac{2}{n^2}}{\sin \frac{2}{n^2}}
\end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
\boxed{\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =
1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| Just use Cauchy's rule...
$$
\lim_{x \to 0} \dfrac{2\sqrt{1-x}-\sqrt[3]{8-x}}{x}=
\lim_{x\to 0} \left(-\frac{1}{\sqrt{1-x}}+\frac{1}{3(8-x)^{2/3}} \right)= -1 + \frac{1}{12}=-\frac{11}{12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
How to solve unfactorable trinomial for x? I have to solve the trinomial $2x^2+4x+1 = 0$, which should equal $-1 \pm \frac{1}{2}\sqrt{2}$
So far, my steps are:
$$2x^2+4x=-1$$
$$2(x^2+2x)=-1$$
$$x^2+2x=\frac{-1}{2}$$
This is where I get stuck. I can take $\frac{-1}{2}$ to the left side, but it's still unfactorable.
How can I solve this equation?
| The general procedure known as completing the square may be executed for any quadratic polynomial
$q(x) = x^2 + \alpha x + \beta \in F[x], \tag 1$
where $F$ is any field with
$\text{char}(F) \ne 2; \tag 2$
if we seek the roots of
$q(x) = x^2 + \alpha x + \beta = 0, \tag 3$
we write this equation as
$x^2 + \alpha x = - \beta; \tag 4$
then in the light of (2) the quantity
$\left( \dfrac{\alpha}{2} \right )^2 = \dfrac{\alpha^2}{4} \in F \tag 5$
is well-defined; we may add it to each side of (4) to obtain
$\left ( x + \dfrac{\alpha}{2} \right )^2 = x^2 + \alpha x + \dfrac{\alpha^2}{4} = \dfrac{\alpha^2}{4} - \beta; \tag 6$
if now
$\exists \gamma \in F, \; \gamma^2 = \dfrac{\alpha^2}{4} - \beta, \tag 7$
then we may write
$x + \dfrac{\alpha}{2} = \pm \gamma, \tag 8$
whence
$x = -\dfrac{\alpha}{2} \pm \gamma, \tag 9$
where we note that (2) implies
$\gamma \ne -\gamma, \; \gamma \ne 0, \tag{10}$
and so the roots (9) are distinct unless $\gamma$ vanishes. It is easy to walk these steps back and show that $x$ as in (9) satisfies (3).
In the present case we have, after dividing the quadratic equation
$2x^2 + 4x + 1 = 0 \tag{11}$
by $2$,
$x^2 + 2x + \dfrac{1}{2} = 0, \tag{12}$
we apply the above with $\alpha = 2$ and $\beta = 1/2$ and we see that
$(x + 1)^2 = x^2 + 2x + 1 = \dfrac{1}{2}, \tag{13}$
and the result
$x = -1 \pm \sqrt{\dfrac{1}{2}} = -1 \pm \dfrac{1}{2} \sqrt 2 = -1 \pm \dfrac{\sqrt 2}{2} \tag{14}$
immediately follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=K\frac{\arccos e}{\sqrt{1-e^{2}}}$ $$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=K\frac{\arccos e}{\sqrt{1-e^{2}}}, (e^{2}\lt1)$$
Find value of $K$ ?
I have solved till this step .
$$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=\frac{1}{2}
\int_{0}^{\pi}\frac{ \pi dx}{1+ e \sin x}$$
| $$F(a)=\int_0^\pi\frac{xdx}{1+a\sin x}$$
Somehow, you have shown that
$$F(a)=\frac\pi2\int_0^\pi\frac{dx}{1+a\sin x}$$
With which CAS seems to agree. We exploit symmetry:
$$F(a)=\pi\int_0^{\pi/2}\frac{dx}{1+a\sin x}$$
Then we preform the substitution $t=\tan\frac{x}2$, to see that
$$F(a)=2\pi\int_0^1\frac{1}{1+a\frac{2t}{t^2+1}}\frac{dt}{t^2+1}$$
So that
$$F(a)=2\pi\int_0^1\frac{dt}{t^2+2at+1}$$
Then we consider the integral
$$I(x;a,b,c)=\int\frac{dx}{ax^2+bx+c}$$
Complete the square on the bottom
$$I(x;a,b,c)=\int\frac{dx}{a(x+\frac{b}{2a})^2+g}$$
Where $g=c-\frac{b^2}{4a}$. I leave it as a challenge to you to use the trig sub $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$ to see that
$$I(x;a,b,c)=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}$$
So we see that
$$F(a)=2\pi\left[I(1;1,2a,1)-I(0;1,2a,1)\right]$$
$$F(a)=\frac{2\pi}{\sqrt{1-a^2}}\left(\arctan\sqrt{\frac{1+a}{1-a}}-\arctan\frac{a}{\sqrt{1-a^2}}\right)$$
Which works for $a^2<1$. Amazingly this simplifies down to
$$F(a)=\frac{\pi\arccos a}{\sqrt{1-a^2}}$$
So we have that $K=\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\int_{-\infty}^{\infty} \frac{\cos (\theta) e^{\theta y}}{2 \cosh(\pi y/2)} y dy=\tan (\theta)$. I'm not a hardcore mathematician. I have exhausted all integration trick I learned in calculus including change of variables, integration by parts. That cosh on denominator is just very much in the way. How one should approach an integral like this.
| Here is a real method.
Let
$$I = \int_{-\infty}^\infty \frac{\cos \theta \, e^{\theta y} \, y}{2 \cosh \left (\frac{\pi y}{2} \right )} \, dy,$$
where $\theta \in (-\pi/2, \pi/2)$.
Now
$$I = \frac{\cos \theta}{2} \int_{-\infty}^0 \frac{y e^{\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy + \frac{\cos \theta}{2} \int_0^\infty \frac{y e^{\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy = I_1 + I_2.$$
In the first of the integrals, enforcing a substitution of $y \mapsto -y$ gives
\begin{align}
I_1 &= -\frac{\cos \theta}{2} \int_0^\infty \frac{y e^{-\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy\\
&= -\cos \theta \int_0^\infty \frac{y e^{-\theta y}}{e^{\frac{\pi y}{2}} + e^{-\frac{\pi y}{2}}} \, dy \tag1\\
&= -\cos \theta \int_0^\infty \frac{y e^{-\theta y - \pi y/2}}{1 + e^{-\pi y}} \, dy\\
&= -\cos \theta \sum_{n = 0}^\infty (-1)^n \int_0^\infty y e^{-(n \pi + \pi/2 + \theta)} \, dy \tag2\\
&= -\cos \theta \sum_{n = 0}^\infty \frac{(-1)^n}{(n \pi + \pi/2 + \theta)^2} \tag3\\
&= -\frac{\cos \theta}{\pi^2} \sum_{n = 0}^\infty \frac{(-1)^n}{(n + \frac{1}{2} + \frac{\theta}{\pi})^2}\\
&= -\frac{\cos \theta}{\pi^2} \sum_{\substack{n = 0\\n \in \text{even}}}^\infty \frac{1}{\left (n + \frac{1}{2} + \frac{\theta}{\pi} \right )^2} - \frac{\cos \theta}{\pi^2} \sum_{\substack{n = 0\\n \in \text{odd}}}^\infty \frac{1}{\left (n + \frac{1}{2} + \frac{\theta}{\pi} \right )^2}\\
&= -\frac{\cos \theta}{4 \pi^2} \sum_{n = 0}^\infty \frac{1}{\left (n + \frac{1}{4} + \frac{\theta}{2 \pi} \right )^2} - \frac{\cos \theta}{4 \pi^2} \sum_{n = 0}^\infty \frac{1}{\left (n + \frac{3}{4} + \frac{\theta}{2 \pi} \right )^2} \tag4\\
&= -\frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left (\frac{\theta}{2 \pi} + \frac{1}{4} \right ) - \psi^{(1)} \left (\frac{\theta}{2 \pi} + \frac{3}{4} \right ) \right ] \tag5
\end{align}
In a similar manner, for the second of the integrals it can be shown that
$$I_2 = \frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) - \psi^{(1)} \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ].$$
Thus
\begin{align}
I &= -\frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left [1 - \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ] + \psi^{(1)} \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ]\\
& \qquad \quad + \frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left [1 - \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) \right ] + \psi^{(1)} \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) \right ]\\
&= -\frac{\cos \theta}{4 \pi^2} \left [\pi^2 \operatorname{cosec}^2 \left (\frac{3 \pi}{4} - \frac{\theta}{2} \right ) - \pi^2 \operatorname{cosec}^2 \left (\frac{\pi}{4} - \frac{\theta}{2} \right ) \right ] \tag6\\
&= -\frac{\cos \theta}{4} \left (\frac{2}{\sin \theta + 1} + \frac{2}{\sin \theta - 1} \right )\\
&= -\frac{\cos \theta}{4} \cdot -\frac{4 \sin \theta}{\cos^2 \theta},
\end{align}
or finally
$$\int_{-\infty}^\infty \frac{\cos \theta \, e^{\theta y} \, y}{2 \cosh \left (\frac{\pi y}{2} \right )} \, dy = \tan \theta,$$
as required.
Explanation
(1) Using $\cosh x = (e^x + e^{-x})/2$.
(2) Geometric expansion.
(3) Integration by parts twice.
(4) Shifting of the indices. Here $n \mapsto 2n$ for $n$ even and $n \mapsto 2n + 1$ for $n$ odd.
(5) Using the series representation for the polygamma function.
(6) Using the reflexion relation for the trigamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric integral involving the fractional part function I have tried the $u$ substitution yet I could not move forwards on this problem. Could you calculate in closed-form this integral?
$$\int_{0}^{\pi/2}\bigg\{\csc(x)\bigg\}\mathrm{dx}$$
| Enforcing the substitution $x\mapsto \arcsin(x)$ followed by $x\mapsto 1/x$ we see that
$$\begin{align}
\int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx&=\int_0^1 \bigg\{\frac1x\bigg\}\frac1{\sqrt{1-x^2}}\,dx\\\\
&=\int_1^\infty \{x\} \frac{1}{x\sqrt{x^2-1}}\,dx\\\\
&=\int_1^\infty \frac{x-\lfloor x\rfloor}{x\sqrt{x^2-1}}\,dx\\\\
&=\sum_{k=1}^\infty \int_{k}^{k+1}\frac{x-\lfloor x\rfloor}{x\sqrt{x^2-1}}\,dx\\\\
&=\sum_{k=1}^\infty \int_{k}^{k+1} \frac{x-k}{x\sqrt{x^2-1}}\,dx\tag1
\end{align}$$
Now, the integral under the summation sign on the right-hand side of $(1)$ can be evaluated in the closed form as
$$\begin{align}
\int_{k}^{k+1} \frac{x-k}{x\sqrt{x^2-1}}\,dx&=(k+1)\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)-k\arctan\left(\frac1{\sqrt{k^2-1}}\right)\\\\
&-\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)\\\\
&+\log(\sqrt{(k+1)^2-1}+(k+1))-\log(\sqrt{k^2-1}+k)
\end{align}\tag2$$
We can easily evaluate the telescoping series
$$\sum_{k=1}^\infty (k+1)\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)-k\arctan\left(\frac1{\sqrt{k^2-1}}\right)=1-\frac\pi2\tag3$$
We can also write the sum $S$ where
$$S=\sum_{k=1}^\infty \left(\log(\sqrt{(k+1)^2-1}+(k+1))-\log(\sqrt{k^2-1}+k)
-\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)\right)$$
as
$$\begin{align}
S&=\lim_{K\to\infty}\left(\log(K+\sqrt{K^2-1})-\sum_{k=2}^K \arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\\\\
&=\log(2)+\lim_{K\to\infty}\left(\log(K)-\sum_{k=2}^K \arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\\\\
&=1+\log(2) -\gamma +\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\tag4
\end{align}$$
Putting $(1)$-$(4)$ together, we find that
$$\int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx=2+\log(2)-\frac\pi2-\gamma-\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\tag5$$
The series in $(5)$ converges quickly with
$$\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\approx -0.0368911$$
Hence,
$$\int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx\approx 0.5082441$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of $\prod^{6}_{r=0}\cos\bigg(\frac{\pi}{21}+\frac{r\pi}{7}\bigg)$
Find value of $$\prod^{6}_{r=0}\cos\left(\frac{\pi}{21}+\frac{r\pi}{7}\right)$$
What I try:
$$\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot \cos\frac{7\pi}{21}\cdot \cos \frac{10\pi}{21}\cdot \cos \frac{13\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{19\pi}{21}$$
$\displaystyle \cos \frac{19\pi}{21} = -\cos \frac{2\pi}{21}$ and $\displaystyle \cos \frac{13\pi}{21}=-\cos \frac{8\pi}{21}$
How do I solve it? Help me, please.
| I believe this is how the problem came into being
Use $\cos(\pi-x)=-\cos x,$
$P=-\prod_{r=0}^6\cos(\pi/21+2r\pi/7),$
Observe that $\cos7(x+2r\pi/7)=\cos7x,$
Now if $\cos(2n+1)y=\cos(2n+1)x$ which is $2^{2n}\cos^{2n+1}x+\cdots+(-1)^n(2n+1)\cos x$
So, the roots of $$2^{2n}\cos^{2n+1}x+\cdots+(-1)^n(2n+1)\cos x-\cos(2n+1)y=0$$ are $\cos(2r\pi/(2n+1)+y)$ where $0\le r\le2n$
So, the product of the roots will be $$\dfrac{\cos7y}{2^{2n}}$$
Here $7y=\pi/3$ and $2n+1=7\iff n=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rigorous or not? I want to prove
$$T(n,k)=\frac{n}{\left\lfloor\frac{n+k+1}{2}\right\rfloor}(T(n-1,k)+T(n-1,k-1))=\binom{n}{k}\binom{n-k}{\left\lfloor\frac{n-k}{2}\right\rfloor}$$
First we know only that
$$T(n,k)=0, n<k$$
$$T(n,k)=0, n<0, k<0$$
$$T(0,0)=1$$
so obviously
$$T(n,0)=\frac{n}{\left\lfloor\frac{n+1}{2}\right\rfloor}T(n-1,0)=\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$$
next we may notice (for $n\geqslant k$)
$$T(n,k)=\frac{n}{\left\lfloor\frac{n+k+1}{2}\right\rfloor}(\frac{\left\lfloor\frac{n-k+1}{2}\right\rfloor}{n}T(n,k)+\frac{k}{n}T(n,k))$$
then
$$T(n,k)=\frac{n}{k}T(n-1,k-1)=\binom{n}{k}T(n-k,0)=\binom{n}{k}\binom{n-k}{\left\lfloor\frac{n-k}{2}\right\rfloor}$$
which agrees with
$$T(n,k)=\frac{n}{\left\lfloor\frac{n-k+1}{2}\right\rfloor}T(n-1,k)$$
Can we say that all steps looks clear? Is there another way to prove it?
| Although there is already a good answer, let me present a different approach.
We can better restate the problem as
$$
\left\{ \matrix{
T(n,k) = 0\quad \left| {\;n < 0\; \vee \;k < 0} \right. \hfill \cr
T(n,k) = \binom{n}{k} \binom{n-k}{\left\lfloor {{{n - k} \over 2}} \right\rfloor} = \hfill \cr
= \left[ {0 = n} \right]\left[ {0 = k} \right] + \left[ {1 \le n} \right]{n \over {\left\lfloor {{{n + k + 1} \over 2}} \right\rfloor }}
\left( {T(n - 1,k) + T(n - 1,k - 1)} \right) \hfill \cr} \right.
$$
where $[condition]$ denotes the Iverson bracket
Let's recall some useful properties of the floor and ceil functions. We have that, for integer $m$
$$
\eqalign{
& m = \left\lfloor {{m \over 2}} \right\rfloor + \left\lceil {{m \over 2}} \right\rceil \cr
& \left\lceil {{m \over 2}} \right\rceil = \left\lfloor {{{m + 1} \over 2}} \right\rfloor \cr}
$$
Calling $B(n,k)$ the binomial expression, we can write it as
$$
B(n,k) = \binom{n}{k} \binom{n-k}{\left\lfloor {{{n - k} \over 2}} \right\rfloor}
= \left[ {0 \le k \le n} \right]{{n!} \over {k!}}{1 \over {\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}}
$$
Then we have
$$
\eqalign{
& {{B(n - 1,k)} \over {B(n,k)}} = \cr
& = \left[ {0 \le k \le n - 1} \right]{{\left( {n - 1} \right)!k!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}
\over {k!\left\lfloor {{{n - 1 - k} \over 2}} \right\rfloor !\left\lceil {{{n - 1 - k} \over 2}} \right\rceil !n!}} = \cr
& = \left[ {0 \le k \le n - 1} \right]{1 \over n}{{\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor !} \over {\left\lfloor {{{n - 1 - k} \over 2}} \right\rfloor !}} = \cr
& = \left[ {0 \le k \le n - 1} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor \cr}
$$
and
$$
\eqalign{
& {{B(n - 1,k - 1)} \over {B(n,k)}} = \cr
& = \left[ {1 \le k \le n} \right]{{\left( {n - 1} \right)!k!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}
\over {\left( {k - 1} \right)!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !n!}} = \cr
& = \left[ {1 \le k \le n} \right]{k \over n} \cr}
$$
So we conclude that
$$
\eqalign{
& {{B(n - 1,k) + B(n - 1,k - 1)} \over {B(n,k)}} = \cr
& = \left[ {0 \le k \le n - 1} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor + \left[ {1 \le k \le n} \right]{k \over n} = \cr
& = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k\left( { \le n - 1} \right) \le n} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor
+ \left[ {0 \le \left( {1 \le } \right)k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]{k \over n} = \cr
& = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]{1 \over n}\left( {\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor + k} \right) = \cr
& = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]\;{{\left\lfloor {{{n + 1 + k} \over 2}} \right\rfloor } \over n} \cr}
$$
and the thesis is demonstrated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Rolling 9 dice probability
Nine fair dice are rolled simultaneously. What is the probability of getting three pairs?
My attempt:
$$P(A) = \frac{\binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times3\times2\times1}{6^{9}}$$
We first choose which three numbers will be pairs, and which dice will be pairs, then we are left with three numbers and three positions for them. Is this correct?
| Here's another approach to counting the numerator:
Take the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9.$ Arrange them in any sequence.
Then pick three of the numbers $1, 2, 3, 4, 5, 6$ and replace $7, 8, 9$ with those numbers in increasing order (that is, $7$ is replaced by the smallest number,
$9$ by the largest).
This produces every possible sequence of nine numbers selected from
$\{1, 2, 3, 4, 5, 6\}$ with three pairs and three singletons, that is, the number of rolls that have exactly three pairs and no other matches, but it produces each such sequence more than once.
In particular, each of the numbers $7, 8,$ and $9$ could originally have appeared either before or after the numbers they eventually are made to duplicate,
so each roll of three pairs is produced exactly eight times.
After correcting for this overcounting, the number of rolls is
$$ \frac{9! \binom63}{8} = \frac52(9!).$$
This turns out to be equal to
$$ \binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times 6 =
\binom{6}{3} \frac{9!7!5!\times6}{(2!7!)(2!5!)(2!3!)}
= \binom{6}{3} \frac{9!}{(2!)^3},$$
confirming your answer for the probability of exactly three pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of the differential equation $(1+y^2-x^2)y'=\frac{1}{x}, y(1)=1$ Solution of the differential equation $(1+y^2-x^2)y'=\frac{1}{x}, y(1)=1$ is the differential equation have solution ? and is this solution is exist?
I am trying to prove by using picard's theorem
here given ODE is $(1+y^2-x^2)y'=\frac{1}{x}$
then $f(x,y)=\frac{1}{(1+y^2-x^2)x}\Rightarrow |f(x,y)|=|\frac{1}{(1+y^2-x^2)x}|$ from here how to processed
| $(1+y^2-x^2)y'=\dfrac{1}{x}$ with $y(1)=1$
$(1+y^2-x^2)\dfrac{dy}{dx}=\dfrac{1}{x}$ with $y(1)=1$
$\dfrac{1}{x}\dfrac{dx}{dy}=1+y^2-x^2$ with $x(1)=1$
Let $u=\dfrac{1}{x^2}$ ,
Then $\dfrac{du}{dy}=-\dfrac{2}{x^3}\dfrac{dx}{dy}$
$\therefore-\dfrac{x^2}{2}\dfrac{du}{dy}=1+y^2-x^2$ with $u(1)=1$
$\dfrac{du}{dy}=2-\dfrac{2(y^2+1)}{x^2}$ with $u(1)=1$
$\dfrac{du}{dy}=2-2(y^2+1)u$ with $u(1)=1$
$\dfrac{du}{dy}+2(y^2+1)u=2$ with $u(1)=1$
I.F. $=e^{\int2(y^2+1)~dy}=e^{\frac{2y^3}{3}+2y}$
$\therefore\dfrac{d\left(ue^{\frac{2y^3}{3}+2y}\right)}{dy}=2e^{\frac{2y^3}{3}+2y}$ with $u(1)=1$
$u=2e^{-\frac{2y^3}{3}-2y}\int e^{\frac{2y^3}{3}+2y}~dy+Ce^{-\frac{2y^3}{3}-2y}$ with $u(1)=1$
$Ce^{-\frac{2}{3}-2}=1$
$C=e^\frac{8}{3}$
$\therefore u=2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}$
$\dfrac{1}{x^2}=2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}$
$x^2=\dfrac{1}{2e^{-\frac{2y^3}{3}-2y}\int_1^ye^{\frac{2y^3}{3}+2y}~dy+e^{\frac{8}{3}-2y-\frac{2y^3}{3}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\mathcal{M}\left(\sin(x)\right)(s) = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)$ using Real Analysis recently I've been investigating Mellin Transforms and this morning solved for case of $\sin(x)$ using Ramunajan's Master Theorem. I was curious if there were any Real based methods to evaluate this integral as in searching around online all proofs seem to rely on Contour Integration.
My approach:
\begin{equation}
\mathcal{M}\left(\sin(x)\right)(s) = \int_0^\infty x^{s - 1}\sin(x)\:dx = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)
\end{equation}
By definiton:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty x^{s - 1}\sin(x)\:dx = \int_0^\infty x^{s - 1}\left[\sum_{m = 0}^{\infty} (-1)^m \frac{x^{2m + 1}}{(2m + 1)!} \right]\:dx \\&= \int_0^\infty x^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(x^2\right)^m}{(2m + 1)!}\:dx
\end{align}
Here we make the substitution $u = x^2$:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty \left(\sqrt{u}\right)^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(u\right)^m}{(2m + 1)!}\frac{du}{2\sqrt{u}} = \frac{1}{2}\int_0^\infty u^{\frac{s - 1}{2}}\sum_{m = 0}^{\infty} (-1)^m \frac{u^m}{(2m + 1)!}\:du \\
&= \frac{1}{2}\int_0^\infty u^{\frac{s + 1}{2} - 1}\sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!}\:du = \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right)
\end{align}
Where
\begin{equation}
g(u) = \sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!} = \sum_{m = 0}^{\infty} \frac{m!}{(2m + 1)!}\frac{(-u)^m}{m!} = \sum_{m = 0}^{\infty} \frac{\Gamma(m + 1)}{\Gamma(2m + 2)}\frac{(-u)^m}{m!}
\end{equation}
We observe that this form enables the use of Ramanujan's Master Theorem. Thus:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right) = \frac{1}{2} \Gamma\left( \frac{s + 1}{2}\right) \cdot \frac{\Gamma\left(- \frac{s + 1}{2} + 1\right)}{\Gamma\left(2\cdot -\frac{s + 1}{2} + 2\right)} \\
&= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)}
\end{align}
Employing Euler's Reflection Formula on the terms in the numerator we arrive at:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)} = \frac{1}{2} \cdot \frac{1}{\Gamma(1 - s)}\cdot \frac{\pi}{\sin\left(\pi \cdot \frac{s + 1}{2}\right)} = \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)}
\end{align}
Now Euler's Reflection Formula:
\begin{equation}
\Gamma(s)\Gamma(1 - s) = \frac{\pi}{\sin(\pi s)} \rightarrow \Gamma(1 - s) = \frac{\pi}{\Gamma(s)\sin(\pi s)}
\end{equation}
Returning to our integral and observing that $\sin\left(\frac{\pi}{2} (s + 1)\right)= \cos\left(\frac{\pi}{2}s\right)$ we arrive at:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)} = \frac{\pi}{2} \cdot \frac{1}{\frac{\pi}{\Gamma(s)\sin(\pi s)} \cdot \cos\left(\frac{\pi}{2}s\right)} = \frac{\Gamma(s)\sin(\pi s)}{2 \cos\left(\frac{\pi}{2} s\right)}
\end{align}
We now use the double angle formula $\sin\left(\pi s\right) = 2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)$ to yield:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{\Gamma(s)\sin\left(\pi s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \frac{\Gamma(s)\cdot 2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \Gamma(s)\sin\left(\frac{\pi}{2} s\right)
\end{align}
As required.
| For $b > 0, \Re(s) > 0$ $$b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx $$
By induction if for some $ b,\Re(b) > 0 $ we have $\forall s, \Re(s) >0,b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx$ then for every $|a/b|< 1, \Re(a+b) > 0$, using the binomial series $$\int_0^\infty x^{s-1} e^{-(a+b)x}dx = \sum_{k=0}^\infty \int_0^\infty x^{s-1}\frac{(-ax)^k}{k!} e^{-bx}dx=\sum_{k=0}^\infty \frac{(-a)^k}{k!} b^{-s-k} \Gamma(s+k)\\ =b^{-s} \Gamma(s)\sum_{k=0}^\infty {-s \choose k} (a/b)^k = (a+b)^{-s} \Gamma(s)$$
From which $b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx $ is true for every $b, \Re(b) > 0$
And hence for $\Re(s) \in (0,1)$
$$\int_0^\infty x^{s-1} \sin(x)dx = \lim_{b \to 0^+} \int_0^\infty x^{s-1} \frac{e^{-(b-i)x}-e^{-(b+i)x}}{2i}dx \\ = \lim_{b \to 0^+} \frac{(b-i)^{-s}-(b+i)^{-s}}{2i} \Gamma(s)= \sin(\pi s/2) \Gamma(s)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-3}^2 \frac{[x]}{[x]^2+2[x]+x^3}dx$ I came across this question which was supposed to be solved in about $1$ or $2$ minutes, but I came across a severe roadblock.
The question was:
Integrate $$\int_{-3}^2 \frac{\lfloor x\rfloor}{\lfloor x\rfloor^2+2\lfloor x \rfloor+x^3}dx$$
I split the integral into $5$ separate ones to substitute a known value of step $x$. But following that I had to use partial fractions with some weird numbers and that made it pretty long. How do I approach this sum?
| Yes you are right we have to split the integral into 5:
$$I_1 = \int_{-3}^{-2}\frac{-3dx}{3 + x^3}$$
$$I_2 = \int_{-2}^{-1}\frac{-2dx}{x^3} = \frac{3}{4}$$
$$I_3 = \int_{-1}^{0}\frac{-dx}{-1 + x^3}$$
$$I_4 = 0$$
$$I_5 = \int_{1}^{2}\frac{dx}{3 + x^3}$$
Separately we try to find the integral
$$\int\frac{dx}{a^3 + x^3}$$
$$= \int(\frac{1}{3a^2(x + a)} + \frac{1}{3a^2}\frac{- x + 2a}{x^2 -ax + a^2})dx$$
$$= \int\frac{1}{3a^2}(\frac{1}{x + a} - \frac{x - \frac{a}{2} - \frac{3a}{2}}{(x - \frac{a}{2})^2 + (\frac{\sqrt{3}a}{2})^2})dx$$
$$= \frac{1}{6a^2}ln\frac{(x + a)^2}{x^2 -ax + a^2} + \frac{1}{\sqrt{3}a^2}tan^{-1}\frac{2x - a}{\sqrt{3}a}$$
Using this formula we have
$$I = I_1 + I_2 + I_3 + I_4 + I_5 = 0.2748 + 0.75 + 0.8356 + 0 + 0.1615 = 2.0219$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sequence converging towards Euler-Mascheroni constant I have been trying to find the limit of the following sequence:
$\lim_{n\to \infty}(\frac{1}{2n+1} + \frac{1}{2n+2}+...+\frac{1}{8n+1})$
Here is my attempt with the Euler-Mascheroni constant:
$(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n+1} + \frac{1}{2n+2}+...+\frac{1}{8n+1}) - (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}) =$
$=\gamma + \varepsilon_{8n+1} + \ln(8n+1) -(\gamma + \epsilon_{2n} + \ln 2n)$
$=\ln\frac{8n+1}{2n}$
Can someone check if this is correct or is there another way to solve this problem?
| I think you can also treat this as a Riemann sum:
$$\sum_{k=1}^{6n+1} \frac{1}{2n+k}=\frac{1}{n}\sum_{k=1}^{6n+1} \frac{1}{2+\frac{k}{n}}\longrightarrow \int_0^{6}\frac{1}{2+x}\,dx=\ln 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find:
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$.
| My usual naive playing.
Suppose the values are
$u, v, w$
and we want to find
$a, d, m, n$
such that
$u = a,
v = a+md,
w = a+nd$.
This is 3 equations in
4 unknowns,
so we expect this to be
underdetermined
and probably have
a one-parameter set of solutions.
Then
$md = v-u,
nd = w-u$
so
$\dfrac{n}{m}
=\dfrac{w-u}{v-u}
=r
$.
Then
$n = mr$
and
$w-v
=(m-n)d
=(m-mr)d
=md(1-r)
$.
For any $d$,
$m
=\dfrac{w-v}{d(1-r)}
$,
$n = mr$,
$a = u$.
For your case,
$u, v, w
=\frac12, \frac13, \frac15
$,
$r
=\dfrac{\frac15-\frac12}{\frac13-\frac12}
=\dfrac{6-15}{10-15}
=\dfrac95
$,
$m
= \dfrac{\frac15-\frac13}{-d\frac45}
= \dfrac{5\cdot-\frac{2}{15}}{-4d}
= -\dfrac{1}{6d}
$,
$n
=mr
=-\dfrac{1}{6d}\dfrac95
=-\dfrac{9}{30d}
=-\dfrac{3}{10d}
$,
$a
=\frac12
$.
Check:
$a+md=
=\dfrac12-\dfrac{1}{6d}d
=\dfrac12-\dfrac16
=\dfrac13
$
and
$a+nd
=\dfrac12-\dfrac{3}{10d}d
=\dfrac12-\dfrac{3}{10}
=\dfrac15
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Determinant of a $3\times 3$ matrix in simplest form. Let $\alpha$ and $\beta $ be fixed non-zero reals and $f(n)=\alpha^n+\beta^n$ with $$A=\begin{pmatrix}
3&1+f(1)&1+f(2)\\1+f(1)&1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4)
\end{pmatrix}.$$ How to find determinant of the matrix $A$ i.e. $|A|,$ in the simplest form? Firstly I thought that matrix may be of a particular form. But its neither circulant matrix nor Vandermonde type or tridiagonal type. Then I tried multilinear property on columns of given matrix as below:
$$\det\begin{pmatrix}
3&1+f(1)&1+f(2)\\1+f(1)&1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4)
\end{pmatrix}=\det\begin{pmatrix}
3&1+f(1)&1\\1+f(1)&1+f(2)&1\\ 1+f(2)&1+f(3)&1
\end{pmatrix} +\det\begin{pmatrix}
3&1+f(1)&f(2)\\1+f(1)&1+f(2)&f(3)\\ 1+f(2)&1+f(3)&f(4)
\end{pmatrix} $$
I further apply multilinear property on columns but didn't get my answer. Please help. Thanks.
| Hint: Consider
$$
\left(\begin{matrix}
1&1&1\\1&\alpha&\beta\\ 1^2&\alpha^2&\beta^2
\end{matrix}\right)\left(\begin{matrix}
1&1&1^2\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2
\end{matrix}\right).
$$ Do you recognize it now?
Possibly more systematic way: Recognizing each column of $A$ is a linear combination of vectors$$
u=\left(\begin{matrix}1\\ 1 \\1 \end{matrix}\right),\quad \ v=\left(\begin{matrix}1\\\alpha\\ \alpha^2\end{matrix}\right),\quad \ w=\left(\begin{matrix}1\\ \beta\\ \beta^2\end{matrix}\right),$$ express the matrix as
$$
A=(u+\ v+ w\ ,\ \ u+\alpha v+\beta w\ ,\ \ u+\alpha^2 v+ \beta^2 w).
$$ Regarding $u,v,w$ as if they were numbers, if necessary,
$$
A=(u,v,w)\left(\begin{matrix}
1&1&1\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2
\end{matrix}\right)=\left(\begin{matrix}
1&1&1\\1&\alpha&\beta\\ 1&\alpha^2&\beta^2
\end{matrix}\right)\left(\begin{matrix}
1&1&1\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2
\end{matrix}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Linear system of equations that arises from recurrence relation. So when solving a third order homogenous recurrence relation I end upp with the task of determening the constants $a,b$ and $c$. The initial conditions give rise to the following system of equations:
$$\left\{
\begin{array}{rcr}
2a+(1+\sqrt{6})b+(1-\sqrt{6})c & = & 0\\
4a+(1+\sqrt{6})^2b+(1-\sqrt{6})^2c & = & 0\\
8a+(1+\sqrt{6})^3b+(1-\sqrt{6})^3c & = & 1
\end{array}
\right.$$
I've verified that this system is correct but now I need to solve it. I find brute force Gaussian elimination quite tedious to do (or maybe there is some trick to make it simple?). Is there any other easier method to solve this without having to spend 3-4 pages of hand written scribble?
The answer is
$$a=-\frac{1}{10}, \ b=\frac{1}{10\sqrt{6}}, \ c=-\frac{1}{10\sqrt{6}}.$$
| I would suggest that you use Cramer's method, i.e. using determinants.
One thing that will help you is to notice the following:
Let $\phi=1+\sqrt6$. Then you have
$$1-\sqrt6 = \frac{(1-\sqrt6)(1+\sqrt6)}{1+\sqrt6}=\frac{-5}{\phi}\ \ \ (*)$$
Put this into the equations, get rid of the fractions and find the determinants. One property that will be useful is that if a row or column of a determinant has a common factor, you can take the factor out in front of the determinant.
After solving the system you will get expressions that contain $\phi$ and you will have to calculate some powers of $\phi$ which is not too hard to do.
Edit: More steps.
By using (*), you can eventually get to the system:
\begin{eqnarray*}
2\phi a+\phi^2 b-5c &=& 0 \\
4\phi^2 a+\phi^4 b+25c &=& 0 \\
2\phi^3 a+\phi^6 b-125c &=& \phi^3
\end{eqnarray*}
Now, the determinants are:
$$D =
\left|\begin{array}{ccc}
2\phi & \phi^2 & -5 \\
4\phi^2 & \phi^4 & 25 \\
8\phi^3 & \phi^6 & -125 \\
\end{array}\right| = 2\phi\cdot\phi^2\cdot(-5)
\left|\begin{array}{ccc}
1 & 1 & 1 \\
2\phi & \phi^2 & -5 \\
4\phi^2 & \phi^4 & 25 \\
\end{array}\right|
$$
This determinant can be calculated using Sarrus' rule.
$$
D_a = -5\phi^2
\left|\begin{array}{ccc}
0 & 1 & 1 \\
0 & \phi^2 & -5 \\
\phi^3 & \phi^4 & 25 \\
\end{array}\right| =-5\phi^5
\left|\begin{array}{cc}
1 & 1 \\
\phi^2 & -5 \\
\end{array}\right|
$$
$$ D_b = -10\phi
\left|\begin{array}{ccc}
1 & 0 & 1 \\
2\phi & 0 & -5 \\
4\phi^2 & \phi^3 & 25 \\
\end{array}\right| = 10\phi^4
\left|\begin{array}{cc}
1 & 1 \\
2\phi & -5 \\
\end{array}\right|
$$
$$ D_c = 2\phi^3
\left|\begin{array}{ccc}
1 & 1 & 0 \\
2\phi & \phi^2 & 0 \\
4\phi^2 & \phi^4 & \phi^3 \\
\end{array}\right| = 2\phi^6
\left|\begin{array}{cc}
1 & 1 \\
2\phi & \phi^2 \\
\end{array}\right|
$$
You should get $$D=-10\phi^4(2\phi^4+\phi^3+5\phi-50)$$
$$D_a = 5\phi^5(\phi^2+5)$$
$$D_b=-10\phi^4(2\phi+5)$$
$$D_c=2\phi^7(\phi-2)$$
The solutions are then $a=D_a/D$, $b=D_b/D$, $c=D_c/D$.
Plugging in the numbers I get the same solution that you gave.
Please let me know if I have made any errors along the way.
| {
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"url": "https://math.stackexchange.com/questions/3116871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $n$ $\in$ $\Bbb Z^+$ such that: $\lfloor\frac{n}{2}\rfloor \cdot \lfloor \frac{n}{3} \rfloor \cdot \lfloor \frac{n}{4} \rfloor = n^2$ Find all the numbers $n$ $\in$ $\Bbb Z^+$ such that:
$$\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2$$
I never worked before with floor function so i'm not completely sure how to solve this. I think (only because $n$ $\in$ $\Bbb Z^+$), i can just multiply (skipping the floor function) and get the answer of $n=24$, but this is floor function so i don't know if there are more solutions.
Any hints?
| You already ofund a solution $n=24$.
For $n<24$, we have
$$ \left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor\le \frac n{24}\cdot n^2<n^2.$$
For $n\ge 30$, we have
$$\begin{align}\left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor -n^2&\ge\frac{n-1}2\frac{n-2}3\frac{n-3}4-n^2\\&=\frac{n^3-30n^2+11n-6}{24}\\&\ge\frac{11n-6}{24}\\&>0.\end{align}$$
Hence we need at most check $25\le n\le 29$.
To simplify checks for these cases, note that $\left\lfloor\frac n2\right\rfloor$, $ \left\lfloor\frac n3\right\rfloor$, $\left\lfloor\frac n4\right\rfloor$ are three distinct integers $\ge 6$. As $29^2$ and $5^4$ cannot be written as product of three distinct integers $\>1$, we can exclude $n=29$ and $n=25$. For $3^6$, the only way to write it as product of three distinct integers is as $3\cdot 9\cdot 27$, but $3<6$, so we can exclude $n=27$ as well.
For $n=26$, we see that only one of the three factors is a multiple of $13$, so this can be ruled out as well. Fially, for $n=28$, $\lfloor \frac n3\rfloor=9$, but $28^2$ is not a multiple of $3$.
| {
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"url": "https://math.stackexchange.com/questions/3119036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$ if $\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac{2}{w}$
If $\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}$, where $a,b,c,d\in\mathcal{R}$ and $\omega$ is a non-real cube root of unity, then prove that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$
As it was asked as a multiple choice question(this expression was one option to pick) I really do not see an easy way to prove this.
$$
\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\
\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\
\frac{1}{a\omega^2+\omega}+\frac{1}{b\omega^2+\omega}+\frac{1}{c\omega^2+\omega}+\frac{1}{d\omega^2+\omega}=2\\
\frac{1}{a\omega+1}+\frac{1}{b\omega+1}+\frac{1}{c\omega+1}+\frac{1}{d\omega+1}=2\\
$$
Is it no coincidence that replacing $\omega$ with other two cube root of unity satisfy the equation ?
| I don't think its that different from @Luis solution, but it seems more convincing to me.
$$
\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\
\implies \frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\
$$
$\omega,\omega^2$ are the solutions of the equation,
$$
\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}\\
\tfrac{(b+x)(c+x)(d+x)+(a+x)(c+x)(d+x)+(a+x)(b+x)(d+x)+(a+x)(b+x)(c+x)}{(a+x)(b+x)(c+x)(d+x)}=\tfrac{2}{x}\\
\frac{4x^4+2x^3\sum a+3x^2\sum ab+x\sum abc}{x^4+x^3\sum a+x^2\sum ab+x\sum abc+abcd}=2\\
\boxed{2x^4+x^3\sum a+0.x^2-x.\sum abc-2abcd=0}
$$
$$
\alpha\beta+\beta\gamma+\gamma\eta+\eta\alpha+\alpha\gamma+\beta\eta=0\quad\&\quad\gamma=\omega,\;\eta=\omega^2\\
\alpha\beta+\omega\beta+\omega^3+\omega^2\alpha+\omega\alpha+\omega^2\beta=0\\
\alpha\beta-\alpha-\beta+1=0\implies \alpha(\beta-1)-(\beta-1)=(\alpha-1)(\beta-1)=0\\
\alpha=1\quad\&\quad\beta=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$,
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
| A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = \frac{dx}{(2x+1)^3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $ Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can simplify this..
$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$
| First of all note that $$\lim_{x\to 0} \frac{1-\cos x\cos 2x}{x^2}=\frac{5}{2}\tag{1}$$ To see why the above equation is true you can observe that the numerator in expression under limit is of the form $1-ab$ where $a, b$ tend to $1$. And we can write $1-ab$ as $$1-a+a(1-b)$$ so that limit in equation $(1)$ is equal to $$\lim_{x\to 0}\frac{1-\cos x} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$$ and thus the desired limit in $(1)$ is $1/2+2=5/2$.
Next we can use the same technique for the limit in question as the numerator is of the form $1-ab$ with $a,b$ tending to $1$. Thus the limit in question is equal to $$\lim_{x\to 0}\frac{1-\cos t} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\tag{2}$$ where $$t=\frac{1-\cos x\cos 2x}{x^2}-\frac{5}{2}\tag{3}$$ The second term in equation $(2)$ evaluates to $2$ and the first term can be written as $$\lim_{x\to 0}\frac{1-\cos t} {t^2}\cdot\left(\frac{t}{x}\right)^2$$ which is same as $$\frac{1}{2}\lim_{x\to 0}\left(\frac{t}{x}\right) ^2$$ Now we have $$\frac {t} {x} =\frac{2-2\cos x\cos 2x-5x^2}{2x^3}\tag{4}$$ and using $\cos 2x=1-2\sin^2x$ and similarly for $\cos x$ we get the numerator in $(4)$ as $$4\sin^2x+4\sin^2(x/2)-5x^2-8\sin^2x\sin^2(x/2)$$ Clearly the last term upon division with $2x^3$ tends to $0$ and hence the limit of fraction in $(4)$ is equal to the limit $$\frac{1}{4}\lim_{u\to 0}\frac{\sin^22u+\sin^2u-5u^2}{u^3}$$ via substitution $x=2u$ and this is the same as $$2\lim_{u\to 0}\frac{\sin 2u-2u}{(2u)^2}\cdot\frac{\sin 2u+2u}{2u}+\frac{1}{4}\lim_{u\to 0}\frac{\sin u-u} {u^2}\cdot\frac{\sin u+u} {u} $$ Since $(\sin u-u) /u^2\to 0$ it follows that the above limit is $0$ and hence the expression in $(4)$ namely $t/x\to 0$. The desired limit is thus $2$.
Note that the above solution is a bit lengthy as it avoids use of L'Hospital's Rule and Taylor series.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $ \sqrt{2}|z|\geq|Re(z)| + |Im(z)|$ I need to prove $\sqrt{2}|z|\geq|Re(z)| + |Im(z)|$
This is what I tried to do :
$Let$ $z=x+iy$ , this gives us $|Re(z)| = |x|$ and $|Im(z)|=|y|$
Also , $|z| = \sqrt{x^2+y^2}$
Then we have $\sqrt{2}\sqrt{x^2+y^2} \geq x +y $
$\Rightarrow$ $\sqrt{2x^2+2y^2} \geq x +y$
$\Rightarrow$ $2x^2+2y^2 \geq (x +y)^2$
$\Rightarrow$ $2x^2+2y^2 \geq x^2 +2xy +y^2$
$\Rightarrow$ $x^2+y^2 \geq 2xy $
Is this correct so far ? How do I proceed from here ?
| Further to greedoid's answer, you really should use $\Leftarrow$ or $\iff$ instead of $\implies$ because we want to show that $a^2+b^2\ge 2ab$, a corollary of $(a-b)^2\ge 0$, implies the desired result $\sqrt{2a^2+2b^2}\ge a+b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding maximum of $y-z$ given quadratic constraint Problem
According to my friend, this is a problem that seems to be solvable by observation.
Try to find the maximum of $(y-z)$ given following two constraints
$$
\begin{aligned}
x+y+z=3\\
x^2+y^2+z^2=9
\end{aligned}
$$
This is a convex QCQP problem and it is solvable by some solver like cvxpy but I am not sure how do I solve this problem by observation?
Except treating it as a QCQP problem, I also tried to eliminate $z$ and finally cast it into maximizing $x+2y-3$ given
$$
x^2+y^2+2xy-3(x+y)=0\tag{1}
$$
the hope is that I factor (1) into terms of $(x+2y-3)$, but I failed to do this.
So does anyone know how to solve this, thank you in advance.
| I'm not sure about using "observation", but here is how I would approach solving your $2$ equations of
$$x + y + z = 3 \tag{1}\label{eq1}$$
$$x^2 + y^2 + z^2 = 9 \tag{2}\label{eq2}$$
You can start to "complete the square" by using \eqref{eq2} - $2 \times$ \eqref{eq1} to get
$$x^2 - 2x + y^2 - 2y + z^2 - 2z = 9 - 2 \times 3 = 3 \tag{3}\label{eq3}$$
Each of $x$, $y$ and $z$ need a term of $1$, so adding this to both sides and then completing the square for each variable gives that
$$\left(x - 1\right)^2 + \left(y - 1\right)^2 + \left(z - 1\right)^2 = 6 \tag{4}\label{eq4}$$
First, note that $y - z = \left(y - 1\right) - \left(z - 1\right)$. Since each term above is squared, the solutions would involve $x - 1 = \pm m$, for some non-negative $m$, and likewise for the other $2$ terms. Also, the maximum values for $\mid y - 1 \mid$ and $\mid z - 1 \mid$ will be reached when $x = 1$ so the first term is $0$ and we then get
$$\left(y - 1\right)^2 + \left(z - 1\right)^2 = 6 \tag{5}\label{eq5}$$
As the MSE question at Maximizing the sum of two numbers, the sum of whose squares is constant shows in several ways, the maximum value of the sum of absolute values is achieved when $\left(y - 1\right)^2 = \left(z - 1\right)^2 = 3$, giving $y - 1 = \sqrt{3}$ and $z - 1 = -\sqrt{3}$, for a maximum value of $y - z$ being $2\sqrt{3}$. Note \eqref{eq1} still holds as $x = 1$, $y = 1 + \sqrt{3}$ and $z = 1 - \sqrt{3}$ gives $x + y + z = 3$, plus \eqref{eq2} also holds, confirming this is a valid solution.
It's important to check the original equations since if we had instead added $2 \times$ \eqref{eq1} to \eqref{eq2} to complete the squares to get $\left(x + 1\right)^2 + \left(y + 1\right)^2 + \left(z + 1\right)^2 = 18$ and then followed the same procedure to set $x = -1$, we would have got $y + 1 = 3$ and $z + 1 = -3$ for $y - z = 6$. However, $x + y + z = -3$ in this case, so \eqref{eq1} wouldn't be true then, and $x^2 + y^2 + z^2 = 21$, so \eqref{eq2} would also fail! Completing the squares by subtracting allows determining the maximum value of $y - z$ most easily as $x = 1$ works in that case to simplify both the resulting equation & to get a maximum result with $x, y, z$ values that still satisfy both of the original $2$ equations.
It's possible this, or something similar, is what your friend was referring to, where if they have enough knowledge & experience doing this sort of thing, then to them it's solved by "observation". However, it's of course best to ask your friend what they specifically meant.
Update: I think I see how this can be solved by "inspection". First, split the RHS value of \eqref{eq1} equally among the $3$ variables to do a similar linear transform of $x = x_1 + 1$, $y = y_1 + 1$ and $z = z_1 + 1$. Now, \eqref{eq1} becomes
$$x_1 + y_1 + z_1 = 0 \tag{6}\label{eq6}$$
Also, the LHS of \eqref{eq2} becomes, including using \eqref{eq6} after grouping,
\begin{align}
x^2 + y^2 + z^2 & = \left(x_1 + 1\right)^2 + \left(y_1 + 1\right)^2 + \left(z_1 + 1\right)^2 \\
& = x_1^2 + 2x_1 + 1 + y_1^2 + 2y_1 + 1 + z_1^2 + 2z_1 + 1 \\
& = x_1^2 + y_1^2 + z_1^2 + 2\left(x_1 + y_1 + z_1\right) + 3 \\
& = x_1^2 + y_1^2 + z_1^2 + 3 \tag{7}\label{eq7}
\end{align}
Thus, \eqref{eq2} is now
$$x_1^2 + y_1^2 + z_1^2 = 6 \tag{8}\label{eq8}$$
The condition to maximize of $y - z = y_1 - z_1$ can now easily be done due to the special condition that allows $x_1 = 0$ in \eqref{eq8} to permit a maximum value where $y_1 = -z_1$ exactly matches the resulting condition in \eqref{eq7} when $x_1 = 0$!
More generally, if the LHS of \eqref{eq1} is $c$ & the LHS of \eqref{eq2} is $d$, the maximum value of $y - z$ would be $2\sqrt{\frac{3d \, - \, c^2}{6}}$, which simplifies to $2\sqrt{3}$ in our case.
| {
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"timestamp": "2023-03-29T00:00:00",
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Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\log\bigg(\frac{1+t^2}{1+3t^2}\bigg)\stackrel{H}{=}\lim\limits_{t\to 0}\frac{1}{2t}\bigg(\frac{2t}{1+t^2}-\frac{6t}{1+3t^2}\bigg)=\lim\limits_{t\to 0}\bigg(\frac{1}{1+t^2}-\frac{3}{1+3t^2}\bigg)=-2$$
It seems the result is the difference between the two functions inside the log, but how this can be possible? Is there another way to compute these type of limits? I mean a calculus way without using theorems.
| Alternatively:
$$\lim_{x\to+\infty} x^2 \ln \left(\frac{x^2+1}{x^2+3}\right)=\\
\lim_{x\to+\infty} \ln \left(\frac{x^2+1}{x^2+3}\right)^{x^2}=\\
\ln \left[\lim_{x\to+\infty} \left[\left(1+\frac{-2}{x^2+3}\right)^{\frac{x^2+3}{-2}}\right]^{\frac{-2x^2}{x^2+3}}\right]=\\
\ln e^{\lim_\limits{x\to+\infty} \frac{-2x^2}{x^2+3}}=\\
\ln e^{-2}=-2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation:
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it:
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1 \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1 \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid.
New attempt.
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1 \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers.
| Hint: We can write the equation as
$$
\left(1+\frac1 x\right)\left(1+\frac1 y\right)\left(1+\frac1 z\right)=2.
$$ Assume without loss of generality $x\le y\le z$ by rearranging. Observe that
$
1+\frac 1 x<2\le \left(1+\frac1 x\right)^3 \implies x\in \{2,3\}
$ and using this information, investigate each case $x=2,3$ and so on.
Addendum, Solution.
*
*For $x=2$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 43$. Then $y>3$ and only $(y,z)=(4,15), (5,9), (6,7)$ work. (Check this using $\frac{z+1}z =\frac{4}{3}\frac{y}{y+1}$.)
*For $x=3$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 32$. Using $y\ge x=3$ and $\frac{z+1}z =\frac{3}{2}\frac{y}{y+1}$, check that only $(y,z)=(3,8), (4,5)$ work.
By rearranging, every solution is a permutation of such solutions with $x\le y\le z$.
| {
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$, so I only have to deal with $\int \frac{\sqrt{x+1}}{x}dx$.
Setting $u=\sqrt{x+1}$, we obtain $du=\frac{1}{2\sqrt{x+1}}dx$, therefore $dx=2udu$ and $x=u^2-1$, so
$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$
$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$
But then $\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$, which is not what I am supposed to obtain. The actual answer is $2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
| No mistake! $log((\frac{\sqrt{x+1}-1}{x})^2)=-2log(\frac{x}{\sqrt{x+1}-1})=-2log(\frac{x(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)})=-2log\frac{x(\sqrt{x+1}+1)}{x}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$
I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?
| Let's write
$$\frac{\ln(1+n+n^3) - 3 \ln(n)}{n(1- \cos(1/n^2))} = \frac{\ln(1 + \frac{1}{n^2} + \frac{1}{n^3})}{n(1- \cos(1/n^2))} = \frac{\frac{1}{n^2} + \frac{1}{n^3} +o(\frac{1}{n^3})}{n (\frac{1}{2n^4} + o(\frac{1}{n^5}))} = \frac{n +1 + o(1)}{\frac{1}{2} + o(\frac{1}{n})} \sim 2n$$
so the limit is $+\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Choosing two constants Let $r=2n+2$ where $n\in\mathbb{N}$
When $r\le 10$, is it possible to choose $\nu >0,\mu >\nu $ such that
$$\max(\frac{1}{6},\frac{1}{8}+\frac{3}{8(r-1)})<\nu<\frac{1}{4}+\frac{1}{2(r-1)}$$
and $$\max(\frac{1}{6},\frac{1}{12}+\frac{\nu}{3})<\mu<\frac{1}{3}+\frac{1}{2(r-1)}-\frac{2\nu}{3}$$
Thanks in advance
| It is easy to check when $r\le 10 $ then $\tfrac{1}{6}\le\tfrac{1}{8}+\tfrac{3}{8(r-1)}$.
If $r=2$ then there is no such $\nu$. Indeed, we have $\tfrac 18+\frac 3{8(r-1)}=\frac 12<\nu$. On the other hand, $\nu<\mu<\frac 13+\frac{1}{2}-\frac {2\nu}{3}$ which implies $\nu<\frac 12$.
If $r\ge 4$ then we have to pick $\tfrac{1}{8}+\tfrac{3}{8(r-1)}<\nu<\tfrac {1}{2}+\tfrac{1}{2(r-1)}$. We have to pick $\mu>\nu$. Since $\nu>\tfrac{1}{8}+\tfrac{3}{8(r-1)}\ge\tfrac 16$ and $\nu>\tfrac 18$, this also assures conditions $\mu>\tfrac 16$ and $\mu>\tfrac 1{12}+\tfrac{\nu}{3}$. So to assure a possibility to chose $\mu$ it remains to satisfy a condition $\nu<\tfrac{1}{3}+\tfrac 1{2(r-1)}-\tfrac{2\nu}{3}$, that is $\nu<\tfrac{3}{5}+\tfrac 3{10(r-1)}$. This can be achieved iff
$\tfrac{1}{8}+\tfrac{3}{8(r-1)}<\tfrac{3}{5}+\tfrac 3{10(r-1)}$, that is $3<19(r-1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding an integral using a table? Am I correct for pattern matching this integral?
I have
$$\int \frac{\sqrt{9x^2+4}}{x^2}dx$$
Does this pattern match with:
$$\int \frac{\sqrt{a^2 + x^2}}{x^2}dx = -\frac{a^2 + x^2}{x} + \ln(x + \sqrt{a^2 + x^2}) + c$$
If I factor out the 9, I get
$$= 3 \int \frac{\sqrt{x^2 + \frac{4}{9}}}{x^2}$$
with $a = \frac{2}{3}$
I get:
$$3 \left( - \frac{\sqrt{\frac{4}{9}+x^2}}{x} + \ln\left(x+\sqrt{\frac{4}{9}+x^2}\right) +c\right)$$
Is this the right track?
Wolfram winds up with a different answer though:
| You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3\ln{(\frac32(x+\sqrt{\frac49 + x^2}))} = 3\ln{(x+\sqrt{\frac49 + x^2})} + 3\ln\frac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
| {
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"timestamp": "2023-03-29T00:00:00",
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $\frac12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + 2Dx^2 + Dx + Ex^3 + 2Ex^2 + E$
= $ \int (A + D + E)x^3 + (3A + B + C + 2D + 2E)x^2 + (2A + 2C + 4B + D)x + (4B + E)$
Turn into matrix, find reduced row echelon form to solve system of equations:
$\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
3 & 1 & 1 & 2 & 2 & 1 \\
2 & 4 & 2 & 1 & 0 & 1 \\
0 & 4 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$
This is where things go wrong, apparently this doesn't reduce down properly :( and I have been relying on RREF to solve systems of equations everytime up until now.
I am also confused where the $x^2 + x + 1$ is supposed to go exactly. I know I put it into the system of equations later but until then, I feel like I kinda just ignored it and left it out of all my work up until then (is that okay?).
Other note: I recognize that this is a proper rational fraction so no long division is nesecary and that this has irreducible factors that are repeated so that is why I split them up into the partial fractions up above in that manner. Did I miss any intermittent steps that made the RREF turn out wrong? I am not sure where I went wrong thus far either
| Hint
Let $x+1=y\iff x=?$
$$\dfrac{x(x+1)+1}{(x+1)^2(x+2)}=\dfrac{y^2-y+1}{y^2(y+1)}=\dfrac1{y+1}-\dfrac1{(y+1)y}+\dfrac1{y^2(y+1)}$$
For the last two terms, replace $1$ with $y+1-y$ in the numerator
| {
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Expression Involving Roots of a Polynomial I'm working on part of a problem, which reads as follows:
Let $\theta$ be a root of the equation $x^{3}+x^{2}+x+2=0$. Express $(\theta^{2}+\theta+1)(\theta^{2}+\theta)$ and $(\theta-1)^{-1}$ in the form $a\theta^{2}+b\theta+c$, where $a,b,c\in\mathbb{Q}$.
My attempt at a solution: I let $f(x)=x^{3}+x^{2}+x+2$; then I was able to show that $f$ is irreducible over $\mathbb{Q}$. Using the division algorithm, we have $f(x)=(x-1)(x^{2}+2x+3)+5$. Since $f(\theta)=0$ by assumption, we evaluate at $x=\theta$ and get $$\begin{align}0=f(\theta)=(\theta-1)(\theta^{2}+2\theta+3)+5&\Longrightarrow \frac{-5}{\theta-1}=\theta^{2}+2\theta+3\\&\Longrightarrow(\theta-1)^{-1}=-\frac{1}{5}\theta^{2}-\frac{2}{5}\theta-\frac{3}{5}.\end{align}$$ My trouble is with finding $(\theta^{2}+\theta+1)(\theta^{2}+\theta)$. I think this is what you do, but I'm not sure: since $0=\theta^{3}+\theta^{2}+\theta+2$, we have $\theta^{3}=-(\theta^{2}+\theta+2)$. Hence, using the fact that $\theta^{4}=\theta\cdot\theta^{3}$, we have $$\begin{align}(\theta^{2}+\theta+1)(\theta^{2}+\theta)&=\theta^{4}+2\theta^{3}+2\theta^{2}+\theta\\
&=-\theta(\theta^{2}+\theta+2)-2(\theta^{2}+\theta+2)+2\theta^{2}+\theta\\
&=-\theta^{3}-\theta^{2}-3\theta-4\\
&=(\theta^{2}+\theta+2)-\theta^{2}-3\theta-4\\
&=-2\theta-2.
\end{align}$$
Does my solution look correct? Thank you in advance for any feedback.
| The systematic way for the first item is to follow what you did for $\theta-1$: compute the remainder of $(x^2+x+1)(x^2+x)$ when divided by $x^3+x^2+x+2$:
$$
(x^2+x+1)(x^2+x) = (x + 1)(x^3+x^2+x+2)+(-2 x - 2)
$$
Therefore,
$$
(\theta^{2}+\theta+1)(\theta^{2}+\theta) = -2\theta-2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A question about the famous paper that finds solutions for $33=x^3+y^3+z^3$ I have a question about the following assertion in this paper, which finds solutions for $33=x^3+y^3+z^3$.
At the bottom of pg 2, it says $$d=|x|-|y|\equiv |z|\mod 3$$
Then it says that for $\epsilon=\{\pm 1\}$, we have
$$x\equiv y\equiv z\equiv \epsilon \mod 3$$
Why is that? It's also possible that $z\equiv 0\mod 3$, and $x\equiv y\mod 3$, right?
| The reason is (partially) given in the part of the sentence which was not quoted:
since every cube is congruent to $0$ or $\pm1 \pmod 9$
More specifically, if $x\equiv a\pmod 3$, where $a$ is $-1,0$ or $1$, then $x^3\equiv a\pmod 9$. This is a very useful restriction, especially if $k\equiv\pm3\pmod 9$.
In particular, if $k = x^3 + y^3 + z^3$ and $k\equiv 3\varepsilon\pmod 9$ with $\varepsilon=\pm1$, the only way to solve the original equation is for all three of $x^3,y^3,z^3$ to be congruent to $\varepsilon\pmod 9$, which means $x,y,z$ are each congruent to $\varepsilon \pmod 3$. Thus we can have $3 \equiv 1 + 1 + 1 \pmod 9$, but it cannot be produced by any other combination of three $-1$s, $0$s and $1$s; similarly for $-3$, which is the relevant value for $k = 33$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$
If
$$\sin(x) - \cos(x) = \frac{1}{3}$$
then determine
$$\sin(x)\cos(x)$$
I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
| Note that
$$\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = -\frac{1}{3\sqrt{2}}.$$
Now, using a double-angle formula:
$$\cos(2x + \pi/2) = 2 \cos^2(x + \pi/4) - 1 = 2 \left( -\frac{1}{3\sqrt{2}}\right)^2 - 1 = -\frac{8}{9}.$$
On the other hand,
$$\cos(2x + \pi/2) = \cos(2x) \cos(\pi/2) - \sin(2x) \sin(\pi/2) = -\sin(2x) = -2 \sin x \cos x.$$
Equating the two, $-2 \sin x \cos x = -\frac{8}{9}$.
In general, this method of using a sum-of-angles formula in reverse is very often useful in analyzing expressions of the form $A \cos x + B \sin x$ when $A$ and $B$ are constants, and is a nice tool to keep available in your toolbox.
| {
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Find all integers of the form $\frac{(x-1)^2(x+2)}{2x+1}$ I made a python program and I have that
\begin{array}{|c|c|}
\hline x & \frac{(x-1)^2(x+2)}{2x+1} \\\hline
-14 & 100 \\\hline
-5& 12 \\\hline
-2&0 \\\hline
-1&4 \\\hline
0&2 \\\hline
1&0 \\\hline
4&6 \\\hline
13&80 \\\hline
\end{array}
And taking very large intervals of integers this seems like the only integers of this form.
However I don't know how can i proved.
Any ideas?
| Assuming $x$ is an integer
as $2x+1$ is odd
$2x+1$ will divide $(x-1)^2(x+2)$
iff $2x+1$ divides $(2x-2)^2(2x+4)=u$
Let $2x+1=y, u=(y-3)^2(y+3)\equiv27\pmod y$
So, the necessary & sufficient condition is: $y=2x+1$ must divide $27$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$
My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
| Given a $4\times 4$ Vandermonde matrix
$$
\left[\begin{matrix}
\color{red}1&\color{red}1&\color{red}1&1\\\color{red}a&\color{red}b&\color{red}c&d\\ a^2&b^2&c^2&\color{blue}{d^2}\\\color{red}{a^3}&\color{red}{b^3}&\color{red}{c^3}&d^3
\end{matrix}\right],
$$ note that the desired determinant is the minor of $d^2$. Considering Laplace expansion, it appears as a minus(-) coefficient of $d^2$ in the Vandermonde determinant
$$\begin{align*}
&\color{red}{(b-a)(c-a)(c-b)}(d-a)(d-b)(d-c)\\=&\color{red}{(b-a)(c-a)(c-b)}(d^3-\color{blue}{(a+b+c)}d^2+\cdots).
\end{align*}$$ Thus we obtain the desired determinant
$$\color{red}{(b-a)(c-a)(c-b)}\color{blue}{(a+b+c)}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $4\sin^2{x}\cos^2{x}-\cos^2{x}=0$
Does anyone know why these answers are wrong?
$$4\sin^2{x}\cos^2{x}-\cos^2{x}=0$$
$(4\sin^2x)(\cos^2x)=\cos^2x$
$4\sin^2x=1$
$\sin x=1/2$
$x=30° , 150°$
Thank you
| $(4\sin^2x)(\cos^2x)=\cos^2x$ implies $\cos x=0$ or $4\sin^2x=1$. For $0\le x<360^\circ$, the former is true for $x=90^\circ,270^\circ$ while the latter is true when $\sin x=\pm1/2$, which in turn is true of $30^\circ,150^\circ,210^\circ,330^\circ$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that the locus of point $N$ is $x^2+y^2=a^2$
Question: A point $P(a\cos\theta,b\sin\theta)$ lies on an ellipse with equation $$\varepsilon:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$The tangent to the ellipse $\varepsilon$ at point $P$ is perpendicular to a straight line $l$ which has passed through its focus and intersected at point $N$. Show that the equation of locus of point $N$ is $x^2+y^2=a^2$.
Suppose the coordinate point of the focus is $F(c,0)$. To find the point $N$ parametrically, I have to deal with the following simultaneous equations.
$$\begin{cases}
y-b \sin \theta=-\dfrac{b}{a} \cot \theta \, (x- a \cos \theta)\\y=\dfrac{b}{a}\tan \theta \,(x-c)
\end{cases}$$
where the first equation is the equation of tangent line at point $P$, and the second equation represents the perpendicular line $l$ that passes through the focus of the ellipse $\varepsilon$ and point $N$.
Solving the above equation for $x$ and $y$ in terms of $a$, $b$ and $c$ is quite complicated and outsmarted because it also involves a new variable $\theta$. Any pretty way to deal with it?
| Preliminaries:-
(1) For the standard ellipse ($\epsilon : \dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$), its foci are $(ae, 0)$ and $(–ae, 0)$; where $e$ is the eccentricity and is related to $a$ and $b$ by $b^2 = a^2(1 – e^2)$.
(2) If $L$ is tangent to $\epsilon$ at $P[\theta]$, then the equation of $L$ is $\dfrac {x \cos \theta}{a} + \dfrac { x \sin \theta}{b} = 1$.
(3) An alternate form of $L$ is $y = mx + c$ for some $m$ and $c$.
(4) Eliminating $c$ from the equations found in (2) and (3), we have $c = \pm \sqrt {m^2a^2 + b^2}$. That is, the equation of $L$ is $L : y – mx = \pm \sqrt {m^2a^2 + b^2}$.
The main part:-
(5) If $N$ is the line that passes thro’ (ae, 0) and perpendicular to $L$, then ….. $N : my + x = ae$.
(6) Adding the Squares of both sides of (4) and (5), we get $(1 + m^2)(x ^2 + y^2) = b^2 + a^2e^2 + m^2a^2$.
After replacing the $b^2$ in (6) by the relation stated in (1), the required result follows when we eliminate $(1 + m^2)$ from both sides of (6).
| {
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Limits: factoring out $x$ from $\lim_\limits{x\to +\infty}\left(\frac{5-x^3}{8x+2}\right)$ So my teacher said that I cannot use arithmetic operation to factor out $x$ from this type of equation, saying that it's because it's composed only by addition and subtraction. But I don't understand clearly, because I get the right answer (according to the book):
$$\lim_{x\to+\infty}\left(\frac{5-x^3}{8x+2}\right)
=\lim_{x\to\infty}\frac{x\times\left(\frac{5}{x}-x^2\right)}{x\times\left(8+\frac{2}{x}\right)}
=\lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}}
=\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)}
=\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}\right)-\lim_\limits{x\to\infty}\left(x^2\right)}{\lim_\limits{x\to\infty}\left(8\right)+\lim_\limits{x\to\infty}\left(\frac{2}{x}\right)}
=\frac{0-\infty}{8+0}
=\frac{-\infty}{8}$$
Applying the infinity property: $\frac{-\infty}{-c}=\infty$
$=-\infty$
Can someone explain to me why I can't factor $x$ out?
| The writing $$
\lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}}
=\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)}
$$
is not so good because $\lim\limits_{x\rightarrow\infty}\left(\frac{5}{x}-x^2\right)$ is not number.
We can say the same words about your next steps.
I think it's better to write the following.
$$\lim_{x\rightarrow\infty}\frac{5-x^3}{8x+2}=\lim_{x\rightarrow\infty}\left(x^2\cdot\frac{\frac{5}{x^2}-x}{8x+2}\right)=-\infty$$
because $$\lim_{x\rightarrow\infty}\frac{\frac{5}{x^2}-x}{8x+2}=\lim_{x\rightarrow\infty}\frac{\frac{5}{x^3}-1}{8+\frac{2}{x}}=-\frac{1}{8}$$ and
$$\lim_{x\rightarrow\infty}x^2=+\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx$, evaluate $\lim\limits_{n\to \infty} nI_n$ If
$$I_n=\int_0^1 \frac{x^n}{x^2+2019}\,\mathrm dx,$$ find $\lim_{n\to \infty} nI_n$. Can somebody help me, please?
I've only found that $$\frac{1}{2020} \le nI_n \le \frac{1}{2019}$$ knowing that $x^2 \in [0,1]$, but it doesn't help to evaluate the limit.
I want a proof without the Arzelà-Ascoli theorem or dominated convergence.
| The term $nx^n$ in the numerator of our integral is almost a derivative: $(x^n)' = nx^{n-1}$, so this suggests the following approach.
Notice that by the product rule,
\begin{align}
\frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg) &= nx^{n-1}\cdot\frac{x}{x^2+2019} + x^n\cdot\frac{\mathrm d}{\mathrm dx}\bigg(\frac{x}{x^2+2019}\bigg) \\
&= \frac{nx^n}{x^2+2019} + x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}
\end{align}
For $0 \leqslant x \leqslant 1$, the second term above is bounded above in absolute value by $|x|^n = x^n$. Integrating and applying the fundamental theorem of calculus gives
\begin{align}
\int_0^1\frac{\mathrm d}{\mathrm dx}\bigg(x^n\cdot \frac{x}{x^2+2019}\bigg)\,\mathrm dx &= n\int_0^1\frac{x^n}{x^2+2019}\,\mathrm dx + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx \\
\leadsto\quad\frac{1}{2020} &= nI_n + \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx.
\end{align}
Hence,
\begin{align}
nI_n = \frac{1}{2020} - \int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx.
\end{align}
By the triangle inequality and the fundamental theorem of calculus again,
\begin{align*}
\bigg|\int_0^1 x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\,\mathrm dx\bigg| &\leqslant \int_0^1\bigg|x^n\cdot \frac{-x^2+2019}{(x^2+2019)^2}\bigg|\,\mathrm dx \\
&\leqslant \int_0^1 x^n\,\mathrm dx \\
&= \frac{1}{n+1} \to 0,\quad\text{as $n\to\infty$.}
\end{align*}
Thus,
$$
\lim_{n\to\infty}nI_n = \frac{1}{2020}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3163280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Irrational integral $\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$ Would anyone be able to verify if this integral is calculated correctly?
$$\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$$
My attempt:
substitute:$\left(t = \sqrt{x}, t^2 = x, 2tdt=dx \right)$
$$
\begin{split}
\int \frac{1}{t} \sqrt{\frac{t-2}{t+2}}\,2t\,dt
&= 2\int\sqrt{\frac{t-2}{t+2}}dt
= 2\int \frac{\sqrt{t-2}}{\sqrt{t+2}}dt \\
&= 2\int \frac{\sqrt{(t-2)(t-2)}}{\sqrt{(t+2)(t-2)}}dt \\
&= 2\int \frac{t-2}{\sqrt{t^2-4}}dt \\
&= \int{\frac{2t-4}{\sqrt{t^2-4}}}dt \\
&= \int \frac{2t}{\sqrt{t^2-4}}dt - 4\int\frac{dt}{\sqrt{t^2-4}} \\
&= \sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C
\end{split}
$$
Substitute back $t = \sqrt{x}$:
result: $\Longrightarrow \sqrt{x-4} - 4\ln{|\sqrt{x} + \sqrt{x-4}|} + C$
| $$...= \color{red}{2}\sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C$$
Anyway, you could take a derivate and check it you self.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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$\int \cos^4{x}dx$ unsolvable with $t = \tan{x}$? I have been told that "universal substitution always works", so I wanted to give it a try on this specific integral.
$\int \cos^4{x}dx$
For some reason it does not work. Please note that I solved this integral in a normal way, just don't understand why the universal substitution does not work.
$\Bigg(t=\tan{x}, \cos^2{x} = \frac{1}{1+t^2},dx=\frac{dt}{1+t^2} \Bigg)$
$\int \cos^4xdx = \int (\cos^2x)^2dx = \int \Big(\frac{1}{1+t^2} \Big)^2 \frac{dt}{1+t^2} = \int \frac{dt}{(1+t^2)(1+t^2)(1+t^2)}$
$\frac{1}{(1+t^2)^3} = \frac{At +B}{1+t^2} + \frac{Ct+D}{(1+t^2)^2} + \frac{Et+F}{(1+t^2)^3}$
$1 = (At+B)(t^4+2t^2+1)+(Ct+D)(1+t^2)+Et+F$
$1 = At^5 + 2At^3 + At + Bt^4 + 2Bt^2 + B + Ct^3 + Ct + Dt^2 + D + Et + F$
Now this boils down to six linear equations:
$0 = A$
$0 = B$
$0 = 2A + C$
$0 = 2B + D$
$0 = A + C + E$
$1 = B + D + F$
Which results in: $A = 0, B = 0, C = 0, D = 0, E = 0, F = 0$ making it unsolvable for me.
Any ideas what went wrong?
| You made a mistake in solving your system of equations. The solution is
$$A = 0, B = 0, C = 0, D = 0, E = 0, F = 1$$
You could have seen this directly by observing that $\frac{1}{(1+t^2)^3}$ is already in the desired form:
$$\frac{1}{(1+t^2)^3} = \frac{0t +0}{1+t^2} + \frac{0t+0}{(1+t^2)^2} + \frac{0t+1}{(1+t^2)^3}$$
You could have skipped this step and go directly to the "reduction formula"
$$\int\frac{dt}{(t^2+1)^m}=\frac{t}{2(m-1)(t^2+1)^{m-1}}+\frac{2m-3}{2m-2}\int\frac{dt}{(t^2+1)^{m-1}}$$
(or integration by parts if you are not familiar with this).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of $a,b \in \Bbb R$ (if exists) such that $-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$ for all $x \in \Bbb R$ Find the values of $a,b \in \Bbb R$ (if exists) such that $$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$$ for all $x \in \Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)\le x^2+ax+b $ $\space$ $\land$ $\space$ $4(x^2+2x+3) \ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
| Let $a,b\in\Bbb{R}$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)\geq0,$$
$$3x^2+(8-a)x+(12-b)\geq0,$$
for all $x\in\Bbb{R}$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)\leq0,$$
$$(8-a)^2-12(12-b)\leq0.$$
Isolating $b$ from both inequalities yields
$$\frac{(a+10)^2}{24}-15\leq b\leq 12-\frac{(8-a)^2}{12}.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532\leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|\leq4\sqrt{\tfrac{34}{3}}.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my Humble request to help me after that.
| For $x=1$ and $y=-1$ we'll get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^3y+y^3x+4\geq2$$ or
$$x^3y+y^3x+2(x^2+xy+y^2)^2\geq0$$ or
$$2x^4+5x^3y+6x^2y^2+5xy^3+2y^4\geq0$$ or
$$2x^4+4x^3y+2x^2y^2+x^3y+2x^2y^2+xy^3+2x^2y^2+4xy^3+2y^4\geq0$$ or
$$(x+y)^2(2x^2+xy+2y^2)\geq0.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/3171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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inequality $(a+b+c+d)(a^3+b^3+c^3+d^3) > (a^2+b^2+c^2+d^2)^2$ I am trying to prove the inequality:
$$(a+b+c+d)(a^3+b^3+c^3+d^3) > (a^2+b^2+c^2+d^2)^2$$ given $a,b,c,d$ are positive and unequal.
starting from LHS
since AM of mth power > mth power of AM
$$(a^3+b^3+c^3+d^3)/4 > ((a+b+c+d)/4)^3$$
multiplying both sides by $(a+b+c+d)$
$$(a+b+c+d)(a^3+b^3+c^3+d^3)/4 > (a+b+c+d)^4/(4^3)$$
$\implies$
$$(a+b+c+d)(a^3+b^3+c^3+d^3) > (a+b+c+d)^4/16$$
now taking expression on the RHS and using AM of mth power > mth power of AM
$$(a^2+b^2+c^2+d^2)/4 > ((a+b+c+d)/4)^2$$
squaring both sides
$$(a^2+b^2+c^2+d^2)^2/16 > (a+b+c+d)^4/(4^4)$$
$$\implies (a^2+b^2+c^2+d^2)^2 > ((a+b+c+d)^4)/16$$
Now I have proved that both LHS and RHS are greater than $$((a+b+c+d)^4)/16$$
still unable to prove LHS > RHS. Please help with this.
| Hint: You will get the left-hand side minus the right-hand side:
$${a}^{3}b+{a}^{3}c+{a}^{3}d-2\,{a}^{2}{b}^{2}-2\,{a}^{2}{c}^{2}-2\,{a}^
{2}{d}^{2}+a{b}^{3}+a{c}^{3}+a{d}^{3}+{b}^{3}c+{b}^{3}d-2\,{b}^{2}{c}^
{2}-2\,{b}^{2}{d}^{2}+b{c}^{3}+b{d}^{3}+{c}^{3}d-2\,{c}^{2}{d}^{2}+c{d
}^{3}
$$ Now you need to combine like terms:
$$a^3b-2a^2b^2+ab^2=ab(a^2-2ab+b^2)$$ and so on.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Convergent value of $\sum\limits^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$ The sum in question is:
$$\sum^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$$
It passes the ratio test:
\begin{align}
&\lim_{n\rightarrow \infty}\frac{(n+1)\left(\frac{5}{6}\right)^{n}}{n\left(\frac{5}{6}\right)^{n-1}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\frac{\left(\frac{5}{6}\right)^{n}}{\left(\frac{5}{6}\right)^{n}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}(1+ \frac{1}{n})\\
=\frac{5}{6} &< 1\Rightarrow \text{convergent}
\end{align}
But now I do not know how to find the convergent value.
| $$f(x)= \frac{1}{1-x}=\sum_{n=0}^\infty x^n\\
f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\\
f'(5/6)=\sum_{n=1}^\infty n(\frac{5}{6})^{n-1}= 36.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}$ I am trying to find $\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}$.
Using L'Hospital's Rule:
since $\frac{d}{dx}(x + 2\sin(x)) = 1 + [0 \cdot \sin(x)] + [2\cdot\cos(x)]=1 +2\cos(x)$
and since $\frac{d}{dx} (x + 10\cos(x)) = 1 + [0 \cdot \cos(x)] + [10 \cdot -\sin(x)]=1-10\sin(x)$
$$\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}= \lim_{x \to 0} \frac{1 +2\cos(x)}{1-10\sin(x)}$$
plug in 0 for x:
Since $\cos(0) = 1$, and $\sin(0) = 0$
$$=\frac{1 + 2(1)}{1 - 10(0)} = \frac{1+2}{1-0} = 3$$
But this is apparently wrong.
| Why so complicated? The numerator goes to $0$, and the denominator goes to $10$, so the limit is $0$. Also you cannot use L'Hospital's rule here since the limit is not one of the indeterminate forms that permit the use of L'Hospital.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| $$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
From $(1)$, $x=\frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(\frac{36-2y}{3})+4y=64 \implies y=6$ and then you can get that $x=24/3=8$
Another Method
From $(1)$, $x=\frac{36-2y}{3}$
From $(2)$, $x=\frac{64-4y}{5}$
But $x=x \implies \frac{36-2y}{3}=\frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) \implies y=6$ and substitute to get $x=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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Tricky integral $\int_{0}^{2}{\frac{\arctan x}{x^2+2x+2}}$ $$\int_{0}^{2}\frac{\arctan{x}}{x^2+2x+2}$$
The solution comes with the substitution $x=\frac{2-t}{1+2t}$. It works perfectly fine, but I wonder, how can I guess something like this?
| Observe that \begin{align}\dfrac{2-t}{1+2\times t }\end{align}leads to think about the following formula, $ab>-1$:
\begin{align}\arctan\left(\frac{a-b}{1+ab}\right)=\arctan\left(a\right)-\arctan\left(b\right)\end{align}
If you perform the change of variable $x=\dfrac{2-t}{1+2\times t }$,
you get something like this:
\begin{align}I&=K\int_0^2\frac{\arctan\left(\frac{2-t}{1+2t}\right)}{x^2+2x+2}dx\\
&=K\int_0^2\frac{\arctan 2 }{x^2+2x+2}dx-K\times I
\end{align}
($K$, a real)
Therefore, if you are facing to, $a>0$, $P$ polynomial:
\begin{align}\int_0^a\frac{\arctan x}{P(x)}dx\end{align}
Try the change of variable $x=\dfrac{a-t}{1+a\times t }$
$\left(t=\dfrac{a-x}{1+a\times x }\right)$
Sometimes, the following formula is helpful.
Let $a>0,b>1$, perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}J&=\int_{\frac{1}{b}}^b \dfrac{\arctan x}{x^2+ax+1}\,dx\\
&=\int_{\frac{1}{b}}^b \dfrac{\arctan\left(\frac{1}{x}\right)}{x^2+ax+1}\,dx\end{align}
Since, for $x>0$,
\begin{align}\arctan x+\arctan\left(\frac{1}{x}\right)=\dfrac{\pi}{2}\end{align}
Therefore,
\begin{align}J&=\dfrac{\pi}{4}\int_{\frac{1}{b}}^b \dfrac{1}{x^2+ax+1}\,dx\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality related with $abcd=(1-a)(1-b)(1-c)(1-d)$ Given that $0<a,b,c,d<1$ satisfying $abcd=(1-a)(1-b)(1-c)(1-d)$. Prove that $$(a+b+c+d)-(a+c)(b+d)\geq 1.$$
First, I have already done a quite similar exercise as below:
"Given that $a^2+b^2+c^2+d^2=1$. Prove that $(1-a)(1-b)(1-c)(1-d)\geq abcd$". The solution is to use the remark of $(a+b-1)^2\geq 0$, which leads to $2(1-a)(1-b) \geq 1-a^2-b^2 \geq 2cd$.
Then, I use the same method for this problem and I have showed that $a^2+b^2+c^2+d^2 \geq 1$. I don't know what to do next with $(a+b+c+d)-(a+c)(b+d)$.
Many thanks!
| Let $\frac{1-a}{a}=x$, $\frac{1-b}{b}=y$, $\frac{1-c}{c}=z$ and $\frac{1-d}{d}=t$.
Thus, $x$, $y$, $z$ and $t$ are positives such that $xyzt=1$ and we need to prove that
$$\sum_{cyc}\frac{1}{1+x}-\left(\frac{1}{1+x}+\frac{1}{1+z}\right)\left(\frac{1}{1+y}+\frac{1}{1+t}\right)\geq1$$ or
$$xz+yt\geq2,$$ which is true by AM-GM:
$$xz+yt\geq2\sqrt{xzyt}=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Draw Blanks approach to solve a probability question of drawing 2 red, 2 blue and 1 green marbles. Q: "There are 6 children at a family reunion, 3 boys and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?" | There are $\binom{5}{2}$ possible positions for the blue balls in the sequence. That leaves three positions in the sequence, of which two must be filled with red balls, so there are $\binom{3}{2}$ positions for the red balls in the sequence. The remaining position in the sequence must be filled with the green ball. Consequently, there are
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 10 \cdot 3 \cdot 1 = 30$$
possible sequences with two blue balls, two red balls, and one green ball.
The probability of obtaining blue, blue, red, red, green in that order is
$$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
The probability of obtaining blue, green, red, red, blue in that order is
$$\frac{5}{11} \cdot \frac{2}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
Notice that the denominators are the same and that the terms in the numerator have just been permuted. Therefore, the probabilities of these sequences are equal. In fact, this is true of each of the $30$ sequences consisting of two blue balls, two red balls, and one green ball.
Therefore, while the probability of any particular sequence is
$$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
the probability that a sequence will contain two blue balls, two red balls, and one green ball can be obtained by multiplying the probability that any particular sequence will appear by the
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 30$$
possible orders in which these balls could appear.
Thus, the probability of obtaining two blue balls, two red balls, and one green ball when five balls are selected from a bag containing five blue balls, four red balls, and two green balls is
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
Since we actually care about which balls are selected rather than the order in which they are selected, we do not need to consider the sequence in which the balls are selected. We just need to find the probability of selecting two of the five blue balls, two of the four red balls, and one of the two green balls when selecting five of the eleven balls in the bag, which is
$$\frac{\dbinom{5}{2}\dbinom{4}{2}\dbinom{2}{1}}{\dbinom{11}{5}}$$
| {
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"question_score": "1",
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How does one subtract from concatenation in order to prove that $4\times 5 + 67 = 45 + 6\times 7$? I noticed that if we get the numbers $4$, $5$, $6$ and $7$, they have an interesting property!
$$4 \times 5 + 67 = 45 + 6 \times 7\tag*{= 87.}$$
I then conjectured that these were the only four consecutive positive integers with this property.
Hmm...
Let the four integers be $n$, $n+1$, $n+2$ and $n+3$. Then we have $$\begin{align} &n(n+1) + (n+2)\| (n+3) \\ =\;&n\| (n+1) + (n+2)(n+3).\tag{$\|\;\small{ \rm stands\;for}\;concatenation$}\end{align}$$ Or $$\begin{align} (n+2)\|(n+3)-n\|(n+1)&=(n+2)(n+3)-n(n+1) \\ &=n^2 + 5n + 6 - n^2 - n \\ &= 4n+ 6 \\ &= 2(2n+3).\end{align}$$
Now this is when I noticed that if $n=4$ then $2(2n+3)=22$. I may be able to reach $22$ like so:
$$\begin{align}(n+2)\|(n+3)-n\|(n+1)&=(n+2-n)\|(n+3-n-1)\\ &=2\|2 \\ &= 22\end{align}$$
If that is true, then this happens: $$\begin{align}22 &= 2(2n+3)\\ \Leftrightarrow \; 11 &= 2n+3 \\ \Leftrightarrow \; \phantom{1}8 &= 2n \\ \\ \therefore \; n &= 4.\end{align}$$ This is my desired result, thus proving the conjecture. But is the equation in the second sandbox true?
How does one subtract from concatenation?
This, in particular, is new to me; I would mostly appreciate a full, explanatory answer.
Thank you in advance.
| We have the following cases:
*
*$n+1,n+3$ both have $k\ge1$ digits; then$$(n+2)||(n+3)-n||(n+1)=10^k(n+2)+(n+3)-10^k(n)-(n+1)\\=2(1+10^k)$$For solutions to your problem in this case,$$2n+3=1+10^k$$You can get infinitely many solutions $\{4,49,499,4999,...\}=\{10^k/2-1:k\in\Bbb N\}$ using different $k$, meaning that the solution is not unique.
*$n+3$ has $k+1$ digits and $n+1$ has $k\ge1$ digits; then $$(n+2)||(n+3)-n||(n+1)=10^{k+1}(n+2)+(n+3)-10^k(n)-(n+1)\\=2+10^k(9n+20)$$The solutions are given by$$2n+3=1+10^{k-1}(45n+100)$$Note that $10^{k-1}(45n+100)\ge45n+100>2n+3$, so there is no solution in this case.
| {
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How do I take the limit of this trigonometric function I was wondering how I could take this limit:
$\lim_{a→1}\frac{\sin(a^4 - 1)}{a^3-1}$
My idea was that if I can get the denominator and the inside of the sin to be the same I can use the sandwich theorem or maybe cancel out the a-1 in the denominator somehow. To go about this I did the following operations:
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a-1)(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a-1)(a+1)(a^2+1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a^4 - 1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)} *\lim_{a\to 1}\frac{(a^2 + 1)(a + 1)}{(a^2 + a + 1)}$$
At this step I tried to use the sandwich theorem to find $\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)}$
However I got $(-\infty,\infty)$ Like so:
$$\lim_{a\to 1} -1 < \sin(a^4-1) < 1$$
$$\lim_{a\to 1} \frac{-1}{a^4 - 1} < \frac{\sin(a^4-1)}{a^4-1} < \frac{1}{a^4 - 1}$$
$$= -\infty < \frac{\sin(a^4-1)}{a^4-1} < \infty$$
So I'm at a loss as to what I can do from here on. Or is what I've done just completely wrong?
| You can just note that
$$
\lim_{a\to 1} \frac{\sin(a^4-1)}{a^3-1} = \lim_{a \to 1}\frac{\sin(a^4-1)}{a^4-1} \cdot \frac{a^4-1}{a^3-1}=\lim_{a\to 1} \frac{\sin(a^4-1)}{a^4-1} \cdot \lim_{a \to 1} \frac{a^4-1}{a^3-1}=1 \times \frac 43= \frac 43
$$
| {
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Sum of divisors of a square I was wondering if there is a nice formula for the number of divisors $d$ of a perfect square ($n^2$), such that $\frac{n}{2} <d < n$. For example, for $n = 12$, the divisors $d$ of $12^2=144$ such that $12/2=6<d<12$ consist of $8$ and $9$, so $f(12) = 2$.
$$$$
Another way to express this is that $12 = 2^2 * 3^1$, so $144=2^4*3^2$. Now we must find the number of tuples $(e_1, e_2)$, such that $6 \lt 2^{e_1} * 3^{e_2} \lt 12$
$$$$
Is there a closed formula for this or at least another way to express this?
| What I'm about to show isn't a formula but rules on how to quickly find the divisors ($d$) between $n/2$ and $n$. For each $d$ $\exists a,b\in\Bbb{N}|2a>b>a,gcd(a,b)=1,a|n,b|n$
$d=\frac{n*a}{b}$
I will show three examples below.
If $n=12$ then $n$'s factors are $1,2,3,4,6, \text{ and } 12$. All $a$ and $b$ must be one of these numbers. All $(a,b)$ pairs that satisfy the inequality ($2a>b>a$) are $(2,3),(3,4),(4,6)$. The gcd of the third pair isn't one, so it is removed.
$$8=\frac{12*2}{3}\quad 9=\frac{12*3}{4}$$
If $n=45$ then $n$'s factors are $1,3,5,9,15,$ and $45$. The three pairs $(3,5),(5,9),(9,15)$ satisfy the inequality. The third pair isn't coprime so it is removed.
$$27=\frac{3*45}{5}\quad 25=\frac{5*45}{9}$$
If $n=60$
Factors of $n$: $1,2,3,4,5,6,10,12,15,20,30,60$
Coprime pairs:
$(2,3),(3,4),(3,5),(4,5),(5,6)$
$$d\in (40,45,36,48,50)$$
While I don't have a general formula for $f(n)$ there is a formula for a specific case. If all factors of $n^2$ less than $n/2$ divides $n$ then the formula for $f(n^2)$ is $$\lceil\frac{\sigma_0(n^2)}{2}\rceil -\sigma_0(n)$$ where $\sigma_0(x)$ is the number of factors of $x$
This formula does work for $12$ and $45$ but not $60$.
| {
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How do I integrate $4\int_0^{\pi/2}\frac{\sec^2(\theta)}{1+2\tan^2(\theta)}\,d\theta$ using symmetry? \begin{align}\int_{-\pi}^\pi \frac{1}{1+\sin^2(\theta)}\,d\theta&=4\int_0^{\pi/2} \frac{1}{1+\sin^2(\theta)}\,d\theta\\\\&=4\int_0^{\pi/2}\frac{\sec^2(\theta)}{1+2\tan^2(\theta)}\,d\theta\\\\&=\left.\left(\frac{\arctan(\sqrt 2 \tan(\theta))}{\sqrt2}\right)\right|_0^{\arctan\pi/2}\\\\&=\sqrt 2\pi\end{align}
I'm having a difficult time understanding how we get from the second to the third line. Apparently, some sort of symmetry is being used.
| U-substitution $u=\sqrt{2}\tan\theta$:
$$
\int\frac{\sec^2\theta}{1+2\tan^2\theta}\,d\theta=
\frac{1}{\sqrt{2}}\int\frac{1}{1+(\sqrt{2}\tan\theta)^2}\frac{d}{d\theta}\left(\sqrt{2}\tan\theta\right)\,d\theta=\\
\frac{1}{\sqrt{2}}\int\frac{1}{1+u^2}\,du=\frac{1}{\sqrt{2}}\arctan{u}+C=\\
\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan\theta)}+C.
$$
If I'm understanding correctly, you're asking how they found the antiderivative of $\int\frac{\sec^2\theta}{1+2\tan^2\theta}\,dx$? That's lines 2 and 3. As far as symmetry is concerned, $f(\theta)=\frac{1}{1+\sin^2\theta}$ is an even function. Graph it and will see that you can integrate it from 0 to $\frac{\pi}{2}$ and quadruple the result.
I'm not sure why it is $\arctan{\pi/2}$ as the upper bound of integration there. That doesn't make much sense. This integral should be treated as an improper integral because $\frac{\sec^2\theta}{1+2\tan^2\theta}$ is not defined at $\frac{\pi}{2}$ (you multiplied the top and bottom by $\frac{1}{\cos^2\theta}$ and that's not defined at $\frac{\pi}{2}$). More specifically, you will jut not be able to plug in $\frac{\pi}{2}$ into the antiderivative because it contains the tangent function and the tangent of $\frac{\pi}{2}$ is not defined:
$$
4\int_{0}^{\frac{\pi}{2}}\frac{\sec^2\theta}{1+2\tan^2\theta}\,d\theta=
4\lim_{b\to\frac{\pi}{2}^-}\int_{0}^{b}\frac{\sec^2\theta}{1+2\tan^2\theta}\,d\theta=\\
\frac{4}{\sqrt{2}}\lim_{b\to\frac{\pi}{2}^-}\arctan{(\sqrt{2}\tan\theta)}\bigg|_{0}^{b}=
\frac{4}{\sqrt{2}}\lim_{b\to\frac{\pi}{2}^-}\left[\arctan{(\sqrt{2}\tan{b})}-\arctan{(\sqrt{2}\cdot 0)}\right]=\\
\frac{4}{\sqrt{2}}\left[\arctan{(+\infty)}-\arctan{(0)}\right]=
\frac{4}{\sqrt{2}}\left(\frac{\pi}{2}-0\right)=\\
\frac{4}{\sqrt{2}}\cdot\frac{\pi}{2}=
\frac{2\pi}{\sqrt{2}}=\frac{\sqrt{2}\cdot 2\pi}{\sqrt{2}\cdot\sqrt{2}}=
\frac{\sqrt{2}\cdot 2\pi}{2}=
\pi\sqrt{2}.
$$
| {
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Find Taylor series of $\sqrt{x}$ centered at $x=4$ and the order 3 Find Taylor series of $\sqrt x$, about $x=4$ and the order 3
I've tried a few timesm but I keep getting a result that does not comply with the answer. Following are the steps I've taken, hopefully I can get a pointer to the factor I'm missing in some of the denominators. I hope all the steps are clear.
Find derivatives of $\sqrt x$
If $f(x) = \sqrt x$, then:
*
*$f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$
*$f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}$
*$f'''(x) = \frac{3}{8}x^{-\frac{5}{2}}$
Write out the Taylor polynomial
$P_3(x) = \sqrt 4 + \frac{1}{2} 4^{-\frac{1}{2}}(x - 4) - \frac{1}{4} 4^{-\frac{3}{2}}(x - 4)^2 + \frac{3}{8} 4^{-\frac{5}{2}}(x - 4)^3$
$P_3(x) = 2 + (\frac{1}{2}) (\frac{1}{\sqrt 4})(x - 4) - (\frac{1}{4}) (\frac{1}{4^{\frac{3}{2}}})(x - 4)^2 + (\frac{3}{8})(\frac{1}{x ^{\frac{5}{2}}})(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{4} \frac{1}{4^{2/2} 4^{1/2}}(x - 4)^2 + \frac{3}{8}\frac{1}{4^{4/2}} \frac{1}{4^{1/2}}(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32}(x - 4)^2 + \frac{3}{256}(x - 4)^3$
But according to the book the series develops like:
$$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2 + \frac{3}{1536}(x - 4)^3$$
So I'm missing some (increasing) factor, I just can't seem to find it. Any hints?
| A Taylor polynomial of degree n centered at $x_0$ is defined as follows:
$$
p_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\\
\frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
You forgot to include the factorials in your Taylor polynomial.
| {
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Can't solve system of linear equations (that need simplification first) I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$\frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12}$$
$$\frac{2y+1}{5} + \frac{x-3y}{4} = \frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12\frac{3x+2}{4} - 12\frac{x+2y}{2} = 12\frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20\frac{2y+1}{5} + 20\frac{x-3y}{4} = 20\frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -\frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
| Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
| {
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Definite integral $\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x= e^{\pi/6} \cos \left( \frac\pi6 \right) - \frac12$
$$
\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x= e^{\pi/6} \cos \left( \frac\pi6 \right) - \frac12
$$
I've spent alot of time moving things around but I can't find a way to actually prove it... Would really appreciate some help.
$$
\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x
\\= \int_0^{\pi/12} \frac{e^{2x}\cos^2(x)}{\frac12 (\cos(x)-\sin(x))^2}{\rm d}x
\\= 2\int_0^{\pi/12}e^{2x} \left( \frac{\cos(x)}{ \cos(x)-\sin(x)} \right)^2{\rm d}x
\\= \int_0^{\pi/12}e^{2x} \left( \frac{1 + \cos(2x)}{ 1-\sin(2x)} \right){\rm d}x
\\= \frac12 \int_0^{\pi/6} e^u \frac{1+\cos(u)}{1-\sin(u)} {\rm d}u
$$
| I think integrating by parts works in most cases where there's a combination of $e^x$ and some other random function and of course we expect an elementary primitive. I will continue what you did:
$$I=\frac12 \int_0^\frac{\pi}{6} e^x \frac{1+\cos x}{\color{red}{1-\sin x}}dx=\frac12 \int_0^\frac{\pi}{6} e^x (1+\cos x)\color{red}{\left(\frac{\cos x}{1-\sin x}\right)'}dx$$
$$=\frac12 e^x (1+\cos x)\left(\frac{\cos x}{1-\sin x}\right)\bigg|_0^\frac{\pi}{6}-\frac12 \int_0^\frac{\pi}{6} e^x \frac{1-\sin x+\cos x}{1-\sin x}\cos xdx$$
$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x\left(\cos x +\frac{{\cos^2 x}}{1-\sin x}\right)dx$$
$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x(\cos x+1+\sin x)dx$$
$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12 e^x(1+\sin x)\bigg|_0^\frac{\pi}{6}$$$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac34 e^\frac{\pi}{6}+\frac12=e^{\large \frac{\pi}{6}}\frac{\sqrt 3}{2}-\frac12 $$
Of course we also have:
$$\int e^x\frac{1+\cos x}{1-\sin x}dx=e^x(1+\cos x)\frac{\cos x}{1-\sin x}-e^x(1+\sin x)+C=\frac{e^x\cos x}{1-\sin x}+C$$
| {
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Combinatorics problem, right solution? We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.
For the remaining two places, I could have $2$ more people of a single profession. This is $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}$ possibilities.
I could also have two people of different professions; a doctor and a laywer, $\binom{5}{1}\binom{3}{1}$; a doctor and an engineer, $\binom{6}{1}\binom{3}{1}$; or an engineer and a laywer $\binom{6}{1}\binom{5}{1}$.
This adds up to $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}+\binom{5}{1}\binom{3}{1}+\binom{6}{1}\binom{3}{1}+\binom{6}{1}\binom{5}{1}=91$ possible committees.
I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.
Thanks in advance!
| Let $A=\{(l,e,d)\in\{1,2,3,4,5,6\}\times\{1,2,3,4,5,6,7\}\times\{1,2,3,4\}\mid l+e+d=5\}$
Then to be found is $$\sum_{(l,e,d)\in A}\binom{6}{l}\binom{7}{e}\binom{4}{d}$$
Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:
*
*$5=3+1+1$
*$5=2+2+1$
*$5=2+1+2$
*$5=1+3+1$
*$5=1+2+2$
*$5=1+1+3$
This provides you a view on set $A$ and shows that the summation has $6$ terms.
| {
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Square root property to solve quadratic equation: $3(x-4)^2=15$ I get $\sqrt{21}$ but solution is $4+-\sqrt{5}$ I am to solve for x using square root property:
$3(x-4)^2=15$
The textbook solution is $4+-\sqrt{5}$ and I am unable to arrive at this. I arrived at $\sqrt{21}$
Here is my working:
$3(x-4)^2=15$
$(x-4)^2=5$ # divide both sides by factor 3
$x^2-16=5$ # multiply out $(x-4)^2$
$x^2 = 21$
$x=\sqrt{21}$
How can I arrive at $x=4+-\sqrt{5}$ # not sure syntax for 'plus or minus' symbol here, just used +-
| Simplifying we get
$$(x-4)^2=5$$ and expanding $$x^2-8x+11=0$$ so $$x_{1,2}=4\pm \sqrt{16-11}$$
| {
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Lagrangian Multipliers Constrained Optimization As far as I understand, Lagrangian multiplier $\lambda$ can take negative and positive values. For the positive values, we find maximum point. For the negative values, we find minimum point.
I think that we cannot find minimum point by using Lagrangian multipliers for this functions:
$f(x, y) = x^2y^2$
$g(x, y) = x^2 + y^2 = 1$
I aim to find minimum and maximum point of $f$ under the condition $g$.
When I apply Lagrangian method, I could find only this solution:
$x = \sqrt\lambda$
$y = \sqrt\lambda$
When I substituted these equalities in $g(x, y)$, I found $\lambda = 1/2$, and $f(x, y) = 1/4$.
What do you think about this ?
There is only positive Lagrangian value, and it makes function $f(x, y)$ produce $1/4$. Is this maximum value or minimum value ? I think it is maximum point.
| Step by step
$$
L = x^2y^2-\lambda(x^2+y^2-1)
$$
so the stationary points (maxima, minima, saddle) are included in the solutions for
$$
\nabla L = 0 =\cases{2xy^2-2\lambda x\\
2y x^2-2\lambda y\\
x^2+y^2-1}
$$
from the first two equations we have $x=0,y=0,y^2=\lambda, x^2 = \lambda$ so substituting into the third we have $2\lambda = 1$ or $\lambda^* = \frac 12$ hence $x^* = \pm\frac{\sqrt 2}{2},\ \ y^* = \pm\frac{\sqrt 2}{2}$ or for any $\lambda$ also $x^* = 0,y^* =\pm 1,\ \ y^*=0,x^*= \pm 1$. Now among those stationary points we should choose according to our needs. Substituting the restriction on the objective function we have
$$
z = x^2(1-x^2),\ \ -1\le x \le 1
$$
so at $x^*=\pm\frac{\sqrt 2}{2}$ we have two maxima and at $x^* = 0, x^*=\pm 1$ we have minima as can be depicted from the following plot
Resuming the stationary points
$$
\begin{array}{ccccc}
x^* & y^* & \lambda^* & x^2y^2 & \text{type}\\
0 & \pm 1 & \text{any} & 0 & \text{min}\\
\pm 1 & 0 & \text{any} & 0 & \text{min}\\
\pm\frac{\sqrt 2}{2} & \pm\frac{\sqrt 2}{2} & \frac 12 & \frac 14 & \text{max}
\end{array}
$$
NOTE
The restriction of $z = x^2y^2$ on the cylinder $x^2+y^2-1=0$ can be depicted in blue in the following plot.
| {
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"url": "https://math.stackexchange.com/questions/3204890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve in $C$ : $P(z)=z^4+2z^3+5z^2+4z+1=0$ where $P(i-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$ Question solve in $C$ :
$P(z)=z^4+2z^3+5z^2+4z+1=0$
where $P(i+-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$
My attempt :
Let $\lambda=i-\frac{1}{2}+i\frac{\sqrt 3}{2}$
$(2\lambda+1)^2=(i(2+\sqrt 3))^2$
$(2\lambda+1)^2+7)^2=(-4\sqrt 3)^2$
$({\lambda}^2+\lambda+2)^2=3$
So we find :
$\lambda^2++2\lambda^3+5\lambda^2+4\lambda+1=0$
But which step !? can be find all root of P(z) !!
| You are given a polynomial $P(x)$ in integer coefficients and an expression
of one of the roots $$\lambda = i - \frac12 + i \frac{\sqrt{3}}{2} = \sqrt{-1} - \frac12 + \frac12 \sqrt{-1}\sqrt{3}$$
which consists of a bunch of square roots of rational numbers.
When you substitute $\lambda$ into $P(x)$, you get $P(\lambda) = 0$. If you follow the expansion of $P(\lambda)$ step by step, you will notice all the square roots in $\lambda$ get squared out, become rational number and disappear in final output. The final expression doesn't depend on the signs of square roots. It only depends on squaring a square root give you back the original number.
This has a very useful implication. If you flip the signs of any of the square roots
in $\lambda$, you get another root!
Since there are two independent square roots, $\sqrt{-1}$ and $\sqrt{3}$ in $\lambda$. Flipping their signs independently give you totally $4$ roots.
$$
\begin{align}
\lambda_{++} &= +i - \frac12 + i\frac{\sqrt{3}}{2}\\
\lambda_{+-} &= +i - \frac12 - i\frac{\sqrt{3}}{2}\\
\lambda_{-+} &= -i - \frac12 - i\frac{\sqrt{3}}{2}\\
\lambda_{--} &= -i - \frac12 + i\frac{\sqrt{3}}{2}
\end{align}$$
Since $P(x)$ is a polynomial of degree $4$, these are all the roots of $P(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$
$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$
$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$
$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$
$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$
$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$
We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$
| Consider $x=a \tan^{1/2}u$ to obtain$$\frac{1}{2|a|}\int_0^{\pi/2}\cos^{-1/2}u\sin^{5/2}udu.$$ Now recall the Beta fucntion.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
How to solve this $3Solve for, x
$$3<x^2-4<x+1 $$
Attempt:
$$7<x^2<x+5 $$
This is difficult.
Solving this equation $$3x-3<6$$ it is easy.
$$3x<9$$
$$x<3$$
| You can see that $3 < x^2 - 4 < x+1$ is equal to $x | 3< x^2 - 4 \cap x | x^2 -4 < x+1 $ and solve for each set
$3 < x^2-4 \leftrightarrow 7 < x^2 \leftrightarrow \sqrt{7} < | x | \leftrightarrow x \in ( - \infty , - \sqrt{7}) \cup ( \sqrt{7} , \infty )$
$ x^2 - 4 < x+1 \leftrightarrow x^2 - 4-(x+1)<0 \leftrightarrow x^2 - x - 5 < 0 \leftrightarrow (x-\frac{1- \sqrt{21}}{ 2 })(x-\frac{1+ \sqrt{21} }{ 2 }) < 0 \leftrightarrow x \in (\frac{1- \sqrt{21}}{ 2 }, \frac{1+ \sqrt{21} }{ 2 })$
You can intersect both solution sets and get the solution of your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solve $7x^2+3x=0$ by completing the square? I am to solve for x: $7x^2+3x=0$
I'm aware that there are multiple approaches to solving a quadratic. In this case, since there is no constant term I decided to go the completing the square route. I know from my textbooks answer that the solutions are $x=0$ and $x=-\frac{3}{7}$.
Here is how far I got:
$7x^2+3x=0$ # want to have leading coefficient 1 not 7
$x^2 + \frac{3}{7}x=0$
take 1/2 of the linear coefficient and then square it:
$\frac{1}{2}*\frac{3}{7}=\frac{3}{14}$
Then square it:
$(\frac{3}{14})^2$ = $\frac{9}{196}$
Add this term to both sides of my equation:
$x^2 + \frac{3}{7}x + \frac{9}{196}=\frac{9}{196}$
This is where I get stuck. Apparently I should be able to factor as a perfect square the left hand side of the equation. Perhaps because I'm working with fractions I cannot see how to do that? How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?
|
How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?
A hint: $n$ is always half of the linear coefficient. You've already calculated that this is $3/14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality $\frac{x^{x^2}}{x^2+y^2}+\frac{y^{y^2}}{y^2+z^2}+\frac{z^{z^2}}{z^2+x^2}\geq \frac{3}{2}$ with $xyz=1$ I'm proud to present the work of my last week end and I think it's hard to prove :
Let $x,y,z>0$ such that $xyz=1$ then we have :
$$\frac{x^{x^2}}{x^2+y^2}+\frac{y^{y^2}}{y^2+z^2}+\frac{z^{z^2}}{z^2+x^2}\geq \frac{3}{2}$$
I try the following inequality to prove my statement :
$$\frac{x^{x^2}}{x^2+y^2}+\frac{y^{y^2}}{y^2+z^2}+\frac{z^{z^2}}{z^2+x^2}\geq \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\geq \frac{3}{2}$$
But it doesn't work .
I try also to apply the well-know inequality :$e^x\geq x+1$ but it gives nothing consequent. So I'm a bit lost with this inequality ...
Any help is appeciated .
Thanks
| The hint.
Prove that:
$$x^{x^2}\geq 2x^3-4x^2+3x.$$
The rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$ Given $$c=\min(a,b,c)~, \quad a+b+c=3 \\
P=(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 )~,$$ I have to prove that $$P+11abc \le 12~.$$
I started with
$$b^2-bc+c^2 \le b^2 \quad \text{and} \quad c^2-ca+a^2 \le a^2~,$$
which led to
$$P \le a^2 b^2 (a^2-ab+b^2 ) \le \frac{4}{9} \left[\frac {(a+b+c)^2}{3}
\right]^3 \le \frac{4}{9} \left[ \frac {(a+b+c)^2}{3} \right]^3=12~.$$
Then I'm stuck.
| The variables should be non-negatives, otherwise the inequality is wrong: $(a,b,c)=(4,0,-1).$
Let $a=\min\{a,b,c\}\geq0$, $b=a+u$, $c=a+v$.
Thus, we need to prove that
$$\prod_{cyc}(a^2-ab+b^2)+\frac{11abc(a+b+c)^3}{27}\leq\frac{4(a+b+c)^6}{243}$$ or
$$567(u^2-uv+v^2)a^4+36(19u^3+3u^2v+3uv^2+19v^3)a^3+$$
$$+18(11u^4+35u^3v-6u^2v^2+35uv^3+11v^4)a^2+$$
$$+18(4a^5+u^4v+10u^3v^2+10u^2v^3+uv^4+4v^5)a+$$
$$+(u^2+11uv+v^2)(2u-v)^2(u-2v)^2\geq0,$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}{(1+\tan x)^2} \,dx$$
| $\displaystyle \int\frac{sin(2x)dx}{(sinx+cosx)^{2}}$
$sin(2x)=(sinx+cosx)^{2}-1$
$\displaystyle sinx+cosx=\sqrt{2}cos(x-\frac{\pi}{4})$
$\displaystyle y=x-\frac{\pi}{4}\Rightarrow dy=dx$
$\displaystyle \int\frac{sin(2x)dx}{(sinx+cosx)^{2}}=\int(1-\frac{1}{2cos^{2}y})dy$
$\displaystyle tan(y)=z\Rightarrow dz=\frac{dy}{cos^{2}y}$
$\displaystyle\int(1-\frac{1}{2cos^{2}y})dy=y-\frac{z}{2}+c $
$\displaystyle \int\frac{sin(2x)dx}{(sinx+cosx)^{2}}=x+\frac{1}{2}\tan{(x-\frac{\pi}{4})}+c_{1}\\\left( c_{1}=c-\frac{\pi}{4} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find a limit for $\frac{({1+x})^\frac{1}{x} - e}{x}$ as x tends to 0 I have done it thus far:
$$\lim_{x \to 0}\frac{{(x+1)}^\frac{1}{x}-e}{x} = \bigg[\frac{0}{0}\bigg] = \frac{((x+1)^\frac{1}{x}-e)'}{x'}=({x+1})^\frac{1}{x} \cdot \left(\frac{\ln(x+1)}{x}\right)' = \\({x+1})^\frac{1}{x} \cdot \frac{\left(\frac{x}{x+1}-(\ln(x+1)\right)}{x^2} = ({x+1})^\frac{1}{x} \cdot \left(\frac{\frac{1}{x+1}}{x} - \frac{\ln(x+1)}{x^2}\right) $$
I don't know what to do next.
Also could someone elaborate as to why when I have to find a derivative for $\frac{\ln(x+1)}{x}$ I need to use the quotient rule, but when I first derived the fraction that is given I could derive numerator and denominator separately?
| With binomial expansion
$$\begin{align}
(1+x)^{\frac1x}-e
&=1+\frac1x\frac{x}{1!}+\frac1x(\frac1x-1)\frac{x^2}{2!}+\frac1x(\frac1x-1)(\frac1x-2)\frac{x^3}{3!}+\cdots-\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots\right)\\
\frac{(1+x)^{\frac1x}-e }{x}&=
-\dfrac{1}{2!}-\dfrac{3-2x}{3!}-\dfrac{6-11x+6x^2}{4!}+\cdots\\
&\to-\dfrac{1}{2!}\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots\right)\hspace{1cm}\text{as}\hspace{0.3cm} x\to0.\\
&=-\dfrac12e
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from the question, I tried:
Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.
But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
| $$\begin{eqnarray*}
N=\sqrt{3}+\sqrt{7}+\sqrt{21} & \Rightarrow & N-\sqrt{21}=\sqrt{3}+\sqrt{7} \\
& \Rightarrow & {{N}^{2}}+21-2N\sqrt{21}=10+2\sqrt{21} \\
& \Rightarrow& \sqrt{21}=\frac{{{N}^{2}}+11}{2+2N} \\
\end{eqnarray*}$$
So $\sqrt{21}$ is rational, which is a contradiction
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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Find exact value of $\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$ How can I compute the series
$$\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$$
I just guess using half-angle formula to compute this series,
but I can't do any approaches.
How should I do to solve this infinite sum?
| $$\sum_{n=1}^\infty \frac1{2^n}\tan\left(\fracπ{2^{n+1}}\right) = 2\frac d {dπ}\sum_{n=1}^\infty \ln\left(\sec\left(\fracπ{2^{n+1}}\right)\right) = 2\frac d {dπ}\ln\left(\frac1{\prod_{n=1}^\infty\cos\left(\fracπ{2^{n+1}}\right)}\right)$$ But $$\prod_{n=1}^\infty\cos\left(\fracπ{2^{n+1}}\right) = \cos\left(\fracπ4\right)\cos\left(\fracπ8\right)\cos\left(\fracπ{16}\right)\cos\left(\fracπ{32}\right)... = \frac{\sqrt2}2\frac{\sqrt{2+\sqrt2}}2\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2... = \frac2π$$
(This is Viète's formula.) Then our original sum is $$2\frac d {dπ}\ln\left(\frac1{\frac2π}\right) = 2\frac d {dπ}\ln\left(\fracπ2\right) = 2\frac{\frac12}{\fracπ2} = \frac2π$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
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Solving $3$x $6$-sided dice of different colours to find the probability of events An experiment consists of $3$ fair, different coloured dice being rolled. The dice are $6$-sided and the sides show numbers $1,\dots,6$. Let $A$ be the event that none of the dice shows numbers $1$ and $2$, and let $B$ be the event that all dice show an odd number.
A) What is the probability of $A$?
B) What is the probability of $B$?
C) What is the probability of $A$ intersecting $B$?
I've solved this question by finding the total number of possible outcomes:
$|S| = 6^3 = 216$
The results were way too long for the marks given which makes me question the method I used for these solutions.
I ended up with:
A) $P(A) = \frac{64}{216} = \frac{8}{27} $
B) $P(B) = \frac{26}{216} = \frac{13}{108}$
C) $P(A \cap B) = \frac{8}{216} = \frac{1}{27}$
| So to think about $A$ for the first role we have $\frac{4}{6}$ possibility as we only want $3,4,5,6$ and this is the same for the other $3$ dices, so $(\frac{4}{6})^3= \frac{8}{27}$.
Next for $B$ we have have the odd numbers to be $1,3,5$ so the possibilities are $\frac{3}{6}$ for each dice, so overall again we cube for three dices so $(\frac{3}{6})^3= \frac{1}{8}$.
Finally we have odd and not one or two, so the number on one dice can only be $3,5$ so with 3 dices: $(\frac{2}{6})^3= \frac{1}{27}$.
But your answers are correct hope this helps :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help trying to find the coefficient in a generating function expansion Find the coefficient of ${x^{20}}$ in the expansion of the generating function $g(x) = \frac{5{(1-x^5)^7}}{(1-x)^{2}}$
I broke the function into two components: $5{(1-x^5)^7}$ and $\frac{1}{(1-x)^2}$
Because I'm looking for the ${x^{20}}$ coefficient, I have 5 terms:
$(a_0 \cdot b_{20}) + (a_5 \cdot b_{15}) + (a_{10} \cdot b_{10}) + (a_{15} \cdot b_5) + (a_{20} \cdot b_0)$
which gives me:
$20+2-1 \choose 20$ $-$ $7 \choose 1$ $15+2-1 \choose 15$$ $+$ $$7 \choose 2$$10+2-1 \choose 10$$ $-$ $$7 \choose 3$$45+2-1 \choose 5$$ $+$ $$7 \choose 4$$0+2-1 \choose 0 $
I believe that I multiply the $5$ in the first polynomial to the coefficient I find, but my answer comes out to be $(5 \cdot (-35))=-175$ which I don't believe is possible. Is this the correct answer?
| Yes you are correct. We have that
$$(1-x^5)^7=1-7x^5+21x^{10}-35x^{15}+35x^{20}+o(x^{20}).$$
and
$$\frac{1}{(1-x)^{2}}=\sum_{n=0}^{\infty} (n+1)x^n.$$
Therefore
$$[x^{20}]\frac{(1-x^5)^7}{(1-x^2)^{2}}=1\cdot (20+1)-7\cdot (15+1)+21\cdot (10+1)-35\cdot (5+1)+35\cdot 1=-35$$
So the desired coefficient is $5\cdot (-35)=-175$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do.
I would really appreciate any help
| It is equivalent to the problem: how many ways can $5$ candies be distributed among $17$ children? Mathematically, it is:
$$x_1+x_2+\cdots +x_{17}=5, 0\le x_i\le 5, i=1,2,...,17.$$
The partitions of $5$ are:
$$\begin{align}\{5\} &\Rightarrow P(17,1)=\frac{17!}{16!}=17\\
\{4,1\} &\Rightarrow P(17,2)=\frac{17!}{15!}=272\\
\{3,2\}&\Rightarrow P(17,2)=\frac{17!}{15!}=272\\
\{3,1,1\}&\Rightarrow \frac{P(17,3)}{2!}=\frac{17!}{2\cdot 14!}=2040\\
\{2,2,1\}&\Rightarrow \frac{P(17,3)}{2!}=\frac{17!}{2\cdot 14!}=2040\\
\{2,1,1,1\}&\Rightarrow \frac{P(17,4)}{3!}=\frac{17!}{6\cdot 13!}=9520\\
\{1,1,1,1,1\}&\Rightarrow \frac{P(17,5)}{5!}=\frac{17!}{120\cdot 12!}=6188\\
\end{align}$$
Hence, the sum is: $20,349$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that every group of order 15 is abelian? I had seen this proof at many places, but everywhere sylows theorem is used.
So is their any way to solve it without using sylows theorem?
| Let $G$ be a group of order $15$. There exist an element $a$ of order $5$ and an element $b$ of order $3$. Now we consider $b^{-1}ab$, and we have $(b^{-1}ab)^5 = 1$. So $o(b^{-1}ab)\mid 5$, which implies $o(b^{-1}ab) = 1$ or $5$. It cannot be $1$, otherwise $ab = b$ and so $a = 1$, a contradiction. Thus $o(b^{-1}ab) = 5$.
We claim that $b^{-1}ab = a^m$ for some $m$. Otherwise $\langle a\rangle$, $\langle b^{-1}ab\rangle$, $\langle bab^{-1} = b^{-2}ab^2\rangle$ are three subgroups of $G$ of order $5$, each two of them intersect trivially. So we have $$G = \{1,a,\dots,a^4,b^{-1}ab,\dots,(b^{-1}ab)^4,b^{-2}ab^2,\dots,(b^{-2}ab^2)^4,b,b^2\}.$$
However, you may check that $ba$ is not in this set.
To check $ba\ne a^k,ba^kb^{-1},b^k$ is easy, we now check $ba\ne b^{-1}a^kb$.
For $k = 1$, we have $aba^{-1}=b^2=b^{-1}$, then $a^2b=ba^2$. Hence $b,a^2$ commutative and thus $b,a = (a^2)^3$ commutative.
For $k = 2$, $bababa = b^{-1}ab$ and so $b^{-1}ab = (ba^2ba)^2$, we have an element of order $10$, a contradiction.
For $k=3$, $baba = b^{-1}ab$ and so $(b^2ab^2)^2 = a$.
For $k = 4$, then $a^{-1}b^{-1} = (ba)^{-1} = b^{-1}ab$ and so $(ab^{-1})^2 = 1$.
Note that $o(b) = 3$, and so $m^3\equiv 1\pmod 5$. The only solution of that equation is $m\equiv 1\pmod 5$. Therefore, $b^{-1}ab = a$. Now $ab$ is of order $15$, $G = \langle ab\rangle$ is cyclic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find Jordan Decomposition of $\left(\begin{smallmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{smallmatrix}\right)$ over $\mathbb{F}_5$
Find the Jordan decomposition of
$$
A := \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}
\in M_3(\mathbb{F}_5),
$$
where $\mathbb{F}_5$ is the field modulo 5.
What I've done so far
The characteristic polynomial is
\begin{equation}
P_A(t) = (4 - t)(1-t)(3-t) - (1-t)
= -t^3 + 8t^2-18t+1
\equiv 4t^3 + 3t^2 + 2t + 1 \mod5.
\end{equation}
Therefore, $\lambda = 1$ is a zero of $P_A$, since $4+3+2+1 = 10 \equiv 0 \mod 5$.
By polynomial division one obtains
$$
P_A(t)
= (t + 4)(4t^2 + 2t + 4)
= (t + 4)(t + 4) (4t + 1)
\equiv 4 (t + 4)^3
$$
Therefore $\lambda = 1$ is the only eigenvalue of $A$.
To find the eigenspace we calculate the kernel of $A + 4 E_3$ and obtain
$$
\text{span}\left( \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right)
$$
Since $(A + 4 E_3)^2 = 0$, the kernel of $(A + 4 E_3)^2$ is the whole space.
Now, I choose $v := (1, 0, 0) \in \text{ker}(A + 4 E_3)^2$ such that $v \not\in \text{ker}(A + 4 E_3)$.
We calculate $(A + 4E)v = (3, 0, 1)$ and then
$$
(A + 4E)
\begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},
$$
but the zero vector can't be a basis vector of our Jordan decomposition.
Have I made a mistake in my calculations?
| Your calculations are fine. However, by definition of kernels, an element of the kernel of $(A+4E)^2$ vanishes when applying $A+4E$ to it twice, so you should not be surprised. You just made the wrong conclusion. The eigenvector $(3,0,1)$ together with the generalized eigenvector $(1,0,0)$ form part of a Jordan basis giving you a Jordan block of size $2$. All you need to do is add another eigenvector which is linear independent to $(3,0,1)$, for example $(1,1,2)$.
Then
$$
\begin{pmatrix}
3 & 1 & 1 \\
0 & 0 & 1 \\
1 & 0 & 2
\end{pmatrix}^{-1}
\begin{pmatrix}
4 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 3
\end{pmatrix}
\begin{pmatrix}
3 & 1 & 1 \\
0 & 0 & 1 \\
1 & 0 & 2
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $\alpha^{\frac13}+\beta^{\frac13}$
If $f(x)=x^2-5x+8, f(\alpha)=0$ and $f(\beta)=0$ then find the value of $\alpha^{\frac13}+\beta^{\frac13}$
$$\alpha+\beta=5$$
$$\alpha \beta=8$$
$$\alpha^{\frac 1 3}=\frac 2 {\beta^{\frac 1 3}}$$
what should I do next?
| Since $f(\alpha)=0$ and $f(\beta)=0$, so $\alpha, \beta$ are solutions of the given equation $f(x)=x^2-5x+8=0$.
So we say that $\alpha=\frac{5 + i \sqrt{7}}{2}$, $\beta=\frac{5 - i \sqrt{7}}{2}$.
Also $\alpha + \beta= 5$, $\alpha \beta =8$
Let $\alpha^{\frac{1}{3}}=a$ and $\beta^{\frac{1}{3}}=b$
Now $(a+b)^3 =a^3 + b^3+3ab(a+b)\implies (\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})^3= \alpha + \beta + 3\alpha^{\frac{1}{3}}\beta^{\frac{1}{3}}(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}}) \implies (\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})^3 =5+6(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})\implies m^3-6m-5=0$ (taking $(\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}})=m$)
Now $m^3-6m-5=0 \implies (m+1)(m^2-m-5)=0\implies m=-1, \frac{1\pm \sqrt{21}}{2}$
Hence values of $\alpha^{\frac{1}{3}}+\beta^{\frac{1}{3}}$ are $-1, \frac{1\pm \sqrt{21}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$. Prove that $x_{n+1} < x_{n}$ Let $$x_{n+1} = \frac{1}{2}(x_{n} + \frac{a}{x_{n}})$$
Prove that $x_{n+1} < x_{n}$ for $a \geq 0$.
Hint: Let the initial guess satisfy $x_{1} > \sqrt{a}$
I am stuck at how to begin this. I would like to use an induction proof, but there is no simple way for me to relate the base case and begin. That is I can't even establish: $$x_{2} < x_{1}$$
How would I do this since there is no given initial term? Or is that a mistake and an initial term should be present?
| $x_{n+1}
= \frac{1}{2}(x_{n} + \frac{a}{x_{n}})
$
so
$x_{n+1}^2
= \frac14(x_{n}^2+2a + \frac{a^2}{x_{n}^2})
$
so
$x_{n+1}^2-a
= \frac14(x_{n}^2+2a + \frac{a^2}{x_{n}^2})-a
= \frac14(x_{n}^2-2a + \frac{a^2}{x_{n}^2})
= \frac14(x_{n}-\frac{a}{x_{n}})^2
$.
Therefore
$x_{n+1}^2 \ge a
$.
Now we can compare
$x_{n+1}$ and
$x_m$.
$x_{n+1}^2-x_n^2
= \frac14(-3x_{n}^2+2a + \frac{a^2}{x_{n}^2})
= -\frac14(3x_{n}^2-2a - \frac{a^2}{x_{n}^2})
$.
Since,
for $n \ge 2$,
$x_n^2 \ge a$,
$\frac{a^2}{x_{n}^2}
\le a$
so
$2a + \frac{a^2}{x_{n}^2}
\le 3a
\le 3x_n^2
$
so
$3x_{n}^2-2a - \frac{a^2}{x_{n}^2}
\ge 0
$
so
$x_{n+1}^2-x_n^2
\le 0
$
so
$x_{n+1}
\le x_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Contour integration problem with sin and cos so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$
I thought the sensible thing to do here would be to substitute in $z=e^{ix}$ and take a contour integral around the unit circle, call this path $\ast$ so that my integral becomes $$\frac{1}{2i}Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)$$
Then, letting $f(z)=\frac{1-z^2}{az+bz^2}$, I thought the function had simple poles at $z=0$ with residue $\frac{1}{a}$ and another simple pole at $z=\frac{-a}{b}$ with residue $\frac{a}{b^2}-\frac{1}{a}$ and thus I get the that $$Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)=2i(2\pi i)\frac{a}{b^2}=-4\pi(\frac{a}{b^2})$$ which is not the answer given, that is: $=\frac{2\pi}{b^2}[a-\sqrt{a^2-b^2}]$,but I can't work out why.
Any help appreciated, thank you in advance.
| We suppose $b\ne 0$, and get rid of it by force, use rather instead of $a,b$ only the variable $c=a/b$.
(After some edits in the OP we have indeed $0<b<a$, thus $1<c$.)
We will assume $c$ real, $|c|>1$, so that there is no zero in the denominator. (And the integral does not diverge.)
Sooner or later we will have to fight against the two roots of $z^2+2c+1$, we denote them by $U,V$, their product is $UV=1$, and thus we may and do assume $|U|<1$, $|V|>1$.
Let $C$ be the unit circle centered in the origin of the complex plane, then formally using $z=e^{ix}$, $\frac 1{iz}\; dz=dx$ we have:
$$
\begin{aligned}
J
&=
\int_0^{2\pi}\frac{\sin^2x}{a+b\cos x}\;dx
\\
&=
\frac 1b\int_0^{2\pi}\frac{\sin^2x}{\cos x+c}\;dx
\\
&=
\frac 1b
\int_C
\frac{
\left(\frac 1{2i}\left(z-\frac 1z\right)\right)^2}
{\frac 12\left(z+\frac 1z\right)+c}\;\frac 1{iz}\;dz
\\
&=
\frac i{2b}
\int_C
\frac{
(z^2-1)^2}
{z^2(z^2+2cz+1)}\;dz\ .
\\[3mm]
&\qquad\text{ The partial fraction decomposition is:}
\\
\frac{
(z^2-1)^2}
{z^2(z^2+2cz+1)}
&=
\frac{
(z^2-1)^2}
{z^2(z-U)(z-V)}
\\
&=
1+\frac 1{z^2}+\frac{U+V}z
-\frac {U-V}{z-V}
-\frac {V-U}{z-U}
\ .
\\[3mm]
&\qquad\text{ Only the residues in $0,U$ contribute, so...}
\\
J&=2\pi i\cdot
\frac i{2b}
\cdot[(U+V)+(U-V)]
\\
&=2\pi \frac {-U}b
=2\pi \frac {c-\sqrt{c^2-1}}b
\ .
\end{aligned}
$$
The last expression corresponds to the expected answer from the OP, recalling that $c=a/b$.
Sage check for the partial fraction decomposition:
sage: var('U,z');
sage: V = 1/U
sage: EE = (z^2-1)^2 / z^2 / (z-U) / (z-V)
sage: EE.partial_fraction(z)
-(U^2 - 1)/(U*z - 1) - (U^2 - 1)/((U - z)*U) + (U^2 + 1)/(U*z) + 1/z^2 + 1
sage: bool( _ == (1 + 1/z^2 + (U+V)/z - (V-U)/(z-U) - (U-V)/(z-V) ) )
True
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| A purely "trig-bashing" algebraic way:
We have
\begin{align*}
&\sin\left(B-\frac\pi4\right)+\sin\left(C-\frac\pi4\right)\\
&=2\sin\left(\frac{B+C}2-\frac\pi4\right)\cos\frac{B-C}2\\
&=2\sin\left(\frac\pi4-\frac{A}2\right)\cos\frac{B-C}2.
\end{align*}
So we want to prove
$$
2\cos\frac{B-C}2-1>0
$$
when $A<\pi/4$, $B,C$ acute.
Note that we have $\lvert B-C\rvert$ is at most $A$ (only in the degenerate case when $B$ or $C$ is $\pi/2$), so
$$
2\cos\frac{B-C}2-1>2\cos\frac{A}2-1>2\cos\frac\pi8-1>0
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solve the following system of equations - (7).
Solve the following system of equations (over the reals). $$\large \left\{ \begin{aligned} (x + y)^2 &= xy + 3y - 1\\ x + y &= \frac{x^2 + y + 1}{x^2 + 1}\end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{aligned} (x + y)^2 - 1 &= y(x + 3)\\ x + y - 1&= \frac{y}{x^2 + 1}\end{aligned} \right. \implies \frac{y}{x^2 + 1}\cdot (x + y + 1) = y(x + 3)$$
$$\left[ \begin{aligned} y &= 0\\ x + y + 1 &= (x^2 + 1)(x + 3) \end{aligned} \right.$$
Plugging in $y = 0$ into the first equation, we have that $x^2 = -1$, which is incorrect for $\forall x \in \mathbb R$.
Then $x + y + 1 = (x^2 + 1)(x + 3) \implies y = x^3 + 3x^2 + 2$.
And I am done with my life.
| Let $\sigma=x^2+1$, note that $\sigma y\ne 0$. The first equation is equivalent to: $$\sigma+y(x+y)-3y=0\tag1$$ and the second: $$x+y-1=\frac{y}\sigma\tag2$$
$$(1)\iff x+y=3-\frac{\sigma}{y}.$$ Replacing this last one into $(2)$ yields: $$2=\frac{y}\sigma+\frac{\sigma}y\iff (y-\sigma)^2=0.$$ Therefore $y=\sigma=x^2+1=2-x$. I let you take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$
Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$
I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain
$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$
I am stuck here. Any hints on solving the R.H.S will be appreciated.
| Try to find what is $$\sin\frac{5\pi}{12}$$
If you get it $$\frac{\sqrt3+1}{2\sqrt2}$$ then you did correctly and you know what to do next using the general definition of
$$\sin x=\sin y $$ yields what you know it !
you can find $$\sin\frac{5\pi}{12}$$ using the formula for $$\sin \frac{x}{2}$$ by taking $x=\frac{5\pi}{6}$ in degrees which is equivalent to 75 degrees
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Continued fraction $[0;2,6,10,14,...,2(2n-1)] = \frac{e-1}{e+1}$ I would like to ask about the following relation, I wonder how to reach it.
\begin{equation}
K_{n=1}^{\infty} \frac{1}{2(2n-1)} = \frac{1}{2+\frac{1}{6+\frac{1}{10+\cdots}}} = \frac{e-1}{e+1} \approx 0.46
\end{equation}
Thank you in advance
| One way you may derrive this, is noting that $\text{tanh}(z)=\cfrac{e^{2z}-1}{e^{2z}+1}$. Thus we find that $\text{tanh}(\frac{1}{2})=\frac{e-1}{e+1}$.
And that the non-simple continued fraction of $\text{tanh}(z)$ is given by:
$$\text{tanh}(z)=\frac{z}{1+\frac{z^2}{3+\frac{z^2}{5+\cdots}}}$$
From this you find that
$$\cfrac{e^{1}-1}{e^{1}+1}=\text{tanh}(\frac{1}{2})=\frac{\frac{1}{2}}{1+\frac{\frac{1}{4}}{3+\frac{\frac{1}{4}}{5+\cdots}}}=\frac{1}{2+\frac{\frac{1}{2}}{3+\frac{\frac{1}{4}}{5+\cdots}}}=\frac{1}{2+\frac{1}{6+\frac{\frac{1}{2}}{5+\cdots}}}=[0,2,6,10,14,\cdots,2(2n-1)]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Examine differentiability of $f$ Examine the differentiability of $f$:
$$f(x,y) = \begin{cases}\displaystyle \frac{x^3}{x^2+y^2} & (x,y) \neq (0,0) \\\\ 0 & (x,y) = (0,0) \end{cases}\:\:. $$
Let's check if $f$ is continuous: let $(x,y) \longrightarrow (0,0)$. Then
$$ \left |\frac{x^3}{x^2+y^2}\right| \le |x|\cdot \left|\frac{x^2}{x^2+y^2}\right | \le |x| \longrightarrow 0, $$
so this is ok. Now let's check the continuity of partial derivatives:
\begin{align} \frac{\partial f}{\partial y}(x,y) &= -\frac{2 x^3 y}{\left(x^2+y^2\right)^2}, \\
\frac{\partial f}{\partial x}(x,y) &= \frac{3 x^2}{x^2+y^2}-\frac{2 x^4}{\left(x^2+y^2\right)^2}. \end{align}
But I am not sure what should I do now?
| Suppose that $f$ is differentiable in $(0,0)$. Then there are a linear map $\mathcal{L}$ and a function $h$, such that
$$f(h_1, h_2) = f(0, 0) + \mathcal{L}(h_1, h_2) + h(h_1, h_2)$$
and
$$\lim_{(h_1,h_2) \to (0,0)} \frac{h(h_1, h_2)}{\sqrt{h_1^2 + h_2^2}} = 0.$$
Notice that $f(0,0) = 0, \ f(h_1, 0) = h_1$ and $f(0, h_2) = 0$, so if there would exist such $\mathcal{L}$ and $h$, it must be $\mathcal{L}(h_1, h_2) = h_1$ for all $(h_1, h_2)$.
Since $f(h_1, h_1) = \frac{1}{2} h_1$, it follows $f(h_1, h_1) - f(0, 0) - \mathcal{L}(h_1, h_1) = - \frac{1}{2}h_1 = h(h_1, h_1)$ and therefore we have
$$\lim_{(h_1,h_1) \to (0,0)} \frac{h(h_1, h_1)}{\sqrt{h_1^2 + h_1^2}} = \lim_{(h_1,h_1) \to (0,0)} - \frac{h_1}{2\sqrt{2h_1^2}} = - \frac{1}{2\sqrt{2}} \neq 0,$$
which is a contradiction. This implies, that $f$ is not differentiable in $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Equation of Normal? I got a question that I need help on:
Q1: The curve C has the equation $2x+3y^2+3x^2y=4x^2$.
The point P on the curve has coordinates $(-1,1).$
(a): Find the gradient of the curve at P.
So I did: $\frac{d}{dx}(2x+3y^2+3x^2y=4x^2$
$2+6y\frac{dy}{dx}+6xy+3x^2\frac{dy}{dx}=8x$
$\frac{dy}{dx}=\frac{8x-2}{6y+6xy}$
Using the coordinates (-1,1)
$\frac{dy}{dx}=\frac{1}{2}$
Q2: Hence find the equation of the normal to curve C at P, giving your answer in the form of $ax+by+c=0$.
I just don't understand what they want me to do, "normal to the curve C at P?"
| $2x+3y^2+3x^2y = 4x^2$
$2+6yy'+6xy+3x^2y'=8x$
$y' = \frac{8x-2-6xy}{6y+3x^2}$
$y'|_P = \frac{-8-2+6}{6+3} =\frac{-4}{9} $
Slope of the normal at $P$, $\ \ $ $m = -\frac{1}{y'} = +\frac{9}{4}$
So the equation of the normal to the curve at $P\equiv(-1,1)$ is,
$y - 1 = m(x-(-1))$
$y - 1 = \frac{9}{4}(x+1)$
$4y-4 = 9x + 9$
$$9x-4y+13=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find length (requires quadratic equation) There is an equation $l = 4\sqrt{(\frac{w^2+d^2}{4})+(h-k)^2}$
Following values are given
$l = 50$
$w = 12$
$d = 10$
$k = 3$
What is $h$?
Following choices are
$14$
$10$
$11$
$12$
$13$
My answers come nothing near these!
Here is what I am doing, please assist
TL;DR: $b = 27.75, b=8.245$
$50 = 4\sqrt{(\frac{w^2+d^2}{4})+(h-k)^2}$
$50 = 4\sqrt{(\frac{12^2+10^2}{4})+(h-3)^2}$
$50 = 4\sqrt{(\frac{244}{4})+h^2-6h+9^2}$
$50 = 4\sqrt{61+h^2-6h+9}$
$50 = 4\sqrt{70+h^2-6h}$
Divide both sides by $4$
$12.5 = \sqrt{70+h^2-6h}$
Raise both sides by power of $2$
$156.25 = 70+h^2-6h$
$0 = 70+h^2-6h - 156.25$
$0 = h^2 - 6h - 86.25$
Now, solve for $h$ with quadratic equation
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\frac{-(-6)\pm\sqrt{(-6)^2-4(-86.25)}}{2}$
$\frac{36\pm\sqrt{36+345}}{2}$
$\frac{36\pm\sqrt{381}}{2}$
$\frac{36 \pm 19.51}{2}$
$b = 27.75, b=8.245$
| From $$\frac{-(-6)\pm\sqrt{(-6)^2-4(-86.25)}}{2}$$
to $$\frac{36\pm\sqrt{36+345}}{2}$$
You have written $36$ for $-(-6)$. If you fix that, what you will find is $h_1 = -6.7596$ and $h_2 = 12.76$. I don't know if question wants you to round it up but with given values, these are the values that $h$ can take.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx$
Prove that
$$\sf I=\int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx=\frac{\pi}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{9\sqrt{3}}-\frac{19}{18}\zeta(3).$$
I have thought about the integral from above after I saw this similar integral and I believe changing the sign to have $\sf 1+x+x^2$ might get us a nice closed form.
So I started using the following formula: $$\sf 2ab=(a+b)^2-a^2-b^2$$
$$\sf \Rightarrow 2I=\int_0^1\frac{\ln^2(1+x^2+x^4)}{x}dx-\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$
Using in the first integral the substitution $\sf x^2\rightarrow x $ gets us:
$$\sf \int_0^1\frac{\ln^2(1+x^2+x^4)}{x}dx=\frac12\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx$$
$$\sf \Rightarrow I=-\frac14\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\frac12\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$
Well, now we only need to find:
$$\sf I(a)=\int_0^1\frac{\ln^2(1+ax+x^2)}{x}dx $$
Then set $a=1$ and $a=-1$. Of course I tried to use Feynman's trick:
$$\sf I'(a)=2\int_0^1\frac{\ln(1+ax+x^2)}{1+ax+x^2}dx$$
But quickly gave up as it doesn't look promising.
Another way might be to let $\sf x+\frac12=\frac{\sqrt 3}{2}t$ in order to get:
$$\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\int_\frac{1}{\sqrt 3}^\sqrt 3 \frac{\ln^2\left(\frac34(1+t^2)\right)}{t-\frac{1}{\sqrt 3}}dt$$
But well.. I would appreciate some help!
Update. In the meantime I found something a conjecture:
$$\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)$$
| As shown in the question we have:
$$\sf I=-\frac14\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\frac12\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$
For the first integral we can write:
$$\sf (a-b)^2=a^2-b^2-2b(a-b);\ a=\ln(1-x^3),b=\ln(1-x)$$
$$\sf \Rightarrow \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\int_0^1 \frac{\left(\ln(1-x^3)-\ln(1-x)\right)^2}{x}dx$$
$$\sf =\color{blue}{\int_0^1 \frac{\ln^2(1-x^3)}{x}dx}-\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx$$
$$\sf \overset{\color{blue}{x^3\to x}}=-\frac23\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=-\frac43\zeta(3)-2J$$
Note also that:
$$\sf \int_0^1 \frac{\ln^2(1-x)}{x}dx=\int_0^1 \frac{\ln^2 x}{1-x}dx=\sum_{n=1}^\infty \int_0^1 x^{n-1}\ln^2 x\, dx=2\sum_{n=1}^\infty \frac{1}{n^3}=2\zeta(3)$$
The latter integral can be found here:
$$\sf J=\int_0^1 \frac{\ln(1-x)\ln(1-x+x^2)}{x}dx=-\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)+\frac{2\pi^3}{27\sqrt 3}-\frac13\zeta(3) $$
$$\Rightarrow \boxed{\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)}$$
The second integral can be found here:
$$\boxed{\sf\int_0^1 \frac{\ln^2(1-x+x^2)}{x}dx=-\frac{4\pi}{9\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{8\pi^3}{27\sqrt{3}}+\frac{22}{9}\zeta(3)}$$
Combining the boxed results yields:
$$\boxed{\sf \int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx=\frac{\pi}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{9\sqrt{3}}-\frac{19}{18}\zeta(3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_3^4 \sqrt {x^2-3x+2} \, dx$ using Euler's substitution
Calculate $\int_3^4 \sqrt {x^2-3x+2}\, dx$ using Euler's substitution
My try:
$$\sqrt {x^2-3x+2}=x+t$$
$$x=\frac{2-t^2}{2t+3}$$
$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$
$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$
$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$ However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.Can you help me?P.S. I must use Euler's substitution because that's the command.
| You can first observe that
$$
\sqrt{x^2-3x+2}=\frac{1}{2}\sqrt{4x^2-12x+8}=\frac{1}{2}\sqrt{(2x-3)^2-1}
$$
so with $2x-3=t$, you get
$$
\frac{1}{4}\int_3^5\sqrt{t^2-1}\,dt
$$
Now use the Euler substitution $\sqrt{t^2-1}=t-u$, so $t^2-1=t^2-2tu+u^2$ and $t=\frac{u^2+1}{2u}$. Thus
$$
2t=u+\frac{1}{u},\qquad 2\,dt=\left(1-\frac{1}{u^2}\right)\,du=\frac{u^2-1}{u^2}\,du
$$
and the integral becomes
$$
\frac{1}{8}\int_{3-2\sqrt{2}}^{5-2\sqrt{6}}\frac{u^2-1}{u^2}\left(\frac{u^2+1}{2u}-u\right)\,du=
-\frac{1}{16}\int_{3-2\sqrt{2}}^{5-2\sqrt{6}}\left(u-\frac{2}{u}+\frac{1}{u^2}\right)\,du
$$
Not nice, but less ugly. Check the computations, please.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integral of $\frac{1}{\sqrt{2-x^2}}$ I know that the integral of $\frac{1}{\sqrt{1-x^2}} = arcsin(x)$, but what is the the integral of $\frac{1}{\sqrt{2-x^2}}$? Is this how you do it?
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2}(1-\frac{x^2}{\sqrt{2}})} = \frac{1}{\sqrt{2}} \arcsin\left(\frac x{\sqrt{2}}\right)$$
| By considering the general integral $$\int \frac{dx}{\sqrt{a^2-x^2}}$$ and applying the substitution $$x=a\sin u$$
You can calculate that $$ \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Comparison Test for Series I am trying to prove that the series below converges by the comparison test.
\begin{align}
\sum_{n=1}^\infty\frac{2n^3+5^n+5\log(n)}{13-n+8^n}
\end{align}
How do I show that:
\begin{align}
\frac{2n^3+5^n+5\log(n)}{13-n+8^n} \le 16\left(\frac{5}{8}\right)^n
\end{align}
| Note for n>1:
$$
\dfrac{2n^3 + 5^n + 5\log(n)}{13-n+8^n} \leq \dfrac{2\cdot 5^n + 5^n + 5^n}{\frac{8^n}{2}} = 8\bigg( \frac{5}{8} \bigg)^n
$$
and note that $\frac{5}{8} < 1$ so the series will converges by geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
| Following @paulinho's suggestion, we want to write $\frac{1}{u^2(u-1)(u+1)^3}$ as a sum of partial fractions. The rest we can build by repeatedly using how to write the reciprocal of a quadratic with partial fractions. Note that $$\frac{1}{(u-1)(u+1)}=\frac12\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\\\implies\frac{1}{(u-1)(u+1)^2}=\frac14\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}\right)\\\implies\frac{1}{(u-1)(u+1)^3}=\frac18\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}-\frac{4}{(u+1)^3}\right)\\\implies\frac{1}{u^2(u-1)(u+1)^3}=\frac18\left(\frac{1}{u^2(u-1)}-\frac{1}{u^2(u+1)}-\frac{2}{u^2(u+1)^2}-\frac{4}{u^2(u+1)^3}\right).$$You can do the rest yourself with such observations as$$\frac{1}{u^2(u\pm 1)}=\pm\frac{1}{u}\left(\frac{1}{u}-\frac{1}{u\pm 1}\right)=\pm\frac{1}{u^2}\mp\frac{1}{u}\pm\frac{1}{u\pm 1},\\\frac{1}{u^2(u+1)^2}=\frac{1}{u^2}-\frac{2}{u}+\frac{2}{u+1}+\frac{1}{(u+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.