Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}$ for $a+b+c+d=1$
Let $a,b,c$, and $d$ be four positive reals satisfying $a+b+c+d=1$. Show that
$$abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}.$$
I tried the inequality between $27abc$ and $(a+b+c)^3$ but it didn't help me
| We rewrite the inequality as :
$$f(a,b,c,d)=-\left(\ln\left(a\right)+\ln\left(b\right)+\ln\left(c\right)+\ln\left(d\right)+\ln\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{176}{27}\right)\right)\geq -\ln\left(1/27\right)$$
We use the equal variable method corollary 1.5 as :
$$a+b+c+d=1$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=constant$$
$$f\left(x\right)=-\ln\left(x\right)$$ define :
$$g\left(x\right)=f'\left(\frac{1}{\sqrt{x}}\right)$$
then we have :
$$g''(x)>0$$
So the $f(a,b,c,d)$ is minimal for $d\geq c\geq b \geq a>0$ and $d=c=b=x$
Remains to show the inequality for $x\in[0,1/3)$ :
$$x^{3}\left(1-3x\right)\left(\frac{3}{x}+\frac{1}{1-3x}-\frac{176}{27}\right)-1/27\leq 0$$
Or :
$$1/27(3x-1)(4x-1)^{2}(11x+1)\leq 0$$
Can you end now ?
Adding remark for another proof :
For $0<a\leq b\leq c \leq d$, $a+b+c+d=1$ and $b=c=0.25$ we have the inequalities :
$$\left(abcd\left(\frac{1}{a}+\frac{9\left(1+\frac{5\left(a-0.25\right)^{2}}{1+a}\right)}{1-a}\right)-\frac{1}{27}-\frac{176}{27}abcd\right)\geq abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}abcd$$
Wich is a one variable inequality .
Now and again fixing $b=c=0.25$ and introducing the variable $u,v>0$ such that $a+u+v+d=1=a+b+c+d$ and $0<a\leq b\leq c \leq d$ we have :
$$abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}abcd\geq auvd\left(\frac{1}{a}+\frac{1}{u}+\frac{1}{v}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}auvd$$
using the previous inequality .
Now it misses one equality case so to be continued...
Edit 01/09/2022 :
We have the following inequalities for $0<a<0.3\leq b \leq c \leq d$ and $a+b+c+d=1$ :
$$0\geq \left(abcd+\frac{a^{2}\left(0.25-a\right)^{2}}{2\left(2-4a\right)}\right)\left(\frac{1}{a}+\frac{9}{\left(d+c+b\right)}-\frac{176}{27}\right)-\frac{1}{27}\geq abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{176}{27}\right)-\frac{1}{27}$$
And we can use the same strategy fixing two new variable $u,v$
| {
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"url": "https://math.stackexchange.com/questions/4521003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
4 & 3\\
-3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
\dfrac{11}{2} & \dfrac{9}{2}\\
-\dfrac{9}{2} & -\dfrac{7}{2}
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
7 & 6\\
-6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
\dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
-\dfrac{3n}{2} & -\dfrac{3n}{2}+1
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$ we get $ A^{2022} = \begin{bmatrix}
3034 & 3033\\
-3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?
| COMMENT.-$A$ has a double eigenvalue $\lambda=1$, is not diagonalizable and its Jordan matrix decomposition is $$A=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$ so we have
$$A^{2022}=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}^{2022}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $r,s,t$ are roots of the cubic equation $x^3+bx^2+cx+d=f(x)$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b, c, d$ Let $r,s,t$ are roots of the cubic equation $f(x) = x^3+bx^2+cx+d$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b,c,d$.
Is there any clever way to solve it?
We know $r+s+t=-b, rs+st+rt=c, rst=d.$ I also have the feeling that $D$ is actually defined as a discriminant of the cubic so there must be a discriminant of a matrix or some tricky way to prove that this is in particular ${\displaystyle b^{2}c^{2}-4c^{3}-4b^{3}d-27d^{2}+18bcd\,.}$ I am just not getting the answer without making my hands dirty i.e. expanding both sides.
| Remark: A solution using middle school skill.
We have
$$(r-s)(r-t)(s-t) = r^2s + s^2t + t^2r - r^2t - t^2s - s^2 r = A - B$$
where
$$A := r^2s + s^2t + t^2r, \quad B := r^2t + t^2s + s^2 r.$$
We have
$$[(r-s)(r-t)(s-t)]^2 = (A - B)^2 = (A + B)^2 - 4AB.$$
We have
\begin{align*}
A + B &= r^2(s + t) + t^2(r + s) + s^2(t + r)\\
&= (r^2 + t^2 + s^2)(s + t + r) - (r^3 + t^3 + s^3).
\end{align*}
We have
$$
AB = rst(r^3 + t^3 + s^3) + (r^3s^3 + s^3t^3 + t^3r^3) + 3r^2s^2t^2.
$$
We have $r + s + t = -b, ~ rs + st + tr = c, ~ rst = -d$.
We have
$$r^2 + t^2 + s^2 = (r + s + t)^2 - 2(rs + st + tr) = b^2 - 2c. \tag{1}$$
Using the known identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$, we have
\begin{align*}
r^3 + t^3 + s^3 &= 3rst + (r + s + t)(r^2 + t^2 + s^2 - rs - st - tr)\\
&= -3d - b (b^2 - 3c)\\
&= -3d - b^3 + 3bc. \tag{2}
\end{align*}
Using the known identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$, we have
\begin{align*}
&r^3s^3 + s^3t^3 + t^3r^3\\
=\,& 3rs\cdot st \cdot tr + (rs + st + tr)[(rs)^2 + (st)^2 + (tr)^2 - rs\cdot st - st\cdot tr - tr\cdot rs]\\
=\,&3(rst)^2 + (rs + st + tr)[(rs + st + tr)^2 - 3rst(r + s + t)]\\
=\,& 3d^2 + c(c^2 - 3bd)\\
=\,& 3d^2 + c^3 - 3bcd. \tag{3}
\end{align*}
Using (1)-(3), we have
$$A + B = (b^2 - 2c)(-b) - (-3d - b^3 + 3bc)
= 3d - bc$$
and
$$AB = -d(-3d - b^3 + 3bc) + (3d^2 + c^3 - 3bcd) + 3d^2 = 9d^2 + db^3 - 6bcd + c^3. $$
Thus, we have
\begin{align*}
(A + B)^2 - 4AB &= (3d - bc)^2 - 4(9d^2 + db^3 - 6bcd + c^3)\\
&= -4b^3d + b^2c^2 + 18bcd - 4c^3 - 27d^2
\end{align*}
which gives
$$[(r-s)(r-t)(s-t)]^2 = -4b^3d + b^2c^2 + 18bcd - 4c^3 - 27d^2.$$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the convergence of $\sum_{n=1}^\infty (1-cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n})$ I'm having trouble determining the convergence of the series:
$\sum_{n=1}^\infty (1-cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n})$
I used $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ and then tried the comparison test for $0 \le (1-cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n}) \le (\sqrt{n+1}-\sqrt{n})$ but I can't conclude anything even If I know $\sum_{n=1}^\infty (\frac{1}{\sqrt{n+1}+\sqrt{n}})$ diverges.
I thought about trying the limit test but I couldn't find a sequence to conclude the convergency.
I also know $\sum_{n=1}^\infty 1-cos(\frac{1}{n})$ converges as $1−cos(\frac{1}{n})$ behaves like $\frac{C}{n^2}$ for large values of n
Any guidance, tip or resolution would be highly aprecciated, thank you.
| Use :
$$1-\cos(\frac{1}{n}) \sim_{+\infty} \frac{1}{2n^2} $$
and
\begin{align}
\sqrt{n+1} - \sqrt{n} &= \sqrt{n} (\sqrt{1+\frac{1}{n}} -1) \\
&\sim_{+\infty} \sqrt{n} \times \frac{1}{2n} \\
&= \frac{1}{2\sqrt{n}}\\
\end{align}
Thus
$$ (1-\cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n}) \sim_{+\infty} \frac{1}{4 n^{\frac{5}{2}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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help with mathematical induction exercise the instructions say: "Consider n and $a_1<a_2<...<a_n$ natural numbers, $n\ge1$. Prove that $$(\sum_{k=1}^n a_k)^2 \le \sum_{k=1}^n a_k^3$$"
this is how I proceeded:
induction base: n = 1 $\implies a_1^2 \le a_1^3$ which is always true, since $a_1$ is a natural number
inductive hypothesis: I assume $(\sum_{k=1}^n a_k)^2 \le \sum_{k=1}^n a_k^3$ is true for $n\in\mathbb N$
inductive step: I verify that $(\sum_{k=1}^{n+1} a_k)^2 \le \sum_{k=1}^{n+1} a_k^3$
$(\sum_{k=1}^n a_k + a_{n+1})^2 \le \sum_{k=1}^{n} a_k^3+a_{n+1}^{3}$
$(\sum_{k=1}^n a_k)^2 + a_{n+1}^2 +2a_{n+1}\sum_{k=1}^n a_k \le \sum_{k=1}^{n} a_k^3+a_{n+1}^3$
$(\sum_{k=1}^n a_k)^2 \le \sum_{k=1}^n a_k^3$ is the inductive hypothesis, therefore I only have to verify that $a_{n+1}^2 +2a_{n+1}\sum_{k=1}^n a_k \le a_{n+1}^3$.
This is where I'm having trouble. anyways, I rewrote the this equation this way:
$2\sum_{k=1}^n a_k \le a_{n+1}^2-a_{n+1}$
since both quantities on the sides of the inequality are positive numbers, I can raise them to the square:
$(\sum_{k=1}^n a_k)^2 \le \frac{a_{n+1}^4-2a_{n+1}^3 + a_{n+1}^2}{4}$
$(\sum_{k=1}^n a_k)^2 \le \frac{a_{n+1}^2(a_{n+1}^2 -2a_{n+1}+1}{4}$
$(\sum_{k=1}^n a_k)^2 \le \frac{a_{n+1}^2(a_{n+1} -1)^2}{4}$
I think I have to somehow prove that
$\frac{a_{n+1}^2(a_{n+1} -1)^2}{4} \ge \sum_{k=1}^n a_k^3$
but I don't know how. thank you for the help :)
| As you pointed out, you need to prove that $2\sum_{k=1}^n a_k \leq a_{n+1}^2-a_{n+1}$. And you have the hypothesis $a_1<\dots<a_{n+1}$.
Then we have that:
$2\sum_{k=1}^n a_k\leq 2\sum_{k=1}^n (a_{n+1}-k) = 2na_{n+1}-(n+1)n$, since $a_{n+1-k} \leq a_{n+1}-k$.
So, it suffices to prove that $2na_{n+1}-(n+1)n\leq a_{n+1}^2-a_{n+1}$, that is:
$(2n-a_{n+1}+1)a_{n+1}\leq (n+1)n$, and because of our hypothesis, we know that $a_{n+1}\geq n$. If $a_{n+1} = n$, there is nothing to be done, and if $a_{n+1}>n$, then there is $c>0$ such that $a_{n+1} = n+c$. In this case we have:
$(2n-a_{n+1}+1)a_{n+1} = (2n-n-c+1)(n+c) = (n+1-c)(n+c) =$
$= (n+1)(n+c) - c(n+c) = (n+1)n + c(n+1-(n+c)) =$
$(n+1)n + c(1-c) \leq (n+1)n$, as we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the integral $\int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? This integral is very similar to the one solved in this answer, but unfortunately the technique used there of splitting up the integral does not seem to yield obviously tractable integrals (at least for Mathematica!).
Edit
This integral arose from a research problem in the theory of orthogonal polynomials, which involved finding the equilibrium measure for a particular weight function, $w(x) = \exp(-|x| \ln(1+|x|))$. The extra factor of 1 in the argument of the logarithm ensures smoothness of this weight function near zero.
Here is where I can get to using the method from the linked post.
\begin{align}
f(t) &= \int_{0}^{1} \dfrac{s \ln(1+s) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s,\\
&= \underbrace{\int_{0}^{1} \dfrac{s \ln(1+s) - s \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s}_{\equiv I_{1}(t)} + \underbrace{\int_{0}^{1} \dfrac{s \ln(1+t) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s}_{\equiv I_{2}(t)}.
\end{align}
For $I_{2}(t)$, we have
\begin{align}
I_{2}(t) &= \ln(1+t) \int_{0}^{1} \dfrac{s - t}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s,\\
&= \ln(1+t) \int_{0}^{1} \dfrac{1}{\sqrt{1-s^{2}}(s + t)} \, \mathrm{d} s,\\
&= \dfrac{2 \ln(1+t)}{\sqrt{1-t^{2}}} \mathrm{arctanh}{\sqrt{\dfrac{1-t}{1+t}}}.
\end{align}
For $I_{1}(t)$, change variables to $x = \sqrt{1-s^{2}}$, and define $b \equiv \sqrt{1-t^{2}}$. Then we have
\begin{equation}
I_{1} = \int_{0}^{1} \dfrac{\ln(1+\sqrt{1-x^{2}}) - \ln(1+\sqrt{1-b^{2}})}{b^{2} - x^{2}} \, \mathrm{d} x.
\end{equation}
But whereas in the linked post this integral was solvable by Mathematica, in this case apparently we have no such luck.
| Continue with
\begin{align}
& \int_{0}^{1} \dfrac{s \ln(1+s) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s\\
= &\ \underset{=J}{\int_{0}^{1} \dfrac{s \ln(1+s) - s \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s} + {\int_{0}^{1} \dfrac{s \ln(1+t) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s}\\
=&\ J + \frac{2 \ln(1+t)}{\sqrt{1-t^{2}}} \mathrm{arctanh}{\sqrt{\dfrac{1-t}{1+t}}}
\end{align}
and evaluate $J$ below with the variable changes
$ x^2=\frac{1-s}{1+s}$ and $ y^2=\frac{1-t}{1+t}$
\begin{align}
J &= \int_{0}^{1} \frac{s\ln\frac{1+s}{1+t}}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s
= \int_{0}^{1} \frac{\ln\frac{1+s}{1+t}}{2\sqrt{1-s^{2}} }\left( \frac 1{s-t} + \frac1{s+t} \right){d} s\\
&=\frac{1+y^2}2\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{y^2-x^2}
+ \frac{\ln\frac{1+y^2}{1+x^2}}{1-y^2x^2}\ dx\\
\end{align}
where
\begin{align}
\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{y^2-x^2}dx
&= \frac1y \Re\bigg( \text{Li}_2\left(\frac{y+1}{y-i}\right)-\text{Li}_2\left(\frac{y-1}{y+i}\right) \bigg) \\
\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{1-y^2x^2}dx
&= \frac1y \Re\bigg( \text{Li}_2\left(\frac{1+y}{1-iy}\right)-\text{Li}_2\left(\frac{1-y}{1+iy}\right)\bigg)- \frac1y \ln y \ \ln\frac{1-y}{1+y}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all natural numbers $n$ such that the sum of the squares of the positive divisors of $n$ is equal to $n^2+2n+37$. If $f(n)$ is the sum of the squares of the positive divisors of n,then find all natural numbers $n$ such that $f(n)=n^2+2n+37$. I tried it first by substituting few values of $n$ but that doesn't work I think. So, then I consider two cases where one is that $n$ is prime but it is not possible. Another is that $n$ is composite and here I got stuck. Any help would be appreciated.
| Let us call the function $f$ as $\sigma_2$ since that's standard notation. Then, we have $\mathbb N\ni n\mapsto \sigma_2(n)=\sum_{d\mid n}d^2$. We start with a few lemmas and propositions some of which are easy enough so I will leave the proofs to you.
Lemma 1 $\sigma_2$ is multiplicative, that is, if $\gcd(a,b)=1$ then $\sigma_2(ab)=\sigma_2(a)\sigma_2(b)$.
Lemma 2 $n$ is prime $\iff$ $\sigma_2(n)=1+n^2$
Proof. The only if part is simple. If $n$ is a prime, then it's only divisors are $1,n$ and hence $\sigma_2(n)=1^2+n^2=1+n^2$. For the other direction, let $p$ be the smallest prime factor of $n$. We write $n=mp$ and for the sake of contradiction, assume $m>1$. Clearly $m\ge p$. In fact the equality is strict because if $m=p$, then $1+p^4=\sigma_2(n)=\sigma_2(p^2)=1+p^2+p^4$ which is absurd. Thus $m>p$. Now, $$1+m^2p^2=\sigma_2(mp)\ge 1+m^2+p^2+m^2p^2\Rightarrow m^2+p^2\le 0$$ which is not possible and thus we have reached a contradiction, and hence $m=1$, completing the proof. $\square$
For ease of writing, let $h(n)=n^2+2n+37$.
Proposition 3 If $\sigma_2=h$, then $n\ne p^2$ for any prime $p$.
Proposition 4 Let $p,q$ be primes with $q>p$ and $\sigma_2=h$. Then $n=pq\Rightarrow q=p+6$.
Proof. Simple calculation. $$p^2q^2+2pq+37=h(pq)=\sigma_2(pq)=1+p^2+q^2+p^2q^2$$ $$\Rightarrow (p-q)^2=36\Rightarrow q=p+6\ \ \ \ \ \ \square$$
Proposition 5 Let $p>3$ be a prime and $\sigma_2=h$. Then $n=p(p+6)\Rightarrow p+6$ is a prime.
Proof. We note that $\gcd(p,p+6)=1$ because if not, then $p=2,3$ but $p>3$. Then $$\sigma_2(p(p+6))=\sigma_2(p)\sigma_2(p+6)=(1+p^2)\sigma_2(p+6)$$
Also
\begin{align*}
\sigma_2(p(p+6))=h(p(p+6))&=p^2(p+6)^2+2p(p+6)+37\\&=p^2(p+6)^2+2p(p+6)+36+1\\&=p^2(p+6)^2+2p(p+6)+(p+6-p)^2+1\\&=p^2(p+6)^2+p^2+(p+6)^2+1\\&=(p^2+1)((p+6)^2+1)
\end{align*}
Thus $\sigma_2(p+6)=1+(p+6)^2$ and hence by lemma 2, we are done. $\square$
The preparations are complete. The trick in the proof of Lemma 2 is going to help again. Let $p$ be the smallest prime factor of $n$, and let $n=mp$. We observe that $m>p$ as $m=p$ does not work by proposition 3, and if $m<p$, then $p$ is not the smallest prime factor of $n$. Then since $\sigma_2=h$, $$m^2p^2+2mp+37=\sigma_2(mp)\ge 1+m^2+p^2+m^2p^2\Rightarrow (m-p)^2\le 36\Rightarrow m-p\le 6$$
and hence
$$p+6\ge m>p$$
We note that the upper bound is tight by proposition 4. Then $n=p(p+k),\ 1\le k\le 6$.
Theorem 5 The only $n$ that works are $n=p(p+6),\ p\ge 3$ prime.
Remark. The above theorem along with proposition 5 essentially says that if $\sigma_2(n)=h(n)$, then $n=pq$ where $(p,q)$ is a sexy prime pair.
Proof. We can check that for $p=2$, this doesn't work. Also, for $p=3$, only $3(3+6)=27$ works (12 cases to check, not that difficult). Then let $p>3$.
Since primes are odd, $n=p(p+k),\ k=1,3,5$ doesn't work as then the smallest prime factor of $n$ is 2, and thus $p=2\not\ge 3$. Also, by proposition 5, $n=p(p+6)$ works with $p+6$ prime. Thus the remaining cases are $n=p(p+k),\ k=2,4$. Then we note that for both these cases $p+2,\ p+4$ have to be prime respectively, because if not, then some $p'$ prime divides $p+2$ (p+4 respectively), and then $p'<p$ which is a contradiction to the minimality of $p$.
Then by proposition 4, both these cases do not work, and hence we are done. $\square$
An interesting observation is that $\sigma_2(n)=n^2+2n+(4k^2+1)$ holds only for $n=pq,\ q=p+2k$. The proof is essentially the same.
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$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$ $$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$$
I am given solution for this definite integral is $\frac{\pi}{2}\left(\ln\frac{\sqrt2+3}{\sqrt2+1}\right)$. Any idea or approach you would use to solve this?
| \begin{align}
&\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}\ dx\\
=& \int_{0}^{\infty}\int_{\sqrt2-1}^{\sqrt2+1}\frac{x}{x^2+y^2}\frac{x}{x^2+4}\ dy \ dx\\
=& \ \frac\pi2 \int_{\sqrt2-1}^{\sqrt2+1}\frac1{y+2}dy= \frac{\pi}{2}\ln\frac{\sqrt2+3}{\sqrt2+1}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If there are only two lattice points $(n, n+1),(n+1, n)$ on the first circle quadrant with radius $r$, then $r^2$ is prime. Conjecture
For lattice points $(x,y) \in \mathbb{Z}^{+}$, define a circle at the origin
$x^2+y^2=r^2$
Let $r = \sqrt{2 n^2 + 2 n + 1}$
where $n$ is odd, $n>3$.
If there are only two lattice points
$(n,\ n+1),(n+1, \ n)$
on the first quadrant arc, then $\large r^2$ is prime.
Examples
Let $n=7$, then $r^2=113$ is prime.
Let $n=9$, then $r^2=181$ is prime.
Question
Is there a way to prove that conjecture?
| We have the equation
$$n^2 + (n+1)^2 = m \tag{1}\label{eq1A}$$
for positive integers $n$ and $m$. Since $\gcd(n,n+1) = 1$, then $\gcd(n,m)=\gcd(n+1,m)=1$ as well. Next, consider if there's a prime factor $p \equiv 3 \pmod{4}$ which divides $m$. Then $n$ has a multiplicative inverse modulo $p$, call it $n^{-1}$, so \eqref{eq1A} becomes
$$(n+1)^2 \equiv -n^2 \pmod{p} \; \to \; (n^{-1}(n+1))^2 \equiv -1 \pmod{p} \tag{2}\label{eq2A}$$
Thus, $-1$ would be a quadratic residue modulo $p$. However, as shown in If p $\equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p., this is not possible. Thus, all prime factors of $m$ are congruent to $1$ modulo $4$.
Next, the sum of squares function for $k=2$ states that
The prime factorization $n=2^{g}p_{1}^{f_{1}}p_{2}^{f_{2}}\cdots q_{1}^{h_{1}}q_{2}^{h_{2}}\cdots$, where $p_{i}$ are the prime factors of the form $p_{i}\equiv 1\pmod {4}$ and $q_{i}$ are the prime factors of the form $q_{i}\equiv 3\pmod {4}$ gives another formula
$\qquad r_{2}(n)=4(f_{1}+1)(f_{2}+1)\cdots$, if all exponents $h_{1},h_{2},\cdots$ are even. If one or more $h_{i}$ are odd, then $r_{2}(n)=0$.
Note $r_2(n)$ is the number of representations of $n$ as a sum of $2$ squares, and their $n$ is the $m$ in \eqref{eq1A}. Since $m$ is odd, then $g = 0$. As explained earlier, all $h_i = 0$. Also, since there are $2$ additional lattice points in each of the other $3$ quadrants, there's a total of exactly $8$ lattice points. Thus, $r_2(m) = 8$, so $f_1 = 1$ and $f_i = 0$ for all $i \gt 1$. This gives $m = r^2 = p_1$, i.e., it's a prime.
| {
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Proving using Mathematical Induction from my discrete math class This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.
This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer."
This is my attempt to prove it.
Step 1: Base case $(n=1)$
$\sum_{k=1}^1\frac{1}{(1)^2}\le2-\frac{1}{1}$
$1\le1$
Step 2: Induction hypothesis
Assume that $\sum_{k=1}^n\frac{1}{k^2}\le 2-\frac{1}{n}$ is true for $n=c$
Then, $\sum_{k=1}^c\frac{1}{k^2}\le 2-\frac{1}{c}$
Step 3: prove that it is true for $n=c+1$
$\sum_{k=1}^{c+1}\frac{1}{k^2}$ $\le$ $2-\frac{1}{c}+\frac{1}{(c+1)^2}$
$\le 2+\frac{-(c+1)^2+c}{c(c+1)^2}$
$\le 2+\frac{-(c^2+2c+1)+c}{c(c+1)^2}$
$\le 2+\frac{-c^2-2c-1+c}{c(c+1)^2}$
$\le 2-\frac{c^2+c+1}{c(c+1)^2}$
$\le 2-\frac{c^2+c}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{c(c+1)}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
I'm Stuck on this step.
| It is very straightforward induction. You just add
$\dfrac{1}{(n+1)^{2}}$ to both sides of $P(n)$ and you get
$2-\dfrac{1}{n}+\dfrac{1}{(n+1)^{2}}$ must be less than $2-\dfrac{1}{n+1}$ .
Making all the simplifications we get
$\dfrac{n+2}{(n+1)^{2}}<\dfrac{1}{n}$ which is obviously true. So we get
$P(n+1)$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$? So this was part of a bigger induction proof problem, but I'm stuck at what I think is the last step. I need to show that
$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$, for $n \ge 1$
Would it be enough to show that the limit as it approaches positive infinity is both 1, and that since the inequality is true for $n=1$, and $1/2 < \sqrt{1/2}$, the function on the right is always greater? I'm pretty sure this isn't the way to go about it but I'm really stuck.
| Although this is a little indirect, it keeps calculation to a
minimum.
For any positive integer $m,$ we have
$$
(m - 1)(m + 1) = m^2 - 1 < m^2,
$$
therefore
$$
\frac{m - 1}{m^2} < \frac1{m + 1},
$$
therefore
$$
\left(\frac{m - 1}m\right)^2 = \frac{(m - 1)^2}{m^2} \leqslant
\frac{m - 1}{m + 1} = 1 - \frac2{m + 1} < 1 - \frac2{m + 2} =
\frac{m}{m + 2},
$$
therefore
$$
\frac{m - 1}m < \sqrt{\frac{m}{m + 2}}.
$$
Let $n$ be a non-negative integer.
Taking $m = 2n + 2,$ and simplifying, we get
$$
\frac{2n + 1}{2n + 2} < \sqrt{\frac{2n + 2}{2n + 4}}
= \sqrt{\frac{n + 1}{n + 2}} = \frac{\sqrt{n + 1}}{\sqrt{n + 2}}.
$$
| {
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Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm 1$, but this implies $a=0$ and this is not always correct. Then I realized that,
$$x^4-2x^2+1=(x^2-1)^2$$
is a perfect square. So, I tried to write the original equation as
$$ax^4-x^3+x+\bigg(x-\frac 1x\bigg)^2=0$$
$$x^2\bigg(ax^2-x+\frac 1x\bigg)+\bigg(x-\frac 1x\bigg)^2=0$$
But I failed again. I couldn't spot the palindromic property.
| The coefficient $a$ should be something difficult to deal with, so isolating $a$ might help.
We know that $x=0$ is not a root, so we can divide by $x^6$ to have
$$a=\frac{x^5-x^4-x^3+2x^2-1}{x^6}$$
$$a=\frac 1x-\frac{1}{x^2}-\frac{1}{x^3}+\frac{\color{red}2}{x^4}-\frac{1}{x^6}$$
Now, we can focus on the RHS which has no $a$, and we are happy if the RHS is of the form $uX^2+vX+w$. The coefficient $\color{red}2$ reminds me of $A^2+\color{red}2AB+B^2=(A+B)^2$. So, we may have $\bigg(\dfrac 1x\pm \dfrac{1}{x^3}\bigg)^2$, then we note that
$$a=\bigg(\frac 1x-\frac{1}{x^3}\bigg)-\bigg(\frac 1x-\frac{1}{x^3}\bigg)^2$$
Letting $X=\dfrac 1x-\dfrac{1}{x^3}$, we have
$$X^2-X+a=0\implies \frac 1x-\frac{1}{x^3}=X=\frac{1\pm\sqrt{1-4a}}{2}$$
and finally get
$$\frac{1\pm\sqrt{1-4a}}{2}x^3-x^2+1=0$$
| {
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Diophantine Equation $a^3+b^3=22c^3$ Found this equation in a book on the Riemann Hypothesis. I've found solution without proof on MathStack asking for an algorithm for all the solutions. I'm trying to find a proof for the smallest positive integer solutions. I think I'm close, but not sure how to proceed.
Solve $a^3+b^3=22c^3$ over the positive integers. Seeking a smallest solution we need $a$,$b$, and $c$ all to be pairwise relatively prime. If the sum of two numbers is even, then those numbers are either both even or both odd. By the above argument then, $a$ and $b$ must both be odd. It follows that their sum is even.
$a^3+b^3=(a+b)(a^2+b^2-ab)$. By 25 trials, it can be shone that $a^2+b^2-ab$ is never divisible by 11. It follows that $a+b$ must be, since the right hand side of the equation is. So $a+b$ is divisible by 22.
$a+b=22k$ for some integer $k$.
$22c^3=a^3+(22k-a)^3=a^3+22^3k^3-3a(22k)^2+ 3a^2(22k)-a^3$
$3ka^2-66ak^2+(22^2k^3-c^3)=0\implies a= \frac{66k^2\pm\sqrt{66^2k^4-12k(22^2k^3-c^3)}}{6k}$
Introduce $n$ as $66^2k^4-12k(22^2k^3-c^3)=36k^2n^2$
It follows that $c^3=k(11^2k^2+3n^2), a=11k+n, b=11k-n$. For $a$ and $b$ to be relatively prime then so are $k$ and $n$. If $z^3=xy$ and $gcd(x,y)=1$, then $x$ and $y$ are also perfect cubes.
So if $3\nmid k$, both factors of $c^3$ are relatively prime. This means $\exists m$ so that $k=m^3$. It then follows that $c^3=m^3(11^2m^6+3n^2)$ and both factors are perfect cubes.
On the other hand, if $3|k$, then $9|k$ and $3|c$. After $c=3p, k=9q$ we have:
$27p^3=11^2\cdot 9^3q^3+3n^2\cdot 9q\implies p^3=q(27\cdot11^2q^2+n^2)$ By previous arguments $q$ and $n$ are relatively prime. It follows that :
$p^3=m^3(3\cdot(33)^2m^6+n^2)$ where both factors are perfect cubes.
This might be a red-herring, but whether $k$ is divisible by 3 or not, we end up with an equation of the form $z^3=3x^2+y^2$. Multiply both sides by 27 and you get $(3z)^3=(9x)^2+3(3y)^2$. You get another triplet that satisfies the equation. Given one solution of this form we have infinitely many others. Not sure where to proceed from here.
Not sure if this is the smallest, but $a=17299, b=25469, c=9954$ will do the job, but I'm not sure how these numbers were arrived at.
| COMMENT.-Apropos of Tomita's comment, I want to add that it is not necessary to go to the (more popularly known) Weierstrass form and that the curve $x^3+y^3=nz^3$ is also in their own right an elliptic curve in which the formulas of sum are given as follows:
If $A=(X_1,Y_1,Z_1)\ne B=(X_2,Y_2,Z_2)$ then $A+B=(X_3,Y_3,Z_3)$ where $$X_3=X_1Z_1Y_2^2-X_2Z_2Y_1^2 \\Y_3=Y_1Z_1X_2^2-Y_2Z_2X_1^2\\Z_3=X_1Y_1Z_2^2-X_2Y_2Z_1^2$$ and if $A=B=(X,Y,Z)$ (case of tangent, not a chord) $$X_3=-Y(2X^3+Y^3)\\Y_3=X(X^3+2Y^3)\\Z_3=Z(X^3-Y^3)$$
►I want to add that having deduced the point of the elliptic curve $x^3+y^3=22$ given by the O. P. and shown in the figure above (I want to believe that it has been calculated by him) is very meritorious and that it is clear that he does not have the tools to know if said point is minimum (i.e. it is a generator of the corresponding additive group). To this end, in the classic paper “The diophantine equation $ax^3+by^3+cz^3=0$, Acta Math. (Stockolm) 85 $(1951)\space p. 203-362$” written by E. S. Selmer, there is a very laborious table (calculated without a computer!) in which there are just over the first hundred values of cube-free integers, $n$, representable as sum of two cubes of rationals. This table also shows the range of the curves and then you can see if the point given by Turloc The Red is minimum or not (in Tomita's comment, it can also be deduced that the point in question is indeed a generator).
| {
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Rotating a Matrix into a Pauli Spin Matrix Suppose we are given the matrix
$$
M=\begin{pmatrix}
z & x-iy\\
x+iy & -z
\end{pmatrix}
$$
constrained by $x^2+y^2+z^2=1.$ Note that by the constraints, the eigenvalues of $M$ are given by $\pm1$. My question is, how do we find a unitary matrix $R$ such that
$$
R^{\dagger}MR=\sigma_z:=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}?
$$
We know such an $R$ exists because $M$ and $\sigma_z$ are unitarily equivalent. This might be a duplicate from:
Find a matrix for a unitary transform between matrices or prove that there is none,
but I still don't know how to construct $R.$
| The eigenvalues of $M$ are found by calculating the characteristic polynomial
$$ \det(M-\lambda I_2)=\det\begin{pmatrix}z-\lambda & x-yi \\ x+yi & -z-\lambda\end{pmatrix}=\lambda^2-(x^2+y^2+z^2). $$
Thus, $\lambda=\pm1$ since $x^2+y^2+z^2=1$. Now, what about the eigenvectors? For $\lambda=1$,
$$ \underbrace{\begin{pmatrix} z-1 & x-yi \\ x+yi &-z-1 \end{pmatrix}}_{M-\lambda I_2}\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} $$
yields the eigenvector $\begin{pmatrix} u \\ v\end{pmatrix}\!=\!\begin{pmatrix}1+z \\ x+yi\end{pmatrix}$ or $\begin{pmatrix}x-yi \\1-z\end{pmatrix}$of squared norm $2(1\pm z)$.
(A quick solution to $au+bv=0$ is $(u,v)=(-b,a)$ or $(b,-a)$.)
Similarly for $\lambda=-1$, we get $\begin{pmatrix}x-yi \\ -1-z\end{pmatrix}$ or $\begin{pmatrix}-1+z \\ x+yi\end{pmatrix}$ again of squared norms $2(1\pm z)$.
Therefore we get the unitary matrix
$$ Q \,=\, \frac{1}{\sqrt{2(1+z)}}\begin{pmatrix}1+z & x-yi \\ x+yi & -1-z\end{pmatrix} \,=\, \frac{1}{\sqrt{2(1-z)}}\begin{pmatrix} x-yi & -1+z \\ 1-z & x+yi\end{pmatrix} $$
which satisfies $MQ=Q\Lambda$ with $\Lambda=\mathrm{diag}(1,-1)=\sigma_z$. This is because $Q$'s columns are eigenvectors of $M$, the columns of $MQ$ are just $M$ applied to the columns of $Q$, and the columns of $Q\Lambda$ with $\Lambda$ diagonal are just the columns of $Q$ scaled by the diagonal elements of $\Lambda$. Thus the spectral decomposition $M=Q\Lambda Q^\dagger$ (where $Q^\dagger=Q^{-1}$ because $Q$ is unitary), which we may rewrite as $Q^\dagger MQ=\sigma_z$.
| {
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Trignometry problem: If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$ If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$
What I did:
$\sin^2\theta + 3\cos\theta = 2$
$3\cos\theta - 1 = 1 - \sin^2\theta$
$3\cos\theta - 1 = \cos^2\theta$
$\cos^3\theta + \frac{1}{\cos^3\theta} = (\cos^2\theta + 1)^3 + \frac{1}{(\cos^2\theta + 1)^3}$
Then I'll have to take LCM and it would be bigger. Can anyone suggest an easy way to solve this problem?
| Let $c=\cos\theta$. From what you found, we have
$$c+\frac{1}{c}=3.$$
Then,
$$c^3+\frac{1}{c^3}=(c+\frac{1}{c})^3-3(c+\frac{1}{c})=3^3-3\times 3=18.$$
| {
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Find $\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x} $ So I am having trouble finding this limit:
$$\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x}$$
The problem is I can't use the derivative of the composition of two functions nor can I use other techniques like l'Hôpital's theorem.
I tried numerous techniques to calculate this limit but in vain so if you have any simple idea that is in the scope of my knowledge ( I am a pre-calculus student ), please do let me know without actually answering the question.
| In this answer I will use the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to derive the limit
\begin{align}
\lim_{x \to 0} \frac{\sin \left( \pi\sqrt{\cos x} \right)}{x}&=\lim_{x \to 0} \frac{\sin \left(\pi - \pi \sqrt{\cos x} \right)}{x}\\
&= \lim_{x \to 0} \frac{\sin \left(\pi\left(1 - \sqrt{\cos x}\right) \right)}{\pi\left(1 - \sqrt{\cos x}\right) }\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\
&=\lim_{x \to 0}\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\
&= \lim_{x \to 0} \frac{\pi\left ( 1 - \cos x \right) }{x \left (1 + \sqrt{\cos x}\right)}\\
&= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos x }x\\
&= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos^2 x }{x\left ( 1 + \cos x\right) }\\
&= \frac{\pi}4 \lim_{x \to 0} \frac{sin^2 x}{x}\\
&= \frac{\pi}4 \lim_{x \to 0} \sin x \\
&= 0
\end{align}
| {
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The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$. The following question is taken from JEE practice set.
The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$.
My Attempt:
$$D\le0\\ \implies 9a^2-4b\le0\\ \implies9a^2-36\le4b-36\\ \implies9(a-2)(a+2)\le4(b-9)$$
Now, $a+2\gt4\implies 9(a-2)(a+2)\gt36(a-2)$
Also, $b\gt b-9\implies4b\gt4(b-9)$
Thus, $4b\gt9(a-2)+a+2)\gt36(a-2)$
Therefore, $\frac{b}{a-2}\gt9$
The answer given is $18$. How to do this?
| We get from $9a^2\le4b$ that $a,b$ and $\frac b{a-2}$ are all positive. It is clear that raising $b$ does not violate the inequality but increases $\frac b{a-2}$, so for any $a$ we minimise $\frac b{a-2}$ by setting $b=\frac94a^2$.
Once that is fixed we then rewrite
$$\frac b{a-2}=\frac94\cdot\frac{a^2}{a-2}=\frac94\left(a+2+\frac{4}{a-2}\right)$$
and differentiate to get $$\frac94\left(1-\frac4{(a-2)^2}\right)=0\implies a=4$$
Hence the minimum of $\frac b{a-2}$ is $\frac94\cdot\frac{4^2}{4-2}=18$.
| {
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How to factorize $\frac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$? How to factorize $\dfrac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$?
Which trigonometric identities to use?
I'm stuck when it comes to $\tan(2x)+\tan(x)$. I don't know which identity to use to turn it into the product.
I was thinking of just transforming them to sines and cosines, but also doesn't get me anywhere.
Thanks for help in advance.
| First note the following identities:
\begin{align}
\cos (3x) &= 4 \cos ^3 (x) - 3 \cos (x) \\
\sin(2x) &= 2 \sin(x) \cos(x) \\
\cos(2x) &= 2 \cos^2(x) - 1
\end{align}
Substituting into the expression yields
\begin{multline}
\frac{\cos(3x) - \cos(x)}{\tan(2x) - \tan(x)} = \frac{4 \cos(x)(\cos^2(x) - 1)}{\frac{2 \sin(x) \cos(x)}{2 \cos^2(x) - 1} - \frac{\sin(x)}{\cos(x)}} = \frac{4 \cos^2(x)(\cos^2(x) - 1)(2\cos^2(x) - 1)}{2 \sin(x) \cos^2(x) - \sin(x)(2\cos^2(x) - 1)} \\= 4 \frac{\cos^2(x)}{\sin(x)}(\cos^2(x) - 1)(2\cos^2(x) - 1).
\end{multline}
Now, substituting back in $2 \cos^2(x) - 1=\cos(2x)$ and $\cos^2(x) - 1 = -\sin^2(x)$ yields
\begin{multline}4 \frac{\cos^2(x)}{\sin(x)}(\cos^2(x) - 1)(2\cos^2(x) - 1) = -4 \cos^2(x) \sin(x) \cos(2x)\\ = -2\cos(x)\sin(2x)\cos(2x) = -\cos(x)\sin(4x)\end{multline}
where in the last two equalities I have used $2 \sin(x) \cos(x) = \sin(2x)$.
| {
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How to solve the equation for $x$ where $x^{(x-1)^2} = 2x +1 $ How to solve for $x$ in the following equation
$$ x^{(x-1)^2} = 2x +1$$
Guessing $(x-1)^2=2$ is a pretty good guess and it works giving $x = \sqrt{2} +1 $
But I want a non-guess solution. I tried thinking in terms of logarithm and binomial expansion but couldn't go far.
| I'm assuming $x$ is non negative, since otherwise $x^{(x-1)^2}$ might not be well defined over the reals.
If $x<1$, then $x^{(x-1)^2}<1$, but $2x+1\ge 1$, so that can't yield a solution. Now suppose $x\ge 1$.
Let $y = (x-1)^2$, so $x^y = 2x+1 =$ $x^2+2-y$, therefore
$$x^y - x^2 = 2-y$$
Because $x \ge 1$, we have $x^y \le x^2$ if $y<2$ and $x^y \ge x^2$ if $y>2$, i.e.,
*
*If $y<2$, $x^y - x^2$ is non positive but $2-y$ is positive.
*If $y>2$, $x^y - x^2$ is non negative but $2-y$ is negative.
Hence $y = 2$ is the only candidate for a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\frac{1}{2})^n = 0$ to prove that $\lim_{n\to\infty} a_n = 0$.
My attempt: $a_n$ is a decreasing sequence because $a_n < \frac{1}{2}a_{n-1}$, for all $n \ge 2$. And we can continue: $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n}$; $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n} <\frac{1}{2}a_{n-1}$. Then, $(\frac{1}{2})^na_{n+1} < (\frac{1}{2})^na_{n} <(\frac{1}{2})^na_{n-1}$ is also true. I see that the squeeze theorem is applicable but I am not sure how to proceed from here. Please advise.
| Testing
Note that for all $n\ge 1$ we have
\begin{align*}
a_n\le \frac{1}{2}a_{n-1}\le \dots \le \left(\frac{1}{2}\right)^{n-1}a_{1}
\end{align*}
Also note that for all $x>0$ exists $m\in \mathbb{N}$ such that $\frac{1}{m}<x$, with this we can conclude that for all $\varepsilon>0$ exists $N\in \mathbb{N}$ such that
\begin{align*}
|a_n|\le \left(\frac{1}{2}\right)^{n-1}a_1<a_1\varepsilon
\end{align*}
for all $n\ge N$. In other words
\begin{align*}
\lim_{n \to \infty}a_n=0
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^\frac13+\dfrac23x\left(x^2-1\right)^{-\frac23}(x^2-1)(x^2+1)\\=\dfrac23x\left(x^2-1\right)^\frac13\left(6x^2+x^2+1\right)\\=\dfrac23x\left(x^2-1\right)^\frac13(7x^2+1)$$ The given answer is $$y'=\dfrac{2x(7x^4-6x^2-1)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$ I don't see my mistake...
| $$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(\frac{\left(x^2-1\right)^{\frac{2}{3}}}{\left(x^2-1\right)^{\frac{2}{3}}}\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}+\frac{2}{3}}\left(7x^2+1\right)\left(\frac{1}{\left(x^2-1\right)^{\frac{2}{3}}}\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)\left(7x^2+1\right)\left(\frac{1}{^{\sqrt[3]{\left(x^2-1\right)^2}}}\right)$$
$$=\frac{2x}{3\sqrt[3]{\left(x^2-1\right)^2}}\left(x^2-1\right)\left(7x^2+1\right)$$
$$=\frac{2x\left(x^2-1\right)\left(7x^2+1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
$$=\frac{2x\left(7x^4+x^2-7x^2-1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
$$=\frac{2x\left(7x^4-6x^2-1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
| {
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what comes to mind is finding $(a + b + c + d)^3$ and subtracting whatever we don't need to get $a^3 + b^3 + c^3 + d^3$. However,
\begin{equation*}
(a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3
\end{equation*}
There is simply no good way to calculate $6abc + 6abd + 6acd + 6bcd $ without expanding everything.
| Let $\omega_1, \ldots, \omega_p$ be the $p$th roots of unity (including $1$), where $p$ is prime. Note that:
$$\omega_1^k + \ldots + \omega_p^k = \begin{cases} p & \text{if } p \mid k \\ 0 & \text{otherwise.} \end{cases}$$
Why? The group of $p$th roots of unity are isomorphic to $\Bbb{Z} / p\Bbb{Z}$, and the map $x \mapsto x^k$ is equivalent to the map $x \mapsto kx$ on $\Bbb{Z} / 5 \Bbb{Z}$. It's well know that this homomorphism is trivial when $p \mid k$, and injective (and thus surjective) otherwise. So, when $p \not\mid k$, we are summing the $p$th roots of unity, which is $0$.
So, if we have a polynomial $q$, we can evaluate $q(\omega_1) + \ldots + q(\omega_p)$ relatively easily. We get:
$$q(\omega_1) + \ldots + q(\omega_p) = p(q_0 + q_p + q_{2p} + \ldots),$$
where $q_i$ is the coefficient of $x^i$, and the sum on the right hand side is necessarily finite, due to the finite degree of $q$.
In your case, where $p = 5$, we have $a^3, b^3, c^3, d^3, (20 + 13)^3$ are (not necessarily respectively) $q(\omega_1), \ldots, q(\omega_5)$, where
$$q(x) = (20x^2 + 13x)^3.$$
We only care about the coefficients of $x^5$ and $x^0$, the latter of which is clearly $0$, and the former is $3 \cdot 20^2 \cdot 13$. So,
$$a^3 + b^3 + c^3 + d^3 + 33^3 = 15 \cdot 20^2 \cdot 13,$$
hence the answer is $15 \cdot 20^2 \cdot 13 - 33^3$
| {
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"timestamp": "2023-03-29T00:00:00",
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A Question on which solution is false and where (with logarithms) This question was put to me yesterday during a meeting with one of my highschool school teacher who taught me mathematics.
The question:
Find the domain of the function $f:x \to$ $log_\frac{1}{10} \frac{2x+1}{x+2}$.From this, solve the inequality $f(x) \ge log_\frac{1}{10} \frac{1}{x}$.
Solution for the domain of the function:
Since $ \frac{2x+1}{x+2} = \frac{2x+4-3}{x+2}$ = 2- $\frac{3}{x+2} \ge 0$,then $x \in $$(-\infty,-2) \cup (-\frac{1}{2},\infty):=S$
Here comes the funny part, my teacher offered two solutions for solving the inequality:
solution 1 (which is also similar to mine) :
With the premise of $x \in S = D_f$, we have that
$log_\frac{1}{10} \frac{2x+1}{x+2} \ge log_\frac{1}{10}\frac{1}{x}$
$\iff$ $\frac{2x+1}{x+2} \le \frac{1}{x}$ $\land$ $\frac{1}{x} \gt 0$
$\iff$ $\frac{2(x+1)(x-1)}{x(x+2)} \le 0$ $\land$ $x \gt 0$
$\iff$ $x-1 \le 0 \land x \gt 0$
$\iff$ $x \le 1 \land x \gt 0$
Hence $x \in (0,1]$
Solution 2(which I personally think that there's something wrong with this solution) :
Consider $f(x) \ge log_\frac{1}{10} \frac{1}{x}$
Choose y such that $f(y) = log_\frac{1}{10} \frac{1}{x}$ $\land$ $x \gt 0$
so
$\begin{cases} f(x) \ge f(y)\\log_\frac{1}{10} (2-\frac{3}{y+2})=log_\frac{1}{10}\frac{1}{x}\\ x \gt 0 \land y\in S\end{cases}$
$\begin{cases} x \le y ,&\text{(because $f$ is decreasing)} \\ 2-\frac{3}{y+2}=\frac{1}{x} ,&\text{(take the $antilog={0.1}^{x}$ of both sides)} \\ x \gt 0 \land y\in S\end{cases}$
$\begin{cases} x+2\le y+2\\ y+2 =\frac{3x}{2x-1}\\ x \gt 0 \land y\in S\end{cases}$
sub the second equation to the first inequality and we will get
$x+2 \le \frac{3x}{2x-1} \land x,2x-1 \gt 0$
$(x+2)(2x-1)-3x \le 0 \land x\gt \frac{1}{2} $
$2(x-1)(x+1) \lt 0 \land x\gt \frac{1}{2} $
$(x-1)\lt 0 \land x\gt \frac{1}{2} $
Hence $ \frac{1}{2} \lt x \le 1 $
Which solution is wrong and where?
| The second solution is wrong because $f$ is only decreasing on each of the two intervals $(-\infty,-2)$ and $(-\frac12,\infty)$ but not on their union.
So, a large forgotten part of the solution is when $y<-2$ (and $x>0$, and $y=\frac{2-x}{2x-1}$), i.e. $0<x<\frac12.$
Moreover, the value $x=\frac12$ is also forgotten because there is no $y$ such that $\frac{2y+1}{y+2}=2.$
| {
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Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|<\delta$. So, $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|=|K||\frac{x(K^2+y^2)-y(K^2+x^2)}{(K^2+y^2)(K^2+x^2)}|\leq |\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}|\leq|\frac{xK^2-yK^2+xy^2}{(K^2+y^2)(K^2+x^2)}|$ I'm a bit stuck here on how to relate this inequality to $|x-y|$. Can someone help out? Thanks!
| Since you started your proof attempting to use the $\epsilon$-$\delta$ definition here is a hint. One can proceed in the following way:
*
*If $K=0$, then $f(x)=0$ for all $x$ and the proof is trivial.
*If $K\neq 0$, then we can express:
$$f(x)=\frac{x/K}{1+x^2/K^2}$$.
The $\epsilon$-$\delta$ proof is a bit laborious, but my best guess is that
$$\delta(\epsilon,y)=\frac{-(\frac{y^2}{K^2}+\frac{1}{{|K|}})+\sqrt{\frac{y^4}{K^4}+\frac{2y^2}{{|K|}^3}+\frac{1+4\epsilon y}{K^2}}}{2|y|/K^2}$$
should do the trick. Do you see why?
Details:
Let $|x-y|<\delta(\epsilon,y)$. Then:
$$
\begin{align}
|f(x)-f(y)|&=\left|\frac{x-y}{K}+\frac{xy}{k^2}(y-x)\right|\\
&\leq \left|\frac{x-y}{K}\right|+\left|\frac{xy}{K^2}(y-x)\right|\\
&=\frac{\left|x-y\right|}{\left|K\right|}\left|1-\frac{xy}{K}\right|\\
&\leq\frac{\left|x-y\right|}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\
&<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\
&<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|\delta(\epsilon,y)y+y^2\right|}{\left|K\right|}\right)\\
&\leq\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\delta(\epsilon,y)|y|+y^2}{\left|K\right|}\right)\\
&=\delta^2(\epsilon,y)\frac{y}{K^2}+\delta(\epsilon,y)\left(\frac{y^2}{K^2}+\frac{1}{\left|K\right|}\right)\\
&=\epsilon \\
\end{align}
$$
| {
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Finding a relationship between coefficients to solve a cubic polynomial From E.J Barbeau's Polynomials, the question states:
Find the relationship between $p$ and $q$ in order that the equation $x^3 + px + q$ may be put into the form $x^4 = (x^2 + ax + b)^2$. Hence, solve the equation $8x^3 - 36x + 27 = 0$.
I have no clue where to start on this. Could anyone assist me with this problem? The answer ($p^3 + 8q^2 = 0$) seems rather random and I don't understand how one could have obtained such a relationship.
| You want $x^3+px+q=0$ to be equivalent to $x^4=(x^2+ax+b)^2$. Notice that:
$$x^4=(x^2+ax+b)^2 \iff (x^2)^2-(x^2+ax+b)^2=0$$
$$\iff (x^2-(x^2+ax+b))(x^2+(x^2+ax+b))=0 \iff (ax+b)(2x^2+ax+b)=0$$
$$\iff 2ax^3+a^2x^2+abx+2bx^2+abx+b^2=0$$
$$\iff 2ax^3+(a^2+2b)x^2+2abx+b^2=0 \iff x^3+\frac{a^2+2b}{2a}x^2+bx+\frac{b^2}{2a}=0$$
In order for this to be equivalent to $x^3+px+q=0$, we must have equal coefficients:
$$\begin{cases}
\frac{a^2+2b}{2a}=0 \implies a^2+2b=0\\
p=b \implies b=p\\
q=\frac{b^2}{2a} \implies a=\frac{b^2}{2q} \implies a=\frac{p^2}{2q}
\end{cases}
$$
Plugging in the last two equations into the first one yields:
$$(\frac{p^2}{2q})^2+2p=0 \implies \frac{p^4}{4q^2}+2p=0 \implies p^4+8pq^2=0 \implies\fbox{$p^3+8q^2=0$}$$
Now for solving $8x^3-36x+27=0 \iff x^3-\frac{9}{2}x+\frac{27}{8}$, we can see that our condition is satisfied, hence we can turn the equation into one of the form $x^4=(x^2+ax+b)^2$. The above equations enable us to quickly find $a$ and $b$:
$$a=\frac{p^2}{2q}=\frac{\frac{9^2}{2^2}}{2\cdot\frac{27}{8}}=3$$
$$b=p=-\frac{9}{2}$$
So our equation becomes:
$$x^4=(x^2+3x-\frac{9}{2})^2 \implies x^2=\pm (x^2+3x-\frac{9}{2})$$
For the $+$ case we get:
$$x^2=x^2+3x-\frac{9}{2} \implies 3x=\frac{9}{2} \implies x_1=\frac{3}{2}$$
For the $-$ case we get:
$$x^2=-x^2-3x+\frac{9}{2} \implies 2x^2+3x-\frac{9}{2}=0 \implies x_{2,3}=\frac{-3\pm 3\sqrt{5}}{4}$$
To conclude, the solutions for $8x^3-36x+27=0$ are:
$$\fbox{$x_1=\frac{3}{2} \space \text{and} \space x_{2,3}=\frac{-3\pm 3\sqrt{5}}{4}$}$$
| {
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Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?
| Let $a\geq 0$ and $$f(a)=\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x}x^a dx$$
You're really looking for the value of $f(0)$.
Note that
$$\begin{split}
f^\prime(a) &= \int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)}x^a dx\\
&= \int_0^1 (x^{m-1}-x^{n-1})\left( \sum_{k\geq 0} (-1)^k x^{kp}\right)x^a dx\\
&= \sum_{k\geq 0} (-1)^k\int_0^1 (x^{kp+a+m-1} -x^{kp+a+n-1}) dx\\
&= \sum_{k\geq 0} (-1)^k\left( \frac 1 {kp+a+m}-\frac 1 {kp+a+n}\right)
\end{split}$$
Integrating back w.r.t. $a$, and noting that $\lim_{a\rightarrow+\infty}f(a)=0$ gives
$$\begin{split}
f(a) &= \sum_{k\geq 0} (-1)^k \ln\left( \frac{kp + a + m}{kp + a + n}\right)\\
&=\ln\left ( \frac{a+m}{a+n} \cdot \frac{p+a+n}{p+a+m}\cdot \frac{2p+a+m}{2p+a+n}\dots\right)
\end{split}
$$
Evaluating at $a=0$:
$$\int_0^1\frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x} dx= \ln\left ( \frac{m}{n} \cdot \frac{p+n}{p+m}\cdot \frac{2p+m}{2p+n}\dots\right)$$
| {
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How do I calculate ln(12) approximate value to two decimals? (edit) The problem I've been assigned literally says just to "Calculate ln(12) without a calculator to two decimals using ln(1+x) series." I've been trying to figure out what series I'm dealing with here and don't seem to find an answer anywhere.
| When $x> 1/2$ we can use the following series for the logarithm:
$$\ln(x) = \sum_{k = 1}^{+\infty} \dfrac{1}{k} \left(\dfrac{x-1}{x}\right)^k = \dfrac{x-1}{x} + \dfrac{1}{2}\left(\dfrac{x-1}{x}\right)^2 + \dfrac{1}{3} \left(\dfrac{x-1}{x}\right)^3 + \ldots$$
Plugging $x = 12$ you get
$$\ln(12) \approx \dfrac{11}{12} + \dfrac{1}{2} \cdot \dfrac{121}{144} + \dfrac{1}{3}\cdot \dfrac{1331}{1728} = \dfrac{11}{12} + \dfrac{121}{288} + \dfrac{1331}{5184} = \dfrac{8261}{5184} = 1.593(..)$$
Where as $\ln(12) = 2.48$ so you can see this may not be the best way for it converges slowly.
There is another series for the logarithm as pointed out by RobJohn, that works for $x > 0$, that is:
$$\ln(x) = 2\sum_{k = 1}^{+\infty} \dfrac{1}{2k-1} \left(\dfrac{x-1}{x+1}\right)^{2k-1} = 2\left(\dfrac{x-1}{x+1} + \dfrac{1}{3}\left(\dfrac{x-1}{x+1}\right)^3 + \dfrac{1}{5}\left(\dfrac{x-1}{x+1}\right)^5 + \ldots \right)$$
Applying it here gives
$$\ln(12) \approx 2\left(\dfrac{11}{13} + \dfrac{1}{3}\dfrac{11^3}{13^3} + \dfrac{1}{5}\dfrac{11^5}{13^5} + \ldots\right) = \frac{12640826}{5569395} = 2.26(...)$$
Which is rather better.
Add some extra terms to get a two decimal precision. Calculations are easy.
| {
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Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fractions too. What happens when the power of $x$ is 4?
My thoughts are to do integration by parts (product rule)
$$\int u v' dx = uv - \int u' v dx$$
But I am unclear what do substitute $u$ and $v$ to
| Integrate by parts as follows
\begin{align}
\int \frac{3}{(x^2 +5)^2}dx &=\int \frac3{10x}d\left(\frac{x^2}{x^2+5}\right)\\
&= \frac{3x}{10(x^2+5)}+\frac3{10}\int\frac{1}{x^2+5}dx\\
&= \frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac x{\sqrt5}+C
\end{align}
| {
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Proving a combinatorial identity having sums of fractions of binomial coefficients Recently I was solving a probability question, and I encountered a summation that I was unable to figure out. I put it on Wolfram Alpha, and it returned an unexpectedly simple solution. The answer I mention is here, and it is correct as I have checked the solution independently using a different method (the second method in the answer). The identity in question is as follows:
$$\sum_{n=1}^{r+1}\frac{\binom{r}{n-1}}{\binom{b+r}{n}}=\frac{b+r+1}{b(b+1)}$$
The linked answer definitely gives a probabilistic proof for the same, but I would very much like a direct proof. Any kind of method is acceptable that is different from what has been done in the answer. Thank you in advance!
| We seek to evaluate
$$\sum_{q=1}^{r+1} {r\choose q-1} {b+r\choose q}^{-1}
= \sum_{q=0}^r {r\choose q} {b+r\choose q+1}^{-1}.$$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We obtain
$$\sum_{q=0}^r {r\choose q} (q+1)
[z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-q-1}
\\ = [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\sum_{q=0}^r {r\choose q} (q+1) (z-1)^{-q}.$$
We get two pieces the first is
$$[z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\left[1+\frac{1}{z-1}\right]^r
\\ = [z^b] \log\frac{1}{1-z} (z-1)^{b-1}
= {b\choose 1}^{-1} = \frac{1}{b}.$$
The second is
$$r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\sum_{q=1}^r {r-1\choose q-1} (z-1)^{-q}
\\ = r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-2}
\sum_{q=0}^{r-1} {r-1\choose q} (z-1)^{-q}
\\ = r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-2}
\left[1+\frac{1}{z-1}\right]^{r-1}
\\ = r [z^{b+1}] \log\frac{1}{1-z} (z-1)^{b-1}
= r \frac{1}{2} {b+1\choose 2}^{-1}
= \frac{r}{b(b+1)}.$$
Collecting we find
$$\frac{1}{b} + \frac{r}{b(b+1)}
= \frac{b+r+1}{b(b+1)}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
A discrete difference computation I am attempting to compute the following discrete difference:
$$\Delta[(3x+2)^2]$$ where $\Delta[f(x)] = f(x+1)-f(x)$
Method 1:
\begin{align*}
\Delta[(3x+2)^2] &= [3(x+1)+2]^2 - (3x+2)^2 \\
&=(3x+5)^2 - (3x+2)^2 \\
&= (9x^2+30x+25)-(9x^2+12x+4) = 18x+21
\end{align*}
Method 2:
Define $f(x)^{(k)} = f(x)f(x-1)\cdots f(x-[k-1]).$ It is known1 that $\Delta[(ax+b)^{(k)}] = ka(ax+b)^{(k-1)}.$ Using the fact that $x^2 = x^{(2)} +x$ :
\begin{align*}
\Delta[(3x+2)^2] &= \Delta[(3x+2)^{(2)} + (3x+2)^{(1)}] \\
&= \Delta[(3x+2)^{(2)}] + \Delta[(3x+2)^{(1)}]\\
&= 6(3x+2) + 3 = 18x+15
\end{align*}
I can't see why I'm getting two different answers. A direct expansion indicates the first calculation is correct. Can anyone help me find the flaw in the second computation?
*
*See e.g. Theorem 5 of Richardson's An Introduction to the Calculus of Finite Differences.
| Using
$$ (a x + b)^2 = (a x + b)(a (x-1) + b) + a \, (a x + b) = (a x + b)^{(2)} + a \, (a x + b)^{(1)}$$
then
\begin{align}
\Delta (a x + b)^{2} &= \Delta (a x + b)^{(2)} + a \, \Delta (a x + b)^{(1)} \\
&= 2 \, a \, (a x + b)^{(1)} + a \, a \, (a x + b)^{(0)} \\
&= 2 \, a \, (a x + b) + a^2 \\
&= a \, (2 a \, x + 2 b + a).
\end{align}
Setting $a = 3$ and $b = 2$ gives $\Delta (3 x + 2)^2 = 3 \, (6 x + 7)$.
Method 3:
\begin{align}
\Delta (a x + b)^{2} &= \Delta (a^2 \, x^2 + 2 a b \, x + b^2) \\
&= a^2 \, \Delta (x^2) + 2 a b \, \Delta (x) + b^2 \, \Delta (1) \\
&= a^2 \, (2 x + 1) + 2 a b \, (1) \\
&= a \, (2 a \, x + 2 b + a).
\end{align}
This is the same as the other results.
| {
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Convert $z=-1$ to polar form? I have an issue when wanting to convert $z=-1$ to the polar form.
Indeed, assuming that $z$ is of the form $z=a+ib$, I start by finding $z=(-1)+(0)i$, thus $a=-1$ and $b=0$.
I know that the polar form of $z$ is $z=re^{i\theta}$, where $r=\sqrt{a^2+b^2}$, thus $r=\sqrt{(-1)^2}=1$.
Then, I need to find $\theta$. I know that $\theta=\operatorname{arg}(z)=\arctan(\frac ba)=\arctan(\frac {0}{-1})=0$.
Therefore, my final answer would be $z=1\times e^{i0}=1$, which is incorrect, as WolframAlpha tells me the correct answer is $z=e^{i\pi}=-1$. I understand that I messed up when finding $\theta=0$ instead of $\theta=\pi$.
Any help is appreciated.
Thank you
| Your attempt is good, but your formula for the argument $\theta=\operatorname{arg}(z)=\arctan\left(\frac{b}{a}\right)=\arctan\left(\frac {0}{-1}\right)=0$ only works in special cases (when
$a > 0$). Actually the formulas are:
$$
\begin{align*}
z &= a + b \cdot \mathrm{i}\\
z &= r \cdot \operatorname{cis}(\theta) = r \cdot e^{\theta \cdot \mathrm{i}}\\
\\
r &= |z| = \sqrt{a^{2} + b^{2}}\\
\theta &= \arg(z) = \operatorname{arctan2}\left( b, ~a \right)\\
\\&\text{with}\\\\
\operatorname{arctan2}\left( b, ~a \right) &=
\begin{cases}
\arctan\left({\frac {b}{a}}\right) \qquad\quad~~~ \text{for } a > 0\\
\arctan\left({\frac {b}{a}}\right) + \pi \qquad \text{for } a < 0 \quad b > 0\\
\pi \qquad\qquad\qquad\quad~~~ \text{for } a < 0 \quad b = 0\\
\arctan\left({\frac {b}{a}}\right) - \pi \qquad \text{for } a < 0 \quad b < 0\\
\frac{\pi}{2} \qquad\qquad\qquad\quad~~ \text{for } a = 0 \quad b > 0\\
-\frac{\pi}{2} \qquad\qquad\qquad~~~ \text{for } a = 0 \quad b < 0\\
\end{cases} + 2 \cdot k \cdot \pi
\end{align*}
$$
If you use that you'll get:
$$
\begin{align*}
z &= -1\\
z &= -1 + 0 \cdot \mathrm{i}\\
z &= \sqrt{\left( -1 \right)^{2} + 0^{2}} \cdot e^{\arctan2\left( 0, ~-1 \right) \cdot \mathrm{i}}\\
z &= 1 \cdot e^{(\pi + 2 \cdot k \cdot \pi) \cdot \mathrm{i}}\\
z &= e^{\left( \pi + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}}\\
z &= e^{\pi \cdot \mathrm{i}} \tag{if k = 0}\\
\end{align*}
$$
The formula $e^{\pi \cdot \mathrm{i}} = -1$ is also known as Euler's formula, named after Leonhard Euler.
However, if you don't want to calculate the angle $\theta$, you can also consider the complex number $z = a + b \cdot \mathrm{i} = x + y \cdot \mathrm{i}$ as a vector (versor) with $z ~\widehat{=} \left[\begin{matrix} a\\ b \end{matrix}\right] = \left[\begin{matrix} x\\ y \end{matrix}\right]$ and draw this in a coordinate system. Measure the angle from the positive a-axis / x-axis to the vector (versor) with a protractor. If the angle you are measuring is in degrees, to get the result in radians you can use $\theta = \text{angle}^{\circ} = \frac{\text{angle}}{180} \cdot \pi$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Area enclosed between the roots of a quadratic Let $f(x)=ax^2+bx+c$
If $f(x)$ has roots $\alpha$ and $\beta$, what is the area enclosed by $f(x)$ and the $x$-axis between $x=\alpha$ and $x=\beta$ in terms of $a,b$ and $c$?
It is also given that $\alpha>\beta.$
If $a=1$, then I thought this might be easier since you get:
$$A=\int_\beta^\alpha{(x^2-(\alpha+\beta)x+({\alpha}{\beta})x) dx}$$
But even after evaluating this, I still wasn't even able to find an answer in terms of the coefficients.
I've been at this problem for a while now, and I would love some help. Any ideas?
| Have you ever seen Archemedes' proof of the quadrature of a parabola?
https://en.wikipedia.org/wiki/Quadrature_of_the_Parabola
Archimedes proved that any secant through a parabola intersecting at points $\alpha, \beta$ form a region with area $\frac{a(\alpha - \beta)^3}{6}$
But using calculus...
$y = ax^2 + bx + c$
let's put this parabola into vertex form.
$y = a(x + \frac {b}{2a})^2 + c - \frac {b^2}{4a}$
Save this for later...
$\int_\beta^\alpha ax^2 + bx + c\ dx$
$\beta = \frac {-b - \sqrt {b^2 - 4ac}}{2a}\\
\alpha = \frac {-b + \sqrt {b^2 - 4ac}}{2a}$
let's perform a u-substitution. $u = x + \frac {b}{2a}, du = dx$
$\displaystyle a\int_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}} u^2 - \frac {b^2 - 4ac}{4a^2}\ du$
$a(\frac 13 u^3 - \frac {b^2 - 4ac}{4a^2}u)|_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}}$
$a(\frac 23 (\frac {b^2 - 4ac}{4a^2})^\frac 32 - 2(\frac {b^2 - 4ac}{4a^2})^\frac 32) = -\frac {(b^2 - 4ac)^\frac 32}{6a^2}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$ So I need to evaluate $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$. Here's what I have thus far:
\begin{align*}
\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz&= \frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{5}{3}- \sin z}dz\\
&=\frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{10i-3e^{i z}+3e^{-i z}}{6i}} \\
&=2 i \int_0^{2 \pi} \frac{1}{10i - 3e^{iz}+3e^{-iz}}dz\\
&= 2 i \int_0^{2 \pi} \frac{1}{10i - 3e^{iz}+3e^{-iz}} \frac{e^{iz}}{e^{iz}}dz\\
&=2 \int_0^{2 \pi} \frac{1}{10iu-3u^2+3} du&& \text{where $u=e^{iz}$, $du=ie^{iz}dz$}\\
&= -2 \int_0^{2 \pi} \frac{1}{3(u+3i)(u+\frac{i}{3})} du
\end{align*}
but I get stuck here, Am I on the right path? Also I don't think that
$$-3(u+3i)(u+\frac{i}{3})=10iu-3u^2+3$$
Im off by a negative for the $10iu$. If possible, when helping use the formula for $\sin z$:
$$\frac{e^{iz}-e^{-iz}}{2i}.$$
Thanks in advance. (PS- why would someone downvote this? Its clearly showing I have work shown and have made an attempt at a solution!)
| Let $t=\tan(\frac{z}{2})$, $\sin z=\frac{2t}{t^2+1}$ and $dz=\frac{2dt}{t^2+1}$ and thinking the integral from $-\pi$ to $\pi$, we get
$$\int_{-\infty}^{\infty}\frac{2}{5}\frac{1}{(t-\frac{3}{5})^2+(\frac{4}{5})^2}dt=\left.\frac{1}{2}\arctan\left(\frac{5t-3}{4}\right)\right\rvert_{-\infty}^{\infty}=\frac{\pi}{2}$$
| {
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} |
If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$
If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$
We can express $1+x^2$ as
$$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$
since $x + y + z = xyz$ implies $1+ x^2 + xy + xz = 1 + x^2yz$. Thus, our desired inequality boils down to
$$\sqrt{\frac{(x+z)(x+y)}{yz}} + \sqrt{\frac{(y+z)(y+x)}{xz}} + \sqrt{\frac{(z+x)(z+y)}{xy}} \ge 6$$
which looks more complicated. Perhaps I am not headed in the right direction? I'd appreciate any hints or solutions. Thank you!
| Your idea gives a simple solution.
Indeed, by AM-GM twice we obtain: $$\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(x+y)}{xyz}}\geq3\sqrt[3]{\frac{8xyz}{xyz}}=6.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponential equations: Why is this method wrong? While solving exponential equations with my Y10 students, they found that you can break down a number into powers with the same base and add one $a^0$ and just use exponents, but never more than one power of one. They tried this out of laziness to do variable change and based on the fact that $a^{f(a)}=a^{g(a)} \rightarrow f(a) = g(a)$. Below is the part of the solution that contains the incorrect step, but why does it work?
$$
3^{x+1}-3^{-x+2}=26
$$
$$
26 = 27 - 1 = 3^3 - 3^0
$$
$$
3^{x+1}-3^{-x+2}=3^3-3^0
$$
$$
!!\rightarrow(x+1)-(-x+2)=3-0 \leftarrow!!
$$
$$
2x-1=3
$$
$$
x = \frac{4}{2}=2
$$
Why does this work?
| It works because the equation is actually involving a variation of the hyperbolic sine function $\sinh(x)=\frac{e^x-e^{-x}}{2}$ which is known to be invertible.
If $a$ is a positive real number different from $1$, let $\sinh_a(x)=\frac{a^x-a^{-x}}{2}$ (I will call it the "hyperbolic sine of base $a$"). Your equation is equivalent to
$$
\begin{array}{lcl}
& & \frac{3^{x+1}-3^{-x+2}}{2\times 3^{3/2}} = \frac{3^3-3^0}{2\times 3^{3/2}} \\
& \iff & \frac{3^{x-1/2}-3^{-(x-1/2)}}{2} = \frac{3^{3/2}-3^{-3/2}}{2} \\
& \iff & \sinh_3\left(x-\frac{1}{2}\right) = \sinh_3\left(\frac{3}{2}\right) \\
\end{array}
$$
$\sinh_a$ is easily shown to be one-to-one (for example because $sh_a(a)=\sinh(x\ln (a))$ and $\sinh$ is one-to-one), so $x-\frac{1}{2}=\frac{3}{2}$ which means that $x=2$.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Step
Let n = 1
⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$
I then proved that the remainder is 0 using polynomial long division.
$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0
∴ s(1) is true
Assumption Step
Assume that s(m) is true
⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial
Inductive Step
To prove that s(m+1) is true
⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial
⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)
⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)
However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!
| Let $$x^{6m+2}-x^{6m+1}+1=\lambda(x^2-x+1)$$
$$x^{6m+8}-x^{6m+7}+1=S$$
$$S=x^6(x^{6m+2}-x^{6m+1})+1$$
$$S=x^6(\lambda(x^2-x+1)-1)+1$$
$$S=x^6\lambda(x^2-x+1)-x^6+1$$
Now dividing $S$ by $x^2-x+1$ yields
$$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{1-x^6}{x^2-x+1}$$
$$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{(1+x-x^3-x^4)(x^2-x+1)}{x^2-x+1}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Prove that $2\cdot 3^x +1= p^y$ has no solution Prove that the Diophantine equation of $3$ variables $(x,y,p)$
$$2\cdot 3^x +1= p^y$$
has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.
I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradiction. But I cannot prove there is no solution for the case $y = 2k+1$.
I try to apply the technique in this answer (equation $7^x = 3 \cdot 2^y +1$) but it doesn't work as $p$ is not known.
Any help would be greatly appreciated!
| We have, as the O.P. noted, the exponent of the prime $p$ is odd so
$$2\cdot3^x=(p-1)(p^{2y}+p^{2y-1}+\cdots+p+1)$$ the second factor having $y+1$ terms is odd so $2$ divides exactly $p-1=2m$, with $m$ odd. It follows
$$3^x=m(p^{2y}+p^{2y-1}+\cdots+p+1)\Rightarrow m=3^a \text { and } \boxed{p=2\cdot 3^a+1} $$ then
$$3^{x-a}=(2\cdot 3^a+1)^{2y}+(2\cdot 3^a+1)^{2y-1}+\cdots+(2\cdot 3^a+1)+1$$ which implies $$3^{x-a}\equiv {1+1+\cdots+1}=2y+1\pmod{3^a}\iff3^{x-a}=2y+1+3^aX$$ then
$2y+1=3^ak$.
Back to the proposed equation we have $$2\cdot3^x+1=(2\cdot3^a+1)^{3^ak}$$ and similarly as above $$2\cdot3^x=2\cdot3^a[(2\cdot3^a)^{3^ak-1}+\cdots+(2\cdot3^a)+1]\Rightarrow 3^{x-a}\equiv 1\pmod{3^a}\iff3^{x-a}=1+3^aY$$ Thus
$$3^a(3^{x-2a}-Y)=1\Rightarrow 3^a=1\Rightarrow a=0\Rightarrow p=3$$
We are done.
| {
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"question_score": "6",
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$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$ with one-variable calculus solution Prove that $$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$$
My textbook says that we need to note that the integral can be transformed into $\int_L (\sqrt{x^2+y^2}\mathrm{d}x+y\ln(\sqrt{x^2+y^2}+x)\mathrm{d}y)$, where $L: y=\sin x$, $0\leq x\leq\pi$. We denote $\sqrt{x^2+y^2}$ by $P(x,y)$, and $y\ln(\sqrt{x^2+y^2}+x)$ by $Q(x,y)$. And then it is not hard to compute $\int_l P\mathrm{d}x+Q\mathrm{d}y$, where $l:(0\leq x\leq\pi\wedge y=0)$, as well as $\int_{l+L} P\mathrm{d}x+Q\mathrm{d}y$ (by Green's formula), (all the directions here are omitted, but it should make no ambiguity). Hence the question is done. I think this method is hard to observe. Do we have a way merely apply theorems of one-variable calculus? I have observed that $\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x})=\cos x\sin x(2\ln\sin x-\ln(x+\sqrt{x^2+\sin^2x}))$, and $\cos x\sin x\ln\sin x=(\sin x\ln\sin x)(\sin x)'$, hence its indefinite integral is computable. I do not know if this observation would help.
| Using the integration by parts, you have
\begin{eqnarray}
&&\int(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x\\
&=&\int\sqrt{x^2+\sin^2x}dx+\frac12\int_{0}^{\pi}\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}\sin^2x\\
&=&\int\sqrt{x^2+\sin^2x}dx+\frac12\bigg(\sin^2x\ln(x+\sqrt{x^2+\sin^2x})-\int\frac{\sin^2x(1+\frac{x+\sin x\cos x}{\sqrt{x^2+\sin^2x}})}{x+\sqrt{x^2+\sin^2x}}\mathrm{d}x\bigg).
\end{eqnarray}
Now combining these two integrals and simplifying will give the primitive.
| {
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How to evaluate this integral containing a Bessel function: $\int_1^2 x^{-2} J_2(x)\,dx$? $$\int_1^2 x^{-2} J_2(x)\,dx$$
I can't solve this integration problem. I want to develop the expression using Derivation of the Bessel function and Recurrence relations between the Bessel functions and get the answer.
| You can evaluate in terms of hypergeometric functions. Recall that
$$J_{\nu}(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{-z^2}{4}\right)$$
Therefore
$$x^{-2}J_2(x)=\frac{1}{x^2}\frac{(x/2)^2}{\Gamma(2+1)}~{}_0F_1\left(;2+1~;~\frac{-x^2}{4}\right) \\ =\frac{1}{8}~{}_0F_1\left(;3~;~\frac{-x^2}{4}\right)$$
Now, with the substitution $$-x^2/4=z\\ \mathrm dx=-\mathrm iz^{-1/2}~\mathrm dz$$ we get
$$\int x^{-2}J_2(x)\mathrm dx=\frac{-1}{8}\int {}_0F_1\left(;3~;~\frac{-x^2}{4}\right)\mathrm dx \\ =\frac{-\mathrm i}{8}\int z^{1/2-1}~{}_0F_1\left(;3~;~z\right)~\mathrm dz$$
Which from here we know has the primitive
$$=\frac{-\mathrm i}{8}\frac{z^{1/2}}{1/2}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~z\right)$$
Putting back our substitution $z=-x^2/4$ this is
$$=\frac{-\mathrm i}{8}(1/2)^{-1}\sqrt{\frac{-x^2}{4}}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right) \\ =\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)$$
I.E, that is
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)\right)=x^{-2}J_2(x)$$
Hence
$$\int_1^2 x^{-2}J_2(x)\mathrm dx=\frac{1}{4}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~-1\right)-\frac{1}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-1}{4}\right)\\ \approx 0.102617$$
Confirmed by Mathematica:
Mathematica says
$${}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg|~z\right)\\=\frac{2~I_0(2\sqrt z)\cdot\big(-1+4z+2\pi z~L_1(2\sqrt z)\big)}{3z}-\frac{2~I_1(2\sqrt z)\cdot\big(-1+2z+2\pi z^{3/2}L_0(2\sqrt z)\big)}{3z^{3/2}}$$
Where
$$I_\nu(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{z^2}{4}\right)$$
Is a Modified Bessel function
and
$$L_\nu(z)=\frac{z~(z/2)^\nu}{\sqrt{\pi}~\Gamma(\nu+3/2)}~{}_1F_2\left(\begin{matrix}1 \\ \frac{3}{2},\nu+\frac{3}{2}\end{matrix}~\bigg|~\frac{z^2}{4}\right)$$
Is a modified Struve function.
But IMO this doesn't qualify as "simpler".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\int\frac{x\arctan x}{x^4+1}dx$ in a practical way I need to evaluate the following indefinite integral for some other definite integral
$$\int\frac{x\arctan x}{x^4+1}dx$$
I found that
$$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{4}-2I(1)$$
where
$$I(t)=\int_0^1\frac{x\operatorname{Ti}_2(tx)}{x^4+1}dx$$
taking derivatives
$$tI'(t)=\int_0^1\frac{x\arctan(tx)}{x^4+1}dx$$
$$(tI')'=\int_0^1\frac{x^2}{(t^2x^2+1)(x^4+1)}dx$$
The answer for the integral is the integrand in my indefinite integral plus some stuff I can integrate on my own, which is where my problem comes from. In reality, I need
$$\int\frac{1}{t}\left(\int\frac{t\arctan{t}}{t^4+1}dt\right)dt$$
But it's fine if I just get an answer for the inside integral.
Back to the indefinite integral. I wanted to use partial fractions
$$\int\frac{x\arctan x}{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}dx$$
We would get four integrals of the form $\int\frac{\arctan{x}}{x+c}dx$. Using the logarithmic definition of arctangent, we get 8 integrals of the form $\int\frac{\ln{(a+bx)}}{x+c}$. That's 8 dilogarithms. I need this for another calculation, and having 8 dilogarithms is too numerous to be practical to work with. I want another way to evaluate this integral.
WAs answer
TO CLARIFY
I am fine with dilogarithms
| I am not sure that you could avoid dilogarithms except if you accept an infinite summation of Gaussian hypergeometric functions.
$$I=\int \frac{x }{x^4+1}\tan ^{-1}(x)\,dx$$ Using integration by parts
$$I=\frac{1}{2} \tan ^{-1}(x) \tan ^{-1}\left(x^2\right)-\frac 12\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ looks a bit more pleasant.
Using
$$\tan ^{-1}\left(x^2\right)=\sum_{n=0}^\infty\frac{(-1)^n}{2 n+1}x^{4n+2}$$
$$J=\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ $$J=\sum_{n=0}^\infty (-1)^n\,\frac{x^{4 n+3}}{(2n+1)(4 n+3)}\,\,\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)$$
Notice that
$$x^{4 n+3}\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)=-(4n+3)\tan ^{-1}(x)+P_{4n+1}(x)$$ where the first polynomials are
$$\left(
\begin{array}{cc}
0 & 3 x \\
1 & \frac{7 x^5}{5}-\frac{7 x^3}{3}+7 x \\
2 & \frac{11 x^9}{9}-\frac{11 x^7}{7}+\frac{11 x^5}{5}-\frac{11
x^3}{3}+11 x \\
3 & \frac{15 x^{13}}{13}-\frac{15 x^{11}}{11}+\frac{5
x^9}{3}-\frac{15 x^7}{7}+3 x^5-5 x^3+15 x \\
4 & \frac{19 x^{17}}{17}-\frac{19 x^{15}}{15}+\frac{19
x^{13}}{13}-\frac{19 x^{11}}{11}+\frac{19 x^9}{9}-\frac{19
x^7}{7}+\frac{19 x^5}{5}-\frac{19 x^3}{3}+19 x \\
\end{array}
\right)$$
| {
"language": "en",
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} |
How to find the constant $C$ such that $f(x)\geq Cx$ Problem :
Define for strictly positive $x$ :
$$f\left(x\right)=\left(\prod_{k=1}^{\operatorname{floor}\left(x\right)}\left(1+\sum_{n=1}^{k}\frac{1}{k\cdot2^{n}}\right)\right)$$
Does there exists a constant $C$ such that :
$$f(x)\geq Cx$$
I think definitely yes and $C\simeq 0.6516...$
Currently Symbolic calculator don't find it .
As side notes we have :
$$-\ln\left(2\right)+\sum_{k=1}^{\infty}\frac{1}{k\cdot2^{k}}=0$$
$$\frac{1}{2}\left(1-3\ln\left(\frac{4}{3}\right)\right)+\left(+\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)\cdot2^{1+2k}}\right)=0$$
and so on .
Edit 26/12/2022 :
we have with a computer :
$$\left|\exp\left(1-\prod_{k=1}^{1000}\left(1-\frac{1}{2^{k}(k+1)}\right)\right)-\sqrt{2}-\frac{1}{2^{9}}-\frac{1}{2^{11}}-\frac{1}{2^{14}}-\frac{1}{2^{16}}+\frac{1}{2^{18}}-\frac{1}{2^{20}}-\frac{1}{2^{21}}-\frac{1}{2^{24}}+\frac{1}{2^{27}}+\frac{1}{2^{30}}-\frac{1}{2^{33}}+\frac{1}{2^{36}}+\frac{1}{2^{38}}+\frac{1}{2^{39}}-\frac{1}{2^{41}}\right|<4*10^{-13}$$
I don't know if we can go like this at infinity .
Question :
How to show the existence and does $C$ admits a closed form ?
Thanks a lot .
| The claim is that $f(x)\ge Cx$ for any $x>0$, where
$$
C= \prod_{k=1}^\infty\left( {1 - \frac{1}{{2^k (k + 1)}}} \right) =0.651649356319290\ldots,
$$
as given by @metamorphy in the comments. Indeed
\begin{align*}
\log f(x) & = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {1 + \sum\limits_{n = 1}^k {\frac{1}{{k \cdot 2^n }}} } \right)} = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right)}
\\ & = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {\frac{{k + 1}}{k}} \right)} + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {\frac{{k + 1}}{k}} \right)} \right]}
\\ & = \log (\left\lfloor x \right\rfloor + 1) + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]}
\\ & \ge \log x + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]}
\\ & \ge \log x + \sum\limits_{k = 1}^\infty {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]}
\\ & =\log x+ \sum\limits_{k = 1}^\infty {\log \left( {1 - \frac{1}{{2^k (k + 1)}}} \right)}
\\ & = \log x + \log C,
\end{align*}
since
$$
\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right) \le 0
$$
for any $k\ge 1$. It is readily seen that
$$
\lim_{x\to +\infty}\frac{f(x)}{x}=C,
$$
i.e., $C$ is the best possible such constant.
Note that $C(\left\lfloor x \right\rfloor+1)$ is a better approximation (and lower bound) to $f(x)$ than $Cx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Probability on Number Theory Problem: Suppose that $a,b,c \in \{1,2,3,\cdots,1000\}$ are randomly selected with replacement. Find the probability that $abc+ab+2a$ is divisible by $5$.
Answer given from the worksheet: $33/125$
My answer: $\frac{641}{3125}$
Attempt: Since $abc+ab+2a = a(bc+b+2)$, either $a \equiv 0 \pmod{5}$ or $bc+b+2 \equiv 0 \pmod{5}$. The first case, which is $a \equiv 0 \pmod{5}$, has a probability of $1/5$. Now on the other case,
$$bc+b+2 \equiv 0 \pmod{5} \Rightarrow b(c+1) \equiv 3 \pmod{5}$$
happens when $b\equiv 1$ and $c \equiv 2$, $b \equiv 2$ and $c \equiv 3$, $b \equiv 3$ and $c \equiv 0$, or $b \equiv 4$ and $c \equiv 1$, total of $4$ solutions. So, this case has a probability of $4 \cdot\frac{1}{5^4} = \frac{4}{625}$. By Inclusion-exclusion principle, the final probability should be $\frac{1}{5} + \frac{4}{625} - \frac{1}{5} \cdot \frac{4}{625} = \frac{641}{3125}.$
This is apparently not the same from the given answer in the worksheet. Where did I go wrong?
| You've made a good attempt. The only mistake I can determine is that, modulo $5$, there are only $5$ possibilities for each of $b$ and $c$, so there are $5^2 = 25$ total potential combinations, not $5^4$. Thus, your $4 \cdot\frac{1}{5^4} = \frac{4}{625}$ should be $4 \cdot\frac{1}{5^2} = \frac{4}{25}$ instead. The final probability using your methodology (i.e., the inclusion-exclusion principle) would then be
$$\frac{1}{5} + \frac{4}{25} - \frac{1}{5} \cdot \frac{4}{25} = \frac{41}{125} \tag{1}\label{eq1A}$$
However, this is also different from what the worksheet states. It seems the problem statement has an issue (e.g., part of it is incorrect, it's missing something, etc.), the worksheet answer is incorrect, or we're both missing something. I assume no explanation is provided regarding how the worksheet's answer was determined.
Also, note that regarding how to calculate this, I would instead consider the $2$ events of $a \equiv 0 \pmod{5}$ and $a \not\equiv 0 \pmod{5}$. The first one has a $\frac{1}{5}$ chance, with $b$ and $c$ being anything so their probability is $1$, giving an overall probability of $\frac{1}{5}$, as you determined. Otherwise, there's a $\frac{4}{5}$ chance that $a \not\equiv 0 \pmod{5}$, with a $\frac{4}{25}$ probability that $b$ and $c$ will be appropriate values. This then gives the probability to be
$$\frac{1}{5} + \frac{4}{5} \cdot \frac{4}{25} = \frac{41}{125} \tag{2}\label{eq2A}$$
i.e., the same as in \eqref{eq1A}. That's because $\frac{4}{25} - \frac{1}{5} \cdot \frac{4}{25} = \frac{5}{5} \cdot \frac{4}{25} - \frac{1}{5} \cdot \frac{4}{25} = \frac{5-1}{5}\cdot \frac{4}{25} = \frac{4}{5} \cdot \frac{4}{25}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
| Graph the integrand and observe that there is "as much area above as below the x-axis". This leads to the conjecture that the value is zero, and also gives a strategy for proving it using a symmetry argument.
First observe that the function is periodic with period $\pi$. You can confirm this by showing that $f(x + \pi) = f(x)$ for all $x$. Since your integral covers two periods we can say that your integral is twice the integral over $[-\frac{\pi}{4}, \frac{3\pi}{4}]$.
Then observe that the graph of $f$ is symmetric with respect to the point $(\frac{\pi}{4},0)$. You can confirm this by showing that $f(\frac{\pi}{4} - x) = - f(\frac{\pi}{4} + x)$.
Thus the integral over $[-\frac{\pi}{4}, \frac{3\pi}{4}]$ is equal to zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$ Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$
substituting $y=ux$ so that $y' = u + x\frac{du}{dx}$"
$u + x\frac{du}{dx} = \frac{2x^2u}{ux^2-x^2} = \frac{2u}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{2u}{u^2-1} - u = \frac{2u}{u^2-1} - \frac{u(u^2-1)}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{-u^3+u}{u^2-1}$
Separating variables...
$\rightarrow \frac{u^2-1}{-u^3+u}du = \frac{1}{x}dx$
Now, from here we can factor the top and bottom of the LHS as so...
$\frac{u^2-1}{-u^3+u} = \frac{(u+1)(u-1)}{-u(u+1)(u-1)} = \frac{-1}{u}$
And I'm thinking this is where I'm going wrong... Is there a reason that this simplification is not valid? I'm thinking that maybe I shouldn't simplify and that I should use partial fraction's instead?
The answer is supposed to be $3yx^2 - y^3 = k$
Edit: I continued as so:
$\frac{u^2-1}{3u-u^3} = \frac{A}{u} + \frac{B}{u-\sqrt{3}} + \frac{C}{u + \sqrt{3}}$.
I got $A = \frac{1}{3}$ by multiplying both sides by $u$ and then taking the limit as $u \rightarrow 0$. A similar method led to $B = \frac{1}{3}$ and $C = \frac{1}{3}$. Thus I arrived at:
$\frac{u^2-1}{3u-u^3} = \frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})}$.
And so:
$(\frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})})du = \frac{1}{x}dx$
Integrating and then exponentiating both sides:
$(u(u+\sqrt{3})(u-\sqrt{3}))^{\frac{1}{3}} = xk$
$(u(u^2-3))^{\frac{1}{3}} = xk$
$u = \frac{y}{x}$ and so:
$(\frac{y}{x}((\frac{y}{x})^2-3))^{\frac{1}{3}} = xk$
This leads to:
$(\frac{y}{x})^3 - \frac{3y}{x} = x^3k$
... Which is still not correct!!!
I haven't done ODE's quite a long time... I've been flying through Schaum's outline on the topic and I've done plenty of problems like this one... But i can't get this one correct!! help appreciated.
| HINT
According to the substitution $y = ux$, it results that
\begin{align*}
y' = \frac{2xy}{y^{2} - x^{2}} & \Longleftrightarrow y' =\frac{2(y/x)}{(y/x)^{2} - 1}\\\\
& \Longleftrightarrow xu' + u = \frac{2u}{u^{2} - 1}\\\\
& \Longleftrightarrow xu' = \frac{2u}{u^{2} - 1} - u\\\\
& \Longleftrightarrow xu' = \frac{3u - u^{3}}{u^{2} - 1}
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to merge odd series and even series of hypergeometric function of Legendre polynomials into one hypergeometric function? On the Wolfram MathWorld page of Legendre Differential Equation, Legendre polynomials are represented as
$$
P_l(x) = c_n
\begin{cases}\begin{align*}
&_2F_1\left(-\frac{1}{2}(l), \frac{1}{2}(l + 1); \frac{1}{2}; x^2\right) && \text{for even } l\\
&x_2F_1\left(\frac{1}{2}(l + 2), \frac{1}{2}(1 - l); \frac{3}{2}; x^2\right) && \text{for odd } l \\
\end{align*}\end{cases}\tag{1}
$$
I also know Legendre polynomials can be written in this single hypergeometric function.
$$
{}_2F_1\left(l + 1, -l; 1; \frac{1 - x}{2}\right) \tag{2}
$$
How to merge (1) to get (2)?
| A quadratic transform for the Gaussian hypergeometric function exists which links (1) and (2):
\begin{align}
{ }_2 F_1\left(a, b ; \frac{a+b+1}{2} ; z\right)&= \\
&\frac{\sqrt{\pi} \Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{b+1}{2}\right)}{ }_2 F_1\left(\frac{a}{2}, \frac{b}{2} ; \frac{1}{2} ;(2 z-1)^2\right)+\\
&+\frac{2 \sqrt{\pi}(2 z-1) \Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a}{2}\right) \Gamma\left(\frac{b}{2}\right)}{ }_2 F_1\left(\frac{a+1}{2}, \frac{b+1}{2} ; \frac{3}{2} ;(2 z-1)^2\right)
\end{align}
Here, by choosing $a=-l,b=l+1,z=(1-x)/2$
\begin{align}
{}_2F_1\left( -l,l+1; 1; \frac{1 - x}{2}\right)&=\frac{\sqrt{\pi}}{\Gamma\left((1-l)/2\right) \Gamma\left(1+l/2\right)}{ }_2 F_1\left(-l/2, (1+l)/2 ;1/2 ;x^2\right)+\\
&+2x\frac{ \sqrt{\pi}}{\Gamma\left(-l/2\right) \Gamma\left((1+l)/2\right)}{ }_2 F_1\left((1-l)/2, 1+l/2 ; 3/2 ;x^2\right)
\end{align}
Now, due to the $\Gamma\left((1-l)/2\right)$ and $\Gamma\left(-l/2\right)$ in the denominators of the coefficients, the second (the first) term vanishes when $l$ is even (odd).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Grouping the real and imaginary terms in the proof of Euler's formula using the MacLaurin series The MacLaurin series:
\begin{align}
\sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots
\\\\
\cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots
\\\\
e^z&=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots
\end{align}
Substitute $z=ix$ in the last series:
\begin{align}
e^{ix}&=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots
\\\\
&=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots
\\\\
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)
\\\\
&=\cos x+i\sin x
\end{align}
My question is how do we group $1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots$ and $+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$ or how do we apply the fact that if a series is absolutely convergent, then every rearrangement of
the series converges to the same sum? It seems like $1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$ corresponds to $\mathbb{N}\times\mathbb{N}$.
| If $\sum_{n=1}^{\infty}a_n = L$, then so is $\sum_{n=1}^{\infty}b_n=L$ for $b_n = a_{n/2}$ if $n$ is even, and $0$ if $n$ is odd. This is because $$\sum_{n=1}^{2N}b_n = \sum_{n=1}^{2N+1}b_n = \sum_{n=1}^{N}a_n\to L$$ as $N\to\infty$. Hence from the convergence of series for cosine and sines, we have \begin{align*}&\left(1-\frac{x^2}{2}+\frac{x^4}{4!}-\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right) \\ &= \left(1+0-\frac{x^2}{2}+0+\frac{x^4}{4!}+\cdots\right)+\left(0+ix+0-\frac{ix^3}{3!}+0+\frac{ix^5}{5!}+\cdots\right)\\ &= 1+ix-\frac{x^2}{2}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}+\cdots\\ &= e^{ix}\end{align*} by just summing two series like $\sum(a_n+b_n) = \sum a_n+\sum b_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
| Alternative approach:
Let $~r~$ denote $\displaystyle \log_{4n} 40\sqrt{3}.$
Then
$$(4n)^r = 40\sqrt{3}, ~~(3n)^r = 45 \implies $$
$$\left[\frac{4}{3}\right]^r = \left[\frac{4n}{3n}\right]^r = \frac{40\sqrt{3}}{45} = \frac{8\sqrt{3}}{9} \implies $$
$$9 \times 4^r = 8\sqrt{3} \times 3^r \implies $$
$$\frac{2^{2r}}{8} = \frac{\sqrt{3} \times 3^r}{9} \implies $$
$$2^{2r - 3} = 3^{r - (3/2)} = \sqrt{3}^{2r-3}. \tag1 $$
The only way that this is possible is if $(2r - 3) = 0 \implies r = (3/2).$
Checking this candidate value for $r$ gives
$$4^{3/2} \times n^{3/2} = 40\sqrt{3} \implies n^{3/2} = 5\sqrt{3}.$$
Checking this further,
$$3^{3/2} \times 5\sqrt{3}$$
does in fact equal $45$.
So, the sole candidate value for $r$, works.
Therefore,
$$n^{3/2} = 5\sqrt{3} \implies n^3 = \left[5\sqrt{3}\right]^2 = 75.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Properties of eigenvectors $$A :=\begin{pmatrix}
5 &−6 &0\\
2 &−2& 0 \\
0 &0 &2 \\
\end{pmatrix} $$
Determine an eigenvector of the matrix that is perpendicular to the vector $\mathbf v = (1, −4, 1).$
I think I need to first determine the eigenvalues and eigenvectors of $A$. Then I can use that a linear combination of eigenvectors of eigenvalue $λ$ is again an eigenvector of eigenvalue $λ$. How can I solve it?
| From the definition of the eigenvector $\mathbf v$ corresponding to the eigenvalue $λ$ we have
$$A\mathbf v=λ\mathbf v$$
Then:
$$A\mathbf v-λ\mathbf v=(A-λI)\mathbf v=0$$
Equation has a nonzero solution if and only if
$$\det(A-λI)=0$$
$$\det(A-λI)=\begin{vmatrix} 5-λ& -6& 0 \\ 2& -2-λ& 0\\ 0& 0& 2-λ\end{vmatrix} $$$$=-λ^3+5λ^2-8λ+4=-(λ-1)(λ^2-4λ+4)=-(λ-1)(λ-2)^2=0$$
$\lambda_1=1$ and $\lambda_2=2$ with multiplicity $2$ are the eigenvalues.
For each eigenvalue, look for the associated eigenvector. $( A' − I ) \mathbf X = \mathbf{0}$. If $\lambda_1=1$ then $$\begin{pmatrix}
4& -6& 0 \\
2& -1& 0\\
0 & 0 & 1
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0 \\
0\\
0
\end{pmatrix}$$
Solving the system you will find a vector of components, for example, $\mathbf v_1=(x_1,y_1,z_1)\ne \mathbf 0$. If $\mathbf v_1\cdot \mathbf v =0$ then $\mathbf v_1 \perp \mathbf v$. Same steps for $\lambda_2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show positivity of a function of two variables in the unit square. Let $$
f(x,y) = x^3 (1 + y + y^2) + y^2 \Big[x^2 (5 + 2 y) + x (-6 - 4 y + y^2) + (1 + 3 y + y^2)\Big]
$$
Show that $f(x,y) \ge 0 $ for $0\le x \le 1$ and $0\le y \le 1$.
Numerical evaluations seem to support the claim. This is a cubic function in $x$ with one negative coefficient in the linear term, all other coefficients are positive. Since $f(x=0) > 0$, the last two terms guarantee positivity for $0 < x < 1/6$. The last three terms show a quadratic function which has its minimum always for $x^* \in [0 \quad 1]$, namely at $x^* = \frac{6 + 4 y - y^2}{10 + 4 y }$ ; however, for small $y$, the value of the sum of the three last terms is negative at $x^*$. How to continue?
| We have
\begin{align*}
f &= (x+1)y^4 + (2x^2 - 4x + 3)y^3 + (x^3 + 5x^2 - 6x + 1)y^2 + x^3y + x^3\\
&\ge (x+1)y^4 + (2x^2 - 4x + 3)y^4 + (x^3 + 5x^2 - 6x + 1)y^2 + x^3y^2 + x^4 \tag{1}\\
&= (2x^2 - 3x + 4)y^4 + (2x^3 + 5x^2 - 6x + 1)y^2 + x^4\\
&\ge (2-x)^2 y^4 + (2x^3 + 5x^2 - 6x + 1)y^2 + x^4\tag{2}\\
&\ge 2\sqrt{(2-x)^2y^4\cdot x^4} + (2x^3 + 5x^2 - 6x + 1)y^2\tag{3}\\
&= y^2(3x-1)^2\\
&\ge 0.
\end{align*}
Explanations:
(1): $(2x^2 - 4x + 3)y^3 \ge (2x^2 - 4x + 3)y^4$ and $x^3y \ge x^3y^2$ and $x^3 \ge x^4$.
(2): $2x^2-3x+4 \ge (2-x)^2$.
(3): AM-GM.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A $n \times n$ linear system Solve the linear system
$
\begin{array}{ c c c c c c c c c c c c c c }
& & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 2 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1)\\
x_{1} & & & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 4 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 2)\\
x_{1} & + & x_{2} & & & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 6 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 3)\\
\vdots & & \vdots & & \vdots & & \ddots & & \vdots & & \vdots & & \vdots & \ \ \ \ \ \ \ \ \ \ \ \\
x_{1} & + & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & & & = & 2n & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n)
\end{array}
$
Attempt
I am not confident yet with my linear algebra, so I tried approaching this problem by adding all of the equations, which gives us
$
\begin{align*}
(n-1)(x_1+x_2+x_3+\ldots+x_{n-1}+x_n)=2+4+6+\ldots+2n \tag*{}
\end{align*}
$
since $x_i$ ($i \in \mathbb{N}, i \leq n$) always appear in every equation except the $i$-th equation. The right-hand side is the sum of first $n$ positive even number, which can be simplified further as
$
\begin{align*}
(n-1)(x_1+x_2+x_3+\ldots+x_{n-1}+x_n) &= n(n+1)
\\[.5em]
x_1+x_2+x_3+\ldots+x_{n-1}+x_n &= \dfrac{n(n+1)}{n-1} \tag*{}
\end{align*}
$
Subtracting respectively Eq. (1), Eq. (2), up to Eq. (n) from Eq. (A), gives us
$
\begin{align*}
x_1 &= \dfrac{n(n+1)}{n-1} - 2 = \dfrac{n(n+1) - 2(n-1)}{n-1} = \dfrac{n^2 - n + 2}{n-1} \tag*{}\\
x_2 &= \dfrac{n(n+1)}{n-1} - 4 = \dfrac{n(n+1) - 4(n-1)}{n-1} = \dfrac{n^2 - 3n + 4}{n-1} \\
\vdots \\
x_n &= \dfrac{n(n+1)}{n-1} - 2n = \dfrac{n(n+1) - 2n(n-1)}{n-1} = \dfrac{3n-n^2}{n-1}
\end{align*}
$
However, I'm rather concerned with the denominator ($n-1$) which causes division by zero when $n = 1$. Did I do a mistake, or maybe did I misunderstand something in the process? And what would be the correct approach to this problem? Thanks.
Note: I found the problem here.
| When $n=1$, you have just a single equation, but the left side is an empty sum ($0$) and the right side is $2$. Thus for $n=1$ your system devolves to a contradiction, and with that you should expect the division to break down.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
Assume that positive numbers a, b, c, x, y, z satisfy $cy+bz =a;
az + cx = b$$bx + ay = c$.
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
I've tried appling A.M.-G.M. inequality but it didn't work. Then applied jensen inequality assuming, $f(x)=\frac{x^2}{1+x}$
As the function is convex for $x>0$,
$$f\left(\frac{x+y+z}{3}\right)\leq \frac{f(x)+f(y)+f(z)}{3}$$
The thing we need to prove is $$f\left(\frac{x+y+z}{3}\right)=1/6$$ but it's not. How to prove it?
| All things are same as that of answer given by Calvin Lin upto $x= \frac{b^2+c^2-a^2}{2bc}$ and similarly for $y,z$.
Clearly we can see that $b^2+c^2>a^2$ ($x$ is positive) and same for $b,c$.
We get an acute angled triangle of sides $a,b,c$ with $x=\cos A, y=\cos B, z=\cos C$. Let the function which to prove is $f(\Delta)$
Now we have to find minimum value of $$f(\Delta)= \frac{\cos^2 A}{1+\cos A}+\frac{\cos^2 B}{1+\cos B}+\frac{\cos^2 C}{1+\cos C}$$
For $A+B+C=\pi$
Now let $u=\cot A, v=\cot B, w=\cot C$
It can be easily proved that,
$$\frac{\cos^2 A}{1+\cos A}=u^2- \frac{u^3}{\sqrt{(u+v)(u+w)}} $$
Hint: Use $uv+vw+wu=1$ for a triangle.
By A.M.$\geq$G.M.,
$$(u+v)+(u+w) \geq 2 \sqrt{(u+v)(u+w)}$$
Or,
$$\frac{1}{u+w}+ \frac{1}{u+v} \geq \frac{2}{\sqrt{(u+v)(u+w)}} $$
Rewriting the above inequality in a different way gives,
$$u^2- \frac{u^3}{\sqrt{(u+v)(u+w)}} \geq u^2-\frac{u^3}{2}\left(\frac{1}{u+w}+ \frac{1}{u+v}\right)$$
Or,
$$\frac{\cos^2 A}{1+\cos A}\geq u^2-\frac{u^3}{2}\left(\frac{1}{u+v}+ \frac{1}{u+w}\right)$$
Adding this inequality for $\cos A, \cos B, \cos C$ gives,
$$f(\Delta)\geq \frac{uv+vw+wu}{2}= \frac{1}{2} $$
One value of $(u,v,w)$ for which to hold equality is $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.( Although equality can hold at other places as well, as described by Calvin Lin)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
AM-GM & Minimization Proof I want to prove that for all $x, y > 0$, $$\cfrac{x+y}{2} \geq \sqrt{xy}$$
Particularly, I want to show that the minimum of $(x+y)/2$ is exactly $\sqrt{xy}$.
This is my attempt:
$\textbf{Proof}$ (Contradiction). Assume if $x, y > 0$, then $(x+y)/2 < \sqrt{xy}$. But
\begin{align*}
x+y &< 2\sqrt{xy} \\
x^2+2xy+y^2 &< 4xy \\
x^2-2xy+y^2 &< 0 \\
(x-y)^2 &< 0
\end{align*}
gives us a contradiction since for all $x, y > 0$, we know $(x-y)^2 > 0$. Therefore there exists no positive reals $x$ and $y$ such that $(x+y)/2 < \sqrt{xy}$, or $\min\bigg(\cfrac{x+y}{2}\bigg) = \sqrt{xy}$. $\qquad \square$
| Instead of talking about minimum, we may say that the inequality
$$\frac{x+y}{2} \geq \sqrt{xy}$$
holds for any $x, y>0$ and the equality holds for $x=y.$
We may prove the inequality directly by starting with $$
\begin{aligned}
& x+y-2 \sqrt{x y}=(\sqrt{x}-\sqrt{y})^2 \geqslant 0 \\
\Rightarrow \quad & x+y \geqslant 2 \sqrt{x y} \\
\Rightarrow \quad & \frac{x+y}{2} \geqslant \sqrt{x y}
\end{aligned}
$$
By the way, we can say that the minimum value $x+y-2\sqrt{xy} $ is $ 0$ for positive values of $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.
| Extraneous roots can be introduced by squaring. We can avoid squaring by multiplying by the conjugate.
The equation
$$\sqrt{x + 1} - \sqrt{1 - x} = 1$$
imposes the restrictions that $x + 1 \ge 0 \implies x \geq -1$ and $1 - x \ge 0 \implies x \leq 1$. Therefore, we know that any valid solution must satisfy $-1 \leq x \leq x$.
If we multiply both sides of the given equation by $\sqrt{x + 1} + \sqrt{1 - x}$, we obtain
$$\sqrt{x + 1} + \sqrt{1 - x} = 2x$$
This imposes the additional constraint that $x \geq 0$. Hence, any valid solution must satisfy $0 \leq x \leq 1$.
We now have the system of equations
\begin{align*}
\sqrt{x + 1} - \sqrt{1 - x} & = 1\\
\sqrt{x + 1} + \sqrt{1 - x} & = 2x
\end{align*}
Adding the equations gives
$$2\sqrt{x + 1} = 1 + 2x$$
Squaring both sides of the equation yields
\begin{align*}
4(x + 1) & = 1 + 4x + 4x^2\\
4x + 4 & = 1 + 4x + 4x^2\\
3 & = 4x^2\\
\frac{3}{4} & = x^2\\
\frac{\sqrt{3}}{2} & = |x|\\
\pm \frac{\sqrt{3}}{2} & = x
\end{align*}
Since we require that $0 \leq x \leq 1$, we discard the solution $x = -\sqrt{3}/2$.
Direct calculation shows that $x = \sqrt{3}/2$ is a valid solution, so the solution set is $S = \{\sqrt{3}/2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618757",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 6
} |
How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
| Again focusing on $x^4-2x^3+3x^2-x+1$, write this as the sum of $f(x) = x^4 - 2x^3 + 3x^2$ and $g(x) = -x + 1$. $f(x) = x^2(x^2 - 2x + 3) = x^2 ((x - 1)^2 + 2)$ which is always nonnegative.
That leaves $g(x)$ which is negative for $x \ge 1$. However, we just need to show that $f(x) > x$ in this domain, which can be shown as $f(x) = (x^2-x)^2 + x^2 \cdot 2 \ge x^2 \cdot 2 \ge 2x > x$. Thus from the first part, $f(x)$ and $g(x)$ are individually nonnegative for all real $x \le 1$, and from the last part, $f(x) + g(x)$ $> x + (1 - x) = 1$ for all real $x > 1$.
Hence proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
How to solve a system of equations over a finite field? I need to solve a system of equations over $\mathbb{Z}_{11}$.
My system is:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 8 \\
7x + 6y + 8z = 10 \\
10x + 3y + 4z = 6
\end{array}
\right.
$$
In matrix form:
$$
\begin{bmatrix}
2 & 5 & 1\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix} =
\begin{bmatrix}
8\\
10\\
6\\
\end{bmatrix}
$$
after $R_1*6$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}=
\begin{bmatrix}
4\\
10\\
6\\
\end{bmatrix}
$$
$R_2-7*R_1$ and $R_3-10*R_1$:
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 5 & 10\\
0 & 0 & 1\\
\end{bmatrix}=
\begin{bmatrix}
4\\
4\\
10\\
\end{bmatrix}
$$
$R_2*9$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{bmatrix}=
\begin{bmatrix}
4\\
3\\
10\\
\end{bmatrix}
$$
$$\left\{
\begin{array}{l}
z = 10 \\
y = 3 - 2z = 3 - 20 = -17=5 \\
x = 4 - 8y - 6z = 4 - 8*5 - 6*10 = -96 = 3
\end{array}
\right.$$
But after substitution found values in the initial system I have an error:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 2*3 + 5*5+10= 41 = 8 \\
7x + 6y + 8z = 7*3 + 6*5 + 8*10 = 131 = 10 \\
10x + 3y + 4z = 10*3 + 3*5 + 4*10= 85 = 8 \neq 6
\end{array}
\right.
$$
Do I have an arithmetic error or the whole way is wrong? Is there an easier and faster way to solve it?
| Since the determinant of your matrix $A$ is $221$, it is non-zero (actually equal to $1$) in the field $\Bbb F_{11}$. Hence the inverse exists, so that $Ax=b$ has the unique solution
$$
b=A^{-1}x=\begin{pmatrix} 0 & 5 & 1\cr 8 & 9 & 2\cr 5 & 0 & 10 \end{pmatrix}\cdot \begin{pmatrix} 8 \cr 10 \cr 6\end{pmatrix} =
\begin{pmatrix} 1 \cr 1 \cr 1\end{pmatrix}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Deduce limit of sequence from another sequence We have a sequence $(a_n)$ defined by the formula $$a_n = \int_0^1 (1 - x^2)^{\frac n2} \mathrm dx$$ We are also asked to prove, which I managed to do, the following recursion formula $$a_n = \frac{n + 3}{n + 2} \cdot a_{n + 2}$$ We are then asked to deduce with this information the limit of the series $$\frac 21 \cdot \frac 23, \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45, \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \frac65 \cdot \frac 67 \dots$$ But I cannot seem to find a relation between these sequences. I am quite sure that the limit should be $\frac \pi 2$, which is $2 \cdot a_1$, but I am not really sure if this is a coincidence and, in case it's not, why.
If someone could me maybe help to figure out how to proceed, I would be very thankful!
| Hint. Note that $0\le (1-x^2)^{(n+2)/2} \le (1-x^2)^{(n+1)/2} \le (1-x^2)^{n/2}\le 1$ when $0\le x \le1$. This gives $0 \le a_{n+2} \le a_{n+1} \le a_n \le 1$. So, $\dfrac{a_{n+2}}{a_n} \le \dfrac{a_{n+1}}{a_n} \le 1$. Can you simplify $\dfrac{a_{n+2}}{a_n}$? What can you say about the limit $\lim_{n\to \infty} \frac{a_{n+1}}{a_n} $? How can you use this limit for the given infinite product?
Edit. I came up with a full solution.
As above, $ \frac{a_{n+2}}{a_n} = \frac{n+2}{n+3}\le \frac{a_{n+1}}{a_n} \le 1 $, and taking limit yields $\lim \frac{a_{n+1}}{a_n} = 1$.
$$a_{2k}= \left( \frac{2k}{2k+1}\right) a_{2k-2} =
\left( \frac{2k}{2k+1}\right) \left( \frac{2k-2}{2k-1}\right) a_{2k-4} = \dots = \frac{(2k)(2k-2)\dots 2 }{(2k+1)(2k-1) \dots 3} a_0,$$
$$a_{2k-1}= \left( \frac{2k-1}{2k}\right) a_{2k-3} =
\left( \frac{2k-1}{2k}\right) \left( \frac{2k-3}{2k-2}\right) a_{2k-3} = \dots = \frac{(2k-1)(2k-3)\dots 3 }{(2k)(2k-2) \dots 4} a_1.$$
We can write the second equation as
$$\frac{a_{2k-1}}{2} = \frac{(2k-1)(2k-3)\dots 3 \cdot 1 }{(2k)(2k-2) \dots 4 \cdot 2} a_1.
$$
(This adjustment makes the multiplicants in numerators and denominators to be all $k$ in the fractions $a_{2k}$ and $a_{2k-1}/2$.)
So taking quotient yields
\begin{align*}
\frac{2 \cdot a_{2k}}{a_{2k-1}} &= \frac{(2k)(2k-2)\dots 2 }{(2k+1)(2k-1) \dots 3} \cdot
\frac{(2k)(2k-2) \dots 4 \cdot 2}{(2k-1)(2k-3)\dots 3 \cdot 1 }\cdot
\frac{a_0}{a_1}\\
& = \frac{(2k)(2k)}{(2k+1)(2k-1)} \dots \frac{2\cdot 2}{3\cdot 1} \frac{a_0}{a_1}.
\end{align*}
By taking limit $k\to\infty$, we have
$$ L\cdot \frac{a_0}{a_1} = 2,$$
$$ L = \frac{2a_1}{a_0} = \frac{\pi}{2},$$
where $L$ is the desired infinite product. This proves that the infinite product converges, and the limit is $\pi/2$.
| {
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Construct equation of ellipse, if focuses are $F_1 (-1, -1), F_2 (1, 1)$, height (perpendicular to $F_1 F_2$) $b = 1$ First of all, I know the right answer. It's $2x^2 - 2xy + 2y^2 - 3 = 0$.
What is unknown to me, is how to obtain that answer.
I've tried to make equation like $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We know $b$.
$c$ is half-distance between $F_1$ and $F_2$, so $c = \sqrt{2}$ (full distance $ = \sqrt{8} = 2\sqrt{2}$)
According to some formulas, $b = \sqrt{a^2 - c^2}$, hence $a^2 = b^2 + c^2 = 1^2 + (\sqrt{2})^2 = 3$
So equation "should be" the next one: $\frac{x^2}{3} + \frac{y^2}{1} = 1$
But it's a wrong equation. Obviously, this ellipse has been rotated by 45 degrees. Some posts hinted me that I should put $(a_0 x + a_1 y)^2$ in numerator.
Unfortunately, I don't know how to "rotate" ellipse in such form of equation. Perhaps, I need somehow to come up with more generalized form of equation
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, but still I don't know to identify the coefficients
Anyway, I would be glad to perceive any help on your behalf.
| If $\left(\begin{matrix}X\\Y\end{matrix}\right)$ is the image of the vector $\left(\begin{matrix}x\\y\end{matrix}\right)$ after rotation by $45^o$, then $$\left(\begin{matrix}X\\Y\end{matrix}\right)=M\left(\begin{matrix}x\\y\end{matrix}\right)$$
Where $$M=\left(\begin{matrix}\cos 45&-\sin 45\\\sin 45&\cos 45\end{matrix}\right)$$
$$\implies \left(\begin{matrix}x\\y\end{matrix}\right)=M^{-1}\left(\begin{matrix}X\\Y\end{matrix}\right)$$
Where $$M^{-1}=\left(\begin{matrix}\cos 45&\sin 45\\-\sin 45&\cos 45\end{matrix}\right)$$
Hence $$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{matrix}\right)\left(\begin{matrix}X\\Y\end{matrix}\right)$$
So, taking the equation of the ellipse you found, i.e. $x^2+3y^2=3$, substitute into this $$x=\frac{1}{\sqrt{2}}X+ \frac{1}{\sqrt{2}}Y$$ and $$y=-\frac{1}{\sqrt{2}}X+ \frac{1}{\sqrt{2}}Y$$ and you get the equation of the required rotated ellipse $$2X^2-2XY+2Y^2=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the number of squares modulo a composite number $143$ I want to know how many classes of modulo $143$ are squares. This is equivalent to find how many $a$ can make $$\begin{align*}
x^2 \equiv a \pmod{143} && (1)
\end{align*}$$ has solutions
On the other hand, I know that $143=11 \times 13$ and $(11,13)=1$, so $$\begin{align*}
x_0 \text{ is a solution of (1) } \iff x_0 \text{ is a solution of} \begin{cases}
x^2 \equiv a \pmod{11} \\ &&(2) \\ x^2 \equiv a \pmod{13}
\end{cases}
\end{align*}$$ That is $a$ is a square modulo $143 \iff a$ is a square modulo $11$ and is a square modulo $13$
It's not hard for me to find that $$1^2=1, \ 2^2=4,\ 3^2=9,\ 4^2=5,\ 5^2=3 \pmod{11}$$ are squares modulo $11$ and $$1^2=1,\ 2^2=4,\ 3^2=9,\ 4^2=3,\ 5^2=12,\ 6^2=10 \pmod{13}$$ are squares modulo $13$ . Based on this information, I'm a little confused to compute how many squares there exist modulo $143$
I think at least, $a=1,4,9$ are squares $\pmod{143}$ , since this time in $(2)$ , plug in $a=1,4,9$ we can pick $x=1,2,3$ , these numbers satisfiy $(2)$ , so $x=1,2,3$ should also satisfy $x^2 \equiv 1,4,9 \pmod{143}$
Also, I find that if $a$ modulo $11$ is $5$ and modulo $13$ is $3$, then $x=4$ is a solution of $(2)$; By some computation, I find the number satisfies this is $16$; similarly, if $a$ modulo $11$ is $3$ and modulo $13$ is $12$, then $x=5$ is a solution of $(2)$. I find the number satisfies this is $25$.
They are $1,4,9,16,25$ all squares modulo $143$. If they are, is the number of squares of modulo $143$ just equal to the number of squares of modulo $11$ (that is, the smaller number in $11 \times13$)
Any help on this? Thanks.
| So there should be $6\cdot 7=42.$ These can be gotten under the isomorphism between $$\Bbb Z_{11}×\Bbb Z_{13}$$ and $$\Bbb Z_{143}.$$
For instance, using Bezout, we can map $(x,y)$ to $-7×11×y+6×13×x.$
You can now check that for instance, $(3,12)\mapsto -924+234\equiv-690\equiv-118\equiv25.$ And $(4,0)\mapsto312\equiv26.$
Etc.
It's well-known that modulo a prime $p$, there are $(p+1)/2$ quadratic residues.
This follows from Euler's criterion: $$\left(\frac ap\right)\equiv a^{\frac {p-1}2}\pmod p,$$ since there are $(p-1)/2$ solutions to $x^{\frac{p-1}2}\equiv1$. Plus zero is always a quadratic residue.
| {
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Evaluating $\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}$ I'm trying to evaluate $$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}.$$
I can see the $\frac{0}{0}$ form, so I'll use L'Hôpital's rule. However, I'll eliminate the sine function in the denominator by multiplying the numerator and denominator by $x$. We know
$$\lim\limits_{x\rightarrow0} \frac{x}{\sin x} = 1.$$
The problem reduces to
$$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^4}.$$
Now I'll use L'Hôpital's rule. The problem now becomes
$$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}(-x)+\sin x}{4x^3}.$$
I don't know how to proceed from here onwards. Using L'Hôpital's rule any more complicates the problem. Any ideas would be appreciated.
| The following steps use De l'Hopital rule.
$$\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)}\right)=\lim _{x\to \:0}\left(\frac{-e^{-\frac{x^2}{2}}x+\sin \left(x\right)}{3x^2\sin \left(x\right)+\cos \left(x\right)x^3}\right)=\lim _{x\to \:0}\left(\frac{e^{-\frac{x^2}{2}}x^2-e^{-\frac{x^2}{2}}+\cos \left(x\right)}{-x^3\sin \left(x\right)+6x^2\cos \left(x\right)+6x\sin \left(x\right)}\right)$$
$$=\lim _{x\to 0}\left(\frac{-e^{-\frac{x^2}{2}}x^3+3e^{-\frac{x^2}{2}}x-\sin \left(x\right)}{-x^3\cos \left(x\right)-9x^2\sin \left(x\right)+18x\cos \left(x\right)+6\sin \left(x\right)}\right)=\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}x^4-6e^{-\frac{x^2}{2}}x^2+3e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)-12x^2\cos \left(x\right)-36x\sin \left(x\right)+24\cos \left(x\right)}\right)$$
$$=\frac{e^{-\frac{0^2}{2}}\cdot 0^4-6e^{-\frac{0^2}{2}}\cdot 0^2+3e^{-\frac{0^2}{2}}-\cos \left(0\right)}{0^3\sin \left(0\right)-12\cdot 0^2\cos \left(0\right)-36\cdot 0\cdot \sin \left(0\right)+24\cos \left(0\right)}=\frac 1{12}. \square$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$ In one of my analysis course, we considered the function
$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$$
Then my teacher told us that $f$ had the following behavior
$$
f(x)\sim\begin{cases}
\frac{-1}{4}x\log x & \text{as } x \to 0^+,\\
\frac{2}{x} & \text{as } x \to \infty.
\end{cases}$$
For the case $x \to \infty$ I understand as in this case $e^{-\frac{1}{4}y^2}$ is negligeable and then
$$f \sim x \left[\frac{-1}{2y^2}\right]^\infty_x = \frac{1}{2x}.$$
However I do not have any intuition of the form of
$$\int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy,$$
which should behave as a $\log$. I thought maybe we could expand $e^{-\frac{1}{4}y^2}$ to the first order so that
$$\frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy = \frac{1}{4}\frac{1}{y} + O(y)$$
and by integrating we indeed get a $\log$. However I don't get how exactly we should get this behavior of $f$. Is there someone more skilful than me who could answer to this question ?
| $$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}\,dy$$
Let $y=2 \sqrt{t}$ to make
$$f(x)=\frac 18 x \int^\infty_{\frac {x^2}4} \frac{1-e^{-t}}{t^2}\,dt$$ Using the incomplete gamma function
$$\int t^{-n} e^{-t}\,dt=-\Gamma (1-n,t)$$
wich makes
$$f(x)=\frac{1}{2 x}-\frac{x}{8}\, \Gamma \left(-1,\frac{x^2}{4}\right)$$ Using the expansion for small positive $x$
$$f(x)=-\frac{1}{8} x \,(2 \log (x)+\gamma -1-2\log(2))+O\left(x^3\right)$$
Now, if $x$ is large, you would find
$$f(x)=\frac 1{2x}-\frac{2}{x^3} e^{-\frac{x^2}{4}}+\cdots$$
Just trying
*
*for $x=\frac 1{10}$, the approximation will give $0.0801781$ while the exact value is $0.0801937$.
*for $x=10$, the approximation will give $0.0499999999999722$ while the exact value is $0.0499999999999742$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve the equation $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$? Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$
Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and why? Please provide the explanation to this question. You do not have to solve the problem.
Thank you very much.
| Obviously, both $x=2$ and $x=3$ are solutions.
If $x$ is neither above, then we can cancel $(x-2)(x-3)$ on both sides of the equation, so we get
$$
x-1 = x-4, \text{or} -1=-3,
$$
which is a contradiction.
Therefore we only have two solutions: $x=2$ and $x=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ I was evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ and got an answer which, I think, is not correct as it is different from wolframalpha's answer.
Here's my work:
$$\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1}\\& =\int \frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx + \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2}\\&= \sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C\tag{3}\end{align}$$
Steps:
$(1.)$ Rationalized the numerator.
$(2.)$ Applied linearity.
$(3.)$ The first integral is done by substituting $x^2 - (3/2)^2 = t$ and the second one is inverse hyperbolic cosine.
WolframAlpha shows this. I also tried differentiating both the answers, but still it's different from mine.
|
\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1p}\\& =\int \frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx + \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2p}\\&= \sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C\tag{3p}\end{align}
*
*Your work is correct, but on precisely the interval $(1.5,\infty).$
This is because
*
*step $(1\mathrm p)$ assumes that $(2x+3)\ge0.$
*step $(3\mathrm p)$ assumes that $x\ge0.$
*For the other piece $(-\infty,-1.5]$ of the integrand's domain:
\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int -\frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1n}\\& =\int -\frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx - \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2n}\\&= -\sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(-\frac{2x}{3}\right) + D.\tag{3n}\end{align}
*Then the final answer is the piecewise function combining the above two pieces, noting that they have independent arbitrary constants $C$ and $D.$
This Desmos graph verifies that the red curve (the given integrand) comprises precisely the blue curve (the derivative function of the positive piece $(3\mathrm p)$) and the green curve (the derivative function of the negative piece $(3\mathrm n)$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1}$ as $x \to 2$ I don't know the answer to the problem. None of my "tricks" work: I see no way of factoring it, rationalization on the denominator doesn't seem useful (because I still end up with a zero). The book has not gone over any fancier techniques such as L'Hospital's rule.
Reference. This problem comes from section $2.3$ of Stewart's Calculus. It's problem $60$, an even number, so I don't even know the final answer.
What have I learned? Sometimes we need to rationalize both numerator and denominator.
| The limit has the indeterminate form $0/0$, which can be eliminated if you rationalize both the numerator and denominator.
\begin{align*}
\lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} & = \lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} \cdot \frac{\sqrt{6 - x} + 2}{\sqrt{6 - x} + 2} \cdot \frac{\sqrt{3 - x} + 1}{\sqrt{3 - x} + 1}\\
& = \lim_{x \to 2} \frac{(2 - x)(\sqrt{3 - x} + 1)}{(2 - x)(\sqrt{6 - x} + 2)}\\
& = \lim_{x \to 2} \frac{\sqrt{3 - x} + 1}{\sqrt{6 - x} + 2}\\
& = \frac{1}{2}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Then $T$ is a linear operator. The problem is that can we explicitly obtain the kernel of this linear map. If $A$ is a singular matrix then $X=xy^*$ satisfies $T(X)=0$ where $Ax=0$ and $y^*A=0$. But there exists matrix $A$ for which there is no connection between $A$ and $X$ for example if $$A=\begin{pmatrix}
1&0&0\\0&0&0\\0&1&0
\end{pmatrix} \qquad \text{and} \qquad X=\begin{pmatrix}
0&0&0\\0&1&0\\0&2&1
\end{pmatrix}$$
So how we can obtain the kernel of $T$ completely.
| Kronecker product is the right tool for analysing such issues. Here is how.
Let us begin by writing
$$T(X)=T_A(X)=A^2X+AXA+XA^2\tag{1}$$
in a different way, using the "vec" operator stacking the columns of any $3 \times 3$ matrix into a single $9 \times 1$ column vector :
$$T(\operatorname{vec}(X))=\operatorname{vec}(A^2XI)+\operatorname{vec}(AXA)+\operatorname{vec}(IXA^2) \tag{3}$$
Now, if you are already familiar with Kronecker product, let us recall the fundamental identity :
$$\operatorname{vec}(AXB)=(B^T \otimes A) \operatorname{vec}(X)\tag{4}$$
Using (4) and denoting $\mathbf{X}=\operatorname{vec}(X)$, (3) can be transformed into :
$$T(\mathbf{X})=(I \otimes (A^2))\mathbf{X}+(A^T \otimes A)\mathbf{X}+((A^2)^T \otimes I)\mathbf{X}$$
Otherwise said :
$$T(\mathbf{X})=\underbrace{((I \otimes (A^2))+(A^T \otimes A)+((A^2)^T \otimes I))}_{\mathbf{T}}\mathbf{X}\tag{5}$$
Otherwise said, operator $T$ is completely described by $9 \times 9$ matrix $\mathbf{T}$.
For example, if
$$A=\begin{pmatrix}
0&0&1\\1&0&0\\0&1&0
\end{pmatrix} \ \text{giving} \ A^2=\begin{pmatrix}
0&1&0\\0&0&1\\1&0&0
\end{pmatrix}$$
$\mathbf{T}$ is the sum of the following $9 \times 9$ matrices:
$$\mathbf{T}=\begin{pmatrix}
A^2&0&0\\0&A^2&0\\0&0&A^2
\end{pmatrix}+\begin{pmatrix}
0&A&0\\0&0&A\\A&0&0
\end{pmatrix}+\begin{pmatrix}
0&0&I\\I&0&0\\0&I&0
\end{pmatrix}=\begin{pmatrix}
A^2&A&I\\I&A^2&A\\A&I&A^2
\end{pmatrix}\tag{6}$$
How is all this connected with the initial question, obtaining the kernel of operator $T$ defined by (1) ?
Plainly, by finding the kernel of $\mathbf{T}$, then doing the inverse operation of $vec$, i.e., by "reshaping" all $9 \times 1$ column vectors constituting a basis of this kernel into $3 \times 3$ matrices.
Remark : In initial expression (1), $T(X)$ can be considered as the differential computed in $A$ of function $C$ defined by $C(X)=X^3$. I am almost sure that properties of operator $T$ can be deduced from this fact.
Edit : Having done extensive numerical simulations (using Matlab), I have found pairs :
$$(A,X) \ \text{solutions of eq. } T_A(X)=0$$
(see equ. (1)) where both $A$ and $X$ have full rank.
Here is a particularily interesting instance (see explanation below) :
$$A=\begin{pmatrix}
-1&1&-2\\2&0& \ \ 1\\2&1& \ \ 1
\end{pmatrix}, \ \ X=\begin{pmatrix}
-1&0&-1\\1&1&1\\1&1&0
\end{pmatrix}$$
Please note that $A$ is specific because $A^3=-3I_3$...
A remarkable fact is that for this choice of $A$, the kernel of $9 \times 9$ matrix $\mathbf{T}$ has dimension $6$ ! Moreover, we can exhibit a basis with $6$ elements of this kernel. Indeed, instead of matrix $X=X_1$ above, we could have taken one of the other five matrices :
$$X_2=\begin{pmatrix}
1&2&1\\1&1&-2\\1&0&-2
\end{pmatrix} \ \text{or} \ X_3=\begin{pmatrix}
-2&-1&1\\1&2&-1\\2&2&0
\end{pmatrix} \ \text{or} $$
$$X_4=\begin{pmatrix}
0&0&-1\\1&1&1\\0&1&-1
\end{pmatrix} \ \text{or} \ X_5=\begin{pmatrix}
2&1&0\\-2&-1&1\\0&2&-1
\end{pmatrix} \ \text{or}$$
$$X_6=\begin{pmatrix}
0&-2&1\\-1&-1&1\\0&1&1
\end{pmatrix}, $$
One can check that all the $X_k, \ k=1,2 \cdots 6$ are full-rank matrices and that they constitute a basis of the kernel of $T_A$.
Of course, one can get, by combining these $X_k$, other matrices $X$ belonging to the kernel of $T£ which are rank-2 or even rank-1.
| {
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$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$? I was trying to solve $$\int\frac{x^2}{\sqrt{5-4x^2 }} dx$$ by using the substitution
$$
u=\sqrt{5-4x^2}\\
dx = \frac{\sqrt{5-4x^2}}{-4x}du$$ So the integral can be written as $$\int \left(-\frac{x}{4}\right) du$$ and
$$
x= \begin{cases}
\sqrt{(5-u^2)/4}, & x >0\\
-\sqrt{(5-u^2)/4}, &x<0
\end{cases}
$$
I've rewritten the integral as
$$-\frac18\int\sqrt{5-u^2}du$$
using trig substitution with $$u= \sqrt5\sin(t)$$ we have
$$-\frac5{16}t - \frac5{16} \sin(t)\cos(t)$$ $$t=\arcsin\left(\frac{u^2}{\sqrt5}\right)$$
If we now reverse the substitutions and simplify, I got the final answer as $$-\frac5{16} \arcsin\left(\sqrt{\frac{5-4x^2}5}\right) - \frac{x}8\sqrt{5-4x^2} $$However this differs from the answer in my textbook which is:
$$-\frac{x}{8}\sqrt{5-4x^2}+\frac5{16}\arcsin\left(\frac{2x}{\sqrt5}\right)+C$$
My textbook used different methods that are simpler and easier, but I want to know where I made a mistake.
| Integration by parts
$$
\begin{aligned}
I&=\int \frac{x^2}{\sqrt{5-x^2}} d x \\
& =-\int x d\left(\sqrt{5-x^2}\right) \\
& =-x \sqrt{5-x^2}+\int \sqrt{5-x^2} d x \\
& =-x \sqrt{5-x^2}+\int \frac{5-x^2}{\sqrt{5-x^2}} d x \\
& =-x \sqrt{5-x^2}+5 \int \frac{d x}{\sqrt{5-x^2}}-I \\
& =\frac{1}{2}\left[5 \sin ^{-1}\left(\frac{x}{\sqrt{5}}\right)-x \sqrt{5-x^2}\right]+C
\end{aligned}
$$
| {
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Foci of ellipse My question is given a ellipse of the equation : $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ where $a>b$ then how we can find the coordinates of the foci. I want to find those coordinates without the presuming that the foci exists because most proofs I found online assume the properties of foci to be true and then take some extreme case to find foci's coordinates.
So to rephrase my question: Given a closed curve of the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ prove that there exists two points inside the curve such that if we take any point on the boundary of the curve and join it to those two points then the sum of those lengths will give a fixed constant based on $a$ and $b$ (Assume $a>b$).
Here's my attempt which gave me nothing:
Let's take a point $P$ on the curve as $\left(x,\ b\sqrt{1-\frac{x^2}{a^2}}\right)$
Let those two points be $(-f,0)$ and $(f,0)$ then the sum of lengths from point $P$ becomes
$$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$
Differentiating this wrt to $f$ and equating to $0$ to find the stationary case
$$\dfrac{f+x}{\sqrt{\left(f+x\right)^2+b^2\left(1-\frac{x^2}{a^2}\right)}}+\dfrac{f-x}{\sqrt{\left(f-x\right)^2+b^2\left(1-\frac{x^2}{a^2}\right)}}=0$$
Squaring and simplifying
$$4bfx\left(1-\frac{x^2}{a^2}\right)=0 \implies f =0$$
which is obviously wrong...
Note that in my attempt I too assumed two points, that the foci will be symmetrical and on the major axis, if we can even take these assumptions out that would be amazing. It's just that with these assumptions I was able to at least start somewhere.
P.S.
$\textbf{thanks to the comments by @Blue}$
differentiating $S$ wrt $x$ and equating to zero
$$\dfrac{2\left(x+f\right)-\frac{2b^2x}{a^2}}{2\sqrt{\left(x+f\right)^2+b^2\cdot\left(1-\frac{x^2}{a^2}\right)}}+\dfrac{-\frac{2b^2x}{a^2}-2\left(f-x\right)}{2\sqrt{b^2\cdot\left(1-\frac{x^2}{a^2}\right)+\left(f-x\right)^2}}=0$$
Squaring and simplifying
$$ a^4b^2fx(f^2-(a^2-b^2))=0$$
So,
$$f= \sqrt{a^2-b^2}$$
| *
*Find the eccentricity of the ellipse.
$$e=\sqrt{1-\dfrac{b^2}{a^2}}$$
*Now you have the following expressions to find the coordinates of the foci:
$$\begin{align*}
F_1 &= \left(ae, 0\right) = \left(a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right),\\
F_2 &= \left(-ae, 0\right) = \left(-a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right).
\end{align*}$$
What's even more wonderful, it works for any kind of ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding steady state equations of infinite-dimensional Markov Chain
Suppose we have the transition matrix
$$P=\begin{bmatrix} 0&1&0&0&0&\cdots\\\\ p&0&1-p&0&0&\cdots\\\\
0&p&0&1-p&0&\cdots\\\\ 0&0&p&0&1-p&\cdots\\\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\\\ \end{bmatrix}$$
with $p\in[1/2,1]$. Show that the solution to the steady-state equations are given by
$$\pi_0=\frac{r-1}{2r}$$
$$\pi_j=\frac{(r+1)(r-1)}{2r^{j+1}}\quad j\geq 1$$
where $r = p∕(1 − p)$.
My attempt: We have that
$$P-I=\begin{bmatrix} -1&1&0&0&0&\cdots\\\\ p&-1&1-p&0&0&\cdots\\\\
0&p&-1&1-p&0&\cdots\\\\ 0&0&p&-1&1-p&\cdots\\\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\\\ \end{bmatrix}$$
and so we must solve the equations
$$\pi_0=p\pi_1$$
$$\pi_1=\pi_0+p\pi_2$$
$$\pi_2=(1-p)\pi_1+p\pi_3$$
$$\pi_3=(1-p)\pi_2+p\pi_4$$
$$\vdots$$
and $\pi_0+\pi_1+\pi_2+\cdots=1$.
The structure of the transition matrix is simple, but it's not clear to me how to proceed to arrive at the desired equations.
| A steady-state solution satisfies $P^T \pi = \pi$, where $\pi$ is a column vector of probabilities. Reading off the matrix system gives
$$\begin{cases}
\pi_0 = p \pi_1 \\
\pi_1 = \pi_0 + p \pi_2 \\
\pi_i = (1 - p) \pi_{i - 1} + p \pi_{i + 1} & i \ge 2 \\
\pi_0 + \pi_1 + ... = 1
\end{cases}$$
Of course, we may rewrite the system to depend strictly on indices less than the one of interest:
$$\begin{cases}
\pi_0 = p \pi_1 \\
\pi_1 = \pi_0 + p \pi_2 \\
\pi_i = \frac{1}{p} \pi_{i - 1} + \left( 1 - \frac{1}{p} \right) \pi_{i - 2} & i \ge 3 \\
\pi_0 + \pi_1 + ... = 1
\end{cases}$$
This is essentially a linear recurrence, where
$$\begin{pmatrix}
\pi_{i + 1} \\
\pi_i
\end{pmatrix} = \underbrace{\begin{pmatrix}
\frac{1}{p} & 1 - \frac{1}{p} \\
1 & 0
\end{pmatrix}}_{A} \begin{pmatrix}
\pi_i \\
\pi_{i - 1}
\end{pmatrix}$$
Let's diagonalize the matrix $A$:
$$|A - \lambda I| = \lambda^2 - \frac{1}{p} \lambda - 1 + \frac{1}{p} = 0$$
We find that the two eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{p} - 1$. $\left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right\}$ is a basis for the eigenspace of $\lambda_1$, while $\left\{ \begin{pmatrix} \frac{1}{p} - 1 \\ 1 \end{pmatrix} \right\}$ is a basis for the eigenspace of $\lambda_2$. Solving the first two equations in the above system gives the "initial condition" for the recurrence:
$$\begin{pmatrix}
\pi_2 \\ \pi_1
\end{pmatrix} = \frac{\pi_0}{p} \begin{pmatrix}
\frac{1}{p} - 1 \\ 1
\end{pmatrix}$$
which we find lies exactly in the eigenspace of $\lambda_2$. Therefore, we may say that in general,
$$\begin{pmatrix}
\pi_{i + 1} \\ \pi_i
\end{pmatrix} = \frac{\pi_0}{p} \left( \frac{1}{p} - 1 \right)^{i - 1} \begin{pmatrix}
\frac{1}{p} - 1 \\ 1
\end{pmatrix}$$
or for $i \ge 1$,
$$\pi_i = \frac{\pi_0}{p} \left( \frac{1}{p} - 1 \right)^{i - 1}$$
We now find that the normalization condition is
$$\pi_0 + \pi_1 + ... = \pi_0 \left( 1 + \frac{1}{p} \sum_{j = 0}^\infty \left( \frac{1}{p} - 1 \right)^j \right) = 1$$
Evaluating the geometric series assuming $\frac{1}{p} - 1 < 1 \implies p > \frac{1}{2}$ yields
$$\sum_{j = 0}^\infty \left( \frac{1}{p} - 1 \right)^j = \frac{1}{1 - \left( \frac{1}{p} - 1 \right)} = \frac{p}{2 p - 1}$$
so that
$$1 + \frac{1}{p} \sum_{j = 0}^\infty \left( \frac{1}{p} - 1 \right)^j = \frac{2 p}{2 p - 1}$$
and thus
$$\pi_0 = 1 - \frac{1}{2 p}$$
which you will find to be equivalent to the given expression. Knowing $\pi_0$, all the other probabilities can be found to be
$$\pi_i = \left( \frac{1}{p} - \frac{1}{2 p^2} \right) \left( \frac{1}{p} - 1 \right)^{i - 1}$$
which after some manipulation can be shown to be equivalent to the given expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Divergence of vector-tensor outer product multiplication I have a material derivative of a tensor quantity $\mathbf{S}$.
$$
\frac{\partial \mathbf{S}}{\partial t} + \vec{v} \cdot \nabla \mathbf{S}
$$
I would like to know if the term $\vec{v} \cdot \nabla \mathbf{S}$ can be rewritten as:
$$
\nabla \cdot \left(\vec{v} \otimes \mathbf{S}\right) - \left(\nabla \cdot \vec{v}\right) \mathbf{S} = \vec{v} \cdot \nabla \mathbf{S}
$$
So that the final expression would be:
$$
\frac{\partial \mathbf{S}}{\partial t} + \nabla \cdot \left(\vec{v}\otimes \mathbf{S}\right) - \left(\nabla \cdot \vec{v} \right) \mathbf{S}
$$
I know this works for scalar quantities, but I do not know if it holds for tensor quantities.
Can anyone help me with this?
Best Regards
| Note that following relations hold:
Scalar $\rightarrow$ Vector:
$$
\operatorname{grad} \phi(\mathbf{x}):=\frac{\mathrm{d} \phi(\mathbf{x})}{\mathrm{d} \mathbf{x}}=: \mathbf{w}(\mathbf{x})
$$
Vector $\rightarrow$ Matrix:
$$
\operatorname{grad} \mathbf{v}(\mathbf{x}):=\frac{\mathrm{d} \mathbf{v}(\mathbf{x})}{\mathrm{d} \mathbf{x}}=: \mathbf{S}(\mathbf{x})
$$
Matrix $\rightarrow$ Tensor (3):
$$
\operatorname{grad} \mathbf{T}(\mathbf{x}):=\frac{\mathrm{d} \mathbf{T}(\mathbf{x})}{\mathrm{d} \mathbf{x}}=: \mathrm{U}^3(\mathbf{x})
$$
Tranpose:
$$
\begin{aligned}
\mathbf{I} \otimes \mathbf{I} & =\left(\mathbf{e}_i \otimes \mathbf{e}_i\right) \otimes\left(\mathbf{e}_j \otimes \mathbf{e}_j\right) \\
(\mathbf{I} \otimes \mathbf{I})^{\stackrel{23}{\text{T}}} & =\mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{e}_i \otimes \mathbf{e}_j \\
(\mathbf{I} \otimes \mathbf{I})^{\stackrel{24}{\text{T}}} & =\mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{e}_j \otimes \mathbf{e}_i
\end{aligned}
$$
Gradient:
$$
\begin{aligned}
\operatorname{grad}(\phi \psi) & =\phi \operatorname{grad} \psi+\psi \operatorname{grad} \phi \\
\operatorname{grad}(\phi \mathbf{v}) & =\mathbf{v} \otimes \operatorname{grad} \phi+\phi \operatorname{grad} \mathbf{v} \\
\operatorname{grad}(\phi \mathbf{T}) & =\mathbf{T} \otimes \operatorname{grad} \phi+\phi \operatorname{grad} \mathbf{T} \\
\operatorname{grad}(\mathbf{u} \cdot \mathbf{v}) & =(\operatorname{grad} \mathbf{u})^{\text{T}} \mathbf{v}+(\operatorname{grad} \mathbf{v})^{\text{T}} \mathbf{u} \\
\operatorname{grad}(\mathbf{u} \times \mathbf{v}) & =\mathbf{u} \times \operatorname{grad} \mathbf{v}+\operatorname{grad} \mathbf{u} \times \mathbf{v} \\
\operatorname{grad}(\mathbf{a} \otimes \mathbf{b}) & =\left[\operatorname{grad} \mathbf{a} \otimes \mathbf{b}+\mathbf{a} \otimes(\operatorname{grad} \mathbf{b})^{\text{T}}\right]^{\stackrel{23}{\text{T}}} \\
\operatorname{grad}(\mathbf{T} \mathbf{v}) & =(\operatorname{grad} \mathbf{T})^{\stackrel{23}{\text{T}}} \mathbf{v}+\mathbf{T} \operatorname{grad} \mathbf{v} \\
\operatorname{grad}(\mathbf{T} \cdot \mathbf{S}) & =(\operatorname{grad} \mathbf{T})^{\stackrel{13}{\text{T}}} \mathbf{S}^{\text{T}}+(\operatorname{grad} \mathbf{S})^{\stackrel{13}{\text{T}}} \mathbf{T}^{\text{T}}
\end{aligned}
$$
Divergence:
$$
\begin{aligned}
\operatorname{div}(\mathbf{u} \otimes \mathbf{v})&=\mathbf{u} \operatorname{div} \mathbf{v}+(\operatorname{grad} \mathbf{u}) \mathbf{v} \\
\operatorname{div}(\phi \mathbf{v})&=\mathbf{v} \cdot \operatorname{grad} \phi+\phi \operatorname{div} \mathbf{v} \\
\operatorname{div}(\mathbf{T} \mathbf{v})&=\left(\operatorname{div} \mathbf{T}^{\text{T}}\right) \cdot \mathbf{v}+\mathbf{T}^{\text{T}} \cdot \operatorname{grad} \mathbf{v} \\
\operatorname{div}(\operatorname{grad} \mathbf{v})^{\text{T}}&=\operatorname{grad} \operatorname{div} \mathbf{v} \\
\operatorname{div}(\mathbf{u} \times \mathbf{v})&=(\operatorname{grad} \mathbf{u} \times \mathbf{v}) \cdot \mathbf{I}-(\operatorname{grad} \mathbf{v} \times \mathbf{u}) \cdot \mathbf{I} \\
&=\mathbf{v} \cdot \operatorname{rot} \mathbf{u}-\mathbf{u} \cdot \operatorname{rot} \mathbf{v} \\
\operatorname{div}(\phi \mathbf{T})&=\mathbf{T} \operatorname{grad} \phi+\phi \operatorname{div} \mathbf{T} \\
\operatorname{div}(\mathbf{T} \mathbf{S})&=(\operatorname{grad} \mathbf{T}) \mathbf{S}+\mathbf{T} \operatorname{div} \mathbf{S} \\
\operatorname{div}(\mathbf{v} \times \mathbf{T})&=\mathbf{v} \times \operatorname{div} \mathbf{T}+\operatorname{grad} \mathbf{v} \times \mathbf{T} \\
\operatorname{div}(\mathbf{v} \otimes \mathbf{T})&=\mathbf{v} \otimes \operatorname{div} \mathbf{T}+(\operatorname{grad} \mathbf{v}) \mathbf{T}^{\text{T}} \\
\operatorname{div}(\operatorname{grad} \mathbf{v})&=\mathbf{0} \\
\operatorname{div}\left(\operatorname{grad} \mathbf{v} \pm(\operatorname{grad} \mathbf{v})^{\text{T}}\right)&=\operatorname{div} \operatorname{grad} \mathbf{v} \pm \operatorname{grad} \operatorname{div} \mathbf{v} \\
\operatorname{div} \operatorname{rot} \mathbf{v}&=0 \\
\end{aligned}
$$
Rotation:
$$
\begin{aligned}
\operatorname{rot} \operatorname{rot} \mathbf{v}&=\operatorname{grad} \operatorname{div} \mathbf{v}-\operatorname{div} \operatorname{grad} \mathbf{v} \\
\operatorname{rot} \operatorname{grad} \phi&=\mathbf{0} \\
\operatorname{rot} \operatorname{grad} \mathbf{v}&=\mathbf{0} \\
\operatorname{rot}(\operatorname{grad} \mathbf{v})^{\text{T}}&=\operatorname{grad} \operatorname{rot} \mathbf{v} \\
\operatorname{rot}(\phi \mathbf{v})&=\phi \operatorname{rot} \mathbf{v}+\operatorname{grad} \phi \times \mathbf{v}-\mathbf{v} \otimes \mathbf{u})
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
| Let us consider the polynomial
$$P_n(x):=1-x^2+x^4-x^6+\cdots x^{2n-2}=\frac{x^{2n}+1}{x^2+1}.$$
We have
$$0<\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx<\int_0^1\left(\frac{x^{2n}}{0+1}\right)^2dx=\frac1{4n+1},$$ and the integral can be made as small as desired.
On another hand, the remainder of the long division of $(P_n(x)-1)^2$ by $(x^2+1)^2$ is a binomial $ax^2+b$, with $a,b$ integer. Then
$$\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx=\int_0^1\left(Q(x)+\frac{ax^2+b}{(x^2+1)^2}\right)dx.$$
As
$$\int_0^1\frac{ax^2+b}{(x^2+1)^2}dx=\frac{b-a}4+\frac{b+a}8\pi$$
we can get arbitrarily close rational approximations by a rational integral.
For example, with $P_2:=1-x^2+x^4$ we have
$$\frac{(x^4-x^2)^2}{(x^2+1)^2}=x^4-4x^2+8-\frac{12x^2+8}{(x^2+1)^2}$$ so that by integration
$$0<\frac15-\frac43+8-\frac{5\pi}2+1<\frac19$$ or
$$\frac{698}{225}<\pi<\frac{236}{75}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "129",
"answer_count": 6,
"answer_id": 4
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| I didn't see an answer using the method that I used, so I'm posting an answer here to 'spread the knowledge!'
This can be applied to higher powers such $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.
Solving by the use of Indeterminate Coefficients:
Assume the series$$1+2+3+4+5\ldots+n\tag1$$
Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\ldots\&c\tag2$$
If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\ldots\&c\tag3$$
And subtracting $(3)-(2)$, gives$$\begin{align*} & n+1=B+C(2n+1)\tag4\\n & +1=2Cn+(B+C)\tag5\end{align*}$$
Therefore, $C=\dfrac 12,B=\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\ldots+n=\dfrac n2+\dfrac {n^2}2=\dfrac {n(n+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 20
} |
Lateral surface area by hand I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.
I have come up with my own solution, but I'm not sure if it's right.
Before substituting values back in, I got
$$ \frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right) \text{ from }0\text{ to }\sqrt2$$
Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$
Thanks so much in advance.
| With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$. Now, let
$$\begin{align}
I_1&=\int\tan^2\theta\sec^3\theta d\theta
\\
&=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta
\\
&=\tan\theta(\sec^3\theta-\sec\theta)-\int3\sec^3\theta\tan^2\theta-\sec\theta\tan^2\theta d\theta
\\
&=\tan\theta(\sec^3\theta-\sec\theta)-3I_1+\int\sec\theta\tan^2\theta d\theta
\end{align}$$
(using the integration by parts as you suggested) so that $I_1=\frac{1}{4}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)$. Let
$$\begin{align}
I_2&=\int\sec\theta\tan^2\theta d\theta
\\
&=\sec\theta\tan\theta-\int\sec^3\theta d\theta
\\
&=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta d\theta-\int\sec\theta d\theta
\\
&=\sec\theta\tan\theta-I_2+\log|\sec\theta-\tan\theta|
\end{align}$$
(using integration by parts with $u=\tan\theta$ and $dv=\sec\theta\tan\theta d\theta$) so that $I_2=\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)$. Now,
$$\begin{align}
\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta
&=\frac{\pi}{4}I_1=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)
\\
&=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)\right)
\\
&=\frac{\pi}{16}\left(\tan\theta\sec^3\theta-\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\log|\sec\theta-\tan\theta|\right)
\\
&=\frac{\pi}{16}\left(2x(1+4x^2)^{3/2}-x\sqrt{1+4x^2}+\frac{1}{2}\log\left|\sqrt{1+4x^2}-2x\right|\right)
\end{align}$$
which is similar to what you have, but not the same.
(Of course, doing this by hand and covering a sheet of paper with scribbles to do the two instances of over-and-back integration by parts, I may well have made a mistake along the way, but my result does appear to check by the method described in aschepler's answer.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/5944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $$
\frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots
$$
[Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
"answer_count": 63,
"answer_id": 22
} |
Work out the values of a and b from the identity $x^2 - ax + 144 = (x-b)^2$ How do I solve the following question?
You are given the identity $x^2-ax+144 = (x-b)^2$
Work out the values of $a$ and $b$.
Question appears in AQA 43005/1H.
| Find the Value of $b$
$x^2-ax+144\equiv(x-b)^2$
Expand the right-hand side:
$x^2-ax+144\equiv{x^2-2bx+b^2}$
Find the coefficient $(x=0)$:
$b^2=144$
$\Longrightarrow b=\pm{\sqrt{144}}=\pm12$
Find the Value of $a$
Substitute $b=\pm12$ back into original equation:
$x^2-ax+144\equiv{x^2\pm{24x}+144}$
$\Longrightarrow -ax\equiv\pm{24x}$
Find the coefficient $(x=1)$:
$a=\pm{24}$
Answer
$a=24, b=12$ and $a=-24, b=-12$
Thanks to @J. M. and @Djaian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/9058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve trigonometric equation: $1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$ Dealing with a physics Problem I get the following equation to solve for $\alpha$
$1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$
Putting this in Mathematica gives the result:
$a==2 \text{ArcTan}\left[\frac{1-m}{1+m}\right]$
However I am unable to get this result myself. No matter what I try normal equation transformations or rewriting the equation with the complex e-Function ..., everything fails. Even going the other Direction from Mathematica's Solution to my original equation resulted in nothing sensible.
Any help of how to do do this transformation is very much appreciated.
Thanks in advance
| Added: As explained in the comments, certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of the same trigonometric function. Since all [direct] trigonometric functions of the simple angle can be expressed rationally as a function of the $\tan$ of the half-angle, such a conversion is adequate for these equations.
Since $$\cos \alpha =\frac{1-\tan ^{2}\frac{\alpha }{2}}{1+\tan ^{2}\frac{%
\alpha }{2}}$$
and
$$\sin \alpha =\frac{2\tan \frac{\alpha }{2}}{1+\tan ^{2}%
\frac{\alpha }{2}}$$
your equation
$$m\cos \alpha +\sin \alpha =1$$
is equivalent to
$$m-m\tan ^{2}\frac{\alpha }{2}+2\tan \frac{\alpha }{2}=1+\tan ^{2}\frac{\alpha }{2}.$$
One may set $x=\tan \frac{\alpha }{2}$ ($\alpha =2\arctan x$), and thus get
the quadratic equation
$$\left( 1+m\right) x^{2}-2x+1-m=0.$$
Its solutions are: $x=\frac{1}{m+1}\left( -m+1\right) $ (if $m\neq -1$) or $%
x=1$ (if $m=-1$), which gives
i) If $m\neq -1$,
$$\alpha =2\arctan x=2\arctan \frac{1-m}{m+1},$$
ii) If $m=-1,$
$$\alpha =2\arctan 1=\frac{\pi }{2}.$$
A different technique to solve a linear equation in $\sin \alpha $ and $\cos
\alpha $ is to use an auxiliary angle $\varphi $. If you set $m=\tan \varphi
$, your equation takes the form
$$\sin \alpha +\tan \varphi \cdot \cos \alpha =1$$
or
$$\sin (\alpha +\varphi )=\cos \varphi =\frac{1}{\pm \sqrt{1+\tan ^{2}\varphi
}}=\pm \sqrt{\frac{1}{1+m^{2}}},$$
and obtain
$$\alpha =\pm \arcsin \sqrt{\frac{1}{1+m^{2}}}-\arctan m.$$
Detailed derivation: from $m\cos \alpha +\sin \alpha =1$ and $m=\tan \varphi
$, we get
$$\sin \alpha +\tan \varphi \cdot \cos \alpha =1\iff\sin \alpha +\dfrac{\sin \varphi }{\cos \varphi }\cdot \cos \alpha =1$$
$$\iff\dfrac{\sin \alpha \cdot \cos \varphi +\sin \varphi \cdot \cos \alpha }{\cos
\varphi }=1\iff\dfrac{\sin \left( \alpha +\varphi \right) }{\cos \varphi }=1$$
$$\iff\sin \left( \alpha +\varphi \right) =\cos \varphi .$$
The identity
$$\cos \varphi =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}$$
can be obtained as follows
$$\sin ^{2}\varphi +\cos ^{2}\varphi =1\iff\dfrac{\sin ^{2}\varphi }{\cos ^{2}\varphi }+1=\dfrac{1}{\cos ^{2}\varphi }$$
$$\iff\tan ^{2}\varphi +1=\dfrac{1}{\cos ^{2}\varphi }\iff\cos ^{2}\varphi =\dfrac{1}{1+\tan ^{2}\varphi }.$$
Therefore
$$\sin \left( \alpha +\varphi \right) =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}\iff\alpha +\varphi =\arcsin \left( \pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}\right) $$
$$\iff\alpha +\arctan m=\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) \qquad (m=\tan \varphi,\ \varphi =\arctan m)$$
and finally
$$\alpha =\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) -\arctan m.$$
| {
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Any tricky method to solve this one? The question :
Prove that :$$ \text{ if } y = 2x^2 - 1,\text{ then } \biggl[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5}+ \cdots \biggr]$$ is equal to $$\frac{1}{2} \biggl[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \cdots \biggr]$$
Here I have modified the question, in the actual question (from my paper) there were four other options given,my approach was to reduce the first expression to $\frac{1}{2} \ln \biggl( \frac{y+1}{y-1} \biggr) $ and then trying to check for each options to find the match, now is there any other approach for this one ? Since it took me sometime for checking each options :( and the desired answer is at the last option!
| One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer.
| {
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How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$
how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$
Thank you
| You can use iterative integration of finite differences. The method works as follows. Assume your problem statement is:
$$\sum^n_{r=1} a_r = ?$$
Assume there is a function $S_n$ such that:
$$\sum^n_{r=1} a_r = S_n - S_1 + c$$
By finite difference we have:
$$a_n = \sum^n_{r=1} a_r - \sum^{n-1}_{r=1} a_r = (S_n - S_1 + c) - (S_{n-1} - S_1 + c) = S_n - S_{n-1} = \Delta S_n$$
Now assume we have a guess $T_n$ for $S_n$ such that some error $S'_n$ remains:
$$S_n = T_n + S'_n$$
The error itself can be expressed as a new integral since finite difference distributes over sums. We have:
$$a_n = \Delta S_n = \Delta T_n + \Delta S'_n$$
And hence we have a new sub problem:
$$\sum^n_{r=1} a'_r = \sum^n_{r=1} (a_r - \Delta T_r) = S'_n - S'_1 + c$$
Lets apply the method to the problem at hand, we have:
$$a_r = r\cdot 2^r$$
We can guess:
$$T_n = n\cdot 2^{n+1}$$
We arrive at:
$$a'_r = r\cdot 2^r - (r\cdot 2^{r+1} - (r - 1) \cdot 2^{r-1+1}) = - 2^r$$
We can guess again:
$$T'_n = - 2^{n+1}$$
We arrive at:
$$a''_r = - 2^r - (- 2^{r+1} - - 2^{r-1+1}) = 0$$
So the integration terminated, and the closed form integral is:
$$S_n = T_n + T'_n = n\cdot 2^{n+1} - 2^{n+1}$$
Using this for the sum we get:
$$\sum^n_{r=1} r\cdot 2^r = S_n - S_1 + c = n\cdot 2^{n+1} - 2^{n+1} - (1\cdot 2^{1+1} - 2^{1+1}) + c = 2\cdot (n - 1)\cdot 2^n + c$$
For $c$ we get:
$$\sum^1_{r=1} r\cdot 2^r = 2 = 0 + c$$
Saw this already in the 1980's in computer algebra systems (CAS), simpler than Gosper's method and different scope.
Bye
| {
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How to sum $\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \cdots$ quickly? The Problem:
$$\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \frac{16}{7 \cdot 8 \cdot 9 \cdot 10} + \cdots$$
Any smarter way to solve this may be within a minute or two?
I am adding my solution,If you are a student please don't read-on I find this particular problem to some sort of interesting so just try it once :-)
I started off with trying to find the $T_n$ th term
$$T_n = \frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}$$
$$ = \frac1{24} \times \biggl( \frac1{2n-1} + \frac3{2n+1} - \frac4{2n+2} \biggr)$$
$$ = \frac1{24} \times \biggl[\biggl( \frac1{2n-1} + \frac1{2n+1}\biggr) + 4 \times \biggl(\frac1{2n+1} - \frac1{2n+2} \biggr) \biggr]$$
After this this becomes really easy,
$$ S_\infty = \frac1{24} \times \biggl[( 1 - \frac13 + \frac13 - \frac15 + \frac15 + \cdots ) + 4 \times (\ln 2 - \frac12)\biggr]$$
$$ = \frac16 \log_e 2 - \frac1{24}$$
But as you can see this approach is a bit tedious and it took me some 30 minutes to reduce the $T_n$ to that form. Any other smart approaches?
| Problems like these are usually amenable to partial fraction decompositions.
From my experience with exam problems like these, once we do the partial fractions (which can be done quite quickly as described below), either we can do some sort of a telescoping sum, or write the resulting as a combination of well known series. In some cases, you can quickly and mechanically apply known estimates for well known series and find the sum. Of course, each problem is different and you might have to get creative.
Let us try that approach to your problem.
Step1: Splitting into Partial Fractions
You can try the following which should give you the partial fraction decomposition pretty quickly.
Set
$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{A}{2n-1} + \dfrac{B}{2n+1} + \dfrac{C}{2n+2}$$
Multiplying by $\displaystyle 2n-1$ and setting $\displaystyle n=1/2$ gives
$\displaystyle A = \dfrac{1/2}{2 \times 2 \times 3} = \dfrac{1}{24}$.
Multiply the original by $\displaystyle 2n+1$ and set $\displaystyle n=-1/2$. This gives us $B$.
$\displaystyle B = \dfrac{-1/2}{2 \times -2 \times 1} = \dfrac{1}{8}$
Multiply the original by $\displaystyle 2n+2$ and set $\displaystyle n = -1$. This gives us
$\displaystyle C = \dfrac{-1}{ 2 \times -3 \times -1 } = \dfrac{-1}{6}$.
Thus
$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{1}{24}\left(\dfrac{1}{2n-1} + \dfrac{3}{2n+1} - \dfrac{4}{2n+2}\right)$$
Apparently this is due to Heaviside: http://en.wikipedia.org/wiki/Heaviside_cover-up_method
Step2: Summing the series
Now to find the sum, you can use the following, which does not require any clever algebraic manipulations and so can be done quite quickly.
$$H_n = \sum_{j=1}^{n} \dfrac{1}{j} = \log n + \gamma + \mathcal{O}\left(\dfrac{1}{n}\right)$$
This gives us
$$\sum_{j=1}^{n} \dfrac{1}{2j - 1} = H_{2n} - \dfrac{1}{2} H_n = \log 2 + \frac{\log n}{2} + \gamma/2 + \mathcal{O}\left(\dfrac{1}{n}\right)$$
Thus your sum to $n$ terms is
$$\dfrac{1}{24}\left(\log 2 + \dfrac{\log n}{2} + \gamma/2 + 3\log 2 + 3 \dfrac{\log (n+1)}{2} - 3 + 3 \gamma/2 - 2\log (n+1) - 2\gamma +2\right) + \mathcal{O}\left(\dfrac{1}{n}\right)$$
Since $\log(n+1) = \log n + \mathcal{O}\left(\dfrac{1}{n}\right)$ we get that the sum to n terms is
$$\dfrac{1}{24}(4 \log 2 -1) + \mathcal{O}\left(\dfrac{1}{n}\right)$$
As $\displaystyle n \to \infty$,
the limit is
$$ \dfrac{1}{24}(4 \log 2 - 1)$$
Note: As J.M points out in the comments below, a very general method similar to the above can be found in Abramowitz and Stegun's book:
The technique you used for summing the
series is described in Abramowitz and
Stegun. (Harmonic numbers and
digamma functions are trivially
related).
Note1: As Mike points, out, there is an updated version of the book available here: http://dlmf.nist.gov/
| {
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How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$ While trying to answer this question, I got stuck showing that
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$
The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.
That got me thinking ...
If you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd, i.e. how to simplify nested cubic radicals of the form
$$\sqrt[3]{a+b\sqrt c}$$
| Apply the known denesting formula
$$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$
where $t$ satisfies $t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0$. Take the example $\sqrt[3]{26+15\sqrt{3}}$ in question and solve
$$t^3-\frac{26}5t^2+9t-\frac{26}5=\frac15(t-2)(5t^2-16t+13)=0$$
which yields $t=2$ and the denestation
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$
Other examples are listed below, along with their resolvent equations
\begin{align}
\sqrt[3]{7+5\sqrt{2}}= 1+\sqrt{2}&\>\>\>\>\>\>\>
t^3-\frac{21}5t^2+6t-\frac{14}5 =0,\>\>\>t=1\\
\sqrt[3]{90-34\sqrt{7}}= 3-\sqrt{7}&\>\>\>\>\>\>\>
t^3+\frac{135}{17} t^2+21 t+\frac{315}{17}=0 ,\>\>\>t=3 \\
\sqrt[3]{\frac{99}2+\frac{59}2\sqrt{\frac52}}= 3+\sqrt{\frac52}&\>\>\>\>\>\>\>
t^3-\frac{297}{59} t^2+\frac{15}2 t-\frac{495}{118}=0 ,\>\>\>t=3 \\
\sqrt[3]{25+10\sqrt{5}}= \frac52+\frac12\sqrt{5}&\>\>\>\>\>\>\>
t^3-\frac{15}{2} t^2+15 t-\frac{25}{2}=0 ,\>\>\>t=5\\
\sqrt[3]{70-22\sqrt{7}}= \sqrt[3]{49}\left(1-\frac{\sqrt{7}}7\right)&\>\>\>\>\>\>\>
t^3+\frac{105}{11} t^2+21 t+\frac{245}{11}=0 ,\>\>\>t=7\\
\end{align}
| {
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How to solve $\int\tan^3(x)\,dx$? How to solve $\int\tan^3(x)\,dx$ ? Do I use substitution?
| A simple way out is to write $\tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}$. Now let $\cos(x) = t$ and rewrite $\sin^2(x) = 1-t^2$.
The integral now becomes $\int \frac{\sin^3(x)}{\cos^3(x)} dx$.
$\cos(x) = t \Rightarrow -\sin(x)dx = dt$ and $\sin^2(x) = 1-t^2$.
Note the numerator of the integral $\sin^3(x) dx$ can be written as $\sin^2(x) \times \sin(x) dx = (1-t^2) (-dt) = (t^2-1) dt$
So the integral becomes $\int \frac{t^2-1}{t^3} dt = \int \frac{dt}{t} - \int \frac{dt}{t^3} = \log(t) + \frac{1}{2t^2} = \log(\cos(x)) + \frac{1}{2 \cos^2(x)} = \log(\cos(x)) + \frac{\sec^2(x)}{2}$
| {
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How to find solutions of linear Diophantine ax + by = c? I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?
I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.
$a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?
Thanks,
Chan
| Look can be deceiving. The integer solution to the equation $ax + by = c$ is anything but easy. Please endure a rather long derivation.
To make it more comprehensible let's first solve the equation for y:
\begin{align*}
ax + by = c\\
by = c - ax\\
y = \frac{c - ax}{b}
\end{align*}
To have an integer solution, $y$ must be an integer, and that is if $c - ax$ is a multiple of $b$, or $c - ax = -nb \iff ax = c + nb$. This has the same meaning as $ax \equiv c \: (mod \: n)$.
To continue, we need this Theorem 1:
The congruence $ax \equiv c \: (mod \: n)$ has a solution iff $gcd(a, n) \: | \: c$.
And this Lemma 2:
If $gcd(p, q) = 1$, then $px \equiv r \: (mod \: q)$ has a solution modulo $q$.
To keep this answer manageable, I would like to skip the proof of Theorem 1 and Lemma 2 (which can be found by googling). Just post a question and comment me if you encounter some trouble with them.
Let's define $d = gcd(a, n)$, and continue the derivation:
\begin{align*}
ax \equiv c \: (mod \: n)\\
ax = c + bn\\
\frac{a}{d} x = \frac{c}{d} + \frac{b}{d} n
\end{align*}
Now we want to switch $b$ and $n$ so $\frac{n}{d}$ could be seen more clearly as the modulo and continue it as following:
\begin{align*}
\frac{a}{d} x = \frac{c}{d} + b \frac{n}{d}\\
\frac{a}{d} x \equiv \frac{c}{d} \: (mod \: \frac{n}{d})
\end{align*}
Note that $\frac{a}{d}$ and $\frac{n}{d}$ from our derivation above is the $p$ and $q$ in the Lemma 2 respectively. Also note that as $d$ is $gcd(a, n)$, so $gcd(\frac{a}{d}, \frac{n}{d}) = 1$. Hence by Lemma 2:
\begin{align*}
\frac{a}{d} x \equiv \frac{c}{d} \: (mod \: \frac{n}{d})
\end{align*}
is our solution to equation $ax + by = c$.
As an example, let us solve $6x - 10y = 4 \iff 6x = 4 + 10y \iff 6x ≡ 4 \: (mod \: 10)$. $Gcd(6, 10) = 2$, and $2 \: | \: 4$, so by Theorem 1 that equation has a solution.
From our derivation, the solution is $\frac{6}{2} x = \frac{4}{2} \: (mod \: \frac{10}{2}) \iff 3x = 2 \: (mod \: 5)$.
By Lemma 2, we have a solution $modulo \: 5$. What it means is if we write the solution in $Z_5$, we would have:
\begin{align*}
\bar{3} \bar{x} = \bar{2}\\
\bar{x} = \bar{3}^{-1} \: \bar{2}
\end{align*}
As in $Z_5$, $\bar{3} \: \bar{2} = \bar{1} = \bar{3} \: \bar{3}^{-1}$, so $\bar{3}^{-1} = \bar{2}$ and we have:
\begin{align*}
\bar{x} = \bar{3}^{-1} \: \bar{2}\\
\bar{x} = \bar{2} \: \bar{2} = \bar{4}
\end{align*}
In $Z_5$, $\bar{x} = \bar{4} \iff x \equiv 4 \: (mod \: 5) \iff x = 4 + 5s \iff x = 4, 9, 13, \cdots$. You can check that indeed $x \equiv 4 \: (mod \:5)$ is the solution.
| {
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
| There are many equivalent ways of proving it.
First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.
(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)
Divisibility by $2$:
Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.
Equivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)
Divisibility by $3$:
$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.
$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.
Equivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.
Divisibility by $7$:
Note that $n$ can be either $7k$ or $7k \pm 1$ or $7k \pm 2$ or $7k \pm 3$.
If $n=7k$ or $n=7k \pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.
If $n=7k \pm 2$, then $n^2 = (7k \pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$.
Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
If $n=7k \pm 3$, then $n^2 = (7k \pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$.
Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
Hence, $7|(n^7-n)$.
Equivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.
Therefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\forall n \in \mathbb{N}$.
Hence, $42|(n^7-n)$, $\forall n \in \mathbb{N}$.
| {
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help solving an integral $\int\left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx$ $\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$
I am stuck on this part:
Let $u=\dfrac{x-1}{3-x}~\longrightarrow$ $~~\rm du=\dfrac {2}{(x-3)^2}\,\rm dx,$ which can be represented as
$\rm du=\dfrac{1}{3-x} - \dfrac{1-x}{(3-x)^2}\,\rm dx$
I cannot "see" how to get to this $2$ $\displaystyle\int \: \frac{(u)^\frac{1}{2}}{(u+1)^2} \:\rm dx$
after this part I know how to solve it;
I just wish someone would show me "step by step" this part
It seems it involves some sort of "leap" of thought; or is there a systematic way doing this using basic algebra?
Thanks.
| First note
$$
du=\frac{2}{(x-3)^2}dx
$$
which implies
$$
\frac{(x-3)^2}{2}du=dx.
$$
Also,
$$
u+1=\frac{x-1}{3-x}+\frac{3-x}{3-x}=\frac{2}{3-x}
$$
which implies
$$
(u+1)^2=\left(\frac{2}{3-x}\right)^2=\frac{4}{(x-3)^2}.
$$
Putting it together you have
$$
\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx = \int u^{1/2}\cdot\frac{(x-3)^2}{2}du = 2\int u^{1/2}\cdot\frac{(x-3)^2}{4}du
= 2\int \frac{u^{1/2}}{(u+1)^2}du.
$$
| {
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How does divisibility test using congruence work? In the book, it said:
Let $n = a_{k}10^{k} + a_{k-1}10^{k-1} + a_{k-2}10^{k-2} + ... + a_110 + a_0$
Then, because $10 \equiv 0 \pmod{2}$ it follows that $10^j \equiv 0 \pmod{2^j}$
What congruence property did they use in this case? Is that:
If $a \equiv b \pmod{k_1}$ and $c \equiv d \pmod{k_2}$ then,
$ab \equiv cd \pmod{k_1k_2}$ ?
I saw one property in the book, which is:
$a \equiv b \pmod{k}$ and $c \equiv d \pmod{k}$then,
$ab \equiv cd \pmod{k}$
But I really don't understand how this property relates to the one above it. Any idea?
| You you need to use this property $j$ times, since:
$10 \equiv 0 \mod{2} $ and $10 \equiv 0 \mod{2}$, then $10 \cdot 10 \equiv 10^2 \equiv 0 \mod{2^2}$
You know that $2|10$ so it must be that $2^2 | 10^2$ (factorization). Repeat again:
$10 \equiv 0 \mod{2}$ and $ 10^2 \equiv 0 \mod{2^2}$, then $10 \cdot 10^2 \equiv 10^3 \equiv 0 \mod{2^3}$
So in the end you get:
$10 \equiv 0 \mod{2}$ and $ 10^{j-1} \equiv 0 \mod{2^{j-1}}$, then $10 \cdot 10^{j-1} \equiv 10^j \equiv 0 \mod{2^j}$
| {
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Formula for sum of series $1+(1+x)+(1+x)^2+.... +(1+x)^n$ What is the formula to get sum of series
$$1+(1+x)+(1+x)^2+.... +(1+x)^n,$$
where $n$ is Integer and $x$ is Rational
| Simplified after the comment below by Yuval Filmus.
$$S=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}$$
is the sum of a geometric progression with ratio $1+x$ and $n+1$ terms, the first of which is $1$. The usual way to derive $S$ is to multiply it by the ratio
$$(1+x)S=(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}+(1+x)^{n+1}$$
and subtract from $S$:
$$\begin{eqnarray*}
S-(1+x)S &=&1+\left( (1+x)-(1+x)\right) +\left(
(1+x)^{2}-(1+x)^{2}\right) \\
&&+\ldots +\left( (1+x)^{n}-(1+x)^{n}\right) -(1+x)^{n+1} \\
&=&1-(1+x)^{n+1}.
\end{eqnarray*}$$
Solving for $S$, we get
$$S=\frac{(1+x)^{n+1}-1}{x}\qquad (x\neq 0).$$
For $x=0$ the sum of the series is
$$S=1+1+1^{2}+1^{3}+\ldots +1^{n}=1+n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A congruence in a quadratic number field
If $\displaystyle L = \frac{3+\sqrt{-3}}{2}$,
and if $x\equiv 1\pmod{L}$,
show that $x^3\equiv 1\pmod{L^4}$.
I have already shown that if $x\equiv 1\pmod{L}$,
then $x^3\equiv 1\pmod{L^3}$.
Thanks.
| Let $w = \frac{-1 + \sqrt{-3}}{2}$. Your question is to show that if $x \equiv 1 \pmod{2+w}$, then $x^3 \equiv 1 \pmod{(2+w)^4}$.
By direct expansion, we see that $(2+w)^4 = 9w$. So it suffices to show $x^3 \equiv 1 \pmod{9}$. Now check the following,
*
*$x \equiv 1 \pmod{2+w}$ is equivalent to $x = a+bw$, such that $a+b \equiv 1 \pmod{3}$, where $a,b$ are integers.
*$(a+bw)^3 = a^3+b^3-3ab^2 + 3ab(a-b)w$.
*Check that each coefficient is divisible by 9, by looking at two cases: either a or b are divisible by 3, or that $a \equiv b \equiv -1 \pmod{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\frac{\mathrm d^{100}}{\mathrm d x^{100}}\frac{x^2+1}{x^3-x}=$? $$f(x)=\frac{x^2+1}{x^3-x}$$
$$f^{(100)}(x)=?$$
I tried differnetiating once and twice, but did not see any pattern emerging and can't guess what the 100th derivative should be.
EDIT
so decomposing this as $$f(x)=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}
$$ does the job. Thanks for the hints! (edit: Sivaram has a complete calculation) Although a similar approach would greatly simplify this (next) problem can someone tell me what is wrong with my approach
My usual line of attack is to use Taylor expansion. For example the next problem in the same list asks for the $100^{th}$ derivative of $$\frac{1}{x^2-3x+2}$$ at $x=0$ within 10% relative error.
NOTE:The above is a mistype, the following attempt is for $\frac{1}{x^2+3x+2}$. A better general approach, which is what I was looking for is described in the answer posted below.
I know I can expand in a Maclaurin series $$\frac{1}{x^2+3x+2}=\frac{1}{2} (1+\frac{x^2+3x}{2} + (\frac{x^2+3x}{2})^2 +\cdots)$$
After taking 100 derivatives I would be left to differentiate the following.
$$\frac{1}{2}((\frac{x^2+3x}{2})^{50}+(\frac{x^2+3x}{2})^{51}+\cdots)$$
$$=\frac{1}{2}\left(\frac{\sum_{k=0}^{50}{{50}\choose{k}}3^{50-k} x^{50+k}
}{2^{50}}+\frac{\sum_{k=0}^{51}{{51}\choose{k}}3^{51-k} x^{51+k}
}{2^{51}}+\cdots\frac{\sum_{k=0}^{100}{{100}\choose{k}}3^{100-k} x^{100+k}
}{2^{100}}\right)$$
Because anything on either side of these values would disappear when i take the hundreth derivative at $x=0$ . And it is also easy to sea that I will get exactly one term from each of the sums, so I get an answer,
$$=100!\sum_{k=0}^{50}\frac{3^{2k}}{2^{50+k}}$$
Which is wrong, well because the answer is too huge and Im to find a number within 10%. Can someone tell me where I went wrong, and if there is a cleaner way to approach these problems.
| $$\frac{x^2+1}{x^3-x} = -\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-1} = -x^{-1} + (x+1)^{-1} + (x-1)^{-1}$$
$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$
Hence, the $n^{th}$ derivative is
$$\frac{(-1)^{n+1} n!}{x^{n+1}} + \frac{(-1)^n n!}{(x-1)^{n+1}} + \frac{(-1)^n n!}{(x+1)^{n+1}} = (-1)^n n! \times \left( \frac{-1}{x^{n+1}} + \frac{1}{(x-1)^{n+1}} + \frac{1}{(x+1)^{n+1}}\right)$$
Similarly,
$$\frac{1}{x^2-3x+2} = \frac{1}{x-2} - \frac{1}{x-1}$$
Hence, the $n^{th}$ derivative is
$$\frac{(-1)^n n!}{(x-2)^{n+1}} - \frac{(-1)^n n!}{(x-1)^{n+1}} = (-1)^n n! \times \left( \frac{1}{(x-2)^{n+1}} - \frac{1}{(x-1)^{n+1}}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| I would like to present you a method based on Daners' derivation of $\zeta(2)$ (which I summarised on MathStackexchange in here couple years ago).
We start defining (for all $n \in \mathbb{N}_0$)
\begin{align}
A_n & =\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x,\\
B_n& =\int_0^{\pi/2}x^2\cos^{2n}x\;\mathrm{d}x,\\
C_n& =\int_0^{\pi/2}x^4\cos^{2n}x\;\mathrm{d}x
\end{align}
and let $\beta_n = B_n/A_n$ and $\gamma_n = C_n/A_n$.
The first integral for $A_n$ is well known and follows a reccurence relation
$$A_{n}=\frac{2n-1}{2n}A_{n-1}\tag{1}.$$
By applying per partes on the $A_n$ integral twice, we get:
$$A_n=\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x=x\cos^{2n}x\bigg{|}_0^{\pi/2}-\frac{x^2}{2}(\cos^{2n}x)'\bigg{|}_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}x^2(\cos^{2n}x)''\;\mathrm{d}x$$
The first two terms vanish, so only the integral remains and since $(\cos^{2n}x)''=2n(2n-1)\cos^{2n-2}x-4n^2\cos^{2n}x$, we get for $n\geq 1$:
$$A_n=(2n-1)nB_{n-1}-2n^2B_{n}\tag{2}$$
Similarly, applying per partes twice on $B_n$ instead, we get
$$B_n=\frac16 (2n-1)nC_{n-1}-\frac13 n^2C_{n}\tag{3}$$
Inserting $(1)$ into $(2)$ and $(3)$ and rearranging, we get a following recurence reltions
$$\frac{1\cdot 2}{4}\frac{1}{n^2}=\beta_{n-1}-\beta_n\tag{4}$$
$$\frac{3\cdot 4}{4}\frac{\beta_n}{n^2}=\gamma_{n-1}-\gamma_n\tag{5}$$
Summing these up, we get, by the telescoping property
$$\sum_{l=1}^k\frac{1}{2l^2}=\beta_0-\beta_k \qquad \text{and} \qquad \sum_{k=1}^n\frac{3\beta_k}{k^2}=\gamma_0-\gamma_n. \tag{6}$$
Inserting one sum into another (expressing $\beta_k$), we get
$$3\beta_0\sum_{k=1}^n\frac{1}{k^2} - \frac32\sum_{k=1}^n\frac{1}{k^2}\sum_{l=1}^k\frac{1}{l^2}=\gamma_0 - \gamma_n. \tag{7}$$
Note that in general
$$2\sum_{k=1}^n a_k\sum_{l=1}^k a_l = 2\sum_{1\leq l
\leq k \leq n}a_k a_l = \left(\sum_{k=1}^n a_k\right)^2+\sum_{k=1}^n a_k^2,$$
substituing for $a_k = 1/k^2$ into $(7)$ and rewriting $\sum_{k=1}^n 1/k^2$ as $2\beta_0 - 2\beta_n$ everywhere, we get
$$\bbox[10px,#ffd]{\sum_{k=1}^n \frac{1}{k^4} = 4\beta_0^2 - 4\beta_n^2 - \frac{4}{3}\gamma_0 + \frac{4}{3}\gamma_n.}\tag{8}$$
However, using the inequalities $x^4<(\pi/2)^2 x^2$ and $\frac{2x}{\pi}<\sin x$ valid on $(0,\frac{\pi}{2})$ and by $(1)$, we get
$$0<\gamma_{n-1} <\left(\frac{\pi}{2}\right)^2\beta_{n-1} < \frac{\int_0^{\pi/2}\sin^2x\cos^{2n-2}x}{\int_0^{\pi/2}\cos^{2n-2}x}=\frac{A_{n-1} - A_n}{A_n} = \frac{1}{2n}.$$
Applying the squeeze theorem, we get
$$\lim_{n\rightarrow \infty} \beta_n = \lim_{n\rightarrow \infty} \gamma_n = 0$$
and hence, taking the limit of $(8)$,
$$\zeta_4 = \lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{1}{k^4} = 4\beta_0^2-\frac{4}{3}\gamma_0 = 4\left(\frac13 \frac{\pi^2}{2^2}\right)^2-\frac{4}{3} \frac{1}{5}\frac{\pi^4}{2^4} = \frac{\pi^4}{90}$$
This finishes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality understanding My textbook says that:
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n<e^n
$$
But I do not understand this. Can you please enlighten me?
Edit: How do you show that
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n
$$
| For the first part
note that each term is
$$\begin{align}
\left(\frac{1+k}{k}\right)^k =& \frac{(1+k)^k}{k^k}\\
=&\frac{1}{k}\frac{(1+k)^k}{k^{k-1}}\\
=&\frac{1}{k}\frac{(1+k)^k}{(1+(k-1))^{k-1}}
\end{align}$$
so every denominator cancels nicely with the previous numerator:
$$ \begin{align}\frac{1}{n!}\left(\frac{(1+1)^1}{(1+0)^0}\frac{(1+2)^2}{(1+1)^1}\frac{(1+3)^3}{(1+2)^2}\cdots\frac{(1+n)^n}{n^{n-1}}\right)\\
=\frac{1}{n!}(1+n)^n\end{align}$$
For the second part
note that $f(k) = \left(1+\frac{1}{k}\right)^k < e$ for every $k>0$ since $f(k)$ is a monotonically increasing function and $\lim_{k\to\infty}f(k) = e$, so that the product of any $n$ of them is less than $ee\cdots e = e^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Uniqueness of the infinite expansion in base $b$ If $b>1$ is an integer, is well know that the numbers $x\in (0,1]$, can be written as
$$x = \sum_{k=1}^{\infty} \frac{a_k}{b^k}$$
for some integers $a_k \in \{0,1,\ldots ,b-1\} $. When $x=\frac{1}{b^n}$, for some $n\in \mathbb{N}$, there is two representations. One of them is always $0$ beginning with some index, that is, is finite. How can I prove that the infinite representation is unique?
| If I recall correctly you can prove that the partial sums of the expansion satisfy the following inequality
$$0 \leq x - \sum_{k = 1}^{n} \frac{a_k}{b^k} < \frac{1}{b^n}$$
and from this inequality you can just prove by induction that if you have two expansions for $x$ then they are actually the same by proving that all the "digits" $a_k$ agree.
Edit:
The inequality can be proved as follows:
$$x - \sum_{k = 1}^{n} \frac{a_k}{b^k} = \sum_{k = n+1}^{\infty} \frac{a_k}{b^k} \leq \sum_{k = n + 1}^{\infty} \frac{b - 1}{b^k} = (b - 1) \sum_{k = n+1}^{\infty} \frac{1}{b^k}$$
$$= (b-1) \left ( \frac{1}{b^{n+1}} + \frac{1}{b^{n+2}} + \cdots \right ) = \frac{(b-1)}{b^{n+1}} \left ( 1 + \frac{1}{b} + \frac{1}{b^{2}} + \cdots \right ) = \frac{b-1}{b^{n+1}}\frac{1}{1 - \frac{1}{b}} = \frac{1}{b^n}$$
Then to prove the uniqueness by induction you use the inequality. So suppose that you have two representations for $x$, say
$$ x = \sum_{k = 1}^{\infty} \frac{a_k}{b^k} = \sum_{k = 1}^{\infty} \frac{c_k}{b^k}$$
Then you want to prove that $a_k = c_k$ for every $k \geq 1$. For example for $n = 1$ you have the two inequalities
$$0 < x - \frac{a_1}{b} < \frac{1}{b} \quad 0 < x - \frac{c_1}{b} < \frac{1}{b}$$
Then by subtracting them you get
$$\frac{-1}{b} < \frac{a_1}{b} - \frac{c_1}{b} < \frac{1}{b} \implies -1 < a_1 - c_1 < 1 \implies a_1 = c_1$$
since $a_1 - c_1$ is an integer. I'll leave the inductive step to you.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to reduce congruence power modulo prime? If I have a congruence equation, says
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be
$$x^{14}x - x^7x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^2x - x.x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^3 - x^4 - 4x - 3 \equiv 0 \pmod{7}$$
?
Thanks,
| Not quite. Look for example at the congruence $x^6 \equiv 0 \pmod{7}$. If one assumes that $x^6 \equiv 1$, things go bad. In this case it is easy to spot that there is a problem, but perhaps in a more complicated setting one might miss it.
I would advise using the fact that $x^7 \equiv x \pmod{7}$, basically a variant of Fermat's Theorem that holds always, not just almost always.
| {
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For what $a$ and $b$ is $9x^4-12x^3+28x^2+ax+b$ a perfect square? If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)
| Even though it doesn't affect the final outcome in this problem, I think it's better long-term practice not to assume that the leading term is positive. This emphasizes that there are two square roots to a square polynomial, differing by sign, just like with integers.
Suppose $9x^4-12x^3+28x^3+ax+b=g(x)^2$ for some polynomial $g(x)$. Because the degree of $9x^4-12x^3+28x^3+ax+b$ is 4, we have $4=\deg(g(x)^2)=2\deg(g(x))$, so that $g$ is a degree 2 polynomial.
Write $g(x)=cx^2+dx+e$. Then
$$9x^4-12x^3+28x^3+ax+b=(cx^2+dx+e)^2=(c^2)x^4+(2cd)x^3+(2ce+d^2)x^2+(2de)x+e^2.$$
Thus $c^2=9$, $2cd=-12$, $2ce+d^2=28$, $2de=a$, and $e^2=b$.
$c^2=9$ implies $c=3$ or $c=-3$.
Combined with $2cd=-12$, this implies that $d=2$ or $d=-2$.
Combined with $2ce+d^2=28$, this implies that $e=4$ or $e=-4$.
If $c=3$, then $d=-2$ and $e=4$. If $c=-3$, then $d=2$ and $e=-4$. Thus,
$$g(x)=\pm(3x^2-2x+4),$$
and $a=2de=-16$ and $b=e^2=16$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are a , b and c? $$y = ax^2 + bx + c$$
which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!
My main question is this solvable? I am doubtful?
| This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed using KmPlot. The figure should give you an intuitive idea of how to go about solving.
*
*The Green line is $y=2x+3$.
*The Blue line is $y=x$.
If the line $y=2x+3$ and the parabola $y=ax^{2}+bx+c$ are going to be tangent at a given point then their slopes are equal. Let's find that out. Slope of line $y=2x+3$ is $2$ and we have $$2 = \frac{dy}{dx} = 2ax+1$$ So you have $x=\frac{1}{2a}$. Also we have
\begin{align*}
2x+3 & = ax^{2} + x
\end{align*}
which says that $$2 \times \frac{1}{2a} + 3 = a \times \frac{1}{4a^{2}} + \frac{1}{2a}=\frac{3}{4a}$$ From this we have $$\frac{1}{a} -\frac{3}{4a} = -3 \Longrightarrow a=-\frac{1}{12}$$
This is for the value $a=-\frac{1}{3}$
This is for the value $a=-\frac{1}{7}$.
| {
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Efficient way to find $a$ in $c = 6a\mod n$ Given $c$ and $n$ in $c = 6a\mod n$, how can I find the lowest positive integer value for $a$?
I could find it iteratively by rewriting as $\displaystyle a = \frac{c + xn}{6}$ and increasing $x$ until $a$ is an integer, but I would prefer a more efficient solution. Is there another way to do this?
Edit: Combined two variables.
Here is an example: $3 = 6a\mod 11$
| First, general comments.
The congruence
$$Ax\equiv c \pmod{n}$$
has a solution $x$ if and only if $\gcd(A,n)|c$. If it does, the standard method is to use the extended Euclidean algorithm to find integers $r$ and $s$ such that
$$d = Ar + ns$$
(where $d=\gcd(A,n)$); then multiply by $\frac{c}{d}$ (an integer, since $d|c$) to get
$$c = A\left(\frac{rc}{d}\right) + n\left(\frac{sc}{d}\right).$$
So one solution is $a=\frac{rc}{d}$. Finding the remainder of $\frac{rc}{d}$ modulo $n$ can be done by direct division. If $d\gt 1$, then you still need to make sure that there is no smaller division by comparing with $\frac{n}{d}$, since given any particular solution to $Ax\equiv c\pmod{n}$, then $x\pm\frac{n}{d}$ is also a solution.
The extended Euclidean algorithm is pretty efficient, but even so there are a few tweaks that can be done to make it even more so (e.g., instead of always requiring the remainder to be positive, you can require that the remainder lie between $-\frac{n}{2}$ and $\frac{n}{2}$). The other steps are pretty efficient as well.
Added. If, as you indicate in comments, the modulus is prime, you can also use Gauss's method. Write the fraction you want (in this case, $\frac{c}{A}$), and multiply numerator and denominator by the smallest $k$ such that $Ak\gt n$ (or such that $Ak$ has a small remainder in absolute value modulo $n$); then reduce and repeat until you get to a denominator equal to $1$.
Now, your particular issue.
However, the case of $A=6$ and prime modulus is particularly simple, since every prime other than $3$ is either congruent to $1$ or to $-1$ modulo $6$. So write $n=6k\pm 1$, and use $k$:
For the example you write, note that $11 = 2\times 6 - 1$:
$$a\equiv \frac{3}{6} = \frac{2\times 3}{2\times 6} = \frac{6}{12} \equiv \frac{6}{1} \equiv 6\pmod{11}.$$
For another, say you want to solve
$$7\equiv 6a\pmod{43}.$$
Since $43 = 6\times 7 + 1$, then we have
$$a \equiv \frac{7}{6} =\frac{7\times 7}{7\times 6} \equiv \frac{49}{42}\equiv \frac{6}{-1}= -6 \equiv 37\pmod{43}.$$
Etc.
In summary. For a prime modulus $n\gt 3$, divide $n$ by $6$, and write $n=6k\pm 1$. If $n=6k-1$, then $a\equiv ck\pmod{n}$, so then you just need to divide $ck$ by $n$ to get the remainder. If $n=6k+1$, then $a\equiv -ck$, so find the remainder for $-ck$ instead.
| {
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What is the length of a sine wave from $0$ to $2\pi$? What is the length of a sine wave from $0$ to $2\pi$? Physically I would plot $$y=\sin(x),\quad 0\le x\le {2\pi}$$ and measure line length.
I think part of the answer is to integrate this:
$$
\int_0^{2\pi} \sqrt{ 1 + (\sin(x))^2} \, {\rm d}x
$$
Any ideas?
| \begin{align}
\int_0^{2\pi}\sqrt{1+\cos^2(x)} dx
&= 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx \\
&= 4 \int_0^{\pi/2}\sqrt{1+\dfrac{1+\cos(2x)}2 }dx\\
&= 4 \sqrt{\dfrac{3}2} \int_0^{\pi/2} \sqrt{1+\dfrac{\cos(2x)}3} dx \\
&= 2\sqrt6 \int_0^{\pi/2} \sum_{n=0}^\infty a_n (\dfrac{\cos(2x)}3)^n dx \tag1
\end{align}
where $$\sqrt{1+t}=\sum_{n=0}^\infty a_n t^n, \mbox{ and }\ a_n = \frac{(-1)^{n+1} (2n-3)!!}{n! 2^n}$$
Let $I_n = \int_0^{\pi/2} \cos^n(2x) dx$, then $I_n = 0\ $ if $\ n\ $ is odd, and $ I_n = \dfrac{\pi}2 b_{n/2} $ if $\ n\ $ is even,
where $b_k = \dfrac{(2k-1)!!}{k! 2^k}$.
The equation $(1)
= 2\sqrt6 \sum\limits_{n=0}^\infty \dfrac{a_n}{3^n} I_n
= 2\sqrt6 \sum\limits_{k=0}^\infty \dfrac{a_{2k}}{3^{2k}} \dfrac{\pi}2 b_k
= \sqrt6 \pi \sum\limits_{k=0}^\infty c_k \tag2$
where $c_k = a_{2k} b_k 3^{-2k}
= -\dfrac{(4k-3)!!}{(2k)! 2^{2k}} \dfrac{(2k-1)!!}{k! 2^k} 3^{-2k}
= -\dfrac{(4k-3)!!}{(k!)^2 2^{4k} 3^{2k}}
= -\dfrac{\binom{4k-3}{2k-1}\binom{2k-1}{k}}{2^{6k-2} 3^{2k} k} $.
Note $c_0 = -(-3)!! = -\dfrac{1(-1)(-3)!!}{1(-1)} = 1$.
Additionally, \begin{align}
\int_0^{2\pi}\sqrt{1+\cos^2(x)} dx & = 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx = 4 \int_0^{\pi/2}\sqrt{2-\sin^2(x)}dx\\
& = 4 \sqrt2 \int_0^{\pi/2} \sqrt{1-\dfrac{\sin^2(x)}2} dx = 4 \sqrt2 E\left(\dfrac1{\sqrt2}\right) \tag3
\end{align}
where $$E(k) = \displaystyle \int_0^{\pi/2} \sqrt{1-k^2 \sin^2(x)} dx
= \dfrac{\pi}2 \sum_{n=0}^{\infty} \left(\dfrac{\dbinom{2n}n}{4^n} \right)^2 \dfrac{k^{2n}}{1-2n}$$ and is referred to as the complete elliptic integral of second kind.
$ E\left(\dfrac1{\sqrt2}\right)
= \dfrac{\pi}2 \sum\limits_{n=0}^{\infty} d_n \tag4$
where $d_n
= \left[ \dfrac{(2n-1)!!}{n!2^n} \right]^2 \dfrac{1}{(1-2n)2^n}$.
By $(2)$ and $(3)$, $E\left(\dfrac1{\sqrt2}\right)
= \dfrac{\sqrt3 \pi}4 \sum\limits_{n=0}^{\infty} c_n \tag5$
$(4) = (5)$.
Because $c_n$ and $d_n \to 0$ as $n\to \infty$, the convergence rate of $c_n$ and $d_n$ are given by
\begin{align}
\lim_{n\to\infty} \left| \dfrac{c_n}{c_{n-1}} \right|
&= \lim_{n\to\infty} \dfrac{(4n-3)(4n-5)}{n^2 2^4 3^2} = \dfrac 1 9\
\mbox{ and }\\
\lim_{n\to\infty} \left| \dfrac{d_n}{d_{n-1}} \right|
&= \lim_{n\to\infty} \dfrac{(2n-1)(2n-3)}{n^2 8} = \dfrac 1 2,\
\mbox{ respectively. }
\end{align}
We obtain the values of $(4)$ and $(5)$ by the power series expansion of
$\sqrt{1-\dfrac{\sin^2(x)}2}$ and $\sqrt{1+\dfrac{\cos(2x)}3}$, respectively. Hence using $(5)$, we can obtain accurate estimates faster.
| {
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"timestamp": "2023-03-29T00:00:00",
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Bounded Sequences I came across the following problems during the course of my self-study of real analysis:
Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ is unbounded.
I know a sequence $(x_n)$ is bounded if there exists a positive number $K$ such that $|x_n| \leq K$ for all $n$. So suppose for contradiction that it is bounded. Maybe we can define sequences $a_n = x_n-1$, $b_n = a_n-\frac{1}{2}$, $c_n = b_n- \frac{1}{3} \dots$ and try to come up with a contradiction?
Show that the sequence $(x_n)$ defined by $x_1 = x$, $x_{n+1} = x_{n}+ 1/x_n$ is unbounded.
Suppose for contradiction that $(x_n)$ is bounded by $K$ for all $n$. Then show that there is some $K' < K$ which is also an upper bound?
Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2!}+ \frac{1}{3!} + \dots + \frac{1}{n!}$ is bounded above by $2$.
So there is some relationship between $n!$ and $2^{n-1}$. I think $n! \geq 2^{n-1}$ and we can prove this by induction on $n$? So $x_n \leq 1+1+ \frac{1}{8} + \dots + \frac{1}{2^{n-1}}$?
| *
*The simplest way to show that a sequence is unbounded is to show that for any $K\gt 0$ you can find $n$ (which may depend on $K$) such that $x_n\geq K$.
The simplest proof I know for this particular sequence is due to one of the Bernoulli brothers Oresme. I'll get you started with the relevant observations and you can try to take it from there:
Notice that $\frac{1}{3}$ and $\frac{1}{4}$ are both greater than or equal to $\frac{1}{4}$, so
$$\frac{1}{3}+\frac{1}{4}\geq \frac{1}{4}+\frac{1}{4} = \frac{1}{2}.$$
Likewise, each of $\frac{1}{5}$, $\frac{1}{6}$, $\frac{1}{7}$, and $\frac{1}{8}$ is greater than or equal to $\frac{1}{8}$, so
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7} + \frac{1}{8} \geq \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} = \frac{1}{2}.$$
Now look at the fractions $\frac{1}{n}$ with $n=9,\ldots,16$; compare them to $\frac{1}{16}$; then compare the fractions $\frac{1}{n}$ with $n=17,\ldots,32$ to $\frac{1}{32}$. And so on.
See what this tells you about $x_1$, $x_2$, $x_4$, $x_8$, $x_{16}$, $x_{32}$, etc.
*Your proposal does not work as stated. For example, the sequence $x_n = 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}$ is bounded by $K=10$; but it's also bounded by $K=5$. Just because you can find a better bound to some proposed upper bound doesn't tell you the proposal is contradictory. It might, if you specify that you want to take $K$ to be the least upper bound of the sequence. Even so, it's hard to establish that a sequence is unbounded that way. (Note also that you haven't really defined the sequence very well: it is undefined for $x=0$, though that is the only problem.)
To get you started: Show that if you start the sequence with $-x$ instead of $x$, then you just get the same sequence multiplied by $-1$. That is, if you fix $x\neq 0$, and you let $y_1=-x$, $y_{n+1}= y_n + (1/y_n)$, then $y_k = -x_k$; so the sequence $(x_n)$ is bounded if and only if the sequence $y_k$ is bounded, and so you may assume $x\gt 0$.
Then show that if $0\lt x\lt 1$, and you let $y_1 = \frac{1}{x}$, $y_{n+1} = y_n+(1/y_n)$, then $y_k=x_k$ for $k\geq 2$; so you may assume that $x\geq 1$.
Now you have that the sequence is increasing. If it were bounded, it would converge, say to $L\gt 0$. Then
$$L = \lim_{n\to\infty}x_n = \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\left(x_n + \frac{1}{x_n}\right) = \lim_{n\to\infty}x_n + \frac{1}{\lim_{n\to\infty}x_n} = L+\frac{1}{L}.$$
I think that's a very big problem for $L$...
*Yes, if you can prove that $n!\gt 2^{n-1}$ for all $n\geq 1$, then you can bound your sequence by a sequence of powers of $\frac{1}{2}$; if you can show that sequence is bounded, you'll be done. And, yes, you can prove the inequality in question by induction on $n$. It's very simple to do. But you messed up your computations later (that second $1$ should be a $\frac{1}{2}$). If $n!\geq 2^{n-1}$, then
$$\begin{align*}
x_n &= 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots + \frac{1}{n!} \\
&\leq 1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{n-1}}\\
&= \frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}}\\
&= \frac{2^n-1}{2^{n-1}}\\
& = 2 - \frac{1}{2^{n-1}}.\end{align*}$$
Almost done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
| Another method by using telescoping sum :-
We know $(a+b)^3-a^3-b^3=3ab(a+b)$ , take
$a=k-1 , b=2$ , then $a+b=k+1$ and $(k+1)^3-(k-1)^3-2^3=6(k-1)(k+1)=6k^2-6$ ,
hence $(k+1)^3-(k-1)^3-8+6=(k+1)^3-k^3+k^3-(k-1)^3-2=6k^2$ , taking sum over $k$
from $1$ to $n$ we get , $\sum_{k=1}^n [(k+1)^3-k^3] + \sum_{k=1}^n [k^3-(k-1)^3] -\sum_{k=1}^n 2 = 6 \sum_{k=1}^nk^2$ , the first
sum on the left hand side is telescoping resulting in $(n+1)^3-1$ , the second sum is also telescoping resulting in $n^3-(1-1)^3=n^3$ , and the third sum is simply $2n$ , hence
$6 \sum_{k=1}^nk^2=(n+1)^3-1+n^3-2n=2n^3+3n^2+n=n(n+1)(2n+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| Since the inequality is homogeneous we may WLOG assume that $xyz=1$ and we have to prove the inequality $(x+y+z)^3 \geq 27$ or $x+y+z \geq 3.$ But the last inequality we obtain immediately from $AM \geq GM$ under the assumption $xyz=1.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
| The result is also a consequence of the Chebyshev (sum) Inequality.
This is a quite useful result, not the least for contest problems! It can be proved by using the Rearrangement Inequality, which is also useful to know. For completeness, we state the full result, though only half of it is needed here.
Theorem: (Chebyshev Inequality) Suppose that $a_1 \le a_2 \le \cdots \le a_n$ and $b_1 \le b_2 \le \cdots \le b_n$.
Let $m=a_1b_n + a_2b_{n-1} +\cdots + a_nb_1$ and $M=a_1b_1+a_2b_2+\cdots + a_nb_n$. Then
$$nm \le (a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots +b_n) \le nM$$
with equality only happening when all the $a_i$ are equal or all the $b_i$ are equal.
To apply the result to our problem, we can without loss of generality assume that
$a \le b \le c$. Let $a_1=a$, $a_2=b$, $a_3=c$, $b_1=a^2$, $b_2=b^2$, and $b_3=c^2$.
Then the conditions of Chebyshev's Inequality are met, and $M=a^3+b^3+c^3$. We can then read off our inequality from the Chebyshev Inequality.
Note that the Chebyshev Inequality yields an immediate generalization of the $3$ variable inequality of the question to $n$ variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
} |
finding remainder by dividing a sum Suppose I am dividing 4^30-2^50 by 5.
I do understand that 4^30 will get converted to 2^60
I could also find the series of remainders for 2^x. It comes out to be 2 4 3 1
By that, I could find (4^30) % 5 = 1 and (2^50) % 5 = 4. But how do I find the combined mod?
i.e. [(4^30)-(2^50)] % 5 = ?
Thank you.
P.S. Please note that by %, I mean mod operator.
| If $a\equiv b\pmod m$ and $c\equiv d\pmod m$, then $a\pm c\equiv b\pm d\pmod m$.
Since $4^{2k+1}\equiv 4\pmod 5$ and $4^{2k}\equiv 1\pmod 5$, $4^{30}=4^{2\cdot 15}\equiv 1\pmod 5$.
Also, $2^{4k}\equiv 1\pmod 5$, $2^{4k+1}\equiv 2\pmod 5$, $2^{4k+2}\equiv 4\pmod 5$ and $2^{4k+3}\equiv 3\pmod 5$. Hence $2^{50}=2^{4\cdot 12+2}\equiv 4\pmod 5$.
Then $4^{30}-2^{50}\equiv 1-4\pmod 5\equiv -3\pmod 5$. But, since $-3\equiv 2\pmod 5$, we have $4^{30}-2^{50}\equiv 2\pmod 5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.
Prove that,
$$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$
| There is a theorem$^1$ that is worth knowing (I used it in this answer to this question) that states:
All direct trigonometric functions of $A$ ($2\alpha$) can be expressed rationally in terms of the tangent of $\frac{A}{2}$ ($\alpha$).
Combining $\sin A=2\sin \frac{A}{2}\cdot \cos \frac{A}{2}$ and $\cos ^{2}
\frac{A}{2}+\sin ^{2}\frac{A}{2}=1$, we get (for $\cos \frac{A}{2}\neq 0$)
$$\sin A=\frac{2\sin \frac{A}{2}\cdot \cos \frac{A}{2}}{\cos ^{2}\frac{A}{2}
+\sin ^{2}\frac{A}{2}}=\frac{2\tan \frac{A}{2}}{1+\tan ^{2}\frac{A}{2}};
\qquad (1)$$
and from $\cos A=\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}$, for $\cos \frac{A}{2}\neq 0$,
$$\cos A=\frac{\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}}{\cos ^{2}\frac{A}{2}
+\sin ^{2}\frac{A}{2}}=\frac{1-\tan ^{2}\frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}.
\qquad (2)$$
To prove the identity multiply the LHS numerator and denominator by $1+\tan
^{2}\frac{A}{2}$ and use $(1)$ and $(2)$:
$$\begin{eqnarray*}
\frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=&\frac{1+2\tan \frac{A}{2}-1+\tan
^{2}\frac{A}{2}}{1+2\tan \frac{A}{2}+1+\tan ^{2}\frac{A}{2}} \\
&=&\frac{2\tan ^{2}\frac{A}{2}+2\tan \frac{A}{2}}{2+2\tan \frac{A}{2}}=\tan
\frac{A}{2}.
\end{eqnarray*}$$
--
$^1$ J. Calado, Compêndio de Trigonometria, 1967.
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