Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$ has all roots real Given the equation:
$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$
For what values of $m$ are all the roots real?
I've rewritten the equation as: $x^4-6x^3+11x^2-6x-m=0$
I'm quite sure this is done with Vieta's but didn't really figure out yet what should
I aim to get out of it in order to get $m \in [-1,\frac{9}{16}]$ which is the correct answer .
| Start from $x^4-6 x^3+11 x^2-6 x-m=0$ and substitute $x=z+\dfrac{3}{2}$
we get
$\left(z+\frac{3}{2}\right)^4-6 \left(z+\frac{3}{2}\right)^3+11 \left(z+\frac{3}{2}\right)^2-6 \left(z+\frac{3}{2}\right)-m=0$
Expand and reorder
$z^4-\frac{5 }{2}z^2+\frac{9}{16}-m=0$
substitute $z^2=w$
$w^2-\frac{5 }{2}w^2+\frac{9}{16}-m=0$
$w=\dfrac{1}{4} \left(5\pm 4 \sqrt{m+1}\right)$
to be real the solutions we need $w\ge 0$
$5\pm 4 \sqrt{m+1}\ge 0$
Begin with
$5+4 \sqrt{m+1}\ge 0\to 4 \sqrt{m+1}\ge -5$
verified for $m\ge -1$
Then we solve
$5- 4 \sqrt{m+1}\ge 0$
$4 \sqrt{m+1}\le 5$
$16(m+1)\le 25 \to 16m \le 9\to m \le \dfrac{9}{16}$
the equation has all real solutions if $\quad-1\le m \le \dfrac{9}{16}$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_0^1 \mathrm e^{-x^2} \,\mathrm dx$ using power series? I'm trying to evaluate
$$\int_0^1 \mathrm e^{-x^2} \, \mathrm dx$$
using power series.
I know I can substitute $x^2$ for $x$ in the power series for $\mathrm e^x$:
$$1-x^2+ \frac{x^4}{2}-\frac{x^6}{6}+ \cdots$$
and when I calculate the antiderivative of this I get
$$x-\frac{x^3}{3}+ \frac{x^5}{5\cdot2}-\frac{x^7}{7\cdot6}+ \cdots$$
How do I evaluate this from $0$ to $1$?
| $$\left[ x-\frac{x^3}{3}+ \frac{x^5}{5*2}-\frac{x^7}{7*6}+ \dots \right]_0^1 = \left( 1-\frac{1}{3} + \frac{1}{5 \cdot 2} - \frac{1}{7 \cdot 6} + \dots \right) - 0 = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$$
So $\int_{0}^{1}e^{-x^2}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$, that is, the answer is whatever the series converges to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
| $$ \sqrt{3+\sqrt{13+4\sqrt3}}= \sqrt{3+\sqrt{1+2\times2\sqrt{3}+(2\sqrt{3})^2}}$$
$$=\sqrt{3+\sqrt{(1+2\sqrt{3})^2}}$$
$$=\sqrt{3+1+2\sqrt{3}}$$
$$=\sqrt{(1+\sqrt{3})^2}$$
$$=1+\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a clever one? Quadratic completion?)
\begin{align}
\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}} & = \\
& = (x^2+1)^{-1/2} -x^2*(x^2+1)^{-3/2} \\
& = (x^2+1)^{-1/2} * ( 1 - x^2 *(x^2+1)^{-1}) \\
& = ...
\end{align}
To give a bit more context, I was calculating the derivative of $\frac{x}{\sqrt{x^2+1}}$ in order to use newtons method for approximating the roots.
| Factor from $\dfrac{1}{\sqrt{x^2+1}}$. You will have:
$$\frac{1}{\sqrt{x^2+1}} \bigg(1 - \frac{x^2}{x^2+1}\bigg) = \frac{1}{\sqrt{x^2+1}} \frac{1}{x^2+1} = \frac{1}{(x^2+1)^\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find A and B in this limit Can you find $a$ and $b$? In how many ways can I find them?
$$\lim_{x\to0} \frac{a+\cos(bx)}{x^2}=-8$$
| For
$\lim_{x\to 0}\frac{\ a+\cos bx}{x^2}=-8
$,
note that,
for small $x$,
$\cos x
\approx 1-\frac{x^2}{2}
$.
Therefore
$\dfrac{\ a+\cos bx}{x^2}
\approx
\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
$.
If
$a+1 \ne 0$,
then
$\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
\to \infty$
as $x \to 0$.
Therefore,
to have the limit exist,
we must have
$a+1=0$
or $a = -1$.
The expression then becomes
$\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
=\dfrac{-\frac{(bx)^2}{2}}{x^2}
=-\frac{b^2}{2}
$.
If this is $-8$,
then
$b^2 = 16$,
so
$b = 4$.
Therefore
$a = -1, b=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Better way to reduce $17^{136}\bmod 21$? What I have done:
Note $17\equiv -4$ mod 21, and $(-4)^2 \equiv 5$ mod 21. So $17^{136} \equiv (-4)^{136} \equiv 5^{68}$ mod 21. Also note $5^2 \equiv 4$ mod 21 and $4^3 \equiv 1$ mod 21, so $5^{68} \equiv 4^{34} \equiv (4^3)^{11}\cdot4 \equiv 4$ mod 21. I feel this is rather complicated, and there should be a better way.
| You could also reduce modulo each of the factors of $21$ and then use the Chinese Remainder Theorem to recover the result modulo $21$.
$$
17^{136} \cong 2^{136} \cong (2^2)^{68} \cong 1^{68} \cong 1 \pmod{3}
$$
and
$$
17^{136} \cong 3^{136} \cong 3^{6 \cdot 22 + 4} \cong (3^6)^{22} \cdot 3^4 \cong 1^{22} \cdot 3^4 \cong 3^4 \cong 9^2 \cong 2^2 \cong 4 \pmod{7} .
$$
Solving the system $x \cong 1 \pmod{3}$ and $x \cong 4 \pmod{7}$, we get $x \cong 4 \pmod{21}$. Therefore $17^{136} \cong 4 \pmod{21}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate a limit using l'Hospital rule Evaluate $$\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}$$
I tried to apply l'Hospital rule in order to get the limit to be equal to
$$\lim_{x\to0}\frac{e^x-1}{2(e^x-\frac{x^2}{2}-x-1)^{-\frac{1}{3}}(e^x-x-1)}$$
but the new denominator has an indeterminate form itself and by repeatedly applying l'Hospital rule, it doesn't seem to help... This is where I got stuck.
| You can use $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)$
$$\qquad{\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-x-1}{3((1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))}{3((\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3(x^3(\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3x^{3\times\frac{2}{3}}((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
\frac{\frac{1}{2}}{3(\frac{1}{6})^{\frac{2}{3}}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A question about asymptotic notations with sums. I need to prove that $$ \sum_{k=0}^{n-2017} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}} = \Theta(3^{n})$$
I'm pretty sure it's straightforward to prove that it's $\Omega(3^{n})$ but I'm not sure how to prove the $O(3^{n})$ part. Maybe with using square root and derivative.
| *
*For the upper bound,, note that
$$\begin{align}
\sum_{k=0}^{n-2017} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}}
&\leq \sum_{k=0}^{n} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}}
\leq \sum_{k=0}^{n} \binom{n}{k} \frac{2^{2k}}{\sqrt{k}} \frac{ \sqrt{k}}{2^{k}}
\\&= \sum_{k=0}^{n} \binom{n}{k} 2^k
= (1+2)^n = 3^n
\end{align}$$
giving the upper bound. We used at the beginning the fact that
$$
\binom{2k}{k} \leq \frac{2^{2k}}{\sqrt{3k+1}}\leq \frac{2^{2k}}{\sqrt{k}}.
$$
*For the lower bound, we have that
$$\begin{align}
\sum_{k=0}^{n-2017} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}}
&
\geq \sum_{k=0}^{n-2017} \binom{n}{k} \frac{2^{2k}}{2\sqrt{k}} \frac{ \sqrt{k}}{2^{k}}
\\&= \frac{1}{2}\sum_{k=0}^{n-2017} \binom{n}{k} 2^k
\geq \frac{1}{2}\sum_{k=0}^{n} \binom{n}{k} 2^k - \frac{2017}{2}\binom{n}{n-2017} 2^n\\
&= \frac{1}{2}(1+2)^n - \frac{2017}{2} \binom{n}{2017}2^n
= \frac{3^n}{2} - \Theta(n^{2017}2^n) = \Theta(3^n)
\end{align}$$
giving the lower bound. There, we used the basic result for any constant $k$, $\binom{n}{k} = \Theta(n^k)$ when $n\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing the local maximum or minimum, while the function changes sign infinitely often Please I need a hand in solving this problem:
These 3 functions' values at $0$ are all $0$ and for $x\ne0$, $$f(x)=x^4\sin\frac{1}{x}, \, g(x)=x^4\left(2+\sin\frac{1}{x}\right), \, h(x)=x^4\left(-2+\sin\frac{1}{x}\right)$$
b- Show that $f$ has neither a local maximum nor a local minimum
at $0$, $g$ has a local minimum, and $h$ has a local
maximum.
The derivatives of these functions change sign infinitely often on both sides of $0$. I couldn't use the 1st nor the 2nd derivative test.
| HINT: we have for 1) $$f'(x)=4x^3\sin\left(\frac{1}{x}\right)+x^4\cos\left(\frac{1}{x}\right)\cdot \left(-\frac{1}{x^2}\right)$$
and for 2)$$g'(x)=4x^3\left(2+\sin\left(\frac{1}{x}\right)\right)+x^4\cdot\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$
and 3)$$h'(x)=4x^3\left(-2+\sin\left(\frac{1}{x}\right)\right)+x^4\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$
additionally use that $$|f(x)|\le x^4$$
and for $$f(x)$$ we have a sign Change from minus to plus
$g(x)$ has a sign change from plus to plus
$h(x)$ has a sign change for minus to minus.....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
System of differential equations $dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$ This is the first time I have seen system of differential equations in this form: $$dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
Can you please help me solve it because I don't even know where to start?
| $$\frac{dx}{1}=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
This system looks like to be involved in solving a PDE with the method of characteristics. The PDE should be :
$$\frac{\partial z(x,y)}{\partial x}+(y+z(x,y))\frac{\partial z(x,y)}{\partial y}=x+y+z(x,y)$$
$\underline{\text{If this supposition is true}}$,
unfortunately the boundary conditions are missing in the wording of the question.
$$\frac{dy}{y+z}=\frac{dz}{x+y+z}=\frac{dz-dy}{(x+y+z)-(y+z)}=\frac{dz-dy}{x}$$
A first family of characteristics comes from $\quad \frac{dx}{1}=\frac{dz-dy}{x} \quad\to\quad z-y-\frac{x^2}{2}=c_1$
A second family of characteristics comes from $\quad \frac{dx}{1}=\frac{dy}{y+z}=\frac{dy}{y+(y+\frac{x^2}{2}+c_1)}=\frac{dy}{2y+\frac{x^2}{2}+c_1}$
$\frac{dy}{dx}=2y+\frac{x^2}{2}+c_1 \quad\to\quad y=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}$
$y+\frac{z-y-\frac{x^2}{2}}{2}+\frac{x^2}{4}+\frac{x}{4}+\frac{1}{8}=c_2e^{2x} \quad\to\quad (4z+4y+2x+1)e^{-2x}=8c_2$
The general solution of the PDE is expressed on the form of an implicit equation :
$$\Phi\left((2z-2y-x^2)\:,\:(4z+4y+2x+1)e^{-2x} \right)=0$$
where $\Phi$ is any function of two variables (to be determined according to some boundary conditions).
$\underline{\text{If the above supposition is false}}$ :
Then, $z$ is function of $x$ only, that is $z(x)$ instead of $z(x,y)$.
The system becomes :
$$1=\frac{y'}{y+z}=\frac{z'}{x+y+z}$$
Following the same calculus, the result is :
$z(x)=y+\frac{x^2}{2}+c_1$
$ y(x)=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}$
$$\begin{cases}
y(x)=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x} \\
z(x)=\frac{c_1}{2}+\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, How many solutions are there if n is odd? $X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, $n \in N^*$
How many solutions are there if n is odd?
From the powers of $\begin{pmatrix}3&6\\ 2&4\end{pmatrix}$ I got that $X=\begin{pmatrix}\frac{3}{\sqrt[n]{7^{n-1}}}&\frac{6}{\sqrt[n]{7^{n-1}}}\\ \frac{2}{\sqrt[n]{7^{n-1}}}&\frac{4}{\sqrt[n]{7^{n-1}}}\end{pmatrix}$. But I'm not sure whether this is the only solution... Could I get some hints on how to get this done? Thank you
| $\det(X^n)=(\det X)^n=\det \begin{pmatrix}3&6\\ 2&4\end{pmatrix} = 0$, so $\det X=0$, and $X$ is singular.
$X$ cannot be the zero matrix, so it has two real eigenvalues: $0$ and $a\neq 0$. $X$ is diagonalizable: there is some non-singular $P$ such that $X=P\begin{pmatrix}0&0\\ 0&a\end{pmatrix}P^{-1}$, yielding $X^n=P\begin{pmatrix}0&0\\ 0&a^n\end{pmatrix}P^{-1}$.
Then $trace(X^n)=7=a^n$. Since $n$ is odd, we have $a=7^{1/n}$, hence $trace(X)=7^{1/n}$.
By Cayley-Hamilton, $X^2-7^{1/n}X=0$, that is $X^2=7^{1/n}X$. It's easy to prove by induction that for all $m\geq 1$, $X^m=7^{(m-1)/n}X$.
With $m=n$, $\begin{pmatrix}3&6\\ 2&4\end{pmatrix}=X^n=7^{(n-1)/n}X$, so that $$X=\frac{1}{7^{(n-1)/n}}\begin{pmatrix}3&6\\ 2&4\end{pmatrix}$$
The only solution is the one found by the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof verification: $x^3 + px - q = 0$ has three real roots iff $4p^3 < -27q^2$ I'm trying to prove the problem in the title. I'm not sure if my proof is clear, or even correct. It seems rather long, so I am not sure. Any advice would be great, especially on parts I might need to expand.
Let $f(x) = x^3 + px - q = 0$, then $f'(x) = 3x^2 + p$. Clearly, if $p \geq 0$ then $f'(x) > 0, \forall x$. Since $f(x) < 0$ as x approaches $-\infty$ (can I just state this?) and since $f(x) > 0$ as x approaches $\infty$ and $f'(x) >0, \forall x$ then by the Intermediate Value Theorem there exists only one solution on the interval $(-\infty, \infty)$ if $p \geq 0$.
Therefore, $f(x)$ [Edited: can only have] has three real roots iff $p < 0$. If $p <0$, then $f'(x) = 0$ for $x = \pm \sqrt{\frac{-p}{3}}$. Now, since $f(x) < 0, x \rightarrow -\infty$ and $f'(x) > 0$ for $x < \sqrt{\frac{-p}{3}}$ then by the IVT there exists one real solution on the interval $\left (-\infty, -\sqrt{\frac{-p}{3}}\right)$ iff $f(-\sqrt{\frac{-p}{3}}) > 0$. Likewise, since $f'(x) < 0$ on $\left (\sqrt{\frac{-p}{3}}, -\sqrt{\frac{-p}{3}}\right)$ then by the IVT there exists only one solution on that interval iff $f(-\sqrt{\frac{-p}{3}}) > 0$ and $f(\sqrt{\frac{-p}{3}}) < 0$. Lastly, since $f(x) > 0, x \rightarrow \infty$ and $f(x) > 0$ for $x > \sqrt{\frac{-p}{3}}$ then there exists only one solution on the interval $\left(\sqrt{\frac{-p}{3}}, \infty\right)$ iff $f(\sqrt{\frac{-p}{3}}) < 0$.
Since $f(x)$ has three roots iff there exists the one such root in each interval mentioned above, we require:
$f(\sqrt{\frac{-p}{3}}) < 0$ and $-f(\sqrt{\frac{-p}{3}}) > 0 \implies -2\left(\frac{-p}{3}\right)^{\frac{3}{2}} + q > 0, -2\left(\frac{-p}{3}\right)^{\frac{3}{2}} - q > 0.$ Multiplying the inequalities we get:
$-4\frac{p^3}{27} - q^2 > 0 \implies 4p^3 < -27q^2.$
| I think your statement is wrong.
Try, $p=-3$ and $q=-2$.
We have $$x^3+px-q=x^3-3x+2=(x-1)^2(x+2),$$
which says that the equation $$x^3+px-q=0$$
has three real roots, but $4p^3<-27q^2$ gives $-108<-108$, which is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$
So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$
Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is
simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$
We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.
| Let $\epsilon>0$
$$\left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right|\leq\frac{4n+8-4n-1}{16n+4}=\frac{7}{16n+4} \leq \frac{7}{16n}$$
We have that $\frac{7}{16n} \to 0$
Thus exists $n_0 \in \mathbb{N}$ such that $\frac{7}{16n}< \epsilon, \forall n \geq n_0$
So $\frac{1}{n}<\frac{16\epsilon}{7} \Rightarrow n> \frac{7}{16\epsilon}$
Take $n_0=[\frac{7}{16\epsilon}]+1$ and we have that $$\forall n\geq n_0=[\frac{7}{16\epsilon}]+1 \Rightarrow \frac{7}{16n}<\epsilon \Rightarrow \left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right| < \epsilon $$
Note that $[x]$ is the integer part of $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$? How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}|_{x=-2}=\frac{1}{(3x^2-2)}|_{x=-2}=\frac{1}{10}$$
$$Bx+C=\frac{1}{x+2}|_{x^2-2x=-2}=\frac{1}{(x+2)}\frac{(x-4)}{(x-4)}|_{x^2-2x=-2}=$$
$$=\frac{(x-4)}{(x^2-2x-8)}|_{x^2-2x=-2}=\frac{(x-4)}{(-2-8)}|_{x^2-2x=-2}=-\frac{1}{10}(x-4)$$
Thus:
$$\frac{1}{x^3-2x+4}=\frac{1}{10}\left(\frac{1}{x+2}-\frac{x-4}{x^2-2x+2}\right)$$
$$I=\frac{1}{10}\left(\int_0^2\frac{\ln x}{x+2}dx-\int_0^2\frac{(x-4)\ln x}{x^2-2x+2}dx\right)$$
What should I do next?
| To address your question of how to handle loops when integrating by parts, let $I=\int e^x\sin x \ dx$. Both functions are transcendental. We'll try using $u_1=e^x$ and $dv_1=\sin x \ dx$. These give $du_1=e^x \ dx$ and $v_1=-\cos x$. Thus, $$I=-e^x\cos x+\int e^x\cos x \ dx.$$
Now we have another integral. We'll try by parts again, and it's important we keep the same arrangement as last time (i.e. we need $u_2=du_1$ and $dv_2=v_1$; in this case we used exponential as $u$ and trigonometric as $v$, though we could have dove it the other way as long as we were consistent) otherwise we will just undo our last step. So, $u_2=e^x$ and $dv_2=\cos x \ dx$. These give $du_2=e^x \ dx$ and $v_2=\sin x$. Thus,
$$
\begin{align}
I&=-e^x\cos x+e^x\sin x-\int e^x\sin x \ dx \\
&=e^x(\sin x-\cos x)-I \\
2I&=e^x(\sin x-\cos x) \\
I&=\frac{1}{2}e^x(\sin x-\cos x).
\end{align}
$$
The important step is realising that when you get back to where you started, you can perform algebra to solve for your result.
That said, I do not guarantee this is what's needed here, just thought it might be worth a try and then you asked. So here you go.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With this in mind: I let $x = \sqrt{3} \tan \theta$ .Moving on, $dx = \sqrt{3} (\sec \theta )^2 $. And.....if $\theta$ is an
acute angle, we see in the illustration below that...
$\tan \theta = \frac{x}{\sqrt{3}}$
Then...getting the equivalent form of $\sqrt{x^2 + 3}$ in terms of $\theta$:
$$\sqrt{x^2 + 3}$$
$$ =\sqrt{(\sqrt{3} \tan \theta)^2 + 3}$$
$$= \sqrt{3(\tan \theta)^2 + 3}$$
$$= \sqrt{3((\tan \theta)^2 + 1)}$$
$$= \sqrt{3} \sqrt{(\sec \theta)^2} $$
$$\sqrt{x^2 + 3} = \sqrt{3} \sec \theta $$
Getting the equivalent form of $x^4$ in terms of $\theta$:
$$x^4 = (\sqrt{3} \tan \theta)^4$$
$$ = 9 (\tan \theta)^4 $$
Getting the equivalent form of $dx$ in terms of $\theta$:
$$dx = \sqrt{3} (\sec \theta )^2 d\theta$$
Substituting these equivalent expressions to the integral above, we get:
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{1}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{\sqrt{3} (\sec \theta )^2}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } d\theta$$
$$ = \int \frac{\sec \theta}{9 (\tan \theta)^4 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\tan \theta)^2)^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \frac{1}{9}\int \frac{\sec \theta}{((\sec \theta)^2 - 1 )^2 } d\theta$$
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$$
My problem is....I don't know how to evaluate $\frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$.
I can't go forward. I'm stuck. How to evaluate $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ properly?
| Hint:
You have done a good substitution so here you are
$$\int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta=\dfrac{1}{9}\int\dfrac{\cos^2\theta}{\sin^4\theta}\cos\theta\,d\theta=\dfrac{1}{9}\int\dfrac{1-\sin^2\theta}{\sin^4\theta}\cos\theta\,d\theta$$
now let $\sin\theta=u$ and continue!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$ I was looking for the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$
My work:
Dividing the $\frac{8t^3 +13}{(t+2)(4t^2+1)}$, I get $2 + \frac{-16t^2 -2t + 9}{4t^3 + 8t^2 + t + 2}$ or $2 + \frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$
Using the partial fraction decomposition on the expression $\frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$, the partial fractions of $\frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$ I get
is $\frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 }$
So...I conclude that....
$$\frac{8t^3 +13}{(t+2)(4t^2+1)} = 2 + \frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 } $$
Getting now the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt$.
$$\int \frac{8t^3 +13}{(t+2)(4t^2+1)} dt = \int \left( 2 + \frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 } \right) dt $$
$$ = \int 2 dt + \int \frac{-3}{t+2} dt + \int \frac{-4t + 6}{4t^2 +1 } dt $$
$$ = \int 2 dt + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{6}{4t^2 +1 } dt $$
$$ = 2t - 3ln(t+2) - \frac{1}{2}ln(4t^2 + 1) + 6\arctan (2t) + C$$
$$ = 2t - \left( (ln(t+2)^3) + ln((4t^2 + 1)^{\frac{1}{2}}) \right) + 6\arctan (2t) + C$$
$$ = 2t - \left( ln((t+2)^3(4t^2 + 1)^{\frac{1}{2}}) \right) + 6\arctan (2t) + C $$
$$ = 2t - \frac{1}{2}\left( 2\left( ln((t+2)^3(4t^2 + 1)^{\frac{1}{2}}) \right)\right) + 6\arctan (2t) + C $$
$$ = 2t - \frac{1}{2}ln((t+2)^6(4t^2 + 1)) + 6\arctan (2t) + C $$
But in my book, it says that the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt$ is $2t - \frac{1}{2}ln((t+2)^6(4t^2 + 1)) + 3\arctan (2t) + C$
My answer is so close to that of the book, but I couldn't locate my error. Where did I go wrong?
| You know that
$$
\int \frac{1}{t^2+1}dt=\arctan t + C.
$$
Thus, let $u=2t$, then
\begin{align}
\int \frac{6}{4t^2+1}dt &= \int \frac{6}{u^2+1}\frac{1}{2}du\\
&=\int \frac{3}{u^2+1}du\\
&=3\arctan u + C\\
&=3\arctan 2t + C.
\end{align}
Thus $\int \frac{6}{4t^2+1}dt = 6\arctan (2t) + c$, you found, is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Binomial Expansion - Simple application of the formula I found this one question in an old book:
What is the coefficient of $x^{n+1}$ in the expansion of $(x+2)^n \cdot x^3$?
Answer (according to my book): $(n^2-n) \cdot 2^{n-3}$
Here is my work: Since $\ T_{k+1} = \binom{n}{k}\cdot a^k\cdot b^{n-k}$, we can obtain a general-term for the expansion of $(x+2)^n \cdot x^3$. Letting $a = 2$ and $b = x$, then $ T_{k+1} = (\binom{n}{k}\cdot 2^k\cdot x^{n-k})\cdot x^3$.
So, we can easily notice that $x^{n+1}$ will show-up in the third term of the expansion (first we have $x^{n+3}$, then $x^{n+2}$ and $x^{n+1}$).
But, if we plug-in $k=2$ in $T_{k+1}$, its coefficient will be equal to $(n-3)! \cdot 2$.
Where is my mistake?
| The binomial theorem yields
$$
(x+2)^n = \sum_{k=0}^n \binom nk x^k 2^{n-k},
$$
and so the coefficient of $x^{n+1}$ in $(x+2)^nx^3$ is
$$
\binom n{n-2}2^{n-(n-2)}) = \binom n2 2^2 = 2n(n-1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequalities involving geometry but I can't post a picture yet How do I show that
$$ \frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right) < \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots \quad ?$$
| After moving the odd terms from the LHS to the RHS, we obtain the following equivalent inequality,
$$\frac 12 \left(\frac 1{4^2}+ \frac 1{6^2}+ \frac 1{8^2}+\dots\right) < \left(1-\frac 12\right)\left( \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots\right).$$
Then note that for all positive integer $n$, each term $\dfrac{1}{(2n)^2}$ is less than $\dfrac{1}{(2n-1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Some formula related with factor of (a+b+c+d) I am looking for some math formula
For example
\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}
First one related with factor a+b and the second one related with factor a+b+c
then
How about some formula related with a,b,c,d
i.e., is there are some equation factors into (a+b+c+d)?
|
$$\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}$$
The two relations are not quite "alike" since the second one is symmetric in $\,a,b,c\,$ (i.e. stays invariant if you permute the variables), while the first one is not (both sides change sign). Maybe a better analog would be for the first relation to be written as $\,a^2+b^2+2ab=(a+b)^2\,$.
With that note, the two equalities duplicate the Newton's identities for $\,n=2\,$ and $\,n=3\,$, respectively (where $p_k$ are the $k^{th}$ power sums, and $e_k$ the elementary symmetric polynomials):
$$
\begin{align}
p_2 + 2 e_2 &= e_1 p_1 \\
p_3 - 3 e_3 &= e_1 p_2 - e_2 p_1 = e_1(p_2-e_2)
\end{align}
$$
The next identity for $\,n=4\,$ would then be:
$$
p_4 + 4 e_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = e_1(p_3+e_3) - e_2 p_2
$$
$$
\begin{align}
\iff a^4+b^4+c^4+d^4 + 4 abcd &= (a+b+c+d)(a^3+b^3+c^3+d^3+abc+abd+acd+bcd) \\ &\quad - (a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1 $ is divisible by $ \ \large 2^{n+2} $ .
Answer:
For $ n=1 \ $ we have
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1.
For $ n=2 $ we have
$ \large 3^{2^2}-1=81-1=80 \ \ and \ \ 2^{2+2}=16 \ $
$Also \ \ \ 16 /80 $ .
Thus the statement hold for $ n=2 $ also.
Let the statement hold for $ n=m \ $
Then $ a_m=3^{2^m}-1 \ \ is \ \ divisible \ \ by \ \ b_m=2^{m+2} \ $
We have to show that $ b_{m+1}=2^{m+3} \ $ divide $ \ \large a_{m+1}=3^{2^{m+1}}-1 \ $
But right here I am unable to solve . If there any help doing this ?
Else any other method is applicable also.
| $$\begin{array}{} &3^{2^m} -1 &=x 2^{m+2} & \text{assumed and checked by } m=1 \text{with $x$ odd}\\
&3^{2^{m+1}} -1 &= 3^{2 \cdot 2^m} -1 & \text{general step in induction} \\
& &= (3^{2^m})^2 -1\\
& &= (3^{2^m} -1)(3^{2^m} +1) \\
& &=x 2^{m+2}(3^{2^m} +1) & \text{ with $x$ odd }\\
& &=x 2^{m+2}(3 \cdot 3^{2^m-1} +1) \\
& &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3^{2^m-1}+1)) \\
& &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3+1) \cdot y) & \text{by $3^{2w+1}+1$}=(3+1)\cdot y\\
& &=x 2^{m+2}2 \cdot (3^{2^m-1}+ 2 y) &\text{by factoring out }2\\
& &=x 2^{m+3} z &\text{with $x$ and $z$ odd}\\
\end{array}$$
$$\begin{array}{l}\implies 2^{m+3} \mid 3^{2^{m+1}}-1 & \phantom {\text{yxcyxcyyxcyxcycyxcyc}}& \text{induction successful, proof complete}
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to Solve Non-Homogeneous Recurrence Relations : $r_n = 2\left(r_{n-1} - \binom{n-1}{2}\right) + \binom{n-1}{2}$? $$r_n = 2\left(r_{n-1} - \binom{n-1}{2}\right) + \binom{n-1}{2}$$
which is equal to $$r_n - 2r_{n-1} = -\frac{n^2-3n+2}{2}$$
This given recurrence relation is derived from the question "How many regions line n could make at most in Euclidean plane?"
To solve this relation and make it into non-recurrent form,
I had looked in Wikipedia but I only get to the point of solving the homogeneous part only.
$r_n - 2r_{n-1} = 0$ gives $\alpha \cdot2^n$ as a solution of homogenous part.
How should one deal with non-homogeneous part?
| Using generating functions
$$f(x)=\sum\limits_{n=0}^{\infty}r_nx^n=r_0+r_1x+r_2x^2+\sum\limits_{n=3}^{\infty}r_nx^n=r_0+r_1x+r_2x^2+\sum\limits_{n=3}^{\infty}\left(2r_{n-1}-\binom{n-1}{2}\right)x^n=\\
r_0+r_1x+r_2x^2+2\left(\sum\limits_{n=3}^{\infty}r_{n-1}x^n\right)-\sum\limits_{n=3}^{\infty}\binom{n-1}{2}x^n=\\
r_0+r_1x+r_2x^2+2x\left(\sum\limits_{n=3}^{\infty}r_{n-1}x^{n-1}\right)-x^3\left(\sum\limits_{n=3}^{\infty}\binom{n-1}{2}x^{n-3}\right)=\\
r_0+r_1x+r_2x^2+2x\left(\sum\limits_{n=2}^{\infty}r_{n}x^{n}\right)-x^3\left(\sum\limits_{n=0}^{\infty}\binom{n+2}{2}x^{n}\right)=...$$
which is, according to this
$$...=r_0+r_1x+r_2x^2+2x\left(f(x)-r_0-r_1x\right)-\frac{x^3}{(1-x)^3}$$
or
$$f(x)=r_0+r_1x+r_2x^2+2xf(x)-2xr_0-2r_1x^2-\frac{x^3}{(1-x)^3}$$
$$f(x)(1-2x)=r_0(1-2x)+r_1x(1-2x)+r_2x^2-\frac{x^3}{(1-x)^3}$$
$$f(x)=r_0+r_1x+\frac{r_2x^2}{1-2x}-\frac{x^3}{(1-x)^3(1-2x)}$$
$$f(x)=r_0+r_1x+\frac{r_2x^2}{1-2x}-\left(\frac{1}{1-2x}-\frac{1}{1-x}+\frac{1}{(1-x)^2}-\frac{1}{(1-x)^3}\right)$$
$$f(x)=r_0+r_1x+r_2x^2\left(\sum\limits_{n=0}^{\infty}(2x)^{n}\right)-\left(\sum\limits_{n=0}^{\infty}(2x)^n\right)+\left(\sum\limits_{n=0}^{\infty}x^n\right)-\left(\sum\limits_{n=0}^{\infty}(n+1)x^n\right)+\left(\sum\limits_{n=0}^{\infty}\binom{n+2}{2}x^{n}\right)$$
Now, by comparing powers of $x$ we have
$$r_0=r_0$$
$$r_1=r_1-2+1-2+\binom{3}{2}=r_1$$
$$r_2=r_2-4+1-3+\binom{4}{2}=r_2$$
$$r_3=2r_2-8+1-4+\binom{5}{2}=2r_2-1$$
$$...$$
$$r_n=2^{n-2}r_2-2^n+1-(n+1)+\binom{n+2}{2}$$
Or
$$r_n=2^{n-2}r_2+\binom{n+2}{2}-2^n-n$$
Note I had to consider arbitrary $r_0, r_1, r_2$, especially $r_2$, to avoid troubles with $\binom{n-1}{2}$, which makes sense for $n\geq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Multi variable function and its corresponding range? How do we find the range of
$z=f(x,y)=1/\ln(4−x^2−y^2)$
1) Replace $t=\ln(4−x^2−y^2)$ and $t\in (−\infty,\ln4)$ and there is a DNE at ln('')=0
2) Find range of $1/t$ ; How is this done?
Answers: $(−\infty,0)∪(1/\ln4,\infty)$
| We know the domain of $f(x,y)=\dfrac{1}{\ln(4-x^2-y^2)}$ is
$$\color{blue}{D_f=\{(x,y)\in\mathbb{R}^2:x^2+y^2<4~,~x^2+y^2\neq3\}}$$
and in this domain $0\leq x^2+y^2<4$ so
$$0< 4-x^2-y^2\leq4~,~(x^2+y^2\neq3)$$
the function $\ln x$ is increasing then
$$-\infty< \ln(4-x^2-y^2)\leq\ln4~,~(x^2+y^2\neq3)$$
with reciprocating we have some irregular points for $\dfrac{1}{\ln(4-x^2-y^2)}$ in $x^2+y^2=3$, and except points on this circle we see
\begin{cases}
f(x,y)\in(-\infty,0)~~~\text{for}~~~0< 4-x^2-y^2<3, \\
f(x,y)\in(\dfrac{1}{\ln4},+\infty)~~~\text{for}~~~3< 4-x^2-y^2<4.
\end{cases}
so
$$\color{blue}{R_f=(-\infty,0)\cup(\dfrac{1}{\ln4},+\infty)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$
. Using $y=\frac{1}{4-x}$, we can substitute into equation 2 to get $\frac{1}{4-x}+\frac{1}{z}=1$ (eq. 2')
Together with equation 3, we can eliminate z by inversing equation 3 to be:
$x + \frac{1}{z}=\frac{3}{7}$
Now, we subtract both (eq. 2') and the inversed equation to get
$\frac{1}{4-x} - x = \frac{4}{7}$
However, this is going no where. The final answer should be x=$\frac{3}{2}$, y = $\frac{2}{5}$ and z = $\frac{5}{3}$.
Any tips or help would be greatly appreciated!
| Multiply (1) by $y$ to get:
$$xy + 1 = 4y \implies y = \frac{1}{4-x}$$
Multiply (2) by $z$ to get:
$$yz + 1 = z \implies z = \frac{1}{1-y}$$
Multiply (3) by $x$ to get:
$$zx + 1 = \frac{7z}{3} \implies x = \frac{3}{7-3z}$$
Plug the equation for $z$ into the equation for $x$. This will give you $x$ in terms of $y$. Now plug your equation for $y$ into this equation and solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using Lagrange multiplier to find the shortest distance from the origin to a given Set An exercise of an old exam wants me to find the point with the shortest distance to the origin which is in $M=\{(x,y): x^2y=2, x>0\}$.
So I think the function I have to minimize is $f(x,y)=\sqrt{(x^2+y^2)}$ with the condition $g(x,y)=x^2y-2$
Then Lagrange says that $$\nabla f = \lambda\nabla g \to \begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}\\ \frac{y}{\sqrt{x^2+y^2}}\end{pmatrix} = \lambda\begin{pmatrix}2xy\\x^2 \end{pmatrix}$$ But here is the point I get stuck.
Since the point has to be in $M$ it follows that $y=\dfrac{2}{x^2}$.
so $$\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}\\ \frac{y}{\sqrt{x^2+y^2}}\end{pmatrix} = \lambda\begin{pmatrix}\frac{4}{x}\\x^2 \end{pmatrix}$$ But that does not really help me to get any further. Thanks for your help.
| If I were you, I will use a simpler way. Since $y=\frac{2}{x^2}$, the distance is
$d(x)^2=x^2+\frac{4}{x^4}$. Applying the AM-GM inequality, we have
$$
d(x)^2=\frac{x^2}{2}+\frac{x^2}{2}+\frac{4}{x^4}\geq 3 \sqrt[3]{\frac{x^2}{2}\cdot \frac{x^2}{2}\cdot \frac{4}{x^4}}\geq 3.
$$
The identity holds when $x^6=8$ and hence $x=\sqrt{2}$ (since $x>0$). The shortest distance is $\sqrt{3}$.
If you really want to use Lagrange multiplier, let us consider
$$
F(x,y)=x^2+y^2
$$
under side condition
$$
G(x,y)=x^2y-2.
$$
Then we need to find $(x,y,\lambda)$ such that
$$
0=F_x-\lambda G_x= 2x-2\lambda xy \quad 0=F_y-\lambda F_y= 2y-\lambda x^2
$$
and the side condition.
The first equation implies $\lambda=1/y$ and hence $2y^2=x^2$ by the second one. Since $y=2/x^2$, we know that $x=\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find mistake in solving $\sin 2x=\sin x + \cos x$ I am solving $\sin 2x = \sin x +\cos x $ for $0\le x \le 360$
$$\sin 2x = \sin x +\cos x$$
$$2 \sin x \cos x =\sin x + \cos x$$
$$(\cos x + \sin x) ^{2} - (\sin x)^{2} - (\cos x)^{2} =(\cos x + \sin x)^{2} - 1=\sin x + \cos x$$
Let $\cos x + \sin x = y$
$$y^{2} - 1= y$$
After solving the quadratic gets $1.618$(I think this is not accepted) and $-0.618$
$$y=\cos x + \sin x =\sin 2x = -0.618$$
$$2x=218.17,321.83,578.17,681.87$$
$$x=109.09,160.92,289.09,340.34$$
But the problem is $109.09$ and $340.34$ is not the solution. I didn't purposely square anything to produce extra solution. The two extra solution satisfy $\sin 2x=-0.618$ but not $\cos x + \sin x=-0.618$.Is there any mistake in my calculations, or is there anyplace I introduced accidentally extra solution? Thanks.
| Consider these three equations:
\begin{align}
\sin 2x = \sin x + \cos x \tag{1} \\
(\sin x + \cos x)^2 - 1 = \sin x + \cos x \tag{2} \\
(\sin 2x)^2 -1 = \sin 2x \tag{3}
\end{align}
You were correct in determining that (1) and (2) are equivalent. Also, (1) does imply (3).
However, what you failed to notice is that (3) does not imply (1), so while all solutions to
\begin{align}
y^2 - 1 &= y, \\
y &= \sin x + \cos x
\end{align}
are solutions to (1), the same isn't true for
\begin{align}
y^2 - 1 &= y, \\
y &= \sin 2x.
\end{align}
To illustarte this, here is a example using only basic algebra and not trigonometry, which might look stupid, but is logically analogous to your soliution:
Problem: Solve
$$
x^2 - x = x - 1. \tag{1}
$$
First, we add $1-x$ to both sides to see that this is equivalent to
$$
(x-1)^2 = 0. \tag{2}
$$
Then, we denote $y=x-1$ to solve $y^2 = 0$ it and get $y=0$. So far so good.
Finally, we use $y = x^2 - x$ (This is a stupid way to do this but it illustrates the error.), and solve
$$
x^2 - x = 0,
$$
and find $x=1$ and $x=0$ as the solutions.
Where did we go wrong?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve $\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=4x^2.$ I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welcome!
Solution: The domain for $\arctan{x}$ is $\mathbb{R},$ thus the function $\sin{(\arctan{x})}$ is defined over the entire $\mathbb{R}$. However, the restricting factor comes from the fact that the domain for $\arcsin{x}$ is $[-1,1]$ which implies that the domain for $\tan{(\arcsin{x})}$ is $[-\arctan{1},\arctan{1}]=[-\frac{\pi}{4},\frac{\pi}{4}].$ We can thus conclude that if there exists a solution $x$ to the above equation, then $x\in[-\frac{\pi}{4},\frac{\pi}{4}]$.
First I note that $$\sin{b}=\pm\frac{\tan{b}}{\sqrt{\tan^2{b}+1}}\Rightarrow\sin^2{(\arctan{x})}=\left(\pm\frac{\tan{(\arctan{x})}}{\sqrt{\tan^2{(\arctan{x})}+1}}\right)^2=\frac{x^2}{x^2+1}.$$
Secondly; $$\tan{c}=\pm\frac{\sin{c}}{\sqrt{1-\sin^2{c}}}\Rightarrow \tan^2{\arcsin{x}}=\left(\pm\frac{\sin{(\arcsin{x})}}{\sqrt{1-\sin^2{(\arcsin{x})}}}\right)^2=\frac{x^2}{1-x^2}.$$
So the equation becomes $$\frac{x^2+1-1+x^2}{x^2}=\frac{2x^2}{x^2} =4x^2\Longleftrightarrow x=\pm\frac{1}{\sqrt{2}}.$$
What's left to do now is to show that these $x:$s both lie in the desired interval. So it boils down to showing that $\frac{1}{\sqrt{2}}\leq\frac{\pi}{4}$. This can be done by multiplying the inequality by $\sqrt{2}/\sqrt{2}$: $$\frac{\sqrt{2}}{4}\leq \frac{\pi}{4}\Longleftrightarrow \sqrt{2}\leq \pi,$$
which clearly is true.
| The allowed values for $x$ are in $[-1,1]$, because of $\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\tan\arcsin x$.
(Note: your $x\in[-\pi/4,\pi/4]$ is wrong and the probable cause for the low grade.)
The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\in(0,1)$. For every root we find, also its negative will be a root.
If $y=\sin\arctan x$, then $\arctan x=\arcsin y$ and so $x=\tan\arcsin y$ so
$$
x^2=\frac{\sin^2\arcsin y}{1-\sin^2\arcsin y}=\frac{y^2}{1-y^2}
$$
whence
$$
y^2=\frac{x^2}{1+x^2}
$$
If $z=\tan\arcsin x$, then
$$
z^2=\frac{\sin^2\arcsin x}{1-\sin^2\arcsin x}=\frac{x^2}{1-x^2}
$$
So your equation $\frac{1}{y^2}-\frac{1}{z^2}=4x^2$ becomes
$$
\frac{1+x^2}{x^2}-\frac{1-x^2}{x^2}=4x^2
$$
that is,
$$
2x^2=4x^4
$$
and, since $x\ne0$, $x^2=\frac{1}{2}$. The only positive root is
$$
x=\frac{1}{\sqrt{2}}
$$
which indeed belongs to $(0,1)$. Thus the solutions are
$$
\frac{1}{\sqrt{2}}\qquad\text{and}\qquad{-}\frac{1}{\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What is the probability of getting the sum $26$ when $7$ chips are taken out? Suppose you have a bag in which there are $10$ chips numbered $0$ to $9$. You take out a chip at random, note its number and then put it back. This process is done $7$ times and after that the numbers are added.
What is the probability that the sum is $26$?
The probability of getting any number at any turn is $\frac1{10}$. Also I know that there are $10^7$ different combinations that can be formed from these numbers. But it would be too inefficient to start counting which ones sum up to $26$.
I have really no idea what to do. Can this be done in a faster way using combinatorics? If yes, how?
| The number of ways of drawing 7 chits with sum equal to 26 is the coefficient of $x^{26}$ in the expansion
\begin{align*}
(1+x+x^2+\cdots +x^9)^7 &= \left(\frac{1-x^{10}}{1-x}\right)^7 \\
&=(1-7x^{10} + 21x^{20} - \cdots)\left(1+\binom{7}{1}x + \binom{8}{2}x^2 + \cdots\right)
\end{align*}
Coefficient of $x^{26}$ is
$$\binom{32}{26} - 7\cdot \binom{22}{16} + 21\cdot \binom{12}{6} = 403305$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality on tangent and secant function
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$
| let $\tan\alpha =a, \tan\beta =b, \tan\gamma =c, \tan\delta =d\implies a,b,c,d>0$
Also $\alpha+\beta+\gamma+\delta=180^{\circ}\implies \alpha+\beta=180^{\circ}-(\gamma+\delta)$
$\implies \tan (\alpha+\beta)=-\tan (\gamma+\delta)$
$\implies \frac{a+b}{1-ab}=-\frac{c+d}{1-cd}$
$\implies a+b+c+d=abc+abd+acd+bcd$ ___(i)
Now Consider $(a+b)(a+c)(a+d)=a^3+(b+c+d)a^2+(bc+bd+cd)a+bcd$
$\implies (a+b)(a+c)(a+d)=a^2(a+b+c+d)+(a+b+c+d)$ (Using (i))
$\implies \frac{a^2+1}{a+b}=\frac{(a+c)(a+d)}{a+b+c+d}$
$\implies \frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}=\frac{(a+c)(a+d)+(b+a)(b+d)+(c+a)(c+b)+(d+b)(d+c)}{a+b+c+d}$
$=a+b+c+d$___(ii)
Now by C-S inequality
$\left((a+b)+(b+c)+(c+d)+(d+a)\right)\left(\frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}\right)\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies 2(a+b+c+d)^2\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies \sqrt2\left(\tan\alpha +\tan\beta +\tan\gamma +\tan\delta\right)\ge\sec\alpha +\sec\beta +\sec\gamma +\sec\delta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve differential equation $y''-2y'+3y=\sin x$ using invers operator method.
$$y''-2y'+3y=\sin x$$
$s^2-2s+3=0$
$s=1 \pm i\sqrt {2}$
uisng operator $(D^2-2D+3)y_1=\sin x$
\begin{aligned}
y_1 &= \frac{1}{D^2-2D+3}\sin x \\
&= \Im\left(\frac{1}{D^2-2D+3} e^{ix}\right) \\
&= \Im \left(\frac{1}{i^2-2i+3}e^{ix}\right) \\
&= \Im \left(\frac{1}{2-2i}e^{ix}\right) && (***) \\
&= \frac{1}{2} \Im \left( \frac{1+i}{1+1} e^{ix}\right) && (***)\\
& = \frac{1}{4} \Im \left((1+i)(\cos x+i \sin x)\right) && (***) \\
& = \dots
\end{aligned}
i tried to solve problem using differential operator, but i dont understand what im means here, and also the formula too. i dont understand the star part, can someone explain what is this? is this relate to complex number? and does anyone know useful link about this material? and also i tried to solve using undetermined coef or variation of parameter, but compare to this it really takes lots of time.
2. and for differential operator:L(D) ->L'(D) so how to differentiate a differential operator?
$(D-1)(D-2)^3 => (D-2)^2(4D-5)$ can someone give me hint thanks
| The starred part consists in basic algebraic manipulations of complex numbers, with $i^2 = -1$ and $e^{ix} = \cos x + i \sin x$. Indeed,
\begin{aligned}
\frac{1}{i^2 -2i + 3} & = \frac{1}{2 -2i}\\
& = \frac{1}{2} \frac{1}{1 - i} \frac{1+i}{1+i} \\
& = \frac{1}{2} \frac{1 + i}{2} \, .
\end{aligned}
Now, multiplying by $e^{ix}$ and taking the imaginary part gives the equations of the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\int_{0}^\pi \frac{2\cos 2\theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$ Given that $$\int_{|z|=1|}\frac{z^2}{2z+1} dz = \frac{i\pi}{4}$$,
show $$\int_{0}^\pi \frac{2\cos 2 \theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$$.
I saw the bounds of the latter integral and thought that I should try and parametrize using $z = e^{2i\theta}$ where $\theta \in [0,\pi]$.
This doesn't seem to simplify easily.
i saw this thread: Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
and the top answer says:
$$\begin{align}
\int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\\end{align}$$
which I don't understand.
How does multiplying a half to the integral with contour $|z|=1$ (parametrized by $z = e^{i\theta}, \theta\in [0,2\pi]$) give the LHS?
I tried looking at it by taking the latter integral and using the substitution $u=\pi + \theta$, in hopes that the integrand simplifies to stay the same but it doesn't, so I can't see why the integral with bounds $0,\pi$ is half the integral that would have bounds $0,2\pi$ (since we would use the parametrization $z=e^{i\theta}$).
| With $\theta\to-\theta$
$$
I=\int_0^{-\pi} \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,(-)d\theta=\int_{-\pi}^0 \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta
$$
\begin{align}
2I
&=\int_{-\pi}^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\, d\theta\\
&=\int_{|z|=1} \frac{{\bf Re\,}(2z^2+z^3)}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\
&={\bf Re\,}\int_{|z|=1} \frac{2z^2+z^3}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\
&={\bf Re\,}\dfrac{1}{i}\int_{|z|=1} \frac{z^2}{2z+1}\, dz,
\end{align}
Note that $\overline{z}=\dfrac{1}{z}$ so $\overline{dz}=d\overline{z}=-\dfrac{dz}{z^2}$ and $\dfrac{\overline{dz}}{\overline{iz}}=-\dfrac{dz}{-i\overline{z}z^2}=\dfrac{dz}{iz}$ and also ${\bf Re\,}(z+\overline{z})=z+\overline{z}$. The rest is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answer. Could someone explain the steps?
| As $\alpha+\beta=\dfrac32, \alpha\beta=\dfrac62$
let $y=\dfrac\alpha\beta\iff y+1=\dfrac3{2\beta}\iff\beta=\dfrac3{2(y+1)}$
But as $\beta$ is a root of $$2x^2-3x+6=0$$
$$2\left(\dfrac3{2(y+1)}\right)^2-3\left(\dfrac3{2(y+1)}\right)+6=0$$
As $y+1\ne0,$ multiply both sides by $\dfrac{2(y+1)^2}3$ to find $$0=3-3(y+1)+4(y+1)^2=4y^2+5y+4$$
By symmetry, we can surmise that the same equation will be reached if we start with $y=\dfrac\beta\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| Using the general form of quadritic equation, $x^2 -(a+b) x + ab$,
we get the values of $a+b$ and $ab$.
Now, the expression $a^3+b^3$ can be reduced to $(a+b)^3 -3ab(a+b)$.
Substitute the value of $a+b$ and $ab$ in the above equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to find minimum and maximum value of x, if x+y+z=4 and $x^2 + y^2 + z^2 = 6$? I just know that putting y=z, we will get 2 values of x. One will be the minimum and one will be the maximum. What is the logic behind it?
| $z=4-x-y$
$x^2+y^2+(4-x-y)^2-6=0$
Differentiate wrt $x$ and $y$
$2x-2(4-x-y)=0$
$2y-2(4-x-y)=0$
Gives $x=y=4/3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Show that $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$ If $\Delta$ is the area of a triangle with side lengths a, b, c, then show that: $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$. Also show that equality occurs in the above inequality if and only if a = b = c.
I am not able to prove the inequality.
| One has $\Delta = \frac{1}{4}\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ (See this link).
Moreover, one has $(a+b-c)(b+c-a) \leq (\frac{a+b-c + b+c-a}{2})^2 = b^2$
So $[(a+b+c)(b+c-a)(c+a-b)(a+b-c)]^2 \leq a^2b^2c^2$.
Thus, $\Delta \leq \frac{1}{4}\sqrt{abc(a+b+c)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all roots of $2x^3+16$ in $\mathbb C$
Find all roots of $p(x) = 2 x^3 + 16$ in $\mathbb C$.
I found my answers to be x = 2, -1+i$\sqrt{3}$, -1+i$\sqrt{3}$.
But when I put the expression into Symbolab, it gives me -2, 1+i$\sqrt{3}$, 1+i$\sqrt{3}$ as roots of p(x) in C.
Can someone explain where I went wrong?
This is how I did it;
First, I rewrite p(x) and get
$2x^3+16=0$
$2x^3=-16$
$x^3=-8$ ------ (1)
Express (1) in Euler form
$x=re^{\theta i}$
$x^3=r^3e^{3\theta i}$
Let w = -8
$ |w| = \sqrt{(-8)^2} = 8$
$\theta = tan^{-1}(0) = 0$
$w = 8e^{0\theta}$
Now I have
$r^3e^{3\theta i} = 8e^{0\theta}$
Equate the modulus and argument
$r^3 = 8$
$r = 2$
$3\theta = 0+2\pi k$ ------- for k $\in Z$
$\theta = {2\pi k \over 3}$
Now I have
$ x = 2(cos {2\pi k \over 3} + i sin {2\pi k \over 3})$
Calculating x by substituting k = 0,1,2
k = 0, x = 2
k = 1, x = 2(${-1 \over 2} + i {\sqrt{3} \over 2}$) = -1+i${\sqrt{3}}$
k = 2, x = 2(${-1 \over 2} - i {\sqrt{3} \over 2}$) = -1-i${\sqrt{3}}$
So all roots of p(x) in C are 2, -1-i${\sqrt{3}}$, -1+i${\sqrt{3}}$
.
.
.
I will be very appreciated if someone can help me out.
| Another way you can calculate this is to use De Moivre's theorem, which states that if $z = r(\cos \theta + i\sin \theta)$, the $n$th roots of $z$ are $$r^{1/n}\left(\cos\frac{\theta+2\pi k}{n} + i \sin \frac{\theta+2\pi k}{n}\right)$$
From $(1)$, which is $x^3=-8$, we find that $z = 8(\cos \pi + i \sin \pi)$, so $r=8$, $\theta=\pi$ (because $8$ and $-8$ are opposite to each other), and $n=3$ (we are finding the $3$rd roots of $z$). Substituting these values into the formula, we have:
First root: $8^{1/3}(\cos \frac{\pi+2\pi *0}{3} + i \sin \frac{\pi+2\pi *0}{n}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 1+i\sqrt{3}$
Second root: $8^{1/3}(\cos \frac{\pi+2\pi *1}{3} + i \sin \frac{\pi+2\pi *1}{3}) = 2(\cos\pi+i \sin \pi) = -2$
Third root: $8^{1/3}(\cos \frac{\pi+2\pi *2}{3} + i \sin \frac{\pi+2\pi *2}{3}) = 2(\cos \frac{5\pi}{3} + i\sin\frac{5\pi}{3}) = 1-i\sqrt{3}$
Use the fact that $n$th roots of $z$ have $n$ complex roots. Have we have found all the complex $3$rd roots of $8$ yet?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Conditional probability of multivariate gaussian I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.
Let $\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda ) $ , where
$$\begin{align}
\mu &= \begin{pmatrix}
1 \\
1
\end{pmatrix}
\end{align}
$$
$$
\begin{align}
\Lambda &= \begin{pmatrix}
3 \quad 1\\
1 \quad 2
\end{pmatrix}
\end{align}
$$
We are tasked with computing: $P(X_1 \geq 2 \mid X_2 +3X_1=3)$
I here begin by doing a transformation,
$$ \mathbf{Y} = (Y_1, Y_2)', \qquad Y_1 = X_1, \qquad Y_2 = X_2 + 3X_1$$
We now are interested in the probability,
$$P(Y_1 \geq 2 \mid Y_2 = 3)$$
Since we can write that $\mathbf{Y = BX}$, it follows that,
$$\mathbf{Y} \in \mathcal{N}(\mathbf{B\mu, B\Lambda B')})$$
where
$$\mathbf{B}= \begin{pmatrix}
1 \quad 0\\
3 \quad 1
\end{pmatrix} \rightarrow \quad \mathbf{B \mu} = \begin{pmatrix}
1 \\
4
\end{pmatrix}, \quad \mathbf{B\Lambda B'}= \begin{pmatrix}
1 \quad 0\\
3 \quad 1
\end{pmatrix} \begin{pmatrix}
3 \quad 1\\
1 \quad 2
\end{pmatrix} \begin{pmatrix}
1 \quad 3\\
0 \quad 1
\end{pmatrix} = \begin{pmatrix}
3 \quad 10\\
10 \; \; 35
\end{pmatrix}$$
We thereafter know that we can obtain the conditional density function by,
$$
f_{Y_1\mid Y_2 = 3} (y_1) = \frac{f_{Y_1,Y_2}(y_1, 3)}{f_{Y_2}(3)} \tag 1
$$
The p.d.f. of the bivariate normal distribution,
$$f_{Y_1, Y_2}(y_1, y_2) = \frac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} e^{\frac{1}{2(1-\rho^2)}(\frac{(y_1 - \mu_1)^2}{\sigma_1^2} - \frac{2 \rho (y_1 - \mu_1)(y_2 - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y_1 - \mu_1)^2}{\sigma_2^2})} $$
The marginal probability density of $Y_2$,
$$f_{Y_2}(y_2) = \frac{1}{\sqrt{2\pi} \sigma_2} e^{-(y_2 - \mu_2)^2 / (2\sigma_2^2)}$$
Given that,
$$\sigma_1 = \sqrt{3}, \quad \sigma_2 = \sqrt{35}, \quad \rho = \frac{10}{\sigma_1 \sigma_2 } = \frac{10}{\sqrt{105}} $$
we are ready to determine (1). However, the resulting expression, which I then need to integrate as follows,
$$
Pr(Y_1 \geq 2 \mid Y_2 = 3) = \int_2^\infty f_{Y_1\mid Y_2 = 3} (y_1) \, dy_1
$$
becomes quite ugly, making me unsure whether I've approached the problem in the wrong way?
Thanks in advance!
| The covariance between $X_1 + \lambda (3 X_1 + X_2)$ and $3 X_1 + X_2$ is $10 + 35 \lambda$, therefore if we take $\lambda = -2/7$, we get
$$\operatorname{P}(X_1 \geq 2 \mid 3 X_1 + X_2 = 3) =
\operatorname{P} \left(
X_1 -\frac 2 7 (3 X_1 + X_2 - 3) \geq 2 \mid 3 X_1 + X_2 = 3 \right) = \\
\operatorname{P} \left( X_1 -\frac 2 7 (3 X_1 + X_2 - 3) \geq 2 \right),$$
and $X_1/7 -2 X_2/7 \sim \mathcal N(-1/7, 1/7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving $n\times n$ determinant using triangular shape I have just started learning to solve nth order determinants by getting it into the triangular shape ( in this way the determinant is equal to the multiple of main or additional diagonal + the determination of the sign ). I have solved a couple of easy ones, but got stuck on this one ( which seems easy, but however I manipulate the rows or columns I can't get it into the triangular shape ).
$$
\begin{vmatrix}
5 & 3 & 3 & \cdots & 3 & 3 \\
3 & 6 & 3 & \cdots & 3 & 3 \\
3 & 3 & 6 & \cdots & 3 & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
3 & 3 & 3 & \cdots & 6 & 3 \\
3 & 3 & 3 & \cdots & 3 & 6 \\
\end{vmatrix}
$$
If someone could give the solution and explain the crucial steps to solving this, it would be very appreciated.
| First, you can subtract the last line from the others which gives you :$$\begin{vmatrix}
2 & 0 &\ldots&0&-3 \\
0 & 3 &\ddots&\vdots& \vdots \\
\vdots &\Large{0}&\ddots&0& \vdots \\
0&\ldots&0&3&-3\\
3&\ldots&\ldots&3&6
\end{vmatrix}$$
Then subtract $\frac{3}{2} L_{1}$ from $L_{n}$:
$$\begin{vmatrix}
2 & 0 &\ldots&0&-3 \\
0 & 3 &\ddots&\vdots& \vdots \\
\vdots &\Large{0}&\ddots&0& \vdots \\
0&\ldots&0&3&-3\\
0&3&\ldots&3&\frac{3}{2}
\end{vmatrix}$$
Finally, subtract $L_{i}$ for every $i \in [2,n-1]$ from $L_{n}$:
$$\begin{vmatrix}
2 & 0 &\ldots&0&-3 \\
0 & 3 &\ddots&\vdots& \vdots \\
\vdots &\Large{0}&\ddots&0& \vdots \\
0&\ldots&0&3&-3\\
0&\ldots&\ldots&0&\frac{3}{2}+3(n-2)
\end{vmatrix}=\begin{vmatrix}
2 & 0 &\ldots&0&-3 \\
0 & 3 &\ddots&\vdots& \vdots \\
\vdots &\Large{0}&\ddots&0& \vdots \\
0&\ldots&0&3&-3\\
0&\ldots&\ldots&0&3(n-\frac{3}{2})
\end{vmatrix}$$
These operations don't affect the determinant so you can apply the formula for a triangular shaped matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}\tag2$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={A\over 2^n-1}+{B\over 2^{n+1}-1}+
{C\over 2^{n+2}-1}\tag3$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 3(2^n-1)}-{1\over 2^{n+1}-1}+
{2\over 3(2^{n+2}-1)}\tag4$$
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}={1\over 9}\tag5$$
| Hint: We can express
$$\frac{1}{2^n-1}=\sum_{k=1}^{\infty} \frac{1}{2^{kn}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Can someone tell me what should i do next? Should I use the inequality between Arithmetic and Geometric mean? Problem 1: Let $a,b,c> 0$
$ab+3bc+2ca\leqslant18$
Prove that:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$.
I started on this way:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$
$\frac{3bc+2ac+ab}{abc}\geqslant 3 $
$\frac{ab+3bc+2ac}{abc}\geqslant 3 \times \frac{abc}{3}$
$\frac{ab+3bc+2ca}{3}\geqslant abc$
| By Holder:
$$18\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)^2\geq(ab+3bc+2ac)\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)\left(\frac{2}{b}+\frac{1}{c}+\frac{3}{a}\right)\geq$$
$$\geq\left(\sqrt[3]{ab\cdot\frac{3}{a}\cdot\frac{2}{b}}+\sqrt[3]{3bc\cdot\frac{2}{b}\cdot\frac{1}{c}}+\sqrt[3]{2ac\cdot\frac{1}{c}\cdot\frac{3}{a}}\right)^3=162$$
and we are done!
The Holder inequality for three sequences is the following.
Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $c_1$, $c_2$,..., $c_n$, $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that:
$$(a_1+a_2+...+a_n)^{\alpha}(c_1+c_2+...+c_n)^{\gamma}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}c_1^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+\left(a_2^{\alpha}b_2^{\beta}c_2^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+...+\left(a_n^{\alpha}b_n^{\beta}c_n^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}\right)^{\alpha+\beta+\gamma}.$$
It follows from convexity of $f(x)=x^k$, where $k>1$.
In our case $\alpha=\beta=\gamma=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the volume between $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral So the volume of $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral shall look similar to $$\int^2_0\int^{y=?}_{y=?}\int^{4-x^2-y^2}_{4-2x} dz dy dx$$ but how do I find the limits on $y$?
| The point is that you are integrating over a domain in which $4-2x<4-x^2-y^2$. In that domain,
$$
-2x < -x^2-y^2\\
2x>x^2+y^2\\
y^2 < 2x -x^2 \\
-\sqrt{2x-x^2} < y < +\sqrt{2x-x^2}
$$
So the integral is
$$
\int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (4- x^2 -y^2 -(4-2x) )dy\,dx = \int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (2x- x^2 -y^2 )dy\,dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of terms in the Multinomial Expansion $(x+\frac{1}{x}+x^2+\frac{1}{x^2})^n$ I am aware that there is a formula to calculate the number of terms in a multinomial expression $(x_1+x_2+x_3+...x_r)^n$, i.e. $^{n+r-1}C_{r-1}$. However, this is in the case when the terms $x_1, x_2, x_3 ... x_r$ are different variables. In my case, the variables are the same; i.e. x, raised to different powers. Can someone please point me in the right direction?
|
We obtain
\begin{align*}
\color{blue}{\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)}&=\frac{1}{x^{2n}}(1+x+x^2+x^3)^n\\
&=\frac{1}{x^{2n}}\left(1+x+x^3\left(1+x\right)\right)^n\\
&=\frac{1}{x^{2n}}(1+x)^n(1+x^3)^n\\
&\color{blue}{=\frac{1}{x^{2n}}\sum_{j=0}^n\binom{n}{j}x^j\underbrace{\sum_{k=0}^n\binom{n}{k}x^{3k}}_{n+1\text{ terms}}}\tag{1}
\end{align*}
Let's have a look at the three factors in (1).
*
*The rightmost sum contains $n+1$ pairwise different terms with exponents $3k,0\leq k\leq n$.
*The leftmost factor $\frac{1}{x^{2n}}$ does not change the number of different terms as it is just a shift of each exponent of $x$ by $-2n$.
*Now we assume $n\geq 2$ and analyse the left sum. A multiplication with the first three terms $\binom{n}{0},\binom{n}{1}x,\binom{n}{2}x^2$ results in $3(n+1)$ terms
\begin{align*}
\sum_{j=0}^n a_kx^{3k\color{blue}{+0}},\sum_{j=0}^n b_kx^{3k\color{blue}{+1}},\sum_{j=0}^n c_kx^{3k\color{blue}{+2}}
\end{align*}
which gives a sum of increasing powers of $x^k, 0\leq k\leq 3(n+1)$. Additionally we observe that whenever $n(>2)$ is increased by $1$, the overall number of terms is increased by $1$.
We conclude: The number of different terms in the expansion of $\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)^n$ is
\begin{align*}
\color{blue}{1}&\qquad \color{blue}{n=0}\\
\color{blue}{4}&\qquad \color{blue}{n=1}\\
3(n+1)+(n-2)=\color{blue}{4n+1}&\qquad \color{blue}{n\geq 2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculus of variation: Lagrange's equation A particle of unit mass moves in the direction of $x$-axis such that it has the Lagrangian $L= \frac{1}{12}\dot x^4 + \frac{1}{2}x \dot x^2-x^2.$ Let $Q=\dot x^2 \ddot x$ represent a force (not arising from a potential) acting on the particle in the $x$-direction. If $x(0)=1$ and $\dot x(0)=1$, then the value of $\dot x$ is
*
*some non-zero finite value at $x=0$.
*$1$ at $x=1$.
*$\sqrt 5 $ at $x=\frac{1}{2}$
*$0$ at $x=\sqrt \frac{3}{2} $
My Attempt: The Lagrange's equation is
$$ \frac{d}{dt}\left (\frac{\partial L}{\partial \dot x}\right )-\frac{\partial L}{\partial x}=Q \tag{1} $$
where
$$ \frac{\partial L}{\partial x}=\frac{\partial }{\partial x}\left(\frac{1}{12}\dot x^4 + \frac{1}{2}x \dot x^2-x^2\right) =\frac{1}{2} \dot x^2-2x$$
and
$$\frac{\partial L}{\partial \dot x}=\frac{1}{3}\dot x^3 +x \dot x $$
Hence equation $(1)$ becomes
\begin{aligned}
\frac{d}{dt}\left( \frac{1}{3}\dot x^3 +x \dot x \right)-\frac{1}{2} \dot x^2+2x &=\dot x^2 \ddot x \\ \Rightarrow\; \frac{d}{dt}\left( \frac{1}{3}\dot x^3\right) +\frac{d}{dt}\left(x \dot x \right) &=\dot x^2 \ddot x+\frac{1}{2} \dot x^2-2x \\ \Rightarrow\; \dot x^2 \ddot x+\frac{d}{dt}(x \dot x ) &=\dot x^2 \ddot x+\frac{1}{2} \dot x^2-2x \\ \Rightarrow\; \frac{d}{dt}(x \dot x ) &=\frac{1}{2} \dot x^2-2x
\end{aligned}
I don't know how to solve further and where to use the given values of $x(0) $ and $\dot x(0)$
| well, continue
\begin{align}
\frac{d}{dt}(x \dot x ) &=\frac{1}{2} \dot x^2-2x \\
\implies \dot{x}^2+x \ddot{x} &= \frac{1}{2}\dot{x}^2-2x \\
\implies 0 &= x \ddot{x}+\frac{1}{2}\dot{x}^2+2x \\
&\vdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove a 3 parameter integral identity I have stumbled upon the following identity:
$$\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt$$
It appears to work for all $b \in \mathbb{R}$, $c \geq 0$ and $|z|<1$.
The identity is related to the integral representation of ${_2F_1}$, the hypergeometric function, however I don't have a nice proof of it.
How can we prove this identity, and are the conditions on the parameters I listed correct?
| We want to show
\begin{eqnarray*}
\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt.
\end{eqnarray*}
Make the substitution $t=\frac{1}{u}$ into the third integral ($dt=-\frac{du}{u^2}$)
\begin{eqnarray*}
\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt = -\int_{\infty}^1 \frac{(1+\frac{1}{u})^{b-c-2}}{u^c(1-z+\frac{1}{u})^b} \frac{du}{u^2} \\
=\int_1^{\infty} \frac{(1+u)^{b-c-2}}{(1+(1-z) u)^b} du.
\end{eqnarray*}
Now combine this with the second integral & the RHS becomes
\begin{eqnarray*}
\int_0^{\infty} \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt.
\end{eqnarray*}
Now make the substition $w=\frac{t}{1+t}$ (so $dw=\frac{dt}{(1+t)^2}$ and $t=\frac{w}{1-w}$) and we get the LHS.
\begin{eqnarray*}
\int_0^{\infty} \frac{(1+t)^{b-c}}{(1+(1-z) t)^b} \frac{dt}{(1+t)^2} &=& \int_0^{1} \frac{(1+\frac{w}{1-w})^{b}}{(1+(1-z) \frac{w}{1-w})^b} \frac{1}{(1+\frac{w}{1-w})^c}dw \\
&=&\int_0^1 \frac{(1-w)^c}{(1-z w)^b} dw.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:
For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$
I attempted to rewrite and simplify the left hand side as:
$$ \sum_{k = 1}^m \frac{\frac{(2m-2)!!}{(2m-2k)!!}}{\frac{(2m-1)!!}{(2m-2k-1)!!}} = \sum_{k = 1}^m \frac{(2m-2)!! (2m-2k-1)!!}{(2m-2k)!! (2m-1)!!} = \frac{\left[2^{m-1} (m-1)! \right]^2}{(2m-1)!} \sum_{k=1}^m \frac{(2m-2k-1)!!}{(2m-2k)!!} $$
But I do not seem to be getting anywhere. Does anyone see how to prove the statement?
| $$\scriptsize\begin{align}
&\;\;\;\frac 1{2m-1}+\frac {2m-2}{(2m-1)(2m-3)}+\frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)}+\cdots+\frac {(2m-2)!!}{(2m-1)!!}\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\left(1+\frac 43\left(1+\frac 21\right)\right)\right)\right)\cdots+\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\left(1+\frac 43\cdot \color{blue}3\right)\right)\right)\cdots+\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\cdot \color{blue}5\right)\right)\cdots+\right)\right)\\\\
&\qquad\qquad \vdots\qquad\qquad\qquad\qquad\qquad \;\unicode{x22F0} \\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\cdot \color{blue}{(2m-5)}\right)\right)\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\cdot \color{blue}{(2m-3)}\right)\\\\
&=\frac 1{2m-1}\cdot \color{blue}{(2m-1)}\\\\
&=\color{red}{\large 1}\\\\\end{align}$$
In alternative notation, starting from the innermost parenthesis,
$$\scriptsize\begin{align}
a_1\;\;&=1+\frac 21&&=3\\
a_2\;\;&=1+\frac 43a_1&&=5\\
a_3\;\;&=1+\frac 65 a_2&&=7\\
&\qquad \vdots &\qquad \\
a_n\;\;&=1+\frac {2n}{2n-1}a_{n-1}&&=2n+1\\
&\qquad \vdots &\qquad \\
a_{m-1}&=1+\frac {2m-2}{2m-3}\cdot {(2m-3)}&&=2m-1\\
S\;\;\;&=\frac 1{2m-1}a_{m-1}&&=1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Closed form of the elliptic integral $\int_0^{2\pi} \sqrt{1+\cos^2(x)}\,dx $ I want to prove the closed form shown in Wikipedia for the arc length of one period of the sine function.
Source of wikipedia
$$\int_0^{2\pi} \sqrt{1+\cos^2(x)} \ dx= \frac{3\sqrt{2}\,\pi^{\frac32}}{\Gamma\left(\frac{1}{4}\right)^2}+\frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{2\pi}}$$
Could someone offer some demonstration for this statement?
| $$\int_0^{2\pi} \sqrt{1+\cos^ 2 x} dx = 4 \int_0^1 \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} dx = 4\int_0^1 \frac{1+x^2}{\sqrt{1-x^4}} dx $$
Now use the beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$
I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm missing?
Thanks in advance
| We know that, $\sin^2\theta+\cos^2\theta = 1$ and
$a^2-b^2=(a-b)(a+b$
then
\begin{split} \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} &= &\frac{\color{red}{\sin^2\theta+\cos^2\theta} +\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \qquad\quad\\\\&=& \frac{\color{red}{\sin^2\theta+(-i\cos\theta)(i\cos\theta)} +\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \\
\\ &=&\frac{\color{red}{[\sin^2\theta- (i\cos\theta)^2]}+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \\\qquad~~~\qquad\\&=&\frac{\color{red}{(\sin\theta +i\cos\theta)(\sin\theta- i\cos\theta)}+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \\\\&=&(\sin\theta +i\cos\theta)\frac{1+\sin\theta- i\cos\theta}{1+\sin\theta-i\cos\theta} \\&=&(\sin\theta +i\cos\theta) \end{split}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Find two arithmetic progressions of three square numbers I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:
\begin{align}
\ & a^2 +r = b^2 \\
& b^2 +r = c^2 \\
& a^2 +c^2 = 2\,b^2 \\
\end{align}
and
\begin{align}
\ & d^2 +r = e^2 \\
& e^2 +r = f^2 \\
& d^2 +f^2 = 2e^2 \\
\end{align}
where $a,b,c,d,r \in \Bbb N$.
Here is an example that almost works:
\begin{align}
\ & 23^2 +41496 = 205^2 \\
& 205^2 + 41496 = 289^2 \\
& 23^2 +289^2 = 2\,(205)^2 \\
\end{align}
and
\begin{align}
\ & 373^2 + 41496 = 425^2 \\
& 425^2 + 41496 = \color{#C00000}{222121} \\
& 23^2 + \color{#C00000}{222121} = 2\,(205)^2 \\
\end{align}
where the difference is $41496$, but the last element isn't a square number.
I can't find an example of two progressions with three numbers and the same common difference. Could you demonstrate that such progressions are nonexistent using reductio ad absurdum to this statement?
| There are infinitely many solutions to the system,
\begin{align}
\ & a^2 +r_1 = b^2 \\
& b^2 +r_1 = c^2 \\
& a^2 +c^2 = 2b^2 \\
\hline
\ & d^2 +r_2 = e^2 \\
& e^2 +r_2 = f^2 \\
& d^2 +f^2 = 2e^2 \\
\end{align}
with $\color{blue}{r_1=r_2}$. Eliminating $r_1$ between the first two equations (and similarly for $r_2$), one must solve the Pythagorean-like,
$$a^2+c^2=2b^2\\ d^2+f^2=2e^2$$
which has solution,
$$a,b,c = p^2 - 2q^2,\; p^2 + 2p q + 2q^2,\; p^2 + 4p q + 2q^2\\ d,e,f = r^2 - 2s^2,\; r^2 + 2r s + 2s^2,\; r^2 + 4r s + 2s^2$$
Hence,
$$r_1 = -a^2+b^2 = -b^2+c^2 = 4 p q (p + q) (p + 2 q)\\ r_2 = -d^2+e^2 = -e^2+f^2 = 4 r s (r + s) (r + 2 s)$$
Thus one must solve,
$$p q (p + q) (p + 2 q) = r s (r + s) (r + 2 s)$$
This is essentially the same equation in this post, hence one solution (among many) is,
$$p,\;q = 2 n (m + 6 n),\; m (m + 4 n)\\
\;r,\;s = m (m + 2 n),\; 4 n (m + 3 n)$$
For example, let $m,n = 1,1$, then,
$$a,b,c = 146, 386, 526\\ d,e,f = 503, 617, 713\\ r_1=r_2= 127680$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Differential equations for chemical reaction $\mathrm{A + 2B \to 3C}$
In a chemical reaction $\mathrm{A + 2B \to 3C}$, the concentrations $a(t)$, $b(t)$ and $c(t)$ of the three substances A, B and C measure up to the differential equations
$$
\begin{align}
\frac{da}{dt} &= -rab^2\tag{1}\\
\frac{db}{dt} &= -2rab^2\tag{2}\\
\frac{dc}{dt} &= -3rab^2\tag{3}
\end{align}
$$
with $r > 0$ and begin condition $a(0) = 1$ and $b(0) = 2$.
Show that $b(t) - 2a(t) = 0$ .
Here is my solution, but it is not right. Any help would be great.
First equation
$$
\begin{align}
\int\frac{da}{a} &= -rb^2\int dt\\
\ln(a) &= -rb^2t + C\\
a(t) &= e^{-rb^2t+ C}\\
a(0) &= 1 &&\to &1 &= e^{0 + C}\\
\ln(1) &= C &&\to &C &= 0\\
a(t) &= e^{-rb^2t}
\end{align}
$$
Second equation
$$
\begin{align}
\frac{db}{dt} &= -2rab^2\\
\int \frac{db}{b^2} &= -2ra\int dt\\
b &= \frac{1}{2rat + C}\\
b(0) &= 2 \quad \to \quad 2 = \frac{1}{C} \quad \to \quad C = \frac{1}{2}\\
b(t) &= \frac{2}{ 2rat + 1}
\end{align}
$$
But now $b(t) - 2a(t) \ne 0$. Where I am making mistake? Any tip will be enough.
| Notice that, by your equations,
$\dfrac{d(2a - b)}{dt} = 2\dfrac{da}{dt} - \dfrac{db}{dt} = -2rab^2 -(-2rab^2) = 0; \tag 1$
hence $2a(t) - b(t)$ is constant. Now
$2a(0) - b(0) = 2(1) - 2 = 0; \tag 2$
the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain any calculus methods.
So I conjugated the parenthesizes two by two, and I got:
\begin{align} f(x)&=\left(x^2-(a+b)^2\right)\left(x^2-(a-b)^2\right) \end{align}
Can somebody help me to take this problem further?
| By your work
$$f(x)=x^4-2(a^2+b^2)x^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-(a^2+b^2)^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-4a^2b^2.$$
Now, we see that
$$f(x)\geq-4a^2b^2$$ and the equality occurs for $x^2=a^2+b^2$, which is possible.
Thus, $$\min_{\mathbb R}f=-4a^2b^2.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to integrate this kind of functions? I was wondering how to evaluate $$\int\frac{sin^4 x}{cos^7 x}dx$$
I tried the usual method of writing the expression in terms of powers of $tan(x)$ and $sec(x)$, but nothing useful came out of it.
My attempt
$$\int\frac{sin^4 x}{cos^7 x}dx$$$$=\int({tan^4x}) ({sec^3x})dx$$$$=\int(tan^4x)(sec{x})(sec^2x)dx$$$$=\int(t^4)({\sqrt{t^2+1}})dt$$
I haven't got any further yet.
My generalized question
How to evaluate $$\int(sin^mx)(cos^nx)dx$$
where $$m,n\in \mathbb{Q}$$ and $(m+n)$ is a negative odd integer.
| \begin{align*}
\int{\frac{sin^{4}x}{cos^{7}x}}\,dx &= \int{tan^{4}x \cdot sec^{3}x}\,dx\\
&= \int{(sec^{2}x - 1)^2 \cdot sec^3{x}}\,dx\\
&= \int{sec^{7}x}\,dx -2\int{sec^{5}x}\,dx + \int{sec^{3}x}\,dx
\end{align*}
Now,
\begin{align*}
I_{2n+1} &= \int{sec^{2n+1}x}\,dx = \int{sec^{2n-1}x\cdot \sec^{2}x}\,dx\\
&= sec^{2n-1}x\int{sec^{2}x}\,dx - \int{((2n-1)sec^{2n-2}x\cdot \sec{x}\cdot\tan{x}\int{sec^{2}x}\,dx})\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)\int{sec^{2n-1}x\cdot tan^{2}x}\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)\int{sec^{2n+1}x}\,dx + (2n-1)\int{sec^{2n-1}x}\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)I_{2n+1} + (2n-1)\int{sec^{2n-1}x}\,dx
\end{align*}
\begin{align*}
&\Rightarrow \quad 2nI_{2n+1} = sec^{2n-1}x\cdot \tan{x} + (2n-1)\int{sec^{2n-1}x}\,dx\\
&\Rightarrow \quad I_{2n+1} = \frac{1}{2n}\left(sec^{2n-1}x\cdot \tan{x} + (2n-1)\int{sec^{2n-1}x}\,dx\right)
\end{align*}
We know, $\int{\sec{x}}\,dx = \log{\vert\sec{x}+\tan{x}\vert} + C$.
Now $n = 1, 2, 3$, give respectively, the expressions of $I_3, I_5, I_7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.
First, I tried to simplify the proof statement but I got an even more complicated: $$a^4b^4+b^4c^4+a^4c^4> a^2b^3c^3+b^2c^3a^3+a^3b^3c^2$$
Then I used Power mean inequality on $\dfrac{1}{a},\dfrac{1}{b},\dfrac {1}{c}$ but that wasn't useful here.
Finally, I tried to solve it using AM-HM inequality but couldn't.
What would be an efficient method to solve this problem? Please provide only a hint and not the entire solution since I wish to solve it myself.
| AM-GM helps!
$$\sum_{cyc}\frac{ab}{c^3}=\frac{1}{4}\sum_{cyc}\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)\geq\frac{1}{4}\sum_{cyc}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}\right)=\sum_{cyc}\frac{1}{c}.$$
Done!
Without $cyc$ we can write the solution so:
$$\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}=$$
$$=\frac{1}{4}\left(\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{2bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{2ca}{b^3}\right)\right)\geq$$
$$\geq\frac{1}{4}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{bc}{a^3}\right)^2\cdot\frac{ab}{c^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{ca}{b^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ab}{c^3}}\right)=$$
$$=\frac{1}{c}+\frac{1}{a}+\frac{1}{b}.$$
The same trick gives also a proof by Holder:
$$\sum_{cyc}\frac{ab}{c^3}=\sqrt[4]{\left(\sum_{cyc}\frac{ab}{c^3}\right)^2\sum_{cyc}\frac{bc}{a^3}\sum_{cyc}\frac{ca}{b^3}}\geq\sum_{cyc}\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}=\sum_{cyc}\frac{1}{c}.$$
Turned out even a bit of shorter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proving equivalence of norms in $\mathbb{R}^2$
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2.$$
My idea:
$$
k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2
\\
\Leftrightarrow
k\sqrt{|x|^2+|y|^2} \leq \sqrt{x^2+2xy+3y^2} \leq K\sqrt{|x|^2+|y|^2}
$$
Now how can I find $k,K$? The problem is these constants must be minimal.
Can someone give me a tip?
Thx
| Here's a version that's more explicitly geometric, but whose underlying mathematics resemble Roberto's matrix diagonalization.
Rewrite the norm in rotated coordinates $(x', y')$, where $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$. We'll choose $\theta$ at our convenience—specifically, we'll choose it such that the coordinate axes of the rotated coordinate system align with the axes of the unit ellipse $||(x, y)||_* = 1$, thus making the $xy$ terms in the norm vanish. In this case, \begin{align*}|| (x, y)||_*^2 =& x'^2 \cos^2 \theta \tag{$x^2$} - 2x' y' \sin \theta \cos \theta + y'^2 \sin^2 \theta \\ &+ 2 (x'^2 - y'^2) \sin \theta \cos \theta + 2x' y' (\cos^2 \theta - \sin^2 \theta) \tag{$2xy$} \\
&+ 3x^2 \sin^2 \theta + 6x' y' \sin \theta \cos \theta + 3y^2 \cos^2 \theta \tag{$3y^2$} \end{align*}
The coefficient of $2x' y'$ is $4 \sin \theta \cos \theta + 2 \cos^2 \theta - 2 \sin^2 \theta = 2 \sin (2\theta) + 2 \cos (2\theta)$. So we can make this term disappear by choosing $\theta = -\pi/8$, for which $$\begin{align*} \sin^2 \theta &= \frac{2 - \sqrt{2}}{4} \\ \cos^2 \theta &= \frac{2 + \sqrt{2}}{4} \\ \sin \theta \cos \theta &= -\frac{\sqrt{2}}{4}\end{align*}$$
We thus have
$$ \begin{align*}
|| (x', y')||_*^2 &= x'^2 (\cos^2 \theta + 2 \sin \theta \cos \theta + 3 \sin^2 \theta) + y'^2 (\sin^2 \theta - 2 \sin \theta \cos \theta + 3 \cos^2 \theta) \\
&= (2 - \sqrt{2}) x'^2 + (2 + \sqrt{2}) y'^2\\ \end{align*}$$
Meanwhile, the ordinary Euclidean norm, invariant under rotation, is $$|| (x', y')||_2^2 = x'^2 + y'^2.$$ The path from here should be evident.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How should one proceed in this trigonometric simplification involving non integer angles? The problem is as follows:
Find the value of this function
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$
when $\omega=33^{\circ}{20}'$ and $\phi=56^{\circ}{40}'$.
Thus,
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right)^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right)^{2}$$
By solving the power I obtained the following:
$$A= \cos^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +\sin^{2}\frac{\omega}{2} +\sin^{2}\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}$$
I noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:
\begin{align}
A &= \cos^{2}\frac{\omega}{2} +\sin^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +\sin^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\
&= 1+1+2\cos\frac{\omega}{2}cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}
\end{align}
Since the latter terms are another way to write prosthapharesis formulas I did the following:
\begin{align}
\cos\alpha +\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\
\cos\alpha -\cos\beta &= -2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \\
\alpha+\beta &= \omega \\
\alpha-\beta &= \phi
\end{align}
By solving the system I found: $\alpha=\frac{\omega+\phi}{2}$ and $\beta=\frac{\omega-\phi}{2}$.
Therefore by inserting these into the problem:
\begin{align}
A &= 1+1 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\
&= 2 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} -\left(-2\sin\frac{\omega}{2}\cos\frac{\phi}{2}\right) \\
&= 2 +\cos\frac{\omega+\phi}{2} +\cos\frac{\omega-\phi}{2} -\left(\cos\frac{\omega+\phi}{2} -\cos\frac{\omega-\phi}{2}\right)
\end{align}
By cancelling elements,
$$A=2 +2\cos\frac{\omega-\phi}{2}$$
However I'm stuck at trying to evaluate these values:
\begin{align}
\omega &= 33^{\circ}{20}'\; \phi=56^{\circ}{40}' \\
\omega-\phi &= \left(33+\frac{20}{60}\right) -\left(56+\frac{40}{60}\right) = -23-\frac{20}{60}
\end{align}
Therefore,
$$A=2+2\cos\left(\frac{-23-\frac{20}{60}}{2}\right).$$
However the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?
| Hint: You did a miscalculation:
$$A=\left (\cos\frac{\omega}{2}+\cos\frac{\phi}{2} \right )^{2}+\left (\sin\frac{\omega}{2}-\sin\frac{\phi}{2} \right )^{2}$$
$$=\cos ^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin ^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2 \frac{\phi}{2}$$
$$=2+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}$$
$$=2+2\left(\cos\frac{\omega}{2}\cos\frac{\phi}{2}-\sin\frac{\omega}{2}\sin\frac{\phi}{2}\right)$$
$$=2+2\cos\left(\frac{\omega}{2}+\frac{\phi}{2}\right)$$
In the last step I used $\cos(x+y)=\cos x\cos y-\sin x \sin y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sqrt{\frac{1+x}{1-x}}}\right)'}_{=D}dx.$$
So, the integrand $D$ remains to simplify:
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2} \quad \quad (1).$$
Setting $a=\frac{1+x}{1-x}$ for notation's sake I get
$$D=x\cdot\frac{1}{1+a}\cdot\frac{1}{2\sqrt{a}}\cdot\frac{1}{(x-1)^2}=\frac{x}{(2\sqrt{a}+2a\sqrt{a})(x^2-2x+1)},$$
and I get nowhere. Any tips on how to move on from $(1)?$
NOTE: I don't want other suggestions to solutions, I need help to sort out the arithmetic to the above from equation (1).
| Use change of variable
$$\theta=\arctan\sqrt{\frac{1+x}{1-x}}\in[0,\frac{\pi}{2}).$$
Then we have
$$x=\frac{\tan^2\theta-1}{\tan^2\theta+1}=\sin^2\theta-\cos^2\theta=-\cos2\theta.$$
Therefore
\begin{align}
\int\arctan\sqrt{\frac{1+x}{1-x}}dx&=-\int\theta\,d\cos 2\theta=-\theta\cos 2\theta+\int\cos 2\theta d\theta\\
&=-\theta\cos 2\theta+\frac{1}{2}\sin 2\theta+C\\
&=x\arctan\sqrt{\frac{1+x}{1-x}}+\frac{1}{2}\sqrt{1-x^2}+C.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Strange symmetry regarding sum $\sum_{n=0}^\infty\frac{n^ne^{-bn}}{\Gamma(n+1)}$ and integral $\int_{0}^\infty\frac{x^xe^{-bx}}{\Gamma(x+1)}dx$ One can show by computation the following for $b>1$
$$\sum_{n=0}^\infty\frac{n^ne^{-b n}}{\Gamma(n+1)}=\frac{1}{1+W_{\color{blue}{0}}(-e^{-b})},\tag{1}$$
(here one assumes that the term with $n=0$ is understood as the limit $\lim_{n\to 0}$ and is equal to $1$) and
$$\int_{0}^\infty\frac{x^xe^{-b x}}{\Gamma(x+1)}dx=\boldsymbol{\color{red}{-}}\frac{1}{1+W_{\color{red}{-1}}(-e^{-b})}.\tag{2}$$
$W_0$ and $W_{-1}$ are different branches of the Lambert W function. One can see that this formulas look similar. I considered them in the hope of obtaining a function for which sum equals integral:
$$
\sum_{n=0}^\infty f(n)=\int_0^\infty f(x) dx.
$$
$(1)$ is the consequence of Lagrange inversion and the integral arises in the probability distribution theory, namely the Kadell-Ressel pdf (see also this MSE post).
Question 1. Can anybody explain the symmetry between $(1)$ and $(2)$ without resorting to direct calculation?
Question 2. Is it possible to alter $(1)$ and $(2)$ to obtain a nice function for which sum equals integral?
If $b=1$ then there is the Knuth series
$$
\sum_{n=1}^\infty\left(\frac{n^ne^{-n}}{\Gamma(n+1)}-\frac1{\sqrt{2\pi n}}\right)=-\frac23-\frac1{\sqrt{2\pi}}\zeta(1/2),\tag{3}
$$
and the "Knuth integral"
$$
\int_0^\infty\left(\frac{x^xe^{-x}}{\Gamma(x+1)}-\frac1{\sqrt{2\pi x}}\right)dx=-\frac13.\tag{4}
$$
Again we see there is a discrepancy.
Question 3. Is it possible to modify the term $\frac1{\sqrt{2\pi x}}$ in $(3)$ and $(4)$ so that the series and the integral agree?
Edit. Of course by mounting some additional terms and parameters one can come up with a formula that technically answers question 2 or 3. What is meant as nice in question 2 might be difficult to formulate explicitly. It is best illustrated by formulas in this MSE post.
|
Question 2. Is it possible to alter (1) and (2) to obtain a function
for which sum equals integral?
A simpler form for $z\in[0,\mathrm{e}^{-1})$:
\begin{align}
\sum_{n=0}^\infty
\frac{(z\,n)^n}{\Gamma(n+1)}
&=
\frac1{1+\operatorname{W}_{0}(-z)}
\tag{1}\label{1}
,\\
\int_0^\infty
\frac{(z\,x)^x}{\Gamma(x+1)}\,dx
&=-\frac1{1+\operatorname{W}_{-1}(-z)}
\tag{2}\label{2}
.
\end{align}
For some $u\in\mathbb{R}$ consider
\begin{align}
\sum_{n=0}^\infty \frac{u}{(n+1)^2}
&=\frac{u\pi^2}6
\tag{3}\label{3}
,\\
\int_0^\infty \frac{u}{(x+1)^2}\,dx&=u
\tag{4}\label{4}
.
\end{align}
Let's add \eqref{3} and \eqref{4}
to \eqref{1} and \eqref{2}, respectively:
\begin{align}
\sum_{n=0}^\infty
\left(
\frac{(z\,n)^n}{\Gamma(n+1)}
+\frac{u}{(n+1)^2}
\right)
&=
\frac1{1+\operatorname{W}_{0}(-z)}
+\frac{u\pi^2}6
\tag{5}\label{5}
,\\
\int_0^\infty
\left(
\frac{(z\,x)^x}{\Gamma(x+1)}
+\frac{u}{(x+1)^2}
\right)
\,dx
&=-\frac1{1+\operatorname{W}_{-1}(-z)}
+u
\tag{6}\label{6}
.
\end{align}
From the right hand sides of \eqref{5} and \eqref{6} for any $z\in[0,\mathrm{e}^{-1})$
we have
\begin{align}
u&=
-6\frac{2+\operatorname{W_0}(-z)+\operatorname{W_{-1}}(-z)}{(\pi^2-6)(1+\operatorname{W_0}(-z))(1+\operatorname{W_{-1}}(-z))}
\end{align}
such that the pair $(z,u)$ satisfies \eqref{5}=\eqref{6}.
For example,
\begin{align}
z&=\tfrac12\ln2
,\quad\operatorname{W_0(-z)}=-\ln2,\quad\operatorname{W_{-1}(-z)}=-2\ln2
,\\
&\sum_{n=0}^\infty
\left(
\frac{(n\ln2)^n}{2^n\Gamma(n+1)}
-
\frac{6(2-3\ln2)}{
(\pi^2-6)(1-\ln2)(1-2\ln2)(n+1)^2
}
\right)
\\
=&
\int_{0}^\infty
\left(
\frac{(x\ln2)^x}{2^x\Gamma(x+1)}
-
\frac{6(2-3\ln2)}{
(\pi^2-6)(1-\ln2)(1-2\ln2)(x+1)^2
}
\right)
\\
=&
\frac{\pi^2(\ln2-1)+6(2\ln2-1)}{
(\pi^2-6)(\ln2-1)(2\ln2-1)
}
\approx 1.549536
.
\end{align}
Edit
Similarly,
\begin{align}
&\sum_{n=0}^\infty
2^{-n}
\left(
\frac{(n\ln2)^n}{\Gamma(n+1)}
+
\frac{\ln2\,(3\ln2-2)}{
(\ln2-1)(2\ln2-1)^2
}
\right)
\\
=&
\int_{0}^\infty
2^{-x}
\left(
\frac{(x\ln2)^x}{\Gamma(x+1)}
+
\frac{\ln2\,(3\ln2-2)}{
(\ln2-1)(2\ln2-1)^2
}
\right)
\\
=&
\frac{2(\ln2)^2-1}{
(\ln2-1)
(2\ln2-1)^2
}
\approx 0.8537740
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Integrate $f(x)=\sqrt{x^2+2x+3}.$ Completing the square and letting $t=x+1$, I obtain $$\int\sqrt{(x+1)^2+2} \ dx=\int\sqrt{t^2+2}\ dt.$$
Letting $u=t+\sqrt{t^2+2},$ I get
\begin{array}{lcl}
u-t & = & \sqrt{t^2+2} \\
u^2-2ut+t^2 & = & t^2+2 \\
t & = & \frac{u^2-2}{2u} \\
dt &=& \frac{u^2+2}{2u^2}du
\end{array}
Thus the integral becomes
$$F(x)=\int \left(u-\frac{u^2-2}{2u}\right)\left(\frac{u^2+2}{2u^2}\right) \ du = \int \left(\frac{u^2+2}{2u}\right)\left(\frac{u^2+2}{2u^2}\right) \ du =\int\frac{u^4+4u^2+4}{4u^3} \ du.$$
This integrand is nicely divided into
\begin{array}{lcl}
F(x) & = & \frac{1}{4}\int u \ du+\int \frac{1}{u} \ du+\int \frac{1}{u^3}=\frac{u^2}{8}+\ln{|u|}-\frac{1}{2u^2}+C \\
& = & \frac{(t+\sqrt{t^2+2})^2}{8}+\ln{|t+\sqrt{t^2+2}|}-\frac{1}{2(t+\sqrt{t^2+2})^2}+C \\
\end{array}
And finally in terms of $x$:
$$F(x)=\frac{(x+1+\sqrt{x^2+2x+3})^2}{8}+\ln{|x+1+\sqrt{x^2+2x+3}|}-\frac{1}{2(x+1+\sqrt{x^2+2x+3})^2}+C.$$
The answer in the book is:
$$F(x)=\frac{1}{2}\left((x+1)\sqrt{x^2+2x+3}+2\ln{|x+1+\sqrt{x^2+2x+3}}|\right)+C.$$
Can anyone help me identify where I missed what?
| Here is how I would work it.
$t = \sqrt 2 \tan \theta\\
dt = \sqrt 2 \sec^2 \theta$
$\int 2\sec^3 \theta\\
\sec\theta\tan\theta + \ln [\sec\theta+\tan\theta]+C\\
\frac {1}{2} t\sqrt {t^2 + 2} + \ln \frac 12 (t+\sqrt{t^2 + 2}+ C\\
\frac {1}{2} (x+1)\sqrt {x^2 + 2x + 3} + \ln [x+1 + \sqrt {x^2 + 2x + 1}]+ C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$
Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$ for $x,y,z \in \mathbb{Z}$.
We must have $x^3+y^3 = 1006d$ and $x^3+z^3 = 1001d$ where $d$ is an integer. This means that $x^3+y^3 \equiv 0 \pmod{1006}$ and $x^3+z^3 \equiv 0 \pmod{1001}$. Rearranging the equation we also get $1001x^3+1001y^3 = 1006x^3+1006z^3$ so $$1001y^3 = 5x^3+1006z^3.$$ How can we continue?
| Well, at least we have some solutions, like
$$\frac{669^3 + 337^3}{669^3 + 332^3} = \frac{1006}{1001}$$
This should be studied with elliptic curves.
May assume $x$, $y$, $z$, rationals, and then may even assume $x=1$. We get
$$\frac{y^3+1}{z^3+1} = \frac{1006}{1001}\\
z^3 +1 = \frac{1001}{1006}(y^3 + 1)$$
This is an elliptic curve, and it has an (easy) rational point ( does not give yet a solution to the problem) $(-1,-1)$. Now, take the tangent to the curve at this point and consider the intersection with the curve. It will be another rational point $(y,z) = (\frac{337}{669}, \frac{332}{669})$.
I lack expertise in elliptic curves, so will leave it here.
$\bf{Added:}$ Took the tangent line to the curve at the last point and intersected it with the curve. Got another point
$$(y,z) = (-11901775977431/50258598213909 , -13216294942936/50258598213909) $$
I guess the question now is whether our elliptic curve has finitely many rational points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$ Question:
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$
My attempts:
*
*Here's the expanded form:$$x^4-6x^3+12x^2-12x+4=0$$
*I've plugged this into several online "math problem solving" websites, all claim that "solution could not be determined algebraically, hence numerical methods (i suppose the quartic formula?) were used"
*I've substituted $y=x^2+2$ but then the $6x$ remains to prevent me from solving.
*I've tried factorizing in other ways, I've tried finding simple first solutions but they are actually not simple so I couldn't find them.
*Factoring into circles and hyperbola to arrive at a geometric solution, by estimating the roots from the graph
For reference, the roots are: (credits to wolframalpha)
$$2+\sqrt{2}, 2-\sqrt{2},1+i, 1-i$$
| $$(x^2+2)^2+8x^2=6x(x^2+2)$$
$$(x^2+2)^2-6x(x^2+2)+8x^2=0$$
Let $x^2+2=U, x=V$. Then
$$U^2-6UV+8V^2=0$$
Then
$$\left(\frac UV\right)^2-6\left(\frac UV\right)+8=0$$
Then
$\frac UV=2$ or $\frac UV=4$
$\frac {x^2+2}{x}=2$ or $\frac {x^2+2}{x}=4$
$x^2-2x+2=0$ or $x^2-4x+2=0$
$$x=2\pm \sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given equations for the side-lines of a parallelogram, why are these the equations for the diagonal-lines? My book, for a parallelogram $ABCD$ with sides as
$$\begin{align}
AB&\;\equiv\; a\phantom{^\prime}x+b\phantom{^\prime}y +c\phantom{^\prime}=0 \\
BC&\;\equiv\; a^\prime x +b^\prime y +c^\prime=0 \\
CD&\;\equiv\; a\phantom{^\prime}x+b\phantom{^\prime} y +c^\prime=0 \\
DA&\;\equiv\; a^\prime x +b^\prime y +c\phantom{^\prime}=0
\end{align}$$
wrote equation of diagonals:
$$AC\;\equiv\; (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0$$
$$BD\;\equiv\; (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0$$
I don't understand why. Please help.
| As Michael Rozenberg has written, they should be
$$\small BD\equiv (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0\tag1$$$$\small AC\equiv (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0\tag2$$
why the equations of diagonals (taking AC for instance) is for the entire line AC and not just points A and C?
$(1)$ can be written as
$$aa'x^2+ab'xy+acx+a'bxy+bb'y^2+bcy+ca'x+cb'y+c^2-aa'x^2-a'bxy-a'c'x-ab'xy-bb'y^2-b'c'y-ac'x-bc'y-c'^2=0,$$
i.e.
$$(a+a')(c-c')x+(b+b')(c-c')y+(c-c')(c+c')=0\tag3$$
Suppose that $c=c'$. Then, the equations of $AB$ and $CD$ are the same, which is impossible. So, we have $c\not=c'$.
So, dividing the both sides of $(3)$ by $c-c'$ gives
$$(a+a')x+(b+b')y+c+c'=0\tag4$$
Suppose that $a+a'=0$ and $b+b'=0$. Then, the equation of $DA$ is $ax+by-c=0$. So, the line $DA$ is parallel to the line $AB$, which is impossible. So, we have $a+a'\not=0$ or $b+b'\not=0$.
It follows that $(4)$, i.e. $(1)$ is the equation of the line $BD$ (assuming that you've already known that $B,D$ satisfy the equation).
Also, $(2)$ can be written as$$(a-a')x+(b-b')y=0\tag5$$
Suppose that $a-a'=0$ and $b-b'=0$. Then, the equations of $AB$ and $DA$ are the same, which is impossible. So, we have $a-a'\not=0$ or $b-b'\not=0$.
It follows that $(5)$, i.e. $(2)$ is the equation of the line $AC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Odd number proof
Prove that for every odd number $n$, it holds that $n^2+3$ is not not divisible
by $8$.
My idea:
Let $n=2k+1$ for $k \in \mathbb{N}$, which implies
$$n^2+3=(2k+1)^2+3=4k^2+4k+1+3=4k^2+4k+4$$
How can I conclude that I cannot divide $4k^2+4k+4$ by $8$?
| Just for fun, a slightly different approach:
If $n^2+3$ is divisible by $8$, then so is $(n-4)^2+3=n^2+3-8n+16$, hence, by induction, $8$ divides $n^2+3$ for at least one $n$ between $-2$ and $2$. But $0^2+3=3$, $(\pm1)^2+3=4$, and $(\pm2)^2+3=7$ are not divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.
| Note that $1$ and $\sqrt{2}$ is a basis for $\mathbb{Q}(\sqrt{2}).$
Hence from
$$3=a^2+b^2+2\sqrt{2}ab$$
We have $a^2+b^2=3$ and $2ab=0$.
From there, you should be able to see a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one
| $S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +... $
Take 1/4 common from the terms whose denominator is even.
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} ( 1 + \frac{1}{4} + \frac{1}{9} + ... ) $
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} S$
$ 1 + \frac{1}{9} + \frac{1}{25} + ...= \frac{3}{4} S = (3/4)(\pi^2 /6) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine constant c so that g(x,y) is continuous at every point $$g(x,y)=\begin{cases} \frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2} & \text{if} & (x,y) \neq (0,0) \\
c & \text{if} & (x,y) = (0,0) \end{cases}$$
Should I set the first function equal to c and then solve using polar coordinates?
| Hint. By using polar coordinates one gets, for $(x,y)\ne (0,0)$,
$$
g(x,y)=\frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2}=\frac{r^3\cos^3 \theta+r^3\cos\theta \sin^2\theta +2r^2}{r^2},\quad r\ne0,
$$ that is
$$
g(x,y)=r\cos^3 \theta+r\cos\theta \sin^2\theta +2,\quad r\ne0,
$$ then this tends to $2$ as $r \to 0^+$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a good approach? $0<|(x^5)<(x^2)^2|=|x|$ so $|x^5+y^5|<|(x^2+y^2)^2|$ so therefor
$$0< \dfrac{|x^5+y^5|}{|(x^2+y^2)^2|}<1$$ then multiply both sides with $\ln(x^2+y^2)$. and take the $$\lim_{(x,y)\to (0,0)} \ln(x^2+y^2).$$ and since it is not $0$ but its negative infinity the limit doesn't exist by the squeeze theorem. which is not the right answer.
| Hint. Note that by letting $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, we have that as $(x,y)\to(0,0)$ then $\rho\to 0$ and
$$0\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right|\leq
\frac{(|\rho\cos\theta|^5+|\rho\sin\theta|^5)|\ln(\rho^2)|}{\rho^4}
\leq \frac{\rho^5(1+1)|\ln(\rho^2)|}{\rho^4}={4\rho|\ln(\rho)|}.$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Inequality proof as a part of calculus lesson As part of a calculus lesson I was required to prove that:
(1) if $\ |x-3| < \frac{1}{2},\ $ then $\ \bigg|\displaystyle{\frac{\sin(x^2 -8x+15)}{4x-7}}\bigg| < \frac{1}{2}$
So, by using $|\sin(t)| \le |t|,$ I can prove that:
$$\bigg|\frac{\sin(x^2 -8x+15)}{4x-7}\bigg| \le \frac{5}{6}|x-3|$$
After proving this inequality I assume that the initial requirement is proved, and therefore I'm done, is that it or am I missing something?
Thanks for your assistance :)
| Because, $x^{2}-8x+15=(x-3)(x-5)\ $, you can start saying that:
First,
$$ \vert{\sin(x^{2}-8x+15)}\vert\leq {x^{2}-8x+15} $$
and then you start like this
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert=\frac{\vert{\sin(x^{2}-8x+15)}\vert}{\vert{4x-7}\vert}\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}$$
Suppose that $\ \displaystyle{\vert{x-3}\vert<\frac{1}{2}}\ $, then
$$ -\frac{1}{2}<x-3<\frac{1}{2}$$
$$ 3-\frac{1}{2}<x<3+\frac{1}{2} $$
$$ \frac{5}{2}<x<\frac{7}{2} $$
and with the last condition you can bound $\displaystyle{\frac{1}{\vert{4x-7}\vert}}$ and $\vert{x-5}\vert$. In fact,
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow \frac{5}{2}-5<x-5<\frac{7}{2}-5 $$
this means
$$ -\frac{5}{2}<x-5<-\frac{3}{2} $$
and the absolute value function is decreasing in the negatives real numbers:
$$ \frac{3}{2}<\vert{x-5}\vert<\frac{5}{2}, $$
so you have $$ \vert{x-5}\vert<\frac{5}{2}. $$
Also,
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow 10<4x<14 \Rightarrow 3<4x-7<7$$
so $\vert{4x-7}\vert>3$ and this implies:
$$ \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}. $$
Then you can multiply both inequalities below:
$$ \vert{x-5}\vert<\frac{5}{2} \qquad \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}$$
to get:
$$ \frac{\vert{x-5}\vert}{\vert{4x-7}\vert}<\frac{5}{6}. $$
So,
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}<\frac{5}{6}\vert{x-3}\vert<\frac{5}{6}\cdot\frac{1}{2}=\frac{5}{12}<\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
$$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$
i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.
I don't get the reasoning completely.
| We show a more general fact:
There is no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=d^2$ where $d$ is a non-negative integer.
We first note that $d\leq b\leq 9$. Since $b^2-4ac=d^2$, we have that
$$P(x):=ax^2+bx+c=a\left(x-\frac{-b+d}{2a}\right)\left(x-\frac{-b-d}{2a}\right)$$
Assume that $P(10)=\overline{abc}=p$ where $p$ is some prime of three digits. Then
$$4ap=\left(20a+b-d\right)\left(20a +b+d\right)$$
which implies that $p$ divides $\left(20a+b-d\right)$ or $\left(20a+b+d\right)$.
Now $1\leq a\leq 9$, $0\leq b\leq 9$, and
$$0<\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 9+9+9\leq 198.$$
Hence $a=1$, and
$$\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 1+9+9\leq 38.$$
Therefore $p>100$ can not divide any of them and we have a contradiction!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the limit of the complex function. $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)}.$$
I have simplified this limit to this extent :
$$e^{ \lim_{x\to a} \left(\left(1- \frac{x}{a}\right){\left(\tan \frac{\pi x}{2a}\right)}\right)}$$
I don't know how to simplify the limit after that. please help.
| \begin{align*}
\log\left(2-\frac{x}{a}\right)^{\tan(\pi x/2a)}&=\left(\tan\frac{\pi x}{2a}\right)\left(\log\left(2-\frac{x}{a}\right)\right)\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\log\left(1+\left(1-\dfrac{x}{a}\right)\right)}{\cos\left(\dfrac{\pi}{2}\left(\dfrac{x}{a}-1\right)+\dfrac{\pi}{2}\right)}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\left(1-\dfrac{x}{a}\right)-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)^{2}+\cdots}{\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)\right)^{3}+\cdots}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{1-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)+\cdots}{\dfrac{\pi}{2}-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\right)^{3}\left(1-\dfrac{x}{a}\right)^{2}+\cdots},
\end{align*}
taking limit as $x\rightarrow a$, then $\log\left(2-\dfrac{x}{a}\right)^{\tan(\pi x/2a)}\rightarrow\dfrac{2}{\pi}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Frobenius method series solution Q: $x^2y^{''}-(x^2+2)y=0$ [1]
Solving using frobenius
$y=\sum_{n=0}^{\infty}a_nx^{x+r}$ [2]
$ y'=\sum_{n=0}^{\infty}(n+r)a_nx^{x+r-1}$ [3]
$ y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}$ [4]
inserting [2,4] into [1]
$x^2 \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}-(x^2+2)\sum_{n=0}^{\infty}a_nx^{x+r}$ [5]
Multiplying through but the cofficent I get
$\sum_{n=0}^{\infty}(n+r-1)(n+r)a_n x^{n+r}-\sum_{n=0}^{\infty}a_nx^{x+r+2}-2\sum_{n=0}^{\infty}a_nx^{n+r}$
setting n=0 I solve for r to get the values of $r=2 , r=-1$
Now this is where my confusion comes in
so for r=2 I rearnage the series like so
$x^2(\sum_{n=0}^{\infty}[(n+r-1)(n+r)a_n -2a_n]x^{n}-\sum_{n=0}^{\infty}a_nx^{n+2})=0$ [6]
So now what I did I want all x's to be of the same power so I did the following:
$k=n+2$
$k-2=n$
$\sum_{k=0}^{\infty}[(k+1)(k+2)a_n -2a_k]x^k-\sum_{k=2}^{\infty}a_{k-2}x^k=0$ [9]
expanding on the series $\sum_{k=0}^{\infty}[(k+1)(k+2)a_n - 2a_n]x^k$ so I can get the staring point of the series in the same postion I did the following
$k=0, 2a_0-2a_0=0 \rightarrow a_0=0$
$k=1, 6a_{1}-2a_{1} \rightarrow a_1=0$
$a_0=a_1=0$ is because [9] is equal to $0$
so now I have the equation
$\sum_{k=2}^{\infty}[(k+1)(k+2)a_n - 2a_n-a_{k-2}]x^k=0$
So I now have a recurrence relation of:
$(k+1)(k+2)a_n - 2a_n-a_{k-2}=0$
$a_n=\frac{a_{k-2}}{k(k+3)}$
Now in my book solution it has the solution of
$a_2=\frac{a_0}{2\cdot 5}$ $a_3=0$ $a_4=\frac{a_0}{2\cdot 5\cdot 4\cdot 7}$
but how is it $a_0$? I know I have gone somewhere wrong but just cant see where.
| Your approach is fine. There is just a small mistake which causes the problem. In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
Let's consider again the series equation
\begin{align*}
\sum_{k=0}^\infty[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k=0\tag{1}
\end{align*}
We consider the coefficient of $x^0$
\begin{align*}
[x^0]:\qquad 2a_0-2a_0=0\qquad \text{resp.}\qquad 0=0
\end{align*}
We observe this equation provides no information at all. So $k=0$ can't be usefully used (this was the mistake) and $a_0$ is left unspecified. This becomes plausible, when we rearrange the equation (1) somewhat. We obtain
\begin{align*}
\sum_{k=0}^\infty&[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k\\
&=\sum_{k=0}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{1}\\
&=\sum_{k=1}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{2}\\
&=0
\end{align*}
Comment:
*
*In (1) we simplify $(k+1)(k+2)a_k-2a_k$ and observe the coefficient $a_0$ vanishes in the left-hand series since $(k+3)k=0$ if $k=0$.
*In (2) we can therefore start the left series with $k=1$.
From (2) we obtain the expected relations:
\begin{align*}
[x^1]:\qquad &&4a_1&=0&\qquad &\color{blue}{a_1=0}\\
[x^2]:\qquad &&10a_2-a_0&=0&\qquad &\color{blue}{a_2=\frac{a_0}{10}}\\
[x^3]:\qquad && 18a_3-a_1&=0&\qquad &\color{blue}{a_3=0}\\
[x^4]:\qquad &&28a_4-a_2&=0&\qquad &\color{blue}{a_4=\frac{a_2}{28}=\frac{a_0}{280}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Three couples sit at random in a line of six seats, probability that no couple sits together? If three married couples (so 6 people) sit in a row of six seats at random, what is the probability that no couples sit together?
Another way to think about it (couples are AB, CD, and EF)
| There are $6!$ possible seating arrangements. From these, we must exclude those in which one or more couples sit in adjacent seats.
There are three ways to select a couple who sit in adjacent seats. That gives us five objects to arrange, the couple and the other four people. The objects can be arranged in $5!$ ways. The couple that sits together can be arranged internally in $2!$ ways. Hence, there are
$$\binom{3}{1}5!2!$$
seating arrangements in which a couple sits in adjacent seats.
However, if we subtract these seating arrangements from the total, we will have subtracted too much since we have counted seating arrangements in which two couples sit together twice, once for each way we could designate one of the couples as the couple that sits in adjacent seats. Since we only want to subtract such couples once, we must add them back.
There are $\binom{3}{2}$ ways to select two couples that sit together. That gives us four objects to arrange, the two couples and the two other people. The objects can be arranged in $4!$ ways. Each of the two couples that sit in adjacent seats can be arranged internally in $2!$ ways. Hence, there are
$$\binom{3}{2}4!2!2!$$
seating arrangements in which two couples sit together.
When we subtracted arrangements in which a couple sits together, we counted seating arrangements in which all three couples sit together three times, once for each way we could have designated one of those couples as the couple that sits together. When we added arrangements in which two couples sit together, we counted seating arrangements in which all three couples sit together three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the three couples as the ones that sit together. Therefore, we have not excluded seating arrangements in which all three couples sit together at all.
There are $3!$ ways to arrange three couples. Each couple can be arranged internally in $2!$ ways. Hence, the number of seating arrangements in which all three couples sit together is
$$\binom{3}{3}3!2!2!2!$$
By the Inclusion-Exclusion Principle, the number of seating arrangements of the three couples in which no couples sit together is
$$6! - \binom{3}{1}5!2! + \binom{3}{2}4!2!2! - \binom{3}{3}3!2!2!2!$$
The probability that no couple sits together is
$$\frac{6! - \dbinom{3}{1}5!2! + \dbinom{3}{2}4!2!2! - \dbinom{3}{3}3!2!2!2!}{6!} = 1 - \frac{\dbinom{3}{1}5!2! - \dbinom{3}{2}4!2!2! + \dbinom{3}{3}3!2!2!2!}{6!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} = \frac{24}{24}=1 $$
Consider the following example using a different method:
$$ \lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0}\frac{\frac{x^4}{x^4}}{\frac{x^4}{x^4}+\frac{x^2}{x^4}} = \lim_{x \rightarrow 0} \frac {1}{1 +\frac{1}{x^2}} = \frac {1}{1+\infty} = \frac{1}{\infty}=0 $$
The graph here clearly tells me the limit should be $0$, but why does l'Hôpital fail?
| After doing derivative one more time you get $12x^2 +2 $ which is not $0$ when $x$ goes to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Prove 11 does not divide $3^{3k-1}+5*3^k$ for any odd k. First I did an induction proof that it does work for even k. Then I started the proof as so.
Suppose there exists a k of the form 2n+1, s.t 11 divides $3^{3k-1}+5*3^k$.
After some algebra I can arrive at this point $5*2^{6n-1}+3(2^{6n-1}+5*3^{2n})$
Since I proved separately this works for even k, the right side sum if of that form, $(2^{6n-1}+5*3^{2n})$ is divisible by 11. So I believe if I can somehow prove that 11 does not divide any power of 2, I would have finished the proof. However I don't know how to do that.
Someone may have to fix the tags as I'm not entirely sure what is appropriate here, sorry.
| $3^{3k-1}+5\times 3^k=3^{k-1}(3^{2k}+15)=3^{k-1}(x^2+15)$ with $x=3^k$
Also $x^2+15\equiv x^2+4\pmod{11}$
$\begin{array}{l}
k=0: & 3^0\equiv 1\pmod{11} & x^2+4\equiv 5\pmod{11}\\
k=1: & 3^1\equiv 3\pmod{11} & x^2+4\equiv 13\equiv 2\pmod{11}\\
k=2: & 3^2\equiv 9\pmod{11} & x^2+4\equiv 85\equiv 8\pmod{11}\\
k=3: & 3^3\equiv 5\pmod{11} & x^2+4\equiv 29\equiv 7\pmod{11}\\
k=4: & 3^3\equiv 4\pmod{11} & x^2+4\equiv 20\equiv 9\pmod{11}\\
k=5: & 3^5\equiv 1\pmod{11} & \text{and it cycles there...}\\
\end{array}$
So $x^2+4$ is never a multiple of $11$ for any $x=3^k$, and since $3^{k-1}$ is neither divisible by $11$ we have our result for any $k\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$
I try to use this well known: if $a>b>0,c>0$,then we have
$$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
$$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
so we have
$$\sum_{k=2}^{n}\dfrac{1}{3^k-1}<\sum_{k=2}^{n}\dfrac{2}{3^k}=\dfrac{1}{3}$$but this is big than $\dfrac{1}{5}$,so how to prove inequality (1)
| Since $3^k > 6$ for $k \geq 2$, $$\frac{1}{3^k-1}<\frac{1}{3^k-\frac{3^k}{6}}=\frac{6}{5}\cdot\frac{1}{3^k}$$ so $$\sum_{k=2}^{n}\dfrac{1}{3^k-1}<\frac{6}{5}\sum_{k=2}^{n}\dfrac{1}{3^k}<\frac{6}{5}\sum_{k=2}^\infty\dfrac{1}{3^k}=\dfrac{1}{5}$$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
show that $\sum_{k=1}^{n}(1-a_{k})<\frac{2}{3}$ Let $a_{1}=\dfrac{1}{2}$, and such $a_{n+1}=a_{n}-a_{n}\ln{a_{n}}$,show that
$$\sum_{k=1}^{n}(1-a_{k})<\dfrac{2}{3}$$
My attemp: let $1-a_{n}=b_{n}$,then we have
$$b_{n+1}=b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}<\cdots<(b_{1})^{2^{n}}=\dfrac{1}{2^{2^n}}$$
where use $\ln{(1+x)}<x,x>-1$
so
$$\sum_{k=1}^{n}(1-a_{k})<\sum_{k=1}^{n}\dfrac{1}{2^{2^{k-1}}}?$$
But $$\sum_{k=1}^{+\infty}\dfrac{1}{2^{2^{k-1}}}=0.816\cdots$$big than$\frac{2}{3}$,so this inequality How to prove it?
| We can prove that $0< a_{n}< 1$ inductively by making use of the graph$:\quad y= x\left ( 1- \ln x \right ).$ We let
$$b_{n}:= 1- a_{n}, \left \{ b_{n} \right \}_{n= 1}^{\infty}\Leftrightarrow b_{1}= \frac{1}{2}, b_{n+ 1}= b_{n}+ \left ( 1- b_{n} \right )\ln\left ( 1- b_{n} \right )$$
Well$,\quad a_{n+ 1}- a_{n}= -a_{n}\ln n> 0,$ then $b_{n+ 1}< b_{n}\Rightarrow 0< b_{n}< b_{1}= \frac{1}{2},$ according to inequality
$$\ln\left ( 1- x \right )< -x,\quad x\in\left ( 0, 1 \right )$$
So$,\quad n> 2$
$$\Rightarrow b_{n}= b_{n- 1}+ \left ( 1- b_{n- 1} \right )\ln\left ( 1- b_{n- 1} \right )< b_{n- 1}- \left ( 1- b_{n- 1} \right )b_{n- 1}= b_{n- 1}^{2}\Rightarrow b_{n}< b_{2}^{2^{n- 2}}=$$
$$= \left ( \frac{1- \ln 2}{2} \right )^{2^{n- 2}}\Rightarrow\sum_{k= 1}^{n}b_{k}< \sum_{k= 1}^{n}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}< \sum_{k= 1}^{\infty}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}= 0.56\cdots< \frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$ Prove that:
$$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$
I know only this method:
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+....=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-...$
But, unfortunately, I could not a hint.
| My method :
$$\left\{ 1-\frac {1}{2}-\frac {1}{4}-...-\frac {1}{128} \right\}+\left\{ \frac {1}{3}-\frac {1}{6}- \frac{1}{12}-...- \frac{1}{192}\right\}+\left\{\frac {1}{5}-\frac{1}{10}-\frac{1}{20}-...- \frac{1}{160}\right\}+...+\left\{ \frac{1}{99}-\frac{1}{198}\right\}+\left\{ \frac{1}{101}+\frac{1}{103}+\frac{1}{105}+...+\frac{1}{199}\right\}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Help solving variable separable ODE: $y' = \frac{1}{2} a y^2 + b y - 1$ with $y(0)=0$ I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve:
$$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$
The solution is given as
$$y(x) = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma},$$ with $\Gamma = \sqrt{b^2 + 2a}$
I am really getting stuck at this exercise and would love to have someone show me how this solution is derived.
One thing I did find out is that this ODE is variable separable. That is,
$$y' = g(x)h(y) = (1) \cdot (\frac{1}{2} ay^2 + by - 1),$$ and therefore the solution would result from solving
$$\int \frac{1}{\frac{1}{2} ay^2 + by - 1} dy = \int dx + C,$$
where C is clearly zero because $y(0) = 0$.
I am now getting stuck at solving the left integral. Could anyone please show me the steps?
UPDATE
So I came quite far with @LutzL solution, however my answers seems to slightly deviate from the solution given above. These are the steps I performed (continuing from @LutzL's answer):
You complete the square $\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}$ and use this to inspire the change of coordinates $u=ay+b$ leading to
$$
\int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2}
$$
and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you perform a partial fraction decomposition for
$$
\frac{2Γ}{u^2-Γ^2}=-\frac{1}{u+Γ}+\frac{1}{u-Γ}
$$
and find the corresponding logarithmic anti-derivatives,
$$
\ln|u-Γ|-\ln|u+Γ|=Γx+c,\\ \frac{u-Γ}{u+Γ}=Ce^{Γx},\ C=\pm e^c
$$
which you now can easily solve for $u$ and then $y$.
Given that $y(0) = 0$ we have $u(y(0)) = u(0) = b$ and therefore the final equation becomes
$$\frac{u(0)-Γ}{u(0)+Γ}= \frac{b-Γ}{b+Γ}=Ce^{Γ\cdot0} = Ce^{Γ \cdot 0} = C$$
Now by first isolating $u$ I get
$$u - \Gamma = u C e^{\Gamma x} + \Gamma C e^{\Gamma x} \Rightarrow \\
u \left( 1 - C e^{ \Gamma x} \right) = \Gamma \left( 1 + C e^{\Gamma x} \right) \Rightarrow \\
u = \frac{\Gamma \left( 1 + C e^{\Gamma x} \right)}{\left( 1 - C e^{ \Gamma x} \right)}$$
Now substituting u and C gives
$$ay + b= \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right)}{\left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)} \Rightarrow \\
y = \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right) - b \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}{a \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}$$
Now using the fact that $\Gamma = \sqrt{b^2 + 2a} \Rightarrow a = \frac{(\Gamma + b)(\Gamma - b)}{2}$ we get that
$$y = \frac{2 \left( \Gamma \left( 1 + \frac{b-\Gamma }{b+\Gamma } e^{\Gamma x} \right) - b \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right) \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\
= \frac{2 \left( (\Gamma - b) + (\Gamma + b) \frac{b-\Gamma }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\
= \frac{2 \left( (\Gamma - b) - (\Gamma + b) \frac{\Gamma - b }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)}$$
Now cancelling the terms $(b + \Gamma)$ and $(b - \Gamma)$ wherever possible and multiplying denominator and nominator by -1 gives
$$y = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(\Gamma + b) \left( \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} - 1 \right)} $$
So clearly, I got the nominator right, but I can not seem to get the denominator to equal $(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma$. Can someone rescue me and show me what I did wrong?
Maybe it helps if I say that x is always positive?
| You complete the square $\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}$ and use this to inspire the change of coordinates $u=ay+b$ leading to
$$
\int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2}
$$
and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you perform a partial fraction decomposition for
$$
\frac{2Γ}{u^2-Γ^2}=-\frac{1}{u+Γ}+\frac{1}{u-Γ}
$$
and find the corresponding logarithmic anti-derivatives,
$$
\ln|u-Γ|-\ln|u+Γ|=Γx+c,\\ \frac{u-Γ}{u+Γ}=Ce^{Γx},\ C=\pm e^c
$$
which you now can easily solve for $u$ and then $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Divergence of reciprocal of primes, Euler On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
& {} = \sum_{p}\frac{1}{p}+ \sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p
^3} + \sum_{p}\frac{1}{4p^4} + \cdots \\
& {} = \left( \sum_p \frac{1}{p} \right) + K
\end{align}
And then Wikipedia says that $K<1$, without any explanation. How do we know that $$\sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p
^3} + \sum_{p}\frac{1}{4p^4} + \cdots$$ is equal to a constant $K<1$?
| Hint :
$$\sum_p \frac {1}{p^s} < \sum_{n=1}^{\infty} \frac 1{n^s} = \zeta(s) $$
$\zeta(s)$ is Riemann Zeta function, and $\zeta(s)$ converges for each $s \in \Bbb R, s>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve differential equation $y'=\frac{3x-y+4}{x+y}$ ÊDIT: Found the second mistake (failed by calculating $u_{1,2}$)!
I might have made a mistake, however, I am not able to detect it. Here we go:
$$y'=\frac{3x-y+4}{x+y}, \quad y(1)=1$$
1.) set $x = X+a$ and $y=Y+b$, so $$\frac{dY}{dX} = \frac{3X-Y+(3a-b+4)}{X+Y+(a+b)}$$
2.) Choose $a,b$ with $a=-1$ and $b=1$, then the differential equation is:
$$\frac{dY}{dX} = \frac{3X-Y}{X+Y} = \frac{3-\frac{Y}{X}}{1+\frac{Y}{X}}$$
3.) Now substitution: $Y = uX$, so $Y' = u+ Xu'$ and we have
$$\frac{dY}{dX} = \frac{3-u}{1+u} = u+Xu'$$
4.) Solve this, ending with:
$$\frac{1+u}{-u^2-2u+3} du = \frac{dX}{X}$$
5.) Solving this by integration, ending with:
$$-\frac{1}{2} \ln (-u^2-2u+3) = \ln(X) + \ln(C)$$
$$-u^2-2u+3 = \exp(-2(\ln(CX))) = e^{\ln((CX)^{-2})} = \frac{1}{(CX)^2}$$
$$-u^2-2u+3-\frac{1}{(CX)^2} = 0$$
6.) So I get $$u_{1,2} = -1 \pm \sqrt{16-\frac{4}{(CX)^2}}$$
Because of the substitution and by having chosen $Y=y-1$ and $X=x+1$ it follows that
$$y-1 = -x-1 \pm (x+1)\cdot \sqrt{16-\frac{4}{(C(x+1))^2}}$$
Finally:
$$y(x) = -x \pm \sqrt{16(x+1)^2-\frac{4}{C^2}}$$
7.) With the initial values I get
$$C=\frac{1}{\sqrt{15}}$$
8.) However, my result is not true for the differential equation I started with, since left hand side and right hand side are not equal.
Thanks for any advice/hint :)
| Your mistake started from 6 (I'm writing $c=C^2$ here)
$$ u^2 + 2u = 3 - \frac{1}{cX^2} $$
$$ (u+1)^2 = 4 - \frac{1}{cX^2} $$
$$ u= -1 \pm \sqrt{4-\frac{1}{cX^2}} $$
or
$$ \frac{y-1}{x+1} =-1 \pm \sqrt{4-\frac{1}{c(x+1)^2}} $$
Using the condition $x = 1, y = 1$ we get
$$ -1 \pm \sqrt{4-\frac{1}{4c}} = 0 $$
Only the plus sign satisfies
$$ c = \frac{1}{12} $$
Final solution
$$ y(x) = 1 + (x+1)\left(-1 + \sqrt{4-\frac{12}{(x+1)^2}} \right) $$
Or
$$ y(x)= -x + 2\sqrt{x^2+2x-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ I've had some hard time determining Galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ because I didn't know exactly how to compute the order of the elements. See here for the computation of the orders and the order of the group.
However, I'm arriving at two contradictory conclusions. I get that the order of the group is 6 and that all the non-identity elements are of order 2. However, I know this cannot be the case since there are only the diedric group and the cyclic group of order 6 and in both you find elements of order three.
What am I doing wrong?
\begin{array}{|c|c|c|c|}
\hline
notation & \sqrt[3]{3} & \sqrt{-3} & order \\ \hline
Id & \sqrt[3]{3} & \sqrt{-3} & 1\\ \hline
& \sqrt[3]{3} & -\sqrt{-3} & 2\\ \hline
& \omega \sqrt[3]{3}& \sqrt{-3}& 2\\ \hline
& \omega \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
& \omega^2 \sqrt[3]{3}& \sqrt{-3}& 2\\ \hline
& \omega^2 \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
\end{array}
I do the computations considering that $\omega = -\frac{1}{2}+\frac{\sqrt{-3}}{2}$. So for instance if I compute the order of the isomorphism that sends $\sqrt[3]{3} \mapsto \omega^2 \sqrt[3]{3}$ and $\sqrt{-3} \mapsto -\sqrt{-3}$ I observe that two applications of this isomorphism on $\sqrt{-3}$ will give the identity and for the other element I have the following chain $\sqrt[3]{3} \mapsto \omega^2 \sqrt[3]{3} \mapsto \omega \omega^2 \sqrt[3]{3}$ observing that $\omega^2 \mapsto \omega$.
Edit
Let me update the table as I go along:
\begin{array}{|c|c|c|c|}
\hline
notation & \sqrt[3]{3} & \sqrt{-3} & order \\ \hline
Id & \sqrt[3]{3} & \sqrt{-3} & 1\\ \hline
& \sqrt[3]{3} & -\sqrt{-3} & 2\\ \hline
& \omega \sqrt[3]{3}& \sqrt{-3}& 3\\ \hline
& \omega \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
& \omega^2 \sqrt[3]{3}& \sqrt{-3}& 3\\ \hline
& \omega^2 \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
\end{array}
So I conclude the group is diedric of 6 elements. Thanks everyone.
| Wouldn't it be simpler to remark that by definition, $\omega$ is a primitive cubic root of $1$, so that your field $K$ is just the splitting field of the polynomial $X^3 - 3$ ? As such, $K/\mathbf Q$ is normal, with Galois group $G$ isomorphic to the permutation group of the roots, so $G\cong S_3 \cong D_6$, generated by the transposition $\tau:\omega \to \omega^2, \sqrt [3] 3 \to \sqrt [3] 3$ and the 3-cycle $\sigma: \sqrt [3] 3 \to \omega \sqrt [3] 3, \omega \to \omega$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\frac{\pi}{3}\le \iint_D \left(x^2+(y-2)^2\right)^{-1/2}\,dx\,dy\le \pi$ where $D$ is the unit disc.
How to show that $$\frac{\pi}{3}\le \iint_D \frac{dx\ dy}{\sqrt{x^2+(y-2)^2}}\le \pi$$ where $D$ is the unit disc centered at the origin?
I was trying to integrate it using polar coordinates but got stuck as using polar coordinates I get $$\int_{\theta=0}^{2\pi}\int_{r=0}^1 \frac{r dr d\theta}{\sqrt{r^2-2r \sin\theta+4}}$$
I have no idea on how to solve it further.
Also, I think we don't need to evaluate the integral and do some computations to show that it lies between $\pi/3$ and $\pi$.
Can someone please help me out?
Thanks
| We have $D=\{(x,y)\ |\ x^2+y^2\leq 1\}$. Clearly $x^2+(y-2)^2 \geq 1$. So we get:
\begin{align}
\iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \iint_D dx dy = \pi
\end{align}
For the other inequality we note that the function $f(x,y)=x^2 + (y-2)^2$ can only have its maximum on the boundary of $D$ (why?). Let $\phi(x,y)=x^2+y^2-1$, so that $D=\{(x,y)\ |\ \phi(x,y)=0\}$. By Lagrange Multiplier method we must solve:
\begin{align}
\begin{cases}
\nabla f(x,y) + \lambda \nabla \phi(x,y) = 0 \\
\phi(x,y)=0\\
\nabla \phi(x,y) \neq 0
\end{cases}
\end{align}
for some $\lambda\in\mathbb{R}$. Solving this is simple and yields $(x,y)=(0,-1)$ is the maximum of $f(x,y)$ on $D$. So $f(x,y)=x^2+(y-1)^2 \leq f(0,-1) = 9$. We get finally:
\begin{align}
\iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \geq \iint_D \frac{1}{\sqrt[]{9}}dx dy = \frac{\pi}{3}
\end{align}
So:
\begin{align}
\frac{\pi}{3} \leq \iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \pi
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Completing squares with three variables. I want to complete the squares for this polynomial
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))+(\frac{1}{2}(y+\frac{3}{2}z))^2-(\frac{1}{2}(y+\frac{3}{2}z))^2-4yz+2y^2-z^2$$
Then things get kinda messy from here and I get totally lost from then on, could anyone help me out factoring this? And telling me if there is an eaay and non-confusing way to solve it?
| We can do it simply using $(a \pm b )^2 = a^2 \pm 2ab + b^2$
$$2x^2 + 2y^2 -z^2 + 2xy +3xz - 4yz $$
$$ x^2 + y^2 + 2xy + x^2 +2(x)(\frac{3}{2}z)+(\frac{3}{2}z)^2 -(\frac{3}{2}z)^2+y^2 -2(y)(2z)+(2z)^2 -(2z)^2-z^2$$
$$(x+y)^2 +(x+\frac{3}{2}z)^2 +(y-2z)^2 - (\frac{\sqrt{29}}{2}z)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
| $$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$
$$ =\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4} $$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}\right) .\frac{2}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{n+4-n}{n+4}\right).\frac{1}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \right) $
and
$A= \left( \begin {array}{ccc} 0&1&0\\ 0&0&1 \\ 0&0&0\end {array} \right) $ work,
but how can I find all the matrices?
| Note that $A^4=0$. Thus all eigenvalues of $A$ must be $0$ thus its Jordan normal form has one of the following forms
$$
A_1=\left( \begin {array}{ccc} 0&0&0\\ 0&0&0 \\ 0&0&0\end {array} \right) \text{ or }
A_2=\left( \begin {array}{ccc} 0&1&0\\ 0&0&0 \\ 0&0&0\end {array} \right)
\text{ or }
A_3=\left( \begin {array}{ccc} 0&1&0\\ 0&0&1 \\ 0&0&0\end {array} \right)
$$
The first two options are immediately disqualified because their square is $0$. Thus $A$ must be similar $A_3$, i.e. $A= CA_3C^{-1}$ for an invertible C. Now $B = A^2 = C A_3^2 C^{-1} = C B C^{-1}$, thus $B$ and $C$ have to commute. Thus your space of solutions to $A^2 = B$ is given by
$$\{CA_3C^{-1}\vert C \in Gl(\mathbb{R},3), [B,C] = 0\}$$
By solving the linear equation $[B,C] = 0$ you see $C$ satisfies $C \in Gl(\mathbb{R},3), [B,C] = 0$ if and only if it is of the form
$$C=\left( \begin {array}{ccc} \lambda&*&*\\ 0&\mu&* \\ 0&0&\lambda\end {array} \right) $$
for $\lambda,\mu \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.
So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.
We shall find the CDF of $Z$ and then differentiate it.
$F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \ dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
How do I do this? Or otherwise is there different more elegant way to get the required pdf?
| $X$ and $Y$ are standard Cauchy hence $\frac{X+Y}{2}$ is standard Cauchy (what easily can be seen using characteristic functions)
So the pdf of $\frac{3}{2}Z$ is also $$f(t) = \frac{1}{\pi}\frac{1}{1+t^2}$$
Hence the distribution of $Z$ is: $$f(z) = \frac{d}{dz}P(Z \le z) = \frac{d}{dz}F\left(\frac{3}{2}Z \le \frac{3}{2}z\right) = \frac{3}{2}f\left(\frac{3}{2}z\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit :
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Using Maclaurin series for $\cos x $:
$$\cos x=1-\frac{x^2}{2!}\quad \text{(We'll stop at order 2)}$$
This leads to :
$$1-\sqrt{1+x^2}\cos x=1-(1+\frac{x^2}{2})(1-\frac{x^2}{2})=\frac{x^4}{4}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4$$
Using Maclaurin series for $\sin x $:
$$\sin x=x-\frac{x^3}{3!}\quad \text{(We'll stop at order 3)}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4 = \left(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}}\right)^4$$
Thus $$\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\frac{\frac{x^4}{4}}{(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}})^4}=\frac{x^4(1-\frac{x^2}{2})^4}{4(x-\frac{x^3}{3!})}=\frac{(x(1-\frac{x^2}{2}))^4}{4(x-\frac{x^3}{3!})}=\frac{1}{4}(\frac{x-\frac{x^3}{2}}{x-\frac{x^3}{3!}})=\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})
$$
Then $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\lim_{x\rightarrow 0}\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})=\frac{1}{4}.$$
But my book says the solution is $\frac{1}{3}$
Where have I done wrong?
Help appreciated!
| Without using L'Hospital & Taylor's Expansion,
$$=\dfrac{1-(1+x^2)(1-\sin^2x)}{1+\cos x\sqrt{1+x^2}}\cdot\dfrac{\cos^4x}{\sin^4x}$$
$$=\dfrac{x^2\sin^2x+(\sin x-x)(\sin x+x)}{x^4}\cdot\dfrac{\cos^4x}{\left(\dfrac{\sin x}x\right)^4(1+\cos x\sqrt{1+x^2})}$$
Now $\dfrac{(\sin x-x)(\sin x+x)}{x^4}=\left(\dfrac{\sin x}x+1\right)\cdot\dfrac{(\sin x-x)}{x^3}$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$
| It's hard to read your handwriting but it looks like you have the right idea but were sloppy in your execution and made so distributive error.
Assume if $n=k$ then statement is true and
$2^{3k} - 3^k = 5P$ for some integer $P$. (Always a good idea to specify what a variable is whenever you introduce it.)
$2^{3k+1} -3^{k+1}= 2*2^{3k} - 3$.
Theres two errors here: $2^{3(k+1)} \ne 2^{3k+1}$ and ... why did the $3^{k+1}$ turn into a $3$.
You should have:
$2^{3(k+1)} - 3^{k+1} = 2^{3k + 3} - 3^{k+1} = 8*2^{3k} - 3*3^{k}$
Can you go from there?
$8*2^{3k} - 3*3^{k} = 5*2^{3k} + 3*2^{3k} - 3*3^k = 5*2^{3k} + 3(2^{3k} - 3^k) = 5*2^{3k} + 3(5P) = 5(2^{3k} + 3P)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Proving that all the real roots of Hermite polynomials are in $(-\sqrt{4n+1}, \sqrt{4n+1})$ The Hermite polynomials are given by:
$H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$
There is the proof that all the roots are real: https://math.stackexchange.com/a/104875/504137.
And I know the fact that they all are bounded i.e. all the roots lie in $(-\sqrt{4n+1}, \sqrt{4n+1})$.
But how to prove that?
| It doesn't look like there are many questions on this topic on Math.SE, so just for fun let's use some tricks from matrix analysis to slightly improve the bound $\sqrt{2n-2}$ in Jack's answer.
Let $A = [a_{ij}]$ be a real $n \times n$ matrix. Define $|A| = [|a_{ij}|]$. We will write $A \geq 0$ if all $a_{ij} \geq 0$. Let $\rho(A)$ be the absolute value of the largest eigenvalue of $A$.
Theorem. Let $A$ and $B$ be real $n \times n$ matrices. If $B - |A| \geq 0$ then $\rho(A) \leq \rho(|A|) \leq \rho(B)$.
This is theorem 8.1.18 in Horn & Johnson's Matrix Analysis.
Theorem. The eigenvalues of the $n \times n$ matrix
$$
\begin{pmatrix}a&b&&&\\c&a&b&&\\&c&\ddots&\ddots&\\&&\ddots&\ddots&b\\&&&c&a\end{pmatrix}
$$
are
$$
\lambda_k = a + 2\sqrt{bc} \cos\left(\frac{k\pi}{n+1}\right), \qquad k=1,\ldots,n.
$$
This is a known fact about tridiagonal Toeplitz matrices. See, e.g., this PDF.
Using J.M.'s matrix from Jack's answer, the matrix
$$
\begin{pmatrix}0&\sqrt{\frac{n-1}2}&&&\\\sqrt{\frac{n-1}2}&0&\sqrt{\frac{n-1}2}&&\\&\sqrt{\frac{n-1}2}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix} -
\begin{pmatrix}0&\sqrt{\frac12}&&&\\\sqrt{\frac12}&0&\sqrt{\frac22}&&\\&\sqrt{\frac22}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix}
$$
is $\geq 0$ for $n \geq 2$, so, by the two theorems above,
$$
\max_k |\zeta_k| \leq \sqrt{2n-2} \cos\left(\frac{\pi}{n+1}\right).
$$
To get a lower bound we can note that when $n$ is even the characteristic polynomial of the matrix
$$
\begin{pmatrix}
0&\sqrt{\frac12}\\
\sqrt{\frac12}&0&0\\
&0&0&\sqrt\frac32\\
&&\sqrt\frac32&0&0\\
&&&0&0&\sqrt\frac52\\
&&&&\sqrt\frac52&0&\ddots\\
&&&&&\ddots&\ddots&\sqrt\frac{n-1}2\\
&&&&&&\sqrt\frac{n-1}2&0
\end{pmatrix}
$$
is
$$
\prod_{k=0}^{(n-2)/2} \left(\lambda^2 - \frac{2k+1}{2}\right),
$$
and when $n$ is odd the characteristic polynomial of the matrix
$$
\begin{pmatrix}
0&0\\
0&0&\sqrt\frac22\\
&\sqrt\frac22&0&0\\
&&0&0&\sqrt\frac42\\
&&&\sqrt\frac42&0&0\\
&&&&0&0&\ddots\\
&&&&&\ddots&\ddots&\sqrt\frac{n-1}2\\
&&&&&&\sqrt\frac{n-1}2&0
\end{pmatrix}
$$
is
$$
-\lambda \prod_{k=0}^{(n-1)/2} \left(\lambda^2 - k\right).
$$
Since these two matrices are $\leq$ the Hermite matrix, we get the lower bound
$$
\max_k |\zeta_k| \geq \sqrt{\frac{n-1}{2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to figure out if there is an actual horizontal tangent without a graph There is this practice problem that asks to determine the points at which the graph of $y^4=y^2-x^2$ has a horizontal tangent.
So I did implicit differentiation to find that
$$\displaystyle\frac{dy}{dx} = \frac{-x}{2y^3-y}$$
To find the horizontal tangent, I set $\frac{dy}{dx}=0$ and solved for $x$:
$$\begin{align}
\displaystyle\frac{dy}{dx} &= 0 \\
\frac{-x}{2y^3-y} &= 0 \\
-x &= 0 \\
x &= 0
\end{align}$$
Then I substituted $x=0$ into the equation of the curve:
$$\begin{align}
y^4&=y^2-(0)^2 \\
0 &= y^4 - y^2\\
0 &= y^2(y+1)(y-1) \\
y&=-1,\,0,\,1
\end{align}$$
I concluded that the points $(0,0)$, $(0,-1)$, and $(0,1)$ were the points with a horizontal tangent.
However, when I graphed this using Desmos, it turns out that the point at $(0,0)$ did not look like it has horizontal tangent.
Graph of y^4=y^2-x^2
How would I have been able to figure this out without graphing it?
| If
$y^4=y^2-x^2
$,
then
diffing implicitly,
$4y^3y'
=2yy'-2x
$
or
$2y^3y'
=yy'-x
$.
If $y' = 0$,
then
$x = 0$.
Putting this in,
$y^4 = y^2$
so the possible values are
$y = 0, \pm 1$.
At $x=y=0$,
suppose $y' = c$.
For small $x$ and $y$,
$y \approx cx$
so
$c^4x^4 \approx c^2x^2-x^2$
or,
dividing by $x^2$,
$c^4 x^2 \approx c^2-1$.
Since
$c^4x^2$ is small,
$c^2-1$ must also be small
so
$c^2 \approx 1$
so
$c \approx \pm 1$.
This means that,
at $(0, 0)$,
$y' = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$ The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I noticed that this is the sum of the reciprocals of the triangle numbers.
Let $t_n = \frac{n(n+1)}{2}$ denote the $n$-th triangle number. Then the question is basically asking us to evaluate
\begin{align}
\sum_{n=1}^{2015} \frac {1}{t_n} & = \sum_{n=1}^{2015} \frac {2}{n(n+1)}\\
& = \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1}
\end{align}
Here's where my first question arises. Do you just have to know that $\frac {2}{n(n+1)} = \frac{2}{n}-\frac{2}{n+1}$? In an exam situation it would be very unlikely that someone would be able to recall that if they had not done a question like this before.
Moving on:
\begin{align}
\sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} & = \left(\frac{2}{1}-\frac{2}{2}\right) +\left(\frac{2}{2}-\frac{2}{3}\right) + \ldots +\left(\frac{2}{2014}-\frac{2}{2015}\right) +\left(\frac{2}{2015}-\frac{2}{2016}\right)\\
&= 2 - \frac{2}{2016} \\
& = \frac {4030}{2016} \\
& = \frac {2015}{1008}
\end{align}
And I'm not sure if this is right. How does one check whether their summation is correct?
|
And I'm not sure if this is right. How does one check whether their summation is correct?
Replace "2015" with "10", do it by hand, and see if your results match with your general formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad. Solutions presented: $(x,y)=(0.625 a^2,0.612372 a^2)$ and $(x,y)=(0.625 a^2,-0.612372 a^2)$. I tried first to make the substitution $u=x+y$ and $v=x-y$, noticing that $x^2+y^2=0.5((x+y)^2+(x-y)^2)$ but could not go far using that route. Then I moved to squaring both equations, hoping to get a solution, but without success.
Hints and answers are appreciated. Sorry if this is a duplicate.
| Hint:
Let $u=\sqrt{x-y},v=\sqrt{x+y}$. The system now reads
$$u+v=a,\\\sqrt{\frac{u^4+v^4}2}-uv=a^2$$
Raising the first equation to the fourth power,
$$a^4=u^4+v^4+4uv(u^2+v^2)+6u^2v^2.$$
Then using $u^4+v^4=2(a^2+uv)^2$ and $u^2+v^2=a^2-2uv$, you get an equation in $uv$, which simplifies:
$$a^4=2(a^2+uv)^2+4uv(a^2-2uv)+6u^2v^2.$$
$uv=-\dfrac{a^2}8$.
When $uv$ is known, $u^2+v^2=a^2-2uv$ gives you $2x$, and $x^2-u^2v^2$ gives you $y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Smallest possible value of expression involving greatest integer function If $a, b, c \gt 0$ then what is the smallest possible value of
$$\left[\frac{a+b}{c}\right]+ \left[\frac{b+c}{a}\right] + \left[\frac{c+a}{b}\right]$$
where $[.]$ denotes greatest integer function.
I tried using the AM GM inequality at first but it was not useful. I also tried adding 1 to each bracket and then subtracting 3 from overall to get the same numerator in each bracket. But this too wasn't useful. I don't have much practice of solving the question involving greatest integer function. Somebody please tell me how to deal with this question.
| Wolog $a \le b \le c$
$[\frac {b+c}a] \ge [\frac {a+a}a] =2$
And $[\frac {a+c}b] \ge [\frac {a + b}b] \ge [\frac bb] = 1$.
So you can not get less than $3$
If $\frac {a+b}{c} < 1$ then $c > a+b$ and $\frac {c+a}b > \frac {2a + b}b= \frac {2a}b + 1$. So If $\frac {c+a}b < 2$ then $\frac {2a}b < 1$ so $b > 2a$. So $\frac {b+ c}a > \frac {2a + a+b}a > \frac {5a}a = 5$.
In other words if $[\frac {a+b}{c}]=0$ and $[\frac {c+a}b] = 1$ then $[\frac {b+c}b] \ge 5$. So you most certainly can not get $3$ as an answer.
So the answer is $4$ or greater.
Can we get four?
Well, if $a = b = c$ then $\frac {a+b}c= \frac {a+c}b=\frac {b+c}a =2$
Well if we can get $\frac {a+b}c$ and $\frac {a+c}b$ just under $2$ and $\frac {b+c}a$ just over $2$ that should do it.
Let $a= .9; b = 1; c= 1.1$ then
$[\frac {a+b}c]=[\frac {1.9}{1.1}] = 1$ and $[\frac {a+c}b]= [\frac 21]$
....oops..... we'll have to shave just a whisker off.
Let $a = .9; b=1; c=1.09$ then $[\frac {a+b}c]=[\frac {1.9}{1.09}]=1$ and
$[\frac {a+c}b]= [\frac {1.99}1]=1$ (barely!)$
And $[\frac {b+c}{a}] = [\frac {2.09}{.9}] =2$.
So $[\frac {a+b}c]+[\frac {a+c}b]+[\frac {b+c}{a}]=4$.
$4$ is the smallest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.
2.) Then $\cos(x) = \frac{1}{3}\sin(x)$
3.) $\sin^2 (x) + \left(\frac{1}{3}\sin(x)\right)^2 = 1$ //Pythagorean identity substitution.
4.) $\left(\frac{4}{3}\sin(x)\right)^2 = 1$ //Combining like terms
5.) $\frac{16}{9}\sin^2(x) = 1$ //Square the fraction so I can move it later.
6.) $\sin^2(x) = \frac{1}{\frac{16}{9}}$ //Divide both sides by $\frac{16}{9}$
7.) $\sin^2(x) = \frac{1}{1} * \frac{9}{16} = \frac{9}{16}$ //divide out the fractions
8.) $\sin(x) = \pm \frac{3}{4}$ //square root both sides.
So $\sin(x) = \pm \frac{3}{4}$ but the book says this is wrong. Any help is much appreciated.
| If $\tan(x)=3$, then $\tan^2(x)=9$. This means that $\frac{\sin^2x}{1-\sin^2x}=9$. So, $\sin^2(x)=\frac9{10}$; in other words (at least if we're on the first quadrant), $\sin(x)=\frac3{\sqrt{10}}$.
Your error lies in item 4: $\sin^2(x)+\left(\frac1{3\sin(x)}\right)^2=\frac{10}9\sin^2(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$ I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$.
So I pose the following problem:
Given $a \in \mathbb{R}-\{0\}$, find $x$ such that $\dfrac{dy}{dx}=0$ where $y=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$
As in most maxima/minima problems, we first (implicitly) differentiate it and set to $0$ to give $$\boxed{\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x} \tag{1}$$ I have no idea how to continue from here. I thought about taking logarithms, but it appears to me that the double $\ln$ in the term $\dfrac{a^2x^2-x+a}{ax}$ would only make the equation worse.
(For the simplest case when $a=1$, the problem is easy: $x=1$ and it is a point of inflexion).
Let's try setting each of the terms to $0$:
Case $1$: $\left(\frac{a^2x^2-x+a}{ax}\right)^x=0$
$ \hspace{1cm}$ This is only possible when the fraction is zero; that is, solving $a^2x^2-x+a=0$ to get $$x=\frac{1\pm\sqrt{1-4a^3}}{2a^2}$$
Case $2$: $\left(\frac{ax^2-a^2x+1}{ax}\right)^a=0$
$ \hspace{1cm}$ This gives $$\begin{align}ax^2-a^2x+1=0&\implies a^2x^2+a=a^3x\\&\implies\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(\dfrac{a^3x-x}{ax}\right)^x=\left(\dfrac{a^3-1}{a}\right)^x=0\end{align}$$
$ \hspace{1cm}$ so for equality between LHS and RHS, we must have $a=1$. However, the equation
$ \hspace{1cm}$ $ax^2-a^2x+1=0$ has no real solutions for such $a$; hence we reach a contradiction.
Case $3$: $\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=0$
$ \hspace{1cm}$ We have $x=\pm \dfrac1a$. Now LHS is $0$, and $$\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(1-\dfrac1a+a\right)^{\frac1a} \neq 0$$
$ \hspace{1cm}$ for $a \in \mathbb{R} - \{\phi\}$, where $\phi$ is the golden ratio.
$ \hspace{1cm}$ Suppose that $a = \phi$. Then $x$ is forced to be $-\dfrac1a=-\dfrac2{1+\sqrt5}$, since $ax^2-a^2x+1=0$
$ \hspace{1cm}$ (undefined) when $x=\dfrac 1a$. This is impossible, since $y$ is only defined when $x>0$ for this
$ \hspace{1cm}$ value of $a$!
Case $4$: $\ln\left(\frac{a^2x^2-x+a}{ax}\right)+\frac{a(ax^2-1)}{a^2x^2-x+a}=0$
$ \hspace{1cm}$ This is impossible from cases $1$ and $3$.
UPDATE: I have provided a partial answer to my question, now with $x$ removed from it.
Any hints on how to solve $(3)=(4)$ are welcome.
Here, on MathOverflow: https://mathoverflow.net/questions/302105/on-finding-the-critical-points-of-fx-leftx-a-frac1ax-righta-left-fra
| A note about Case $1$ with $\,a:= 2^{-\frac{2}{3}}\,$ and $\,\displaystyle x\to 2^{\frac{1}{3}}\,$ .
Left side $\,=0\,$ for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\, ax^2-1=0\,$ and $\,\displaystyle ax^2-a^2x+1=\frac{3}{2}\ne 0\,$ .
Right side $\,=0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>1\,$:
$\,\displaystyle \lim_{z\to +0 \\x>0} z^x\ln z=0\,$ and therefore $\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \ln \frac{a^2x^2-x+a}{ax} \to 0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>0\,$
$\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \frac{a(ax^2-1)}{a^2x^2-x+a} = \left(a^2x^2-x+a\right)^{x-1} (ax^2-1) x^{-x} a^{1-x} = 0\,$
for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\,x-1>0\,$ and $\,\displaystyle a^2x^2-x+a=0\,$ and $\, ax^2-1=0\,$ .
If you like to work with recursions for e.g. $\,a>0,\,a\neq 1\,$
it can make sense to choose $\,\displaystyle z:=x+\frac{1}{ax}\,$ so that you get
$\displaystyle x=f_{1,2}(z):=\frac{z}{2}\pm\sqrt{(\frac{z}{2})^2-\frac{1}{a}}\,$ and $\,\displaystyle y=(z-a)^a-\left(az-\frac{1}{a}\right)^x\,$ .
We get $\,\displaystyle \frac{dy}{dz}=(z-a)^{a-1} - \left(az-\frac{1}{a}\right)^x \left(\frac{x}{z-\frac{1}{a^2}}+\frac{1}{2-\frac{z}{x}}\ln\left(az-\frac{1}{a}\right)\right)$
and a possible recursion with $\displaystyle z_0>\max\left(a;\frac{2}{\sqrt{a}};\frac{1}{a^2}\right)\,$ could be
$\displaystyle z_{n+1}=a+\left(\left(az_n-\frac{1}{a}\right)^{f(z_n)} \left(\frac{f(z_n)}{z_n-\frac{1}{a^2}}+\frac{1}{2-\frac{z_n}{f(z_n)}}\ln\left(az_n-\frac{1}{a}\right)\right)\right)^{\frac{1}{a-1}}\,$
with $\,f(z)\in\{f_1(z);f_2(z)\}\,$ .
For time reasons I haven't checked this recursion, sorry. But it's a try to simplify the calculations.
Note:
Because of $\,\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\,$ with $\,\displaystyle \frac{dy}{dx}:=0\,$ we have $\,\displaystyle \frac{dy}{dz}=0\,$ or $\,\displaystyle \frac{dz}{dx}=0\,$ .
$\displaystyle \frac{dz}{dx}=0\,$ means $\,ax^2-1=0\,$ which leads directly to my comment about Case $1$ .
This strengthens the claim that the substitution $\,\displaystyle z:=x+\frac{1}{ax}\,$ makes sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?
How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?
I first started with the case $x=1/2$ to try out. We see that $n^2(1/2)^n(1/2)=n^2/2^{n+1}$ which goes to $0$ as $n$ goes to infinity since $n^2 < 2^{n+1}$.
Now I need to generalize. If I take the derivative of $f_n(x)$ then I get, after simplifying,
$$f_n'(x)= n^2x^{n-1}(n-(n+1)x).$$
So, we know $x=0$ and $x=n/(n+1)$ are critical points.
I plotted the graph and I see that as $n$ gets larger, the maximum of the function on $[0,1]$ grows and gets closer to $x=1$.
| For $0 < x < 1$ we have $y = 1/x - 1 > 0$ and $ x = 1/ (1 + (1/x - 1))= 1/(1+y).$
Thus,
$$n^2 x^n = \frac{n^2}{(1 + y)^n}.$$
Using the binomial expansion $(1+y)^n = 1 + ny + \frac{1}{2}n(n-1)y^2 + \frac{1}{6}n(n-1)(n-2)y^3 + \ldots,$we have
$$0 \leqslant n^2 x^n = \frac{n^2}{(1 + y)^n} < \frac{n^2}{n(n-1)(n-2)}\frac{6}{y^3},$$
and $\lim_{n \to \infty} n^2x^n = 0$ by the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. Of course, if we prove this stronger statement we are done. Any idea?
| We have that $$a^4+b^4=(a+b)(a^3-a^2b+ab^2-b^3)+2b^4$$ so if $a+b$ is a factor of $a^4+b^4$ it is also a factor of $2b^4$ and (by symmetry) $2a^4$
If the highest common factor of $a$ and $b$ is $y$ so that $a=py$ and $b=qy$ we find that $(p+q)y$ is a factor of $2q^4y^4$. Now $p+q$ can have no factor in common with $q$ by construction, so $p+q|2y^4$. We find an easy solution by setting $p+q=y$.
You might want to think about constructing a counterexample before going further, by tightening things up a bit.
If we want to be tight against a constraint we might try $y^4=\frac {p+q}2$. With $y=2$ this would give $p+q=32$. Then $a=2p$ and $b=2q$ and $a^4+b^4=16 (p^4+q^4)$ and this is divisible by $p+q=32$ because $p$ and $q$ have the same parity. But now put $p=1, q=31$ with $a=2, b=62$ and $a^2+b^2=4+3844=3848$ is not divisible by $32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Compute the sum fast. How can I compute the following sum in the fastest way possible?
$y = 1 + x + ... + {x}^{{n}^{3}}=\sum_{i = 0}^{n}{x}^{{i}^{3}}$
I wrote that $n^3 - (n-1)^3 = 3n^2-3n+1$, but so far it does not help a lot.
| This is a modified version of my answer here, which was the same question only with square exponents rather than cubed ones.
You have already calculated $x^{(n+1)^3} = x^{n^3}\cdot x^{3(n+1)^2-3(n+1)+1}$ (I shifted the indices by one to conform with my other answer). We also have $x^{3(n+1)^2-3(n+1)+1}=x^{3n^2-3n+1}\cdot x^{6n}$. You could then use the following recursion:
*
*Get $x^{n^3}, x^{3n^2 - 3n +1}$ and $x^{6n-6}$ from previous iteration
*Calculate $x^{6n} = x^{6n-6}\cdot x^6$
*Calculate $x^{3(n+1)^2 - 3(n+1) + 1} = x^{3n^2 - 3n +1}\cdot x^{6n}$
*Calculate $x^{(n+1)^3} = x^{n^3}\cdot x^{3(n+1)^2 - 3(n+1) + 1}$
*Add $x^{(n+1)^3}$ to your summation variable
*Send $x^{(n+1)^3}, x^{3(n+1)^2 - 3(n+1) + 1}$ and $x^{6n}$ to the next iteration
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$
Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$
I am getting:
$$\sin(x) = x - \frac{x^3}{3!} + R_4(x)$$
where $R_4(x) = \frac{\cos(c)x^5}{5!}$ for some $c$ between $0$ and $x$
I want to prove $R_4(x)\geq 0$ to arrive at the result $x-x^3/3 \leq \sin(x)$.
for:$$0 \leq x \leq \pi/2 \Rightarrow 0<c<\pi/2 \Rightarrow R_4(x)\geq 0$$ but for: $$-\pi/2 \leq x < 0 \Rightarrow -\pi/2 < c < 0 \Rightarrow R_4(x) < 0$$
How can I proceed with this ? There are many cases that I need to check
| The aim is to show that:
$$ x - \frac{x^3}{3!} \leq \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + ...\leq x $$
For $sin x\leq x$ it suffices to use MVT:
$$\cos c =\frac{\sin x- \sin 0}{x-0}=\frac{\sin x}{x}\implies -1\leq \frac{\sin x}{x}\leq \implies \frac{\sin x}{x}\leq 1 \implies sin x\leq x$$
For $x - \frac{x^3}{3!}\leq \sin(x) $ see here
Proof for $\sin(x) > x - \frac{x^3}{3!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the shortest distance from a line to circle while their equations are given Consider a line $L$ of equation $ 3x + 4y - 25 = 0 $ and a real circle $C$ of real center of equation $ x^2 + y^2 -6x +8y =0 $
I need to find the shortest distance from the line $L$ to the circle $C$.
How do I find that?
I am new to coordinate geometry of circles and line.
And I noted the slope of $L$ to be $\frac{-A}{B} = \frac{-3}{4} $ which means the line is inclined to $ -37° $ with $+x axis$ And circle centered at $ (3,-4) $ and of radius $5$ units.
By diagram, its difficult to figure out. Can we figure out easily by diagram or there is an algebraic way which is good for this?
| Hint:
Any point on the circle can be set as $$P(3+5\cos t,5\sin t-4)$$
The distance of this point from $L$ will be $$\dfrac{|3(3+5\cos t)+4(5\sin t-4)-25|}{\sqrt{3^2+4^2}}$$
$3(3+5\cos t)+4(5\sin t-4)-25=5(3\cos t+4\sin t)-32$
Now $-\sqrt{3^2+4^2}\le3\cos t+4\sin t\le\sqrt{3^2+4^2}$
$\iff-5\cdot5-32\le5(3\cos t+4\sin t)-32\le5\cdot5-32$
$\iff7\le|3(3+5\cos t)+4(5\sin t-4)-25|\le57$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Ratio of an inscribed circle's tangent to original square In the diagram, the circle is inscribed within square $PQRS,$ $\overline{UT}$ is tangent to the circle, and $RU$ is $\frac{1}{4}$ of $RS.$ What is $\frac{RT}{RQ}$?
| Suppose that the circle touches $RS$, $TU$ and $RQ$ at $X$, $Y$ and $Z$ respectively.
If $RQ=4$, $RU=1$. Let $RT=x$.
Then $TU=TY+YU=TZ+XU=2-x+1$.
So $1^2+x^2=(3-x)^2$ and thus $1+x^2=9-6x+x^2$.
$\displaystyle x=\frac{4}{3}$.
$\displaystyle \frac{RT}{RQ}=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find out all solutions for the system Given the system
$$ \left[
\begin{array}{ccc|c}
x_1&x_2&x_3&k\\
x_1&x_2&kx_3&1\\
x_1&kx_2&x_3&1\\
kx_1&x_2&x_3&1\\
\end{array}
\right] $$
I tried to solve this...It looks simple but I found a problem at the end...
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
1&1&k&1\\
1&k&1&1\\
k&1&1&1\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
0&0&k-1&1-k\\
0&k-1&0&1-k\\
k-1&0&0&1-k\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
0&0&1& \frac{1-k}{k-1}\\
0&1&0& \frac{1-k}{k-1}\\
1&0&0& \frac{1-k}{k-1}\\
\end{array}
\right] $$
Finally,
$$ \left[
\begin{array}{ccc|c}
1&0&0& -1\\
0&1&0& -1\\
0&0&1& -1\\
0&0&0&k+3\\
\end{array}
\right] $$
There is no way to get infinitely many solutions.
$$$$
However, I tried to put the system into other online calculator...
Click me$$$$
There is infinitely many solutions when k=4 or other numbers. What's wrong of my work...
| The system has solutions only if
$\det(A|B)=0$, otherwise $\text{rank }(A|B)=4>\text{rank }(A)$
that is $k^4-6 k^2+8 k-3=0\to (k-1)^3 (k+3)=0$
for $k=-3$ and $k=1$
for $k=-3$ we get the system
$\begin {cases}
x+y+z=-3\\
x+y-3 z=1\\
x-3 y+z=1\\
-3x+y+z=1\\
\end{cases}
$
which has solution $(-1,-1,-1)$
for $k=1$ we get
$\begin {cases}
x+y+z=1\\
x+y+ z=1\\
x+y+z=1\\
x+y+z=1\\
\end{cases}
$
that is
$x+y+z=1$ which gives infinite solutions
$(t,u,1-t-u)$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$
| Hint: Your expression equals
$$x^2[(1+1/x^2)^{1/2} - (1+1/x^3)^{1/3}].$$
Now use the fact that $(1+h)^p = 1 + ph +o(h)$ as $h\to 0.$ (This fact is equivalent to the statement that the derivative of $(1+x)^p$ at $x=1$ is $p.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Subsets and Splits