Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Asymptotics of $\Sigma x^n/n^p$ I am looking for asymptotics of
$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$
in terms of $N$, where $x>1$ and $p$ is a positive number.
So I am looking for a $O(f(N))$ type result. For example what is the approximate value of
$\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is probably something like $\frac{5^N}{N^2} (a+\frac{b}{N}+\frac{c}{N^2}+\cdots)$.
| Just playing around to see what happens.
If $x > 1$
then
$\begin{array}\\
\sum_{n=1}^N \dfrac{x^n}{n^p}
&=x^N\sum_{n=1}^N \dfrac{x^{n-N}}{n^p}\\
&=x^N\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(N-n)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(1-n/N)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} x^{-n}(1-n/N)^{-p}\\
&=\dfrac{x^N}{N^p}\left(1+\dfrac1{x(1-1/N)^p}+\dfrac1{x^2(1-2/N)^p}+...\right)\\
\end{array}
$
To see how (or if)
this converges,
look at the ratio of
consecutive terms.
$\begin{array}\\
\dfrac{x^{-n-1}(1-(n+1)/N)^{-p}}{x^{-n}(1-n/N)^{-p}}
&=\dfrac{(N-n)^p}{x(N-n-1)^{p}}\\
&=\dfrac{1}{x(1-1/(N-n))^{p}}\\
&\approx\dfrac{1}{x(1-p/(N-n))}
\qquad\text{for small } n\\
&\approx\dfrac{1}{x}(1+p/(N-n))\\
&\approx\dfrac{1}{x}(1+p/(N(1-n/N))\\
&\approx\dfrac{1}{x}(1+(p/N)(1+n/N))\\
&=\dfrac{1}{x}(1+p/N+pn/N^2)\\
&\approx\dfrac{1}{x}(1+p/N)\\
\end{array}
$
This is $1$ when
$x(1-1/(N-n))^{p}
=1$
or
$1-1/(N-n)
=x^{-1/p}
$
or
$N-n
=\dfrac{1}{1-x^{-1/p}}
$.
Using the approximate ratio of
$\approx\dfrac{1+p/N}{x}
$,
the approximate expansion is
$\dfrac{x^N}{N^p}\left(1+\dfrac{1+p/N}{x}+\dfrac{(1+p/N)^2}{x^2}+...\right)
$.
That's as far as I can take it now.
Hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/2201643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$ $$\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$$
I tried everything substitution,beta gamma function everything but I am unable to convert this into a standard form so How do I solve this problem.
| $$\int^{\pi}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx + \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos x}} \mathrm dx = 2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx$$
Then use the formula
$$B(x,y) = 2\int^{\pi/2}_0 (\sin t)^{2x-1} (\cos t)^{2y-1} dt$$
For $2y-1 = 0$ and $2x-1 = -\frac{1}{2}$ we have $x = \frac{1}{4}$ and $y = \frac{1}{2}$
$$2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$$
Remark
It seems the question was edited
Define the incomplete elliptic integral of the first kind as
$$F(\phi,k) = \int^{\phi}_0 \frac{\mathrm dx}{\sqrt{1-k^2\sin^2 x}}$$
Suppose that $0 < \phi \leq \frac{\pi}{4}$ and $k=\sqrt{2}$
$$F(\phi,\sqrt{2}) = \int^{\phi}_0 \frac{\mathrm dx}{\sqrt{\cos(2x)}} = \frac{1}{2}\int^{\pi/2}_{\pi/2-2\phi} \frac{\mathrm dx}{\sqrt{\sin(x)}} $$
$$F(\phi,\sqrt{2})= \frac{1}{2}\int^{\pi/2}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}}-\frac{1}{2}\int^{\pi/2-2\phi}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}}$$
Hence we have
$$\int^{\pi/2-2\phi}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F(\phi,\sqrt{2})$$
For the case $\phi = \pi/8$
$$\int^{\pi/4}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F(\pi/8,\sqrt{2})$$
Or using the other representation
$$\int^{\pi/4}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F\left(\frac{\pi}{8} \left. \right|2 \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Let $\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)$
We need prove $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$
We have: $\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]$
By C-S we have: $\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}$
$\Rightarrow 2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)$
$\Rightarrow LHS^2\ge 2(x+y+z+2\sqrt {xy}+2\sqrt {yz}+2\sqrt {xz})=2(\sqrt{x}+\sqrt{y}+\sqrt{z})=RHS^2$
Can do other way ?
| Apply the popular inequality $\sqrt{\dfrac{a^2+b^2}{2}}\ge \dfrac{a+b}{2}$, with $a = \sqrt{2x}, b = \sqrt{2y}$, and do the same for the other two pairs $(b,c)$ and $(c,a)$, and add all $3$ inequalities to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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To find factor of a polynomial equation
One of the factors of $4x^2+y^2+14x-7y-4xy+12$ is equal to
*
*$2x-y+4$
*$2x-y-3$
*$2x+y-4$
*$2x-y+3$
Step $1$:
$4x^2+y^2-4xy$ can be simplified as $(2x-y)^2$
Step $2$:
$14x-7y$ can be simplified as $7(2x-y)$
and finally
$(2x-y) (2x-y+7) + 12$
I can able to factor to this extent only. however can't able to arrive at the answer.
The answer is given in the book. it states that $4x^2+y^2+14x-7y-4xy+12$ is product of $(2x-y+3)$ and $(2x-y+4)$ I am in need of steps
| You can do $2x-y=k$ and then
$$k^2+7k+12=(k+3)(k+4)$$
and then you get
$$(2x-y+3)(2x-y+4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Exponential map of Beltrami-Klein model of hyperbolic geometry In the Betrami-Klein model of hyperbolic geometry, geodesics are represented as straight lines. Hence the exponential map of a tangent vector $\mathbf{v}$ at a point $\mathbf{p}$ is $\mathbf{p} + \lambda \mathbf{v}$, where $\lambda$ is a scalar that depends on $\mathbf{p}$ and $\mathbf{v}$. For example, suppose $\mathbf{p} = 0$. Then the exponential map is
$$ \exp_\mathbf{p}(\mathbf{v}) = \hat{\mathbf{v}} \tanh \lvert \mathbf{v} \rvert $$
Notice that $\tanh \lvert \mathbf{v} \rvert < 1$ always, since we cannot reach the circle at infinity with any finite $\mathbf{v}$.
Now consider the diagram below:
where $\mathbf{q} = \mathbf{p} + \lambda \mathbf{v}$. The hyperbolic distance between $\mathbf{p}$ and $\mathbf{q}$ must be equal to the length of the tangent vector, that is, $d(\mathbf{p}, \mathbf{q}) = \lvert \mathbf{v} \rvert$. This hyperbolic distance is
$$ d(\mathbf{p}, \mathbf{q}) = \frac{1}{2} \log \frac{\lvert\mathbf{p} - \mathbf{r_1}\rvert \lvert\mathbf{q} - \mathbf{r_2}\rvert}{\lvert\mathbf{p} - \mathbf{r_2}\rvert \lvert\mathbf{q} - \mathbf{r_1}\rvert}$$
where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the two ideal points intersected by the straight line connecting $\mathbf{p}$ and $\mathbf{q}$. We can find these ideal points as follows:
\begin{align}
1 &= \mathbf{r} \cdot \mathbf{r} \\
&= (\mathbf{p} + \mu \mathbf{v}) \cdot (\mathbf{p} + \mu \mathbf{v}) \\
&= \mathbf{p} \cdot \mathbf{p} + 2 \mu \mathbf{p} \cdot \mathbf{v} + \mu^2 \mathbf{v} \cdot \mathbf{v}
\end{align}
Hence
\begin{align}
\mu &= \frac{-2 \mathbf{p} \cdot \mathbf{v} \pm \sqrt{(2 \mathbf{p} \cdot \mathbf{v})^2 - 4 (\mathbf{v} \cdot \mathbf{v})(\mathbf{p} \cdot \mathbf{p} - 1)}}{2 \mathbf{v} \cdot \mathbf{v}} \\
&= \frac{-\mathbf{p} \cdot \mathbf{v} \pm \sqrt{(\mathbf{p} \cdot \mathbf{v})^2 - (\mathbf{v} \cdot \mathbf{v}) (\mathbf{p} \cdot \mathbf{p} - 1)}}{\mathbf{v} \cdot \mathbf{v}}
\end{align}
so that $\mathbf{r} = \mathbf{p} + \mu \mathbf{v}$ for each root. Hence we have
\begin{align}
\lvert \mathbf{v} \rvert &= \frac{1}{2} \log \frac{\lvert\mathbf{p} - (\mathbf{p} + \mu_1 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_2 \mathbf{v})\rvert}{\lvert\mathbf{p} - (\mathbf{p} + \mu_2 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_1 \mathbf{v})\rvert}
\\
&= \frac{1}{2} \log \frac{\lvert\mu_1 \mathbf{v}\rvert \lvert(\lambda - \mu_2) \mathbf{v}\rvert}{\lvert\mu_2 \mathbf{v}\rvert \lvert(\lambda - \mu_1) \mathbf{v}\rvert}
\\
&= \frac{1}{2} \log \frac{\lvert\mu_1\rvert \lvert\lambda - \mu_2\rvert}{\lvert\mu_2\rvert \lvert\lambda - \mu_1\rvert}
\end{align}
or equivalently
\begin{align}
\left\lvert \frac{\lambda - \mu_2}{\lambda - \mu_1} \right\rvert = \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\frac{\lambda - \mu_2}{\lambda - \mu_1} = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\lambda - \mu_2 = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} (\lambda - \mu_1)
\\
\lambda \left( 1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \right) = \mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\lambda = \frac{\mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}}{1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}}
\end{align}
We pick the root $\lambda > 0$. Is this result correct? Can it be simplified? Can anyone find a source for it?
| I did some calculation and so and if
$P = (p_x,p_y)$ and $Q=(p_x+v_x,p_y+v_y)$ are two points on a chord the ideal points (points on the unit circle are:
$$I_1 = \left ( \frac{v_y \ c + v_x \sqrt{v_x^2 +v_y^2 -c^2}}
{v_x^2 +v_y^2} \ , \
\frac{-v_x \ c + v_y \sqrt{v_x^2 +v_y^2 -c^2} }{v_x^2 +v_y^2} \right) $$
$$I_2 = \left ( \frac{v_y \ c - v_x \sqrt{v_x^2 +v_y^2 -c^2}}
{v_x^2 +v_y^2} \ , \
\frac{-v_x \ c - v_y \sqrt{v_x^2 +v_y^2 -c^2} }{v_x^2 +v_y^2} \right) $$
with $ c =p_x v_y - v_x p_y $
I don't know if this will help you and also I don't know how to write this as a formula of complex numbers (but will make a seperate question for that :)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+y))\sin (\frac{1}{2}(x-y))}{(x+y)(x-y)}\\
&=-\dfrac{2 \sin y}{2y}\dfrac 12\\
&=-\dfrac{\sin y}{2y}\end{align}
My question: Is $-\dfrac{\sin y}{2y}$ the final answer or can it be calculated further as $-\dfrac12$?
I also try different route:
Let $p=x-y$ so $x=p+y$ and $p\rightarrow 0$ as $x \rightarrow y$. Hence,
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{p\rightarrow 0}\dfrac{\cos (p+y)-\cos y}{p^2+2py+y^2-y^2}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))\sin (\frac{1}{2}p)}{p(p+2y)}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))}{(p+2y)} \dfrac{\sin (\frac{1}{2}p)}{p}\\\\
&=-\dfrac{2\sin y}{2y} \dfrac{1}{2}\\\\
&=-\dfrac{\sin y}{2y}\end{align}
Okay, so that left me with the same result. What is the correct final answer, $-\dfrac{\sin y}{2y}$ or $-\dfrac 12$?
Thanks
| Your answer is correct. A simpler approach: let $f(x)= \cos x$. Then
$\dfrac{\cos x-\cos y}{x^2-y^2}=\dfrac{f( x)-f(y)}{x-y}* \dfrac{1}{x+y} \to f'(y)* \dfrac{1}{2y}$ for $x \to y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2
$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$
I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and delighted if someone could give some hints about how to atack this kind of problem... what is the thinking behind? Is it just about intuition? Or is it about test all sort of algebraic manipulation?
| Try that
$$\lim_{x \to 2}\frac{\sqrt{x^2+12}-4}{2 - \sqrt{x^3-4}} = \lim_{x \to 2 }\frac{(\sqrt{x^2+12}-4)(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})}{(2 - \sqrt{x^3-4})(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})} = \lim_{x \to 2}\frac{(x^2-4)(2 + \sqrt{x^3-4})}{(8-x^3)(\sqrt{x^2+12}+4)}$$
Then you can use that the limit of a product is a product of the limits.
$$=\underbrace{\lim_{x\to 2}\frac{2+\sqrt{x^3-4}}{(4 + \sqrt{x^2+12})}}_{=1/2}\lim_{x \to 2}\frac{(x^2-4)}{(8-x^3)} = -\frac{1}{2}\underbrace{\lim_{x \to 2}\frac{x+2}{x^2+2x+4}}_{=1/3} = -\frac{1}{6}$$
Where, after using $8-x^3 = - (x^3-8)$ and $x^2-4=(x-2)(x+2)$ we noted that
$$(x^3 - 8) = (x - 2)(x^2 + 2x + 4)$$
In general you are correct in the way of treat this kind of problems. This type of manipulations of polynomials are very useful: when you get a polynomial such as $x^3-8$ and it is easy to see that $x=2$ is a root you can divide by $x-2$ and lower the order of the polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is every integer odd $p$ not divisible by $3$ a sum of difference of powers of $2$ and $3$ or vice versa? Does every odd integer $p$ not divisible by $3$ have the form
$2^a+3^b$, $2^a-3^b$, and or $3^b-2^a$ for integers $a$ and $b$?
Please show proofs if possible. Thanks for help.
| No. $53$ is a counterexample.
There are no integers $a,b$ such that $53=2^a+3^b$ since $$2^a\not=53-3^0,53-3^1,53-3^2,53-3^3$$
Suppose that there are integers $a,b$ such that $53=2^a-3^b$.
It is easy to see that $a\ge 2$ and $b\ge 2$.
We have
$$2\equiv (-1)^a-0\pmod 3\implies \text{$a$ is odd}$$
Also, we have
$$1\equiv 0-(-1)^b\pmod 4\implies \text{$b$ is odd}$$
Here, let $a=2a'+1,b=2b'+1$ where $a',b'$ are positive integers.
Now
$$3\equiv 2\cdot 4^{a'}-3\cdot 9^{b'}\equiv 2(-1)^{a'}-3(-1)^{b'}\equiv 0,1,4\pmod 5$$
which is a contradiction.
Suppose that there are integers $a,b$ such that $53=3^b-2^a$.
It is easy to see that $a\ge 1$ and $b\ge 1$.
We have
$$2\equiv 0-(-1)^a\pmod 3\implies \text{$a$ is even}$$
Also, we have
$$1\equiv (-1)^b-0\pmod 4\implies \text{$b$ is even}$$
So, letting $a=2a',b=2b'$ where $a',b'$ are positive integers, we have
$$3\equiv 9^{b'}-4^{a'}\equiv (-1)^{b'}-(-1)^{a'}\pmod 5\implies \text{$b'$ is odd, $a'$ is even}$$
Letting $a=4k,b=4m+2$ where $k$ is a positive integers and $m$ is a non-negative integer, we have
$$5\equiv 9\cdot 81^m-16^k\equiv 9\cdot 1^m-0\equiv 9\pmod{16}$$
which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ is a perfect cube then consider $d>1$ then there is a $p$ that is prime and $p \mid d$.We have :
$p\mid a-b \Rightarrow p \mid (2b)^3 \Rightarrow p \mid 2b \Rightarrow p\mid 2$ or $p\mid b$
If $p\mid b$ then also $p\mid a$ (as $p\mid a-b$ holds). Then we will get to $p \mid 1$ because:
$ p \mid a^2+b^2+ab+a+b,p \mid a^2+b^2+ab+a+b+1 \Rightarrow p \mid 1$
which is clearly wrong then we have $p\mid 2$ so $p=2$ means $d=2^k$ where $k$ is a natural number including $0$.In the case $d=1$ we have the right result.So assemble $k \ge 1$.Because $2 \mid a-b$ we can conclude that $a,b$ have the same parity @Ghartal showed in his answer that if $a,b$ are both even we have a right result but if $a,b$ are both odd we don't.So maybe we have to prove $a,b$ can,t be both odd.
| Let us consider the Diophantine equation
$$(a-b)(a^2+ab+b^2+a+b+1) = 8b^3.\tag1$$
Let $\ \gcd(a,b) = d.$
$\text{If }\mathbf {d > 1,}\text{ then }a=md,\quad b=nd,\quad \gcd(m,n) = 1,$
$$(m-n)\left((m^2+mn+n^2)d^2 + (m+n)d + 1\right) = 8d^2n^3.\tag2$$
${\text{If }\mathbf{2\mid d},\ \text{then}}$
$$2\mid m-n,\quad 2\nmid (m^2+mn+n^2)d^2 + (m+n)d + 1,$$
$$d\mid m-n,\quad \gcd(d, m^2+mn+n^2)d^2 + (m+n)d + 1) = 1,$$
so
$$m-n=\pm8d^2,\quad (m^2+mn+n^2)d^2 + (m+n)d + 1 = \pm n^3,$$
$$a-b = \pm8d^3 = (\pm2d)^3.$$
$\text{If }\mathbf {(d > 2)\wedge (d\nmid2),}\text{ then the case } (2\nmid m+n)\text{ contradicts with (2), so }$
$$2\nmid m,\quad 2\nmid n,\quad (m-n=\pm 2d^2) \vee (m-n = \pm4d^2),$$
$$\left[\begin{align}
&\begin{cases}
m = n\pm2d^2\\
4d^6 - 3d^2n(n\pm2d^2) + 2d(n\pm d^2) + 1 = \pm4n^3
\end{cases}\\
&\begin{cases}
m = n\pm4d^2\\
16d^6 - 3d^2n(n\pm4d^2) + 2d(n\pm2d^3) + 1 = \pm2n^3,
\end{cases}
\end{align}\right.$$
$$\left[\begin{align}
&\begin{cases}
m = n\pm2d^2\\
\mp4n^3 - 3d^2n^2 + 2d(1\mp3d^3)n + 4d^6 \pm2d^3 + 1 = 0
\end{cases}\\
&\begin{cases}
m = n\pm4d^2\\
\mp2n^3 - 3d^2n^2 + 2d(1\mp6d^2)n + 16d^6 \pm4d^3 + 1 = 0,
\end{cases}
\end{align}\right.$$
and both of systems leads to the same equation without solutions
$$d^2n^2+1\equiv0\pmod4.$$
$\text{If }\mathbf {d = 1,}\text{ then the proof of the previous case is valid. Solutions also do not.}$
Thus, the issue equation has the integer solutions only in the case $\mathbf{2\nmid \gcd(a,b)},$
when $\mathbf{a-b}\ $ is a perfect cube.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
} |
Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1 = -\frac{3}{2}\sec \theta $$
$$2\sec^2 \theta -2 = {-3}\sec \theta $$
$$2\sec^2 \theta +3\sec \theta - 2 = 0 $$
Is this factoring correct?:
$$(2\sec\theta+4)(\sec\theta-1) = 0 $$
$$2\sec\theta+4 =0$$
$$2\sec\theta = -4 $$
$$\sec \theta = -2$$
$$(2\pi/3), (4\pi/3) $$
$$\sec\theta - 1 = 0$$
$$\sec\theta = 1 $$
$\sec\theta=1$ would evaluate to $0$, for some reason that is an invalid answer? (according to my assignment)
So is $\ (2\pi/3), (4\pi/3) $ the full answer?
| Why factor? It's fine, but in general when faced with something like
$$2\sec^2 \theta +3\sec \theta - 2 = 0$$
you could set $u=\sec\theta$ and then use the quadratic formula.
$$2u^2 +3u - 2 = 0$$
$$u = \frac{-3 \pm \sqrt{9-4(2)(-2)}}{2(2)} = \frac{-3 \pm 5}{4} = \frac12 , -2$$
Then solve for possible $\theta$ from $\sec\theta=1/2$ and $\sec\theta=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ ∵ For any positive integer $n$
$n^2-2(n+1)^2+(n+2)^2$
= $n^2-2n^2-4n-2+n^2+4n+4=2$
∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$
= $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$
= $1+(2+2+...+2)+9998^2-2×9999^2$
I do not know how many lots of two will be inside the aforementioned parenthesis.
| There would be $((9995-2)/3)+1$ lots of 2.
Following;
$1+2×(((9995-2)/3)+1)+9998^2-2×9999^2$
$2×3332+(9998^2-9999^2)+(1-9999^2)$
$6664+(9998-9999)(9998+9999)+(1-9999)(1+9999)$
$6664-19997-99980000$
= $-99993333$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Closed Form of Series $ \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $. Problem:
A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $\Large1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a, b,$ and $c$ are integers. Find $a+b+c.$
How would I begin to do this? Any easy theorem/trick?
| HINT:
$$\dfrac1{3^{2^m}-1}-\dfrac1{3^{2^m}+1}=\dfrac2{3^{2^{m+1}}-1}$$
$$\dfrac2{3-1}-S=\dfrac2{3-1}-\dfrac2{3+1}-\left(\dfrac{2^2}{3^2+1}+\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)$$
$$=\dfrac{2^2}{3^2-1}-\dfrac{2^2}{3^2+1}-\left(\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$
Prove:
$$\begin{align}
\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt]
\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7} &=\phantom{-}24 \tag{2}
\end{align}$$
| Let $\cos\frac{2\pi}{7}=x$, $\cos\frac{4\pi}{7}=y$ and $\cos\frac{4\pi}{7}=z$.
Hence,
$$x+y+z=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2};$$
$$xy+xz+yz=\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=$$
$$=\frac{1}{2}\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=-\frac{1}{2}$$ and
$$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8}.$$
Thus,
$$\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=$$
$$=\frac{x^3y^3+x^3z^3+y^3z^3}{x^3y^3z^3}=\frac{(xy+xz+yz)^3-3(x+y+z)(xy+xz+yz)xyz+3x^2y^2z^2}{x^3y^3z^3}=$$
$$=\frac{\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)\cdot\frac{1}{8}+3\left(\frac{1}{8}\right)^2}{\left(\frac{1}{8}\right)^3}=-88$$ and
$$\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=$$
$$=\frac{x^2y^2+x^2z^2+y^2z^2}{x^2y^2z^2}=\frac{(xy+xz+yz)^2-2(x+y+z)xyz}{x^2y^2z^2}=$$
$$=\frac{\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)\cdot\frac{1}{8}}{\left(\frac{1}{8}\right)^2}=24$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The quotient of the euclidean division $\text{Let } n,\ a_1,\ a_2, \ldots, \ a_n \text{ be natural numbers, such that:} $
$$n = q_1a_1 + r_1, 0\le r_1<a_1$$
$$q_1 = q_2a_2 + r_2, 0\le r_2<a_2$$
$$\vdots$$
$$q_{k-1} = q_ka_k + r_k, 0\le r_k<a_k$$
I have to prove that
$$\begin{cases}
\hfill q_k = \left\lfloor\frac{n}{\prod_{i=1}^{k} a_i}\right\rfloor \\
\hfill \left\lfloor \frac{n}{\prod_{i=1}^{k} a_i} \right\rfloor = \left\lfloor\frac{\left\lfloor\frac{n}{\prod_{i=1}^{k-1} a_i}\right\rfloor}{a_k}\right\rfloor\\
\end{cases}$$
Could someone give me some hints how to prove that without using induction?
|
Proof $\mathbb{1}:$
$\boxed{\left\lfloor \frac{n}{\prod_{i=1}^{k} a_i} \right\rfloor = \left\lfloor\frac{\left\lfloor\frac{n}{\prod_{i=1}^{k-1} a_i}\right\rfloor}{a_k}\right\rfloor}$
Note that $\bigg\lfloor\frac{[x]}{m}\bigg\rfloor=\bigg\lfloor\frac{x}{m}\bigg\rfloor$ when $m$ is positive integer as in this case. Why? Because $x=\lfloor x \rfloor+f$ for some $0\leq f<1$ which means $\bigg\lfloor\frac{x}{m}\bigg\rfloor=\bigg\lfloor\frac{[x]}{m}+\frac{f}{m}\bigg\rfloor$ and since $\frac{f}{m}$ is positive and less than 1, it equals $\bigg\lfloor\frac{[x]}{m}\bigg\rfloor$.
Replace $x$ with $\bigg\lfloor\frac{n}{\prod_{i=1}^{k-1}a_i}\bigg\rfloor$ and $m$ with $a_k$ and be done.
Proof $2:$
$\boxed{q_k=\Bigg\lfloor\frac{n}{\prod_{i=1}^{k}a_i}\Bigg\rfloor}$
From the statement $n=q_1a_1+r_1~;~ 0 \leq r_1 < a_1$ , we get that $\bigg\lfloor\frac{n}{a_1}\bigg\rfloor=\bigg\lfloor q_1+\frac{r_1}{a_1}\bigg\rfloor$. Since $\frac{r_1}{a_1} < 1,$ $\bigg\lfloor\frac{n}{a_1}\bigg\rfloor=q_1$
Now see, by the same arguments, $\bigg\lfloor\frac{q_1}{a_2}\bigg\rfloor=q_2$. From what we did prove in the first part (Proof 1), we get $\bigg\lfloor\frac{q_1}{a_2}\bigg\rfloor=\Bigg\lfloor\frac{\bigg\lfloor\frac{n}{a_1}\bigg\rfloor}{a_2}\Bigg\rfloor=\bigg\lfloor\frac{n}{a_1a_2}\bigg\rfloor$.
Proceed in this manner to get $q_k=\Bigg\lfloor\frac{n}{a_1a_2a_3 \cdots a_k}\Bigg\rfloor=\Bigg\lfloor\frac{n}{\prod_{i=1}^{k}a_i}\Bigg\rfloor$
Q.E.D
| {
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"url": "https://math.stackexchange.com/questions/2219398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Compute the mean for the $f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$ using probability distribution. How I compute the mean for the $$f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$$ using probability distribution.
My attempt$$\mu_x=E(X)=\int_{-\infty}^{\infty}xf(x)dx\\= \int\limits_{1}^{2}\,x\cdot\dfrac {3}{2}\sqrt {x-1}\;dx\\= \dfrac {3}{2} \left [ \int\limits_1^2x\sqrt {x-1}\,dx \right]$$
Now, don't understand what to do. Thanks in advance.
| Now, use integration by parts formula$$\displaystyle\int u\; dv=u\;v-\int v\;du\\\text{Let }u=x,\;\;dv=\sqrt{x-1} \; dx\\du=dx,\;\;v=\dfrac{2}{3}(x-1)^{3/2}\\ = \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \int \dfrac{2}{3}(x-1)^{3/2} \,dx\right]^2_1\\ =\left[ \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \dfrac{2}{3}×\dfrac{2}{5} (x-1)^{5/2} \,\right]\right]^2_1\\= \left[x(x-1)^{3/2} - \dfrac{2}{5} (x-1)^{5/2}\right]^2_1\\ = 2- \dfrac{2}{5}\\\mu=\dfrac{8}{5} = 1.6 $$
go there for better explanation on probability distribution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix inversion question with different dimensions
Let $A=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]$ and $B=\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]$ verify the matrix equation $X=AX+B$
*
*What are the dimensions of $X$?
*Find the matrix $X$. Is the matrix you found the unique solution of the equation?
Can you please discuss how to answer this type of question, I already know that $X$ has to be $3\times 2$ and this is clear through the equation given $X = AX + B$
Please discuss the second part of the question about how we could determine Matrix $X$
| We know $X=\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]$ so we can say
\begin{align}X&=AX+B\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0\\-a&-b\\-c&-d\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}4&1\\5-a&2-b\\6-c&3-d\end{matrix}\right]
\end{align}
Therefore we have the following equalities
\begin{align}a&=4\\
b&=1\\
c&=5-a=1\\
d&=2-b=1\\
e&=6-c=5\\
f&=3-d=2\end{align}
and thus $$X=\left[\begin{matrix}4&1\\1&1\\5&2\end{matrix}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possible which makes sense since you can't take a small element of the form $(x^2+a^2)$. How can I then solve this integration without using trigonometry?
| Following your idea we get $du=2xdx$ so :
$$I=\int (x^2+a^2)^{-3/2} dx=\int\frac{2xdx}{2x(x^2+a^2)^{3/2}}=\int \frac{du}{2\sqrt{u^4-a^2u^3}}=\frac{1}{2}\int\frac{du}{u^{3/2}\sqrt{u-a^2}} $$
Let $v=\frac{1}{u-a^2}$, we get $dv=-\frac{1}{(u-a^2)^2}$ :
$$I=-\frac{1}{2}{\displaystyle\int}\dfrac{1}{\left(a^2v+1\right)^\frac{3}{2}}\,\mathrm{d}v$$
Let $w=a^2v+1$ then $dw=a^2$ :
$$I=-\frac{1}{2}\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{a^2}}}{\displaystyle\int}\dfrac{1}{w^\frac{3}{2}}\,\mathrm{d}w=\dfrac{1}{a^2\sqrt{w}}$$
And finally :
$$I=\dfrac{1}{a^2\sqrt{\frac{x}{x-a^2}}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given $f(n)$ how to find the maximum value of $f(n^5)$? A friend of mine recently asked me this question.
For a positive integer $n$, a function $f(n)$ is defined as:
$f(n)=$ sum of digits in $n$.
given $f(n)=5$ find the maximum value of $f(n^5)$.
I tried solving this problem by putting random values, but my friend gave me a hint that the answer is greater than 100. Now I am completely lost because I couldn't find a single $n$ for which the value of $f(n^5)$ comes out to be greater than 100. Is there a proper way to solve it?
| Consider the terms generated by $(a+b+c+d+e)^5$. There will be $5$ of the form $a^5$ which will have a coeffiecient of $1$ & each will therefore give a contribution of $1$, there will be $20$ of the form $a^4 b$ which will have a coeffiecient of $5$ & will therefore give a contribution of $5$, etc ... terms whose coefficients are a multiple of $10$ will only give that multiple. The following table summarises the possible contributions
\begin{eqnarray*}
\begin{array}{|c|c|c|c|c|}
\hline
form & multiplicity & coefficient & contribution & tot \\ \hline
a^5 & 5 & 1 & 1 & 5 \\ \hline
a^4b & 20 & 5 & 5 & 100 \\ \hline
a^3b^2 & 20 & 10 & 1 & 20 \\ \hline
a^3bc & 30 & 20 & 2 & 60 \\ \hline
a^2b^2c & 30 & 30 & 3 & 90 \\ \hline
a^2bcd & 20 & 60 & 6 & 120 \\ \hline
abcde & 1 & 120 & 3 & 3 \\ \hline
\end{array}
\end{eqnarray*}
Totting up the tot ... we make the best $f(n^5)=\color{red}{398}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum of the serie $\sum_{n=1}^{\infty}\frac{1}{4n^3-n}$ I am trying to calculate the sum of this infinite series
$$\sum_{n=1}^{\infty}\frac{1}{4n^3-n}.$$
I only know that
$$\frac{1}{4n^3-n}=-\frac{1}{n}+ \frac{1}{2n+1} +\frac{1}{2n-1}.$$
Can you help me, please?
thanks.
| $$\sum_{n=1}^{N}\frac{1}{4n^3-n}=-\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N} \left(\frac{1}{2n+1} +\frac{1}{2n-1}\right)\\=-\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=0}^{N} \frac{1}{2n+1}-1-\frac{1}{2N+1}\\= -\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=1}^{2N+1}\frac{1}{n}-2\sum_{n=1}^{N}\frac{1}{2n}-1-\frac{1}{2N+1}$$
Now use $\sum_{n=1}^{N}\frac{1}{n}=\ln(N)+\gamma+o(1)$.
$$\sum_{n=1}^{N}\frac{1}{4n^3-n}=-2\ln(N)+2\ln(2N+1)-1-\frac{1}{2N+1}+o(1)\\=2\ln\left(\frac{2N+1}{N}\right)-1-\frac{1}{2N+1}+o(1)\xrightarrow[N\rightarrow\infty]{}2\ln(2)-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How I find the angle 'a'.
Please let me know if you can prove the angle a is 150°.
| Let $\triangle ABC$ satisfy $\angle BAC=96^\circ$, $\angle ABC = \angle ACB=42^\circ$. Let $CK$ be a line with K on $AB$ such that $\angle BCK=18^\circ$ and $A,K$ are on the same side of $BC$. Let $F$ be the point on $BC$ such that $\angle AFC=60^\circ$. Let $AF$ meet $KC$ at $D$. Our goal is to prove that $\angle DBC=30^\circ$ and hence $A,B,C,D$ are the four points depicted in your graph.
First, let $E$ be the point such that $BE//CD$ and meets $AF$ at $E$. notice that $$\angle BAE=\angle AFC-\angle ABC=60^\circ-42^\circ=18^\circ=\angle DCF$$
$$\angle ABE = \angle ABC+\angle CBE=\angle ABC + \angle KCF=60^\circ=\angle DFC$$
Hence, we have $\triangle DCF \sim \triangle EAB$.
Notice that $\angle CAD = \angle BAC-\angle BAE= 96^\circ-18^\circ=78^\circ = \angle DFC + \angle DCF = \angle ADC$, so $\triangle ACD$ is an isosceles triangle and $AC=CD$.
With $\triangle DCF \sim \triangle EAB$, $AC=CD$, and $BE//CD$ we have
$$\frac{BF}{BE}=\frac{CF}{CD}=\frac{CF}{AC}=\frac{CF}{AB}=\frac{DF}{BE}$$
Hence, $BF=DF$ and $\angle DBF = \angle BDF = \frac{1}{2}\cdot \angle DFC = 30^\circ$.
Therefore, we now have proved that our constructed $\triangle ABC$ is in fact the same as the given graph with the two lines of angles $30^\circ$ and $18^\circ$!
Hence, $a=\angle BDA = 180^\circ-\angle BDF = 150^\circ$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\bigg)$$
| Hint 1:
$$\sqrt{1-4xt}-\sqrt{1-4x}={4x(1-t)\over\sqrt{1-4xt}+\sqrt{1-4x}}$$
which suggests the change of variable $u=4xt$.
Hint 2: Letting $1-u=w^2$ (and abbreviating $\sqrt{1-4x}$ to $a$), we have
$$\int{du\over a\sqrt{1-u}(\sqrt{1-u}+a)}={1\over a}\int{-2wdw\over w(w+a)}$$
Now use partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove by Induction $1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$ Prove by induction that the following equality holds true for all n that's an element of a natural number.
$$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$$
My work:
Base Case: $n = 1$
l.s = 2
r.s = 2
True
Induction Hypothesis: Assume for some $k$ that's an element of a natural number, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ k(3k-1) = k^2(k+1)$$
Now show that, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ (k+1)(3(k+1)-1) = (k+1)^2((k+1)+1)$$
$$(k+1)(3k +2) = (k+1)^2(k+2)
$$
$$3k^2 + 5k+ 2 = k^3 + 4k^2 + 5k + 2
$$
$$0 = k^3 + 4k^2-3k^2 + 5k - 5k + 2 - 2
$$
$$0 = k^3 + k^2
$$
by Induction Hypothesis, $k^3 + k^2 = k(3k-1)$
I know you're not supposed to start off with what you are trying to prove/show but if I reverse this whole process, wouldn't that be a correct proof? Is there a faster way?
Thanks
| The claim is as follows:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
So consider the base case, that is, when $n=1$
Then:
$$\sum_{k=1}^{1}k(3k-1)=1(3(1)-1)=2 \ \checkmark$$
Assume the claim holds for the $n^{th}$ case, that is:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
Then show it holds for the $(n+1)^{th}$ case:
$$\sum_{k=1}^{n+1}k(3k-1)=(n+1)^2(n+2)$$
So consider the following:
$$\sum_{k=1}^{n+1}k(3k-1)=\sum_{k=1}^{n}k(3k-1)+(n+1)(3(n+1)-1)=\sum_{k=1}^{n}k(3k-1)+(n+1)(3n+2)$$
Substituting our assumption, we have:
$$=n^2(n+1)+(n+1)(3n+2)=(n+1)(n^2+3n+2)=(n+1)(n+1)(n+2)=(n+1)^2(n+2)$$
Thus the $(n+1)^{th}$ case holds, given the $n^{th}$ case.
Therefore by induction, the original claim holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find area of triangle given one angle Question: In triangle ABC, all three side lengths are integers and angle A = 60 degrees. The point D lies on the side AC so that BC = BD and DC = 2. Find the smallest possible area of the triangle ABC.
I know that because BC=BD, one of the inner triangles is isosceles. However, I do not understand how this knowledge can help me solve the problem.
| The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us another Diophantine constraint. Draw the altitude $BH$ from $B$ onto $AC$:
Since $\triangle BCD$ is isosceles, $H$ is the midpoint of $CD$, so $CH = 1$; by looking at the right triangle $\triangle BHC$, we get $\cos \angle C = \frac{CH}{BC} = \frac1a$. The Law of Cosines tells us that $$c^2 = a^2 + b^2 - 2b.$$
Substituting the first equation into the second yields $$c^2 = (b^2 + c^2 - bc) + b^2 - 2b \implies bc = 2b^2 - 2b \implies c=2(b-1),$$ and substituting this value of $c$ back into the first equation tells us that $$a^2 = b^2 + (2b-2)^2 - b(2b-2) \implies a^2 - 3(b-1)^2 = 1,$$ which is a Pell equation. The simplest solution to $x^2-3y^2=1$ is $x=2,y=1$, and all other solutions are obtained by taking the coefficients of $1$ and $\sqrt3$ in $(2 + \sqrt3)^k$ for some $k$.
If we use $x=2$ and $y=1$, then this gives us $a=b=c=2$, which is certainly a solution to the problem: in this case, $\triangle ABC$ is equilateral and $D=A$. This triangle has area $\sqrt 3$.
If you don't like this solution, because it's a somewhat degenerate case, then $(2+\sqrt3)^2 = 7+4\sqrt3$ yields $x=7$ and $y=4$ as the next solution. In this case, $a=7$, $b=5$, and $c=8$, which is the triangle I used for the diagram. By Heron's formula, $\triangle ABC$ has area $10 \sqrt 3$.
One of these, depending on whether you find the equilateral solution acceptable or not, is the solution with the smallest area, but we can find infinitely many triangles with integer sides that satisfy the conditions, by taking higher powers of $2+\sqrt 3$ to solve Pell's equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer.
Manually doing fist elements gives
$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$
I'm trying to restate it as
$$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$
so that I cut off the beginning of the series and it's tail.
| $\displaystyle \sum\limits_{k=2n+1}^{4n-1} \frac{1}{k} - \sum\limits_{k=n+1}^{2n-1} \frac{1}{2k} \approx (\ln 4 – \ln 2) - \frac{1}{2}(\ln 2 – \ln 1) = \frac{\ln 2}{2} $
Explanation (with suitable variables $a,b,n,q,u,v$, e.g. $a\in\mathbb{N}$, $n\to\infty$):
$\displaystyle \prod\limits_{k=1}^{an} (1+\frac{x}{k}) \approx \frac{(an)^x}{\Gamma(1+x)}$
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} (1+\frac{x}{k}) \approx (\frac{(an)^x}{\Gamma(1+x)})/(\frac{n^x}{\Gamma(1+x)}) = a^x $
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} (1+\frac{x}{qk})^q \approx a^x $ , $\hspace{2cm} q\to\infty$ :
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} \exp(\frac{x}{k}) \approx a^x$ , $\hspace{2cm} log$ :
$=> \hspace{1cm}\displaystyle \sum\limits_{k=n}^{an} \frac{1}{k} \approx \ln a $
$=> \hspace{1cm}\displaystyle \sum\limits_{k=bn+u}^{an+v} \frac{1}{k} \approx \ln a - \ln b \enspace\enspace$ with constants $\enspace|u|,|v|\ll n $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to explain for my daughter that $\frac {2}{3}$ is greater than $\frac {3}{5}$? I was really upset while I was trying to explain for my daughter that $\frac 23$ is greater than $\frac 35$ and she always claimed that $(3$ is greater than $2$ and $5$ is greater than $3)$ then $\frac 35$ must be greater than $\frac 23$.
At this stage she can't calculate the decimal so that she can't realize that $(\frac 23 = 0.66$ and $\frac 35 = 0.6).$
She is $8$ years old.
| In a first step you can observe with the child that
$$
\frac{2}{3} = 1-\frac{1}{3}
\quad \text{ and } \quad
\frac{3}{5} = 1 - \frac{2}{5}.
$$
In words, you are looking at what is remaining of the pie when you have taken either one third or two fifths.
Then, you can show that cutting a third into two pieces gives two sixths, hence $\frac{1}{3}= \frac{2}{6}$ and then
$$
\frac{2}{3} = 1-\frac{2}{6}
\quad \text{ and } \quad
\frac{3}{5} = 1 - \frac{2}{5}.
$$
You can then conclude by the fact that sharing fairly between more people leads to smaller shares, thus $\frac{2}{6} < \frac{2}{5}$. Hence, the comlplement to two shares is larger:
$$
\frac{2}{3} = 1-\frac{2}{6}
>
\frac{3}{5} = 1 - \frac{2}{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 29,
"answer_id": 25
} |
Evaluate $\int_0^{\infty} \frac{1}{1+x^6} dx$ using Contour Integral My attempt:
I am considering a contour C consisting of upper half circle $|z|=R$ and the real axis from $-R$ to $R$ and finding $\int_C \dfrac{1}{z^6+1}dz$ around it.
First, I find the poles of this function, which are given by $z^6+1=0$
The poles are all simple poles, $z=\pm i,\dfrac{\sqrt{3}}{2} \pm i\dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \pm i\dfrac{1}{2} $
Only the 3 poles $z= i,\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2},-\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2} $ lie within the contour, so I find the residues at these poles and then apply residue theorem to find $\int_C \dfrac{1}{z^6+1}dz$
Residue at $z=i$ comes out to be $\dfrac{1}{6i}$, at $z=\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2}$ comes out to be $\dfrac{1}{3i(1-\sqrt3 i)}$ and at $z=\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2}$ is $\dfrac{-1}{6i}$
So from Cauchy's theorem I get $\int_C \dfrac{1}{z^6+1}dz=2\pi i \times $sum of the residues= $\dfrac{2\pi}{3(1-\sqrt3 i)}$
Taking $R \to \infty$ and using $f(z)$ is an even function, I get my final answer as: $\int_0^{\infty} \frac{1}{1+x^6} dx=\dfrac{\pi}{3(1-\sqrt3 i)}$
Is this solution correct? Is there a shorter or better way to do this? Thank you.
| You have miscalculated the residues at $\dfrac{\pm\sqrt{3} + i}{2}$. Note that if $\zeta$ is a simple zero of $f$, then the residue of $\dfrac{1}{f}$ at $\zeta$ is $\dfrac{1}{f'(\zeta)}$, so here for $f(z) = z^6 + 1$, we have the residues
$$\frac{1}{f'(\zeta)} = \frac{1}{6 \zeta^5} = \frac{\zeta}{6\zeta^6} = - \frac{1}{6}\zeta.$$
Thus the integral over $C$ is - for $R > 1$ -
$$-\frac{2\pi i}{6} (\zeta_1 + \zeta_2 + \zeta_3) = -\frac{\pi i}{3}\biggl(\frac{\sqrt{3} + i}{2} + i + \frac{-\sqrt{3}+ i}{2}\biggr) = -\frac{\pi i}{3}\cdot 2i = \frac{2\pi }{3}.$$
Letting $R \to +\infty$, the integral over the semicircle tends to $0$, whence
$$\int_{-\infty}^{+\infty} \frac{dx}{x^6+1} = \frac{2\pi}{3}.$$
The integral over $[0,+\infty)$ is half of that by parity.
Note that the integrand is real on the real axis, so the integral needs to be real too, by that you could have seen that you must have made a mistake somewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-3xy+y^2$$
$$B(x,y,z) = 2xy+yz-3xz$$
$$C(x,y,z,t) = x^2+y^2+2xy-xt+2yt$$
For $A$ I did $A(x,y) = x^2-3xy+y^2 = x^2-2xy+y^2-xy = (x-y)^2-xy$ which is indefinite.
For $B(x,y) = 2xy+yz-3xz$ I remembered that $(x+y+z)^2 = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2$ so $2xy+yz-3xz = (x+y+z)^2 -x^2-y^2-z^2-yz+xz$ which won't help anything.
For $C(x,y,z,t)$ there's too much terms, how should I do it?
| There is an algorithm for solving $P^T A P = D,$ where $A$ is some given symmetric matrix, $D$ is diagonal, while $\det P = 1.$ See reference for linear algebra books that teach reverse Hermite method for symmetric matrices
In this case we get
$$
P =
\left(
\begin{array}{rr}
1 & \frac{3}{2} \\
0 & 1
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{rr}
1 & 0 \\
\frac{3}{2} & 1
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -\frac{3}{2} \\
-\frac{3}{2} & 1
\end{array}
\right)
\left(
\begin{array}{rr}
1 & \frac{3}{2} \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
1 & 0 \\
0 & -\frac{5}{4}
\end{array}
\right)
$$
Since the determinant is $1,$ it is fairly quick to find the inverse of $P.$ Let us call it
$$
Q =
\left(
\begin{array}{rr}
1 & -\frac{3}{2} \\
0 & 1
\end{array}
\right)
$$
The outcome is
$$ Q^T D Q = A. $$ The diagonal entries of $D$ and the rows of $Q$ give
$$ \left(x - \frac{3}{2} y \right)^2 -\frac{5}{4} y^2 \; \; \; = \; \; \; x^2 - 3xy + y^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Complex numbers involving roots of unity Let $z\in \mathbb{C}$ and $n\in \mathbb{N}, n \geq 1$. Solve the following equation: $$z+z^2+\dots+z^n=n|z|^n$$
Obviously, $(z,n)=(0,n)$ and $(1,n)$ are solutions, $\forall n \geq 1$. Considering $z \neq 0,1$ I tried to use $$z+z^2+\dots+z^n=z(1+z+\dots+z^{n-1})=z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})$$
where $\epsilon=\cos \frac{2\pi}{n}+i\sin \frac{2\pi}{n}$, such that $$z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})=n|z|^n$$
I then tried to take the norm on both sides and obtain some sort of inequalities, but didn't manage to get somewhere.
| Expanding on the work of @WB-man by looking at the solutions graphically.
$n=2$
$$
f(z) = z+ z^2 -2 \left| z\right|^{2}
$$
Find the roots of the real and imaginary components. Solve
$$
\begin{align}
\text{Re } f &= x - x^2 - 3 y^2 = 0, \\
\text{Im } f &= y + 2 x y2 = 0, \\
\end{align}
$$
The solution for the real component is plot in $\color{blue}{blue}$, imaginary component in $\color{red}{red}$.
The solutions are the intersection of the $\color{blue}{blue}$ and $\color{red}{red}$.
$n=3$
$$
f(z) = z + z^2 + z^3 - 3 \left| z\right|^{3}
$$
Solve
$$
\begin{align}
\text{Re } f &= -3 x^2 \sqrt{x^2+y^2}-3 y^2 \sqrt{x^2+y^2}+2 x^2+x-2 y^2 = 0, \\
\text{Im } f &= y + 4 x y = 0, \\
\end{align}
$$
$n=4$
$$
f(z) = z + z^2 + z^3 + z^{4} - 4 \left| z\right|^{4}
$$
Solve
$$
\begin{align}
\text{Re } f &= x + 2 x^2 - 3 x^4 - 2 y^2 - 14 x^2 y^2 - 3 y^4 = 0, \\
\text{Im } f &= y + 4 x y + 4 x^3 y - 4 x y^3 = 0, \\
\end{align}
$$
$n=5$
$$
f(z) = z + z^2 + z^3 + z^{4} + z^{5} - 5 \left| z\right|^{5}
$$
Solve
$$
\begin{align}
\text{Re } f &= x^7+x^6-21 x^5 y^2+x^5-15 x^4 y^2+x^4+35 x^3 y^4-10 x^3 y^2+15 x^2 y^4-6 x^2 y^2-7 y^6 \sqrt{x^2+y^2}-21 x^2 y^4 \sqrt{x^2+y^2}+2 x^2-7 x^6 \sqrt{x^2+y^2}-21 x^4 y^2 \sqrt{x^2+y^2}-7 x y^6+5 x y^4+x-y^6+y^4-2 y^2 = 0, \\
%
\text{Im } f &= 5 x^4 y+4 x^3 y-10 x^2 y^3-4 x y^3+4 x y+y^5+y = 0, \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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} |
If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 +q$, then find the value of $p$ and $q$ If $x^2 - 3x +2$ is a factor of $x^4-px^2+q$ then find the value of $p$ and $q$.
My attempt:
$$x^2-3x+2$$
$$x^2-2x-x+2$$
$$x(x-2)-1(x-2)$$
$$(x-1)(x-2)$$
How do I proceed further?
P.S: Edit after Deepak's comment!
| EDIT1
Factors of $ (x^2 - 3 x + 2) $ are $(x-1),(x-2)$ and each factor produces a zero remainder. Each factor zero remainder requires that
$$ 1-p+q = 0, \, 16-4p+q = 0 $$
Solving
$$ p=5, q=4 $$
and original expression was
$$ x^4 - 5 x^2 + 4 =(x-1)(x+1)(x-2)(x+2)$$
BTW the answer is same even if problem is set in any of six ways..
$$(x^2-3x+2),(x^2-x-2),..,..,..$$
is a factor of $ x^4-px^2+ q $ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\int \arcsin(3x - 4x^3)dx$
$$\int \arcsin(3x - 4x^3)dx$$
I think the best idea here is to use integration by parts and remove the arcsine from the integral.
$$\int \arcsin(3x - 4x^3)dx = x\arcsin(3x - 4x^4) - \int {3x(1-4x^2) \over \sqrt{1- (3x - 4x^3)^2}} dx$$
Now to integrate $\displaystyle \int {3x(1-4x^2) \over \sqrt{1- (3x - 4x^3)^2}} dx$ I tried substituting $u = 3x - 4x^3$, but then I get $\displaystyle \int {x\over \sqrt{1- u^2}} du$.
Now I need to express $x$ in terms of $u$ that is I need to solve $4x^3 - 3x + u = 0$, for which I get a hopeless answer.
Any ideas how to solve this integrand ?
| Another approach:
Let, $x=\sin t\implies dx=\cos t\ dt$
\begin{align*}
\int\arcsin\left(3x-4x^3\right)\ dx&=\int\arcsin\left(3\sin t-4\sin^3t\right)\cos t\ dt\\
&=\int\arcsin(\sin3t)\cos t\ dt\\
&=\int3t\cos t\ dt\\
&=3\left[t\int\cos t\ dt-\int\left\{\dfrac{d}{dt}(t)\cdot \left(\int\cos t\ dt\right)\right\}\right]\\
&=3\left[t\cdot \sin t+\cos t\right]+c\\
&=3[\arcsin x\cdot x+\cos(\arcsin x)]+c\\
&=3[x\cdot\arcsin x+\cos(\arcsin x)]+c\\
&=3x\arcsin x+3\cos\left[\arccos\sqrt{1-x^2}\right]+c\\
&=3x\arcsin x+3\sqrt{1-x^2}+c.
\end{align*}
or, you can use direct inverse formula: $\arcsin\left(3x-4x^3\right)=3\arcsin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the Saddle Points of $f(x,y) = x^3 +3xy^2-3x^2-3y^2+4$ I started by finding the partial derivatives :
$$ f_x(x,y) = 3x^2+3y^2-6x,$$
$$f_y(x,y) = 6xy -6y$$
but I cant seem to find the critical points from here.
| To find critical points, we solve:
$$\begin{cases}
3x^2 + 3y^2 - 6x &=& 0 \\
6xy - 6y &=& 0
\end{cases}$$
From the second equation, we obtain $x=1$ or $y=0$.
Putting $x=1$ to the first equation gives $y=\pm1$.
Putting $y=0$ to the first equation gives $x=2$ or $x=0$.
By the Second partial derivative test:
$$H(x,y) = \begin{pmatrix}f_{xx}(x,y) &f_{xy}(x,y)\\f_{yx}(x,y) &f_{yy}(x,y)\end{pmatrix} = \begin{pmatrix}6x-6 & 6y\\6y &6x-6\end{pmatrix}$$
Then:
$$D(x,y) = \det(H(x,y)) = (6x-6)^2 - (6y)^2$$
*
*$D(1,-1) = 0^2 - (-6)^2 = -36$
*$D(1,1) = 0^2 - (6)^2 = -36$
*$D(0,0) = (-6)^2 - (0)^2 = 36$
*$D(2,0) = (6)^2 - (0)^2 = 36$
Thus $(1,-1)$ and $(1,1)$ are two saddle points.
(created using Geogebra)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$
Any help is appreciated. I'm entering a challenge and can't reach the solution.
| Alternative approach: notice that
$$t^4+t^2+1 = (t^2+1)^2-t^2 = (t^2+t+1)(t^2-t+1)$$
so
$$\frac{t^4+t^2+1}{t^2+t+1} = t^2-t+1$$
so either
$$t^2+t+1=0$$
or
$$t^2-t+1=13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$
Given$$x+y=2010,\\(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942,$$
find the value of $\sqrt[3]{x}+\sqrt[3]{y}$.
My attempts:
Let $u=\sqrt[3]{x} , v=\sqrt[3]{y}$.
$(u+v)(u^2-uv+v^2)=2010 \tag{1}$
$(u+v)(uv+11u+11v+121)=2942
\tag{2}$
$(1)+(2)$:
$$(u+v)(u^2+v^2+11u+11v+121)=4952$$
How to proceed?
| From what you've done, we have that
$$(1)\iff 2010=(u+v)^3-3uv(u+v)\tag3$$
and that
$$(2)\iff uv(u+v)=2942-11(u+v)^2-121(u+v)\tag4$$
From $(3)(4)$, we have
$$2010=(u+v)^3-3(2942-11(u+v)^2-121(u+v)),$$
i.e.
$$t^3+33t^2+363t-10836=0\tag5$$
where $t=u+v$.
Now checking if a divisor of $10836=2^2\times 3^2\times 7\times 43$ is a solution, we see that $t=12$ is a solution to have
$$(5)\iff (t-12)(t^2+45t+903)=0$$
where $t^2+45t+903\gt 0$ since $45^2-4\cdot 1\cdot 903\lt 60^2-4\times 900=0$.
It follows from this that $$\sqrt[3]{x}+\sqrt[3]{y}=\color{red}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
| Square to remove the absolute value: if $x\neq 2$,
$$\left|\frac{3x+1}{x-2}\right|<1\iff (3x+1)^2<(x-2)^2\iff8x^2+10x-3<0.$$
Now this quadratic polynomial has reduced discriminant $\Delta'=5^2+24=49$, whence the roots $-3/4$ and $1/4$, and it is negative between the roots, so the solutions are
$$-\frac32<x<\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$ What is real root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$?
Some remarks:
*
*I saw this quintic in wikipedia
*Real root is given $x=1+{\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$ in wikipedia.
*I used the transformation $x=y+1$ (Tschirnhaus transformation) and $y^5 + 20 y^3 + 20 y^2 + 30 y + 10=0$. (We can remove the term of degree four.)
*Therefore, we have to solve $x^5 + 20 x^3 + 20 x^2 + 30 x + 10=0$ and we have to find $x={\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$.
*But, I want to know how to solve this without plugging it in and verifying an already known root. Can the depressed quintic be solved? Or does one need to use another method to solve this polynomial?
| First off, I don't know of a systematic way to solve it. But suppose one has the magic inspiration to effect the substitution $x=y+2\,$, then the equation in $y$ turns out to be: $$y^5 + 5 y^4 + 30 y^3 + 90 y^2 + 135 y + 81 = 0$$
At this point, the equation became easy to solve. The ratios between symmetric coefficients are $90/30=3^1\,$, $135/5=3^3\,$, $81/1=3^4\,$, suggesting the substitution $y=3z\,$, which gives: $$3 z^5 + 5 z^4 + 10 z^3 + 10 z^2 + 5 z + 1=0 \;\;\iff\;\; 2z^5 + (z+1)^5=0$$
The latter has the obvious real solution $1+ \cfrac{1}{z}=-\sqrt[5]{2}\,$ which can be rationalized as: $$z = \frac{-1}{1+\sqrt[5]{2}} \cdot \frac{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4}{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4} = \frac{-1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4}{1+2} $$
Reverting back to $x=3z+2$ gives $x=1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding maximum value of $x^2y$ given following condition.
Find the maximum value of $x^2y$ given the constraint that : $$x+y+\sqrt{2x^2+2xy+3y^2} =k ~\text{(constant)}~ \text{and} ~~ x,y \in \mathbb{R^+}$$ (Answer is in terms of '$k$')
I tried using AM-GM , $$\frac x2+\frac x2+y \ge 3\Bigg(\frac{x^2y}{4}\Bigg)^{\frac13}$$
And the same thing : $$\frac{2x^2}{3}+\frac{2x^2}{3}+\frac{2x^2}{3}+xy+xy+ 3y^2 \ge 6\Bigg(\frac{8}{3} (x^2y)^4 \Bigg)^{\frac 16}$$
But when I solved for upper bound of $x^2y$ using these two inequalities, my answer didn't match. Where am I wrong? Can you please help by giving a correct way of solving this question?
Thanks.
| I think you mean that the variables are positives.
By the given we obtain:
$$2x^2+2xy+3y^2=(k-x-y)^2$$ or
$$x^2+2y^2+2k(x+y)=k^2$$ or
$$(x+k)^2+2\left(y+\frac{k}{2}\right)^2=\frac{5k^2}{2}$$ or
$$2(x+k)^2+(2y+k)^2=5k^2.$$
Now, by AM-GM and Holder we obtain:
$$5k^2=2(x+k)^2+(2y+k)^2\geq3\sqrt[3]{\left((x+k)^2(2y+k)\right)^2}\geq$$
$$\geq3\sqrt[3]{\left(\left(\sqrt[3]{2x^2y}+k\right)^3\right)^2}=3\left(\sqrt[3]{2x^2y}+k\right)^2.$$
Thus, $$\sqrt[3]{2x^2y}+k\leq k\sqrt{\frac{5}{3}}$$ or
$$x^2y\leq\frac{k^3}{2}\left(\sqrt{\frac{5}{3}}-1\right)^3.$$
The equality occurs for $x=2y$, which gives the answer:
$$\frac{k^3}{2}\left(\sqrt{\frac{5}{3}}-1\right)^3.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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User remainder theorem to estimate $ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2} $ within $\frac{2}{5}$ I am trying to use the alternate series test to estimate:
$ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2} $
The question asks:
By the alternating series test, which of the following is known to be an estimate of L to within an error of $\frac{2}{5}$?
choices: 10, 9, 12, 11, 31/3.
In doing the work, I did not get an answer. The first terms of the series are:
12 - 3 + $\frac{4}{3} - \frac{3}{4} + \frac{12}{25} - \frac{1}{3} + \frac{12}{49}$. Based on the remainder theorem for alternating series, I need to add up the first 5 terms, which gives me a value of 10.06333 and the remainder should be less than the absolute value of the 6th term (1/3). However, my answer does not match any of the multiple choice possibilities above. Did I do something wrong?
| In order to estimate
$$ \eta(2) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} $$
we may quote Euler. He noticed (in his third or fourth proof of the Basel problem, $\approx 1741$) that we may simply exploit the trigonometric integral
$$ \int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx = \frac{1}{2}\arcsin^2(1) = \frac{\pi^2}{8}. $$
Since the Taylor series of $\arcsin(x)$ and $\frac{1}{\sqrt{1-x^2}}$ in a neighbourhood of the origin are well-known, we have:
$$\begin{align*}
\frac{\pi^2}{6}=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\,dx&=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}}\,dx\\
&=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx
+\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\int_0^1x^{2n}\frac{x}{\sqrt{1-x^2}}\,dx\\
&=\frac{4}{3}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\left[\frac{(2n)!!}{(2n+1)!!}\right]\\
&=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\\
&=\frac{4}{3}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}\right)\\
&=\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2).\end{align*}$$
As a consequence, $\eta(2)=\frac{1}{2}\zeta(2)=\color{red}{\large\frac{\pi^2}{12}}$ can be approximated through the Taylor series of the squared arcsine:
$$ \eta(2)=\frac{3}{2}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}} $$
Since the terms of the last series decay pretty fast to zero, by considering just the first five terms of such series we get an approximation of $\eta(2)$ with an error term less than $10^{-4}$:
$$ \eta(2)=\frac{\pi^2}{12}\approx\color{red}{\frac{9211}{11200}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $6(M+m)$ Let $q$, $r$ are unit vectors such that $\vec p = \vec q \times \vec p + \vec r$. $M$ is the maximum and $m$ is the minimum value of $[\vec p \vec q \vec r]$, then we have to find the value of $6(M+m)$.
Should $\vec p$ be zero? As in $\vec p = \vec q \times \vec p + \vec r$, $p$ cannot be equal to its perpendicular.
|
$$\mathbf{p}=\mathbf{q} \times \mathbf{p}+\mathbf{r} \tag{1}$$
$\mathbf{p} \cdot \mathbf{p}$ gives
\begin{align}
p^2 &= \mathbf{p} \cdot \mathbf{r} \\
p^2 &= p\cos \theta
\end{align}
$$p=\cos \theta \tag{2}$$
$\mathbf{p} \cdot \mathbf{q}$ gives
$$\mathbf{p} \cdot \mathbf{q}=\mathbf{q} \cdot \mathbf{r}=\cos \phi \tag{3}$$
$\mathbf{p} \cdot \mathbf{r}$ gives
\begin{align}
\mathbf{p} \cdot \mathbf{r} &= 1-\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} \\
\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} &= 1-\mathbf{p} \cdot \mathbf{r} \\
&= 1-p^2 \\
&= 1-\cos^2 \theta
\end{align}
$$\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} = \sin^2 \theta \tag{4}$$
$\mathbf{p} \times \mathbf{q}$ gives
\begin{align}
\mathbf{p} \times \mathbf{q} &=
(\mathbf{q} \times \mathbf{p}) \times \mathbf{q}+
\mathbf{r} \times \mathbf{q} \\
&= q^2 \mathbf{p}-(\mathbf{p} \cdot \mathbf{q}) \, \mathbf{q}+
\mathbf{r} \times \mathbf{q} \\
&= \mathbf{p}-(\mathbf{p} \cdot \mathbf{q}) \, \mathbf{q}+
\mathbf{r} \times \mathbf{q} \\
\mathbf{p} \times \mathbf{q} \cdot \mathbf{r}
&= \mathbf{p} \cdot \mathbf{r}-
(\mathbf{p} \cdot \mathbf{q})(\mathbf{q} \cdot \mathbf{r})
\end{align}
$$\mathbf{p} \times \mathbf{q} \cdot \mathbf{r}=
\cos^2 \theta-\cos^2 \phi \tag{5}$$
Equating $(4)$ and $(5)$,
\begin{align}
\sin^2 \theta &= \cos^2 \theta-\cos^2 \phi \\
1-\cos^2 \theta &= \cos^2 \theta-\cos^2 \phi \\
\cos^2 \theta &= \frac{1+\cos^2 \phi}{2} \\
\cos^2 \theta-\cos^2 \phi &= \frac{1-\cos^2 \phi}{2} \\
\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} &= \frac{\sin^2 \phi}{2} \\
m &= 0 \\
M &= \frac{1}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?
Thanks!
| The second (finite) series is a particularization of the first (with a shifted index):
$$
1 + x + x^{2} + x^{3} + \cdots + x^{n} + O(x^{n+1})= \frac{1}{1 - x}
$$
means: There exists constant $C > 0$ and $\delta > 0$ such that if $|x| < \delta$, then
$$
\left|\frac{1}{1 - x} - (1 + x + x^{2} + x^{3} + \cdots + x^{n})\right| \leq C|x|^{n+1}.
\tag{1}
$$
By contrast, the second (with $n$ replaced by $n + 1$) reads
$$
1 + r + r^{2} + r^{3} + \cdots + r^{n}
= \frac{r^{n+1} - 1}{r - 1}
= \frac{1 - r^{n+1}}{1 - r}
= \frac{1}{1 - r} - \frac{r^{n+1}}{1 - r}.
\tag{2}
$$
Equation (2) gives the specific form of the $O(x^{n+1})$, namely $\dfrac{x^{n+1}}{1 - x}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify sum over trigonometric functions I need to prove that assuming $N>2$,
$$\frac{1}{N}\sum_{k=1}^{N}\sin\left(\frac{2k}{N}\pi\right)\sin\left(\frac{2\pi}{N}(k-x)\right) = \frac{1}{2}\cos\left(\frac{2\pi}{N} x\right).$$
It should be achievable solely by using trigonometric identities but I am stuck and can't find a way to prove it.
| $$(\sin \frac{2\pi k}{n})(\sin \frac{2\pi (k-x)}{n})=\frac{1}{2}(\cos\frac{2\pi x}{n}-\cos\frac{2\pi (2k-x)}{n})$$
Using $$\sin A \sin B=\frac{1}{2} . \cos (A-B) - \frac{1}{2}.\cos (A+B)$$
Now, $$\frac{1}{n} \sum^n_{k=1} .\frac{1}{2}\cos\frac{2\pi x}{n}-\frac{1}{2}.\frac{1}{n} \sum^n_{k=1}\cos\frac{2\pi (2k-x)}{n}$$
For the second term's credibility to be $0$, You may follow these methods here: Method 1
This should leave you with:
$$\frac{1}{2}\cos \frac{2\pi x}{n}$$
Did this help or should I provide more?
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.
I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent series, I have the following:
$$f(z)=\left(\frac{z+1}{z}\right)\left(\frac{1}{z-2}\right)=\left(\frac{z+1}{z^2}\right)\left(\frac{1}{1-2/z}\right)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(\frac{1}{1-2/z}\right)$$ and therefore $$f(z)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots\right)=\frac{1}{z}-\frac{1}{z^2}+\frac{2}{z^3}-\frac{4}{z^4}+\cdots$$
so the residue at 0 is 1.
Similarly, $$f(z)=\left(\frac{z+1}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)$$ and so $$f(z)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(1-\frac{z-2}{2}+\frac{(z-2)^2}{4}-\frac{(z-2)^3}{8}+\cdots\right)$$
Thus $$f(z)=\frac{3}{2(z-2)}-\frac{1}{4}+\frac{1}{8}(z-2)-\frac{1}{16}(z-2)^2+\cdots$$
and so the residue at 2 is $\frac{3}{2}$.
So I think $\int_Cf(z)dz=2\pi i(1+\frac{3}{2})= 5\pi i$, but that's not what the book is telling me - the book says the answer should be $2\pi i$. What am I doing wrong?
| You can do this without calculating any residues. For any $R>2,$ Cauchy's theorem shows your integral equals
$$\int_{|z|=R} \frac{z+1}{z^2-2z}\, dz = \int_0^{2\pi} \frac{(Re^{it} + 1)iRe^{it}}{R^2e^{2it} - 2Re^{it}}\, dt.$$
Do a little work to see this equals
$$\tag 1 i\int_0^{2\pi} \frac{1 + e^{-it}/R}{1 - 2e^{-it}/R}\, dt.$$
Remember, this stays the same for any $R>2.$ The integrands in $(1)$ $\to 1$ uniformly on $[0,2\pi]$ as $R\to \infty.$ Hence $(1)\to i\cdot 2\pi = 2\pi i$ as $R\to \infty.$ It follows that $2\pi i$ is the value of your integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the unique vector x in the row space for A, for which Ax = b Given the Vector $b = (9, -6, 3)^T$
$A =\begin{pmatrix}
1 & 2 & 0 & 1\\
-1 & -1 & -1 & 0 \\
1 & 0 & 2 &-1
\end{pmatrix}$
I know that the row space is $ \begin{pmatrix} 1 & 0 & 2& -1\end{pmatrix}$ , $ \begin{pmatrix} 0 & 1 & -1& 1\end{pmatrix}$
Not really sure how to proceed.
| Let us take a vector $\ x=(x_1,x_2,x_3,x_4) $
For us to solve the given problem we need to solve the the system $\ Ax=b$
In our case$\ A=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}$ and $\ b=\pmatrix{9\\-6\\3}$
$\ Ax=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}*\pmatrix{x_1\\x_2\\x_3\\x_4}=\pmatrix{9\\-6\\3}$
After solving this system and taking $\ x_3,x_4$ as free variables you should get that$\ x=(3-2x_3+x_4,3+x_3-x_4,x_3,x_4)=(3,3,0,0)+x_3(-2,1,1,0)+x_4(1,-1.0,1)$
From this we conclude that our vector should be from vector space $\ (3,3,0,0)+span((-2,1,1,0),(1,-1.0,1))$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find roots of $f = x^4 - 2x^3 + x^2 + 4x +a \in \Bbb R[x]$ knowing the real part of one of them I have an algebra exam and I've taken this exercise from some old tests.
Let $f = x^4 - 2x^3 + x^2 + 4x +a \in \Bbb R[x]$. Find all $a \in \Bbb R$ for which $f$ has a root $z \in \Bbb C - \Bbb R$ such that the real part of $z$ equals $1$.
I know that the sum of all roots should equal 2 in this polynomial, and if there's one complex root it's conjugate should also be a root. Also the sum of a complex and it's conjugate its $2 Re(z)$ so $z + \overline{z} = 2$ and the sum of the other roots should equal $0$. So if $x_1$ and $x_2$ are the other roots of $f$, $x_1 = -x_2$.
The product of all roots equals $a$ as well.
On the other hand I can form a polynomial that will divide $f$ using $z$ and $\overline z$, i.e. $g(x)= x^2 - 2 Re(z)x + |z|^2$.
I've tried evaluating both $f(x)$ and $g(x)$ with some values:
$f(1)=4+a$
$g(1)=1 - 2Re(z)+1+b^2$
$f(0)=a$
$g(0)=1+b^2$
I think that the answer should be somewhere close this path but I'm stuck... suggestions? thanks!
| The sum of the two other (possibly complex) roots of $f$ is $0$ so we have two options:
*
*The two other roots are pure-imaginary and conjugate. In this case, $f$ is divisible by $x^2 + u^2$ (and the roots are $\pm iu$) for some $u \in \mathbb{R}$.
*The two other roots are real. In this case, $f$ is divisible by $x^2 - u^2$ (and the roots are $\pm u$).
Hence, we should have the factorization
$$ (x^2 - 2x + |z|^2)(x^2 \pm u^2) = x^4 - 2x^3 + x^2 + 4x + a. $$
Expanding the left hand side, we get
$$ x^4 - 2x^3 + (|z|^2 \pm u^2)x^2 \mp 2u^2x \pm |z|^2 u^2 = x^4 - 2x^3 + x^2 + 4x + a.$$
Comparing coefficients, we get $u^2 = 2$ so $u = \pm \sqrt{2}$ and $|z|^2 - u^2 = |z|^2 - 2 = 1$ so $|z| = \sqrt{3}$ and so $a = -|z|^2u^2 = -6$. The final factorization is
$$ x^4 - 2x^3 + x^2 + 4a - 6 = (x^2 - 2x + 3)(x^2 - 2) = \\(x - \sqrt{2})(x + \sqrt{2})(x - (1 + i\sqrt{2}))(x - (1 - i\sqrt{2})). $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area. Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
| The area between $y=4-x^2$ and $y=0$ is $\int_{-2}^{2}(4-x^2)\,dx = \frac{32}{3}$.
We have to find some constant $b\in(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $\frac{16}{3}$. That leads to the equation
$$ \int_{-\sqrt{4-b}}^{\sqrt{4-b}}(4-b-x^2)\,dx = \frac{16}{3} $$
and ultimately to $b=\color{red}{2(2-\sqrt[3]{2})}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the nature of the triangle Consider a triangle $ABC$ for which $\tan \frac{A}{2} = \frac{a}{b + c}$, where $a = BC, \: b = AC, \:c = AB$.
Then, the triangle is:
A) equilateral
B) right triangle with $A = \frac{\pi}{2}$
C) right triangle with $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}$
D) acute
E) obtuse
All I could do was to eliminate "$A$", which is pretty obvious.
| Option C) is correct by trig simplification.
By Sine Rule $a$ is proportional to $\sin A $ with circumcircle diameter as constant of proportionality.
Using shorthand for $ ( s= \sin, c=\cos )$
$$ \tan A/2 = \dfrac{s_{A/2}}{c_{A/2}}= \dfrac{s_A}{s_B+s_C}= \dfrac{2 s_{A/2}c_{A/2}}{2 s_{(B+C)/2}c_{(B-C)/2}} \tag1 $$
since
$$ \frac{(A+B+C)}{2}=\pi/2,\tag2 $$
$$ \cos A/2 = \sin (B+C)/2 \tag3 $$
Plug into 1) RHS, it simplifies to
$$ \dfrac{s_{A/2}}{c_{(B-C)/2}} \tag4 $$
LHS
$$ \tan A/2 = \dfrac{s_{A/2}}{c_{A/2}} \tag5 $$
In RHS of 4) and 5) numerators are same, so equate denominators
$$ \cos A/2 = \cos \frac{B-C}{2} \tag6 $$
For inverse cosine function equality we should have either
$$ \frac{A}{2} = + \frac{B-C}{2} \tag7$$
or
$$ \frac{A}{2} = - \frac{B-C}{2} \tag8 $$
And further on comes to either $B= \pi/2 $ or $C= \pi/2 $ making it a right triangle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the Diophantine Equation $ax^2 + bx + c = dy^2 + ey + f$? As a fun topic for personal research, I am trying to determine everything I can about Pythagorean triples for which all three integers are one less than a perfect square (please don't say anything about this if there is a known solution, I want to try to figure it out myself).
Here is my work so far;
Let $a^2 + b^2 = c^2$ be a primitive triple (that is, $a,b,c$ are pairwise relatively prime). Then the objective is to find a positive integer $g$ so that for some positive integers $x,y,z$;
$$ag = x^2 - 1$$
$$bg = y^2 - 1$$
$$cg = z^2 - 1$$
This is equivalent to solving $\dfrac{x^2-1}{a} = \dfrac{y^2-1}{b} = \dfrac{z^2-1}{c} \in \mathbb{Z}^+$.
Now, consider the problem of solving for $x$. The equation given for $x$ can be written $x^2 \equiv 1 \pmod{a}$. Let $\alpha$ be a solution to this equation in the reduced residue system mod $a$. Then we have a family of solutions to the Diophantine equation $ag = x^2 - 1$ given by $x = at + \alpha, t \in \mathbb{N}$. This leads to
$$\dfrac{x^2-1}{a} = \dfrac{(at+\alpha)^2-1}{a} = at^2 + 2\alpha t + \dfrac{\alpha^2-1}{a}$$
Now, doing the same thing for $y$ and $z$, we can reduce the problem to finding $t,m,n$ such that
$$at^2 + 2 \alpha t + \dfrac{\alpha^2-1}{a} = bm^2 + 2 \beta m + \dfrac{\beta^2-1}{b} = cn^2 + 2 \gamma n + \dfrac{\gamma^2-1}{c}$$
In case it helps, I found a pretty nice way to factorize these
$$at^2 + 2 \alpha t + \dfrac{\alpha^2-1}{a} = a \bigg{(} t + \dfrac{\alpha+1}{a} \bigg{)} \bigg{(} t + \dfrac{\alpha-1}{a} \bigg{)}$$
with the equivalent factorizations holding for the other expressions as well.
where $\alpha^2 \equiv 1 \pmod{a}, \beta^2 \equiv 1 \pmod{b}, \gamma^2 \equiv 1 \pmod{c}$. So my question is whether there is any method for solving an equation like this, even solving the equations pairwise, in the general case or for any special properties of $a,b,c,\alpha,\beta,\gamma$.
Here are all the solutions of which I am aware;
When $a,b,c = 33,56,65$, we can set $x,y,z = 10,13,14$. This yields the equation $99^2 + 168^2 = 195^2 \implies (3a)^2 + (3b)^2 = (3c)^2$.
When $a,b,c = 133,156,205$, we have $x,y,z = 265, 287, 329$. This yields the equation $70224^2 + 82368^2 = 108240^2 \implies (528a)^2 + (528b)^2 = (528c)^2$.
Thanks in advance!
| Legendre gave a general solution to the equation
$$
ax^2+bxy+cy^2+dx+ey+f=0.
$$
Your equation/system can certainly be put in this form, and then the parameterization given by Legendre would provide you with the answers you seek.
| {
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Integrate $\int_{0}^{\pi}\frac{\sin(x)}{(\sin^{2} (x) + k\cos^{2}(x))^{1/2}}\,dx$ I am faced with a dilemma. I have the following integral:
$$I(g,b) = \int_{0}^{\pi} \frac{\sin(\theta)}{(\sin^2(\theta) + \frac{g}{b}\cos^2(\theta))^{1/2}} d\theta = \int_{0}^{\pi} \frac{\sin(\theta)}{(1 - (1-\frac{g}{b})\cos^2(\theta))^{1/2}} d\theta$$
I know that if $g=b=1$:
$$I(1,1) = \int_{0}^{\pi} \frac{\sin(\theta)}{(\sin^2(\theta) + \cos^2(\theta))^{1/2}} d\theta = \int_{0}^{\pi} \sin(\theta) d\theta = \big[-\cos(\theta)\big]_{0}^{\pi} =2$$
However if I solve the integral by substitution;
$$u= \big( 1-\frac{g}{b}\big)^{\frac{1}{2}} \cos(\theta) $$
$$du= -\big( 1-\frac{g}{b}\big)^{\frac{1}{2}} \sin(\theta) d\theta$$
$$-\frac{du}{\big( 1-\frac{g}{b}\big)^{\frac{1}{2}} \sin(\theta)}= d\theta$$
when
$\theta=0 $ ; $u = \big( 1-\frac{g}{b}\big)^{\frac{1}{2}} =u_0$
$\theta=\pi $ ; $u = -\big( 1-\frac{g}{b}\big)^{\frac{1}{2}} =u_{\pi}$
$u_0 = - u_{\pi}$
Thus the integral becomes:
$$I(g,b) = \int_{-u_{\pi}}^{u_{\pi}}- \frac{\sin(\theta)}{(1 - u^2)^{\frac{1}{2}}}\frac{du}{\big( 1-\frac{g}{b}\big)^{\frac{1}{2}} \sin(\theta)} = -\frac{1}{\big( 1-\frac{g}{b}\big)^{\frac{1}{2}} }\int_{-u_{\pi}}^{u_{\pi}} \frac{du}{(1 - u^2)^{\frac{1}{2}}} $$
Now, depending on whether the minus sign at the fron of the expresison is taken within the integral or not we can see that this is either arrccos or arcsin.
$$\int \frac{du}{(1-u^2)^{\frac{1}{2}}} = \arccos(u)$$
$$\int \frac{-du}{(1-u^2)^{\frac{1}{2}}} = \arcsin(u)$$
If g=b=1:
For arcsin:
$$I(1,1) = -\frac{1}{(1-\frac{1}{1})^{\frac{1}{2}}} \bigg(\arcsin(-(1-\frac{1}{1})^{\frac{1}{2}} ) -\arcsin((1-\frac{1}{1})^{\frac{1}{2}} )\bigg) =\frac{1}{(1-\frac{1}{1})^{\frac{1}{2}}}\bigg(2\arcsin((1-\frac{1}{1})^{\frac{1}{2}} ) \bigg) = -\frac{2}{0} \arcsin(0)= -\frac{0}{0} $$
For arccos:
$$I(1,1) = \frac{1}{(1-\frac{1}{1})^{\frac{1}{2}}} \bigg( \arccos(-(1-\frac{1}{1})^{\frac{1}{2}} ) -\arccos((1-\frac{1}{1})^{\frac{1}{2}} ) \bigg)= \frac{1}{0} \bigg(\arccos(0)-\arccos(0)\bigg)= \frac{\bigg(\pi-\pi\bigg)}{0} =\frac{0}{0}$$
This has been solved in published work of others (whom are now dead and thus cannot be asked), but without any steps other than an additional function being $ E(g,b)=E(1,1)= 1-I(1,1) = -1$.
Thus $$I(1,1) = 2$$.
I(1,1) MUST be 2, but I do not see how the arcsin and arcos integrals are "wrong", now how to solve this integral.
In general g is between 0 and 1.
b is any real positive number greater than g.
EDIT:
The solution to the $I(g,b)$ integral has to be directly applicable for all g values between 0 and 1 including 0 and 1.
| Case 1: $ \ $ Let $k=g/b>1$. The indefinite integral is
$$
\int{\sin t\over\sqrt{\sin^2 t + k \cos^2t}} dt = -\frac{\log(\sqrt{2(k - 1)} \cos t + \sqrt{(k - 1) \cos2t + k + 1})}{\sqrt{k - 1}} + C
$$
$$
= -{\sinh^{-1}(\sqrt{k-1} \cos t)\over\sqrt{k-1}} + C_1.
$$
The definite integral from $0$ to $\pi$ is
$$
\int_0^{\pi}{\sin t\over\sqrt{\sin^2 t + k \cos^2t}} dt =
{\log(\sqrt{k}+\sqrt{k-1}) - \log(\sqrt{k}-\sqrt{k-1})\over\sqrt{k-1}}
$$
$$
= {2 \sinh^{-1}\sqrt{k - 1}\over\sqrt{k - 1}}.
$$
This is not directly applicable to the case $k=g/b=1$; however,
$$
\lim_{k\to1} {2 \sinh^{-1}\sqrt{k - 1}\over\sqrt{k - 1}}=2
$$
as expected.
Case 2: $ \ $ Now consider $0<k<1$. We have
$$
\int{\sin t\over\sqrt{\sin^2 t + k \cos^2t}} dt =
-\frac{\arcsin(\sqrt{1-k} \cdot\cos t)}{\sqrt{1-k}}
+ C
$$
and the definite integral from $0$ to $\pi$ is
$$
\int_0^{\pi}{\sin t\over\sqrt{\sin^2 t + k \cos^2t}} dt =
{2 \arcsin\sqrt{1-k}\over\sqrt{1-k}}=
{2 \arccos\sqrt{k}\over\sqrt{1-k}}.
$$
Again, as $k\to1$, the limit is $2$ as expected:
$$
\lim_{k\to1} {2 \arcsin\sqrt{1-k}\over\sqrt{1-k}} = 2.
$$
Case 3: $ \ $ For $k=0$, the definite integral is equal to $\pi$ (because the integrand is $1$).
Case 4: $ \ $ For $k=1$, the definite integral is equal to $2$ (as already explained in the question).
| {
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$
But how to find $\alpha^6+\beta^6$ ?
| Hint: Here $\alpha^2=2\alpha -4$ and same for $\beta$, as they are the roots of the equation $x^2-2x+4=0$.
Now calculate the value of ${\alpha}^6$ and ${\beta}^6$ in terms of $\alpha^2 and \beta^2$. Then substitute their value again.
| {
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Certain properties of numbers such that $n \mid 2^n+1$ A natural number $n>1$ is called good if$$n \mid 2^n+1.$$ For example, $n=3$ is good, as $3 \mid 2^3+1=9$. Prove that if $N_1$ and $N_2$ are good, then:
*
*$\mathrm{lcm}(N_1,N_2)$ and $\gcd(N_1,N_2)$ are good,
*$N_1\cdot N_2$ is good.
This seems pretty difficult for me. Any hints?
| You can prove "good" ness pretty easily without knowing much more about $N_1 $ and $N_2$, but I thought I'd share the properties that make numbers "good", and thus you can see that your problem becomes trivial.
We are going to go case by case where we assume something about $n$, take it to its conclusion, and then sew together the cases to get the overall answer.
Case 1: n is 1
1 divides every integer, and 21+1=3 is an integer. n = 1 is part of the solution set.
Case 2: n is even
If $n$ is even, every multiple of $n$ must also be even. But $2^n + 1$ is clearly odd, so no even $n$ will divide it. So $n$ cannot be even.
Case 3: If n is a power of an odd prime
Let $n=p^k$ where p is an odd prime. If $n$ divides $2^n + 1$, so does $p$
$2^n+1=2^{p^k} + 1 = 2^{p^{k-1}*p} + 1 = (2^{p^{k-1}})^p + 1$
$2^{p^{k-1}$ is relatively prime to $p$ (hint, both $p$ and $2$ are primes). Let $2^{p^{k-1}}$ = a
By Fermat's Little Theorem, we know that $a^p \equiv a\mod p$
So, $(2^{p^{k-1}})^p + 1 \equiv 2^{p^{k-1}} + 1\mod p)$
We can keep reducing this modulo $p$ and reducing the powers of $p$ in the exponent until we have only 1, so we get $2^p + 1 \equiv 2 + 1 \equiv 3\mod p$
This means that $2^n + 1 \equiv 3\mod p$ . But $p$ divides $2^n + 1$, so $p$ must divide 3 as well, i.e. $p = 3$
So if n is the power of an odd prime, it must be a power of 3.
Does this mean that all powers of 3 satisfy our condition? We don’t know yet. We are currently only gathering “necessary” conditions, not “sufficient”.
Case 4: If n is a product of powers of distinct odd primes
This is the last and most general case, as all odd numbers are a products of powers of odd primes. If $n$ is a product of prime powers, then let $p$ be least such prime that divides $n$.
Let $n = mp^k$ , where $m$ is a product of powers of primes larger than $p$, and $k$ is some integer.
Similar to case 3, we can see that $2^n + 1 = 2^{mp^k} + 1 = (2^{mp^{k-1}})^p + 1 \equiv 2^{mp^{k-1}} + 1\mod p$
We can keep reducing this modulo $p$ and pick off powers of p until we get $2^m + 1$
We have thus shown that if $p$ divides $2^n + 1$, it then divides $2^m + 1$
$2^m \equiv -1\mod p$, so $2^{2m} \equiv 1\mod p$ so $4^m \equiv 1\mod p$
Let $q$ be the order of $4\mod p$ (i.e. the lowest integer satisfying $4^q \equiv 1\mod p$). We know 4 has an order as it is relatively prime to any odd prime. We know $q$ divides $m$ as that’s a property of the order.
Also, as $q$ is the order of $4$ mod $p$ , $q$ must also divide $\phi(p)$, i.e. $q$ must also divide $p - 1$.
However, $q$ is clearly larger than $p - 1$ . $q$ is a divisor of $m$, and all divisors of $m$ are at least as big as their prime factors, all of which are larger than $p$ by design.. So $q$ cannot divide $p - 1$. This gives us our coveted contradiction.
So n cannot be a product of powers of distinct odd primes.
Cases 2 and 4 are eliminated.
So we are left with two cases
*
*$n = 1 $
*$n = 3^k$ for some integer $k$
This is really just one case ($3^0 = 1$), but we’ll get to that later. These $3^k$ powers are just candidates.
How do we prove that every power $3^k$ divides $2^{3^k} + 1$ ?
We can try proving this by induction. We know that for $k = 1$, $2^3 + 1 = 9$ which is divisible by $3^1$.
Let $2^{3^k} + 1 = a*3^k$, $a,k \in N $
All these exponents can make one feel wonky, but we can observe that $2^{3^k}$ can be turned into $2^{3^{k+1}}$ by cubing.
So cubing both sides, we get:
$2^{(3^k)(3)} + 3*2^{(3^k)(2)} + 3*2^{3^k} + 1 = (a^3)(3^{3k})$
$2^{3^{k+1}} + 1 = (a^3)(3^{3k}) - 3*2^{3^k}(2^{3^k} + 1) $
We know that $3^{3k}$ is clearly divisible by $3^{k+1}$ as $3k > k + 1$.
So let $(a^3)(3^{3k}) = b.3^{k+1}$
$2^{3^{k+1}} + 1 = b.3^{k+1} - 2^{3^k}*3*a*3^k$
$2^{3^{k+1}} + 1 = b.3^{k+1} - 2^{3^k}*a*3^{k+1}$
The RHS is clearly divisible by $3^{k+1}$, so we have our proof.
The answer is: all numbers $n=3^k$ where $k \geq 0$ satisfy the property “$n$ divides $2^n + 1$”
From this point, proving your statements about good numbers is trivial.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the coefficient of $x^2y^2z^4$ in the expansion of $(x-2y+z)^8$. I need help solving this question. How do I start?
Attempt at solution
*The term that contains $x^2y^2z^4$ is ${8 \choose 4} (-2y)^2(z)$ ?
*
| We can also iteratively apply the binomial theorem in order to find the coefficients. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.
We obtain
\begin{align*}
[x^2y^2z^4](x-2y+z)^8&=[x^2y^2z^4]\sum_{j=0}^8\binom{8}{j}x^j(-2y+z)^{8-j}\tag{1}\\
&=[y^2z^4]\binom{8}{2}(-2y+z)^6\\
&=\binom{8}{2}[y^2z^4]\sum_{j=0}^6\binom{6}{j}(-2y)^jz^{6-j}\tag{2}\\
&=\binom{8}{2}\binom{6}{2}(-2)^2\\
&=1680
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem and select the coefficient of $x^2$.
*In (2) we again apply the binomial theorem and select the coefficient of $y^2z^4$.
| {
"language": "en",
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Finding the value of $a$ and $b$ for the curve $y=ax^2+\frac{b}{x}$ given $\frac{dy}{dx}=-5$ at $(2,-2)$. The curve $y= ax^2 + \frac{b}{x} $ has a gradient of $-5$ at the point $(2,-2)$. Find the value of $a$ and $b$.
These are my workings -
$$y= ax^2 + \frac{b}{x} \tag{1}$$
Sub $x=2 , y=-2$
$$8a+b=-4 \tag{2}$$
$$\frac{dy}{dx}\ = 2ax - bx^{-2} $$
When $x=2$ , $\frac{dy}{dx}\ = -5 $
$$-5 = 2ax - bx^{-2} \tag{3}$$
Here, I have two unknown constants $a$ and $b$. How should I find $1$ of the constant first? Or had I made a mistake earlier? Thanks !!
| You can easily obtain simultaneous equations from both $y$ and $\frac{dy}{dx}$. You are generally on the right track (You've found an equation using $y=ax^2+\frac{b}{x}$). However, for the derivative on equation $(3)$, you seem to have forgotten to substitute $x=2$. If you had substituted it, you would have gotten the simultaneous equations required. Below is the approach I would use:
Using the curve and substituting the known values at the point $(2,-2)$.
$$y=ax^2+\frac{b}{x} \implies -2=a\cdot 2^2+\frac{b}{2}$$
You evaluated your derivative correctly. Substituting the known values at the point $(2,-2)$ gives:
$$\frac{dy}{dx}=2ax-\frac{b}{x^2} \implies -5=2a\cdot 2-\frac{b}{2^2}$$
This gives you the system of equations:
$$\begin{cases} 4a+\frac{b}{2}=-2 \\ 4a-\frac{b}{4}=-5 \end{cases} \iff \begin{cases} 16a+2b=-8 \\ 16a-b=-20 \end{cases}$$
You can subtract one of the equations from the other to eliminate $a$. This gives:
$$16a+2b-(16a-b)=-8-(-20) \iff 2b+b=12$$
Can you solve for $b$, then find the corresponding value of $a$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer is $[-2;0) $. Should there be a different approach?
| If we differentiate, $\frac d{dx}(\sqrt{x+2}-\sqrt{x+3})=\frac{1}{2}\Big(\frac1{\sqrt{x+2}}-\frac1{\sqrt{x+3}}\Big)>0$. So, provided $x\geq-2$, the LHS is continuous and increasing with $x$ and it will be equal to the RHS when $x=0$. Therefore it is less than the RHS when $-2\leq x<0$.
To see it is increasing without calculus, note that
$$\sqrt{x+2}-\sqrt{x+3}=\frac{-1}{\sqrt{x+2}+\sqrt{x+3}},$$
and the denominator is clearly increasing with $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260395",
"timestamp": "2023-03-29T00:00:00",
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How to calculate $14^{2017^{2017}} \mod 60$? $14^{2017^{2017}} \mod 60$
So, I know that I should begin with decomposing $60$ to the prime factors, which are: $ 3, 2^2, 5$, now I should calculate $14^{2017^{2017}} mod$ all three of these prime factors.
My question is, what is the easiest way of calculating $14^{2017^{2017}} \mod 3$?
| Sketch:
Since $14\equiv 2\pmod 3$,
$$14^{2017^{2017}}\equiv 2^{2017^{2017}}\pmod 3.$$
Now, we know that the powers of $2$ satisfy
\begin{align*}
2^1=2&\equiv 2\pmod 3\\
2^2=4&\equiv 1\pmod 3\\
2^3=8&\equiv 2\pmod 3\\
2^4=16&\equiv 1\pmod 3\\
2^5=32&\equiv 2\pmod 3
\end{align*}
and so on.
In other words, since $2^2\equiv 1\pmod 3$, we only need to compute the power modulo $2$. Since $2017^{2017}\equiv 1^{2017}\equiv 1\pmod 2$, we know that the power on $2$ is odd. Hence
$$
14^{2017^{2017}}\equiv 2^{2017^{2017}}\equiv 2^1\equiv 2\pmod 3.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$? Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$ ?
This is a homework problem, so if someone could just give me an outline/hint of the method used to solve this it would be much appreciated.
Thanks
| To prove the solution is $x=1,y=2,z=3$ we can write the constrained problem as
$min f(y,z)$ where $f(y,z) = \frac{108}{y^2 z^3} + y + z$.
Solving for the minimum of $f(y,z)$ we get:
$$
\frac{\partial f}{\partial y} = -2 \frac{108}{y^3 z^3} +1 = 0
$$
$$
(yz)^3 = 216
$$
$$
yz = 6
$$
$$
\frac{\partial f}{\partial z} = -3 \frac{108}{y^2 z^4} +1 = 0
$$
$$
(yz^2)^2 = 324
$$
$$
yz^2 =18
$$
Solving the pair of quadratic equations we get $y=2$ and $z=3$, and hence $x=1$. So the minimum value of the sum is $6$.
| {
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Find $\lim_{n \to \infty} M(n)/n$ where $M(n)$ is the largest integer satisfying a binomial inequality Let $M (n) $ be the largest integer such that ${m \choose n-1}>{m-1 \choose n} $, then what is the value of $\lim _{n \to \infty} \frac {M (n)}{n} $?
Attempt: After solving and simplifying the given condition, I get a quadratic in $m$ as $m^2-m (1+3n)+n^2-n < 0$. Solving for $m$ we have $m=\frac {(3n+1) \pm \sqrt {5n^2+10n+1}}{2} $ thus I get two values for limit as $\frac{3 \pm \sqrt {5}}{2} $ . As both are positive I can't discard any value straight forward. So what should be done to discard one value or are both values the answer? Note that there can be more than one correct answer.
| Note that from the condition we get the ratio of the two binomials to be $\frac{mn}{(m-n+1)(m-n)}$ so our condition simplifies to $(m-n+1)(m-n) - mn <0$. That is $m^2 - 3mn + m - n + n^2< 0$. (note that this is different to your condition). We can simplify this to $m^2 - m(3n-1) + n^2 - n < 0$. This quadratic has roots $$\alpha,\beta = \frac{3n-1 \pm \sqrt{5n^2 - 2n+1}}{2}$$ where $\alpha$ corresponds to the root larger root and $\beta$ the smaller one. (Note that this differs from your roots as well).
It is clear that $m$ must lie between $\beta < m < \alpha$. Since $\alpha - \beta= \sqrt{5n^2 - 2n +1} > 1$ then there is at least one integer between $\beta$ and $\alpha$. So $M(n)$ is the greatest integer in this interval. That is $\alpha - 1 \leq M(n) \leq \alpha$ or equivalently $\frac{\alpha - 1}{n} \leq \frac{M(n)}{n} \leq \frac{\alpha}{n}$.
Now we squeeze $$\lim \frac{\alpha - 1}{n} \leq \frac{\lim M(n)}{n} \leq \lim \frac{\alpha}{n}$$
But since $$\lim \frac{\alpha}{n} = \lim\frac{3 - \frac{1}{n} + \sqrt{5 - \frac{2}{n} + \frac{1}{n^2}}}{2} = \frac{3+\sqrt{5}}{2}$$ and $\lim \frac{\alpha - 1}{n} = \lim \frac{\alpha}{n}$ we squeeze $M(n) \to \frac{3+\sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that the determinant represents a straight line. Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$, then show that the equation
represents a straight line.
Expansion of determinant is tedious. I can see that matrix is symmetric but not able to use this fact. Could someone help me with this?
| \begin{eqnarray*}
Det\left[
\begin{array}{ccc}
ax-by-c&bx+ay&cx+a \\
bx+ay&-ax+by-c&cy+b\\
cx+a&cy+b&-ax-by+c \\
\end{array}
\right]=(x^2+y^2+1)(a^2+b^2+c^2)(ax+by+c).
\end{eqnarray*}
The first two factors are positive so $\color{red}{ax+by+c=0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^k \leq 5^k$ is true and I want to prove that implies $k+1$. Using the inductive hypothesis I multiply both sides by $4$ to get this:
$$4 \cdot 2^{k} + 4\cdot 4^{k} \leq 4 \cdot 5^{k}$$
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
I will use again the induction hypothesis, this time I'll multiply both side by $5$ to get:
$$5 \cdot (2^{k} + 4^{k}) \leq 5^{k+1}$$
I can say that $2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k})$ and $4 \cdot 5^{k} \leq 5^{k+1}$ so I concatenate them:
$$2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k} \leq 5^{k+1}$$
However this doesn't feel right. I'm assuming that $5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k}$ which there's no way I can be sure about. At this point I'm stuck since the whole reasoning seems wrong.
| It can be done much easier: assume $2^k+4^k\le 5^k$, and see
\begin{align}
5^{k+1}&=5\cdot5^k\\
&\ge 5\cdot 2^k+5\cdot4^k\\
&\ge2\cdot 2^k+4\cdot4^k\\
&=2^{k+1}+4^{k+1}\\
\end{align}
Completing the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Solving the differential equation $y'' + y = 8\cos(x) \cos(2x)$ I'm solving the differential equation $y'' + y = 8\cos(x) \cos(2x)$
I started to find the homogeneous solution:
We search the zeros of the associated polynomial:
$$r^2 + 1 = 0$$
This yields $r = i$ or $r = -i$
Hence, the homogeneous solution is:
$$y_h = e^0(c_1\cos(x) + c_2\sin(x)) + e^0(c_3\cos(-x) + c_4\sin(-x))$$
And by rewriting this, we find:
$$y_h = c_1\cos(x) + c_2\sin(x)$$
I have trouble making a suggestion for the particular solution. I would suggest something like:
$$y_p = (A\cos(x) + B\sin(x))(C\cos(2x) + D\sin(2x))$$
but the answer my book gives is :
$$y = x_1\cos(x) + (c_2 + 2x)\sin(x) - 1/2\cos(3x)$$
Where does the $\cos(3x)$ come from? This makes me wonder my suggestion won't work.
Any help will be appreciated.
| The C.F.$=c_1\cos x+c_2\sin x$,
By inverse operator rule: When
\begin{align*}
P.I.&=\dfrac{1}{f(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2,\text{ when }[f(-a^2)\neq0]\\
&=x\dfrac{1}{f'(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2,\text{ when }[f(-a^2)=0,\ f'(-a^2)\neq0]\\
&=x^2\dfrac{1}{f''(D^2)}\cos(ax+b)\ \text{ put }D^2=-a^2,\text{ when }[f'(-a^2)=0,\ f''(-a^2)\neq0]\\
&=\cdots
\end{align*}
So the P.I. is given by (using inverse operator rules),
\begin{align*}
P.I.&=\dfrac{4}{D^2+1}(\cos3x+\cos x)\\
&=4\left[\dfrac{1}{D^2+1}\cos3x+\dfrac{1}{D^2+1}\cos x\right]\\
&=4\left[\dfrac{1}{-3^2+1}\cos3x+x\cdot\dfrac{1}{2D}\cos x\right]\\
&=4\left[-\dfrac{1}{8}\cos3x+\dfrac{x}{2}\cdot\dfrac{1}{D}\cos x\right]\\
&=4\left[-\dfrac{1}{8}\cos3x+\dfrac{x}{2}\sin x\right]\\
&=2x\sin x-\dfrac{1}{2}\cos3x.
\end{align*}
So, $y=C.F.+P.I.=c_1\cos x+c_2\sin x+2x\sin x-\dfrac{1}{2}\cos3x$
or, $y=c_1\cos x+(c_2+2x)\sin x-\dfrac{1}{2}\cos3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ For real numbers $a, b, c, d >0$ I have to prove that $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$ always holds. I tried using the mean inequalities for multiple numbers but I always get in trouble with the different orders of the two roots. And in case I make an estimate that can handle the orders of the roots, it is too strong. Can anyone help?
| Use this inequality:
$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ equivalent $\frac{1}{4}(a+b+c+d) \geq \sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ Now just prove$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq \frac{1}{4}(a+b+c+d)$ equivalent $4(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$ equivalent $3(a^2+b^2+c^2+d^2)\geq 2(ab+ac+ad+ bc + bd + cd)$ equivalent $(a-b)^2 + (a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 + (c-d)^2 \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
The value of infinite product At a first look I observed that every term of this product is less than 1 so the limit will also be less than 1 but clueless about how to find limit .
$$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$
| Note that
$$\ln\left(\frac{n^3-1}{n^3+1}\right)=\ln\left(\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right)\\=\ln(n-1)-\ln(n+1)+\ln((n+1)^2-(n+1)+1)-\ln(n^2-n+1).$$
Hence, by taking the logarithm of the partial product we obtain a telescoping sum,
$$\ln\left(\prod_{n=2}^N \frac{n^3-1}{n^3+1} \right)=\sum_{n=2}^N \ln\left(\frac{n^3-1}{n^3+1}\right)\\=\ln(1)+\ln(2)-\ln(N)-\ln(N+1)+\ln(N^2+N+1)-\ln(3)\\
=\ln(2/3)+\ln\left(\frac{N^2+N+1}{N(N+1)}\right)\stackrel{N\to\infty}{\to} \ln(2/3).$$
Thus the infinite product yields $2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Halving a vector and halving an angle
An object is subjected to two equal forces along two different directions. If the magnitude of one of them is halved, the angle which the new resultant makes with the other component force is also halved. What is the angle between the forces?
Well I'm not sure whether the vector notation in the second figure is right or not – if wrong please correct it (I'm learning vectors).
| Let's set
$$
\left\{
\begin{array}{ll}
\vec{p} &= p \cos{\alpha} \hat{i} + p \sin{\alpha}\hat{j} \\
\vec{p^*} &=p \cos{\alpha} \hat{i} - p \sin{\alpha}\hat{j} \\
\end{array}
\right.
$$
$\Rightarrow \vec{p} + \vec{p^*} = 2p \cos{\alpha}\hat{i}$. But
$\frac{1}{2}\vec{p} + \vec{p^*} = \frac{3}{2}p \cos{\alpha} \hat{i} - \frac{1}{2} p \sin{\alpha} \hat{j} = \vec{p}_N$.
Now, it was defined that the angle between $\vec{p}_N$ and $\vec{p^*}$ is $\frac{\alpha}{2}$:
$$
\cos{\frac{\alpha}{2}} = \frac{\vec{p}_N \cdot \vec{p^*}}{\left|\vec{p}_N \right| \left|\vec{p^*} \right|} = \frac{\frac{3}{2}p^2 \cos^2{\alpha} + \frac{1}{2}p^2 \sin^2{\alpha}}{p^2 \sqrt{ \frac{9}{4}\cos^2{\alpha} + \frac{1}{4} \sin^2{\alpha} }} = \frac{ \frac{1}{2} + \cos^2{\alpha} }{\sqrt{ \frac{1}{4} + 2 \cos^2{\alpha}}} = \frac{ 1 + 2\cos^2{\alpha} }{\sqrt{ 1 + 8 \cos^2{\alpha}}}
$$
We also note that $\cos{2x} = 2\cos^2{x} - 1 \Rightarrow \cos{\frac{\alpha}{2}} = \sqrt{ \frac{\cos{\alpha} + 1}{2} }$. Inserting this result into the previous equation, and taking the notation $\cos{\alpha} = \eta$, we obtain the equation
$$
\frac{\eta+1}{2} = \left( \frac{1+ 2\eta^2}{\sqrt{1+8\eta^2}} \right)^2
=\frac{(1+ 2\eta^2)^2}{1+8\eta^2} = \frac{4\eta^4 + 4\eta^2 +1}{8\eta + 1}
$$
$$
\Rightarrow 8\eta^3 + 3\eta^2 + \eta +1 = 8\eta^4 + 8 \eta^2 +2
$$
$$
\Rightarrow 8\eta^4 - 8\eta^3 - \eta + 1= 0
$$
The solutions to this equation are $\eta = 1$ and $\eta= \frac{1}{2}$, giving $\alpha = 0$ (trivial) or $\alpha = 60^{\circ}$. The angle between the forces is therefore $2\alpha = 120^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $\tan{x}$'s and I don't know how to go to $t$, let alone express $\cos{x}$ in $t$.
| Note that $$\sin^{2}\left(\frac{x}{2}\right) = \frac{1-\cos x}{2},\quad \cos^{2}\left(\frac{x}{2}\right) = \frac{1+\cos x}{2},$$
so $$\tan^{2}\left(\frac{x}{2}\right) =\frac{1-\cos x}{1+\cos x}.$$
Setting $$t = \tan \left(\frac{x}{2}\right)$$
we get
\begin{align*}
t^{2} &= \frac{1-\cos x}{1+\cos x}\\ \implies t^{2} + t^{2}\cos x &= 1 - \cos x\\ \implies (1+t^{2})\cos x &= 1-t^{2}\\
\implies \cos x &= \frac{1-t^{2}}{1+t^{2}}
\end{align*}
*Edited based on Arnaldo's suggestion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve for xy in terms of a and b $$x^2 + xy + y^2 = a$$
$$x + y = b$$
$xy = ?$
I tried this and did this:
$xy = a - x^2 - y^2$
$xy = a - (x^2 + y^2)$
$xy = a - (x + y)(x - y)$
$xy = a - b(x - y)$
At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.
Any help would be appreciated.
| $x^2 + xy + y^2 = a$ ....(1)
$x + y = b$
Squaring both sides,
$x^2 + y^2 + 2xy = b^2$
$x^2 + y^2 = b^2 - 2xy$
Put value of $x^2 + y^2$ in equation (1),
$b^2 - 2xy + xy = a$
$b^2 - xy = a$
$xy = b^2 - a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\
-3y^2-4y=5x^2+24x+26 \Leftrightarrow \\
???$$
What do I do next?
| Naive geometric solution:
The locus of points whose distances are a fixed ratio (not equality) from two given points is a circle. Call the points $A$ and $B$. The circle's center lies on the (extended) line $AB$.
If you want all the points twice as far from $A$ as from $B$, then the two points where its circumference intersects line $AB$ are $\frac13$ of the way from $B$ to $A$, and at distance $d(AB)$ on the other side of $B$ (opposite from $A$). The center is the midpoint between these points, at distance $\frac13 d(AB)$ on the far side of $B$ from $A$. That makes the radius of the circle $\frac{2}{3}d(AB)$.
In this case, $A$ is at $2i$ and $B$ is at $-3$. That puts the center at the point $2i(1-t)-3t$ for $t=\frac{4}{3}$, and it makes the radius $\frac{2}{3}|2i+3|$. I.e., the center is at $-4-\frac23i$, and the radius is $\frac{2\sqrt{13}}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Teach me how to transform integrals Trig for $\int _{0}^{\pi }\sin ^{2}\left( \psi \right) \sin \left( m\psi \right) d\psi $ How to transform the integral
$$\int _{0}^{\pi }\sin ^{2}\left( \psi \right) \sin \left( m\psi \right) d\psi $$
to
$$\int _{0}^{\pi }\left( \dfrac {1} {2}-\dfrac {1} {2}\cos 2\psi \right) \sin m\psi d\psi $$
What is the general method you need to solve trig questions like this. How are do you know which identities to use and which ones should you always have memorised to derive this.
| The general approach
Considering trigonometric integrals there are a myriad of different techniques to solve them. Sometimes integration by parts is the best option, other times a clever substitution or a trigonometric identity saves the day. Sadly, the only way to know which one to use is to experiment and solve as many problems as possible.
My toolbox
In my toolbox I only store three basic identities when it comes to integrating sine and cosine.
$$
\begin{align*}
1 & = (\sin x)^2 + (\cos x)^2 \\
\cos(A+B) & = \cos(A)\cos(B) - \sin(A) \sin(B) \\
\sin(A+B) & = \cos(A) \sin(B) + \cos(B) \sin(A)
\end{align*}
$$
Everything else I can derive from these pretty simple formulas. Of course as mentioned earlier I know how to use integration by parts, and use substitutions as well.
The problem at hand
Let's look at your problem and use a fairly clever trick. I start with defining the following pair of integrals
$$
I = \int_0^{\pi} (\sin \psi)^2 \sin (m\psi)\, \mathrm{d}\psi
\quad \text{and} \quad
J = \int_0^{\pi} (\cos \psi)^2 \sin (m\psi)\,\mathrm{d}\psi
$$
Notice how simple we can solve the following integral
$$
I + J = \int_0^\pi (\sin^2 \psi + \cos^2\psi) \sin (m\psi)\, \mathrm{d}\psi =
\begin{cases}
\cfrac{1 - \cos m\pi}{m} & \text{if} \qquad m \neq 0 \\
0 & \text{if} \qquad m = 0
\end{cases}
$$
Similarly, the integral $J - I$ can be evaluated fairly easy with the use of
$\cos^2\psi - \sin^2\psi = \cos 2\psi$ and
$$
\cos ax \sin bx = \frac{\sin(a x + b x) - \sin(a x - b x)}{2}\,.
$$
This can be derived by taking the difference between $\sin(A+B)$ and $\sin(A-B)$ and then using the addition formula for sine, written at the top. Thus,
$$
\begin{align*}
J - I & = \int_0^\pi (\cos^2\psi - \sin^2\psi )\sin (m\psi) \,\mathrm{d}\psi \\
& = \int_0^\pi \cos (2\psi) \sin (m\psi)\, \mathrm{d}\psi \\
& = \frac{1}{2}\int_0^\pi \sin (2 + m)\psi - \sin(2 - m)\psi \,\mathrm{d}\psi \\
& = \begin{cases}
\cfrac{m}{m^2 - 4}(1 - \cos m\pi) & \text{if} \qquad m^2 \neq 4 \\
0 & \text{if} \qquad m^2 = 4
\end{cases}
\end{align*}
$$
Assuming $m \neq 0$ and $m^2 \neq 2$ we then have the following system of equations
$$
\begin{align*}
J - I & = \cfrac{1 - \cos m\pi}{m} \\
J + I & = \cfrac{m}{m^2 - 4}(1 - \cos m\pi)
\end{align*}
$$
Solving this set of equations yields
$$
I = \frac{2}{m} \frac{\cos(m\pi) - 1}{m^2 - 4}
$$
Doing the same in the cases $m^2 = 4$ and $m = 0$, just give $I = 0$, thus
$$ \int_0^\pi (\sin \psi)^2 \sin(m\psi) \,\mathrm{d}\psi
= \begin{cases} \cfrac{2}{m} \cfrac{\cos(m\pi) - 1}{m^2 - 4} & \text{if} \qquad m\neq -2,0,2 \\ 0 & \text{if} \qquad m = -2, 0, 2 \end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral inequality with two increasing functions Let $f,g:[0,1]\rightarrow\mathbb{R}^+$ be increasing functions such that $f\leq g$. Is there a constant $c>0$ (independent of $f,g$) for which there exists some $r\geq 0$ (possibly dependent on $f,g$) such that
$$\int_{x:f(x)\leq r\leq g(x)}g(x)dx+\int_{f(x)\geq r}f(x)dx\geq c\int_0^1g(x)dx ?$$
As an example, let $g(x)=x$ and $f(x)=x^2$. The integral $\int_0^1g(x)dx$ is $\frac{1}{2}$. On the left-hand side, for fixed $r$, the first integral is from $x=r$ to $x=\sqrt{r}$ and amounts to $\frac{1}{2}(r-r^2)$. The second integral is from $x=\sqrt{r}$ to $x=1$ and amounts to $\frac{1}{3}(1-r\sqrt{r})$.
Suppose $g(x)=x^t$ and $f(x)=x^s$ for $s\geq t\geq 1$. Then the right-hand side is $\frac{1}{t+1}$. The left-hand side is
$$\frac{r^{\frac{t+1}{s}}}{t+1}-\frac{r^{\frac{t+1}{t}}}{t+1}+\frac{1}{s+1}-\frac{r^{\frac{s+1}{s}}}{s+1}.$$
| $$
\newcommand {\diff} {\mathrm d}
\newcommand{\spc}[1] {\quad #1 \quad}
$$
For convenience, rewrite the inequality as $\max\limits_r \frac A B \ge c$, where
$$A = \int\limits_{f(x) \leq r \leq g(x)} g(x)\; \diff x + \int\limits_{f(x) \geq r} f(x)\; \diff x, \quad B = \int\limits_0^1g(x)\; \diff x$$
For $f(x) = \frac 1 {1 + s(1 - x)}$ and $g(x) = \frac 1 {1 + t(1 - x)}$, $s > t > 1$:
\begin{align*}
\frac A B
& \spc = \left( \int\limits_{1 + \frac 1 t \left( 1 - \frac 1 r \right)}^{1 + \frac 1 s \left( 1 - \frac 1 r \right)} \frac {\diff x} {1 + t(1 - x)} + \int\limits_{1 + \frac 1 s \left( 1 - \frac 1 r \right)}^1 \frac {\diff x} {1 + s(1 - x)} \right) \Bigg/ \int\limits_0^1 \frac {\diff x} {1 + t(1 - x)} \\\\
& \spc = -\; \frac {\ln \Big( \frac t s + \left( 1 - \frac t s \right) r \Big) + \frac t s \ln r} {\ln (1 + t)} \qquad \bigg| \qquad r \in \left[ {{\frac 1 {1 + t}}, 1} \right] \\
\end{align*}
We can exclude values of $r$ below $g(0) = \frac 1 {1 + t}$, as these are not good cadidates for maximizing $\frac A B$, for mostly obvious reasons. Thus, $\max\limits_r \frac A B$ occurs at one of $r = \frac 1 {1 + t}$, $r = 1$, or some value of $r$ such that $\frac \diff {\diff r} \frac A B = 0$.
\begin{align*}
\frac \diff {\diff r} \frac A B
& \spc = -\; \frac 1 {\ln (t + 1)} \left( \frac {s - t} {t + (s - t) r} + \frac t {sr} \right) \spc = 0 \qquad \bigg| \qquad r = \frac {t^2} {t^2 - s^2} < 0 \\
\end{align*}
The solution for $\frac \diff {\diff r} \frac A B = 0$ is spurious, as it lies outside $\left[ {{\frac 1 {1 + t}}, 1} \right]$. Thus, we need only consider $r = \frac 1 {1 + t}$ and $r = 1$ for $\max\limits_r \frac A B$:
\begin{align*}
\frac A B \bigg|_{r = 1} & \spc = 0 \\
\frac A B \bigg|_{r = \frac 1 {1 + t}}
& \spc = -\; \frac {\ln \Big( \frac t s + \left( 1 - \frac t s \right) \frac 1 {1 + t} \Big) + \frac t s \ln \frac 1 {1 + t}} {\ln (1 + t)}
\spc = \frac t s + 1 -\; \frac {\ln \left(\frac {t^2} s + 1 \right)} {\ln (t + 1)} \\
\end{align*}
Obviously, $\max\limits_r \frac A B = \frac A B \big|_{r = \frac 1 {1 + t}} = \frac t s + 1 -\; \frac {\ln \left(\frac {t^2} s + 1 \right)} {\ln (t + 1)}$.
If we put $k = \frac t s$, $k \in (0, 1)$,
\begin{align*}
\max \frac A B & \spc = 1 + k -\frac {\ln (k t + 1)} {\ln (t + 1)}
\spc \to k - \frac {\ln k} {\ln t} \spc \to k \qquad \bigg| \qquad t \to \infty
\end{align*}
We can choose $k$ arbitrarily close to 0, and then, by the above, choose $t$ such that $\max\limits_r \frac A B$ will be arbitrarily close to $k$, for the corresponding pair of $f(x) = \frac 1 {1 + \frac t k (1 - x)}$ and $g(x) = \frac 1 {1 + t(1 - x)}$. Hence, we can choose $f$ and $g$ such that $\max\limits_r \frac A B$ will be arbitrarily close to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the point on the graph of $f (x) = \sqrt x$ closest to the point $(3,0)$ $(x,y)=(x,\sqrt{x})$
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$
$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$
The question is: Should I put those x-values in g(x) or the orginal graph,$
y=\sqrt{x}$. To me, it's logical to put it into g(x), but in an example it was reverse
| You want to minimize the function
$$
f(x)=\|(x,\sqrt{x})-(3,0)\|^2=\|(x-3,\sqrt{x})\|^2=(x-3)^2+(\sqrt{x})^2=x^2-5x+9
$$
The function $f$ reaches its minimum value at
$$
h=-\dfrac{-5}{2(1)}=\dfrac{5}{2}
$$
and the minimum value is
$$
k=f(h)=\dfrac{11}{4}
$$
Hence the point of $y=\sqrt{x}$ closest to the point $(3,0)$ is the point
$\left(\dfrac{5}{2},\sqrt{\dfrac{5}{2}}\right)$, and its distance to the point $(3,0)$ is $\sqrt{k}=\dfrac{\sqrt{11}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all the points on the cone $z^2=x^2+4y^2$ that are the closest to the point $(0,0,c)$
From all the points on the cone $z^2=x^2+4y^2$ find the closest to the point $(0,0,c)$. State explicitly the minimal distance. $c$ is a constant.
Lagrange multipliers can be used here.
Let the constraint function $g=x^2+4y^2-z^2$.
Let the minimization function $f=(x-0)^2+(y-0)^2+(z-c)^2$ as function of finding the distance in between 2 points ion 3d space. We're also going to keep $f$ in squares so we don't have to get the factor out $z$ from the ellipse equation.
Then we have:
$$
\nabla g=\langle2x, 8y,-2z \rangle\\
\nabla f=\langle 2x, 2y,2(z-c)\rangle
$$
So:
$$
\begin{cases}
2x=k\cdot2x\\
2y=k\cdot 8y\\
2(z-c)=k\cdot(-2z)
\end{cases}
$$
We have that $k=\frac{1}{4}$ then $x=0$ and $z=\frac{4c}{5}$.
If we plug those values into the ellipse equation we get that:
$$
c^2=\frac{125y^2}{4}\Rightarrow c=\pm\frac{\sqrt{125}y}{2}
$$
So the points would be $(0,y,\frac{\sqrt{125}y}{2})$ and $(0,y,-\frac{\sqrt{125}y}{2})$.
First I think something is wrong with my calculations (although I double checked it). Also I'm not sure what does this mean that we have 2 points and how am I supposed to know which one is the minimum.
| Minimizing the squared distance,
$$\begin{array}{ll} \text{minimize} & x^2 + y^2 + (z-c)^2\\ \text{subject to} & z^2 = x^2 + 4 y^2\end{array}$$
Let the Lagrangian be
$$\mathcal L (x,y,z,\lambda) := x^2 + y^2 + (z-c)^2 + \lambda (x^2 + 4 y^2 - z^2)$$
Taking the partial derivatives and finding where they vanish, we obtain
$$\begin{array}{rl} (1 + \lambda) \, x &= 0\\ (1 + 4\lambda) \, y &= 0\\ (1 - \lambda) \, z &= c\\ x^2 + 4 y^2 - z^2 &= 0\\\end{array}$$
We have two cases to consider.
$\color{blue}{\boxed{\lambda = -1}}$
In this case, $y = 0$ and $z = \frac c2$. The value of $x$ is given by
$$x = \pm \sqrt{ z^2 - 4 y^2 } = \pm \frac c2$$
Hence, we have the two points
$$(x,y,z) = \left( \pm \frac c2, 0, \frac c2 \right)$$
whose squared distance from $(0,0,c)$ is $\frac{c^2}{2}$.
$\color{blue}{\boxed{\lambda = - \frac 14}}$
In this case, $x = 0$ and $z = \frac{4c}{5}$. The value of $y$ is given by
$$y = \pm \frac 12 \sqrt{ z^2 - x^2 } = \pm \frac{2c}{5}$$
Hence, we have the two points
$$(x,y,z) = \left( 0, \pm \frac{2c}{5}, \frac{4c}{5} \right)$$
whose squared distance from $(0,0,c)$ is $\frac{c^2}{5} < \frac{c^2}{2}$. This is the minimal squared distance.
Example
Let $c = 5$. Hence, the points on the cone closest to $(0,0,5)$ are $(x,y,z) = ( 0, \pm 2, 4 )$. Here is a plot of the cone and the line segment whose endpoints are $(0,0,5)$ and $( 0, 2, 4)$
The length of the line segment is $\sqrt 5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taylor expansion of composite functions Suppose one has a function of the form: $$f(x)=\sqrt{1+a\cos^{2}(x)}$$ where $a$ is some parameter.
Firstly, what is the correct method for Taylor expanding such a composite function? Should one just repeatedly use the chain rule $(f\circ g)'(x)=f'(g(x))g'(x)$ to determine the coefficients of the Taylor expansion, such that it has the form: $$(f\circ g)(x)=(f\circ g)(x_{0})+(f\circ g)'(x_{0})(x-x_{0})+\frac{1}{2}(f\circ g)''(x_{0})(x-x_{0})^{2}+\cdots$$ or is there a different method?
Secondly, although not technically a Taylor expansion, if the parameter $a<<1$ is it reasonable to approximate the function as $$\sqrt{1+a\cos^{2}(x)}\approx 1+\frac{a}{2}\cos^{2}(x)$$ i.e. naively plugging in $a\cos^{2}(x)$ to the Taylor expansion for $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\mathcal{O}(x^{3})$?
| $$f(x)=\sqrt{1+a - a\sin^{2}(x)} = \sqrt{1+a}\sqrt{1 - {a\over a+1}\sin^2x}$$
$$ = \sqrt{1+a}\left(1 - {1\over2}{a\over a+1}\sin^2x + {1\over2!}{1\over2}\left({1\over2}-1\right)\left(-{a\over a+1}\right)^2\sin^4x\right) +O(x^6)$$
$$ = \sqrt{1+a}\left(1 - {1\over4}{a\over a+1}(1-\cos2x) - {1\over32}{a^2\over (a+1)^2}(1-2\cos 2x+\cos^22x)\right)+O(x^6)$$
$$ = \sqrt{1+a}\left(1 - {1\over4}{a\over a+1}(1-\cos2x) - {1\over64}{a^2\over (a+1)^2}(2-4\cos 2x+ 1+\cos4x)\right)+O(x^6)$$
$$ = \sqrt{1+a} - \sqrt{1+a}\left({1\over4}{a\over a+1} + {1\over16}{a^2\over (a+1)^2}\right)(1-\cos2x) +\sqrt{1+a} + {1\over64}{a^2\over (a+1)^2}(1-\cos4x)+O(x^6)$$
$$ = \sqrt{1+a} - \sqrt{1+a}\left({1\over4}{a\over a+1} + {1\over16}{a^2\over (a+1)^2}\right)\left({4x^2\over2!}-{16x^4\over4!}\right)$$
$$ + {\sqrt{1+a}\over64}{a^2\over (a+1)^2}\left({16x^2\over2!}-{256x^4\over4!}\right)+O(x^6)$$
$$ = \sqrt{1+a} - {\sqrt{1+a}\over4}{a\over a+1}x^2 + {\sqrt{1+a}\over4!}\left({4a\over a+1} - {3a^2\over (a+1)^2}\right)x^4+O(x^6).$$
| {
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"url": "https://math.stackexchange.com/questions/2289116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the locus of $|z-2i|=3|z+3|$ I got as far as
$$|z-2i|=3|z+3| \Leftrightarrow \\
(\ldots) \Leftrightarrow \\
x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\
x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\
-8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\
8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\
y^2+\frac{1}{2}y-x^2-\frac{27}{4}x=\frac{85}{8}$$
Then I tried to complete the square:
$$y^2+\frac{1}{2}y = 0\Leftrightarrow (y+\frac{1}{4})^2-\frac{1}{16}$$
$$-x^2-\frac{27}{4}x=0 \Leftrightarrow -(x+\frac{27}{8})^2+\frac{729}{64} = 0 \Leftrightarrow (x+\frac{27}{8})^2-\frac{729}{64} = 0$$
And so the equation becomes:
$$(y+\frac{1}{4})^2-\frac{1}{16}+(x+\frac{27}{8})^2-\frac{729}{64}=\frac{85}{8} \Leftrightarrow \\
(y+\frac{1}{4})^2+(x+\frac{27}{8})^2=\frac{85}{8}+\frac{1}{16}+\frac{729}{64} $$
According to my book the center is $-\frac{27}{8}-\frac{1}{4}i$ which I got right but the radius is $\frac{\sqrt{117}}{8}$.
What went wrong?
| i have $$|x+iy-2i|=3|x+3+iy|$$ from here we get
$$\sqrt{x^2+(y-2)^2}=3\sqrt{(x+3)^2+y^2}$$
can you finish now?
| {
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Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL? I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?
| Let $a$ be the sidelength of the triangle $AKL$. Then $K$ and $L$ are the intersection points between the circle $\omega$ centered in $A$ with radius $a$ and $BC$ and $CD$ respectively. Since both the circle $\omega$ and the square $ABCD$ are symmetric with respect to the line $AC$ then $|BK| = |DL|$.
Let $b = |BK|$. Then by applying the Pythagorean Theorem on $ABK$ we get
$$a^2 = 1 + b^2$$
and by applying it on $KCL$ we get
$$a^2 = 2(1-b)^2 = 2 - 4b + 2b^2$$
Subtracting the two equations and solving for $b$ we get
$$b = 2 \pm \sqrt{3}.$$
Note that $0 < b < 1$ so we have to discard the option $2+\sqrt{3}$. Then
$$b = 2 - \sqrt{3}.$$
Now
$$|AKL| = |ABCD| - |ABK| - |KCL| - |ALD| = 1 - \dfrac{b}{2} - \dfrac{(1-b)^2}{2} - \dfrac{b}{2} = \dfrac{1-b^2}{2}$$
$$ = \boxed{2\sqrt{3} - 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve a system of linear equations subject to Boolean constraints Say I have an equation on the form
$$c_1^{\top} = c_2^{\top} X$$
where $c_1, c_2 \in \mathbb N^n$ and $X \in \{0,1\}^{n \times n}$ is a square matrix. How do I solve for $X$?
| (let's call $c_1 = a$ and $c_2 = b$ for the rest of the question)
Let's say $x \in \mathbb{R}^{n\times x}$ (with $x_{ij}\in \{0,1\}$) and $a,b\in\mathbb{R}^n$. This makes:
$$ \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}^T = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}^T \begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\
x_{21} & x_{22} & \cdots & x_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
x_{n1} & x_{n2} & \cdots & x_{nn} \end{bmatrix} $$
Thus
$$ \begin{array}{rcl}
a_1 & = & b_1 x_{11} + b_2 x_{21} + \cdots + b_n x_{n1} \\
a_2 & = & b_1 x_{12} + b_2 x_{22} + \cdots + b_n x_{n2} \\
&\vdots& \\
a_n & = & b_1 x_{1n} + b_2 x_{2n} + \cdots + b_n x_{nn}
\end{array}$$
It is easy to show that no unique sollution exists. Let's give a counter example:
take $a = \begin{bmatrix}4 & 1 & 3\end{bmatrix}$ and $b = \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}$. This gives us:
$$ \begin{array}{rcl}
4 & = & 1 x_{11} + 2 x_{21} + 1 x_{31} \\
1 & = & 1 x_{12} + 2 x_{22} + 1 x_{32} \\
3 & = & 1 x_{13} + 2 x_{23} + 1 x_{33}
\end{array}$$
Do you see that different solutions are:
$$ \begin{bmatrix} 4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\
1 & 0 & 1 \\
1 & 0 & 1 \end{bmatrix} $$
As well as:
$$ \begin{bmatrix} 4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 0 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{bmatrix} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that surfaces $x + 2y – lnz + 4 = 0$ and $x^2 - xy – 8x + z + 5 = 0$ are tangent at $(2,-3,1)$. $x + 2y – ln(z) + 4 = 0$
$x^2 - xy – 8x + z + 5 = 0$
$\nabla [1,2,\frac{-1}{z}] $
$\nabla [2x-y-8,-x,1] $
$\nabla(P0) [1,2,-1] $
$\nabla(P0) [-1,-2,1] $
I've stuck at this point and i don't know what to do next.
| Two surfaces $S_1$ and $S_2$ are tangent at a point $P$ if and only if
$P \in S_1 \cap S_2, \tag{1}$
i.e., $P$ lies in each of $S_1$, $S_2$; and
$T_PS_1 = T_PS_2, \tag{2}$
that is, the tanget planes to $S_1$ and $S_2$ at $P$ are the same. Taking $S_1$ to be the surface
$f_1(x, y, z) = x + 2y - \ln z + 4 = 0, \tag{3}$
and $S_2$ to be
$f_2(x, y, z) = x^2 - xy - 8x + z + 5 = 0, \tag{4}$
we first verify
$P = (2, -3, 1) \in S_1 \cap S_2 \tag{5}$
by showing
$f_1(P) = f_2(P) = 0; \tag{6}$
we have
$f_1(2, -3, 1)= 2 + 2(-3) -\ln 1 + 4 = 2 - 6 + 4 = 0, \tag{7}$
and
$f_2(2, -3, 1) = 2^2 - 2(-3) - 8(2) + 1 + 5 = 4 + 6 - 16 + 1 + 5= 0; \tag{8}$
thus
$P \in S_1 \cap S_2. \tag{9}$
To see that the tangent planes to $S_1$ and $S_2$ at $P$ coincide, we may show the normal vectors to each surface at $P$ are collinear. We do this by calculating $\nabla f_1$ and $\nabla f_2$ at $P$, viz.
$\nabla f_1 = (1, 2, -\dfrac{1}{z}), \tag{10}$
$\nabla f_2 = (2x - y - 8, -x, 1); \tag{11}$
thus
$\nabla f_1(P) = (1, 2, -1) \tag{12}$
and
$\nabla f_2(P) = (2(2) - (-3) - 8, -2, 1) = (-1, -2, 1); \tag{13}$
we see that
$\nabla f_2(P) = -\nabla f_1(P). \tag{14}$
Since the normals to $S_1$ and $S_2$ are collinear at the point of intersection $P$, their tangent planes are one and the same there; the surfaces are indeed tangent at $P$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the Limit with Series Expansion Problem
Evaluate the following using series expansion.
$$ \lim_{x \rightarrow 0} \frac{\cosh{x}-\cosh{2x}}{x \cdot \sinh{x} }$$
I really don't know what I should do. If I expand $\cosh$ and $\sinh$ normally, (they'd look something like $\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n)!}$) I get a fraction with two infinite sums with the limit $0$ on both top and bottom and if I expand it using Taylor, I have to use $x \neq 0$ and I get a sum with infinite summands.
I don't have anymore ideas and any help would be appreciated. Thank you in advance.
| As other answers said, you are correct and the limit is easy to find using this way.
However, the same procedure can be used to see how is approached the limit. Just use one extra term in the expansions $$\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cosh(2x)=1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right)$$ $$\sinh(x)=x+\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{\left(1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right) \right)-\left(1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right) \right) } { x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right)}$$
$$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{-\frac{3 x^2}{2}-\frac{5 x^4}{8}+O\left(x^6\right) }{ x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right) }$$ Simplify by $x^2$ and use long division to finally get $$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=-\frac{3}{2}-\frac{3 x^2}{8}+O\left(x^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Alternatively note that $$\begin{align}x^5 + x + 1 &= x^5 - x^2 + x^2 + x + 1\\ & =x^2(x^3-1) + \color{red}{x^2+x+1} \\ & = x^2(x-1)\color{red}{(x^2+x+1)} + \color{red}{x^2+x+1} \\& =\color{blue}{(x^3-x^2+1)}\color{red}{(x^2+x+1)} \end{align}$$
where we used the well known identity $x^3 - 1 = (x-1)(x^2+x+1)$ in the third equality.
Meta: whilst the first step may seem arbitrary and magical, it is natural to want to insert a term like $x^2$ or $x^3$ into the equation in order to get some traction with factorising $x^5 + x^k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Why is $x^2+x$ the same as writing $(x+0.5)^2-0.25$? I find it extremely weird that a translation vertically by a variable should cause the graph to move both vertically and horizontally. Also, why 0.5 and 0.25?
| Consider a more general case: ($a > 0$)
\begin{align}
a \, x^2 + b \, x &= (\sqrt{a})^2 x^2 + 2 \cdot \frac{b}{2 \, \sqrt{a}} \cdot \sqrt{a} x + \frac{b^2}{4 \, a} - \frac{b^2}{4 \, a} \\
&= \left(\sqrt{a} x + \frac{b}{2 \, \sqrt{a}}\right)^2 - \frac{b^2}{4 \, a}.
\end{align}
In this example the values of $(a,b)$ are $(1, 1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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The sum of the coefficients of $P(x)$ of 4th degree given that $P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod{x^2-x-2}\equiv 141 \pmod{x+2}$ Let there be a polynomial $P(x)$ of 4th degree. The remainder of $P(x)$ divided by both $2x^2+4$ and $x^2-x-2$ is equal to $x-1$. So:
$$P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod{x^2-x-2}$$
Given that $P(x)\equiv 141 \pmod{x+2}$, I'm asked to calculate the sum of the coefficients of $P(x)$.
I don't have any idea how I should go about solving the question, any help would be appreciated. Thanks.
| Hint 1:$$P\left( x \right) =a\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1\\ P(x)\equiv 141{ \quad \left( mod\quad x+2 \right) \Rightarrow }P\left( -2 \right) =141\\ P\left( -2 \right) =48a-3=141\\ 48a=144\\ a=3\\ P\left( x \right) =3\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1\\ $$
Hint 2
$P(1)=$sum of the coefficients of $P(x)$.
| {
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Determine exact values using angle sum and difference identity (trig) Ok so we have just learnt the basic angle sum and difference identities:
$$\begin{array}{l}\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B\\\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B\\\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\end{array}
$$
The question i have been given is determine the exact value of $\tan {15^0}$ degrees.
So i assume you use the tan identity, sub in
$$\tan \left( {{{45}^0} - {{30}^0}} \right) = \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan 30}}
$$
After simplifying it down as far as possible i end up with:
$$\frac{{\left( {3\left( {3 - \sqrt 3 } \right)} \right)}}{{\left( {3\left( {3 + \sqrt 3 } \right)} \right)}}
$$
However the correct answer is: ${2 - \sqrt 3 }$
I'm not sure if iv'e done something wrong or are completely off track, but could someone please explain (as simply as possible) how to achieve this answer.
| Here you go,
$$\begin{align}\tan \left( {{{45}^0} - {{30}^0}} \right) &= \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan {{30}^0}}}\\\\
&= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \left( 1 \right)\left( {\frac{1}{{\sqrt 3 }}} \right)}}\\\\
&= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\ \\
&= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}\\\\
&= \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}\\\\
&= \frac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\\\\
&= \frac{{4 - 2\sqrt 3 }}{2}\\\\
&= 2 - \sqrt 3 \end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the limit to infinity of this function with summation? I need someone with knowledge of limits to infinity and limits of summations to please work out the following:
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} $$
For context, I want to determine if the following sequence approaches $\frac{2}{3}$:
$$ \frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{11}{16}, \frac{21}{32}, \frac{43}{64}, \frac{85}{128}, \frac{171}{256}, \frac{341}{512}, . . . $$
Thank you =)
| Your sequence does not converge. Firstly $\sum_{j=0}^{i-1} (-1)^j 2^j$ is sum of geometric series with quotient -2 and is equal to $\frac{1}{3} (1-(-2)^i)$.
We have
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} =
\lim_{i\to \infty} \frac{2^{i}}{2^{i+1}} + \lim_{i\to \infty} \frac{ 1-(-2)^{i}}{3\times 2^{i+1}} = \frac{1}{2} + \lim_{i\to \infty} \frac{ 1}{3\times 2^{i+1}}+ \lim_{i\to \infty} \frac{ -(-2)^{i}}{3\times 2^{i+1}} = \frac{1}{2} -\frac{1}{6} \lim_{i\to \infty} (-1)^i$$
Since the last does not converge, the limit does not converge.
Actually the sequence with $i-$th term $a_i = \frac{1}{2} -\frac{1}{6}(-1)^i$ has two subsequences, that converge to $1/3$ and $2/3$, respectively:
$$
\lim_{i\to \infty}a_{2i} = \frac{2}{3}, \qquad \lim_{i\to \infty}a_{2i-1} = \frac{1}{3}\cdot
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Express $\frac {a^2+(a+b)^2}{a(a+b)}$ in terms of $x$ and $y$. Given that ${ab}=x$ and ${a+b}=y$. Express $\dfrac{a^2+(a+b)^2}{a(a+b)}$ in terms of $x$ and $y$.
I try many way still cannot succeed. Please use some elementary way to help me to solve this. As I am not familiar with MathJax, so if exists any mistake please forgive me. Thanks.
| As suggested in the comments, $a$ and $b$ are the roots of $t^2-yt+x=0$. We get this from: $$a=\frac xb\implies\frac xb+b=y\implies b^2-yb+x=0$$ and similarly for $a$ leads to the same equation.
Solving this gives $$t=\frac{y\pm\sqrt{y^2-4x}}{2}$$
Wlog, let $a$ be the root with the $+$ sign and $b$ be the root with the $-$ sign. Then $$\frac ya=\frac{2y}{y+\sqrt{y^2-4x}}=\frac{2}{1+\sqrt{1-4\frac {x}{y^2}}}$$
$$\frac {a^2+(a+b)^2}{a(a+b)}=\frac{a^2+y^2}{ay}=\frac ay+\frac ya=\frac{1+\sqrt{1-4x/y^2}}{2}+\frac{2}{1+\sqrt{1-4x/y^2}}$$
This can be simplified to $$\frac{4y^2+4x-(3y^2-4x)\sqrt{1-4x/y^2}}{8x}$$
(I think - that $3$ looks suspicious so this could be worth checking).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the power series representation of the function I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$
| Note that the geometric series is valid also in the complex field
$$
\frac{1}
{{1 - z}} = \sum\limits_{0\, \leqslant \,n} {z^{\,n} } \quad \left| \begin{gathered}
\;z \in \;\mathbb{C}\; \hfill \\
\,\left| z \right| < 1 \hfill \\
\end{gathered} \right.
$$
and that the partial fraction decomposition can be operated also in the complex field,
so that in your case we can write
$$
\begin{gathered}
f(z,\omega ) = \frac{1}
{{\left( {2z - 3} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} = \hfill \\
= \frac{1}
{{2\left( {z - 3/2} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} = \hfill \\
= - \frac{1}
{{3\left( {3/2 - \omega } \right)\left( {3/2 + \omega } \right)\left( {1 - 2/3\,z} \right)}} + \frac{1}
{{4\omega ^2 \left( {3/2 - \omega } \right)\left( {1 - z/\omega } \right)}} + \frac{1}
{{4\omega ^2 \left( {3/2 + \omega } \right)\left( {1 - \left( { - z/\omega } \right)} \right)}} \hfill \\
\end{gathered}
$$
and the expression for $a_n$ follows easily.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Every odd number is of the form $2n+1=(n+1)^2-n^2=(n+1)^2-n^2+0^2$ and every even integer is of the form $2n=(n+1)^2-n^2-1^2$.
Now the second part:
Let, $6=x^2+y^2$ i.e $z=0$.. Now, $x,y\ne1$ as if one of them is one the other one will be $\sqrt{5}$. $x,y\ne2$ as if one of them is $2$ other one will be $\sqrt{2}$. We don't need to check it for $3$ as the squares of integers is always positive.
Now, assume that $6=x^2-z^2$ i.e $y=0$ then $6=(x+z)(x-z)$ . Since $6=-1*-6=6*1=2*3$ and sum of any these two pairwise factors is always odd. While sum of $x+z,x-z$ is even($2x$). A contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Note that
\begin{align*}
(\sqrt{-3-2i}+\sqrt{-3+2i})^2 &= (-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3+2i) \\
&= -6 + 2\sqrt{13}\\
&= 2(\sqrt{13} - 3).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\frac{x}{1-x^2}=x+x^3+x^5+\cdots$$
$$\frac{x^2}{1-x^4}=x^2+x^6+x^{10}+\cdots$$
$$\cdots$$
Add them all and get the answer. Unfortunately, I havn't got a direct method to calculate it. Appreciate for your help!
| $$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}.(1-x^{2^{n-1}})}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}\frac{-(1-x^{2^{n-1}})+1-x^{2^{n}}}{(1-x^{2^n}).(1-x^{2^{n-1}})}
$$
$$
\sum_{n=1}^{\infty}(\frac{1}{1-x^{2^{n-1}}}-\frac{1}{1-x^{2^n}})
$$
Now consider upto $k$ terms then this can be written as
(second term of $nth$ expression cancelled by first term of $(n+1)th$ expression) ,
$$
\frac{1}{1-x}-\frac{1}{1-x^{2^k}}
$$
For $k\to\infty$ and $x\in(0,1)$,
$$
\frac{1}{1-x}-1
$$
Which is equal to
$$
\frac{x}{1-x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
} |
Find matrices $P$ and $Q$ to check if matrices $A$ and $B$ are equivalent( $ Q^{-1}AP = B$ )? $ \begin{align}&A=\begin{pmatrix}
6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\
12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\
\end{pmatrix} \\\\
& B = \begin{pmatrix}
4 & -4 & 0 & -8 & -12 & 8 & -4 & -8 & 4 & -12 \\
-9 & -3 & -6 & 3 & -6 & 9 & -3 & -3 & -6 & 3\\
\end{pmatrix}
\end{align}$
I need to find out if $A$ and $B $ are equivalent or not.
To do that I have rearranged the definition of matrix equivalence:
\begin{align} Q^{-1}AP & = B \\\\ AP & = QB \end{align}
So now I have
$
\begin{pmatrix}
6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\
12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\
\end{pmatrix} \cdot
\begin{pmatrix}
x_{1,1} & \cdots & x_{1,2} \\
\vdots & & \vdots \\
x_{10,1}& \cdots & x_{10,2}
\end{pmatrix} =
\begin{pmatrix}
y_{1,1} & \cdots & y_{1,2} \\
\vdots & & \vdots \\
y_{10,1}& \cdots & y_{10,2}
\end{pmatrix} \cdot
\begin{pmatrix}
4 & -4 & 0 & -8 & -12 & 8 & -4 & -8 & 4 & -12 \\
-9 & -3 & -6 & 3 & -6 & 9 & -3 & -3 & -6 & 3\\
\end{pmatrix}
$
Question: How can I solve that big equation system with two variables $x$ and $y$ and get the matrices $P$ and $Q$?
Is there an easier (and hopefully more safe) way to solve that task?
| If you only want to know if $A$ and $B$ are equivalent or not, you only need to show whether they have the same rank. A short explanation of this is that applying invertible square matrix on the left is just applying multiple elementary row operations, and applying invertible square matrix on the right is just applying multiple elementary column operations.
For this case, since there are only two rows in the matrices, the fastest way is to look at their column spaces.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Definite integral of $\frac{1}{\sqrt{\tan x}}$ i would like to ask you for help to evaluate $$\int_{0}^{\pi/2}\frac{1}{\sqrt{\tan x}}dx$$
I didn't find an appropriate substitution.
Thanks for help.
| Hint
As in the previous answer, let
$$z=\sqrt{\tan(x)}\implies x=\tan ^{-1}\left(z^2\right)\implies dx=\frac{2 z}{z^4+1}\,dz$$ This makes $$\int\frac{dx}{\sqrt{\tan x}}=2\int \frac{dz}{z^4+1}$$ Now $$z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt 2 z+1)(z^2-\sqrt 2 z+1)$$ Now, using partial fraction decomposition $$\frac 1{z^4+1}=\frac 1{(z^2+\sqrt 2 z+1)(z^2-\sqrt 2 z+1)}$$ $$\frac 1{z^4+1}=\frac 1{2\sqrt 2}\left(\frac{z+\sqrt{2}}{z^2+\sqrt{2} z+1}-\frac{z-\sqrt{2}}{z^2-\sqrt{2} z+1}\right)$$ which seems to lead to quite standard integrations leading to some $\tan^{-1}(.)$ and $\log(.)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3}
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies a_{n+1} = \frac{4} {2 - \frac{3} {a_n} }
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????
| If you have proved that the sequence has a limit $l$, then $l$ can be calculated by the following:
$$\lim_{n\to \infty}2a_{n+1}a_n = \lim_{n\to \infty}(4a_n + 3a_{n+1}) $$
and hence
$$2l^2=4l+3l$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiate $y = x^{\sqrt{x}}$ (Simplification)
I understand the gist of the problem. Just the simplification confused me.
How did $\sqrt{x}/x$ turn into $1/\sqrt{x}$?
Why does that circled 1 turn into a 2?
| (1) $\displaystyle \sqrt{x}\cdot\frac{1}{x}=\sqrt{x}\cdot\frac{1}{(\sqrt{x})^2}=\frac{1}{\sqrt{x}}$.
(2) $\displaystyle \frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2}{2}\cdot\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2}{2\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2+\ln x}{2\sqrt{x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic) I tried using the following approach:
$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$
Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,
$$x=(2222^5+5555^2)k,\ k\in N$$
$$x=(1111^2)(32\cdot1111^3+25)k$$
The term in parentheses was found to be $54165190296027657$, which is divisible by $7$. Thus $7 | x$.
Another method (edit):
We can also write $$2222^{5555} = (7\times 317+3)^{5555} = 7p+3^{5555}$$
Similarly $$5555^{2222}=(793\times7+4)^{2222} = 7q+4^{2222}$$
So we have $$x=7(p+q)+(3^5+4^2)r$$
Here we have $3^5+4^2$ is divisible by $7$. So $x$ is also divisible by $7$.
But is there a simple way without multiplication of large numbers? Thank you!
| We have
\begin{align}
2222^5 + 5555^2&=2222^5 +5555^5-5555^5+ 5555^2\\
&=2222^5 +5555^5-5555^2( 5555^3-1)\\
\end{align}
$2222^5 +5555^5$ is divisible by $2222+5555=7(1111)$ as $5$ is odd.
$$5555^3-1=(7\times793+4)^3-1=7p+4^3-1=7p+63$$
is divisible by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| This is a variation of Michael Rozenberg's answer,
$$\cos x+\sin x=\sqrt2\cos\left(\dfrac\pi4-x\right)$$
Set $\dfrac\pi4-x=y\iff x=?$
$2\sin x\cos x=\sin2x=\sin2\left(\dfrac\pi4-y\right)=\cos2y=2\cos^2y-1$
So, we have $$0=4(2\cos^2y-1)-2\sqrt3\cos y+1=8\cos^2y-2\sqrt3\cos y-3$$ which is a quadratic equation in $\cos y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\frac{c}{a}$
$α+δ+β+δ=\frac{-2B}{A}$
⇒ $α+β+2δ =\frac{-2B}{A}$
$(α+δ)(β+δ)=\frac{C}{A}$
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$
$(α+β)^2=\frac{4b^2}{a^2}$
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)
$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)
From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
| WLOG, $A=a=1$ (otherwise you can divide the trinomials by their leading coefficient).
Then
$$(x+\delta)^2+2B(x+\delta)+C=x^2+2(\delta+B)x+\delta^2+2B\delta+C=x^2+2bx+c,$$ and
$$b^2-c=(\delta+B)^2-(\delta^2+2B\delta+C)=B^2-C.$$
| {
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"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Is it true: $\sum_{n=1}^\infty \frac{a_n}{b_n}$&$\sum_{n=1}^\infty \frac{b_n}{c_n}$ are irrational=>$\sum_{n=1}^\infty \frac{a_n}{c_n}$ is irrational Assume that the:
$$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ and $$\sum_{n=1}^\infty \frac{b_n}{c_n}$$
are convergent and irrational, then if
$$\sum_{n=1}^\infty \frac{a_n}{c_n}$$ is convergent
should it also be irrational then for the integer $a_n$, $b_n$ and $c_n$?
I assume this is false, but can not find any counterexample.
| Another example with all terms positive is
$$ a_n = 2n+1, \qquad b_n = n(n+1)(2n-1), \qquad c_n = n(n+1)(n+2)(n+3)(2n+1). $$
Then we can prove that
$$ \sum_{n=1}^{\infty} \frac{a_n}{b_n} = \frac{1}{3} + \frac{8}{3}\log 2, \qquad \sum_{n=1}^{\infty} \frac{b_n}{c_n} = \frac{9}{15} - \frac{8}{15}\log 2, \qquad \sum_{n=1}^{\infty} \frac{a_n}{c_n} = \frac{1}{18}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximal and minimal value of a function $4x^2-4xy+y^2$ at the border $x^2+y^2=25$. So my first thought was to find the critical points of the $4x^2-4xy+y^2$ and I calculated.
$$F_{x}=8x-4y,\quad F_{y}=2y-4x$$
and I found out that critical point are satisfying $x=2y$.
Then I put then to equation $x^2+y^2=25$.
And I calculated that $x=\sqrt5,y=2\sqrt5$, and $x=-\sqrt5$, $y=-2\sqrt5$.
But I don't know if they are maximum or minimum.
My last idea was to to put $x^2=25-y^2 $ to the function $4x^2-4xy+y^2$ and then calculate the derivative. But I do not know if it is a good solution.
| $x =u + 2v\\
y = 2u - v$
$4x^2 -4xy + y^2 = 4(u^2 + 4uv + 4v^2) - 4(2u^2 -3uv - 2v^2) + (4u^2 - 4uv + v^2) = 25 v^2$
$x^2 + y^2 = (u^2 + 4uv + 4v^2) + (4u^2 - 4uv + v^2) = 5u^2 + 5v^2 = 25$
Minimize $25v^2$ constrained by: $5u^2 + 5v^2 = 25$
$v = 0, u = \pm \sqrt 5\\
(x,y) = (\sqrt 5,2\sqrt 5), (-\sqrt 5, -2\sqrt5)$
Maximize $25v^2$ constrained by: $5u^2 + 5v^2 = 25$
$v = \pm \sqrt 5, u= 0\\
(x,y) = (2\sqrt 5, -\sqrt 5), (-2\sqrt 5, \sqrt 5)\\
4x^2 -4xy +y^2 = 125$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
To compute $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz$ where $\mathcal{C}$ is the unit circle in $\mathbb{C}$
Let $\mathcal{C}$ denote the unit circle in $\mathbb{C}$ centred at the origin taken anticlockwise. Compute the value of the integrtal $$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz$$
I don't recon we can use the residue theorem to compute this integral because the integrand is not analytic anywhere.
I then tried substituting $z=e^{i\theta}$ and integrating it the traditional way but cant make much of the integrated $|1+e^{i\theta}+e^{2i\theta}|^2$.
Any help would be much appreciated. Thanks.
| We want the integral
$$\frac{1}{2\pi i}\int_C |1+z+z^2|^2dz=\frac{1}{2\pi}\int_0^{2\pi}|1+z+z^2|^2z~d\theta$$
since $z=e^{i\theta}$.
Now consider the absolute value portion,
$$
\begin{align}
|1+z+z^2|^2
&={(1+z+z^2)(1+z+z^2)^*}\\
&={(1+z+z^2)(1+z^{-1}+z^{-2})}\\
&={(1+z+z^2)(1+z^{-1}+z^{-2})\frac{z^2}{z^2}}\\
&={\left(\frac{1+z+z^2}{z} \right)^2}\\
&={\left(\frac{1}{z}+1+z \right)^2}\\
&={(1+2\cos\theta)^2}\\
\end{align}
$$
We can return to solve the integral
$$
\begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2|^2dz
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta)^2(\cos\theta+i\sin\theta)~d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta)^2\cos\theta~d\theta\\
&=2
\end{align}$$
This is in agreement with our own numerical solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain. Find the set of values of $a$ for which the function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain.
For $f(x)$ to be increasing,$f'(x)=3(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^2+5\ge 0$
So,$(1-\frac{\sqrt{21-4a-a^2}}{a+1})\ge 0$ gives $(-\infty,-5)\bigcup(2,\infty)$ but the answer in my book is $[-7,-1)\bigcup[2,3]$
| The domain gives $a^2+4a-21\leq0$ and $a\neq-1$, which gives $-7\leq a\leq3$ and $a\neq-1$.
Now, $1-\frac{\sqrt{21-4a-a^2}}{a+1}=0$ for $a=2$ and use the intervals method.
Maybe the mistake was in solving of the equation:
$$a+1=\sqrt{21-4a-a^2}.$$
We have $a>-1$ and $21-4a-a^2\geq0$, which gives $-1<a\leq3$.
Now, after squaring of the both sides we obtain $(a-2)(a+5)=0$, which gives $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots, Then $p+2q+r$ is
If $f(x)=x^3+px^2+qx+r$ has three distinct non negative integer roots and
$(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots,
Then $p+2q+r$ is
$\bf{Attempt:}$ Assuming $f(x) = x^3+px^2+qx+r = (x-a)(x-b)(x-c)$,
where $a,b,c \geq 0$ and $a,b,c \in \mathbb{Z}$
Then $f(x^2-4x+7) = (x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r $
$=(x^2-4x+7-a)(x^2-4x+7-b)(x^2-4x+7-c)$
Could some help me how to solve it , thanks
| $$
f(x^2-4x+7)=((x-2)^2+(3-a))((x-2)^2+(3-b))((x-2)^2+(3-c))
$$
has no real roots so $3-a$, $3-b$, and $3-c$ are at least $1$. Hence
$$
\{a,b,c\}=\{0,1,2\}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$ We know that $$\DeclareMathOperator{\arccot}{arccot}\arctan(-x)=-\arctan(x)$$ $$\arccot(-x)=\pi-\arccot(x)$$ $$\cot(-x)=-\cot(x)$$ $$\tan(-x)=-\tan(x)$$So I have figured out that $$\arctan(\tan(-\frac{3\cdot \pi}{5}))=\arctan(-\tan(\frac{3\cdot \pi}{5}))=-\arctan(\tan(\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5},$$ $$\arccot(\cot(-\frac{3\cdot \pi}{5}))=\arccot(-\cot(-\frac{3\cdot \pi}{5}))=\pi-\arccot(\cot(\frac{3\cdot \pi}{5}))=\pi-\frac{3\cdot \pi}{5}.$$Coming from these, I have solved the problem as follows:$$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\arccot(\cot(-\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5}+\pi-\frac{3\cdot \pi}{5}=-\frac{6\cdot \pi}{5}
+\pi=-\frac{\pi}{5}.$$But the book says it is incorrect. Could somebody point out where I have made a mistake?
| The function $\arctan(\tan x)$ returns a “normalized” $x$, that is the unique angle $x'\in(-\pi/2,\pi/2)$ so that $\tan x=\tan x'$.
Since $-3\pi/5<-\pi/2$, but $2\pi/5=-3\pi/5+\pi\in(-\pi/2,\pi/2)$, you have
$$
\arctan\Bigl(\tan\Bigl(-\frac{3\pi}{5}\Bigr)\Bigr)=\frac{2\pi}{5}
$$
Do similarly for the arccotangent, but with the “normalization” in $(0,\pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergent, non-geometric, infinite sum as a fraction? How would you go about finding the sum of an infinite series that is known to be convergent, but is not geometric?
The particular example I am interested in is as follows: $\sum_{j=0}^{\infty}\left [(6j+1)(\frac{1}{2})^{6j+2}\right ]$.
The only catch to this is that I need to find the answer in fractional form. I've looked online and can't find anything useful relating to non-geometric series' or even to finding the fractional answer.
| We have
\begin{eqnarray*}
\sum_{j=0}^{\infty} x^j =\frac{1}{1-x}.
\end{eqnarray*}
Differentiate this
\begin{eqnarray*}
\sum_{j=0}^{\infty} j x^{j-1} =\frac{1}{(1-x)^2}.
\end{eqnarray*}
Your sum can be rewitten as
\begin{eqnarray*}
\frac{6}{256}\sum_{j=0}^{\infty} j \left( \frac{1}{64} \right)^{j-1} +\frac{1}{4}\sum_{j=0}^{\infty} \left( \frac{1}{64} \right)^j & = & \frac{3}{4} \frac{1}{(1-\frac{1}{64})^2} +\frac{1}{4} \frac{1}{(1-\frac{1}{64})} \\
&=& \frac{32}{1323} + \frac{16}{63} = \color{red}{\frac{368}{1323}}.
\end{eqnarray*}
EDIT : In more detail ...
\begin{eqnarray*}
S= \sum_{j=0}^{\infty} (6j+1) \left( \frac{1}{2} \right)^{6j+2} = 6 (\frac{1}{2})^2 \sum_{j=0}^{\infty}j \left( \left(\frac{1}{2} \right)^6 \right)^j +(\frac{1}{2})^2 \sum_{j=0}^{\infty} \left( \left(\frac{1}{2} \right)^6 \right)^j
\end{eqnarray*}
Now $(1/2)^6=1/64$ and the first sum needs the exponent to decrease by $1$ ... so
\begin{eqnarray*}
S= \frac{6}{4 \times 64}\sum_{j=0}^{\infty} j \left( \frac{1}{64} \right)^{j-1} +\frac{1}{4}\sum_{j=0}^{\infty} \left( \frac{1}{64} \right)^j & = & \frac{3}{2 \times 64} \frac{1}{(1-\frac{1}{64})^2} +\frac{1}{4} \frac{1}{(1-\frac{1}{64})} \\
& = & \frac{3}{2 \times 64} \left(\frac{64}{63}\right)^2 +\frac{1}{4} \frac{64}{63} \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$
Where $\beta(n)$ is Beta dirichlet function
$(1)$ becomes
$$ln\left({2\over \pi}\right)+{1\over n\Gamma(n)}\sum_{n=1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}{x^{n-1}\over e^x+e^{-x}}\mathrm dx\tag2$$
How can we show that the closed form for $(1)$ is correct?
| Let us consider the two sums separately:
$$
S_1 = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)}{n} \\
S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n}
$$
For $S_1$, let us first consider the related sum, for which $S_1$ is the limiting value $S(1)$:
$$
S(x) = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)x^n}{n} \\
S'(x) = \sum_{n=1}^\infty (-1)^{n-1}\beta(n)x^{n-1} = \sum_{n=1}^\infty (-1)^{n-1}x^{n-1}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n} \\
= \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \sum_{n=1}^\infty \left( \frac{-x}{2k+1} \right)^{n-1} = \sum_{k=0}^\infty \frac{1}{2k+1+x} = \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right)
$$
Where $\Phi$ is the Lerch Transcendent, and $|x|<1$. Now, upon integration, we can recover $S_1$:
$$
S_1 - S(0) =\int_0^1 S'(x) dx = \int_0^1 \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right)dx = \int_\frac{1}{2}^1 \Phi (-1, 1, x)dx \\
= \lim_{s\to0}\int_\frac{1}{2}^1 \Phi (-1, 1+s, x)dx = \lim_{s\to0} \left[ \frac{-1}{s}\Phi (-1, s, x) \right|_\frac{1}{2}^1 \\
= \lim_{s\to0} \frac{\Phi \left(-1, s, \frac{1}{2}\right) - \Phi(-1, s, 1)}{s} = \lim_{s\to0} \frac{2^s \beta(s) - \eta(s)}{s} \\
=^H \lim_{s\to0} \ \ln(2) 2^s \beta(s) + 2^s \beta'(s) - \eta'(s) = \frac{1}{2} \ln(2) + \ln \frac{\Gamma^2(\frac{1}{4})}{2 \pi \sqrt{2}} - \frac{1}{2} \ln \frac{\pi}{2} = \ln \frac{\sqrt{2 \pi}}{\Gamma^2(\frac{3}{4})}
$$
And as $S(0) = 0$, this is all $S_1$. For $S_2$:
$$
S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n} = \ln \prod_{n=1}^\infty \left( \frac{n+1}{n} \right)^{(-1)^{n-1}} \\
= \lim_{N \to \infty} \ln \prod_{n=1}^N \frac{n^2}{(n - \frac{1}{2})(n + \frac{1}{2})} = \lim_{N \to \infty} \ln\frac{\pi \Gamma^2(N+1)}{2\Gamma(N + \frac{1}{2})\Gamma(N + \frac{3}{2})} = \ln \frac{\pi}{2}
$$
The sum therefore equals:
$$
S_1-S_2=\ln \frac{2\sqrt2}{\sqrt{\pi} \ \Gamma^2(\frac{3}{4})}
$$
as predicted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
I have tried many different substitutions to no avail. Any help is appreciated.
| Another way, using trigonometric identities
$$\sin{2x} = 2 \cdot \sin{x} \cdot \cos{x}$$
$$1 - \cos{x} = 2 \cdot \sin^2\frac{x}{2}$$
is as follows:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
$$= \int\dfrac{dx}{\sin x-(1 - \cos x)}$$
$$= \int\dfrac{dx}{2 \cdot \sin\frac{x}{2} \cdot \cos\frac{x}{2}-(2\cdot sin^2\frac{x}{2})}$$
$$= \int\dfrac{dx}{2 \cdot \sin\frac{x}{2}\cdot \big(\cos\frac{x}{2}-sin\frac{x}{2}\big)}$$
$$= \frac{1}{2}\int\dfrac{cosec\frac{x}{2} \cdot dx}{\big(\cos\frac{x}{2}-sin\frac{x}{2}\big)}$$
$$= \frac{1}{2}\int\dfrac{cosec^2\frac{x}{2} \cdot dx}{\big(\cot\frac{x}{2}-1\big)}$$
$$= \int\dfrac{cosec^2\frac{x}{2} \cdot \frac{dx}{2}}{\big(\cot\frac{x}{2}-1\big)}$$
$$ = -ln\bigg(cot\bigg(\frac{x}{2}\bigg) - 1\bigg) + C$$
Hence
$$\int\dfrac{dx}{\sin x+\cos x-1}= \boxed{- ln\bigg(cot\bigg(\frac{x}{2}\bigg) - 1\bigg) + C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.