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Fermat's infinite descent for finding the squares that sum to a prime Fermat's theorem on sum of two squares states that an odd prime $p = x^2 + y^2 \iff p \equiv 1 \pmod 4$
Applying the descent procedure I can get to $a^2 + b^2 = pc$ where $c \in \mathbb{Z} \gt 1$
I want $c = 1$, so how do I proceed from here? How do I apply the procedure iteratively?
Example:
$$
p = 97
$$
$$97 \equiv 1 \pmod 4 \implies \left(\frac{-1}{97}\right) = 1 \implies x^2 \equiv -1 \pmod {97}$$ has a solution
$$x^2 + 1 \equiv 0 \pmod {97}$$
$$x^2 + 1 = 97m$$
We find an $x,m$ that solves the equation.
$$x = 75, m = 58$$
Now, we pick an $a,b$ such that $\frac{-m}{2} \leq a,b \leq \frac{m}{2}$
$$a \equiv x \pmod m = 17$$
$$b \equiv y \pmod m = 1$$
Observations:
*
*$ a^2 + b^2 \equiv x^2 + 1 \equiv 0 (\mod m)$
*$ (a^2 + b^2) = mc$
*$ (x^2 + 1) = mp$
Plugging in $a,b,m$ for 2, we get $c = 5$
By this identity, we know that
$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$
**$(a^2 + b^2)(x^2 + 1^2) = (ax + b)(a - bx) = m^2pc$
Dividing ** by $m^2$, $pc = (\frac{ax+b}{m})^2 + (\frac{a-bx}{m})^2$
Plugging in $a,b,m,p,c$ we get that $22^2 + (-1)^2 = 97*5$
So we have two squares that add up to 5 times our $p$. How do we turn the 5 into a 1? What is the next step in the descent?
| $$22^2+1=97\times 5$$
Use the idendity $$(2a+b)^2+(a-2b)^2=5a^2+5b^2$$
The numbers $a,b$ obvioulsy solve the equation $a^2+b^2=97$ , if
$$2a+b=22\ ,\ a-2b=1$$
Solving this system gives $a=9\ ,\ b=4$
| {
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"url": "https://math.stackexchange.com/questions/1525113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve this first order nonlinear PDE? Given
$u=x{{u}_{x}}+y{{u}_{y}}+\frac{1}{2}\left( u_{x}^{2}+u_{y}^{2} \right)$ , find a solution with $u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)$ .
Not confortrable with my solution as follows. Please help.
Standard Charpit's method leads to
$\frac{dp}{0}=\frac{dq}{0}=\frac{du}{xp+{{p}^{2}}+qy+{{q}^{2}}}=\frac{dx}{x+p}=\frac{dy}{y+q}.$
$\Rightarrow p=a,q=b\Rightarrow u=ax+by+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$
$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right).$
$\Rightarrow {{b}^{2}}=\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax.$
$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+y\sqrt{\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax}\text{ where }a\text{ is an arbitary constant}.$
Solution alternative (for Clairaut's form) :
$dz=pdx+qdy$
$z=ax+by+c\text{ }.$
$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+c\text{ }\Rightarrow \text{ }c=\frac{1}{2}\left( 1-{{x}^{2}} \right)-ax\text{ }.$
$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+by.$
| Hint It is in Clairaut's form. Can you see?
$u=ax+by+\frac{1}{2}(a^2+b^2)$
then use the initial condition to eliminate $b$
| {
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"url": "https://math.stackexchange.com/questions/1528048",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
This is a 1989 ARML problem. One, ugly way to solve this is:
Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to see that $18$ works.
What's a nicer way?
Hint
Difference of squares
| Multiply first and last term and middle terms and take 1 on RHS. $(x^2+3x)(x^2+3x+2)=379^2-1^2$ so substitute $x^2+3x=y$ you will get a simple quadratic ie $y(y+2)=380\times 378$ which are also the factors . Get the value of $y$ and and then resubstituting $y=x^2+3x$ you will get the value of $x$ Hope you can take it from here to find $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of the triangle determined by the line $x+y=3$ and the bisector of angle between the lines $x^2-y^2+2y=1$ What is the area of the triangle formed by the lines $x+y=3$ and angle bisectors of pair of straight lines $x^2-y^2+2y=1$ . I found the intersection point of these equations $(1,2)$ but not getting any idea of how to find the area . Any way to solve this problem. Thanks!
| First, observe that just like @Nicholas said, the equation $\,x^2-y^2+2y=1\,$ defines two lines:
\begin{align}
x^2-y^2+2y=1 \iff x^2 = (y-1)^2 \implies
\begin{cases} l_1: & y = x + 1 \\ l_2: & y = -x + 1 \end{cases}
\end{align}
The slope of the first one is $\,\dfrac{\pi}{4} = 45º,\,$ and for the second one the slope is $\,\dfrac{3\pi}{4} = 135º,\,$ i.e. they are perpendicular.
Equations of bisecting lines can be found by adding and subtracting equations of original lines:
\begin{align}
\begin{cases}b_1:&x=0\\b_2:&y=1\end{cases}
\end{align}
Points of intersection of these lines with each other and with the line $\,l_0: \ \; y = 3-x\,$ are
\begin{align}
A & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ x = 0\end{cases}
&\implies&& A &= \big(\,0,\,3\,\big) \\
B & := b_1 \cap b_2 &\iff& &&\begin{cases}x = 0\\y=1\end{cases}
&\implies&& B &= \big(\,0,\,1\,\big) \\
C & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ y=1\end{cases}
&\implies&& C &= \big(\,2,\,1\,\big) \\
\end{align}
Observe that $\;b_1\perp b_2\;$ so that $\;\triangle\, ABC\;$ is right triangle.
Therefore the area $\,S\,$ of $\;\triangle\, ABC\;$ is just a half of product of legs $\, AB\,$ and $\,BC\,$:
\begin{align}
S_{\triangle ABC} = \dfrac{1}{2}\,\big\|\left(0,\,1\right) - \left(0,\,3\right)\big\|\, \big\|\left(2,\,1\right) - \left(0,\,1\right)\big\| = 2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove that this matrix is nonsingular? How can I prove that $
\begin{vmatrix}
\frac{1}{(x_1-y_1)^2} & \frac{1}{(x_1-y_2)^2} & \cdots & \frac{1}{(x_1-y_n)^2} \\
\frac{1}{(x_2-y_1)^2} & \frac{1}{(x_2-y_2)^2} & \cdots & \frac{1}{(x_2-y_n)^2} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{(x_n-y_1)^2} & \frac{1}{(x_n-y_2)^2} & \cdots & \frac{1}{(x_n-y_n)^2}
\end{vmatrix} \ne 0
$?
For $x_i$ and $y_i$ are all distinct.
| Even if all the $x_i$ and $y_j$ are distinct, this is not true.
For instance, if $x_1=2/3$, $x_2=2$, $y_1=1$, $y_2=0$, we have
$$
\begin{vmatrix}\frac1{(x_1-y_1)^2}&\frac1{(x_1-y_2)^2}\\\frac1{(x_2-y_1)^2}&\frac1{(x_2-y_2)^2}\end{vmatrix}
=\begin{vmatrix}
\frac1{(2/3-1)^2}&\frac1{(2/3-0)^2}\\
\frac1{(2-1)^2}&\frac1{(2-0)^2}
\end{vmatrix}
=\begin{vmatrix}
9&9/4\\
1&1/4
\end{vmatrix}=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to simplify $ 2 \sqrt{2}\left(\sqrt{9-\sqrt{77}} \right) $ How to simplify
$$
2 \sqrt{2}\left(\sqrt{9-\sqrt{77}} \right)
$$
so that it has no nested radicals?
This question is same as that already posted but with a different point of view.
| $$
2\sqrt{2}\sqrt{9-\sqrt{77}}=2\sqrt{2}\left( \sqrt{\frac{9+2}{2}}-\sqrt{\frac{9-2}{2}}\right)=2\sqrt{11}-2\sqrt{7}
$$
By the formula:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
that can easily verified and works well when $a^2-b$ is a perfect square.
You can see my answer to Denesting a square root: $\sqrt{7 + \sqrt{14}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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integrate sin(x). OK. I have a doubt with this: I know $-\cos(x) + k =\int \sin x\,dx$ but doing
$\sin(x)=2\sin\frac{x}{2}\cos\frac{x}{2}$ I get $\int \sin x\,dx = \int 2\sin\frac{x}{2}\cos\frac{x}{2}\,dx $ if $ u = \sin \frac{x}{2}$ then $du = \cos \frac{x}{2} \frac{dx}{2}$ then $$\int \sin x\,dx = \int 2\sin \frac{x}{2}\cos \frac{x}{2}\, dx =4\int \sin \frac{x}{2} \cos\frac{x}{2}\frac{dx}{2}=4\int u\,du= 4\frac{u^2}{2}= 2u^2 = 2\sin^2\frac{x}{2}$$ then if $\theta = \frac{x}{2} \rightarrow \cos 2\theta =-2\sin^2(\theta)+k \rightarrow \frac{\cos(2\theta)}{2}=\cos^2 \theta -1+\frac{k}{2} $ if $k=0$, then $$\frac{\cos 2\theta}{2}=\cos^2 \theta-1.$$ Now, why isn't it a trigonometric identity? or Is it? Because I think that I found one.
| You can simply use
$$\sin(x)=\cos(\pi/2-x)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_\limits{x \to 0} (1 +\sin^2 x)^{\frac{1}{\ln(\cos x)}}$ $$\lim_{x \to 0} (1 + \sin^2 x)^{\frac{1}{\ln(\cos x)}}$$
I evaluated $\sin$ and $\cos x$ but what can be done with $\ln\left(1-\frac{x^2}{2}\right)$ or $\ln\left(\frac{2 - x^2}{2}\right)$?
Assume that L'Hopital is forbidden but you can use asymptotic simplifications like big and small $o$ notations and Taylor series.
| You can write the function as
$$(1 + \sin^2 x)^{ \frac{1}{\sin^2 x} \frac{\sin^2x}{\ln(\cos x)}}$$
Further
$$\frac{\sin^2x}{\ln(\cos x)}=\frac{x^2+o(x^2)}{\ln(1-\frac{x^2}{2}+o(x^2))}=\frac{x^2+o(x^2)}{-\frac{x^2}{2}+o(x^2)}\to-2$$
And
$$(1 + \sin^2 x)^{ \frac{1}{\sin^2 x} } \to e$$
Hence...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\displaystyle \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx+\int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx$
Evaluation of $\displaystyle \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx+\int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx\;,$ Now Put $x=t^{12}\;,$ Then $dx = 12t^{11}dt$
So we get $$I = 12\int\frac{t^{11}}{t^4+t^3}dt = 12\int \frac{t^8}{1+t}dt = 12\int\frac{(t^8-1)+1}{1+t}dt$$
So we get $$I = 12 \int (1+t+t^2+t^3+....+t^7)dt+12\ln |1+t|$$
and $\displaystyle J = \int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx\;,$ Now put $x=u^6\;,$ We get $dx = 6u^5dt$
So we get $$ J = \int\frac{\ln(1+u)}{u^3+u^2}\cdot 6u^5dt = 6\int \frac{u^2\ln(1+u)}{1+u}du$$
Now How can I solve Integral $J\;,$ after that
Help required
Thanks
| Let $v=\ln(1+u)$ so $u=e^v-1$ and $dv=\frac{du}{1+u}$.
So $$J=6\int (e^v-1)^2 v dv$$
Or
$$J=6\int (e^{2v}-2e^v+1)v dv$$
$$J=6\int ve^{2v}-2ve^v+v dv$$
Which shouldn't be hard. Break this into three integrals, use integration by parts for the first two and the third is easy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$3x + 1$ is even iff $5x-2$ is odd I'm asked to prove 'Let $x \in Z$. $3x + 1$ is even iff $5x-2$ is odd'. I have the following proof techniques in my toolbox: trivial/vacuous proofs (not so relevant in this case), direct proof and proof via the contrapositive.
1) If $3x + 1$ is even, then $5x- 2$ is odd.
2) If $5x - 2$ is odd, then $3x + 1$ is even.
Theorem 1a: An odd number multiplied by an odd number yields an odd number.
Proof via direct proof. Let $n, m \in Z$. Assume n and m are odd numbers, so they can be written as $n = 2a + 1$ and $m = 2b + 1$ for any $a,b \in Z$.
Therefore $mn$ = $(2a + 1)(2b + 1)$ = $4ab + 2a + 2b + 1$ = $2(2ab + a + b) + 1$. Since $2ab + a + b + 1$ is an integer, $2(2ab + a + b) + 1$ is odd.
Theorem 1b: Adding and odd number to an odd number yields an even number.
Proof via direct proof. Let $n, m \in Z$. Assume n and m are odd numbers so they can be written as $n = 2a + 1$ and $m = 2b + 1$ for any $a,b \in Z$.
Therefore $m + n$ = $(2a + 1) + (2b + 1)$ = $2a + 2b + 2$ = $2(a + b + 1)$. Since $a + b + 1$ is an integer, $2(a + b + 1)$ is an even integer.
Theorem 1: If $3x + 1$ is even, then $5x- 2$ is odd.
Proof via direct proof. We know via 1a and 1b that for $3x + 1$ to be even $x$ has to be odd. Since $x$ is odd, $x = 2k + 1$ for any $k \in Z$.
Therefore $5(2k + 1) - 2$ = $10k + 5 - 2$ = $10k + 3$ = $10k + 2 + 1$ = $2(5k + 1) + 1$. Since $5k + 1$ is an integer, $2(5k + 1) + 1$ is an odd integer.
Theorem 2a: Subtracting an even number from an odd number yields an odd number.
Proof via direct proof. Assume $m,n \in Z$, $m$ is odd, $n$ is even, then: $m = 2a + 1$ and $n = 2b$.
$m - n$ = $(2a + 1) - 2b$ = $2a - 2b + 1$ = $2(a - b) + 1$. Since $a-b$ is an integer, $2(a - b) + 1$ is an odd integer.
Theorem 2: If $5x - 2$ is odd, then $3x + 1$ is even.
We know via 1a and 2a that for $5x - 2$ to be odd, $x$ must be odd.
So, $x = 2k + 1$ for $k \in Z$. Therefore:
$3(2k + 1) + 1$ = $6k + 4$ = $2(3k + 2)$. Since $3k + 2$ is an integer, $2(3k + 2)$ must be an even integer. $\blacksquare$
Since this is the first time I've needed to establish intermediate results I was wondering if this approach is correct.
| If they have the same parity, then their sum is divisible by two.
but you have $(3x+1)+(5x-2)=8x-1=2(4x-1)+1$ odd. so they need to have different parities.
| {
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Solving the 4th Degree Equation $x^4- 8\sqrt{3}x^2 - 16 = 0$ I'm learning radical simplification and our teacher gave us this equation to solve:
$$x^4-8\sqrt{3}x^2-16=0$$
She told us to consider $y=x^2$ to transform the equation into a quadratic equation, which we can solve. However, when I apply the quadratic formula to the equation:
$$y^2-8\sqrt{3}y-16=0$$
I get:
$$y={8\sqrt{3}\pm\sqrt{\sqrt{192}-64}\over 2}\equiv y=4\sqrt{3}\pm8$$
And after that I get stuck. Our teacher solved another equation in class and she transformed the result of the quadratic formula into the square of a binomial, so that you're able to square root it and get the value of $x$, however, I haven't been able to transform $y=4\sqrt{3}\pm8$ into a square of a binomial. Maybe there's another way around it, but I can't seem to find it.
Any help would be greatly appreciated, thanks in advance! :)
| $$x^4-8\sqrt{3}x^2-16=0\Longleftrightarrow$$
Substitute $y=x^2$:
$$y^2-8\sqrt{3}y-16=0\Longleftrightarrow$$
$$y^2-8\sqrt{3}y=16\Longleftrightarrow$$
$$y^2-8\sqrt{3}y+16\cdot 3^{2\cdot\frac{1}{2}}=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$
$$\left(y-4\sqrt{3}\right)^2=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3^{2\cdot\frac{1}{2}}}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{64}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm 8\Longleftrightarrow$$
$$y=\pm 8+4\sqrt{3}\Longleftrightarrow$$
$$x^2=\pm 8+4\sqrt{3}\Longleftrightarrow$$
$$x=\pm\sqrt{\pm 8+4\sqrt{3}}$$
So the solutions are:
$$x_1=\sqrt{8+4\sqrt{3}}=\sqrt{2}+\sqrt{6}\approx 3.86370$$
$$x_2=-\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{2}-\sqrt{6}\right)\approx -1.0325i$$
$$x_3=\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{6}-\sqrt{2}\right)\approx 1.0325i$$
$$x_4=-\sqrt{8+4\sqrt{3}}=-\left(\sqrt{2}+\sqrt{6}\right)\approx -3.86370$$
| {
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I need integrate this $\int_{} \frac{1}{\sqrt{1-z^2}-z} dz$ I don't know how to integrate this $\int_{} \frac{1}{\sqrt{1-z^2}-z} dz$ I tried with suspstitution $ t=\sqrt{1-z^2}-z $ but it doesn't work. Please help!
| $$\int\frac{1}{\sqrt{1-z^2}-z}\space\text{d}z=$$
Substitute $z=\sin(u)$ and $\text{d}z=\cos(u)\space\text{d}u$. Then $\sqrt{1-z^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\sin^{-1}(z)$:
$$\int\frac{\cos(u)}{\cos(u)-\sin(u)}\space\text{d}u=$$
$$\int\frac{\sec^2(u)}{\sec^2(u)-\sec^2(u)\tan(u)}\space\text{d}u=$$
$$\int\frac{\sec^2(u)}{1-\tan(u)+\tan^2(u)-\tan^3(u)}\space\text{d}u=$$
Substitute $s=\tan(u)$ and $\text{d}s=\sec^2(u)\space\text{d}u$:
$$\int\frac{1}{-s^3+s^2-s+1}\space\text{d}s=$$
$$\int\left(\frac{s+1}{2(s^2+1)}-\frac{1}{2(s-1)}\right)\space\text{d}s=$$
$$\int\frac{s+1}{2(s^2+1)}\space\text{d}s-\int\frac{1}{2(s-1)}\space\text{d}s=$$
$$\frac{1}{2}\int\frac{s+1}{s^2+1}\space\text{d}s-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
$$\frac{1}{2}\int\left(\frac{s}{s^2+1}+\frac{1}{s^2+1}\right)\space\text{d}s-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
$$\frac{1}{2}\int\frac{s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
Substitute $p=s^2+1$ and $\text{d}p=2s\space\text{d}s$:
$$\frac{1}{4}\int\frac{1}{p}\space\text{d}p+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
$$\frac{\ln\left|p\right|}{4}+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
$$\frac{\ln\left|p\right|}{4}+\frac{\tan^{-1}\left(s\right)}{2}-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
$$\frac{\ln\left|p\right|}{4}+\frac{\tan^{-1}\left(s\right)}{2}-\frac{1}{2}\int\frac{1}{s-1}\space\text{d}s=$$
Substitute $w=s-1$ and $\text{d}w=\text{d}s$:
$$\frac{\ln\left|p\right|}{4}+\frac{\tan^{-1}\left(s\right)}{2}-\frac{1}{2}\int\frac{1}{w}\space\text{d}w=$$
$$\frac{\ln\left|p\right|}{4}+\frac{\tan^{-1}\left(s\right)}{2}-\frac{\ln\left|w\right|}{2}+\text{C}=$$
$$\frac{\ln\left|s^2+1\right|}{4}+\frac{\tan^{-1}\left(s\right)}{2}-\frac{\ln\left|s-1\right|}{2}+\text{C}=$$
$$\frac{\ln\left|\tan^2(u)+1\right|}{4}+\frac{\tan^{-1}\left(\tan(u)\right)}{2}-\frac{\ln\left|\tan(u)-1\right|}{2}+\text{C}=$$
$$\frac{\ln\left|\tan^2\left(\sin^{-1}(z)\right)+1\right|}{4}+\frac{\tan^{-1}\left(\tan\left(\sin^{-1}(z)\right)\right)}{2}-\frac{\ln\left|\tan\left(\sin^{-1}(z)\right)-1\right|}{2}+\text{C}=$$
$$\frac{1}{4}\left(-\ln\left|1-2z^2\right|+2\tanh^{-1}\left(\frac{z}{\sqrt{1-z^2}}\right)+2\sin^{-1}(z)\right)+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3\tan^2x+6\tan x+11}{1+\tan ^2x}dx=\frac{k\pi+\lambda}{6}$ If $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3\tan^2x+6\tan x+11}{1+\tan ^2x}dx=\frac{k\pi+\lambda}{6}$$,then the value of $(k+\lambda)$ is
$$(A)10\hspace{1cm}(B)12\hspace{1cm}(C)14\hspace{1cm}(D)16$$
My Attempt:
Let $$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3\tan^2x+6\tan x+11}{1+\tan ^2x}dx.......................(1)$$
Applying $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3\cot^2x+6\cot x+11}{1+\cot ^2x}dx$$
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3+6\tan x+11\tan^2x}{1+\tan ^2x}dx...........(2)$$
Add $(1)$ and $2$ to get
$$2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{14+12\tan x+14\tan^2x}{1+\tan ^2x}dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}14+\frac{12\tan x}{1+\tan ^2x}dx$$
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}7+\frac{6\tan x\sec^2x}{(1+\tan ^2x)^2}dx$$
$$I=\frac{7\pi}{6}+3\int_{\frac{1}{\sqrt3}}^{\sqrt3}\frac{2tdt}{(1+t^2)^2}$$
$$I=\frac{7\pi}{6}+3\int_{\frac{4}{3}}^{4}\frac{dp}{p^2}=\frac{7\pi}{6}-3(4-\frac{4}{3})=\frac{7\pi}{6}-8=\frac{7\pi-48}{6}$$
So $k+\lambda=-41$,but the answer given is $16$.
Where have i gone wrong?Please help me.Thanks.
| Let $$I = \int\frac{3\tan^2 x+6\tan x+11}{1+\tan^2 x}dx = 3\int\frac{1+\tan^2 x}{1+\tan^2 x}dx+\int\frac{6\tan x+8}{1+\tan^2 x}dx$$
So $$I = 3x+6\int \sin x\cos xdx+8\int \cos^2 xdx$$
So $$I = 3x+3\int \sin 2xdx+4\int (1+\cos 2x)dx = 3x-\frac{3}{2}\cos 2x+4x+2\sin 2x$$
So we Get $$I = 7x-\frac{3}{2}\cos 2x+2\sin 2x$$
Now $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3\tan^2 x+6\tan x+11}{1+\tan^2 x}dx = \frac{7\pi}{6}-\frac{3}{2}\left[-\frac{1}{2}-\frac{1}{2}\right]+2\left[0\right]=\frac{7\pi}{6}+\frac{3}{2}$$
So we Get $$ = \frac{7\pi+9}{6} = \frac{k\pi+\lambda}{6}$$
So we get $$k+\lambda = 7+9 = 16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Canonical Complete Residue System $\pmod 7$ Find all solutions in the canonical complete residue system modulo $7$ of the congruence
$$3x \equiv 2 \pmod 7$$
I know that there is a unique solution, because $\operatorname{gcd}(3, 7)=1$ which is obviously a divisor of $2$. However, I don't exactly where I need to go from there in order to get to the general solution.
| To solve the congruence $3x \equiv 2 \pmod{7}$, we can apply the extended Euclidean algorithm.
$$7 = 2 \cdot 3 + 1$$
Solving for $1$ yields
$$1 = 7 - 2 \cdot 3$$
Hence,
$$3 \cdot -2 \equiv 1 \pmod{7}$$
Multiplying both sides of the congruence by $2$ yields
$$3 \cdot -4 \equiv 2 \pmod{7}$$
Since $-4 \equiv 7 - 4 \equiv 3 \pmod{7}$, we obtain
$$3 \cdot 3 \equiv 2 \pmod{7}$$
Hence, $x \equiv 3 \pmod{7}$.
Check: If $x \equiv 3 \pmod{7}$, then $3x \equiv 3 \cdot 3 \equiv 9 \equiv 7 + 2 \equiv 2 \pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How does one get from $p+3=3k+2$ then $2^{p+3} \equiv 4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$ How does one get from $p+3=3k+2$ then $2^{p+3}\equiv4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$.
I am just starting with modular arithmetic so any help would be greatly appreaciated.
Credit to user236182 for original answer, just wanted to know how he got to it
| Given that $p+3=3k+2$ it follows that
$$2^{p+3}=2^{3k+2}=(2^3)^k\cdot2^2=8^k\cdot4\equiv1^k\cdot4=4\pmod7.$$
Plugging this into your next expression yields
$$5\cdot2^{p+3}-31\equiv5\cdot4-31\equiv-11\equiv3\pmod7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Decomposing a composite function: $f(g(x))=\frac{x^4+x^2}{1+x^2}$, then $f$=? We have a composite function $f(g(x))=((x^4+x^2)/(1+x^2))$ and a single function $g(x)=1-x^2$. We want to know $f(1/2)$. From theory we know that $g(x)$ had been plugged in $f(x)$, but how would i find standalone $f(x)$ from here? I thought of dividing the result by $g(x)$, but that isn't true.
| Alternate way to do this is to find an $x$ such that $g(x)=\frac{1}{2}$.
Since $$g(\frac{1}{\sqrt{2}}) = \frac{1}{2}$$
We have $$f(\frac{1}{2})=f(g(\frac{1}{\sqrt{2}}))=\frac{(\frac{1}{\sqrt{2}})^4+(\frac{1}{\sqrt{2}})^2}{1+(\frac{1}{\sqrt{2}})^2} = \frac{\frac{1}{4}+\frac{1}{2}}{1+\frac{1}{2}} = \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim_{x\to 7/4^+}\tfrac{3x}{4x-7}=\infty$ by definition $$\lim_{x\to \frac{7}{4}^+}\frac{3x}{4x-7}=\infty $$
I want to prove that for every $M>0$ exists $\delta$ for which $ 0<x-\frac{7}{4}<\delta $ such that $ f(x)>M $
What I tried:
$$f(x)=\frac{3x}{4x-7}>M\iff\frac{1}{M}>\frac{4x-7}{3x} =* $$
Then I took some $\delta_1 = \frac{1}{4}$ and then:
$$0<x-\frac{7}{4}<\frac{1}{4}\iff \frac{7}{4}<x<2\iff7<4x<8\iff0<4x-7<1 $$
Final proof:
We take arbitrary $M>0 $.
Then we take some $ \delta_{1}=\frac{1}{4}$. Then
$$ 0<x-\cfrac{7}{4}<\cfrac{1}{4}\iff \cfrac{7}{4}<x<2\leftrightarrow\cfrac{21}{4}<3x<6 $$
So $$ f(x)=\cfrac{3x}{4x-7}>\cfrac{5}{4x-7}=\cfrac{5}{4\left(x-\cfrac{7}{4}\right)}=\cfrac{5}{4}\cdot\cfrac{1}{x-\cfrac{7}{4}}>M\iff \cfrac{1}{x-\cfrac{7}{4}}>\cfrac{4M}{5}\iff x-\cfrac{7}{4}<\frac{5}{4M}$$
Then $ \delta_{2}=\frac{5}{4M} $
$$ \delta=\min\{\delta_{1},\delta_{2}\}=\min\left\{\cfrac{5}{4M},\cfrac{1}{8}\right\} $$
| First of all a two-sided limit does not exist!
$$\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{3x}{4x-7}=$$
$$3\left(\lim_{x\to\left(\frac{7}{4}\right)^-}x\right)\left(\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{1}{4x-7}\right)=$$
$$3\left(\frac{7}{4}\right)\left(\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{1}{4x-7}\right)=$$
$$\frac{21}{4}\left(\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{1}{4x-7}\right)=$$
Since $\lim_{x\to\left(\frac{7}{4}\right)^-}\space(4x-7)=0$ and $4x-7<0$ for all $x$ just to the left of $\frac{7}{4}$.
So $\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{1}{4x-7}=-\infty$:
$$\lim_{x\to\left(\frac{7}{4}\right)^-}\space\frac{3x}{4x-7}=-\infty$$
Now you can proof the one from the other side!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Differential equation with separation of the variables I have to slove the following differential equation with separation of the variables (without any initial conditions)
$xy'=\frac{1}{y^2}-y$
This is what I have done so far
$x\frac{dy}{dx}=\frac{1-y^3}{y^2}$
$\int \frac{dx}{x}=\int \frac{y^2 dy}{1-y^3}$
${\rm ln} \ \vert x \vert +c_1=-\frac{1}{3} {\rm ln} \ \vert 1-y^3 \vert +c_2 $
${\rm ln} \ \vert x \vert=-\frac{1}{3} {\rm ln} \ \vert 1-y^3 \vert +c \ $ with $c=c_2-c_1$
How shoud I go on? I don't know what to do with the absolute values. The solution should be $y(x)=(1-\frac{e^{3c}}{x^3})^{1/3}$.
| $$xy'(x)=\frac{1}{y(x)^2}-y(x)\Longleftrightarrow$$
$$x\cdot\frac{\text{d}y(x)}{\text{d}x}=\frac{1}{y(x)^2}-y(x)\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}=\frac{1-y(x)^3}{xy(x)^2}\Longleftrightarrow$$
$$\frac{\frac{\text{d}y(x)}{\text{d}x}\cdot y(x)^2}{1-y(x)^3}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{\frac{\text{d}y(x)}{\text{d}x}\cdot y(x)^2}{1-y(x)^3}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$-\frac{1}{3}\ln\left|-y(x)^3+1\right|=\ln|x|+\text{C}\Longleftrightarrow$$
$$\ln\left|-y(x)^3+1\right|=-3\ln|x|-3\text{C}\Longleftrightarrow$$
$$\ln\left|-y(x)^3+1\right|=\ln\left(\frac{1}{|x|^3}\right)-3\text{C}\Longleftrightarrow$$
$$\left|-y(x)^3+1\right|=e^{\ln\left(\frac{1}{|x|^3}\right)-3\text{C}}\Longleftrightarrow$$
$$\left|-y(x)^3+1\right|=\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$1-y(x)^3=\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow\space\space\vee\space\space 1-y(x)^3=-\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$-y(x)^3=\frac{e^{-3\text{C}}}{|x|^3}-1\Longleftrightarrow\space\space\vee\space\space -y(x)^3=-\frac{e^{-3\text{C}}}{|x|^3}-1\Longleftrightarrow$$
$$y(x)^3=1-\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow\space\space\vee\space\space y(x)^3=1+\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$y(x)=\frac{\sqrt[3]{|x|^3-e^{-3\text{C}}}}{|x|}\space\space\vee\space\space y(x)=-\frac{\sqrt[3]{-1}\sqrt[3]{|x|^3+e^{-3\text{C}}}}{|x|}\space\space\vee\space\space y(x)=\frac{(-1)^{\frac{2}{3}}\sqrt[3]{|x|^3+e^{-3\text{C}}}}{|x|}$$
Where $\text{C}$ is an arbitrary constant
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Express $\frac{\sin 7\theta}{\sin \theta}$ in powers of $\sin \theta$ only By using DeMoivre's theorm express
$$\frac{\sin 7\theta}{\sin \theta}$$
in the powers of Sine only
answer given in the book is
$$7-56\sin ^2\theta+112\sin ^4 \theta-64\sin^6 \theta$$
can any one help to solve the question
| Steps To Carry Out
1) First take a look at this link which is a guide for DeMoivre's formula.
2) Using step 1 show that
$$\sin (7x) = 64\sin \left( x \right)\cos {\left( x \right)^6} - 80\sin \left( x \right)\cos {\left( x \right)^4} + 24\sin \left( x \right)\cos {\left( x \right)^2} - \sin \left( x \right)$$
3) Replace $\cos^2(x)=1-\sin^2(x)$ and obtain
$$\sin (7x) = 7\sin \left( x \right) - 56\sin {\left( x \right)^3} + 112\sin {\left( x \right)^5} - 64\sin {\left( x \right)^7}$$
4) Divide by $\sin(x)$
$${{\sin (7x)} \over {\sin (x)}} = 7 - 56\sin {\left( x \right)^2} + 112\sin {\left( x \right)^4} - 64\sin {\left( x \right)^6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| How do you define a fraction?
The definition I know is that for any number $a \neq 0$ we say the symbol $\frac{1}{a}$ satisfies $a \cdot \frac{1}{a} = 1$. One can show that this is well defined for any $a \neq 0$ and unique. Further, the symbol $\frac{b}{a}$ for any numbers $b,a$ with $a\neq 0$ is defined as $\frac{b}{a}:=b \cdot \frac{1}{a}$.
If $\frac{1}{0}$ would exists, then clearly by definition we would have $$\frac{0}{0}=0 \cdot \frac{1}{0} = 1.$$ However,
the symbol $\frac{1}{a}$ is not definied for $a=0$, because there exists no number $b$ with $0 \cdot b = 1$, so there is no value to which we can assign the symbol $\frac{1}{0}$.
Example of the above definition:
We can compute $\frac{1}{2}$, because we know
$$ 2 \cdot \frac{1}{2} = \frac{1}{2} +\frac{1}{2} =1$$ and since
$$ 2 \cdot 0.5 = 0.5 + 0.5 = 1 $$ it follows that
$$\frac{1}{2} = 0.5$$
holds.
| {
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"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 1
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On $e^{5x}+e^{4x}+e^{3x}+e^{2x}+e^{x}+1$ Define the following,
$$F_2(x) := \frac{1}{2}+\frac{(2x)}{1!} B_2\Big(\tfrac{1}{2}\Big)+\frac{(2x)^2}{2!}B_3\Big(\tfrac{1}{2}\Big)+\frac{(2x)^3}{3!}B_4\Big(\tfrac{1}{2}\Big)+\dots $$
$$\color{brown}{F_3(x)} := \frac{1}{3}+\frac{(3x)}{1!} B_2\Big(\tfrac{1}{3}\Big)+\frac{(3x)^2}{2!}B_3\Big(\tfrac{1}{3}\Big)+\frac{(3x)^3}{3!}B_4\Big(\tfrac{1}{3}\Big)+\dots $$
$$F_4(x) := \frac{1}{4}+\frac{(4x)}{1!} B_2\Big(\tfrac{1}{4}\Big)+\frac{(4x)^2}{2!}B_3\Big(\tfrac{1}{4}\Big)+\frac{(4x)^3}{3!}B_4\Big(\tfrac{1}{4}\Big)+\dots $$
$$\color{brown}{F_6(x)} := \frac{1}{6}+\frac{(6x)}{1!} B_2\Big(\tfrac{1}{6}\Big)+\frac{(6x)^2}{2!}B_3\Big(\tfrac{1}{6}\Big)+\frac{(6x)^3}{3!}B_4\Big(\tfrac{1}{6}\Big)+\dots $$
From Summation of Series (2nd ed) by L.B Jolley, page 26, it seems that,
$$\color{brown}{F_3(x)} \overset{?}= \frac{1}{1+e^x+e^{2x}}\\
\color{brown}{F_6(x)} \overset{?}= \frac{1}{1+e^{x}+e^{2x}+e^{3x}+e^{4x}+e^{5x}}\tag1$$
I assume that $B_n(x)$ are the Bernoulli polynomials. However, when I try to numerically evaluate those two $F_k(x)$ using Mathematica, the LHS does not agree with the RHS.
Questions:
*
*How to interpret $B(n)$ such that $(1)$ holds true?
*And do $F_2(x)$ and $F_4(x)$ evaluate similar to $(1)$?
| Ad Question 1: The polynomials $B_n(x)$ defined in L.B.W. Jolley's book Summation of Series are close relatives of the Bernoulli polynomials defined e.g. in L. Comtet's classic Advanced Combinatorics but not the same.
To differentiate the polynomials defined in L.B.W. Jolley's book from the more commonly defined Bernoulli polynomials we use, we denote them from now on with $\widetilde{B}_n$.
In number (1128) on p. 228 in Jolley's book and in the subsequent section (1129) we find a representation of $\widetilde{B}_n(x)$ as generating series.
\begin{align*}
t\frac{e^{xt}-1}{e^x-1}=\sum_{n\geq 1}n\widetilde{B}_n(x)\frac{t^n}{n!}\tag{1}
\end{align*}
A generating function of the Bernoulli polynomials $B_n(x)$ is given as
\begin{align*}
t\frac{e^{xt}}{e^t-1}=\sum_{n\geq 0}B_n(x)\frac{t^n}{n!}
\end{align*}
Since a generating function of the Bernoulli numbers $B_n=B_n(0)$ is
\begin{align*}
\frac{t}{e^t-1}=\sum_{n\geq 0}B_n(0)\frac{t^n}{n!}
\end{align*}
we obtain the following relationship between Bernoulli polynomials $B_n(x)$ and $\widetilde{B}_n(x)$
\begin{align*}
\widetilde{B}_n(x)=\frac{B_n(x)-B_n(0)}{n}\qquad\qquad n\geq 1
\end{align*}
Ad question 2: The answer is affirmative. The nice formulas of $F_t(x)$ are valid for all $t\in \mathbb{N}$.
On one hand we obtain for $t\geq 1$ by expanding numerator and denominator of $F_t(x)$ with $1-e^x$
\begin{align*}
F_t(x)&=\frac{1}{1+e^x+e^{2x}+\cdots+e^{(t-1)x}}\\
&=\frac{1-e^x}{1-e^{xt}}
\end{align*}
On the other hand note, that the general expression of $F_t(x)$ is
\begin{align*}
F_t(x)=\frac{1}{t}+\sum_{n\geq 1}\widetilde{B}_{n+1}\left(\frac{1}{t}\right)\frac{(tx)^n}{n!}\tag{2}
\end{align*}
We obtain
\begin{align*}
F_t(x)&=\frac{1}{t}+\sum_{n\geq 1}\widetilde{B}_{n+1}\left(\frac{1}{t}\right)\frac{(tx)^n}{n!}\\
&=\frac{1}{t}+\sum_{n\geq 2}\widetilde{B}_n\left(\frac{1}{t}\right)\frac{(tx)^{n-1}}{(n-1)!}\\
&=\frac{1}{t}+\frac{1}{tx}\sum_{n\geq 2}n\widetilde{B}_n\left(\frac{1}{t}\right)\frac{(tx)^{n}}{n!}\tag{3}\\
&=\frac{1}{t}+\frac{1}{tx}\left(tx\frac{e^{\frac{1}{t}tx}-1}{e^{tx}-1}-\widetilde{B_1}\left(\frac{1}{t}\right)tx\right)\tag{4}\\
&=\frac{1}{t}+\frac{1}{tx}\left(tx\frac{e^x-1}{e^{tx}-1}-x\right)\\
&=\frac{e^x-1}{e^{tx}-1}\\
\end{align*}
and the claim follows.
Comment:
*
*In (3) we use the generating series for $n\widetilde{B}_n(x)$ according to (1)
*In (4) note that $\widetilde{B}_1(x)=x$
Epilogue:
Symmetry
The Function $F_t(x)$ and its reciprocal $\frac{1}{F_t(x)}$ admit a nice nearly symmetrical representation with respect to Bernoulli polynomials and their arguments. The following is valid
\begin{align*}
F_t(x)-1&
=\sum_{n\geq 0}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\
\frac{1}{F_t(x)}&
=\sum_{n\geq 0}\frac{B_{n+1}(t)-B_{n+1}(0)}{n+1}\frac{x^n}{n!}
\end{align*}
This holds, since according to (2) we get
\begin{align*}
F_t(x)&=\frac{1}{\sum_{k=0}^{t-1}e^{kx}}
=\frac{1}{t}+\sum_{n\geq 1}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\
&=1+\sum_{n\geq 0}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\
\end{align*}
and
\begin{align*}
\frac{1}{F_t(x)}&=\sum_{k=0}^{t-1}e^{kx}
=\sum_{k=0}^{t-1}\sum_{n\geq 0}\frac{(kx)^n}{n!}
=\sum_{n\geq 0}\left(\sum_{k=0}^{t-1}k^n\right)\frac{x^n}{n!}\\
&=\sum_{n\geq 0}\frac{B_{n+1}(t)-B_{n+1}(0)}{n+1}\frac{x^n}{n!}
\end{align*}
Historical Note
The definition of the Bernoulli numbers $\widetilde{B}_n$ in L.B.W. Jolley's book is taken from the British Association Report from 1877. The first few numbers are
\begin{array}{cccccccccc}
n&0&1&2&3&4&5&6&7&8\\
\hline\\
\widetilde{B}_n&-1&\frac{1}{6}&\frac{1}{30}&\frac{1}{42}&\frac{1}{30}
&\frac{5}{66}&\frac{691}{2730}&\frac{7}{6}&\frac{3617}{510}\\
\\
B_n&0&-\frac{1}{2}&\frac{1}{6}&0&-\frac{1}{30}&0&\frac{1}{42}&0&-\frac{1}{30}
\end{array}
This corresponds for $n\geq 1$ to Comtet's statement in section 1.14
(notation somewhat adapted) ... Bernoulli numbers, denoted by $b_n$ by Bourbaki, are sometimes also defined by:
\begin{align*}
\frac{t}{e^t-1}=1-\frac{1}{2}t+\sum_{n\geq 1}(-1)^{n+1}B_n\frac{t^{2n}}{(2n)!}
\end{align*}
Each $B_n$ is then $>0$ and equals $(-1)^{n+1}B_{2n}$ as a function of our Bournoulli numbers.
| {
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"url": "https://math.stackexchange.com/questions/1558245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$
\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a formula.
| Notice, the series for product up to $n$th term $-3, -4, -5, -6, \ldots -(n+2)$
the series for power of $(6x+8)$ up to $n$th term is $4, 5, 6, \ldots (n+3)$
hence, the $n$th derivative can be generalized as follows $$y^n=4(-3)(-4)(-5)\ldots (-(n+2))(6)^n\frac{1}{(6x+8)^{n+3}}=2(-1)^n\left(1\cdot 2\cdot 3\cdot 4\ldots (n+2)\right)(6)^n\frac{1}{(6x+8)^{n+3}}$$ $$=2(-1)^n(n+2)!(6)^n\frac{1}{(6x+8)^{n+3}}$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y^n=\frac{2(-1)^n(n+2)!6^n}{(6x+8)^{n+3}}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $f(3)$ such that $f(f(x))=x^2-5x+9$ How can one calculate $f(3)$ when $f(f(x))=x^2-5x+9$
I tried this:
$f(f(3))=3$
I'm stuck here.
| $f(f(f(x))) = f(x^2 - 5x +9) = f(x)^2 - 5 f(x) + 9$.
set $x = 3$, then $f(3) = f(3)^2 - 5 f(3) + 9$. then $f(3) = 3$
| {
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} |
Complex Analysis Inequality Involving Exponential I am trying to prove the following result:
For $0<|z|<1$, $\frac{|z|}{4}< |e^z - 1| < \frac{7|z|}{4}$.
My attempt :
First, I converted the problem to (after mult. by 4):
$$|z|< 4|e^z - 1| < 7|z|.$$
Then, I subtracted $|z|$ so that:
$$0< 4|e^z - 1|-|z| < 6|z|.$$
Note that:
$$4|e^z - 1| \leq 4\left(\sum_{i=0}^\infty \frac{|z|^k}{k!}-1\right) \\ \leq|z|(4\frac{4}{1-|z|} -1) \\ = |z|(\frac{3+|z|}{1-|z|}).$$
Which from then we could take the "Sup" over all $\exp|z|$ so that
we get the above is bounded by something like $4 + \exp|z|$.
This is where I get stuck on this. Any pointers would definitely be appreciated.
Thanks.
| From the Taylor series for the exponential function you get
for $0 < |z| < 1$ (with $r := |z|$):
$$
\left| \frac{e^z - 1}{z} - 1 \right | =
\left| \frac{z}{2!} + \frac{z^2}{3!} + \frac{z^3}{4!} + ... \right | \\
\le \frac{r}{2!} + \frac{r^2}{3!} + \frac{r^3}{4!} + ...\\
= \frac{r}{2} \left( 1 + \frac{r}{3} + \frac{r^2}{3\cdot 4} + ...\right) \\ \le
\frac{r}{2} \left( 1 + \frac{r}{3} + \frac{r^2}{3^2} + ...\right)
$$
Since $r <1$, the last expression is less than
$$
\frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + ...\right)
= \frac{1}{2} \frac{1}{1-\frac{1}3} = \frac 34
$$
using the geometric series. It follows that
$$
1 -\frac 34 < \left| \frac{e^z - 1}{z} \right | < 1 + \frac 34
$$
which is the desired inequality.
| {
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Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$) Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$)
The inequality is equivalent to $x^2+(y-25)^2\le 15^2$ that represents the set of points in the circle of radius $15$ and center coordinate $C(0,25)$.
In this set, how to find one complex number which has the largest argument?
| Having posed that $z = x+iy$, then:
$$|z-25i|\le 15 \Rightarrow x^2 +(y-25)^2 \le 15^2.$$
Recall that:
$$\arg(z) =
\begin{cases}
f(y) & x > 0 \\
f(y) + \pi & x<0 \wedge y \ge 0 \\
f(y) - \pi & x<0 \wedge y < 0\\
\frac{\pi}{2} & x = 0 \wedge y > 0 \\
-\frac{\pi}{2} & x = 0 \wedge y < 0 \\
\text{undefined} & x = 0 \wedge y = 0
\end{cases},$$
where
$$f(y) = \arctan\left(\frac{y}{\sqrt{15^2 - (y-25)^2}}\right).$$
Notice that $x = \pm \sqrt{15^2 - (y-25)^2}$.
We want to find the maximum of $\arg(z)$ with respect to $y$. The $\arctan$ is monotonically increasing, then we can work on $f(y)$:
$$\frac{\partial }{\partial y} f(y) = \frac{25(y-16)}{\left(\sqrt{15^2 - (y-25)^2}\right)^3}.$$
Imposing it equal to $0$, you get $y = 16$ and $x=12$.
For $z = 12 + i16$, the argument is:
$$\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) \simeq 0.9273.$$
When $x$ is negative, that is $x = - \sqrt{15^2 - (y-25)^2}$, then we have another candidate $z = -12 + 16i$. In this case $\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) + \pi \simeq 4.0689$.
Last case to consider is $x = 0$ and $y = 40$. Here, we get that $\arg(z) = \frac{\pi}{2} \simeq 1.5708.$
Finally, we conclude that $z = -12+16i$ is the maximum of $\arg(z)$, and $z$ stays in the circle you described.
| {
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Computing $ \int_0^{2\pi} \frac{1}{a^2\cos^2 t+b^2 \sin^2 t} dt \;; a,b>0$.
Using Residue Theorem find $\displaystyle \int_0^{2\pi} \frac{1}{a^2\cos^2 t+b^2 \sin^2 t} dt \;; a,b>0$.
My Try:
So, I am going to use the ellipse $\Gamma = \{a\cos t+i b \sin t: 0\leq t\leq 2\pi\}$.
On $\Gamma$, $z=a\cos t+i b \sin t$, so $|z|^2=z\bar{z}=a^2\cos^2 t+b^2 \sin^2 t$.
Now, $dz=-a\sin t+i b \cos t dt$.
Hence, the integral becomes $\displaystyle \int_\Gamma \frac{dz}{z(iab+(\sin t \cos t)(b^2-a^2))}$. I know that $\displaystyle \int_\Gamma \frac{dz}{z}=2\pi i$. Now, how do I get rid of $\sin t \cos t$ part? I am stuck here. Can somebody please explain how?
| Note that we have for $z=a\cos t+ib\sin t$,
$$\begin{align}
\frac{1}{z}\,dz&=\frac{\bar z}{|z|^2}\,dz\\\\
&=\frac{\left(a\cos t-ib\sin t\right)}{a^2\cos^2 t+b^2\sin^2 t}\,\left(-a\sin t+ib\cos t\right)\,dt\\\\
&=\left(\frac{(b^2-a^2)\sin t\cos t}{a^2\cos^2 t+b^2\sin^2 t}\right)\,dt+i\left(\frac{ab}{a^2\cos^2 t+b^2\sin^2 t}\right)\,dt
\end{align}$$
Therefore, we have
$$\begin{align}\int_0^{2\pi}\frac{1}{a^2\cos^2 t+b^2\sin^2 t}\,dt&=\frac{1}{ab}\text{Im}\left(\oint_C \frac{1}{z}\,dz\right) \tag 1\\\\
&=\frac{2\pi}{ab}
\end{align}$$
where $C$ is the elliptical contour around the origin and we used the Residue Theorem to evaluate the integral on the right-hand side of $(1)$.
| {
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The minimum value of $ f(x) = | x - 1 | + | x - 2 | + | x - 3 | $ is? I don't get it why my solution is wrong :
My solving :
$ f(x) = | x - 1 | + | x - 2 | + | x - 3 | $
When $ x\leq 1 $
$ f(x) = | x - 1 | + | x - 2 | + | x - 3 | = 0 $
= $ 6 -3x $
since $ x\leq 1 $
$ f(x)\leq 3 $ also it clear $ f(x)\geq 0 $
=> The min value is 0 .
I dont have to consider other cases, since 0 is the min value f(x) can take .
Book has given the answer : $ 2 $ .
| We have
\begin{align}
f(x) & =
\begin{cases}
-(x-1)-(x-2)-(x-3) & = -3x+6 & \text{if } x \leq 1\\
(x-1) - (x-2) - (x-3) & = -x+4 & \text{if } x\in[1,2]\\
(x-1) + (x-2) - (x-3) & = x & \text{if } x\in[2,3]\\
(x-1) + (x-2) + (x-3) & = 3x-6 & \text{if }x \geq 3
\end{cases}
\end{align}
Hence, the minimum occurs at $x=2$ and take a value of $2$ as well. The plot is as shown below.
| {
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"source": "stackexchange",
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"answer_count": 8,
"answer_id": 0
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Finding the integer solutions of $y^2 = x^3 - 12$ I tried to find the solution of this equation, or conclude there are none. This i what i found out:
I noticed that $x \neq -1$ mod $3$.
We can write $y^2 + 4 = x^3 - 8 = (x-2)(x^2+2x+4)$
I tried to find a prime divider from the righthand side so i could use legendre on the left hand side, but couldnt solve it. Can someone help me?
| As you've noticed, $y^2+4=(x-2)\left(x^2+2x+4\right)$.
If $x$ is even, then let $x=2k$. Then $y=2m$, so $m^2=2k^3-3$, impossible (use mod $8$, as has been suggested in the comments. $2k^3\equiv \{0,2,6\}\pmod{8}$, but then $m^2\equiv \{5,7,3\}\pmod{8}$, contradiction, because $5,7,3$ are not quadratic residues mod $8$).
If $x$ is odd, then: let $p$ be a prime divisor of either $x-2$ or $x^2+2x+4$. Then $p$ is odd and $p\mid y^2+4$, so $\left(y2^{-1}\right)^2\equiv -1\pmod{p}$, so $p=4t+1$ (by Quadratic Reciprocity).
Therefore all the prime divisors of $x-2$, $x^2+2x+4$ are of the form $4t+1$. Also $x^2+2x+4=(x+1)^2+3>0$ and $y^2+4>0$, so $x-2>0$, so $x-2\equiv 1\pmod{4}$ and $x^2+2x+4\equiv 1\pmod{4}$. The first congruence gives $x\equiv 3\pmod{4}$, but $3^2+2\cdot 3+4\not\equiv 1\pmod{4}$, contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration $\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta}$ by using residues
$$\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta}$$
By substituting $\cos m\theta$ to $\frac{z^m+z^{-m}}{2}$ and $d\theta$ to $\frac{-i}{z}dz$,I get
$$\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta} =\int_C \frac{i(z^6+1)^2}{4z^5(z^2-2)(2z^2-1)}dz$$
where $C$ is a unit circle.
and I found out that isolated singular points of $\frac{i(z^6+1)^2}{4z^5(z^2-2)(2z^2-1)}$ in $C$ are $$z_1=0$$ $$z_2=1/\sqrt2$$ $$z_3=-1/\sqrt2$$
So only what I have to do is calculating their residues and smmanding, and mutiplying $2\pi$.
But as you can see, their order of pole are so high, so that I can't handle it with handwriting. Would be there any simplified way to get
$$\frac{3\pi}{8}$$
for the answer?
| Write $\cos^2{3 \theta} = (1 + \cos{6 \theta})/2$. Evaluate each resulting integral separately. You can eliminate the pole at $z=0$ by writing $\cos{6 \theta} = \operatorname{Re}{\left (e^{i 6 \theta}\right )} $. The rest is straightforward.
To illustrate, the integral is
$$\frac12 \int_0^{2 \pi} d\theta \frac1{5 - 4 \cos{2 \theta}} + \frac12 \operatorname{Re}{ \int_0^{2 \pi} d\theta \frac{e^{i 6 \theta}}{5 - 4 \cos{2 \theta}}}$$
which, when expressed as an integral over the unit circle, we get, for the first integral
$$\frac{i}{2} \oint_{|z|=1} dz \frac{z}{(z^2-2)(2 z^2-1)} $$
and for the second
$$\frac{i}{2} \oint_{|z|=1} dz \frac{z^7}{(z^2-2)(2 z^2-1)} $$
Keep in mind that you will be taking the real part of the second integral when done.
| {
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Solving $(n+1)(n+2)…(n+k)−k = x^2$ Let $n$ and $k$ be positive integers.
Need to find all pairs of $(n,k)$ such that
$$(n+1)(n+2) \cdots (n+k)−k = x^2,$$
where $x^2$ is a perfect square.
| Note: This answer is not mine. It is taken from here. I write it just for completeness and because I have been asked to do it.
Suppose that $k>5$ is composite. Then, it is well known that $k|(k-1)!$. Also, $$k!|(n+1)(n+2) \cdots (n+k),$$ so let $$(n+1)(n+2) \cdots (n+k) = k!t.$$ Then, $k((k-1)!t-1)$ is a perfect square. Since $\text{gcd}(k,(k-1)!t-1) = 1$, therefore, both those numbers must be perfect squares. Let $$(k-1)!t-1 = x^2.$$ Then, $$x^2+1 = (k-1)!t$$ can have no prime factor which is $3\pmod{4}$. So, $k-1 < 3 \implies k \le 3$, contradicting our assumption that $k$ is composite.
If $k=4$ we must have that $(n^2+5n+5)^2 - 5$ is a perfect square, contradiction.
If $k$ is prime, suppose that $k \ge 5$. Let the perfect square in question be $x^2$. Then $$8|5!|k!|(n+1)(n+2) \cdots (n+k),$$ and therefore $$x^2 \equiv -k \pmod{8}.$$ Since $k$ is odd, and any odd square is $1 \pmod{8}$, so $k \equiv 7 \pmod{8}$. Then, $$k-1 = 8m+6 = 2(4m+3),$$ has some prime factor of the form $3 \pmod{4}$. Let this prime be $q$. Then $$q|k!|(n+1)(n+2) \cdots (n+k),$$ so $x^2 \equiv -k \equiv -1 \pmod{q}$, contradicting the fact that $\left( \frac{-1}{q} \right) = -1$.
We are left with the cases $k=1,2,3$. For $k = 1$, $n$ perfect square is a trivial solution. For $k=2$, $n^2+3n$, must be a perfect square. But $$(n+1)^2 < n^2+3n < (n+2)^2$$ for $n > 1$. $n = 1$ is however, a solution. For $k = 3$, $$(n+1)(n+2)(n+3) = x^2+3.$$ At least one of the $3$ terms on the $\text{LHS}$ is $2 \pmod{3}$ and has a $2 \pmod{3}$ prime factor. Suppose that this prime $q \neq $. Then, $q|x^2+3$, is a contradiction since $\left( \frac{-3}{q} \right) = -1$. Otherwise, all $2 \pmod{3}$ prime factors of the LHS are $2$. If $n$ is odd, then $$8|(n+1)(n+3)|x^2+3,$$ contradiction. So $n$ is even and therefore, we can only have $n+2 \equiv 2 \pmod{3}$. Also, since $2^2||x^2+3$, so $v_2(n+2) = 2$. Then, $(n+2)/4 \equiv 2 \pmod{3}$ is odd and therefore, the $\text{LHS}$ does have a $2 \pmod{3}$ factor, contradiction.
| {
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"question_score": "11",
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Compute $\lim_{x\to 0} \frac{x-\sin x}{x-\tan x}$ with L'hopital rule. Compute $\lim_{x\to 0} \frac{x-\sin x}{x-\tan x}$ with L'hopital rule.
Well, I know that I need to use the Quotient rule first $(\frac{f}{g})'$. but then, what do I have to do next?
Edit: I meant by quotient rule: $\frac{f'g - fg'} {g^2}$
| We will prove this without the use of L'Hospital's rule. Let $L= \lim_{x \to 0} \dfrac{x-\sin(x)}{x-\tan(x)}$. We then have
$$L = \lim_{x \to 0} \dfrac{x-\sin(x)}x \cdot \dfrac{x}{x-\tan(x)}$$ Assume $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} = M$ and $\lim_{x \to 0} \dfrac{x-\tan(x)}{x^3} = N$ exists. We then have
$$M = \lim_{x \to 0} \dfrac{3x-\sin(3x)}{(3x)^3} = \lim_{x \to 0} \dfrac{3x-3\sin(x)+4\sin^3(x)}{27x^3} = \dfrac19\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} + \dfrac4{27}\lim_{x \to 0} \dfrac{\sin^3(x)}{x^3}$$
Hence, we obtain that
$$M = \dfrac{M}9 + \dfrac4{27} \implies \dfrac{8M}9 = \dfrac4{27} \implies M = \dfrac16$$
Similarly, we have
\begin{align}
N & = \lim_{x \to 0} \dfrac{3x-\tan(3x)}{(3x)^3} = \lim_{x \to 0} \dfrac{3x-\dfrac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}}{27x^3} = \lim_{x \to 0} \dfrac{3x-3\tan(x)-9x\tan^2(x)+\tan^3(x)}{27x^3(1-3\tan^2(x))}\\
& = \dfrac19 \cdot \lim_{x \to 0}\dfrac{x-\tan(x)}{x^3} \cdot \dfrac1{1-3\tan^2(x)} + \lim_{x \to0}\left(\dfrac1{27}\cdot\dfrac{\tan^3(x)}{x^3} - \dfrac13 \cdot \dfrac{\tan^2(x)}{x^2}\right)\cdot \dfrac1{1-3\tan^2(x)}\\
& = \dfrac{N}9 + \dfrac1{27}-\dfrac13 = \dfrac{N}9 - \dfrac8{27}
\end{align}
Hence, we obtain that
$$N = \dfrac{N}9 - \dfrac8{27} \implies \dfrac{8N}9 = -\dfrac8{27} \implies N = -\dfrac1{3}$$
Hence, $$L = \dfrac{M}N = \dfrac{1/6}{-1/3} = - \dfrac12$$
| {
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$\int\dfrac{\cos^2 x}{1+\sin^2 x}dx$ I have trouble finding the integral of
$$\int_0^{\frac{\pi}{2}}\dfrac{\cos^2 x}{1+\sin^2 x}dx$$
But the answer I'm getting is the negative value of the correct answer. I substituted $\tan x=t$;
| $$\int\dfrac{\cos^2 x}{1+\sin^2x}dx=\int\dfrac{\cos^2 x}{\cos^2x+2\sin^2x}dx=\int\dfrac{dx}{1+2\tan^2x}=\int\dfrac{(2+2\tan^2 x)-(1+2\tan^2 x)}{1+2\tan^2x}dx=\int\dfrac{2\sec^2x}{1+2\tan^2x}dx-\int dx$$
$$=\int\dfrac{d(\tan x)}{\left(\frac{1}{\sqrt{2}}\right)^2+\tan^2 x}-\int dx=\sqrt{2}\tan^{-1}(\sqrt{2}\tan x)-x+C$$
| {
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Limit of $(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}$, where do I get it wrong? I am tasked with evaluating
$$\lim_{(x, y) \to (0, 0)} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}.$$
We can rewrite the expression it as follows:
$$
e^{\ln(1+x^2+y^2)\cdot\frac{1}{x^2+y^2+x y^2}}
$$
Now, my first guess was that this tends to $1$. But that is apparently incorrect, it tends towards $e$, which one can see just by graphing it. That is also what the book says.
Now, we have the "standard limit":
$$\lim_{x \to 0^+} \frac{\ln(1+x)}x = 1$$
So I can kinda see how we can get the power to tend toward $1$ so we get $e^1$, but the expression is not exactly the same, we have $xy^2$ tacked on to the end of the denominator of the fraction. Furthermore, $xy^2$ doesn't have to be positive.
I tried using the substitution $x=\frac{1}{y^2}$, that gives us an expression that tends towards infinity both at $0$ and infinity and is never zero.
Can anyone point out to me what I'm missing. It feels like it should be super-obvious.
| Let $x^2+y^2=r^2 .$ We have $$|x|<a\implies |x y^2|\leq a y^2\leq a(x^2+y^2)=a r^2 .$$ So let $xy^2=r^2(1+b)$ where $b\to 0$ as $r\to 0 .$ Observe that this implies that $x^2+y^2+xy^2>0$ for sufficiently small positive $r.$ So, for all sufficiently small $r>0$ we have $$\log (1+x^2+y^2)^{1/(x^2+y^2+x y^2)}=$$ $$ =\frac {1} {1+b} \frac {1}{r^2}\log (1+r^2)$$ which tends to $1$ as $r\to 0.$
| {
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"url": "https://math.stackexchange.com/questions/1579120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$ Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$
What method to use in order to find the closed form of summation $\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}$?
| METHOD 1: Use the Euler-Maclaurin Formula
While the easier way is perhaps to recognize the limit as a Riemann sum, I thought it might be instructive to present a way forward that used the Euler-Maclaurin Formula. To that end, we write
$$\begin{align}
\sum_{k=1}^n\sqrt{1+\frac kn}&=\int_0^n \sqrt{1+\frac xn}\,dx+\frac12 \left(\sqrt{2}-1\right)+\frac1{24\,n}\left(\frac{\sqrt{2}}{2}-1\right)+O\left(\frac{1}{n^2}\right)\\\\
&=\frac{2n}{3}\left(2^{3/2}-1\right)+\frac12 \left(\sqrt{2}-1\right)+\frac1{24\,n}\left(\frac{\sqrt{2}}{2}-1\right)+O\left(\frac{1}{n^2}\right)
\end{align}$$
Dividing by $n$ and taking the limit produces the expected result!
METHOD 2: Use the Squeeze Theorem
Here we use the squeeze theorem. Note that the sum of interest is bounded as
$$\int_0^n\sqrt{1+\frac xn}\,dx\le \sum_{k=1}^n\sqrt{1+\frac kn}\le \int_1^{n+1} \sqrt{1+\frac kn}\,dx$$
Carrying out the integrals yields
$$\frac{2n}{3}\left(2^{3/2}-1\right)\le \sum_{k=1}^n\sqrt{1+\frac kn}\le \frac{2n}{3}\left(2^{3/2}\left(1+\frac1{2n}\right)-\left(1+\frac1{n}\right)\right)$$
whereupon dividing by $n$ and invoking the Squeeze Theorem yields
$$\lim_{n\to \infty}\frac1n \sum_{k=1}^n\sqrt{1+\frac kn}=\frac23 \left(2^{3/2}-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582557",
"timestamp": "2023-03-29T00:00:00",
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If $\small {x+\sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } = 4}$, solve for $x$. I came across this olympiad algebra problem, asking to solve for $x$:
$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$
Here was my try:
If $$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$$
Then $\quad \sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } =4-x$.
Further, I tried squaring the equation on both sides, but that doesn't seem to solve my problem. Please help.
Thank you.
| Hint: Observe that $:x = (x+1) + (x+2) - (x+3)$. This leads us to letting $a = \sqrt{x+1}, b = \sqrt{x+2}, c = \sqrt{x+3} \Rightarrow a^2+b^2-c^2 + ab+bc+ca = 4, b^2-a^2 = 1 = c^2-b^2 \Rightarrow a^2-1 +ab+bc+ca= 4 \Rightarrow (a+b)(a+c) = 5 \Rightarrow \dfrac{a+c}{b-a} = 5 \Rightarrow a+c = 5b-5a\Rightarrow 6a = 5b-c\Rightarrow 6\sqrt{x+1} = 5\sqrt{x+2} - \sqrt{x+3}$. Can you continue?
| {
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The minimum number of non real roots of the equation $x^4-2x^3+2x^2-x=k$ is?
The minimum number of non real roots of the equation
$x^4-2x^3+2x^2-x=k$ is?
k is any real number.
I plotted this on https://www.desmos.com/calculator/vpfpjwyxz8.It seems that the answer will be 2.But how to solve it manually?
| The derivative of $p(x) = x^4-2x^3+2x^2-x$ is $4 x^3-6 x^2+4 x-1= (2 x-1) (2 x^2-2 x+1)$, $<0$ on $(-\infty, 1/2)$, $0$ at $x=1/2$, and $>0$ on $(1/2, \infty)$. Moreover, $\lim_{x\ \to -\infty} p(x) = \lim_{x\ \to \infty} p(x)= +\infty$. Now, $f(1/2) =-3/16$. Therefore, the number of real solutions of the equation $p(x) = k$ is $0$ if $k < -3/16$, $1$ if $k = -3/16$, and $2$ if $k > -3/16$. Thus, for every $k$ the equation $p(x)=k$ has at least $2$ non-real solutions. Well, in fact we'd better check whether those complex roots are not multiple. The zeroes of $p'(x)$ are $1/2 \pm i/2$ and $p(1/2 \pm i/2) = -1/4$ ( $< -3/16$ of course),so yes, the equation $p(x) = -1/4$ has two non-real roots ( conjugate, each multiplicity $2$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int \frac{8 dx}{3 \cos 2x + 1}$ to arctan rather than log [Edit: It is becoming increasingly likely that the expected answer containing arctan might be a typo from my book, which is transcribed correctly here, so an answer containing log might be correct after all. Sorry about that and thanks for the answers so far, I've learned a lot!]
I'm trying to integrate:
$$
\int \frac{8 dx}{3 \cos 2x + 1} \\
$$
So I tried substituting $x$ for $z$:
$$
z = \tan \frac{x}{2} \\
\cos x = \frac{1 - z^2}{1 + z^2} \\
dx = \frac{2 dz}{1 + z^2} \\
$$
Then I tried replacing $\cos 2x$ with $2 \cos^2 x - 1$ followed by some work:
$$
\int \frac{4(1 + z^2) dz}{z^4 - 4z^2 + 1} \\
$$
If I try substituting $z$ back to $x$, it seems my answer will contain log rather than the expected arctan:
$$
\frac{x}{3} - \frac{5}{6} \arctan (2 \tan \frac{x}{2}) + C \\
$$
Did I take a wrong turn somewhere?
| Notice, the steps of substitution you showed are correct but your final answer is wrong. It shouldn't be in terms of $\arctan$
if you let $z=\tan \frac{x}{2}\implies dx=\frac{2\ dz}{1+z^2}$ then $$\int \frac{8\ dx}{3\cos 2x+1}$$
$$=\int \frac{8\ dx}{3(2\cos^2x-1)+1}=\int \frac{4\ dx}{3\cos^2x-1}$$
$$=\int \frac{4}{3\left(\frac{1-z^2}{1+z^2}\right)^2-1}\frac{2\ dz}{1+z^2}$$
$$=\int \frac{4}{3\left(\frac{1-z^2}{1+z^2}\right)^2-1}$$
$$=\int \frac{4(1+z^2)}{z^4-4z^2+1}\ dz$$
$$=4\int \frac{1+\frac{1}{z^2}}{\left(z-\frac 1z\right)^2-2}\ dz$$
$$=4\int \frac{d\left(z-\frac{1}{z}\right)}{\left(z-\frac 1z\right)^2-(\sqrt 2)^2}$$
$$=\frac{4}{2\sqrt 2}\ln\left|\frac{z-\frac{1}{z}-\sqrt 2}{z-\frac{1}{z}+\sqrt 2}\right|+C$$
substituting back $z=\tan \frac x2$,
$$=\sqrt 2\ln \left|\frac{\tan\frac{x}{2}-\cot\frac{x}{2}-\sqrt 2}{\tan\frac{x}{2}-\cot\frac{x}{2}+\sqrt 2}\right|+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving functional equation $f(x)\cdot f(y)-xy=f(x)+f(y)-1$ As in the title. Substituting $y=x$ we get:
$$ f(x)^2-x^2=2f(x)-1 $$
after rearranging, we get:
$$ f(x)(f(x)-2)=(x+1)(x-1) $$
And I rather cannot assume that e.g. $f(x)=x+1$ and $f(x)-2=x-1$, so what should I do now?
| Note that in your first step you had the following equation:
$$f(x)^2-x^2=2f(x)-1$$
Instead of factoring, if we rearrange all the terms on one side we get:
$$f(x)^2 - 2f(x) - x^2 + 1 = 0$$
Note we have a quadratic, with $f(x)$ acting like our $x$ and the term $1-x^2$ being our constant term.
Solving, we get:
$$f(x) = \frac{2 \pm \sqrt{4 -4(1-x^2)}}{2}$$
$$ = \frac{2 \pm \sqrt{4 - 4 + 4x^2)}}{2}$$
$$ = \frac{2 \pm \sqrt{4x^2}}{2}$$
$$ = \frac{2 \pm 2x}{2}$$
Therefore, we get $2$ solutions for $f(x)$:
$$f(x) = \frac{2 + 2x}{2} = 1+x$$
$$f(x) = \frac{2 - 2x}{2} = 1-x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$ $\displaystyle \int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$
I had been told that $x =\displaystyle \frac{1-y}{1+y}$ solves it, but it takes me back to where I started.
| Using the suggested substitution $x = \dfrac{1-y}{1+y}$, we get
$\dfrac{3(1+x)}{1-2x-x^2} = -\dfrac{3(1+y)}{1-2y-y^2} \ \ \ \ \ \ \ \ \dfrac{1}{1+x^2} = \dfrac{(1+y)^2}{2(1+y^2)} \ \ \ \ \ \ \ \ dx = \dfrac{-2}{(1+y)^2}\,dy$.
and thus,
$I = \displaystyle \int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2}$
$= \displaystyle \int_1^0 \tan^{-1}\left(-\frac{3(1+y)}{1-2y-y^2}\right)\cdot \dfrac{(1+y)^2}{2(1+y^2)} \cdot \dfrac{-2}{(1+y)^2}\,dy$
$= \displaystyle \int_0^1 \tan^{-1}\left(-\frac{3(1+y)}{1-2y-y^2}\right) \frac{dy}{1+y^2}$
$= \displaystyle \int_0^1 \tan^{-1}\left(-\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2}$.
Now, assuming we use a definition of $\arctan$ whose range is $[0,\tfrac{\pi}{2}) \cup (\tfrac{\pi}{2},\pi)$ instead of the usual $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$ then we have $\arctan(A) + \arctan(-A) = \pi$. Hence,
$2I = I + I = $ $\displaystyle \int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} + \int_0^1 \tan^{-1}\left(-\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2}$
$= \displaystyle \int_0^1 \left[\tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right)+\tan^{-1}\left(-\frac{3(1+x)}{1-2x-x^2}\right)\right]\frac{dx}{1+x^2}$
$= \displaystyle\int_0^1 \pi \dfrac{dx}{1+x^2}$
$= \pi \cdot \dfrac{\pi}{4} = \dfrac{\pi^2}{4}$.
Since $2I = \dfrac{\pi^2}{4}$, we have $I = \dfrac{\pi^2}{8}$.
Remark: If we use the usual definition of $\arctan$ whose range is $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$, then we get $I = 0$, which is what Wolfram Alpha suggests as the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation: $\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$ How to solve the equation
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
Can anyone give me some hints?
| HINT: $$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
$$\cos^2(x)+\cos^2(3x)=1-\cos^2(2x)$$
$$\cos^2(x)+\cos^2(3x)=(1-\cos(2x))(1+\cos(2x))$$
$$\cos^2(x)+\cos^2(3x)=(2\sin^2(x))(2\cos^2(x))$$
$$\cos^2(x)+(4\cos^3(x)-3\cos (x))^2=2(1-\cos^2(x))(2\cos^2(x))$$
$$16\cos^6x-20\cos^4x+6\cos^2x=0$$
now, let $\cos^2 x=t$, $$8t^2-10t+3t=0$$
$$t(2t-1)(4t-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Question on circles inscribed in a rectangle In rectangle $ABCD$, $AB = 8$ and $BC = 20$. Let $P$ be a point on $AD$ such that angle $\measuredangle BPC = 90^\circ$. If $r_1, r_2, r_3$ are the radii of the incircles of triangles $APB, BPC$ and $CPD$, what is the value of $r_1 + r_2 + r_3$?
| Given the inradius equation for a right triangle, as hinted by @Joey, just replace the variables:
$$
\begin{eqnarray}
r_1 &=& \frac{AB + AP - BP}{2} \\
r_2 &=& \frac{CD + DP - CP}{2} \\
&=& \frac{AB + (BC - AP) - CP}{2} \\
r_3 &=& \frac{BP + CP - BC}{2} \\
\end{eqnarray}
$$
And then simplify:
$$
\begin{eqnarray}
R &=& r_1 + r_2 + r_3 \\
&=& \frac{AB + AP - BP}{2} +
\frac{AB + (BC - AP) - CP}{2} +
\frac{BP + CP - BC}{2} \\
&=& AB \\
\end{eqnarray}
$$
So the total radii of the incircles is 8.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z$ and $\omega$ are two complex no. and $\theta = \arg\left(\frac{\omega -z}{z}\right)\;,$ Then Max. of $\tan^2 \theta$
If $\omega$ and $z$ are two complex number such that $|\omega| = 1$ and $|z|=10$ and Let $\displaystyle \theta = \arg\left(\frac{\omega -z}{z}\right)$
Then Maximum possible value of $\tan^2 \theta$
$\bf{My\; Try::}$ Let $\omega = =e^{i\alpha} = \cos \alpha+i\sin \alpha$ and $z = 10e^{i\beta}=10(\cos \beta+i\sin \beta)$
Given $\displaystyle \theta = \arg\left(\frac{\omega -z}{z}\right)$
Now $$\displaystyle \frac{\omega-z}{z} = \frac{\omega}{z}-1 = \frac{1}{10}e^{i(\alpha-\beta)}-1 = \frac{1}{10}\left[\cos (\alpha-\beta) +i\sin (\alpha-\beta) \right]-1$$
So we get $$\displaystyle \frac{\omega}{z}-1 = \frac{1}{10}\cos (\alpha -\beta)-1+i\frac{1}{10}\sin (\alpha-\beta)$$
So $$\displaystyle \tan \theta = \frac{\sin (\alpha-\beta)}{\cos(\alpha - \beta)-10}$$
| $f(x)=\frac{sinx}{cosx-10}$ is $2\pi$ periodic and is decreasing in first quadrant and fourth quadrants and increasing in second and third. so obviously max of $tan^2 \theta$ occurs at $x=\frac{\pi}{2}$ Or $x=\frac{3\pi}{2}$ so max is $\frac{1}{100}$
| {
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How to Simplify $-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$? A problem requires finding the value of x, such that
$$-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$$
The site reduces the problem to $\tan(\frac x4) = \sqrt{3}$.
How one should do this?
| Simplify:
$$
\begin{aligned}
-\frac14\sin\frac x4 + \frac{\sqrt3}4\cos\frac x4 &=\frac12\left(-\frac12\sin\frac x4 + \frac{\sqrt{3}}2\cos\frac x4\right)\\
&=\frac12\left(-\sin\frac\pi6\sin\frac x4+\cos\frac\pi6\cos\frac x4\right)\\
&=\frac12\cos(\frac\pi6+\frac x4)
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find an implicit function from this relation? This is from Tenenbaum & Pollard's "Ordinary Differential Equations" book, chapter 1, Exercise 2, problem # 14:
Find the function g(x) that is implicitly defined by the relation:
(1) $\sqrt{x^2 -y^2} + \arccos{(x/y)} = 0, y\neq0$
The answer is $y = g(x) = x$, and I have no idea how they got that.
My attempt:
$\sqrt{x^2 -y^2} + \arccos{(\frac{x}{y})} = 0$
$\arccos{(\frac{x}{y})} = -\sqrt{x^2 -y^2}$
$\cos(\arccos{(\frac{x}{y})}) = \cos(-\sqrt{x^2 -y^2})$
$\cos(\arccos{(\frac{x}{y})}) = \cos(\sqrt{x^2 -y^2})$
$\frac{x}{y} = \cos(\sqrt{x^2 -y^2})$
$y = \frac{x}{\cos(\sqrt{x^2 -y^2})}$
I'm not sure where to go after that...
| Since $\sqrt{x^2-y^2}$ is defined, we have $|y|\leq|x|$.
Since $\arccos(x/y)$ is defined, we have $|y|\geq|x|$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
I was hoping some things would cancel out when I expanded this but nothing. I think using inequalities will help.
| So after reducing the equation the equation will look like that:
$$27\csc^6(x)\sec^2(x)=256$$
OR:
$$\sin^6(x)\cos^2(x)=\frac{27}{256}$$
Then:
$$\sin^6(x)-\sin^8(x)=\frac{27}{256}$$
Or:
$$\sin^8(x)-\sin^6(x)+\frac{27}{256}=0$$
I found that $\sin^2(x)-\frac{3}{4}=0$ is divisible above equation.I there used the features of high degree polynomials.
So the equation will look like in this way:
$$(\sin^2(x)-\frac{3}{4})(\sin^6(x)+\frac{1}{4}\sin^4(x)+\frac{3}{4}\sin^2(x)+\frac{27}{64})=0$$
And:
$$\sin^2(x)-\frac{3}{4}=0$$
The rest is for you.
| {
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"url": "https://math.stackexchange.com/questions/1591141",
"timestamp": "2023-03-29T00:00:00",
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How to compute $\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$? I have a problem with this limit, I dont know what method to use. After several attempts the eventually find was obtained ($\infty$). But in reality the result sought is $0$.
Can you explain the method and the steps used? Thanks.
$$\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$$
| If you naively compute the limit of the base and the exponent, you'll get $1^\infty$ which is undefined. Fortunately, this type of indeterminacy admits a classic approach:
$$\lim\limits_{x \to 0} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2+1}{x+1} -1 \right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2+1 -x -1}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2-x}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left[ \left(1 + \frac{x^2-x}{x+1}\right)^{\frac {x+1} {x^2 - x}} \right] ^{\frac{1}{\sin^2 x} \frac {x^2 - x} {x+1}} .$$
The quantity between square brackets is known to tend to $\rm e$, therefore it remains to compute the limit of the exponent:
$$\lim \limits _{x \to 0} \frac{1}{\sin^2 x} \frac {x^2 - x} {x+1} = \lim \limits _{x \to 0} \frac x {\sin x} \frac 1 {\sin x} \frac {x-1} {x+1} ;$$
the first fraction is known to tend to $1$, the third to $-1$ and the middle one to $\frac 1 {0_+} = \infty$, so the exponent tends to $-\infty$, and thus the whole expression tends to ${\rm e} ^{- \infty} = 0$.
| {
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Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$.
I have done the following:
Let $t=\tan \frac{x}{2}\in ]0,1[$, then using the $t$-formulae the LHS becomes
$$(1+\sec x)(1+\csc x) = \frac{2t+1+t^2}{t(1-t^2)}$$
I could simply it further but then I keep lowering the bound, now I continue as follows:
$$\frac{2t+1+t^2}{t(1-t^2)} = \frac{2}{1-t^2} + \frac{1}{t(1-t^2)}+\frac{t^2-1}{t(1-t^2)}+\frac{1}{t(1-t^2)}$$
Now since $1-t^2$ and $t \in ]0,1[$ then $\frac{1}{t(1-t^2)}> 1$ and so is $\frac{1}{1-t^2}$ then:
$$> 2+1+1-\frac{1}{t}>4$$
Close, but no cigar. Where could I improve the bounding?
| $$(1+\sec x)(1+\csc x) = 1+\sec x+\csc x+\frac{1}{\sin(x)\cos(x)}$$
where $f(x)=\sec x+\csc x$ is a convex function over $\left(0,\frac{\pi}{2}\right)$, with the property that $f(x)=f\left(\frac{\pi}{2}-x\right)$. It follows that $x\in\left(0,\frac{\pi}{2}\right)$ implies $f(x)\geq f\left(\frac{\pi}{4}\right)$. The same holds for $g(x)=\frac{1}{\sin(x)\cos(x)}$, hence it follows that:
$$ \begin{eqnarray*}x\in\left(0,\frac{\pi}{2}\right)\quad\Longrightarrow\quad (1+\sec x)(1+\csc x)&\geq& 1+ f\left(\frac{\pi}{4}\right)+g\left(\frac{\pi}{4}\right)\\&=&3+2\sqrt{2}\\&>&5.\end{eqnarray*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. I have no idea how do this. Please help me.
| Notice that $a x^2 + b x + c = 0$ has solutions $\frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$ by the quadratic formula. Using this result and working backwards, we see that $x = \frac{(c+\sqrt{c^2+4})}{2}$ implies that $ x^2 -c x -1 = 0$. So $x^2 -1 = cx$ then dividing both sides by x, we get $\frac{x^2 - 1}{x}=c$ So $c = x - \frac{1}{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Let $|x|\leq1, |y|\leq 1$. Show that$0 \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy \sqrt{1-x^{2}} \sqrt{1-y^{2}} \leq 1$ I would appreciate if somebody could help me with the following problem
Let $|x|\leq1, |y|\leq 1$. Show that
$$
0 \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy \sqrt{1-x^{2}} \sqrt{1-y^{2}} \leq 1
$$
| We need only note that
\begin{align*}
x^{2}+y^{2} -2x^{2} y^{2}+2xy \sqrt{1-x^{2}} \sqrt{1-y^{2}} &= \left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\,\right)^2\\
&\le \left(\sqrt{(x^2+y^2)(2-x^2-y^2)}\,\right)^2\\
&\le \left(\frac{1}{2}(x^2+y^2 + 2-x^2-y^2)\,\right)^2\\
&=1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculation of the Jordan canonical form Given the matrix $ F= \begin{bmatrix}
3 & -1 & 0 \\
1 & 1 & -2 \\
0 & 0 & 2
\end{bmatrix}$ calculate the Jordan canonical form such that $F = T F_j T^{-1}$.
The characteristic polynomial is $ (\lambda -2)^3= 0$ so the eigenvalue is $\lambda = 2$
The eigenvectors $v$ are given by $(F - \lambda I)v = 0$ so $ \begin{bmatrix}
1 & -1 & 0 \\
1 & -1 & -2 \\
0 & 0 & 0
\end{bmatrix}v = 0 $ so the kernel is $ < \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}>$
Now I need to calculate the kernel of $(F - \lambda I)^2$
$(F - \lambda I)^2=\begin{bmatrix}
1 & -1 & 0 \\
1 & -1 & -2 \\
0 & 0 & 0
\end{bmatrix}*\begin{bmatrix}
1 & -1 & 0 \\
1 & -1 & -2 \\
0 & 0 & 0
\end{bmatrix}=\begin{bmatrix}
0 & 0 & 2 \\
0 & 0 & 2 \\
0 & 0 & 0
\end{bmatrix}$
$(F - \lambda I)^2 v_2 = 0$ so the kernel is $<\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} ><\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} >$
So far my calculations should be right but now come the problems.
The matrix $T $ should be $\begin{bmatrix}
1 & 0 & x_1 \\
0 & 1 & x_2 \\
0 & 0 & x_3
\end{bmatrix}$ where the third column is:
$(F - \lambda I) \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} = \begin{bmatrix}
1 & -1 & 0 \\
1 & -1 & -2 \\
0 & 0 & 0
\end{bmatrix} \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} = \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}$
But this value makes the matrix $T$ not invertible. Where is my mistake?
| If we don't care about finding the change of basis matrix $T$, then we can arrive at the Jordan form quickly.
First, note that the characteristic polynomial of $F$ is
$$
\chi_F(t)=(t-2)^3
$$
This implies that $2$ is the only eigenvalue of $F$.
Next, note that
$$
\dim\DeclareMathOperator{null}{null}\null(F-2\,I)=1
$$
That is, the eigenvalue $2$ has geometric multiplicity one. This means that the Jordan form of $F$ has one Jordan block. Hence the Jordan form of $F$ is
$$
\begin{bmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{bmatrix}
$$
Of course, the change of basis matrix can be computed using @Fabian's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I integrate this expression I looked over on the internet but I don't know hot to integrate this... Can someone give a hint
$$ \int \frac {\sqrt x} {x^2 -x + 1} dx $$
| $$\int\dfrac{\sqrt x dx}{x^2-x+1} =\int\dfrac{2xd\sqrt x}{x^2-x+dx} = \int \dfrac{2u^2du}{u^4-u^2+1}, \text { where } u = \sqrt x.$$
HINT:
$$\dfrac{2u^2}{u^4-u^2+1} = \dfrac{2u^2}{u^4+u^2+1 -3u^2} = $$$$\dfrac{2u^2}{(u^2 -\sqrt 3u +1)(u^2 + \sqrt3 u +1)}$$$$=\dfrac {Au+B}{u^2 -\sqrt 3u +1} + \dfrac {Cu+D}{u^2 + \sqrt 3u +1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simultaneous Equations (Stuck on the algebra)
Question: Solve the following simultaneous equations for real values of x and y
$$
\left\{
\begin{array}{l}
9^{2x+y} - 9^x \times 3^y = 6 \\
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
\end{array}
\right.
$$
What I have attempted; for the first equation
$$
9^{2x+y} - 9^x \times 3^y = 6 \\
9^{2x+y} - 3^{2x} \times 3^y = 6 \\
9^{2x+y} - 3^{2x+y} - 6 = 0
$$
Let $z = 3^{2x+y}$. Then
$$
z^2 - z - 6 = 0
\iff (z-3)(z+2) = 0
\iff z = 3, z \ne -2
$$
where
$$
3^{2x+y} = 3 \iff y = 1-2x.
$$
Now using the second equation I get
$$
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
$$
Substituting $y = 1-2x$
$$
\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3.
$$
Now this is the part I am stuck on , how do I solve for $x$ algebraically?
| You finished with $$\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3$$ Go to natural logarithms (the only ones I know !); this will then write $$\frac{\log (4-2 x)}{\log (x+1)}+\frac{\log (5-x)}{\log (x+1)}=3$$ Multiply by the denominator appearing in the lhs (hoping that $x\neq -1$) and simplify.
So, $$\log (4-2 x)+\log (5-x)=3\log(x+1)$$ that is to say $$(4-2x)\times (5-x)=(x+1)^3$$ Expand and simplify to get $$x^3+x^2+17 x-19=0$$ where $x=1$ is an obvious solution. Perform the long division to get $$x^3+x^2+17 x-19=(x-1)(x^2+2 x+19)=0$$ The quadratic term does not show real solutions (in the complex domain, they would be $(x_{1,2}=-1\pm3 i \sqrt{2})$.
So $x=1$ is your solution and you were about to get it.
Happy New Year !!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proof of an infinite sum using Fourier Series I was revising for my calculus exam and I came across a question that asked to find the Fourier Series of $f(x)=1+x$, on $-1<x<1$, which I did. Which I found to be:
$$f(x) = 1+\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x)$$
I checked this by graphing it and it seemed to hold. However, the latter part of the question asks to evaluate the FS as $x=\frac{1}{2}$, to calculate:
$$\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{2m-1}$$
When I set $x=\frac{1}{2}$, I end up with:
$$\frac{3}{2} = 1+\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n\pi} \sin\bigg(\frac{n\pi}{2}\bigg)$$
$$\frac{\pi}{4} = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} \sin\bigg(\frac{n\pi}{2}\bigg)$$
However I cannot transform the sine function into something else, any help?
| If you look at the sine graph then you'll see that
$$\begin{eqnarray*}
\sin(0) &=& 0 \\ \\
\sin\left(\frac{\pi}{2}\right) &=& +1 \\ \\
\sin\left(\frac{2\pi}{2}\right) &=& 0 \\ \\
\sin\left(\frac{3\pi}{2}\right) &=& -1 \\ \\
\sin\left(\frac{4\pi}{2}\right) &=& 0 \\ \\
\sin\left(\frac{5\pi}{2}\right) &=& +1
\end{eqnarray*}$$
This pattern continues because $\sin(x+2\pi) \equiv \sin(x)$ for all real $x$. What we see is that for all integers $k \ge 1$, we have
$$\sin\left(\frac{n\pi}{2}\right) =
\left\{ \begin{array}{ccc}
(-1)^{k+1} & : & n =2k-1 \\
0 & : & \mathrm{otherwise}
\end{array}\right.$$
This means you can ignore all of the terms in your sequence for which $n$ is even, and only consider the terms for which $n$ is odd. This gives:
$$\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\,\sin\left(\frac{n\pi}{2}\right) &=& \sum_{k=1}^{\infty} \frac{(-1)^{(2k-1)+1}}{(2k-1)}\,\sin\left(\frac{(2k-1)\pi}{2}\right) \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{2k}(-1)^{k+1}}{(2k-1)} \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{3k+1}}{(2k-1)} \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k-1)}
\end{eqnarray*}$$
In the last step, I used the fact that, for all $k \in \mathbb N$,
$$(-1)^{3k+1}=(-1)^{2k}\cdot(-1)^{k+1} = \left[(-1)^2\right]^k \cdot (-1)^{k+1} = [+1]^k\cdot (-1)^{k+1} = (-1)^{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
| If you look at where the $0$'s are in your matrix, it is fairly easy to see that your matrix becomes block-triangular when transformed to (for instance) the ordered basis $[e_1,e_3,e_2,e_4]$, in other words the standard basis but with the second and third vectors interchanged. The change of basis swaps the second and third rows and the second and third columns, and gives you
$$A'=
\pmatrix{5&2&-7&2\\-5&0&-8&3\\0&0&3&-4\\0&0&5&-6}.
$$
Since change of basis does not affect the determinant, and the determinant of a block-triangular matrix is the product of the determinants of the diagonal blocks, you get
$$
\det(A)=\det(A')
=\left|\matrix{5&2\\-5&0}\right|\times\left|\matrix{3&-4\\5&-6}\right|
=10\times2=20.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
calculating a limit of sequence with tan Calculating a limit
$$\lim_{n\longrightarrow +\infty} \dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}$$
thanks.
| First write
$$\tan\left(\frac{\pi n}{2n+1}\right)=\cot\left(\frac{\pi/2}{2n+1}\right)$$
Then, in THIS ANSWER I showed that from basic geometry the cotangent function satisfies the inequalities
$$\frac{\cos(x)}{x}\le \cot(x)\le \frac1x$$
for $0 < x\le \pi/2$.
Therefore, we have
$$\frac{(2n+1)\cos\left(\frac{\pi/2}{2n+1}\right)}{\frac\pi 2 n\left(1+\frac2{n^2}+\frac{1}{n^3}\right)^{1/3}}\le \frac{\tan\left(\frac{\pi n}{2n+1}\right)}{\left(n^3+2n-1\right)^{1/3}}\le\frac{2n+1}{\frac\pi 2 n\left(1+\frac2{n^2}+\frac{1}{n^3}\right)^{1/3}}$$
whereupon invoking the Squeeze Theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\tan\left(\frac{\pi n}{2n+1}\right)}{\left(n^3+2n-1\right)^{1/3}}=\frac{4}{\pi}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Verify Trigonometric Identities $1/\sin50^\circ + √3/\cos50^\circ=4$
I have tried it as:
LHS
$(\cos50+√3 \sin50)/\sin50\cos50$
$(2\cos50+2√3 \sin50)/2\sin50\cos50$
$(2\cos50+2√3 \sin50)/(\sin100)$
Now, whats next??
| \begin{align}
& \frac 1 {\sin50^\circ} + \frac{\sqrt 3}{\cos50^\circ} = 2 \left( \frac {1/2} {\sin50^\circ} + \frac{\sqrt 3/2}{\cos50^\circ} \right) = 2\left( \frac{\sin30^\circ}{\sin50^\circ} + \frac{\cos 30^\circ}{\cos50^\circ} \right) \\[10pt]
= {} & 2\cdot \frac{\sin30^\circ\cos50^\circ + \cos30^\circ\sin30^\circ}{\sin50^\circ\cos50^\circ} = 4\cdot \frac{\sin(30^\circ+50^\circ)}{2\sin50^\circ\cos50^\circ} \\[10pt]
= {} & 4\cdot \frac{\sin(30^\circ+50^\circ)}{\sin(2\cdot50^\circ)} = 4\cdot\frac{\sin80^\circ}{\sin100^\circ}
\end{align}
Now recall why $\sin80^\circ = \sin100^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Locus of intersection of two lines If the tangent at any point P of a circle $x^2 + y^2 = a^2$ meets the tangent at a fixed point A $(a,0)$ in T and T is joined to B , the other end of the diameter through A . Then we have to prove that the locus of intersection of AP and BT is an ellipse whose eccentricity is $\frac{1}{\sqrt{2}}$
I tried ,
Let P$ (h,k)$ on the circle .
The equation of tangent is $hx + ky = a^2 $ and through A it is $x = a$
So the point is $(a , \frac{a^2 - ha}{ k })$
Equation of AP is $( h-a)y = k(x-a)$
Equation of BT is$ ky = (a-h ) ( x+a) $
Then I tried to solve both the equation , but I am not getting the result .
Thank you in advance
|
$$\begin{align}
\triangle CDB \sim \triangle TAB \qquad\to\qquad \frac{y}{a+x} \;=\; \frac{t}{2a} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{2y}{a+x} \\[6pt]
\triangle CDA \sim \triangle OAT \qquad\to\qquad \frac{y}{a-x} \;=\; \frac{\;a\;}{t} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{a-x}{y}
\end{align}$$
$$ $$
$$\frac{2y}{a+x} = \frac{a-x}{y} \quad\to\quad x^2 + 2y^2 = a^2 \quad\to\quad \frac{x^2}{a^2} + \frac{y^2}{a^2/2} = 1$$
$$\to\quad \text{eccentricity }^2 \;=\; \frac{a^2 - a^2/2}{a^2} \;=\; \frac{1}{2} \qquad\square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck with LDU-factorization of a matrix where D should contain zeros I thought that L-D-U- factorization of a square matrix (L=lower triangular factor, D=diagonal factors, U=upper triangular factor) was always possible and meaningful even if I encounter zeros on the diagonal factor D. But my algorithm is not correct in that cases in that the resulting factors do not reproduce the source.
Let $$ M = \begin{bmatrix}
1& 2& 3& 4& 5\\
2& 4& 6& 8& 0\\
3& 6& 9& 2& 5\\
4& 8& 2& 6& 0\\
5& 0& 5& 0& 5
\end{bmatrix}$$
which is just the top-left of the basic $10 \times 10$ multiplication-table (modulo $10$). It has full rank; but in the naive LDU-algorithm one would assign zeros to some entries of D because zeros occur in the top-left element of the intermediate matrices of the iteration. If I do that, I get LDU-components
$$\small L=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 \\
3 & 0 & 1 & 0 & 0 \\
4 & 0 & 0 & 1 & 0 \\
5 & 0 & 0 & 2 & 1 \\
\end{bmatrix}
D= \begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -10 & 0 \\
0 & 0 & 0 & 0 & 20 \end{bmatrix}
U=\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}$$
and if I put them together they don't reproduce the source M:
$$ \text{chk}=L \cdot D \cdot U = \small \begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 6 & 8 & 10 \\
3 & 6 & 9 & 12 & 15 \\
4 & 8 & 12 & 6 & 0 \\
5 & 10 & 15 & 0 & 5
\end{bmatrix}$$
The problem is not, that the matrix-rank of M were not sufficient - Pari/GP gives the inverse and even the diagonalization.
Q: Can this be repaired? Can a meaningful L-D-U-decomposition be given?
Of course, if a general argument exists why and when invertible/diagonalizable matrices cannot be LU or LDU-decomposed I'd like to learn that, too.
| You have to do this by pivoting. In the second step of the LU or LDU decomposition, your matrix becomes
M = \begin{bmatrix}
1& 2& 3& 4& 5\\
0& 0& 0& 0& -10\\
0& 0& 0& -10& -10\\
0& 0& -10& -10& -20\\
0& -10& -10& -20& -20
\end{bmatrix}
You will then need to multiply it by a permutation matrix $P=$
\begin{bmatrix}
1& 0& 0& 0& 0\\
0& 0& 0& 0& 1\\
0& 0& 0& 1& 0\\
0& 0& 1& 0& 0\\
0& 1& 0& 0& 0
\end{bmatrix}
Then your lower triangular matrix $L_P$ becomes $PLP^{-1}$, and the $LDU$ decomposition formula is $PA=L_PDU$. Refer to this article about LU with pivoting.
Your $L$ looks almost correct, but there isn't a $2$ in the 5th row and 4th column. The $U$ should be
\begin{bmatrix}
1& 2& 3& 4& 5\\
0& 1& 1& 2& 2\\
0& 0& 1& 1& 2\\
0& 0& 0& 1& 1\\
0& 0& 0& 0& 1
\end{bmatrix}
And $D=$
\begin{bmatrix}
1& 0& 0& 0& 0\\
0& -10& 0& 0& 0\\
0& 0& -10& 0& 0\\
0& 0& 0& -10& 0\\
0& 0& 0& 0& -10
\end{bmatrix}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\dfrac{x^3+y^3+z^3}{3} \geq xyz + \dfrac{3}{4}|(x-y)(y-z)(z-x)|$
For nonnegative real numbers $x,y,$ and $z$ prove that $$\dfrac{x^3+y^3+z^3}{3} \geq xyz + \dfrac{3}{4}|(x-y)(y-z)(z-x)|.$$
The absolute value sign seems to make this difficult. Should I prove this in two cases? One for $(x-y)(y-z)(z-x) \geq 0$ and one for $(x-y)(y-z)(z-x) < 0$?
| First from AM-GM inequality,
$$
\left(|(x-y)(y-z)(z-x)|\right)^{1/3}
\leq \frac{|x-y|+|y-z|+|z-x|}{3}\leq \frac{2}{3}(x+y+z)
$$
and
\begin{align}
\left(|(x-y)(y-z)(z-x)|\right)^{2/3}
&=\left(|(x-y)^2(y-z)^2(z-x)^2|\right)^{1/3}\cr
&\leq \frac{(x-y)^2+(y-z)^2+(z-x)^2}{3}\cr
&=\frac{2}{3}(x^2+y^2+z^2-xy-yz-xz).
\end{align}
Therefore,
\begin{align}
\frac{x^3+y^3+z^3}{3}-xyz&=\frac{1}{3}(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&\geq \frac{3}{4}|(x-y)(y-z)(z-x)|.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The number of real roots of $x\lfloor x\rfloor+187=\lfloor x^{2}\rfloor+\lfloor x\rfloor$ Find the number of real roots of the equation $x\lfloor x\rfloor+187=\lfloor x^{2}\rfloor+\lfloor x\rfloor$.
I search other methods except checking case by case.
| For
$x[x]+n=[x^{2}]+[x]
$,
the solutions are
$x = m+\frac{n-m}{m}$
for
$n/2 < m \le n$.
This problem is $n=187$.
Let
$[x] = m$
and
$y = x-m$,
$0 \le y < 1$.
$\begin{array}\\
(m+y)m+n
&=[(m+y)^2]+m\\
\text{or}\\
m^2+n+my
&=[m^2+2my+y^2]+m\\
&=m^2+m+[2my+y^2]\\
&=m^2+m+2my\\
\text{or}\\
n
&=m+my\\
\text{or}\\
my
&=n-m\\
\end{array}
$
Since
$0 \le y < 1$,
$y = k/m$
where
$0 \le k < m$.
Therefore
$k
= n-m
$.
Since $0 \le k < m$,
$n \ge m$
and
$n < 2m$,
so
$n/2 < m \le n$.
For any such $m$,
let
$x = m+\frac{n-m}{m}$.
Then
$\begin{array}\\
x[x]+n
&=m(m+\frac{n-m}{m})+n\\
&=m^2+(n-m)+n\\
&=m^2+2n-m\\
\text{and}\\
[x^2]+[x]
&=[(m+\frac{n-m}{m})^2]+[m+\frac{n-m}{m}]\\
&=m^2+2(n-m)+m\\
&=m^2+2n-m\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct
So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$.
Thank you for your help
| Note that
$$\begin{split}
3^{2(k+1)+1} + 2^{(k+1)+2} &= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\
&= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}).
\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Find the nth power of a matrix Let matrix A is
$$A=\left (\begin{array}{rrr}
1&0&0 \\
1&1&0\\
0&0&1
\end{array}\right)$$
How can I find the 30 th power of A..
Is diagonalization possible?
I found the eigen values But cannot continue.
| Notice that
$$
\begin{pmatrix}
1 & 0 \\
a & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
b & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
a+b & 1
\end{pmatrix}
\quad
\text{for all $a,b \in K$},
$$
where $K$ denotes the ground field. Therefore
$$
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}^n
=
\begin{pmatrix}
1 & 0 \\
n & 1
\end{pmatrix}.
$$
Now notice that your matrix is a block matrix of the form
$$
\begin{pmatrix}
1 & 0 & \\
1 & 1 & \\
& & 1
\end{pmatrix}
=
\begin{pmatrix}
A & \\
& B
\end{pmatrix}
$$
with
$$
A =
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\quad\text{and}\quad
B = (1).
$$
Therefore
$$
\begin{pmatrix}
1 & 0 & \\
1 & 1 & \\
& & 1
\end{pmatrix}^n
=
\begin{pmatrix}
A & \\
& B
\end{pmatrix}^n
=
\begin{pmatrix}
A^n & \\
& B^n
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & \\
n & 1 & \\
& & 1
\end{pmatrix}.
$$
PS: Alternatively one can realize that your matrix is an elementary matrix, namely the one adding the fist row to the second. Applying this $n$-times is the same as just adding the first row $n$-times to the second, which is represented by the matrix we calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all positive integers $n$ such that $n+2008$ divides $n^2 + 2008$ and $n+2009$ divides $n^2 + 2009$ I wrote
$$
\begin{align}
n^2 + 2008
&= (n+2008)^2 - 2 \cdot 2008n - 2008^2 + 2008 \\
&= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008^2 + 2008 \\
&= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008 \cdot 2009
\end{align}
$$
to deduce that $n+2008$ divides $n^2 + 2008$ if and only if $n+2008$ divides $2008 \cdot 2009$.
Similarly, I found that $n+2009$ divides $n^2 + 2009$ if and only if $n+2009$ divides $2009 \cdot 2010$.
I can't seem to get anywhere from here. I know that $n=1$ is an obvious solution, and I think it's the only one, although I can't prove/disprove this. I know that any two consecutive integers are coprime but I haven't been able to use this fact towards the solution.
Could anyone please give me any hints?
For the record, this is a question from Round 1 of the British Mathematical Olympiad.
| Here another approach
Since $(n+2008)|( n^{2} + 2008)$ and $(n+2009) |( n^{2} + 2009)$ then
$(n+2008)k = n^2 + 2008$ and $(n+2009)m = n^2 + 2009$ for some $k,m \in \mathbb{Z}$
$\textbf{Case 1:}$ $k=1$ or $m=1$
$(n+2008) = n^2 + 2008$ or $(n+2009) = n^2 + 2009$
in any case leads to $n^2-n=0$ that is $n=1$ and $n=0$
$\textbf{Case 2:}$ $m>k>1$
$$n^2+2009=(n+2009)m = (n+2008)m+m > (n+2008)k + m > n^2+2009$$ which is impossible
$\textbf{Case 3:}$ $k>m>1$
$$n^2+2008=(n+2009-1)k = (n+2009)k - k \ge (n+2009)(m+1) - k = (n+2009)m +(n+2009-k) \ge n^2+2009 $$ also impossible, the last inequality holds because $k< n+2009$, suppose not
$k \ge n+2009 >n+2008 \Rightarrow n^2 + 2008 =(n+2008)k > (n+2008)^2 > n^2 + 2008$ a contradiction
$\textbf{Case 4:}$ $k=m>1$
$$(n+2008)k+1=(n+2009)k \Longrightarrow k =1 \Longrightarrow n=1$$
So the solutions are $\boxed{n=1}$ and $\boxed{n=0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of the power series $\sum\limits_{n=2}^{\infty}\frac{n^2}{(n-1)(n+2)}x^n$ The series is convergent for $|x|<1$ and divergent for $|x|>1$.
I can't find the sum. Integrating three times gives
$$\frac{n^2}{(n^2-1)(n+3)(n+2)^2}x^{n+3}$$
that should have a closed form.
How to find the sum of this series?
| You can write
$$
\frac{n^2}{(n-1)(n+2)}=1+\frac{-n+2}{(n-1)(n+2)}=
1+\frac{1}{3}\frac{1}{n-1}-\frac{4}{3}\frac{1}{n+2}
$$
with partial fraction decomposition. So you reduce the problem to finding the sum of
$$
\sum_{n=2}^{\infty}x^n,
\quad
\sum_{n=2}^{\infty}\frac{x^n}{n-1},
\quad
\sum_{n=2}^{\infty}\frac{x^n}{n+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving a polynomial has only one positive root. Let $m, n, p\in \mathbb R$, $n>0, p>0$. Prove that the following equation has exactly one positive solution:
$$x^5-mx^3-nx-p=0.$$
Here is my attempt: Let $f(x)=x^5-mx^3-nx-p$, $f$ is continuous on $\mathbb R$ and $f(0)=-p<0, \lim_{x\to +\infty} f(x)=+\infty$. This implies, there exists $\xi>0$ such that $f(\xi)=0$ as a consequence of Bolzano-Cauchy theorem. Moreover,
$$f'(x)=5x^4-3mx^2-n.$$
Since $(3m)^2+20n>0$ and $-5n<0$, we can easily see that $f'(x)=0$ has two roots $$x=\pm \frac{3m+\sqrt{9m^2+20n}}{10}.$$
I cannot continue to verify that $f(x)=0$ has only one positive solution?
| Let $f(x)=x^5-mx^3-nx-p$.
$f(0)<0, \lim_{x\to\infty}f(x)=\infty$, so the IVT gives that $f$ has a positive root.
Let's call the least such root $a$, noting thus that $a^5=ma^3+na+p$
Then, for $x\ge 0, f'(x)=5x^4-3mx^2-n$.
Note that $f'(a)=5a^4-(3ma^2+n)>5a^4-3(ma^2+n)=5a^4-3(a^4-\frac{p}{a})=2a^4+\frac{3p}{a}>0$
Furthermore, for $x\ge a$, $f'(x)=5x^4-3mx^2-n\ge x^2(5a^2-3m)-n\ge a^2(5a^2-3m)-n$.
Now:
*
*$a^5=ma^3+na+p\implies a^2=m+\frac{n}{a^2}+\frac{p}{a^3}$
*$\implies a^2(5a^2-3m)=a^2(2m+\frac{5n}{a^2}+\frac{5p}{a^3})\ge 5n$
*$\implies f'(x)\ge5n-n=4n>0$ for $x\ge a$
So the function has a root at $a$, and increases to the right of $a$, giving uniqueness.
| {
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"url": "https://math.stackexchange.com/questions/1607742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Least value of $C$ for which $CT_n+1$ is always a triangular number.
For $n \geq 1$, the $n$th triangular number is defined as $T_n = 1+2+\cdots+n$. Let $C$ be the least positive integer such that $CT_n + 1$ is a triangular number for all positive integers $n$. How many factors does $C$ have?
Attempt
We have that $T_n = \dfrac{n(n+1)}{2}$ and $CT_n+1 = \dfrac{Cn(n+1)}{2}+1 = \dfrac{Cn(n+1)+2}{2} = \dfrac{k(k+1)}{2} \implies Cn(n+1)+2 = k(k+1)$. We must find a $C$ such that for every $n$ there exists an integer $k$ that satisfies the relation. How do I proceed?
| If
$m = \frac{n(n+1)}{2}
$,
$2m = n(n+1)
=n^2+n
$
so
$8m = 4n^2+4n
$
or
$8m+1 = 4n^2+4n+1
=(2n+1)^2
$.
For the reverse way,
if $8m+1$
is a square,
it must be the square
of an odd number,
so
$8m+1 = (2n+1)^2$,
and just do the above in reverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sin{\pi\sqrt{n}}$ Revising for an exam:
Let $a_n = \sin{(\pi\sqrt{n})}.$ Show that:
(i) $a_{n+1} - a_{n} \rightarrow 0$
(ii) The sequence $(a_n)$ is bounded.
(iii) $(a_n)$ does not converge.
My attempt:
(i) ???
(ii) min($\sin(x)$) = -1, max($\sin{x}$) = 1, so $-1 \leq a_n \leq 1, \forall n \in \mathbb{N}$. Thus 1 is an upper bound and -1 is a lower bound.
(iii) $a_n$ has a monotonic subsequence which converges to 1 by the Bolanzo-Weierstrass theorem. Note that the subsequence $a_{n^2}$ converges to 0. Since $0 \neq 1$, $a_n$ does not converge.
| (i) Since $\left|\sin(x)\right|\le\left|x\right|$
$$
\begin{align}
\left|a_{n+1}-a_n\right|
&=\left|\sin\left(\pi\sqrt{n+1}\right)-\sin\left(\pi\sqrt{n}\right)\right|\\[6pt]
&=2\left|\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}2\right)\sin\left(\pi\frac{\sqrt{n+1}-\sqrt{n}}2\right)\right|\\[3pt]
&\le2\cdot1\cdot\frac\pi2\left(\sqrt{n+1}-\sqrt{n}\right)\\[3pt]
&=\frac\pi{\sqrt{n+1}+\sqrt{n}}\\[3pt]
&\le\frac\pi{2\sqrt{n}}\tag{1}
\end{align}
$$
(ii) Since $\left|\sin(x)\right|\le1$, we have
$$
\begin{align}
\left|a_n\right|
&=\left|\sin\left(\pi\sqrt{n}\right)\right|\\
&\le1\tag{2}
\end{align}
$$
(iii) The limit of the subsequence
$$
\begin{align}
\lim\limits_{n\to\infty}a_{n^2}
&=\lim\limits_{n\to\infty}\sin\left(\pi\sqrt{n^2}\right)\\[3pt]
&=\lim\limits_{n\to\infty}0\\[3pt]
&=0\tag{3}
\end{align}
$$
Since
$$
\begin{align}
\left|\,n+\tfrac12-\sqrt{n^2+n}\,\right|
&=\frac{\frac14}{n+\frac12+\sqrt{n^2+n}}\\
&\le\frac1{8n}\tag{4}
\end{align}
$$
and $\cos(x)\ge1-\frac12x^2$, we have that
$$
\begin{align}
\left|a_{n^2+n}\right|
&=\left|\sin\left(\pi\sqrt{n^2+n}\right)\right|\\
&\ge\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\sin\left(\pi\left(n+\frac12\right)\right)\right|\\
&-\left|\sin\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\cos\left(\pi\left(n+\frac12\right)\right)\right|\\
&=\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\right|\\
&\ge1-\frac12\frac{\pi^2}{64n^2}\\[6pt]
&\gt0.9\tag{5}
\end{align}
$$
for $n\ge1$.
If the sequence converged, then the limit must be the limit of the subsequence computed in $(3)$. However, $(5)$ precludes the limit from being $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 1
} |
How do you find the sum: $\sum_{r=1}^6 \tan^2\left(\frac{r \pi}{n}\right)$ How do you find the sum: $$\sum_{r=1}^6 \tan^2\left(\frac{r \pi}{n}\right)$$
I managed to solve this question using complex numbers so I thought I'd share the solution. If you know of any better method of solving it, please do share.
| Let $x=\frac{\pi}{7}:$
$$(\cos x + i \sin x)^7 = \cos 7x + i \sin 7x$$
Since $\sin 7x$ = Imaginary part of $(\cos x + i \sin x)^7$
$$\sin7x = \sum_{r=0}^3 \binom{7}{2r+1}(\cos x)^{7-(2r+1)} \cdot(i\sin x)^{2r+1} $$
We also know that $\sin 7x = 0$
$$0 = \binom{7}{1}(\cos x)^{6} (\sin x)^{1} -\binom{7}{3}(\cos x)^{4} (\sin x)^{3}+\binom{7}{5}(\cos x)^{2} (\sin x)^{5}-\binom{7}{7}(\cos x)^{0} (\sin x)^{7}$$
$$0 = 7(\cos x)^{6} (\sin x)^{1} -35(\cos x)^{4} (\sin x)^{3}+21(\cos x)^{2} (\sin x)^{5}-(\cos x)^{0} (\sin x)^{7}$$
$$0 = 7\tan x - 35\tan^3 x + 21\tan^5 x - \tan^7 x$$
$$0 = \tan^6 x - 21\tan^4 x + 35\tan^2 x -7$$
Since the roots of the above equation are $\tan \frac{\pi}{7},\tan \frac{2\pi}{7},\tan \frac{3\pi}{7},\tan \frac{4\pi}{7},\tan \frac{5\pi}{7},\tan \frac{6\pi}{7}$
and $\tan \frac{\pi}{7}$=$-\tan \frac{6\pi}{7}$, on pairing the roots of the equation, we have:
$$\left(\tan^2 x - \tan^2 \frac{\pi}{7}\right)\left(\tan^2 x - \tan^2 \frac{2\pi}{7}\right)\left(\tan^2 x - \tan^2 \frac{3\pi}{7}\right)=0$$
Hence we have an equation in $\tan^2 x$. The given question requires us to find 2 times the sum of the roots, since
$$\sum_{r=1}^6 tan^2\left(\frac{r \pi}{n}\right)=2\sum_{r=1}^3 tan^2\left(\frac{r \pi}{n}\right)$$
Thus we find the required sum as:
$$2\left(\frac{-b}{a}\right) = 2 \cdot21 =42$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$ Let $x$ and $y$ be two positive numbers:
Prove that $$ \left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right).$$
I answered this one by squaring the two expressions. And therefore finding the difference after squaring the formulas. I don't even know if it's right but I wanted to find another way to answer this question?
| Just notice that the following inequalities are equivalent to each other:
$$
\begin{align*}
\frac {x+y}{x^2+y^2} &\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)\\
\frac {x+y}{x^2+y^2} &\le \frac12 \cdot \frac{x+y}{xy}\\
\frac 1{x^2+y^2} &\le \frac12 \cdot \frac{1}{xy}\\
2xy &\le x^2+y^2
\end{align*}
$$
The last one is a well-known inequality:
*
*Simple algebra question - proving $a^2+b^2 \geqslant 2ab$
*Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$
*Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$
*Show that for all real numbers $a$ and $b$, $\,\, ab \le (1/2)(a^2+b^2)$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Let $a,b,c,d$ be distinct integers such that the equation $(x-a)(x-b)(x-c)(x-d)-9=0$ has an integer root $r$,then find the value of $a+b+c+d-4r.$ Let $a,b,c,d$ be distinct integers such that the equation $(x-a)(x-b)(x-c)(x-d)-9=0$ has an integer root $r$,then find the value of $a+b+c+d-4r.$
As $r$ is the integer root of the equation $(x-a)(x-b)(x-c)(x-d)-9=0$,so $(r-a)(r-b)(r-c)(r-d)=9$
or $(a-r)(b-r)(c-r)(d-r)=9$
We need to find $(a-r)+(b-r)+(c-r)+(d-r)$ but i do not know how to find that.
| $a,b,c$ and $d$ are distinct integers.
So $r-a,r-b,r-c$ and $r-d$ are all distinct integers.
Again, $9=3^2$
Hence the only possible factorisation: $$(r-a)(r-b)(r-c)(r-d)=3\cdot 3\cdot 1 \cdot 1$$ or
$$(a-r)(r-b)(c-r)(r-d)=3\cdot 3\cdot 1 \cdot 1$$
So we have $a=r+3$, $b=r-3$, $c=r+1$, $d=r-1$.
Thus $\color{blue}{a+b+c+d-4r} = \color{red}{0}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Order of growth of sequence $f_{n} = 2f_{n-1} + f_{n-2}$ I'm currently stuck with the following problem.
How do I calculate the order of growth of the following sequence:
$f_{n} = 2f_{n-1} + f_{n-2}$
Assuming that
$f_{0} =1$
and $f_{1} = 1$
I've got the following hints:
they used $(1- 2x - x^2) = (1-Ax)(1-Bx) $
to get the roots of the solution.
How do I get from the squence to $(1-2x-x^2)$ and why do I have to equal it to $(1-Ax)(1-Bx)$
Somehow the solution is $(1+\sqrt{2})^n$
Thanks for your help community
| Process 1:
Start with $1-2x-x^2$. Now, the given, from the problem, is that
\begin{align}
1-2x-x^2 &= (1- a x)(1 - b x) \\
&= 1 -(a+b) x + ab x^2.
\end{align}
Equating coefficients leads to $a + b = 2$ and $ab = -1$. Using this information leads to
\begin{align}
a - \frac{1}{a} &= 2 \\
a^2 - 2 a - 1 &= 0 \\
a &= 1 \pm \sqrt{2}.
\end{align}
By choosing $a = 1 + \sqrt{2}$ then $b$ becomes $b = 1 - \sqrt{2}$.
Since $f_{n}$ increases with $n$ then consider a solution of the form
$$f_{n} = A \, (1+\sqrt{2})^{n} + B \, (1-\sqrt{2})^{n}$$
With $f_{0} = 1$ and $f_{1} = 1$, then it can be found that $A = B = \frac{1}{2}$ and
$$f_{n} = \frac{1}{2} \, \left[ (1+\sqrt{2})^{n} + (1-\sqrt{2})^{n} \right].$$
Now, in terms of growth of $f_{n}$, the following is noticed. It can be determined that $0 \leq 1-\sqrt{2} \leq -1$ and $3 \leq 1 + \sqrt{2} \geq 1$. Taking powers of these values leads to $|(1-\sqrt{2})^{n}| \to 0$ as $n \to \infty$ and $(1+\sqrt{2})^{n} \to \infty$. This leads to
$$f_{n} \approx (1+\sqrt{2})^{n} $$
as $n \to \infty$.
| {
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Induction: Prove that $5^{3n} + 7^{2n-1}$ is divisible by $4$
Prove that $5^{3n} + 7^{2n-1}$ is divisible by $4$ for all $n \in \mathbb{N}$.
*
*For $n=1$,
$\Rightarrow 5^3 + 7^1 \Rightarrow 132 \mid 4$ (which is divisible by $4$)
*Let us assume given equation holds true for $n = m$,
$\Rightarrow 5^{3m} + 7^{2m-1} | 4$
*Now for $n = m+1$,
*
*$5^{3m+3} + 7^{2m+2-1}$
*$5^{3m} \cdot 5^3 + 7^{2m-1} \cdot 7^2$
*$5^{3m} \cdot 125 + 7^{2m-1} \cdot 49$
How do I go ahead from here? I am kind of stuck.
| Use the binomial theorem:
$$5^{3n} + 7^{2n-1}=(4+1)^{3n}+(8-1)^{2n-1}=4a+1^{3n}+8b+(-1)^{2n-1}=4c+1-1=4c$$
| {
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Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq1+4\sqrt{\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
We can prove this inequality by using $b=a+u$, $c=a+v$ and $d=a+w$ after squaring of the both sides.
I am looking for a nice proof. Thank you!
| Let: $ a+b+c+d=1 $
Then the inequality can be rewritten:
$$a\cdot \frac{c+d}{a+b}+b \cdot \frac{d+a}{b+c}+c \cdot \frac{a+b}{c+d}+d \cdot \frac{b+c}{d+a} \ge 4 \cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
$$\frac{a (c+d)^2+c (a+b)^2}{(a+b)(c+d)}+\frac{b (d+a)^2+d (b+c)^2}{(b+c)(d+a)} \ge 2 \cdot \sqrt {\frac{(a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2)}{(a+b)(b+c)(c+d)(d+a)}} \ge$$
$$\ge 4\cdot \sqrt {\frac{abcd}{(a+b)(b+c)(c+d)(d+a)}}$$
$$\Leftrightarrow (a (c+d)^2+c (a+b)^2)(b (d+a)^2+d (b+c)^2) \ge 4abcd$$
$$4 abcd \sum\limits_{cyc} a^2+ \left ( 5 abcd\sum\limits_{cyc}ab+\sum\limits_{cyc}(a^4bc+ab^2d^3+ac^2d^3) \right) + $$
$$+\left ( 4abcd \cdot (ac+bd)+2 (a^3b^2c+ac^3d^2)+2 (b^3c^2d+bd^3a^2) \right) \ge 4abcd\cdot (a+b+c+d)^2= 4abcd$$
| {
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Equality of two binomial coefficient containing expressions
Why is
$$ \begin{align}
&\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
&=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2}\\[6pt] & \end{align} $$ ?
Should'nt it be $$\begin{align}\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}-\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k-1}2^{n-2k}\\[6pt] & \end{align} $$
I know this must be a silly question but I can't understand this...please help me out!
| This is a partial answer
Since
$$\sum_{k=0}^n(-1)^k \binom{n-k-1}{k} 2^{n-2k} = \frac{2^{n+1}\cdot n + 1}{2^n}$$
you get immediately that
$$\begin{align}
2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1} &= \sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}\\
&= \left(\left( \sum_{k=0}^{n}(-1)^k\binom{n-k-1}{k}2^{n-2k} \right) - \left( -1^{n}\binom{n-n-1}{n}2^{n-2n}\right) \right)\\
&= \frac{2^{n+1}\cdot n + 1}{2^n} - \frac{ -1^n \binom{-1}{n}}{2^n}\\
&= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n}\\
\end{align}$$
and
$$\begin{align}
\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} &= 2^{-1} \sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-1}\\
&=2^{-1} \left( \left( \sum_{k=0}^{n-1} (-1)^k\binom{n-k-2}{k}2^{n-2k-1} \right) - \left( (-1)^{n-1}\binom{n-(n-1)-2}{n-1}2^{n-2(n-1)-1} \right)\right)\\
&= 2^{-1} \left( \left( \frac{2^{(n-1)+1}\cdot (n-1) + 1}{2^{n-1}} \right) -\left( -1^{n-1} \binom{-1}{n-1}2^{-n+1} \right) \right)\\
&= \frac{2^n\cdot (n-1) + 1}{2^n} -\left( -1^{n-1} \binom{-1}{n-1}2^{-n} \right)\\
&= \frac{2^n\cdot (n-1) + 1}{2^n} - \frac{-1^{n-1} \binom{-1}{n-1}}{2^n}\\
&= \frac{2^n\cdot (n-1) + 1 +(-1^n)\binom{-1}{n-1}}{2^n}
\end{align}$$
thus you can rewrite the RHS of your statement as
$$\begin{align}
RHS &= 2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} \\
&= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n} - \frac{2^n\cdot (n-1) + 1 +(-1^n) \binom{-1}{n-1}}{2^n}\\
&= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} - 2^n\cdot (n-1) - 1 - (-1^n) \binom{-1}{n-1}}{2^n}\\
&= \frac{2^n(n+1) + (-1^{n-1}) \binom{-1}{n} + (-1^{n-1}) \binom{-1}{n-1}}{2^n}
\end{align}$$
As for the LHS you should easily be able to rewrite it, and check if the statement is true.
| {
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Define $P(x)\in [\mathbb{R}]$ is a polynomial that satisfy $P(x).P(y)=P(\frac{x+y}{2})^{2}-P(\frac{x-y}{2})^{2}$ Define $P(x)\in [\mathbb{R}]$ is a polynomial that satisfy
$P(x).P(y)=P(\frac{x+y}{2})^{2}-P(\frac{x-y}{2})^{2}$ (1)
Some notes may help you.
It is very clear that with y=0 then (1)$\Leftrightarrow P(x).P(0)=0$
We just need to consider $P(x)\neq 0$ so $P(0)=0$.
With $y=-x$ then (1)$\Leftrightarrow P(x)=-P(-x)$
| If $P$ is constant, then it is equal to $0$:
Otherwise:
When $X \to \infty$
$$
\begin{align}
P(X)^2 = a_n^2 X^{2n} + a_na_{n-1} X^{2n-1} + o(X^{2n-1})
\end{align}
$$
Let $y\in\Bbb R$, with $P(y)\neq 0$. When $x\to +\infty$
$$
\begin{align}
P(x+y)^2 - P(x-y)^2 &= a_n^2 (x+y)^{2n} + a_na_{n-1} (x+y)^{2n-1} \\
&- a_n^2(x-y)^{2n} - a_na_{n-1}(x-y)^{2n-1} + o(x^{2n-1})\\
& =a_n^2x^{2n} + a_n^2 (2n)x^{2n-1}y + a_na_{n-1} x^{2n-1} \\
&-a_n^2x^n + a_n^2 (2n)x^{n-1}y - a_na_{n-1}x^{2n-1} + o(x^{2n-1})\\
&=4a_n^2 ny x^{2n-1} + o(x^{2n-1})\\
&\approx 4a_n^2 ny x^{2n-1}
\end{align}
$$
And
$$
\begin{align}
P(2x) P(2y) &\approx P(2y)a_n 2^nx^n \\
4a_n ny x^{2n-1} &\approx P(2y)2^nx^n
\end{align}
$$
Meaning that $2n-1 = n$, so $n=1$ and $4a_1y = P(2y)2$
So $P(2y) = 2a_1y$
$$P(x) = a_1 x $$
Where $a_1$ can be any non zero constant.
| {
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Probability that each number obtained by throwing a die is no smaller than the preceding number
A fair die is thrown 4 times. Find the probability that the each number obtained is no smaller than the preceding number.
*
*If all numbers obtained are same, number of such outcomes $$=\dbinom{6}{1}=6$$
*If 3 numbers obtained are same, number of ways is equal to the number of ways of choosing any two numbers $=\dbinom{6}{2}$. But these two number can be arranged as $x,y,y,y$ or $x,x,x,y$ where $x<y$. So $$2\times\dbinom{6}{2}$$
*If 2 numbers are same, the cases are $a_1<a_2<a_3=a_4$, $a_1<a_2=a_3<a_4$, $a_1=a_2<a_3<a_4$, $a_1=a_2<a_3=a_4$. Number of ways $$=3\times\dbinom{6}{3}+\dbinom{6}{2}$$
*if all 4 numbers are different, $$\dbinom{6}{4}$$
Total number of possibilities $=6^4$.
Even though I got the correct answer, is there any shorter method?
| Nope, that seems likely to be the most concise way to do it.
Count ways to pick $n\in\{1,2,3,4\}$ unique numbers, and to arrange them with $n-1$ "$>$" signs, to meet the criteria.
$$\begin{array}{|l:l|} \hline
\rm a{=}a{=}a{=}a & \dbinom 3 0 \dbinom 6 1
\\\hdashline\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \dbinom 3 1\dbinom 6 2
\\\hdashline\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \dbinom 3 2 \dbinom 6 3
\\\hdashline\rm a{>}b{>}c{>}d & \dbinom 3 3 \dbinom 6 4
\\ \hline\end{array}$$
$$\dfrac{\sum_{n=1}^4 \dbinom{3}{n-1}\dbinom{6}{n}}{6^4} = \dfrac{\dbinom 6 1 + 3\dbinom 6 2 + 3 \dbinom 6 3 + \dbinom 6 4}{6^4}$$
| {
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Find if $\sqrt[4]{x^4+y^4}, \sqrt{x^4+y^4}$ are differentiable in $(0,0)$
Find if
$$f(x,y)=\sqrt[4]{x^4+y^4}$$
$$g(x,y)=(f(x,y))^2$$
are differentiable in $(0,0)$.
well, $g(x)$ is clearly $\sqrt{x^4+y^4}$, so I guess the answer will be similar to $f(x)$.
$f_x=x^3/(x^4+y^4)^{3/4}$, but what now? $f_x$ has no value in $(0,0)$, so can it be differentiable there?
| Using the definition: in both cases, as $(0,0)$ is a local minimum, the (possible) differential will be zero if it exists:
$$
\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-(0,0)(x,y)}{\|(x,y)\|} =
\lim_{(x,y)\to(0,0)}\frac{\sqrt[4]{x^4+y^4}}{\sqrt{x^2+y^2}} =
\lim_{r\to 0}\frac{r(\cos^4\theta+\sin^4\theta)}r.
$$
($\not\exists$ because depends of the angle)
$$
\lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)-(0,0)(x,y)}{\|(x,y)\|} =
\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}} =
\lim_{r\to 0}\frac{r^2(\cos^4\theta+\sin^4\theta)}r = 0.
$$
And only $g$ is differentiable.
EDIT: clarification of second limit:
$$
\left|\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}\right| =
\frac{r^2(\cos^4\theta+\sin^4\theta)}r \le 2 r = 2\|(x,y)\|\to 0,
$$
and the squeezing theorem proves that the lim is 0.
| {
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How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$
| Note that, by applying the Squeeze theorem you get
\begin{align}
&-1 \le \sin x \le 1\\
&\Rightarrow \lim_{x \to \infty} \frac{-1}{x} \le \lim_{x \to \infty} \frac{\sin x}{x} \le \lim_{x \to \infty} \frac{1}{x}\\
&\Rightarrow 0 \le \lim_{x \to \infty} \frac{\sin x}{x} \le 0\\
&\Rightarrow \lim_{x \to \infty} \frac{\sin x}{x} = 0\\
&\Rightarrow \lim_{y \to 0+} y \sin \frac{1}{y} = 0
\end{align}
so
\begin{align}
\lim_{x\to 0^+} \frac{x^2\sin\frac{1}{x}}{\ln(1+2x)} &= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x^2 \cdot 2x}{\ln(1+2x) \cdot 2x}\\
&= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x}{2}\\
&= 0\\
\end{align}
| {
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How to prove $\sin3θ=3\sinθ-4\sin^3θ$ I was solving an A Level past paper (November 2014 P32) when I stumbled upon this question. It first asks us to expand $\sin(2θ+θ)$ which is easy using the identity $\sin(A+B)=\sin A\cos B+\cos A\sin B$, but what I am not able to do is prove that this is equal to $3\sin θ-4\sin^3 θ$. I tried using the $\sin 2A=2 \sin A \cos A$ identity and the basic $\cos^2θ=1-\sin^2θ$ identity, but that isn't working. I'd really appreciate it if someone would guide me through this question. Thanks in advance!
| We will use the following trigonometric formulas:
\begin{align}
\color{red}{\sin(x+y)\,}&\color{red}{=\sin x\cos y+\sin y\cos x}\\
\color{green}{\sin (2x)\,}&\color{green}{=2\sin x\cos x}\\
\color{blue}{\cos (2x)\,}&\color{blue}{=1-2\sin^2x}\\
\color{magenta}{\cos^2 x\,}&\color{magenta}{=1-\sin^2 x}
\end{align}
So
\begin{align}
\sin(2\theta+\theta)\,&=\color{red}{\sin(2\theta)\cos \theta+\sin \theta\cos(2\theta)}\\
&=\color{green}{2\sin\theta\cos^2\theta}+\sin \theta(\color{blue}{1-2\sin^2\theta})\\
&=2\sin\theta(\color{magenta}{1-\sin^2\theta})+\sin\theta-2\sin^3\theta\\
&=3\sin\theta-4\sin^3\theta
\end{align}
| {
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prove that $-1 \le \frac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$
For the real numbers $a, b, c, d$ prove that $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$
Actually if we let $\vec{u} = (a, b)$ and $\vec{v} = (c, d)$ then by dot product we got $$\begin{split} cosθ & = \dfrac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|} \\ & =\dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \end{split}$$
But $-1 \le cosθ \le 1$ hence $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$
Is it possible to prove the above inequality using only standard inequalities like AM-GM, Cauchy-Swarz, Rearrangement inequality or Jensen inequality?
| The inequality $-1 \leq x \leq 1$ is equivalent to $x^2 \leq 1$.
Use that fact to prove the given inequality, i.e.
just re-work it, raising to the power of 2. And it will be proven.
| {
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Evaluation of $ \int_{0}^{1}\ln\left(\sqrt{1+x}-\sqrt{1-x}\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)dx$
$\bf{My\; Try::}$ Let $$I = \frac{1}{2}\int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)^2dx = \frac{1}{2}\int_{0}^{1}\ln\left(2+2\sqrt{1-x^2}\right)dx$$
So $$I =\frac{1}{2}\ln(2)+\frac{1}{2}\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)\cdot 1dx$$
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\left[x\cdot \ln(1+\sqrt{1-x^2})\right]_{0}^{1}+\frac{1}{2}\int_{0}^{1}\left(\frac{x^2}{1+\sqrt{1-x^2}}\right)\cdot\frac{1}{\sqrt{1-x^2}}dx$$
Now Put $x=\sin \phi\;,$ Then $dx = \cos \phi d\phi$ and changing limits, We get
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(1-\cos \phi\right) d\phi = \frac{\ln 2}{2}+\frac{\pi}{4}-\frac{1}{2}$$
Can we solve it any other way, If yes Then plz explain here
Thanks
| If $y=\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)$, then after solving for $x$ while assuming $x$ is positive, you have $x=e^y\sqrt{1-\frac{e^{2y}}{4}}$.
View the integral as an area. There is a rectangle that is $[0,1]\times[0,\ln\sqrt{2}]$, plus a section above that we can integrate with respect to $y$:
$$\int_{y=\ln\sqrt{2}}^{y=\ln2}e^y\sqrt{1-\frac{e^{2y}}{4}}\,dy$$ which is a candidate for $u$-substitutions with $u=e^{y}/2$:
$$\int_{u=\sqrt{2}/2}^{u=1}2\sqrt{1-u^2}\,du=\left[u\sqrt{1-u^2}+\arcsin(u)\right]_{u=\sqrt{2}/2}^{u=1}=\frac{\pi}{4}-\frac{1}{2}$$
Add this to the rectangle's area of $\ln\sqrt{2}$ to get $$\ln\sqrt{2}+\frac{\pi}{4}-\frac{1}{2}$$
| {
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Finding the equation of a circle from the equation of its tangents Given the equation of a pair of lines
: $36x² - 63xy + 20y² + 54x - 17y - 10 =0.$
If the circle touches one of the lines at (-3,-1) and the other at some point then find the equation of the circle.
I know of one way of doing this is by taking the general equation is the circle and then using $T² = SS1$ and comparing it by the given equation but it is tedious.
So I want to know of a smarter way to do this.
| $36x^2−63xy+20y^2+54x−17y−10=0$
$\implies (12x-5y-2)(3x-4y+5)=0$
$L_{1}$: $12x-5y-2=0$
$L_{2}$: $3x-4y+5=0$
Let $x^2+y^2+2ax+2by+c=0$ be the circle $C$.
Then $T=L_{1} \cap L_2=(1,2)$ and $P=C\cap L_{2}=(-3,-1)$.
Now $x(1)+y(2)+a(x+1)+b(y+2)+c=0$ is the equation of the chord (polar) $L_3$.
Substitute $(-3,-1)$ into $L_{3}$,
$(-3)(1)+(-1)(2)+a(-3+1)+b(-1+2)+c=0$
$-2a+b+c-5=0 \: \cdots \cdots (1)$
$L_{1}, L_{2}$ are equidistant from centre $O(-a,-b)$,
$\displaystyle \left| \frac{-12a+5b-2}{13} \right|=
\left| \frac{-3a+4b+5}{5} \right|$
$5^{2}(-12a+5b-2)^{2}=13^{2}(-3a+4b+5)^{2}$
$(9a-7b-5)(7a+9b+25)=0 \: \cdots \cdots (2)$
Slope of $\displaystyle OP=\frac{-1+b}{-3+a}=-\frac{4}{3}$
$ 4a+3b=15 \: \cdots \cdots (3)$
On solving, $\displaystyle (a,b,c)=
\left( \frac{24}{11}, \frac{23}{11}, \frac{80}{11} \right) \text{ or }
\left( 14, -\frac{41}{3}, \frac{140}{3} \right)$
$C: \left \{
\begin{array}{rcl}
11x^{2}+11y^{2}+48x+46y+80 &=& 0 \\
3x^{2}+3y^{2}+84x-82y+140 &=& 0
\end{array}
\right. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Vandermonde's identity? How to continue? I have:
$$\sum\limits_{k = 1}^{10}k\binom{10}{k}\binom{20}{10-k} = $$
and I know that it doesn't matter if $k = 0$ so it also equals:
$$= \sum\limits_{k = 0}^{10}k\binom{10}{k}\binom{20}{10-k} = $$
and now it really reminds Vandermonde's identity so it's tempting to write:
$$= \binom{30}{10} \cdot \sum\limits_{k = 1}^{10}k = \binom{30}{10} \cdot 10!$$
but it seems wrong because on smaller numbers the equations doesn't hold... What's the right way to continue in order to get an equivalent expression without sigma?
| Ok I found an algebric answer!
$$\begin{align}
&\sum\limits_{k=1}^{10}k\binom{10}{k}\binom{20}{10-k}\\= &\sum\limits_{k=0}^{10}k\binom{10}{k}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}k\cdot\frac{10!}{(k)!(10-k)!}\binom{20}{10-k} \\= &\sum\limits_{k = 0}^{10}\frac{10!}{(k-1)!(10-k)!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(10-k)!}\binom{20}{10-k} \\ = &\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(9 - (k-1))!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \binom{9}{k-1}\binom{20}{10-k} \\ = &10\cdot \sum\limits_{k = 0}^{10}\binom{9}{k-1}\binom{20}{10-k} \\= &10\cdot \sum\limits_{k = 0}^{9}\binom{9}{k}\binom{20}{9-k} \\= &10 \cdot \binom{29}{9}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Why can't you apply the natural logarithm rule to integrate $\int \frac{1}{\sqrt{x}}dx$? I understand that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, or $\frac{1}{x^2} = x^{-2}$, but why wouldn't you be able anyhow to apply the rule for which $\int \frac{1}{x}dx = \ln{|x|} + C$, and have, for example $\int \frac{1}{x^2}dx = \ln{|x^2|} + C$?
| Notice:
*
*$$\frac{1}{\sqrt{x}}=\frac{1}{x^{\frac{1}{2}}}=x^{-\frac{1}{2}}$$
So:
$$\int\frac{1}{x^n}\space\text{d}x=\int x^{-n}\space\text{d}x=\frac{x^{1-n}}{1-n}+\text{C}$$
Set $n=\frac{1}{2}$:
$$\int\frac{1}{\sqrt{x}}\space\text{d}x=\int\frac{1}{x^{\frac{1}{2}}}\space\text{d}x=\int x^{-\frac{1}{2}}\space\text{d}x=\frac{x^{1-\left(\frac{1}{2}\right)}}{1-\left(\frac{1}{2}\right)}+\text{C}=2\sqrt{x}+\text{C}$$
Now if we use the same rule with $n=1$, we get:
$$\int\frac{1}{x^1}\space\text{d}x=\int x^{-1}\space\text{d}x=\frac{x^{1-1}}{1-1}+\text{C}=\frac{x^{0}}{0}+\text{C}$$
$$\color{red}{\text{And we can't divide by}\space0}$$
And if we know $\frac{\text{d}}{\text{d}x}\ln(x)=\frac{1}{x}$:
$$\int\frac{\text{d}}{\text{d}x}\ln(x)\space\text{d}x=\int\frac{1}{x}\space\text{d}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try:}$ For lower bound $$\sum^{99}_{k=1}\frac{1}{(k+1)\sqrt{k}}>\sum^{99}_{k=1}\frac{1}{(k+1)k}=\sum^{99}_{k=1}\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{99}$$
Now I didn't understand how can I solve it, any help?
Thanks
| We have:
$$ \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}\geq \frac{1}{2(k+1)\sqrt{k}}$$
hence:
$$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\leq 2\sum_{k=1}^{99}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)=2-\frac{1}{5}$$
but in a similar way:
$$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\geq \frac{1}{2}+2\sum_{k=2}^{99}\left(\frac{1}{\sqrt{k-1/4}}-\frac{1}{\sqrt{k+3/4}}\right)=\frac{1}{2}+\frac{4}{\sqrt{7}}-\frac{4}{\sqrt{399}}$$
so $\lfloor A \rfloor = \color{red}{1}$. Another great victory for creative telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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Proving number of partitions of $n$ to $3$ parts at most. I have an exercise, to prove that the number of partitions of $n$ to at most $3$ integers is $\frac{(n+3)^2}{12}$ rounded. I tried to prove by induction but I don't know how.
| The number of partitions of $n$ into at most $k$ integers is equal to the number of partitions of $n$ into integers no bigger than $k$. Let's call this number $p(n,k)$. In general $$p(n,k)=p(n,k-1)+p(n-k,k)$$ starting with $p(0,0)=1$ and $p(n,0)=0$ for $n\not=0$ and zero if $n$ or $k$ are negative.
For given $k$, patterns are going to repeat every $k!$ terms, so we can find and show by induction that $p(m,1)=1$ and $p(2m,2)=p(2m+1,2)=m+1$ for $m \ge 0$ and thus
*
*$p(6m+0,3)= 3m^2+3m+1 = \frac{((6m+0)+3)^2}{12} + \frac1{4}$
*$p(6m+1,3)= 3m^2+4m+1 = \frac{((6m+1)+3)^2}{12} - \frac1{3}$
*$p(6m+2,3)= 3m^2+5m+2 = \frac{((6m+2)+3)^2}{12} - \frac1{12}$
*$p(6m+3,3)= 3m^2+6m+3 = \frac{((6m+3)+3)^2}{12} + 0$
*$p(6m+4,3)= 3m^2+7m+4 = \frac{((6m+4)+3)^2}{12} - \frac1{12}$
*$p(6m+5,3)= 3m^2+8m+5 = \frac{((6m+5)+3)^2}{12} - \frac1{3}$
Since all the fractions on the right hand side are smaller than $\frac12$, this leads to the conclusion that you can round and say $$p(n,3)= \left[\frac{(n+3)^2}{12}\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Would this solution of the limit of the sequence be correct? Let's suppose that I have the sequence $a_n = \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \ldots + \frac{n}{n^2}, n \in \mathbb{N}$. And I have to find the limit of the sequence as $n \rightarrow \infty$. Would the below solution be correct?
The sequence $a_n$ can be rewritten as
\begin{align}
a_n &= \sum_{k=1}^{n} \frac{k}{n^2} \\
&= \frac{1}{n^2} \sum_{k=1}^{n} k \\
&= \frac{1}{n^2} \cdot \frac{n(n+1)}{2} \\
&= \frac{n+1}{2n}.
\end{align}
Thus we have
\begin{align}
\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \frac{n+1}{2n} = \lim_{n \rightarrow \infty} \frac{n}{2n} = \frac{1}{2}.
\end{align}
| This seems correct. Personally I would not say $\lim\frac {n+1}{2n}=\lim\frac n{2n}$. Not without explanation at least (e.g. include the step $\lim \frac{n+1}{2n}=\lim \left(\frac{n}{2n}+\frac 1{2n}\right)$). I would rather say $\lim \frac {n+1}{2n}=\lim \frac{1+\frac1n}{2}$. But this is of course a matter of preference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
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} |
simplifying complex fractions when proving inverses with function composition? I'm working with two functions, $f(x)=\frac{x-3}{x+4}$ and $g(x)=\frac{4x+3}{1-x}$.
I need to simplify $f(g(x))$ and its opposite, but i'm not sure of the procedures regarding the more complicated forms like
$$\frac{\frac{4x+3}{1-x}-3}{\frac{4x+3}{1-x}+4}$$
| with the substitution $x \rightarrow g(x)$ the numerator of $f(x)$ becomes:
$$
x-3 \rightarrow \frac{4x+3}{1-x}-3=\frac{7x}{1-x}
$$
and the denominator becomes
$$
x+4 \rightarrow \frac{4x+3}{1-x}+4=\frac{7}{1-x}
$$
so
$$
f(g(x))= \frac{\frac{7x}{1-x}}{\frac{7}{1-x}}= x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$ without L'Hospital's Theorem I've been trying to evaluate$$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$$
I tried:
(a) Rationizing the numerator -> Error
(b) Rationizing the denominator -> Error
(c) Factoring out $x$ -> Error
Finally, I used L'Hospital's Rule and got the answer $3/2$, but that is not what I am supposed to do, for not a single word about this Theorem was mentioned during my lectures.
Is there any other way to solve this limit without this Theorem?
| Here is a same solution:How can I evaluate $\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$ without invoking l'Hôpital's rule?
$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3} * \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}} = lim_{x\to1}{\frac{(x^2-1 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{1}{\sqrt{x^2+3}+2}} = = lim_{x\to1}{\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}} = \frac{\sqrt{1^2+8}+3}{\sqrt{1^2+3}+2} = \frac{6}{4} = \frac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Seeking non-inductive, combinatorial proof of the identity $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ How do you prove
$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$$
without induction?
I'm looking for a combinatorial proof of this.
| We can count the sum $$2^2 + 4^2 + \dots + (2n)^2 = {2n + 2 \choose 3}$$ as follows:
Choose three numbers from $\{1, 2, \dots, 2n+2\}$. If the largest number chosen is odd, notate it $2m+1$ - the other two smaller numbers can be chosen in ${2m \choose 2}$ ways. If the largest number chosen is $2m+2$, the other two numbers can be chosen in ${2m+1 \choose 2}$ ways. From the identity
$${2m \choose 2} + {2m+1 \choose 2} = (2m)^2$$
and summing this for $m = 0, 1, 2, \dots, n$, the even result follows and the general result follows from dividing by two.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ Any ideas on evaluating the definite integral
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx$$
The best numerical approximation I could get is $0.2796245358$.
Is there even a closed form solution?
| Using the Fourier series of $\ln^2(2\sin x):$
$$\ln^2(2\sin x)=\left(\frac{\pi}{2}-x\right)^2+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cos(2nx),\quad 0<x<\pi$$
we get
$$
\int_0^{\frac{\pi}{2}}x\ln^2(2\sin x)\, dx$$
$$=\int_0^{\frac{\pi}{2}} x\left(\frac{\pi}{2}-x\right)^2\, dx+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^{\frac{\pi}{2}} x\cos(2nx)\, dx$$
$$=\frac{15}{32}\zeta(4)+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(\frac{(-1)^n}{4n^2}-\frac{1}{4n^2}\right)$$
$$=\frac{45}{32}\zeta(4)-\frac12\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}$$
$$=\operatorname{Li_4}\left(\frac12\right)-\frac{19}{32}\zeta(4)+\frac78\ln(2)\zeta(3)-\frac14\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2).$$
On the other hand, we can write
$$\int_0^{\frac{\pi}{2}}x\ln^2(2\sin x)\, dx$$
$$=\ln^2(2)\int_0^{\frac{\pi}{2}}x\, dx+2\ln(2)\int_0^{\frac{\pi}{2}}x\ln(\sin x)\, dx+\int_0^{\frac{\pi}{2}}x\ln^2(\sin x)\, dx$$
$$=\frac78\ln(2)\zeta(3)-\frac34\ln^2(2)\zeta(2)+\int_0^{\frac{\pi}{2}}x\ln^2(\sin x)\, dx.$$
Thus,
$$\int_0^{\frac{\pi}{2}} x\ln^2(\sin x)\,dx=\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{19}{32}\zeta(4)+\frac{1}{2}\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Basic algebra problem: $ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $ Basic algebra problem I can't seem to figure out: $$ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $$ $x,y \in \mathbb{R}, x^2 \neq y^2, xy\neq0$.
Now I know the result is: $\frac{xy}{y-x}$, but I am not sure how to get it, I get into a mess like this: $=x+\frac{x^2}{y}-\frac{y^2}{x}-y=\frac{x(xy)+x^3-y^3-y(xy)}{xy}=?$ which doesn't seem to help me much. Halp please.
| First write $\frac1x+\frac1y = \frac{y}{xy}+\frac{x}{xy} = \frac{x+y}{xy}$.
Then write $\frac1{x^2}-\frac1{y^2} = \frac{y^2}{x^2y^2}-\frac{x^2}{x^2y^2} = \frac{y^2-x^2}{x^2y^2}=\frac{(y-x)(x+y)}{x^2y^2}$.
Therefore $$\frac{\frac1x+\frac1y}{\frac1{x^2}-\frac1{y^2}} = \frac{\frac{x+y}{xy}}{\frac{(y-x)(x+y)}{x^2y^2}} = \frac{x+y}{xy} \cdot \frac{x^2y^2}{(y-x)(x+y)} = \frac{(x+y)x^2y^2}{xy(x+y)(y-x)} = \frac{xy}{y-x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a > 0$,$b>0$, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$ If $a$ and $b$ are positive integers, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$
-I played around with numbers and the conditions and it seems that it's common sense that if $a = b$ then $a = b = 1$ or $2$, For example:
If $a = b = 4$, then $\frac{1}{a} + \frac{1}{b}$ would not be an integer.
-Any help is appreciated
| If $\frac1a + \frac1b = \frac{a+b}{ab}$ is an integer, then $a$ and $b$ both divide $a+b$. This means that $a$ divides $(a+b)-a=b$ and that $b$ divides $(a+b)-b=a$. So $a$ and $b$ divide each other, whence they must be equal.
Now if $\frac1a + \frac1b = \frac2a$ is an integer, then $a$ must divide $2$. So $a=1$ or $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$
Prove that for nonnegative $x,y,z$ that $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$
I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.
The reason I am asking this is because it comes from the solution below where it claims it is true.
We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a \le \frac{1}{3}(S_a+S_b+S_c)^2=\frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.
| $\frac{1}{3}(x+y+z)^2 = \frac{2}{3}(xy + zx + yz) + \frac{1}{3}(y^2 + z^2 + x^2) \geq xy + yz + xz.$
since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2\geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) \geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\pi^2$ so close to $10$? Noam Elkies explained why $\pi^2=9.8696...$ is so close to $10$ using an inequality on Euler's solution to Basel problem
$$\frac{\pi^2}{6}=\sum_{k=0}^{\infty} \frac{1}{\left(k+1\right)^2}$$
to form a telescoping series. As a better approximation, taking one more term of this series, $$\pi^2\approx 9.9$$ was obtained.
http://www.math.harvard.edu/~elkies/Misc/pi10.pdf
There is a simple series for $\pi$ whose first term is $3$, so it may be regarded as an explanation why $\pi$ is close to $3$ and a proof that $\pi>3$, because all the remaining terms are positive as well.
$$
\begin{align}
\pi
&=
3\sum_{k=0}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)} \\
&=
3+3\sum_{k=1}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)}
\end{align}
$$
Q: Is there a similar way to show that $\pi^2$ is close to $10$ and have an approximation closer than $9.9$?
| Truncating the following series by Ramanujan that converges to $\pi^2$ from above,
$$\pi^2=10-\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}=10-\frac{1}{8}-\frac{1}{216}-\frac{1}{1728}-\frac{1}{8000}-\frac{1}{27000}-...$$
yields
$$\pi^2<10-\frac{1}{8}=\frac{79}{8}=9.875$$
Both $10$ and $\frac{79}{8}$ are convergents from the continued fraction for $\pi^2$.
From below,
$$\pi^2=\frac{39}{4}+\sum_{k=0}^\infty\frac{4}{((k+1)(k+2)(k+3))^2}=\frac{39}{4}+\frac{1}{9}+\frac{1}{144}+\frac{1}{900}+\frac{1}{3600}+\frac{1}{11025}+...$$
taking the first term only gives
$$\pi^2>\frac{39}{4}+\frac{1}{9}=\frac{355}{36}=9.861...$$
Finally,
$$10-\frac{1}{4}<\frac{355}{36}<\pi^2<\frac{79}{8}<10$$
[EDIT]
These series have a sixth order polynomial in the denominator. The same approximations $\dfrac{39}{4}$ and $10$ are related to simpler series using fourth order polynomials.
$$\begin{align}
\pi^2 &= 9 + \sum_{k=0}^\infty \frac{3}{(k + 1)^2 (k + 2)^2}\\
&=\frac{39}{4} + \sum_{k=1}^\infty \frac{3}{(k + 1)^2 (k + 2)^2}\\
\\
\pi^2 &= \frac{21}{2} - \sum_{k=0}^\infty \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\
&=10 -\sum_{k=1}^\infty \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $\int \frac{dx}{(9+x^2)^2}$
$$\int \frac{dx}{(9+x^2)^2}$$
$x=3\tan\theta$
$dx=\frac{3}{\cos^2\theta}d\theta$
$$\int\frac{\frac{3}{\cos^2\theta}}{(9[1+\tan^2\theta])^2} \,d\theta = \int\frac{\frac{3}{\cos^2\theta}}{(\frac{9}{\cos^2\theta})^2} \, d\theta =\frac{3}{81} \int \cos^2\theta \, d\theta=\frac{3}{162}\int (\cos2\theta+1) \, d\theta$$
$u=2\theta$
$du=2d\theta$
$$\frac{3}{324}\int (\cos u+1) du=\frac{3}{324}(\sin u+u)+c=\frac{3}{324}(\sin2\theta+2\theta)+c$$
I know that $\theta=\arctan(\frac{x}{3})$
How should I continue?
| Hint. You may use
$$
\sin (2\theta)=\frac{2\tan \theta}{1+\tan^2\theta}
$$ and
$$
\theta= \arctan \frac{x}3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\sin(2\theta)$ when $\cot(\theta) + \tan(\theta) = 2.5 $ I have an homework question that goes like:
$\cot(\theta) + \tan(\theta) = 2.5 $ is valid on some angles $\theta$ at section $0 < \theta < \pi/2$.
Find the value of $\sin(2\theta)$. (There is no need to find the value of $\theta$ in order to get the answer).
(a) $0.4 \quad$ (b) $0.5 \quad$ (c) $0.6 \quad$ (d) $0.7 \quad$ (e) $0.8$
An explanation on how to get the correct answer will be helpful.
| \begin{align*}
\cot \theta+\tan \theta
&= \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \\
&= \frac{\cos^{2} \theta+\sin^{2} \theta}{\sin \theta \cos \theta} \\
&= \frac{1}{\sin \theta \cos \theta} \\
&= \frac{2}{\sin 2\theta}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof $\dbinom{n}{0} - \frac{1}{2}\dbinom{n}{1} + \cdots + (-1)^n\frac{1}{2^{n-1}}\dbinom{n}{n-1}$ For all $n \ge 1$,
$$\binom{n}{0} - \frac{1}{2}\binom{n}{1} + \frac{1}{2^2}\binom{n}{2} - \frac{1}{2^3}\binom{n}{3} + \cdots + (-1)^n\frac{1}{2^{n-1}}\binom{n}{n-1} = 0,$$ if $n$ is even
$$\frac{1}{2^{n-1}}, \text{ if } n \text{ is odd}$$
I need to use the binomial theorem in this question.
| Try to use the binomial formula on $(1+(-\frac{1}{2}))^n$
This will give you to the proof of the statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.
Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows;
here.
$$ax^2+2hxy+by^2=0$$
Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides :
$$ax+hy=\pm y\sqrt{h^2-ab}.$$
What should I do next?
| If $ax^2+2hxy+by^2=0$ bisects the coordinate axes then the points $(x,x)$ and $(x,-x)$ belong to these lines. If we consider $(x,x)$ we get $$ax^2+2hx^2+bx^2=0$$ and if we consider $(x,-x)$ we get $$ax^2-2hx^2+bx^2=0.$$ If $x\ne 0$ then it is $$a+b=\pm 2h\implies (a+b)^2=4h^2.$$
Conversely, if $4h^2=(a+b)^2$ then, if $2h=a+b$ we have $ax^2+2hxy+by^2=(x+y)(ax+by)$ and if $2h=-(a+b)$ we have $ax^2+2hxy+by^2=(x-y)(ax+by).$ In any case, $ax^2+2hxy+by^2=0$ contains one of the lines $x+y=0$ or $x-y=0,$ which bisect the coordinate axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Integrating $\int^1_0 \frac{x^2e^{\arctan x}}{\sqrt{x^2+1}}\,dx$ This is a very hard integral that I am trying to solve. I’ve tried many substitutions, integration by parts, but I cannot evaluate this. Are there any other approaches I can take to solve this
$$\int^1_0 \dfrac{x^2e^{\arctan x}}{\sqrt{x^2+1}}\,dx$$
| Let $\displaystyle{I = \int^1_0 \frac{x^2 e^{\tan^{-1} x}}{\sqrt{x^2 + 1}} \, dx}.$
Let $x = \tan u, dx = \sec^2 u \, du.$ Limits: $ x = 1, u = \pi/4$ and $x = 0, u = 0.$ So
\begin{align*}I &= \int^{\frac{\pi}{4}}_0 \frac{\tan^2 u e^u}{\sqrt{\tan^2 u + 1}} \cdot \sec^2 \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u \tan^2 u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u (\sec^2 u - 1) \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u \sec^3 u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 \sec^2 u \cdot e^u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &= \Big{[}\tan u \cdot e^u \sec u \Big{]}^{\frac{\pi}{4}}_0 - \int^{\frac{\pi}{4}}_0 \tan u (e^u \sec u + e^u \sec u \tan u) \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du \quad \mbox{(by parts)}\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \tan^2 u \sec u \, du -\int^{\frac{\pi}{4}}_0 e^u \sec u \, du\end{align*}
Since the second integral here is the integral we started with, we have
\begin{align*}\Rightarrow 2I &= \sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du - \Big{[}e^u \sec u \Big{]}^{\frac{\pi}{4}}_0 + \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du \quad \mbox{(by parts)}\\&=\sqrt{2} e^{\frac{\pi}{4}} - \sqrt{2} e^{\frac{\pi}{4}} + \frac{1}{2}\\ &= \frac{1}{2}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.