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Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(t) = rt + s$, then for a general circle $\mathrm C(x, y) = (x - \alpha)^2 + (y - \beta)^2 - \gamma^2$ we can substitute the the parameters of the line and get
$$\mathrm C(x(t), y(t)) = a t^2 + bt + c = 0,$$
where $a,b,c \in \Bbb R$ and they depend on $p$,$q$.
Now if this quadratic has $2$ roots then the circle intersects the line at $2$ points and same for $1$ and $0$ roots.
If we are given that $\mathrm Z$ is tangent then we can using completing the square and get $(t - d(p, q) ) ^2 + e(p, q) = 0$.
So our condition for the line tangent to the circle would be $e(p, q) = 0$.
Parameterising the three equations,
$\mathbf L^\prime (l) = (2,0) + l (0,1)$, $\mathbf M^\prime(m) = (0, 5) + m(-1, 0)$ and $\mathbf N^\prime(n) = (2 , -1) + n(4,3)$.
Substituting these in the general equation of a circle, $\mathrm C(x, y) = (x - h)^2 + (y - k)^2 - r^2$.
$$\left\{ \begin{align} \mathrm{C(L_x, L_y)} &= (2-h)^2 +(l-k)^2 -r^2 \\ \mathrm{C(M_x, M_y)} &= (m+h)^2 +(5-k)^2 -r^2 \\ \mathrm{C(N_x, N_y)} &= (2+4n-h)^2 +(3n-1-k)^2 -r^2 \end{align} \right.$$
As per the above method,
From first equation we get $r^2 = (2 - h) \iff r = |2 - h|$ and from second equation, $r = |5 - k|$.
Completing the square on the third equation we get,
$$\mathrm{C(N_x, N_y)} = \left(5n + \dfrac{5 - 4h - 3k}{5}\right) + 5 - 4h + 2k - r^2 + h^2 + k^2 - \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
Therefore,
$$ 2 + 2k + h^2 + k^2 = r^2 + 4h + \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
So here we got three equations:
\begin{align} 2 +2k +h^2 +k^2 &= r^2 +4h +\left[1 - \dfrac{4h-3k}{5}\right]^2 \\ r &= |5-k| \\ r &= |2-h|. \end{align}
I don't mind solving the third equation after substituting for $r$ from first in it, it is just a quadratic. What I don't understand is whether I should take $r = + (5 - k)$ or $r = - (5 - k)$ and $r = +(2 - h)$ or $r = - (2 - h)$?
Which two out of those should I take? And why?
| You have by distance from a point to a line formula,
$$|h-2|=|k-5|=\frac{|3h-4k-10|}{5}=r.$$
Consequently, $h=2\pm r$, and $k=5\pm r$. So we have four cases
*
*$h=2+r$, $k=5+r$ implies $|3(2+r)-4(5+r)-10|=|r+24|=5r \implies r=6$. So the equation of the circle is $$\color{red}{(x-8)^2+(y-11)^2=36}.$$
*$h=2+r$, $k=5-r$ implies $|3(2+r)-4(5-r)-10|=|7r-24|=5r \implies r=2,12$. So the equation of the circles are $$\color{red}{(x-4)^2+(y-3)^2=4},$$ and $$\color{red}{(x-14)^2+(y+7)^2=144}.$$
*$h=2-r$, $k=5+r$ implies $|3(2-r)-4(5+r)-10|=|7r+24|=5r$, which does not have any solution such that $r >0$.
*$h=2-r$, $k=5-r$ implies $|3(2-r)-4(5-r)-10|=|r-24|=5r \implies r=4$. So the equation of the circle is $$\color{red}{(x+2)^2+(y-1)^2=16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Asymptotic notation question (logarithms) I am working through some problems and I have come across one I do not understand. Could someone clarify why
$$2x^3 + 3x^2\log(x) + 7x + 1$$
is $O(x^3\log(x))$ for $x>0$?
I guess I am missing some sort of knowledge to be able to answer this question :(
| if f(x) = $2x^3 + 3x^2\log(x) + 7x + 1$, f(x) = $x^3$$(2 + \frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3})$. Then if x approaches the infinity, ($\frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3}$) is negligible comparing to $x^3$.
And, if $f(x)$ = $2x^3 + 3x^3\log(x) + 7x + 1$, f(x) = $x^3log(x)$$(\frac{2}{log(x)} + 3 + \frac{7}{x^2} + \frac{1}{x^3})$. Then if x approaches the infinity, ($3+ \frac{7}{x^2} + \frac{1}{x^3}$) is negligible comparing to $x^3log(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the infinite product $\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$ $$\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$$
I've seen some similar questions asked. But this one is different from all these. Euler product does not apply. One cannot simply factorize $\left(1-\frac{2}{(2n+1)^2}\right)$ since the $\sqrt{2}$ on top will prevent terms from cancelling. Any help will be appreciated!
Note: we are expected to solve this in 2 mins.
| Considering the partial products
$$A_p=\prod_{n=1}^{p} \left(1-\frac{2}{(2n+1)^2}\right)$$ a CAS produced
$$A_p=-\cos \left(\frac{\pi }{\sqrt{2}}\right) \frac{\Gamma
\left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma
\left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma
\left(p+\frac{3}{2}\right)^2}$$ and using Stirling approximetion for large $p$, this leads to $$\log\left(\frac{\Gamma
\left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma
\left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma
\left(p+\frac{3}{2}\right)^2} \right)=\frac{1}{2p}+O\left(\frac{1}{p^2}\right)$$ then $$A_p =-\cos \left(\frac{\pi }{\sqrt{2}}\right)\left(1+\frac{1}{2p}+O\left(\frac{1}{p^2}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answer. Could someone explain the steps?
| $$\alpha'=\frac{\alpha}{\beta} ,\beta'=\frac{\beta}{\alpha}$$ now find sum and product of $\bf new$ roots
$$\quad{S'=\alpha'+\beta'\\p'=\alpha'\times \beta'\\
S'=\alpha'+\beta'=\frac{\alpha}{\beta} +\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{s^2-2p}{p}=\frac{(-\frac{-3}{2})^2-2.\frac{6}{2}}{\frac{6}{2}}\\p'=\alpha'\times\beta'=\frac{\alpha}{\beta} \times\frac{\beta}{\alpha}=1}$$ then put into $$\quad{x^2-s'x+p'=0\\x^2-\frac{-5}{4}x+1=0 \to \times 4 \\4x^2+5x+4=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How to prove $\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}$? $$
\quad{\forall x\in \mathbb{R}:\\
\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}}$$ I want to prove this.I tried to graph it and see whats going on ...https://www.desmos.com/calculator/xgjovvkal6
I also tried to prove it by derivation ,but it become complicated .
Can anybody give me an idea ? I am thankful in advance.
| There is a nice proof for $x\geq0$.
By Minkowski we obtain:
$$\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73}-\sqrt{157}=$$
$$=\sqrt 2\left(\sqrt{x^2} + \sqrt {{x^2} + x + \frac{1}{2}} + \sqrt {{x^2} - 5x + \frac{13}{2}} + \sqrt {{x^2} - 11x + \frac{73}{2}}-\sqrt{\frac{157}{2}}\right)=$$
$$=\sqrt 2\left(\sqrt{x^2} + \sqrt {\left(x+\tfrac{1}{2}\right)^2+\tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{5}{2}\right)^2+ \tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{11}{2}\right)^2+ \tfrac{25}{4}}-\sqrt{\tfrac{157}{2}}\right)\geq$$
$$=\sqrt 2\left(\sqrt{\left(x+x+\tfrac{1}{2}-x+\tfrac{5}{2}-x+\tfrac{11}{2}\right)^2+\left(0+\frac{1}{2}+\frac{1}{2}+\frac{5}{2}\right)^2}-\sqrt{\tfrac{157}{2}}\right)=$$
$$=13-\sqrt{157}>0.$$
We'll prove that for $x\leq0$ our inequality is also true.
Indeed, after replacing $x$ at $-x$ we need to prove that:
$$-\sqrt 2 x + \sqrt {2{x^2}-2x + 1} + \sqrt {2{x^2}+10x + 13} + \sqrt {2{x^2}+ 22x + 73}>\sqrt{157},$$ where $x\geq0.$
But $$\sqrt{2x^2-2x+1}=\sqrt{2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}}\geq\frac{1}{\sqrt2},$$
$$\sqrt{2x^2+10x+13}-\sqrt2x-\sqrt{11}=\frac{2x^2+10x+13-2x^2-2\sqrt{22}x-11}{\sqrt{2x^2+10x+13}+\sqrt2x+\sqrt{11}}>0$$ and $$\sqrt{2x^2+22x+73}\geq\sqrt{73}.$$
Thus, it's enough to prove that:
$$\frac{1}{\sqrt2}+\sqrt{11}+\sqrt{73}>\sqrt{157},$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| Vieta's theorem is the key, like in your previous question. It gives $\alpha+\beta=-\frac{5}{3}$ with no effort and
$$\alpha^3+\beta^3 = (\alpha+\beta)\left((\alpha+\beta)^2-3\alpha\beta\right) = -\frac{5}{3}\left(\frac{25}{9}-4\right) = \color{red}{\frac{55}{27}}$$
with very simple manipulations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do we use EEA and why do we multiply by 4 throughout? I am trying to understand the following example and solution, but I am confused by why we use EEA and why we need to multiply by $4$?
Example. Find a solution to
$$\begin{align}x &\equiv 88 \phantom{1}\mod 6 \\x &\equiv 100 \mod 15 \end{align}$$
Solution 1:
From the first equation we know we want $x − 88 = 6k$ for some integer $k,$ so $x$ is of the form $88+6k$.
From the second equation, we also want $88+6k \equiv 100 \mod 15$, so we want $6k\equiv 12\mod 15$.
Use the extended Euclidean Algorithm to find that $15(1) + 6(−2) = 3.$ Multiply through by $4$ to get $15(4) + 6(−8) = 12.$ Thus $6(−8) \equiv 12 \mod 15$, and we can take $k = −8$. This results in $x = 88+6(−8) = 40$.
| This is how I would solve it.
$x \equiv 88 \pmod{6}$ implies $x = 88 + 6s$ for some integer $s$.
So $x \equiv 100 \pmod {15}$ becomes.
Eliminating $x$, we get
\begin{align}
x &\equiv 100 \pmod {15}\\
88 + 6s &\equiv 100 \pmod {15} \\
6s &\equiv 12 \pmod{15} &(\text{$3$ divides $6,12,$ and $15$})\\
2s &\equiv 4 \pmod 5 & (2^{-1}\equiv 3 \pmod 5) \\
s &\equiv 2 \pmod 5 &(\text{After mult. through by 3})
\end{align}
So $s =2+5t$ for some integer $t$.
Hence
\begin{align}
x &= 88 + 6s \\
x &= 88 +6(2+5t) \\
x &= 100 + 30t
\end{align}
In other words, $x \equiv 100 \pmod{30}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question about analytic geometry (vectors) Let $a , b, c$ be three vectors such that $a+3b+c = o$ where $o$ is $(0,0,0)$ vector. we also know $|a| = 3 , |b| = 4 , |c| =6$. we want to find $a.b + a.c + b.c$
I know we can solve it by saying $a+b+c = -2b$ and then squaring the sides. and at last we get the answer 3/2 . But my problem is that is it even possible for $|a| = 3 , |b| = 4 , |c| =6$ to be such that $a+3b+c = o$ ? I think it is not possible because of triangle inequality. If it is possible, how? and If it is not, why do we get an answer for it?
| Dot the equation with each of the vectors
\begin{eqnarray*}
\mid a \mid^2 +3 a \cdot b +a \cdot c =0 \\
a \cdot b +3 \mid b \mid^2 +b \cdot c =0 \\
a \cdot c +3 b \cdot c+\mid c \mid^2 =0 \\
\end{eqnarray*}
Now add the first and third equations and subtract the second $2(a \cdot b +a \cdot c +b \cdot c)+9+36-48=0$ ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{4x}{x+7}I know how to solve the problem. The reason I post the problem here is to see whether there is a quick approach, rather than a traditional method, to solve the problem.
The problem is: find $x$ that satisfy $\frac{4x}{x+7}<x$
I considered two cases: $x+7>0$ and $x+7<0$, and then went through details to find x. The process took me a few minutes. I wonder whether there is a simple way to find the answer.
| Transform $\frac{4x}{x+7}<x$ into the equivalent $-\frac{x^2+3x}{x+7}=\frac{4x}{x+7}-x<0$ or $\frac{x^2+3x}{x+7}>0$.
This reduces to question to finding the sign of $\frac{x^2+3x}{x+7}$, which can be solved by making a simple table:
$$
\begin{array}{c|ccccc}
& & -7 & & -3 & & 0 & \\
\hline
x & - & - & - & - & -& 0 & + \\
x+3 & - & - & - & 0 & + & + & + \\
x(x+3) & + & + & + & 0 & - & 0 & +\\
x+7 & - & 0 & + & + & + & + & +\\
\dfrac{x^2+3x}{x+7} & - & ! & + & 0 & -& 0 & +
\end{array}
$$
The answer is therefore $(-7,-3) \cup (0, +\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conditional probability of multivariate gaussian I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.
Let $\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda ) $ , where
$$\begin{align}
\mu &= \begin{pmatrix}
1 \\
1
\end{pmatrix}
\end{align}
$$
$$
\begin{align}
\Lambda &= \begin{pmatrix}
3 \quad 1\\
1 \quad 2
\end{pmatrix}
\end{align}
$$
We are tasked with computing: $P(X_1 \geq 2 \mid X_2 +3X_1=3)$
I here begin by doing a transformation,
$$ \mathbf{Y} = (Y_1, Y_2)', \qquad Y_1 = X_1, \qquad Y_2 = X_2 + 3X_1$$
We now are interested in the probability,
$$P(Y_1 \geq 2 \mid Y_2 = 3)$$
Since we can write that $\mathbf{Y = BX}$, it follows that,
$$\mathbf{Y} \in \mathcal{N}(\mathbf{B\mu, B\Lambda B')})$$
where
$$\mathbf{B}= \begin{pmatrix}
1 \quad 0\\
3 \quad 1
\end{pmatrix} \rightarrow \quad \mathbf{B \mu} = \begin{pmatrix}
1 \\
4
\end{pmatrix}, \quad \mathbf{B\Lambda B'}= \begin{pmatrix}
1 \quad 0\\
3 \quad 1
\end{pmatrix} \begin{pmatrix}
3 \quad 1\\
1 \quad 2
\end{pmatrix} \begin{pmatrix}
1 \quad 3\\
0 \quad 1
\end{pmatrix} = \begin{pmatrix}
3 \quad 10\\
10 \; \; 35
\end{pmatrix}$$
We thereafter know that we can obtain the conditional density function by,
$$
f_{Y_1\mid Y_2 = 3} (y_1) = \frac{f_{Y_1,Y_2}(y_1, 3)}{f_{Y_2}(3)} \tag 1
$$
The p.d.f. of the bivariate normal distribution,
$$f_{Y_1, Y_2}(y_1, y_2) = \frac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} e^{\frac{1}{2(1-\rho^2)}(\frac{(y_1 - \mu_1)^2}{\sigma_1^2} - \frac{2 \rho (y_1 - \mu_1)(y_2 - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y_1 - \mu_1)^2}{\sigma_2^2})} $$
The marginal probability density of $Y_2$,
$$f_{Y_2}(y_2) = \frac{1}{\sqrt{2\pi} \sigma_2} e^{-(y_2 - \mu_2)^2 / (2\sigma_2^2)}$$
Given that,
$$\sigma_1 = \sqrt{3}, \quad \sigma_2 = \sqrt{35}, \quad \rho = \frac{10}{\sigma_1 \sigma_2 } = \frac{10}{\sqrt{105}} $$
we are ready to determine (1). However, the resulting expression, which I then need to integrate as follows,
$$
Pr(Y_1 \geq 2 \mid Y_2 = 3) = \int_2^\infty f_{Y_1\mid Y_2 = 3} (y_1) \, dy_1
$$
becomes quite ugly, making me unsure whether I've approached the problem in the wrong way?
Thanks in advance!
| https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions Just apply the result after you obtain the distribution of $\mathbf{Y}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$ Find value of
$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$
I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}$$
$\implies$
$$S+\frac{1}{S}=\frac{\sum_{i=1}^{399}(20+\sqrt{i})+2S_1+\sum_{i=1}^{399}(20-\sqrt{i})+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$ $\implies$
$$S+\frac{1}{S}=\frac{15960+2S_1+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$
where $$S_1=\sum_{i \ne j=1}^{399}\left(\sqrt{20+\sqrt{i}}\right)\left(\sqrt{20+\sqrt{j}}\right)$$ and like wise
$$S_2=\sum_{i \ne j=1}^{399}\left(\sqrt{20-\sqrt{i}}\right)\left(\sqrt{20-\sqrt{j}}\right)$$
Any way to proceed from here?
| Let
$$S_1=\sum_{k=1}^{m^2-1} \sqrt{m+\sqrt{k}}$$
and
$$S_2=\sum_{k=1}^{m^2-1} \sqrt{m-\sqrt{k}}.$$
(your sum is $S_1/S_2$ at $m=20$). Note that
$$\left(\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}\right)^2=2m+2\sqrt{m^2-k},$$
so
$$\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}=\sqrt{2}\sqrt{m+\sqrt{m^2-k}}.$$
Thus
$$S_1+S_2=\sum_{k=1}^{m^2-1} \left(\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}\right) = \sqrt{2}\sum_{k=1}^{m^2-1} \sqrt{m+\sqrt{m^2-k}}.$$
However, under the substitution $k\to m^2-k$, the sum on the RHS is simply $S_1$. Thus,
$$S_1+S_2=S_1\sqrt{2}$$
$$S_2=\left(\sqrt{2}-1\right)S_1$$
$$\frac{S_1}{S_2}=\boxed{1+\sqrt{2}}.$$
As Robert Israel points out, this result is independent of $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find all points on which a function is discontinuous. $ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3+y^3}{x^2+y^2} $ seems pretty hard to analyse.
| $x = r\cos \theta$, $y = r\sin \theta$
instead of $(x,y) \rightarrow (0,0)$ I can now use $r\rightarrow0$
$$\begin{align}
\lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2\cos^2\theta + r^2\sin^2\theta}
&=\, \lim_{r\to0} \frac{r (\cos^3\theta + \sin^3\theta)}{\cos^2\theta + \sin^2\theta} \\
&=\,0
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456976",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer.
Well if this is an integer then $3x-1 \mid 2x^2-3$ so $2x^2-3=3x-1(k)$ such that $k\in \mathbb{Z}$ from here not sure where to go I know that it has no solutions I just can't see the contradiction yet.
| Using the extended Euclidean algorithm in $\mathbb Q[x]$, we get
$$
25 = (-9)(2x^2-3)+(6x+2)(3x-1)
$$
Therefore, if $3x-1$ divides $2x^2-3$, then it divides $25$.
So, $3x-1 \in \{\pm 1, \pm 5, \pm 25 \}$ and there is not much to test.
An easier (but equivalent) argument is $2x^2-3 = (3x-1)q(x) + r$, where $r$ is the value of $2x^2-3$ at $x=1/3$, the root of $3x-1$. This gives $r=-25/9$. The rest follows as above.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\int \frac{x}{x^5-7} dx$? I have tried out many trigonometric substitution like $x=\sin^{\frac{2}{5}}z$. But it did not work.
| Substitute $x=\sqrt[5]{7}u$, and you have
$$
\begin{align}
\int\frac{\sqrt[5]{7}u}{7u^5-7}\,\sqrt[5]{7}du
&=\frac{\sqrt[5]{49}}{7}\int\frac{u}{u^5-1}\,du\\
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{Bu+C}{u^2-2hu+1}+\frac{Du+E}{u^2-2ku+1}\right)\,du\\
\end{align}
$$
where $h=\cos(2\pi/5)=\frac{\sqrt{5}-1}{4}$ and $k=\cos(4\pi/5)=\frac{-\sqrt{5}-1}{4}$. The values of $A,B,C,D,E$ can be found using linear algebra, after recombining the three terms and comparing the coefficients of powers of $u$ in the numerator to that of $\frac{u}{u^5-1}$.
Continuing:
$$
\begin{align}
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{Bu+C}{(u-h)^2+1-h^2}+\frac{Du+E}{(u-k)^2+1-k^2}\right)\,du\\
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{B(u-h)+Bh+C}{(u-h)^2+1-h^2}+\frac{D(u-k)+Dk+E}{(u-k)^2+1-k^2}\right)\,du\\
&=\frac{\sqrt[5]{49}}{7}\left(A\ln\lvert u-1\rvert+\frac{B}{2}\ln\left((u-h)^2+1-h^2\right)+\frac{Bh+C}{\sqrt{1-h^2}}\arctan\left(\frac{u-h}{\sqrt{1-h^2}}\right)\right.\\
&\phantom{{}={}}\left.+\frac{D}{2}\ln\left((u-k)^2+1-k^2\right)+\frac{Dh+E}{\sqrt{1-k^2}}\arctan\left(\frac{u-k}{\sqrt{1-k^2}}\right)\right)+\text{constant}\\
\end{align}
$$
Note $1-h^2=\sin^2(2\pi/5)$ and $1-k^2=\sin^2(4\pi/5)$, which may help some of you needed to carry this through to an explicit end.
The "constant" may take two different values on either side of $u=1$ where the discontinuity is.
The last step would be to back-substitute from $u$ to $x$. Then the "constant" might change values at $x=\sqrt[5]{7}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying a long polynomial equation in (the reciprocal of) the Golden Ratio I'm trying to show that the following equation holds true:
$$4\sigma^{12}+11\sigma^{11}+11\sigma^{10}+9\sigma^9+7\sigma^8+5\sigma^7+3\sigma^6+\sigma^5+\sigma^4+\sigma^3+\sigma^2+\sigma = 1 + 2\sigma$$
where $\sigma$ is the reciprocal of the golden ratio; that is, $\sigma := \frac12(\sqrt{5} - 1)$.
There must be a good way to show this. However, so far all my attempts have failed.
This problem has a real world application. If you are interested in background take a look at A Fresh Look at Peg Solitaire[1] and in particular at Figure 9.
[1] G. I. Bell [2007], A fresh look at peg solitaire, Math. Mag. 80(1), 16–28, MR2286485
| As $\sigma$ verifies the identity $\sigma^2=-\sigma+1$, you can perform the following reductions (the powers of $\sigma$ have been omitted for conciseness):
$$4+11+11+9+7+5+3+1+1+1+1-1-1,\\
7+15+9+7+5+3+1+1+1+1-1-1,\\
8+16+7+5+3+1+1+1+1-1-1,\\
8+15+5+3+1+1+1+1-1-1,\\
7+13+3+1+1+1+1-1-1,\\
6+10+1+1+1+1-1-1,\\
4+7+1+1+1-1-1,\\
3+5+1+1-1-1,\\
2+4+1-1-1,\\
2+3-1-1,\\
1+1-1,\\
0+0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$
is this method correct?
| Since
\begin{align}
\left|\frac{x^3+y^3}{x^2+y^2}\right|&=\left|(x+y)\cdot\frac{x^2-xy+y^2}{x^2+y^2}\right|\\
&=|x+y|\left|1-\frac{xy}{x^2+y^2}\right|\\
&\le|x+y|\left(1+\frac{|xy|}{x^2+y^2}\right)\quad(\because\text{the triangle inequality})\\
&\le|x+y|\left(1+\frac{1}{2}\right)\quad(\because\text{the AM-GM inequality})\\
&=\frac{3}{2}|x+y|
\end{align}
and $\displaystyle \lim_{(x,y)\to(0,0)}\frac{3(x+y)}{2}=0$, $\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^3+y^3}{x^2+y^2}=0$ by the Squeeze Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}\tag2$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={A\over 2^n-1}+{B\over 2^{n+1}-1}+
{C\over 2^{n+2}-1}\tag3$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 3(2^n-1)}-{1\over 2^{n+1}-1}+
{2\over 3(2^{n+2}-1)}\tag4$$
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}={1\over 9}\tag5$$
| You are almost done. Note that the three series on the LHS of your last line are convergent and they can be written as
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$
which is equal to
$$
{1\over 3}\sum_{n=1}^{2}{1\over 2^n-1}-\sum_{n=2}^{2}{1\over 2^{n}-1}+\left(\frac{1}{3}-1+{2\over 3}\right)\sum_{n=3}^{\infty}{1\over 2^{n}-1}=\frac{1}{9}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can someone tell me what should i do next? Should I use the inequality between Arithmetic and Geometric mean? Problem 1: Let $a,b,c> 0$
$ab+3bc+2ca\leqslant18$
Prove that:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$.
I started on this way:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$
$\frac{3bc+2ac+ab}{abc}\geqslant 3 $
$\frac{ab+3bc+2ac}{abc}\geqslant 3 \times \frac{abc}{3}$
$\frac{ab+3bc+2ca}{3}\geqslant abc$
| By the AM-GM inequality, we have $$\left(\frac{ab+3bc+2ac}{3}\right)^3\ge 6{(abc)}^2$$
$$\Rightarrow \frac 1{abc}\ge \sqrt{\frac{3^3.6}{(ab+3bc+2ac)^3}}$$
$$\Rightarrow \frac{ab+3bc+2ac}{abc}\ge \sqrt{\frac {3^3.6}{ab+3bc+2ac}}\ge\sqrt{\frac{3^3.6}{18}}=3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Summing up $3+5+9+17+...$
Find the sum of sum of $3
+5+9+17+...$ till $n$ terms.
Using Method of differences, the sum of the series is
$$\sum\limits_{j=1}^n 2^{j-1}+n$$
I am facing difficulty in evaluating $$\sum\limits_{j=1}^n 2^{j-1}$$. How do I do that?
Now I have $2^0 + 2^1 + 2^2 ... 2^{n-1}$
The sum of this series is : $2^n- 1$ as sum of GP is given by $a(1-r^n)/(1-r)$.
Here $a = 1, r =2$
| Maybe do you mean the following?
$$3+5+9+17+...=\sum_{k=1}^n2^k+n=\frac{2(2^n-1)}{2-1}+n=2^{n+1}-2+n.$$
If so let $S_n=3+5+...+(2^n+1).$
Thus, $$S_n-S_1=\sum_{k=2}^n(S_k-S_{k-1})=\sum_{k=2}^n(2^k+1)=\frac{4(2^{n-1}-1)}{2-1}+n-1,$$
which gives $$S_n=2^{n+1}+n-2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve integral $\int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}$ I need to find:
$$\int\frac{1}{\sin x+ \cos x+\tan x +\cot x}\ dx$$
My attempts:
I have tried the conventional substitutions. I have tried the $\tan(x/2)$ substitutions, tried to solve it by quadratic but nothing has worked so far.
| Let $t=x-\frac\pi4$ and recognize $2\sin x\cos x= 2\cos^2t-1$
$$I= \int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}
=\frac1{\sqrt2}\int \frac{\cos^2t-\frac12}{\cos^3t-\frac12\cos t+\frac1{\sqrt2}}dt
$$
Factorize the denominator ($y=\cos t$)
$$y^3-\frac12y+\frac1{\sqrt2}
=(y+a)\left(y-\frac {a-ib}2 \right) \left(y-\frac {a+ib}2 \right)
$$
where $a$ given below is the real root and $b =\frac12\sqrt{\frac3{\sqrt2a}-\frac12}$
$$a= \bigg(\frac1{2\sqrt2}+\frac16\sqrt{\frac{13}3}\bigg)^{1/3}+ \bigg(\frac1{2\sqrt2}-\frac16\sqrt{\frac{13}3}\bigg)^{1/3}
\overset{\cdot}=1.0758
$$
Then, partially-fractionalize the integrand
\begin{align}
I&= \frac1{3+\sqrt2a}\int \frac{\frac1{\sqrt2}}{\cos t+a}
+\Re \frac{a+\sqrt2 -i \frac{a^2}b}{\cos t +\frac{ib-a}2}\ dt\\
\end{align}
and apply the known integral
$\int \frac{1}{\cos t+p}dt=\frac2{\sqrt{p^2-1}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{p+1}{p-1}}}
$
to obtain
\begin{align}I
&= \frac1{3+\sqrt2a}\bigg(\frac{\sqrt2}{\sqrt{a^2-1}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{a+1}{a-1}}}
+4\Re \frac{a+\sqrt2-i\frac{a^2}b}{\sqrt{(ib-a)^2-4}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{ib-a+2}{ib-a-2}}} \bigg)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Laurent series of algebraic function Consider function
$$f(z)=\frac{1}{\sqrt{a^2-z^2}+b\sqrt{c^2-z^2}}.$$
It has pole
$$z = \frac{\sqrt{b^2 c^2 - a^2}}{\sqrt{b^2-1}}$$.
I wanna find Laurent series of $f$ around singular point.
For $b=1$, I put $f$ in partial fraction to proced.
How to find Laurent series for $b \ne 1$?
| Hint:
$$
\frac{1}{\sqrt{a^2-z^2}+b\sqrt{c^2-z^2}}=\frac{\sqrt{a^2-z^2}-b\sqrt{c^2-z^2}}{a^2-z^2-b^2(c^2-z^2)}=\frac{\sqrt{a^2-z_0^2-(z^2-z_0^2)}-b\sqrt{c^2-z_0^2-(z^2-z_0^2)}}{(b^2-1)(z^2-z_0^2)},
$$
where
$$
z_0 = \frac{\sqrt{b^2 c^2 - a^2}}{\sqrt{b^2-1}}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.
I have a few observations made but not much room for a full solution.
First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.
Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.
From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
| Here is a solution using less machinery than in Will Jagy's answer.
As Daniel Robert-Nicoud noted in comments, the fact that $3^a+7^b\equiv(-1)^a+(-1)^b$ mod $4$ implies $a$ and $b$ must have opposite parity in order for the (even) sum $3^a+7^b$ to be a perfect square. So we seek to show that $(a,b)=(1,0)$ is the only solution in the (odd,even) case and $(a,b)=(2,1)$ is the only solution in the (even,odd) case.
If $a$ is odd and $b=2m$ is even, then we can rewrite $3^a+7^b=n^2$ as $3^a=(n+7^m)(n-7^m)$. This implies $n+7^m$ and $n-7^m$ are each powers of $3$, say $3^c$ and $3^d$ with $c+d=a$. But then $2\cdot7^m=3^c-3^d$, so we must have $d=0$ and $c=a$, since $3\not\mid2\cdot7^m$. One solution to $2\cdot7^m=3^a-1$ is $(a,m)=(1,0)$. It remains to show there are no solutions with $m\gt0$. For this it suffices to note that $3^a\equiv1$ mod $7$ if and only if $6\mid a$, which contradicts the assumption that $a$ is odd.
In the other parity case, if $a=2m$ is even and $b$ is odd, then, along similar lines, we must have $2\cdot3^m=7^b-1$, which has $(m,b)=(1,1)$ as one solution. After checking that there is no solution with $m=0$, it remains to show there are no solutions with $m\gt1$. If there were, then we would have $7^b\equiv1$ mod $9$, which would imply $b\equiv3$ mod $6$ (since $6$ is the order of the multiplicative group of units mod $9$ and $7^3\equiv(-2)^3=-8\equiv1$ mod $9$). Writing $b=3r$ (with $r$ odd, but that doesn't matter), we note that $(7^3-1)\mid(7^{3r}-1)$, so that $7^3-1=2\cdot3^2\cdot19$ divides $2\cdot3^m$, an obvious contradiction.
| {
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Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe I need to differentiate or integrate? But I'm not sure how that might work.
Any ideas? Thanks.
| Hint
$${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.=\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j $$
$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} x^{j+1} \binom n j =\int \sum_{j=0}^{n} (-1)^j x^{j}\binom n j =\int (1-x)^n=- \frac { (1-x)^{n+1}} {n+1} $$
$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j =\int_0^1 (1-x)^n=\frac 1 {n+1}$$
| {
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$\varepsilon$-$N$ proof question
Use the definition of limits to prove $$\lim_{n\to\infty} \frac{n^3-n^2+9}{n^3-3n} =1 $$
So far I got
$$ \left| \frac{n^3-n^2+9}{n^3-3n} -1 \right| < \epsilon, $$
the left hand side is the same as
$$ \left| - \frac{n^2-3n-9}{n(n^2-3)} \right|$$
which we can then drop the absolute value and then make the inequality argument that
$$ \frac{n^2-3n-9}{n(n^2-3)} < \frac{n^2}{n(n^2-3)} =\frac{n}{(n^2-3)}.$$
From here I'm having trouble getting rid of the last $- 3$ to make the argument that $$\frac{1}{n} < \epsilon $$ which then means $\frac{1}{\epsilon} < n$
Should I choose a different $\epsilon$? Is there some alternate way of expressing something that I'm missing. Thanks for the help!
| Since you choose $n$ large anyway, you may assume $n^2 - 3 > \frac{1}{2} n^2$ (that is, you choose $n>\sqrt 2$, or simply $n>2$). Then
$$\frac{n}{n^2 - 3}< \frac{n}{\frac 12 n^2} = \frac{2}{n}.$$
Now we want $\frac 2n<\epsilon$. So we need $n >\frac {2}{\epsilon}$.
Combining the two information, one choose
$$N = \max\{2, \frac{2}{\epsilon}\}. $$
| {
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Basic modular inverses I know how to do modular inverses in a hypothetical sense with the Euclidian method, and have been trying to do the, but I seem to keep getting the incorrect answer.
I'm trying to find the inverse of $\;5\pmod {13}$, for example.
The answer should be 8, but I can't seem to get that. These are my steps:
\begin{align}
13 & = 2(5)+3\\
5 & = 1(3)+2\\
3 & =1(2)+1
\end{align}
\begin{align}
1 & =3-1(2)\\
& =3-1(5-1(3))\\
& = 2(3)-5\\
& = 2(13-2(5))-5\\
& = 2(13)-4(5)-5\\
& = 2(13)-5(5)
\end{align}
I'm sure I'm just misunderstanding the steps, but I don't know how.
Also, can the same method to used to find the inverse of $5\pmod{11}$, since $11=2(5)+1$? I immediately don't know how to proceed from here.
| Ok, after applying the extended Euclidean algorithm applied to $13$ (modulus) and $5$ (the number you want to invert)
you will find that
$$1 = 13\cdot 2 + -5 \cdot 5$$
as stated. But taking this whole equation modulo $13$, we get
$$1 \equiv 13\cdot 2 + -5 \cdot 5 \equiv -5 \cdot 5 \equiv 8 \cdot 5 \pmod{13}$$
using that multiplies of $13$ vanish (are equivalent to $0$) and that $8 \equiv -5 \pmod{13}$ (add $13$ to $-5$). So $8$ and $5$ are each other's inverses in $\mathbb{Z}_{13}$. And indeed $5 \times 8 = 40$ is one plus $39$, a multiple of $13$.
For $5$ modulo $11$ we first write $1$ as a combination of $11$ and $5$:
$$1 = 1\cdot 11 - 2\cdot 5$$ and taking everything modulo $11$ again, the first term vanishes and we get that $-2$ is the inverse of $5$ modulo $11$, but $-2 \equiv 9 \pmod{11}$, so we can also use $9$ if that's more convenient. (and indeed $9\times 5 = 45 = 1 + 4\times 11 \equiv 1 \pmod{11}$ so that works out.
| {
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Find two arithmetic progressions of three square numbers I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:
\begin{align}
\ & a^2 +r = b^2 \\
& b^2 +r = c^2 \\
& a^2 +c^2 = 2\,b^2 \\
\end{align}
and
\begin{align}
\ & d^2 +r = e^2 \\
& e^2 +r = f^2 \\
& d^2 +f^2 = 2e^2 \\
\end{align}
where $a,b,c,d,r \in \Bbb N$.
Here is an example that almost works:
\begin{align}
\ & 23^2 +41496 = 205^2 \\
& 205^2 + 41496 = 289^2 \\
& 23^2 +289^2 = 2\,(205)^2 \\
\end{align}
and
\begin{align}
\ & 373^2 + 41496 = 425^2 \\
& 425^2 + 41496 = \color{#C00000}{222121} \\
& 23^2 + \color{#C00000}{222121} = 2\,(205)^2 \\
\end{align}
where the difference is $41496$, but the last element isn't a square number.
I can't find an example of two progressions with three numbers and the same common difference. Could you demonstrate that such progressions are nonexistent using reductio ad absurdum to this statement?
| $$(a,b,c,d,e,f,r)=(1,29,41,23,37,47,840)$$
satisfies
$$a^2 +r = b^2,\quad b^2 +r = c^2,\quad a^2 +c^2 = 2b^2$$
$$d^2 +r = e^2,\quad e^2 +r = f^2,\quad d^2 +f^2 = 2e^2$$
| {
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Finding the pdf of $Y=X^2-X$ given that the pdf of $X$ is $f_X(x) = 3x^2I_{(0,1)}(x)$
Consider a random variable $X$ having pdf
$$f_X(x) = 3x^2I_{(0,1)}(x)$$
Give the pdf of $Y=X^2−X$ and the value of $E(Y)$.
I have from an inverse calculator that if $Y=X^2−X$ then $X = {1\over2}(1\pm \sqrt{4Y+1})$. We have,
$F_Y(y)=P(Y\leq y)$
$=P(X^2-X\leq y)$
$=P(X\leq{1\over2}(1\pm \sqrt{4Y+1}))$
$=P(({1\over2}(1- \sqrt{4Y+1}) \leq X \leq {1\over2}(1 + \sqrt{4Y+1}))$.
$=F_X({1\over2}(1 + \sqrt{4Y+1}))$ $-$ $F_X({1\over2}(1- \sqrt{4Y+1})$
$=\frac{1}{8}\left(\sqrt{4y+1}-1\right)^3+\frac{1}{8}\left(\sqrt{4y+1}+1\right)^3$
I do not think I am doing this correctly. Based on other similar problems that I am looking at, this was much more tedious in comparison, and so I think I might have made an error. I am new to these kind of problems, so any help would be much appreciated.
| Sketch the graph $Y=X^2-X$, one can see that
$$0 \ge Y \ge -\frac{1}{4}$$
and there are two values of $X$ corresponding to each $Y$, except for $Y=-\frac{1}{4}$.
By transformation method formula
$$f_Y(y)=f_X(x_1)\bigg|\frac{dx_1}{dy}\bigg|+f_X(x_2)\bigg|\frac{dx_2}{dy}\bigg|$$
$$=\frac{3}{8}(1-\sqrt{4y+1})^2\bigg|\frac{d}{dy}(1-\sqrt{4y+1})\bigg|+\frac{3}{8}(1+\sqrt{4y+1})^2\bigg|\frac{d}{dy}(1+\sqrt{4y+1})\bigg|$$
$$=\frac{3}{8}(2+4y-2\sqrt{4y+1})\frac{2}{\sqrt{4y+1}}+\frac{3}{8}(2+4y+2\sqrt{4y+1})\frac{2}{\sqrt{4y+1}}$$
$$=3\frac{1+2y}{\sqrt{1+4y}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a + b + c + bc. My math trainer friend ask my help to look for the shortest possible solution for this problem:
Let $a$, $b$ and $c$ be positive integers such that
$$\left \{\begin{matrix}a + b + ab = 15 \\
b + c + bc = 99 \\
c + a + ca = 399\end{matrix}\right. $$
Find $a + b + c + bc.$
I tried elimination but it took us for about 3 min to do it. This question was intended for 15 seconds only.
| Adding $1$ to each equation, we get that
$$\left \{\begin{matrix}a + b + ab +1 = 16 \\
b + c + bc+1 = 100 \\
c + a + ca+1 = 400\end{matrix}\right. $$
Factoring, we get
$$\left \{\begin{matrix}(b+1)(a+1) = 16 \\
(b+1)(c+1) = 100 \\
(c+1)(a+1) = 400\end{matrix}\right. $$
Dividing equation 2 from equation 3 and multiplying the result with equation 1, we get that $(a+1)^2=64$, meaning $a+1=8$. Equation one gives us $b+1=2$, and equation three gets us $c+1=50$, getting that
$$a+b+c+bc=7+1+49+49=106$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How should one proceed in this trigonometric simplification involving non integer angles? The problem is as follows:
Find the value of this function
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$
when $\omega=33^{\circ}{20}'$ and $\phi=56^{\circ}{40}'$.
Thus,
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right)^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right)^{2}$$
By solving the power I obtained the following:
$$A= \cos^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +\sin^{2}\frac{\omega}{2} +\sin^{2}\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}$$
I noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:
\begin{align}
A &= \cos^{2}\frac{\omega}{2} +\sin^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +\sin^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\
&= 1+1+2\cos\frac{\omega}{2}cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}
\end{align}
Since the latter terms are another way to write prosthapharesis formulas I did the following:
\begin{align}
\cos\alpha +\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\
\cos\alpha -\cos\beta &= -2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \\
\alpha+\beta &= \omega \\
\alpha-\beta &= \phi
\end{align}
By solving the system I found: $\alpha=\frac{\omega+\phi}{2}$ and $\beta=\frac{\omega-\phi}{2}$.
Therefore by inserting these into the problem:
\begin{align}
A &= 1+1 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\
&= 2 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} -\left(-2\sin\frac{\omega}{2}\cos\frac{\phi}{2}\right) \\
&= 2 +\cos\frac{\omega+\phi}{2} +\cos\frac{\omega-\phi}{2} -\left(\cos\frac{\omega+\phi}{2} -\cos\frac{\omega-\phi}{2}\right)
\end{align}
By cancelling elements,
$$A=2 +2\cos\frac{\omega-\phi}{2}$$
However I'm stuck at trying to evaluate these values:
\begin{align}
\omega &= 33^{\circ}{20}'\; \phi=56^{\circ}{40}' \\
\omega-\phi &= \left(33+\frac{20}{60}\right) -\left(56+\frac{40}{60}\right) = -23-\frac{20}{60}
\end{align}
Therefore,
$$A=2+2\cos\left(\frac{-23-\frac{20}{60}}{2}\right).$$
However the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?
| There is a mistake in your expansion of the second bracket.
We should have
$A=\cos^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2\frac{\phi}{2}$
$=2+2(\cos\frac{\omega}{2} \cos\frac{\phi}{2} -\sin\frac{\omega}{2}\sin\frac{\phi}{2})$ .
In the bracket we have the expansion for $\cos(\frac{\omega}{2}+\frac{\phi}{2})$
and fortunately $\omega+\phi=90^\circ$.
Thus $A=2+2\cos(45^\circ)$
$=2+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need Clarification on Trig Substitution $$\int\frac{x}{(1+x^2)^\frac{3}{2}}\ \mathrm dx$$
becomes
$$\sin\theta = \frac{x}{\sqrt{x^2+1}}$$
I am aware that we would use $x = a^2\tan^2\theta$. and that our answer before we plug in from the triangle becomes $\sin\theta$.
I just took a test and got the answer of
$$\sin\theta = \frac{x}{(x^2+1)^\frac{3}{2}}$$
Why does the $\dfrac{3}{2}$ become $\dfrac{1}{2}$ when we are plugging in for $\sin\theta$?
| I'd write $\displaystyle \int \frac 1 {(x^2+1)^{3/2}} \Big( x\, dx\Big) = \int \frac 1 {u^{3/2}} \left( \frac 1 2\,du\right) = \cdots,$ etc.
But if you must use a trigonometric substitution you can do this:
\begin{align}
x & = \tan\theta \\
dx & = \sec^2\theta\,d\theta \\[10pt]
\int \frac x {(1+x^2)^{3/2}} \,dx & = \int \frac{\tan\theta}{\sec^3\theta} \, \sec^2\theta\,d\theta = \int \sin\theta\,d\theta = -\cos\theta + C \\[10pt]
& = \frac{-1}{\sqrt{1+x^2}} + C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integrate $f(x)=\sqrt{x^2+2x+3}.$ Completing the square and letting $t=x+1$, I obtain $$\int\sqrt{(x+1)^2+2} \ dx=\int\sqrt{t^2+2}\ dt.$$
Letting $u=t+\sqrt{t^2+2},$ I get
\begin{array}{lcl}
u-t & = & \sqrt{t^2+2} \\
u^2-2ut+t^2 & = & t^2+2 \\
t & = & \frac{u^2-2}{2u} \\
dt &=& \frac{u^2+2}{2u^2}du
\end{array}
Thus the integral becomes
$$F(x)=\int \left(u-\frac{u^2-2}{2u}\right)\left(\frac{u^2+2}{2u^2}\right) \ du = \int \left(\frac{u^2+2}{2u}\right)\left(\frac{u^2+2}{2u^2}\right) \ du =\int\frac{u^4+4u^2+4}{4u^3} \ du.$$
This integrand is nicely divided into
\begin{array}{lcl}
F(x) & = & \frac{1}{4}\int u \ du+\int \frac{1}{u} \ du+\int \frac{1}{u^3}=\frac{u^2}{8}+\ln{|u|}-\frac{1}{2u^2}+C \\
& = & \frac{(t+\sqrt{t^2+2})^2}{8}+\ln{|t+\sqrt{t^2+2}|}-\frac{1}{2(t+\sqrt{t^2+2})^2}+C \\
\end{array}
And finally in terms of $x$:
$$F(x)=\frac{(x+1+\sqrt{x^2+2x+3})^2}{8}+\ln{|x+1+\sqrt{x^2+2x+3}|}-\frac{1}{2(x+1+\sqrt{x^2+2x+3})^2}+C.$$
The answer in the book is:
$$F(x)=\frac{1}{2}\left((x+1)\sqrt{x^2+2x+3}+2\ln{|x+1+\sqrt{x^2+2x+3}}|\right)+C.$$
Can anyone help me identify where I missed what?
| Compute
\begin{align}
\frac{(t+\sqrt{t^2+2})^2}{8}-\frac{1}{2(t+\sqrt{t^2+2})^2}
&=
\frac{(t+\sqrt{t^2+2})^2}{8}-\frac{(t-\sqrt{t^2+2})^2}{2(t^2-(t^2+2))^2}
\\[6px]
&=
\frac{t^2+2t\sqrt{t^2+2}+(t^2+2)-t^2+2t\sqrt{t^2+2}-(t^2+2)}{8}
\\[6px]
&=\frac{t\sqrt{t^2+2}}{2}
\end{align}
Do you see it, now?
| {
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"url": "https://math.stackexchange.com/questions/2480587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Implicit Differentiation of this function [If
$x^2 + xy + y^3 = 1$,
find the value of
$y'''$
at the point where
x = 1.]1
Am i going the right direction, if so how will know what is the values of y' and y''
| It's easier, I think, to avoid division as much a possible, thus:
With
$x^2 + xy + y^3 = 1, \tag 1$
and $x = 1$,
$1 + y + y^3 = 1; \tag 2$
thus
$y(y^2 + 1) = y^3 + y = 0; \tag 3$
thus, if we assume $y \in \Bbb R$,
$y = 0; \tag 4$
that is, we have
$(x, y) = (1, 0). \tag 5$
We find, taking $d/dx$ of (1),
$2x + y + xy' + 3y^2y' = 0; \tag 6$
with $x = 1$ and $y = 0$,
$2 + y'(1) = 0, \tag 7$
so
$y'(1) = -2; \tag 8$
differentiating (6) yields
$2 + y' + y' + xy'' + 6y(y')^2 + 3y^2y'' = 0, \tag 9$
or
$2 + 2y' + xy'' + 6y(y')^2 + 3y^2 y'' = 0; \tag{10}$
we insert the values of $x, y, y'$:
$2 + 2(-2) + y'' = 0, \tag{11}$
whence
$y''(1) = 2; \tag{12}$
we now differentiate (10) one more time and obtain
$2y'' + y'' + xy''' + 6(y')^3 + 12yy'y'' + 6yy'y'' + 3y^2y''' = 0, \tag{13}$
or
$3y'' + xy''' + 6(y')^3 + 18yy'y'' + 3y^2y''' = 0; \tag{14}$
inserting the values for $x$, $y$, $y'$ and $y''$ yields
$3(2) + y''' + 6(-2)^3 = 0, \tag{15}$
or
$y''' = 42. \tag{16}$
This could obviously go on forever but I'll stop here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.
On the other end of spectrum is the reasonably easy to understand series
$$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$
But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.
A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.
Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.
| For modest requirements of accuracy, we can use a quite basic and easy to understand convergence acceleration technique to transform the Gregory/Leibniz series into something that yields the result with little computation.
If we have an alternating series
$$\sum_{n = 0}^{\infty} (-1)^n a_n$$
where the $a_n$ are slowly converging to $0$, then it takes little guessing to believe that the arithmetic mean of two successive partial sums of the series is a much better approximation to the value of the series than either or the two partial sums. We can write the mean as
$$\sum_{n = 0}^{m-1} (-1)^n a_n + \frac{(-1)^m a_m}{2} = \frac{a_0}{2} + \frac{1}{2} \sum_{n = 0}^{m-1} (-1)^n(a_n - a_{n+1})$$
and see that if $\bigl(a_n - a_{n+1}\bigr)_{n\in \mathbb{N}}$ is a decreasing sequence we immediately have an error bound of $\frac{1}{2}(a_{m} - a_{m+1})$ for this arithmetic mean. If $(a_n)$ is slowly converging, then this is much smaller than either of $a_m$ and $a_{m+1}$. For nice $(a_n)$, this can be iterated.
If we apply that to $a_n = \frac{1}{2n+1}$, we find $a_n - a_{n+1} = \frac{2}{(2n+1)(2n+3)}$, which is decreasing - and still converges slowly to $0$, so we get a significant improvement from then applying the method again. We get
\begin{align}
\frac{\pi}{4} &= \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1} \\
&= \frac{1}{2} + \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\
&= \frac{1}{2} + \frac{1}{6} + 2\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\
&= \frac{1}{2} + \frac{1}{6} + \frac{1}{15} + 6\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)}
\end{align}
and so on, each series converging faster than the previous. Since the terms get more complicated, it is computationally - at least when done by hand - better to start accelerating not right at the beginning of the series, but rather compute a partial sum of the original series, and then use acceleration only on the remainder. Let's say that starting the remainder at $n = 4$ wouldn't need an unreasonable amount of computation, then we get
\begin{align}
\frac{\pi}{4} &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{2n+1} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + 2\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} + 6\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)} \\
&> 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} \\
&> 0.78519
\end{align}
and thus $\pi > 3.14076$.
Of course Machin-like formulæ are far more efficient. But it's not too terrible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
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Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
| For $\epsilon >0$ sufficiently small $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon \Leftrightarrow \left|\dfrac{-7}{9n^2+3}\right|<\epsilon $$ $$ \Leftrightarrow \dfrac{7}{9n^2+3}<\epsilon \Leftrightarrow \frac{7}{\epsilon}-3<9n^2\Leftrightarrow \frac{7-3\epsilon}{9\epsilon}<n^2\Leftrightarrow \sqrt{\frac{7-3\epsilon}{9\epsilon}}<n$$ That is, for every $\epsilon >0$ sufficiently small if we choose $$n_0=\left\lfloor\sqrt{\frac{7-3\epsilon}{9\epsilon}}\right\rfloor +1$$ then $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon $$ if $n\ge n_0$ so, $$\lim_{n\to +\infty}\dfrac{n^2-2}{3n^2+1}=\dfrac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $n=x^2+y^2$ if it is a product of primes congruent 3 mod 4 and 1,2 mod 4 I am stuck trying to show that if
\begin{equation}
n=\prod p_i^{2e_i}\prod q_j^{f_j}
\end{equation}
where each $p_i\equiv3(mod~4)$ and each $q_j\equiv 1,2(mod 4)$ then each $p_i^{2e_i}$ and each $q_j^{f_j}$ is the norm of an element in $\mathbb{Z}[i]$ and thus $x^2+y^2=n$ for some $x,y\in \mathbb{Z}$
I really don't know where to start, so any hint would be so helpful.
| This is a simple model to show the solution.
Consider following product of primes in forms $(4k+1) and (4k+3)$:
$$n=(4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3). . .$$
$$(4 k_1+1)(4 k_2+3)≡ 3 mod 4$$
$$(4 k_3+1)(4 k_4+3)≡ 3 mod 4$$
$$⇒ (4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3)≡ 1 mod 4$$
Or $$(4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3) = 4 k +1$$
It can be seen that in a t term product if number of terms in form of $(4 k +1)$is $(t/2 +1)$ then $n≡1 mod 4$, that is n, due to generalized Fermat theorem, (the sum of two squares theorem)can be written as the sum of two squares in particular cases.
It can be shown that there are infinitely many numbers of form $(4 p +1)$ which can be prime or composite and are the sum of two squares; thus one of squares say x must be even and the other must be odd, suppose:
$x= 2 a ⇒ x^2 = 4 a^2 $
$y= 2 b +1 ⇒ y^2 = 4 b^2 + 4 b + 1$
⇒ $x^2 + y^2=4( a^2 + b^2 + b) +1$
Since a and b can be any natural number then there are infinitely many numbers of form( 4 p +1) which are the sum of two squares. These numbers can be prime like $5417 = 4 . 1354 +1 = 44^2 + 59^2 $( resulted from a=22 and b = 29) or product of two or more numbers, like:
$2009 = 4 . 502 +1= 7 . 7 . 41 = 28^2 + 35^2$ (resulted from a=14 and b=17)
or primes like:
$ 485 = 4 . 121 +1 = 5 . 97 = 14^2 + 17^2$ (resulted from a=7 and b=8)
There is no reason for the lack of infinitely many numbers like 485 which are the product of two or more primes and are the sum of two squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $1+{x\over 2}-{x^2 \over 8}<\sqrt{1+x}$ for all $x>0$? How to prove $1+(1/2)x-(1/8)x^2<\sqrt{1+x}$ for $x>0$ by Taylor Expansions up to $n$ order or Mean Value Theorem? I tried to apply MVT on $\sqrt{1+x}$ and get $\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x$ for $\xi\in (0,x)$. How to do next?
| $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{\left(\sqrt{x+1}\right)'''_{x=\xi}x^3}{6}$$ and since $\left(\sqrt{x+1}\right)'''>0$ for all $x>0$, we are done!
A proof without Taylor series.
If $1+\frac{x}{2}-\frac{x^2}{8}<0$ or $x>2+2\sqrt3$ then it's obvious.
But for $0<x\leq2+2\sqrt3$ it's enough to prove that
$$(x^2-4x-8)^2<64(1+x)$$ or
$$x^3(x-8)<0.$$
Done!
| {
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Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$?
I would greatly appreciate it if people could please explain how one would go about this.
| You can find roots of $z^4+4=0$. For example, one easy way is to multiply the equation by $z^4-4$ to get $z^8 - 16 = 0$, so we conclude that roots of $z^4+4$ are $8$th roots of unity (times $\sqrt 2$) that are not $4$th roots of unity, i.e. the roots are $$\{\omega\sqrt 2,\omega^3\sqrt 2,\omega^5\sqrt 2,\omega^7\sqrt 2\}$$ where $\omega = e^{i\pi/4} = \frac{\sqrt 2} 2+i\frac{\sqrt 2} 2.$ Multiplying by $\sqrt 2$ and rotating in complex plane we get that the roots are $$\{1+i,-1+i,-1-i,1-i\}$$ and finally $$z^4+4=(z-1-i)(z+1-i)(z+1+i)(z-1+i).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.
| At your step
$$3=a^2+2\sqrt 2 ab +2b^2$$, you can rearrange it to be
$$3-a^2-2b^2=2\sqrt 2 ab$$, the left hand side is a rational number while the right hand side is irrational, which you can't do unless a and b are zero, which does not satisfy the first condition.
| {
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"source": "stackexchange",
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Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one
| The usual trick: separate even and odd indices of the sum (things converge absolutely, so you can).
We have $$\frac{\pi^2}{6}
= \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n)^2}
+\sum_{n=0}^\infty \frac{1}{(2n+1)^2}
$$
but
$$
\sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{4n^2}
= \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{4}\frac{\pi^2}{6}
$$
so
$$
\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}\,.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Combinatoric problem with balls
Suppose there are $10$ balls in an urn, $4$ blue, $4$ red, and $2$ green. The balls are also numbered $1$ to $10$. How many ways are there to select an ordered sample of four balls without replacement such that the number B $\geq 0$ of blue balls, the number $R \geq 0$ of red balls, and the number $G \geq 0$ of green balls are all different?
If $|B|,|R|,|G|$ are all different and $|B|+|R|+|G|=4$, then there can only be combinations with the numbers $0,1,3$.
So I listed all of the possibilities :
$310$,$130,031,301$
For $310$ we have $3$ blue balls, $1$ red ball and $0$ green. So we have $(4 \times 3 \times 2) \times (4) $ ways
For $130$ we have $1$ blue balls, $3$ red ball and $0$ green. So we have $(4 ) \times (4 \times 3 \times 2) $ ways
For $031$ we have $0$ blue balls, $3$ red ball and $1$ green. So we have $(4 \times 3 \times 2) \times (2)$ ways
For $301$ we have $3$ blue balls, $0$ red ball and $1$ green. So we have $(4 \times 3 \times 2) \times (2)$ ways
Adding up we get $96+96+48+48=288$
The real answer should be $1152$. What am I doing wrong?
| You've got the color combinations correct: $RBG = [(3,1,0), (1,3,0), (3,0,1), (0,3,1)]$.
For the first color combination, you have $4$ sets of numbers you can draw for the reds, and for those numbers, $6$ orders to draw them in. For the one blue, you have $4$ choices for the number, and $4$ choices for when you draw it relative to the reds. So that's $4 \cdot 6 \cdot 4 \cdot 4 = 384$ ways.
The second color combination is done the same way as the first.
For the third color combination, you have again $4 \cdot 6 = 24$ ways to draw the reds. For the one green, you have $2$ choices for the number, and $4$ choices for when you draw it relative to the reds. So that's $4 \cdot 6 \cdot 2 \cdot 4 = 192$ ways.
The fourth color combination is done the same way as the third.
So we get $384 + 384 + 192 + 192 = 1152$ ways.
| {
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"timestamp": "2023-03-29T00:00:00",
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l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} = \frac{24}{24}=1 $$
Consider the following example using a different method:
$$ \lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0}\frac{\frac{x^4}{x^4}}{\frac{x^4}{x^4}+\frac{x^2}{x^4}} = \lim_{x \rightarrow 0} \frac {1}{1 +\frac{1}{x^2}} = \frac {1}{1+\infty} = \frac{1}{\infty}=0 $$
The graph here clearly tells me the limit should be $0$, but why does l'Hôpital fail?
| You haven't checked whether L'Hopital could be applied each time.
| {
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Modular arithmetic and probability $a$, $b$ and $c$ are chosen from $1$-$999$ with order and replacement, what is the probability that $a^2 + bc$ is divisible by $3$? (each choice is equally likely)
I split it up into 2 cases
$a^2$ has a remainder of $1$ while $bc$ has a remainder of $2$
$a^2$ has a remainder of $1$, $2/3$ of the time, $bc$ has a remainder of $2$ if $b$ and $c$ have $2$ and $1$. $(2/3 \times 1/3 \times 1/3)2$ (because $b$ and $c$ can be interchanged) = $4/27$
$a^2$ has a remainder of 0 while $bc$ has a remainder of $0$
$a^2$ has a remainder of $0$ $1/3$ of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$
$4/27 + 2/9 = 10/27$
Please tell me what i did wrong.
| The error is here:
$a^2$ has a remainder of $0$ 1/3 of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$
You multiplied these numbers, which would be appropriate if you had "...if $b$ and $c$ have remainders of $0$." Since it's or instead, this changes. We can use the inclusion-exclusion principle here:
\begin{align*}
P(b\text{ or } c \text{ has remainder }0 ) &= P(b \text{ has remainder } 0) + P(c \text{ has remainder }0) \\
&- P(b \text{ and } c \text{ have remainder }0) \\
&=1/3 + 1/3 - 1/9 = 5/9.
\end{align*}
| {
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find the value of ''a'' and ''b'' for which function is always increasing
Question Let f(x)= x$^{3}+ax^{2}+bx+5sin^{2}x$ be an increasing
function$\forall$ x $\in$$\mathbb{R}$$ Then$
(a) $a^2 -3b-15$<0
(b) $a^2 -3b-15$>0
(c)$a^2 -3b+15<$0
(d)a>0,b>0
My Approach i tried by making perfect square of the derivative
$\left\{ x+\frac{a}{3}\right\} ^{2}$+$\frac{b}{3}$-$\frac{a}{9}-5\geq$0
$\left\{ sin2x\geq-1\right\} $ $\Longrightarrow$a$^{2}$-3b+45$\leq$0
but it doesn't gives the required result
| When we differentiate the function we get
\begin{align*}
3x^2+2ax+b+10\sin(x)\cos(x)=3x^2+2ax+b+5\sin(2x)
\end{align*}
Now as the function should be increasing, its differential should be everywhere nonnegative. But if it is everywhere nonnegative, so will
\begin{align*}
3x^2+2ax+b-5
\end{align*}
as $\sin(2x)$ is between -1 and 1. So if we now solve
\begin{align*}
3x^2+2ax+b-5=0
\end{align*}
we get
\begin{align*}
x=\frac{-2a\pm \sqrt{4a^2-4\cdot 3(b-5)}}{6}
\end{align*}
As it should be everywhere nonnegative, we are only allowed to have at max 1 solution or no solutions, so the argument squareroot should be negative:
\begin{align*}
4a^2-12b+60\leq 0
\end{align*}
which is equivalent to
\begin{align*}
a^2-3b+15\leq 0
\end{align*}
I'm missing a minussign, but I think this method should work, hopefully it is clear.
| {
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Olympiad inequality
Let $a,b,c,d \in [0,1]$.
Prove
$$ \frac {a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}+abcd\le3$$
Hi. Im a school student so ive been unable to look for ways to apply this problem. I attempted to use AM-GN on 1+a etc but the bound $2\sqrt{a}$ was too large and was easily larger than 3. Thus id like some help solving this problem.
| For $a,b,c,d \in [0,1]$, let
$$
f(a,b,c,d)
=
\frac{a}{1+b}
+
\frac{b}{1+c}
+
\frac{c}{1+d}
+
\frac{d}{1+a}
+
abcd
$$
The goal is to prove $f(a,b,c,d) \le 3$, for all $a,b,c,d \in [0,1]$.
For fixed $b,c,d \in [0,1]$, let $g(a) = f(a,b,c,d)$.
Then
$$g''(a) = \frac{2d}{(a+1)^3}$$
hence $g''(a) \ge 0$, for all $a \in [0,1]$.
It follows that $g$ is convex on $[0,1]$, hence the maximum value of $g$ on $[0,1]$ is achieved at one of the endpoints.
Thus, for fixed $b,c,d$, the function $f(a,b,c,d)$ is maximized at $a=0$ or $a=1$.
The same reasoning applies to each of the variables $b,c,d$, with the other $3$ variables held fixed.
Thus, to prove $f(a,b,c,d) \le 3$, for all $a,b,c,d \in [0,1]$, it suffices to verify that $f(a,b,c,d) \le 3$, for all $a,b,c,d \in \{0,1\}$.
If any of $a,b,c,d$ is zero, the expression
$$
\frac{a}{1+b}
+
\frac{b}{1+c}
+
\frac{c}{1+d}
+
\frac{d}{1+a}
+
abcd
$$
reduces to a sum of $3$ fractions, each of which is at most $1$, hence the sum is at most $3$.
If $a,b,c,d$ are all equal to $1$, the expression
$$
\frac{a}{1+b}
+
\frac{b}{1+c}
+
\frac{c}{1+d}
+
\frac{d}{1+a}
+
abcd
$$
evaluates to exactly $3$.
Thus, the claimed inequality is proved.
| {
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Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be correct to use this and try solving through induction?
| The product $p(p+1)$ of two consecutive numbers is divisible by $2$
because either $p$ or $p+1$ is even.
The product $p(p+1)(p+2)$ of three consecutive numbers is divisible by $3$
because either $p,\ p+1$ or $p+2$ is a multiple of $3$.
$f(n)=n(n^2-1)=(n-1)n(n+1)$ is then divisible by $3$.
Also for $n=2p+1$ odd then $(n-1)(n+1)=(2p)(2p+2)=4p(p+1)$ is divisible by $4\times 2$
So $f(n)$ is divisible by $8\times 3=24$.
| {
"language": "en",
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Show that $|\sin x−\cos x|≤ 2$ for all $x.$ Please help me out in answering this:
Show that $|\sin x−\cos x|≤ 2$ for all $x.$
I don't know where to start and I think we have to use the mean value theorem to show this.
| I have another method for you if you want to use some straight-up calculus optimization. Let us first determine the extrema of $\sin(x)-\cos(x)$. To do this, we take the derivative and set it equal to $0$. In this case, the derivative is $\cos(x) +\sin(x)$, so we want to solve $\cos(x) + \sin(x) = 0$, or $\cos(x) = -\sin(x)$.
So when are $\sin(x)$ and $\cos(x)$ negatives of each other? This occurs precisely when $x = \frac{3\pi}{4}$ or $x= -\frac{\pi}{4}$. This means our extrema are at these values of $x$.
Let's now classify these extrema. First plug in $x = \frac{3\pi}{4}$ to $\sin(x)-\cos(x)$:
$$\sin\bigg(\frac{3\pi}{4}\bigg) -\cos\bigg(\frac{\pi}{4}\bigg) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}.$$
Now plug in $x= -\frac{\pi}{4}$:
$$\sin\bigg(-\frac{\pi}{4}\bigg) -\cos\bigg(-\frac{\pi}{4}\bigg) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}.$$
Thus, we have found that $\sqrt{2}$ is a maxiumum of $\sin(x) -\cos(x)$ and $-\sqrt{2}$ is a minimum. That is to say
$$ -\sqrt{2} \leq \sin(x) -\cos(x) \leq \sqrt{2},$$
for all $x$. But by the definition of the absolute value, this means that
$$ |\sin(x) -\cos(x)| \leq \sqrt{2},$$
and $\sqrt{2} < 2$, so we have
$$ |\sin(x) -\cos(x)| \leq \sqrt{2} < 2.$$
And then we are done.
| {
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Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was having four variables and only one equation. While analyzing my test paper, this is the only problem I (and my friends too) couldn't figure out even after giving this problem several hours. So I came here for some help.
Question : Solve for $a,b,c,d \in \Bbb R$, given that $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$
Since only one equation is given, there must be involvement of making of perfect squares, such that they all add up to $0$. Thus, resulting in few more equations. But how to?
I tried a lot of things, such as making $(a-b)^2 $ by adding the missing terms and subtracting again, but got no success.
Thanks!
| Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$.
With help of a CAS, one can verify
$$\begin{align}
F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)
&= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\
&= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$
If one set $(a,b,c,d)$ to $\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)$, one find
$$\begin{align}
F(a,b,c,d) = 0
&\iff p = p-q = q-r = r-s = s = 0\\
&\iff p = q = r = s = 0
\end{align}
$$
This implies the equation at hand has a unique solution:
$$(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$$
Update
About the question how I come up with this. I first write $F(a,b,c,d)$ as
$$\begin{align}
F(a,b,c,d) &= a^2 + b^2 + c^2 + d^2 - ab - bc - cd - da + d(a-1) + \frac25\\
&= \frac12((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2) + d(a-1) + \frac25
\end{align}\tag{*1}
$$
To simplify the term $d(a-1)$, I introduce $\lambda, \mu$ such that
$$\begin{cases}
d &= \frac12 + \lambda + \mu\\
a &= \frac12 + \lambda - \mu
\end{cases}
\quad\implies\quad d(a-1) = \lambda^2 - \left(\frac12+\mu\right)^2
$$
Now $d-a = 2\mu$ and $(a-b)^2 + (b-c)^2 + (c-d)^2
\ge 3\left(\frac{d-a}{3}\right)^2
= \frac43 \mu^2$.
If one substitute this back into $(*1)$, one find
$$F(a,b,c,d) \ge \frac83\mu^2 + \lambda^2 - (\frac12 + \mu)^2 + \frac25
= \lambda^2 + \frac53\left(\mu - \frac{3}{10}\right)^2$$
In order for $F(a,b,c,d) = 0$, we need
$$\lambda = 0,\quad\mu = \frac{3}{10}
\quad\text{ and }\quad(a-b)^2 + (b-c)^2 + (c-d)^2 = \frac13(d-a)^2$$
The last condition forces $d-c = c-b = b-a = \frac13(d-a)$ and leads to the solution $(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$.
This is a little bit sloppy to describe, so I look at expansion
of $F(a,b,c,d)$ near the solution and obtain a simpler description of $F$ in terms of $p,q,r,s$.
| {
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Splitting field of $x^3-3x+1$ I know there is a trick in that you can use $\omega + \omega ^{-1}$ where $\omega$ is a 9th root of unity but I can't see how this relates. I know this question has already been asked on here but there was no solution given using the trick.
| $x = z + \frac 1z\\
x^3 - 3x + 1 = z^3 + \frac 1{z^3} + 1 = \frac {1}{z^3} (z^3 + e^\frac {2\pi i}{3})(z^3 + e^\frac {-2\pi i}{3})\\
x^3 - 3x + 1 = (x + 2\cos \frac {2\pi}{9})(x + 2\cos \frac {4\pi}{9})(x + 2\cos \frac {8\pi}{9})$
| {
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Completing squares with three variables. I want to complete the squares for this polynomial
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))+(\frac{1}{2}(y+\frac{3}{2}z))^2-(\frac{1}{2}(y+\frac{3}{2}z))^2-4yz+2y^2-z^2$$
Then things get kinda messy from here and I get totally lost from then on, could anyone help me out factoring this? And telling me if there is an eaay and non-confusing way to solve it?
| If you mean writing polynomial as the sum of squares, here is one solution:
$p=2x^2+2y^2-z^2+2xy+3xz-4yz$
polynomial p can be written as following forms:
$p=(x+y)^2+(x+z)^2+(y-2z)^2-5z^2+xz$
$p=(x+y)^2 +(x+2z)^2+(y-2z)^2-5z^2 -xz$
Summing these relations and dividing the result by 2 we get:
$p= (x+y)^2 +\frac{1}{2}(x+z)^2+\frac{1}{2}(x+2z)^2 +(y-2z)^2 -(5^{1/2}z)^2$
| {
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Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$
I have to solve this system of equations \begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}
From the third equation I have done this to manipulate it:
$$ x^3+y^3-z^3=43-3xy \implies (x+y)(x^2-xy+y^2)-z^3=43-3xy\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$
Notice from the first equation $x+y=z+1$ and $(x+y)^2 = z^2+5$
So the above equation reduces to $z^2+5z-38-3xyz=0$
How do I get rid of the $xyz$?
| $$\begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}\Rightarrow z^2+5=z^2+2z+1\Rightarrow z=2$$
Because of $x^3+y^3=(x+y)(x^2-xy+y^2$ it follows $$\begin{cases} x+y=3 \\ x^2+y^2=17 \end{cases}\Rightarrow (x,y)=(-1,4),(4,-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
| What exactly is your problem with that expression? You can't see why the two statements on the left and right sides of the equals sign are equivalent?
\begin{align}\require{cancel}
\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}
&=\frac{\cancel{n}y}{\cancel{n}(n+4)} + \left(\frac{n+4}{n+4}-\frac{n}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \left(\frac{n+4 -n}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \left(\frac{\cancel{n}+4 \cancel{-n}}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \frac{4 }{n+4}\cdot\frac{1}{2}\\
&=\frac{y}{n+4} + \frac{2\cdot 2 }{2(n+4)}\\
&=\frac{y}{n+4} + \frac{\cancel{2}\cdot 2 }{\cancel{2}(n+4)}\\
&=\frac{y}{n+4} + \frac{2}{n+4}\\
&=\frac{y+2}{n+4}
\end{align}
If you want to get from $\frac{y+2}{n+4}$ to $\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$, just trace the steps backwards.
| {
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Why is the limit of $\frac{1}{n}$ is $0$ however the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent? As I recall, according to the test for divergence, if you have a series $\sum_{n=1}^{\infty}a_{n}$ and if the limit of $a_{n}$ is $0$, then the series is convergent. the limit of $\frac{1}{n}$ is $0$. However, if we use the integral test, we realize that the series of $\frac{1}{n}$. I am confused to why it is like this. Am I overlooking something?
| Actually, if $a_{n}$ does not tend to $0$, then $\sum_{n=1}^{\infty}a_{n}$ is divergent. The opposite is not true.
A good example is $a_n = 1/n$. To visualize this, consider the series,
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + ...$$
Now replace all $a_n$ to the closest power of $2$ which is smaller. For example $\frac{1}{3}$ becomes $\frac{1}{4}$. The new series is then,
$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} + ...$$
Notice that if I group the terms, what happens
$$1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + (\frac{1}{16} + \frac{1}{16} + ...$$
Each of the terms add up to half
$$1 + \frac{1}{2} + (\frac{1}{2} ) + (\frac{1}{2}) + (\frac{1}{2} ) + ...$$
From here I'm pretty sure you can see that this series diverges. Now coming to the original series. Since the new series corresponding terms are smaller and this series diverges, so does the series $\sum_{n=1}^{\infty}a_{n}$
| {
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Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \right) $
and
$A= \left( \begin {array}{ccc} 0&1&0\\ 0&0&1 \\ 0&0&0\end {array} \right) $ work,
but how can I find all the matrices?
| You have $A^4=0$, and since it is a $3\times 3$ matrix it follows that $A^3=0$. This means that
$$A\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0\end{pmatrix}=0$$
and
$$\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0\end{pmatrix}A=0$$
Giving that
$$A=\begin{pmatrix}
0&a&b\\
0&c&d\\
0&0&0\end{pmatrix}$$ Now we have
$$A^2=\begin{pmatrix}
0&ac&ad\\
0&c^2&dc\\
0&0&0\end{pmatrix}$$
So $c=0$ and $ad=1$. Thus the general solution is
$$A=\begin{pmatrix}
0&a&b\\
0&0&a^{-1}\\
0&0&0\end{pmatrix}.$$
| {
"language": "en",
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A tangent line to $y=\frac1{x^2}$ cuts the $x$-axis at $A$ and $y$ at $B$, minimize $AB$. The problem:
A tangent line to $y= \frac{1}{x^2}$ intersects the x-axis at the point A and the y-axis at the point B.
What is the length of the shortest such line segment AB?
I know that the graph of $y= \frac{1}{x^2}$ looks like a bell centered around the y-axis.
The function is not equal to zero for x at any point, which is why a tangent line (unknown) will cross the x-axis at one point (A), and the y-axis at another point (B).
The first derivative of $y=\frac{1}{x^2}$ is $\frac{-2}{x^3}$.
I don't know where to go from here. Do I use the first derivative of the function and plug in values A and B as unknown values? I only need guidance on how to begin, and I can solve the rest from there. Thank you.
| Some preliminary theory: If a line tangent to a given function intersects the x-axis at a point A and the y-axis at a point B, then it must form a right triangle with the right angle situated at the origin and the two sides with vertices located at $(A, 0)$ and $(0, B)$ of lengths $A$ and $B$ respectively. The length of the hypotenuse of a right triangle thus formed is the line segment $\overline{AB}$.
First of all, let's find the equation of the line tangent to the function $y=\frac{1}{x^2}$ at an arbitrary point $\alpha$ in general:
$$
y - f(\alpha)=f'(\alpha)(x-\alpha)\implies\\
y - \frac{1}{\alpha^2}=-\frac{2}{\alpha^3}(x - \alpha)\implies\\
y=-\frac{2}{\alpha^3}x+\frac{3}{\alpha^2}
$$
Now, in the problem it is stated that when $x=A$, $y=0$ and when $x=0$, $y=B$. So, plugging in these two pairs of values for $x$ and $y$, we will work out the lengths of the two legs of the right triangle expressed in terms of our arbitrary point $\alpha$.
The length of the side along the x-axis:
$$
0=-\frac{2}{\alpha^3}A+\frac{3}{\alpha^2}\implies\\
A=\frac{3\alpha}{2}
$$
The length of the side along the y-axis:
$$
B=-\frac{2}{\alpha^3}\cdot 0+\frac{3}{\alpha^2}\implies\\
B=\frac{3}{\alpha^2}
$$
Now we can construct a function $L$ which will be the length of the hypotenuse of our imaginary right triangle expressed in terms of our point $\alpha$:
$$
L(\alpha)=\sqrt{A^2+B^2}=\sqrt{\left(\frac{3\alpha}{2}\right)^2+\left(\frac{3}{\alpha^2}\right)^2}=\frac{3\sqrt{\alpha^8+4}}{2\alpha^2}
$$
So, here's what we've got. That function $L$ is a function that can give us any possible value for the length of the line segment $\overline{AB}$ including the one that's the shortest. All we need to do is to plug in the right number for $\alpha$. So, we need to find an $\alpha$ such that when we plug it into the function, it will spit out the functional value that's the smallest of all possible values that the function can yield. In order to do that, we need to find its first derivative, set it equal to zero and solve for $\alpha$. That alpha is that magic number which will give us the shortest hypotenuse.
$$
L'(\alpha)=\left(\frac{3\sqrt{\alpha^8+4}}{2\alpha^2}\right)'=\frac{6\alpha^5}{\sqrt{\alpha^8+4}}-\frac{3\sqrt{\alpha^8+4}}{\alpha^3}
$$
$$
\frac{6\alpha^5}{\sqrt{\alpha^8+4}}-\frac{3\sqrt{\alpha^8+4}}{\alpha^3}=0\\
\alpha=\pm\sqrt[4]{2}
$$
This means that there are two such points. Now, let's plug these points back into the length function and find out the shortest hypotenuses:
$$
L(\sqrt[4]{2})=\frac{3\sqrt{(\sqrt[4]{2})^8+4}}{2(\sqrt[4]{2})^2}=3\\
L(-\sqrt[4]{2})=\frac{3\sqrt{(-\sqrt[4]{2})^8+4}}{2(-\sqrt[4]{2})^2}=3
$$
The lengths as expected are positive values and actually the same.
Answer: the length of the shortest such line segment $\overline{AB}$ is $3$ and there are actually two such line segments.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit :
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Using Maclaurin series for $\cos x $:
$$\cos x=1-\frac{x^2}{2!}\quad \text{(We'll stop at order 2)}$$
This leads to :
$$1-\sqrt{1+x^2}\cos x=1-(1+\frac{x^2}{2})(1-\frac{x^2}{2})=\frac{x^4}{4}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4$$
Using Maclaurin series for $\sin x $:
$$\sin x=x-\frac{x^3}{3!}\quad \text{(We'll stop at order 3)}$$
$$\tan^4x=\left(\frac{\sin x}{\cos x}\right)^4 = \left(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}}\right)^4$$
Thus $$\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\frac{\frac{x^4}{4}}{(\frac{x-\frac{x^3}{3!}}{1-\frac{x^2}{2}})^4}=\frac{x^4(1-\frac{x^2}{2})^4}{4(x-\frac{x^3}{3!})}=\frac{(x(1-\frac{x^2}{2}))^4}{4(x-\frac{x^3}{3!})}=\frac{1}{4}(\frac{x-\frac{x^3}{2}}{x-\frac{x^3}{3!}})=\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})
$$
Then $$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}=\lim_{x\rightarrow 0}\frac{1}{4}(\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3!}})=\frac{1}{4}.$$
But my book says the solution is $\frac{1}{3}$
Where have I done wrong?
Help appreciated!
| Hint. You need a longer expansion of $\sqrt{1+x^2}$ and $\cos(x)$:
$$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\quad,\quad
\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$
Then
\begin{align}\sqrt{1+x^2}\cos(x)&=\left(1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\right)\left(1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\right)\\
&=\left(1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\right)+\frac{x^2}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)-\frac{x^4}{8}\left(1+o(1)\right)+o(x^4)\\
&=1-\frac{x^2}{2}+\frac{x^4}{4!}+\frac{x^2}{2}-\frac{x^4}{4}-\frac{x^4}{8}+o(x^4)\\
&=1-\frac{x^4}{3}+o(x^4).
\end{align}
Can you take it from here?
| {
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Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$
| By I.H. we have $2^{3n}-3^{n}=5k$ for some integer $k$ and thus $2^{3n}=3^{n}+5k$.
Now we have:
$$2^{3(n+1)}-3^{n+1} = 8\cdot 2^{3n}-3^{n+1}= 8(3^{n}+5k)-3^{n+1} = 3^n(8-3)+40k = 5\cdot (3^n+8k)$$
and we are done.
| {
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Write formula for matrix in terms of Fibonacci numbers How could I express this matrix in terms of Fibonacci numbers? It seems like I'd have to use induction once I have a candidate for a formula but I'm unsure of where to start with expressing the matrix in terms of Fibonacci numbers.
Thanks in advance!
Let $T:\mathbb{R^2}\rightarrow \mathbb{R^2}$ be a linear map such that
$$T\left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{c}
y\\
x+ y\\
\end{array}
\right)$$
using the basis $\beta=\{e_1,e_2\}$
$$e_1=\left(
\begin{array}{c}
1\\
0\\
\end{array}
\right),\quad e_2=\left(
\begin{array}{c}
0\\
1\\
\end{array}
\right)$$
Write a formula for the matrix$$ [T^n]_\beta, \forall n\in\mathbb{N}$$ in terms of Fibonacci numbers.
| Let us try to do it for $n = 1$. In that case, $T(e_1) = e_2$ and $T(e_2) = e_1 + e_2$, so this gives the matrix $\begin{pmatrix}0 \quad 1 \\ 1 \quad 1\end{pmatrix}$ for $T$.
If we have to find $T^n$ now, the first thing that we do is to find some elementary powers of $T$. Let's try to find $T^2$:
$$
T^2 = \begin{pmatrix}1 \quad 1\\ 1 \quad 2\end{pmatrix}, T^3 = \begin{pmatrix}1 \quad 2\\ 2 \quad 3\end{pmatrix},T^4 = \begin{pmatrix}2 \quad 3\\ 3 \quad 5\end{pmatrix}
$$
So the pattern, as can be seen clearly, is that $T^n = \begin{pmatrix} F_{n-1} \quad F_n \\ F_{n} \quad F_{n+1}\end{pmatrix}$.
The best we can do is to prove this by induction : Note that $T\begin{pmatrix}F_{n-1} & F_{n}\\ F_{n} & F_{n+1}\end{pmatrix} = \begin{pmatrix} F_{n} & F_{n+1} \\ F_{n-1} + F_n & F_{n} + F_{n+1}\end{pmatrix}$, from where you can conclude.
| {
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Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...$
But, But I'm stuck here.
| A method like this can work:
$$\left( x+1-x\right)×\left(x+1+x \right)×\left((x+1)^2+x^2 \right)×\left( (x+1)^4+x^4\right)×...×\left((x+1)^{64}+x^{64}\right)=(x+1)^{128}-x^{128}$$
| {
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Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I tried rewriting the expression as $$\sum_{i=1}^n \frac{\bar{x_i}-1}{|x_i-1|^2}$$
I then used the fact that $$x^{n+1}-1=(x-1)(x^n+x^{n-1}+...+x+1=0$$ so $x_i$ are the complex nth roots of unity.Using cosine formula I found that $$|x_i-1|^2=2-2\cos(\frac{2i\pi}{n+1})=4(\sin \frac{\pi}{n+1})^2$$
After substituting this I couldn't simplify the resulting expression.Any ideas?Thanks.
| Let's do it without any tricks using standard results from the usual theory of polynomial equations. If $\alpha$ is a root of $f(x) =0$ then $\beta=\alpha-1$ is a root of $f(x+1)=0$. Thus it follows that $y_{i} =(x_{i} - 1)$ are the roots of equation $$(y+1)^{n}+(y+1)^{n-1}+\dots+(y+1)+1=0$$ ie $$y^{n} +(n+1)y^{n-1}+\dots+(1+2+\cdots+n)y+ (n+1)=0$$ ie $$y^{n} +(n+1)y^{n-1}+\dots+\frac{n(n+1)}{2}y+(n+1)=0$$ Further note that if $\alpha$ is a root of $f(x) =0$ then $\gamma=1/\alpha$ is a root of $x^{n} f(1/x)=0$. And therefore $z_{i} =1/y_{i}=1/(x_{i}-1)$ are the roots of $$(n+1)z^{n}+\frac{n(n+1)}{2}z^{n+1}+\dots+(n+1)z+1=0$$ and hence the sum of its roots is $$-\frac{n(n+1)/2}{n+1}=-\frac{n}{2}$$ which answers your first question.
The answer to your second question is based on the fact that the given equation can be written as $$\frac{x^{n+1}-1}{x-1}=0$$ and the roots of the equation $x^{n+1}-1=0$ are given as $$\cos\left(\frac{2k\pi} {n+1} \right)+i\sin\left(\frac{2k\pi}{n+1}\right), \, k=0,1,2,\dots,n$$ The root $x=1$ corresponds to $k=0$ and hence the roots $x_{k} $ of the original equation are given by $$x_{k} =\cos\left(\frac{2k\pi}{n+1}\right)+i\sin\left(\frac{2k\pi}{n+1}\right), \, k=1,2,\dots,n$$ Further we have $$\frac{1}{x_{k}-1}=-i\cdot\frac{1}{2}\cot\left(\frac{k\pi}{n+1}\right)-\frac{1}{2} $$ Since the sum of these roots is $-n/2$ it follows that $$\sum_{k=1}^{n}\cot\left(\frac{k\pi}{n+1}\right)=0$$
| {
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Approximation of exponential function by power series Let $x \in (-\frac{1}{2},\frac{1}{2}), n \in \mathbb{N}$
How can I choose a $n$ that the the inequality is valid?
$$\left|e^x-\sum_{k=0}^n \frac{x^k}{k!}\right| \leq \frac{|e^x|}{10^{16}}$$
My ideas:
Try some values for $n$ and verify the inequality for value greater than $-1/2$ and less than $1/2$ because of the monotony of the exponential function... But I could not find a $n$.
| $$e^x-\sum_{k=0}^n \frac{x^k}{k!}= \sum_{k=n+1}^\infty \frac{x^k}{k!}= \frac{x^{n+1}}{(n+1)!}+ \frac{x^{n+2}}{(n+2)!}+\cdots = x^{n+1} {\left[ \frac{1}{(n+1)!}+ \frac{x}{(n+2)!}+\frac{x^2}{(n+3)!}+\cdots\right]} $$
$$\left|x^{n+1} {\left[ \frac{1}{(n+1)!}+ \frac{x}{(n+2)!}+\frac{x^2}{(n+3)!+\cdots}\right]}+\cdots\right|\le |x|^{n+1} {\left[ \frac{1}{0!}+ \frac{x}{1!}+\frac{x^2}{2!}{+\cdots}\right]}= |x|^{n+1} e^x \le \frac{1}{2^{n+1}}e^x
$$
Now choose $n$ so that $\frac{1}{2^{n+1}}\le 10^{-16}$.
| {
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Rational series for positive integer root of $x>1$ With reference to the method @MarkViola used to prove a formula for finding the square root of $x>1$ in Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$
I have now found a formula for generating a rational series for arbitrary positive integer roots of a number $x>1$:
$$\sqrt[r]{x}=\sum_{n=0}^\infty \binom{rn}{n} \frac{((r-1)n)!}{r^{2n}\,n!\,\prod_{k=1}^n(\prod_{j=1}^{r-2}(rk-j))}\left(1-\frac{1}{x} \right)^n \tag{1}$$
It is quite messy. Can it be simplified? or Is there a simpler way of achieving this?
Update: Using Mathematica I have now managed to simplify my extremely ugly formula to a much more palatable form:
$$\sqrt[r]{x}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n+\frac{1}{r}\right)}{n! \,\Gamma \left(\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n \tag{2}$$
As yet through I don not not how to get from (1) to (2) and I have no general proof for either formula, apart from the specific proof referenced above for $r=2$.
Any thoughts?
Note Added 03/06/2020
For Reference
$$\frac{1}{\sqrt[r]{x}}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n-\frac{1}{r}\right)}{n! \,\Gamma \left(-\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n$$
and
$$\frac{1}{\sqrt[r]{x}}=\sum _{n=0}^{\infty }\binom{r n}{n} \frac{ ((r-1) n )! }{ r^{2 n} \, n! \,(r \,n -1)\prod _{k=1}^n \left(\prod _{j=1}^{r-2} (r k -(r-j)) \right) } \left(1-\frac{1}{x} \right)^n $$
| $\def\peq{\mathrm{\phantom{=}}{}}$Note that for $0 < y < 1$, using the binomial series and $Γ(z + 1) = zΓ(z)$,\begin{align*}
(1 - y)^{-\frac{1}{r}} &= \sum_{n = 0}^∞ \binom{-\frac{1}{r}}{n} (-y)^n = \sum_{n = 0}^∞ \frac{\left( -\dfrac{1}{r} \right) \cdots \left( -\dfrac{1}{r} - n + 1 \right)}{n!} (-y)^n\\
&= \sum_{n = 0}^∞ \left( \dfrac{1}{r} \right) \cdots \left( \dfrac{1}{r} + n - 1 \right) \frac{y^n}{n!} = \sum_{n = 0}^∞ \frac{Γ\left( n + \dfrac{1}{r} \right)}{Γ\left( \dfrac{1}{r} \right)} · \frac{y^n}{n!}.
\end{align*}
Taking $y = 1 - \dfrac{1}{x}$ for $x > 1$,$$
x^{\frac{1}{r}} = \sum_{n = 0}^∞ \frac{Γ\left( n + \dfrac{1}{r} \right)}{n!\,Γ\left( \dfrac{1}{r} \right)} \left( 1 - \frac{1}{x} \right)^n.
$$
Now to prove (1). For any $n \geqslant 1$, since\begin{align*}
&\peq \prod_{k = 1}^n \prod_{j = 1}^{r - 2} (rk - j) = \prod_{k = 1}^n \left( \frac{1}{rk(rk - r + 1)} \prod_{j = 0}^{r - 1} (rk - j) \right)\\
&= \left( \prod_{k = 1}^n \frac{1}{rk(rk - r + 1)} \right) \left( \prod_{k = 1}^n \prod_{j = 0}^{r - 1} (rk - j) \right) = \frac{1}{n!\,r^n} \left( \prod_{k = 1}^n \frac{1}{rk - r + 1} \right) (rn)!,
\end{align*}
then\begin{align*}
&\peq \frac{1}{r^{2n}} \binom{rn}{n} \frac{((r - 1)n)!}{\displaystyle \prod_{k = 1}^n \prod_{j = 1}^{r - 2} (rk - j)} = \frac{1}{r^{2n}} · \frac{1}{n!} \color{blue}{\prod_{k = 0}^{n - 1} (rn - k)} · \frac{\color{blue}{((r - 1)n)!}}{\displaystyle \frac{1}{n!\,r^n} \left( \prod_{k = 1}^n \frac{1}{rk - r + 1} \right) \color{blue}{(rn)!}}\\
&= \frac{1}{r^n} \prod_{k = 1}^n (rk - r + 1) = \frac{1}{r^n} \prod_{k = 0}^{n - 1} (rk + 1) = \left( \dfrac{1}{r} \right) \cdots \left( \dfrac{1}{r} + n - 1 \right) = \frac{Γ\left( n + \dfrac{1}{r} \right)}{Γ\left( \dfrac{1}{r} \right)}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
$$\cot A = 2\\
\frac{\cos A }{ \sin A }= 2\\
\cot B = \frac34\\
\frac{\cos B }{ \sin B} = \frac34$$
What should I do with $A+B+C=180^\circ$?
| You have
$$ \sin C = \sin \big(180^\circ -(A+B)\big) = \sin (A+B) $$
Then, using angle-sum formulas to obtain
$$ \cot (A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} = \frac{2}{11} $$
And
$$ \csc^2 (A+B) = \cot^2 (A + B) + 1 = \frac{125}{121} $$
You know what to do next.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$ The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I noticed that this is the sum of the reciprocals of the triangle numbers.
Let $t_n = \frac{n(n+1)}{2}$ denote the $n$-th triangle number. Then the question is basically asking us to evaluate
\begin{align}
\sum_{n=1}^{2015} \frac {1}{t_n} & = \sum_{n=1}^{2015} \frac {2}{n(n+1)}\\
& = \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1}
\end{align}
Here's where my first question arises. Do you just have to know that $\frac {2}{n(n+1)} = \frac{2}{n}-\frac{2}{n+1}$? In an exam situation it would be very unlikely that someone would be able to recall that if they had not done a question like this before.
Moving on:
\begin{align}
\sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} & = \left(\frac{2}{1}-\frac{2}{2}\right) +\left(\frac{2}{2}-\frac{2}{3}\right) + \ldots +\left(\frac{2}{2014}-\frac{2}{2015}\right) +\left(\frac{2}{2015}-\frac{2}{2016}\right)\\
&= 2 - \frac{2}{2016} \\
& = \frac {4030}{2016} \\
& = \frac {2015}{1008}
\end{align}
And I'm not sure if this is right. How does one check whether their summation is correct?
| For your question about how $\sum{\frac{2}{n(n+1)}}$ is split,
$$\frac{2}{n(n+1)} = \frac{2[(n+1)-n]}{n(n+1)}$$
$$\frac{2[(n+1)-n]}{n(n+1)} = \frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)}$$
$$\frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$
So your question about you would know this in an exam situation, is that these type of question will have a similar type of denominator. Like in this question, if you observe the denominator $n(n+1)$ it can be split into $(n+1)-n$ which is equal to 1. So this makes splitting the fraction into two fractions easier such that the summation could be found out.
The rest of the steps are correct and the summation also probably correct. Hope you understood!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\dfrac{x}{1-x^2}))+(-\dfrac{x^2}2)}{x^4}$$
$$\lim_{x \to 0} \frac{(\dfrac{(1-\cos (\dfrac{x}{1-x^2}))}{(\dfrac{x}{1-x^2})^2})(\dfrac{x}{1-x^2})^2+(-\dfrac{x^2}2)}{x^4}$$
now what ?
| $$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(\frac{x}{1-x^2}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}$$
Now for Taylor's expansion we have that
$$\lim\limits_{x \to 0}\frac{1}{1-x^2}=1-x^2+o(x^2)$$
So
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x(1-x^2+o(x^2)\right)}{x^4}=$$
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}$$
Now recall that for Taylor's expansion
$$\lim\limits_{x \to 0}\cos t=1-\frac{t^2}{2}+\frac{t^4}{24}+o(t^4)$$
where, in our case, $t=(x-x^3)$.
Hence
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-1+\frac{x^2}{2}+x^4-\frac{x^4}{24}+o(x^4)}{x^4}=$$
$$=\lim\limits_{x \to 0}\frac{\frac{23}{24}x^4+o(x^4)}{x^4}=\frac{23}{24}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Getting x in terms of y I have this equation:
$$\dfrac{x}{y} = \dfrac{y-x}{x}$$
How would I separate $x$ and $y$ in $x^2+xy-y^2=0$ ?
| If
$\dfrac{x}{y} = \dfrac{y-x}{x}, \tag 1$
then, assuming $x \ne 0 \ne y$, as we must for the equation to make sense, we may set
$\alpha = \dfrac{x}{y}, \tag 2$
and writing (1) as
$\dfrac{x}{y} = \dfrac{y}{x} - 1, \tag 3$
we find
$\alpha = \alpha^{-1} - 1, \tag 4$
or
$\alpha^2 = 1 - \alpha, \tag 5$
or
$\alpha^2 + \alpha - 1 = 0; \tag 6$
then the quadratic formula yields
$\alpha = \dfrac{1}{2}(-1 \pm \sqrt 5); \tag 7$
since we know no more about $x$ and $y$, the best we can do is, from (1), write
$x = \alpha y, \tag 8$
i.e.,
$x = (\dfrac{1}{2}(-1 \pm \sqrt 5))y; \tag 9$
curiously enough, since
$(\dfrac{1}{2}(-1 + \sqrt 5))(\dfrac{1}{2}(-1 - \sqrt 5)) = -1, \tag{10}$
we may also write
$y = -(\dfrac{1}{2}(-1 \pm \sqrt 5))x. \tag{11}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
find a basis for $R^4$ given the vectors. i need help with this excercise..
find a basis for $R^4$ given the vectors.
$$\begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}\quad \text{and}\quad \begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix}$$
What method do I follow to find the other two vectors?
| Clearly, $v_1$ and $v_2$ are linearly independent, then you can add a new vector $v_3$ that is not in the span of
$$v_1=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}\quad \text{and}\quad v_2=\begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix}$$
Finally add a 4th vector $v_4$ that is not an element in the span of $v_1,v_2,v_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad. Solutions presented: $(x,y)=(0.625 a^2,0.612372 a^2)$ and $(x,y)=(0.625 a^2,-0.612372 a^2)$. I tried first to make the substitution $u=x+y$ and $v=x-y$, noticing that $x^2+y^2=0.5((x+y)^2+(x-y)^2)$ but could not go far using that route. Then I moved to squaring both equations, hoping to get a solution, but without success.
Hints and answers are appreciated. Sorry if this is a duplicate.
| $$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.$$
From first equation,
$$2x + 2\sqrt{x^2 - y^2} = a^2 \tag 2$$
$$(x- a^2/2)^2 = x^2 - y^2$$
$$x^2 + \dfrac{a^4}4 - xa^2 = x^2 - y^2 $$
$$-\dfrac{a^4}4 + xa^2 = y^2 \tag 3$$
Add (2) to second equation
$$ x + \sqrt{x^2 + y^2} = \dfrac32 a^2$$
$$x^2 + \dfrac94a^4 - 3a^2x = x^2 + y^2 $$
$$ 3a^2 x +y^2 -\dfrac94a^4 = 0$$
Substituting from 3
$$ 3a^2 x -\dfrac{a^4}4 + xa^2 -\dfrac94a^4 = 0$$
$$ 4a^2 x -\dfrac{10}4a^4 = 0$$
$$x = \dfrac{10}{16}a^2$$
Substituting back into 3
$$y^2 = -\dfrac{a^4}4 + \dfrac{10}{16}a^4 = \dfrac{6}{16}a^4$$
$$ y = \pm\dfrac{\sqrt{6}}{4}a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$ The question is to evaluate $$\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$$ if $x^2 \neq (n \pi-1) \forall n\in N$
I tried to rewrite the integral as $$\int \frac{x\sqrt{(2\sin(x^2+1))(1-\cos(x^2+1))}}{(2\sin(x^2+1))(1+\cos(x^2+1))}$$
If there would have been a radical sign in denominator too the problem would be quite simple but I couldn't understand how to proceed from here.Any ideas?
| Note that the substitution $x^2+1=y$ and $2x\,\mathrm dx=\mathrm dy$ leads to\begin{align}\frac12\int\frac{\sqrt{2\sin(y)-2\sin(y)\cos(y)}}{2\bigl(\sin(y)+2\sin(y)\cos(y)\bigr)}\,\mathrm dy&=\frac{\sqrt2}4\int\frac{\sqrt{\sin(y)-\sin(y)\cos(y)}}{\sin(y)+2\sin(y)\cos(y)}\,\mathrm dy\\&=\frac{\sqrt2}4\int\frac{\sqrt{\csc(y)-\cot(y)}}{1+\cos y}\mathrm dy.\end{align}Usinge the formulas for $\sin(2\theta)$, $\cos(2\theta)$, and $\cot(2\theta)$, it is not hard to see that this is equal to $\frac{\sqrt2}4\cdot\frac23\tan^{3/2}\left(\frac y2\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| The unpopular limit formula $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1}$$ is your friend here. First divide the numerator and denominator of the expression by $x-7$ and note that the denominator itself becomes a fraction $$\frac{\sqrt[4]{x+9}-2} {x-7}=\frac{t^{1/4}-16^{1/4}}{t-16}$$ where $t=x+9\to 16$. By the limit formula $(1)$ the above fraction tends to $(1/4)16^{-3/4}=1/32$.
Next consider the numerator which has now become $$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}=\frac{u^{1/2}-9^{1/2}}{u-9}-\frac{v^{1/3}-27^{1/3}}{v-27}$$ where $u=x+2\to 9,v=x+20\to 27$. By the limit formula $(1)$ the above expression tends to $(1/2)9^{-1/2}-(1/3)27^{-2/3}=7/54$. The final answer is now $(7/54)/(1/32)=112/27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$ I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$.
So I pose the following problem:
Given $a \in \mathbb{R}-\{0\}$, find $x$ such that $\dfrac{dy}{dx}=0$ where $y=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$
As in most maxima/minima problems, we first (implicitly) differentiate it and set to $0$ to give $$\boxed{\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x} \tag{1}$$ I have no idea how to continue from here. I thought about taking logarithms, but it appears to me that the double $\ln$ in the term $\dfrac{a^2x^2-x+a}{ax}$ would only make the equation worse.
(For the simplest case when $a=1$, the problem is easy: $x=1$ and it is a point of inflexion).
Let's try setting each of the terms to $0$:
Case $1$: $\left(\frac{a^2x^2-x+a}{ax}\right)^x=0$
$ \hspace{1cm}$ This is only possible when the fraction is zero; that is, solving $a^2x^2-x+a=0$ to get $$x=\frac{1\pm\sqrt{1-4a^3}}{2a^2}$$
Case $2$: $\left(\frac{ax^2-a^2x+1}{ax}\right)^a=0$
$ \hspace{1cm}$ This gives $$\begin{align}ax^2-a^2x+1=0&\implies a^2x^2+a=a^3x\\&\implies\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(\dfrac{a^3x-x}{ax}\right)^x=\left(\dfrac{a^3-1}{a}\right)^x=0\end{align}$$
$ \hspace{1cm}$ so for equality between LHS and RHS, we must have $a=1$. However, the equation
$ \hspace{1cm}$ $ax^2-a^2x+1=0$ has no real solutions for such $a$; hence we reach a contradiction.
Case $3$: $\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=0$
$ \hspace{1cm}$ We have $x=\pm \dfrac1a$. Now LHS is $0$, and $$\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(1-\dfrac1a+a\right)^{\frac1a} \neq 0$$
$ \hspace{1cm}$ for $a \in \mathbb{R} - \{\phi\}$, where $\phi$ is the golden ratio.
$ \hspace{1cm}$ Suppose that $a = \phi$. Then $x$ is forced to be $-\dfrac1a=-\dfrac2{1+\sqrt5}$, since $ax^2-a^2x+1=0$
$ \hspace{1cm}$ (undefined) when $x=\dfrac 1a$. This is impossible, since $y$ is only defined when $x>0$ for this
$ \hspace{1cm}$ value of $a$!
Case $4$: $\ln\left(\frac{a^2x^2-x+a}{ax}\right)+\frac{a(ax^2-1)}{a^2x^2-x+a}=0$
$ \hspace{1cm}$ This is impossible from cases $1$ and $3$.
UPDATE: I have provided a partial answer to my question, now with $x$ removed from it.
Any hints on how to solve $(3)=(4)$ are welcome.
Here, on MathOverflow: https://mathoverflow.net/questions/302105/on-finding-the-critical-points-of-fx-leftx-a-frac1ax-righta-left-fra
| I've made some progress on the question, but I'm nowhere close to solving it.
For ease of reading, I will repeat the boxed equation below: $$\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x$$
At the end of my post above, there was a suggestion of setting $$\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=k$$ in the hope that we can find $k$ in terms of $a$, since only this has simple solutions for $x$.
The consequences of this are as follows: $$ax^2+1=a(a+\sqrt[a]k)x\tag{1}$$$$\small x=\frac{a+\sqrt[a]k}2\pm\frac{\sqrt{a^2(a+\sqrt[a]k)^2-4a}}{2a}\implies x^2=\frac{a(a+\sqrt[a]k)^2\pm(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-2}{2a}\tag{2}$$
This means that using $(1)$, $$\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=\frac{a(ax^2-1)}{x(ax^2+1)-a^2x^2}=\frac{ax^2-1}{\sqrt[a]kx^2}=\boxed{\frac a{\sqrt[a]k}-\frac1{\sqrt[a]kx^2}}$$
We now deal with the natural logarithm term on the RHS, using $(1)$. Notice that $$\ln\left(\frac{a^2x^2-x+a}{ax}\right)=\ln\left(\frac{a(ax^2+1)}{ax}-\frac1a\right)=\boxed{\ln\left(a(a+\sqrt[a]k)-\frac1a\right)}$$ Hooray, we've got rid of the $x$! Similarly, we have $$\left(\frac{a^2x^2-x+a}{ax}\right)^x=\boxed{\left(a(a+\sqrt[a]k)-\frac1a\right)^x}$$
The last term next to the logarithmic term is a pain in the neck. However, $(1)$ can still be used. $$\frac{a(ax^2-1)}{a^2x^2-x+a} = \frac{a(ax^2+1)-2a}{a(ax^2+1)-x}=\boxed{\frac{a^2(a+\sqrt[a]k)x-2a}{a^2(a+\sqrt[a]k)-x}}$$
Let's combine all of the boxed terms to see what we currently have: $$k\left(\frac a{\sqrt[a]k}-\frac1{\sqrt[a]kx^2}\right)=\left(\ln\left(a(a+\sqrt[a]k)-\frac1a\right)+\frac{a^2(a+\sqrt[a]k)x-2a}{a^2(a+\sqrt[a]k)-x}\right)\left(a(a+\sqrt[a]k)-\frac1a\right)^x$$
Now we use $(2)$. The steps to the final result are horrible, so I will just give it here. Note that when I took logarithms in the RHS, I used the fact that $\ln(ab^c)=\ln a + c\ln b$. Just hope I haven't made any mistakes: $$\text{LHS}=\frac k{\sqrt[a]k}\left(1-\frac{4a-2}{a^2(a+\sqrt[a]k)^2\pm(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-2}\right)\tag{3}$$ and $$\begin{align}\ln(\text{RHS})&=\frac{a(a+\sqrt[a]k)\pm\sqrt{a^2(a+\sqrt[a]k)^2-4a}}{2a}\ln\left(a(a+\sqrt[a]k)-\frac1a\right)\tag{4}\\\small&+\ln\left(\ln\left(a(a+\sqrt[a]k)-\frac1a\right)+\frac{a^3(a+\sqrt[a]k)^2\pm a^2(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-4}{2a^3(a+\sqrt[a]k)-(a+\sqrt[a]k)\pm\sqrt{a^2(a+\sqrt[a]k)^2-4a}}\right)\end{align}$$ Is it even possible now to rearrange this to make $k$ the subject?
Here is a visualisation of the equation $\text{LHS}=\text{RHS}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548041",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Computing an infinite product Let $a_0=5/2$ and $a_k=a^2_{k-1}-2$ for all $k\geq 1$.The question is to compute $$\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
I tried to calculate few terms.$a_0=5/2$, $a_1=17/4,a_2=257/16$ it seems that $a_k$ is of the form $2^{2^k}+2^{-2^{k}}$ however I am not sure about it.How to proceed without doing much guesswork.Any ideas?
| The explicit formula $a_k=2^{2^k}+2^{-2^{k}}$ is straightforward to prove by induction.
We also have
$$ a_k+1 = a_{k-1}^2-1 = (a_{k-1}-1)(a_{k-1}+1) $$
hence
$$ a_{k-1}-1 = \frac{a_k+1}{a_{k-1}+1} $$
and
$$ \prod_{k=1}^{N}\left(a_k-1\right)=\prod_{k=2}^{N+1}\left(a_{k-1}-1\right)=\frac{a_{N+1}+1}{a_1+1}. $$
Since
$$ \frac{x^{2^M}-y^{2^M}}{x-y} = \left(x^{2^{M-1}}+y^{2^{M-1}}\right)\cdots(x+y) $$
by picking $M=N+1, x=2$ and $y=\frac{1}{2}$ we get
$$a_N\cdot a_{N-1}\cdots a_0 =\frac{2^{2^{N+1}}-2^{-2^{N+1}}}{\frac{3}{2}}$$
and
$$ \prod_{k=0}^{N}\frac{a_k-1}{a_k}=\frac{3}{5}\cdot\frac{\prod_{k=1}^{N}(a_k-1)}{\prod_{k=1}^{N}a_k} = \frac{\left(\frac{3}{2}\right)^2}{a_1+1}\left(1+\frac{1}{2^{2^{N+1}}-2^{-2^{N+1}}}\right).$$
By letting $N\to \infty$ it follows that the wanted product equals $\frac{\left(\frac{3}{2}\right)^2}{a_1+1}= \color{red}{\large\frac{3}{7}}.$
| {
"language": "en",
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"answer_count": 1,
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} |
eigen vector that orthogonal to each other, symmetry matrix symmetry matrix $\left(\begin{array}{ccc} 0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{array}\right) $
one of eigen value is $\lambda_1=2$ and one of eigen vector is $x_1=\left(\begin{array}{ccc} \frac{1}{\sqrt3} \\
\frac{1}{\sqrt3} \\
\frac{1}{\sqrt3} \end{array}\right) $
then i found the two other eigen value is $\lambda_2=-1 $ & $\lambda_3=-1$
but the question want eigen vector that is orthogonal to each other and has magnitude $
|x_2|=|x_3|=1$
here how can i find the two other eigen vector that orthogonal to each other?
using usual computation $x=-y-z, i cant find the vector that is orthogonal to each other and with magnitude 1.
using trial and error?
| \begin{eqnarray*}
\begin{bmatrix}
\frac{2}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\
\end{bmatrix},
\begin{bmatrix}
0 \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \\
\end{bmatrix}
\end{eqnarray*}
Edit: Note that
\begin{eqnarray*}
\begin{bmatrix}
2 \\ -1 \\ -1 \\
\end{bmatrix},
\begin{bmatrix}
-1 \\ 2 \\ -1 \\
\end{bmatrix},
\begin{bmatrix}
-1 \\ -1 \\ 2 \\
\end{bmatrix}
\end{eqnarray*}
are eigenvectors with eigenvalue $-1$ and between them they span a space of dimension $2$, so we need to find a linear combination that is orthogonal the first one
\begin{eqnarray*}
\begin{bmatrix}
2 \\ -1 \\ -1 \\
\end{bmatrix} \cdot ( \alpha
\begin{bmatrix}
2 \\ -1 \\ -1 \\
\end{bmatrix}+ \beta
\begin{bmatrix}
-1 \\ 2 \\ -1 \\
\end{bmatrix}+ \gamma
\begin{bmatrix}
-1 \\ -1 \\ 2 \\
\end{bmatrix} ) .
\end{eqnarray*}
This gives $ 6 \alpha -3 \beta -3 \gamma=0$ and $ \alpha=1, \beta=-2, \gamma=0$ will do, and now you just need to normalise.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Is The Value of This Definite Integral $0$?
Q. Evaluate -$$\int\limits_0^\pi {{{x\,dx} \over {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} $$
My Try -
$$\eqalign{
& Let\,I\, = \,\int\limits_0^\pi {\frac{{x\,dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^\pi {\frac{{\left( {\pi - x} \right)\,dx}}{{{a^2}{{\cos }^2}\left( {\pi - x} \right) + {b^2}{{\sin }^2}\left( {\pi - x} \right)}}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \pi \int\limits_0^\pi {\frac{{dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} \,\, - \,\,I \cr
& \,\,\,\,\,\,\,I = \frac{\pi }{2}\int\limits_0^\pi {\frac{{dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} \cr
& {\text{Now}}\,{\text{divide}}\,{\text{numerator}}\,{\text{and}}\,{\text{denominator}}\,{\text{by}}\,\,{{\text{b}}^{\text{2}}}{\text{co}}{{\text{s}}^{\text{2}}}{\text{x,}}\,\,\,\, \to \,\,\boxed1 \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{\pi }{{2{b^2}}}\int\limits_0^\pi {\frac{{{{\sec }^2}x\,\,dx}}{{\frac{{{a^2}}}{{{b^2}}} + {{\tan }^2}x}}} \cr
& Let\,\,t = \tan x,\,\,dt = {\sec ^2}x\,dx\,\,and\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when\,\,x = 0,\,\,t = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,and\,\,x = \pi ,\,\,t = 0 \cr
& \,\,\,\,\,\,\,I = \frac{\pi }{{2{b^2}}}\int\limits_0^0 {\frac{{dt}}{{\frac{{{a^2}}}{{{b^2}}} + {t^2}}}} = 0 \cr} $$
First Question - $$\eqalign{
& {\text{Infact,}}\,\,{\text{the}}\,{\text{answer}}\,{\text{is}}\,{\text{not}}\,{\text{0}}\,{\text{and}}\,{\text{is}}\,{\text{obtained}}\,{\text{by}}\,{\text{splitting}}\,{\text{the}}\,{\text{integral}}\,{\text{further}}\,{\text{in}}\,step\,\boxed1{\text{.}}\, \cr
& {\text{What}}\,{\text{mistake}}\,{\text{have}}\,{\text{I}}\,{\text{committed}}\,{\text{here?}} \cr} $$
$$\eqalign{
& {\text{Instead,}}\,{\text{in}}\,step\,\boxed1\,{\text{,}}\,\,{\text{if}}\,{\text{we}}\,{\text{divide}}\,{\text{by}}\,{{\text{a}}^{\text{2}}}{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}\,{\text{and}}\,{\text{perform}}\,{\text{similar}}\,{\text{substitution,}} \cr
& = \frac{\pi }{{2{a^2}}}\int\limits_0^\pi {\frac{{\cos e{c^2}x\,\,dx}}{{\frac{{{b^2}}}{{{a^2}}} + {{\cot }^2}x}}} \cr
& Let\,\,t = \cot x,\,\,dt = \cos e{c^2}x\,dx\,\,and\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when\,\,x = 0,\,\,t = not\,defined \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,and\,\,x = \pi ,\,\,t = not\,defined \cr} $$
Second Question -$${\text{So,}}\,{\text{why}}\,{\text{can't}}\,{\text{we}}\,{\text{say}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,{\text{the}}\,{\text{integral}}\,{\text{is}}\,{\text{not}}\,{\text{defined?}}$$
| As @José Carlos Santos points out, using a substitution of $t = \tan x$ on the interval $[0,\pi]$ introduces a discontinuity at the point $x = \pi/2$.
One way this can be avoided, as you correctly identified, is by first changing the interval of integration. If we note that the integrand in the integral
$$I = \int^\pi_0 \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x},$$
is both an even and periodic function with a fundamental period of $\pi$, since for any continuous and bounded function $f$ with period $\mathfrak{a}$
$$\int^{b + \mathfrak{a}}_b f(x) \, dx = \int^\mathfrak{a}_0 f(x) \, dx,$$
where $b \in \mathbb{R}$, we have
\begin{align*}
I &= \frac{\pi}{2} \int^\pi_0 \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}\\
&= \frac{\pi}{2} \int^{-\pi/2 + \pi}_{-\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \qquad \text{(since it is periodic with period $\pi$)}\\
&= \frac{\pi}{2} \int^{\pi/2}_{-\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}\\
&= \pi \int^{\pi/2}_0 \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \qquad \text{(since it is even)}.
\end{align*}
Now your substitution of $t = \tan x$ can be made.
| {
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"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite Integral $\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$ for Logarithm and Algebraic.
Evaluate
$$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$$
I have been thinking for this question quite some time, I've tried all the methods I have learnt but still getting nowhere. Hope that someone can explain it for me. Thanks in advance.
| If you want to avoid series altogether one could always try Feynman's trick of differentiating under the integral sign.
Consider
$$I(a) = \int^1_0 \frac{\ln \left [a(x^2 - x) + 1 \right ]}{x - x^2} \, dx, \quad a \geqslant 0.$$
Note that $I(0) = 0$ and we require $I(1)$.
Differentiating under the integral sign with respect to the parameter $a$ we have
$$I'(a) = - \int^1_0 \frac{dx}{a(x^2 - x) + 1}.$$
Completing the square in the denominator before integrating we have
\begin{align*}
I'(a) &= -\frac{1}{a} \int^1_0 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \left (\frac{1}{2} \sqrt{\frac{4 - a}{a}} \right )^2}\\
&= -\frac{2}{a} \sqrt{\frac{a}{4 - a}} \left \{\tan^{-1} \left [2 \sqrt{\frac{a}{4 - a}} \left (x - \frac{1}{2} \right ) \right ]\right \}^1_0\\
&= -\frac{4}{a} \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}}.
\end{align*}
Thus
$$I(1) = -4 \int^1_0 \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}} \cdot \frac{da}{a}.$$
Letting
$$u^2 = \frac{a}{4 - a},$$
observing that
$$a = \frac{4u^2}{1 + u^2},$$
we have
$$da = \frac{8 u}{(1 + u^2)^2} \, du,$$
while for the limits of integration: $(0,1) \mapsto (0,1/\sqrt{3})$. Thus
$$I(1) = - 8 \int^{1/\sqrt{3}}_0 \frac{\tan^{-1} u}{1 + u^2} \, du.$$
This last integral can be readily found by setting $t = \tan^{-1} u$. Doing so yields
$$I(1) = -8 \int^{\pi/6}_0 t \, dt = -8 \left [\frac{t^2}{2} \right ]^{\pi/6}_0,$$
or
$$\int^1_0 \frac{\ln \left (x^2 - x + 1 \right )}{x - x^2} \, dx = -\frac{\pi^2}{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove the associative la of multiplication for 2x2 matrices, show that (AB)C=A(BC)? Let
$$A= \left[
\begin{array}{cc|c}
a_{11}&a_{12}\\
a_{21}&a_{22}
\end{array}
\right] , B= \left[
\begin{array}{cc|c}
b_{11}&b_{12}\\
b_{21}&b_{22}
\end{array}
\right],C= \left[
\begin{array}{cc|c}
c_{11}&c_{12}\\
c_{21}&c_{22}
\end{array}
\right] $$
I did the exercise, but I have doubts if I have the procedure correctly, if you can tell me if I have an error it would be very helpful. This is my answer:
$$ AB= \left[
\begin{array}{cc|c}
a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\
a_{21}b_{21}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}
\end{array}
\right] $$
$$ (AB)C= \left[
\begin{array}{cc|c}
a_{11}b_{11}C_{11}+a_{12}b_{21}C_{21}&a_{11}b_{12}C_{12}+a_{12}b_{22}C_{22}\\
a_{21}b_{21}C_{11}+a_{22}b_{21}C_{21}&a_{21}b_{12}C_{12}+a_{22}b_{22}C_{22}
\end{array}
\right] $$
$$ BC= \left[
\begin{array}{cc|c}
b_{11}C_{11}+b_{21}C_{21}&b_{12}C_{12}+b_{22}C_{22}\\
b_{21}C_{11}+b_{21}C_{21}&b_{12}C_{12}+b_{22}C_{22}
\end{array}
\right] $$
$$ A(BC)= \left[
\begin{array}{cc|c}
a_{11}b_{11}C_{11}+a_{12}b_{21}C_{21}&a_{11}b_{12}C_{12}+a_{12}b_{22}C_{22}\\
a_{21}b_{21}C_{11}+a_{22}b_{21}C_{21}&a_{21}b_{12}C_{12}+a_{22}b_{22}C_{22}
\end{array}
\right] $$
Therefore (AB)C= A(BC)
| It's wrong, this is the correct solution for $(AB)C$:
$$ (AB)C=\left[
\begin{array}{cc|c}
(a_{11}b_{11}+a_{12}b_{21})C_{11}+(a_{11}b_{12}+a_{12}b_{22})C_{21}&(a_{11}b_{11}+a_{12}b_{21})C_{12}+(a_{11}b_{12}+a_{12}b_{22})C_{22}\\
(a_{21}b_{21}+a_{22}b_{21})C_{11}+(a_{21}b_{12}+a_{22}b_{22})C_{21}&
(a_{21}b_{21}+a_{22}b_{21})C_{12}+(a_{21}b_{12}+a_{22}b_{22})C_{22}
\end{array}
\right] $$
For the matrix $A(BC)$ is the same.
| {
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"answer_count": 1,
"answer_id": 0
} |
Substitution in integral gives $0$ to $0$ bounds $\require{cancel}$
I must find the length of the curve given by those 2 equations:
$x = \cos^3t$
and
$y = \sin^3t$ between $0$ and $2\pi$.
I can find their derivative: $\frac{dx}{dt} = -3\cos^2t\sin t$ and $\frac{dy}{dt} = 3\sin^2t\cos t$.
Then I can apply the formula (results on the right have been acquired with my calculator):
$$L_0^{2\pi} =
\int_0^{2\pi}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}}dt =
\int_0^{2\pi}{\sqrt{\left(-3\cos^2t\sin t\right)^2 + \left(3\sin^2t\cos t\right)}}dt = 6$$
$$L_0^{2\pi} =
\int_0^{2\pi}{\sqrt{9\cos^4t\sin^2t + 9\sin^4t\cos^2t}}\,dt = 6$$
$$L_0^{2\pi} =
3\int_0^{2\pi}{\sqrt{\cos^2t\sin^2t\cdot(\cos^2t + \sin^2t)}}\,dt = 6$$
$$L_0^{2\pi} =
3\int_0^{2\pi}{\sqrt{\cos^2t\sin^2t}}\,dt = 6$$
$$L_0^{2\pi} =
3\int_0^{2\pi}{\cos t\sin t}\,dt = 0\,\,???$$
Let $z = \sin t \Rightarrow dz = \cos t\,dt$
For
$t = 0, z = 0$ and for $t = 2\pi, z = 0$
$$L_0^{2\pi} =
3\int_0^0\cancel{\cos t}\cdot z\cdot\frac{1}{\cancel{\cos t}}\,dz = 3\int_0^0z\,dz = 0$$
$$L_0^{2\pi} = 3\left[\frac{z^2}{2}\right]_0^0 = 0$$
So how does the value of the integral changes only with a square root simplification (multiplying exponents $\frac{1}{2}$ and $2$)?
$\int_0^0$ doesn't make much sense to me but we weren't told how to deal with this kind of situation. How should I proceed to get the good answer?
| The problem is that $\sqrt{\cos^2 t\sin^2 t}$ is only $\cos t \sin t$ when $\cos t \sin t \ge 0$. Otherwise it's $-\cos t \sin t$.
| {
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"answer_count": 1,
"answer_id": 0
} |
If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. Of course, if we prove this stronger statement we are done. Any idea?
| Well,
\begin{align}
&& a+b &\mid a^4 + b^4 \\
&\iff & a+b &\mid a^4 + b^4 - (a+b)^4 + 4ab(a+b)^2 \\
&\iff & a+b &\mid 2a^2b^2 \\
&\iff & a+b &\mid ab\bigl((a+b)^2 - (a^2+b^2)\bigr) \\
&\iff & a+b &\mid ab(a^2+b^2)\,
\end{align}
so for coprime $a,b$ it follows that $a+b \mid a^4+b^4 \iff a+b \mid a^2+b^2$.
Writing $a = gx,\, b = gy$ with $g = \gcd(a,b)$, we see that $a+b \mid a^4 + b^4$ if and only if $x+y \mid g^3(x^2+y^2)$, and $a+b \mid a^2+b^2$ if and only if $x+y \mid g(x^2+y^2)$. So if we find coprime $x,y$ and a $g$ such that $x+y \nmid g(x^2+y^2)$ but $x+y \mid g^3(x^2+y^2)$, we have a counterexample.
Choosing $x = 1,\, y = 7$ and $g = 2$ provides one, $1+7 = 8 \nmid 100 = 2(1+7^2)$, but $8 \mid 400$. So
$$(2+14) \mid 2^4 + 14^4\qquad\text{and}\qquad 2+14 \nmid 2^2 + 14^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
| Let me give a general method which is useful
for this sum $\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}$
we have
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}=\sum_{n=0}^\infty\dfrac{1}{(2n+1)(2n+3)}$
we can use this method
$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
where $\psi$ is digamma function.
now we can write
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}= \frac{1}{4}\sum_{n=1}^\infty\dfrac{1}{(n+\frac{1}{2})(n+\frac{3}{4})}=\frac{1}{2}$
since $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find out all solutions for the system Given the system
$$ \left[
\begin{array}{ccc|c}
x_1&x_2&x_3&k\\
x_1&x_2&kx_3&1\\
x_1&kx_2&x_3&1\\
kx_1&x_2&x_3&1\\
\end{array}
\right] $$
I tried to solve this...It looks simple but I found a problem at the end...
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
1&1&k&1\\
1&k&1&1\\
k&1&1&1\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
0&0&k-1&1-k\\
0&k-1&0&1-k\\
k-1&0&0&1-k\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
0&0&1& \frac{1-k}{k-1}\\
0&1&0& \frac{1-k}{k-1}\\
1&0&0& \frac{1-k}{k-1}\\
\end{array}
\right] $$
Finally,
$$ \left[
\begin{array}{ccc|c}
1&0&0& -1\\
0&1&0& -1\\
0&0&1& -1\\
0&0&0&k+3\\
\end{array}
\right] $$
There is no way to get infinitely many solutions.
$$$$
However, I tried to put the system into other online calculator...
Click me$$$$
There is infinitely many solutions when k=4 or other numbers. What's wrong of my work...
| Let us add the last three equations to gather to get: $$(k+2)(x_1+x_2+x_3) =3$$
Then we use the first one to get, $$(k+2)(k)=3 \implies k=1, -3$$
So, ?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$
| Let's put $x=1/t$ so that $t\to 0^{+}$ and the expression under limit is transformed into $$\frac{\sqrt{1+1/t^{2}}-\sqrt[3]{1+1/t^{3}}}{t}=\frac{\sqrt{1+t^{2}}-\sqrt[3]{1+t^{3}}}{t^{2}}$$ and this can be expressed as a difference of two standard limits $$\lim_{t\to 0^{+}}\frac{(1+t^{2})^{1/2}-1}{t^{2}}-t\cdot \frac{(1+t^{3})^{1/3}-1}{t^{3}}$$ which equals $1/2-0(1/3)=1/2$. The standard limit formula $$\lim_{x\to 0}\frac{(1+x)^{n}-1}{x}=n$$ used above is a direct consequence of the more general limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Residue calculations involving logarithms. I am trying to solve a definite integral using residue theory. At some step, I wish to calculate the residue of the function $\frac{\log{(z)}}{z^3+8}$ at $-2e^{2\pi i/3}$. The log function is defined for $0 < \arg z < 2\pi$.
I'm having difficulty doing this simple looking calculation. Can someone please help me with this?
I have been given a hint that $Res(\frac{\log{(z)}}{z^3+8},z_i)=\frac{-z_i log(z_i)}{24}$. Why and how is this true?
| $$\frac{\log(z)}{z^3+8}= \frac{\log(z)}{(z+2e^{\frac{2i\pi}3})(z^2-2e^{\frac{2i\pi}{3}}z+4e^{\frac{-2i\pi}{3}})}$$
Then the residue at $\displaystyle z=-2e^{\frac{2\pi i}{3}}$ should be:
$$\frac{\log(-2e^{\frac{2\pi i}{3}})}{(-2e^{\frac{2\pi i}{3}})^2-2(-2e^{\frac{2\pi i}{3}})(e^{\frac{2i\pi}{3}})+4e^{\frac{-2i\pi}{3}}}= \frac{\log (-2e^{\frac{2i\pi}{3}})}{4e^{\frac{4i\pi}{3}}+ 4e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}= \frac{\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}= \frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{6i\pi}{3}}+ 4}=\frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{12}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Solving $z^2 - 8(1-i)z + 63 - 16i = 0$ with reduced discriminant $z^2 - 8(1-i)z + 63 - 16i = 0$
I have come across a problem in the book Complex Numbers from A to Z and I do not understand the thought process of the author when solving:
The start to the solution is as follows:
$\Delta$ = discriminant
$\Delta ' = (4 - 4i)^2 - (63 - 16i) = -63 - 16i$
and $r = |\Delta '| = \sqrt{63^{2} + 16^{2}} = 65$, where $\Delta ' = (\frac{b}{2})^{2} - ac$
It is this last part I do not understand. I do not believe this definition for the discriminant has been previously mentioned in the book, and I struggle to see where it has come from.
Thanks
| Since $∆' = -63-16i$
$$ z = \frac{ -b±\sqrt{∆'}}{a} $$
$$ z= 8 - 8i ± \sqrt{-63-16i} $$
Using Quadratic Formula, instead of reduced determinant
$$ z= \frac{8(1-i)±\sqrt{\left[8(1-i)\right]^2-4\cdot1\cdot(63+16i}}{2\cdot1} $$
$$ = \frac{8-8i±\sqrt{\left[64(2i)\right]-\cdot(4×63+4×16i}}{2} $$
$$ = \frac{8-8i±\sqrt{\left[(128i)\right]-(192+64i}}{2} $$
$$ = \frac{8-8i±\sqrt{128i-192-64i}}{2} $$
$$ = \frac{8-8i±\sqrt{192+64i}}{2} $$
$$ = \frac{8-8i±\sqrt{16(12+4i)}}{2} $$
$$ = \frac{8-8i±4\sqrt{4(3+i)}}{2} $$
$$ = \frac{8-8i±8\sqrt{(3+i)}}{2} $$
$$ = \frac{8(1-i)±8\sqrt{(3+i)}}{2} $$
$$ = 4(1-i)±4\sqrt{(3+i)} $$
$$ = 4[(1-i)+\sqrt{(3+i)}], 4[(1-i)-\sqrt{(3+i)}]$$
Are two possible solutions of your equation.
| {
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Sum based on stolz theorem The Sum is as follows:
$$ \lim_{n\to \infty} \frac{1} {\sqrt n} \left( \frac {1} {\sqrt n}+ \frac {1} {\sqrt {n+1}} +...+ \frac{1}{\sqrt{2n}} \right) $$.
I solved as follows using stolz theorem
$$ \lim_{n\to \infty} \frac{x_n-x_{n-1}} {y_n-y_{n-1} } = \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} - \frac {1} {\sqrt{2n-1}}} {\sqrt n - \sqrt{n-1}}$$
where $$ {\{x_n\} = \left( \frac {1} {\sqrt n}+ \frac {1} {\sqrt {n+1}} +...+ \frac{1}{\sqrt{2n}} \right)} $$
and
$$ {\{y_n\} = \sqrt n} $$
I rationalized the denominator and got the answer as zero.
But the solution in the book is given as
$$ \lim_{n\to \infty} \frac{x_n-x_{n-1}} {y_n-y_{n-1} } = \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} + \frac {1} {\sqrt{2n-1}}- \frac{1} {\sqrt{n-1}}} {\sqrt n - \sqrt{n-1}}$$
$$ = 2(\sqrt2-1) $$
how they got the second step that is
$$ \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} + \frac {1} {\sqrt{2n-1}}- \frac{1} {\sqrt{n-1}}} {\sqrt n - \sqrt{n-1}}$$
Please explain.
Thanks in advance.
| You've made a mistake in evaluating $x_n-x_{n-1}$ \begin{align}x_{n}=&&\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}&+\frac1{\sqrt{2n-1}}+\frac1{\sqrt{2n}}\\x_{n-1}=&\frac1{\sqrt{n-1}}&+\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}\end{align}
They didn't.
| {
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Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& = 2^n \cdot 3^n + 2^{3n} \\
\end{align}
\begin{align}
\text{Assume} ~ f(k) & = 6^k + 8^k \\
& = 2^k \cdot 3^k + 2^{3k} \\
f(k+1) & = 2^{k+1} \cdot 3^{k+1} + 2^{3(k+1)} \\
\end{align}
Where do I go from here?
| Because $$6^n+8^n=(6+8)(6^{n-1}-6^{n-2}\cdot8+...-6\cdot8^{n-2}+8^{n-1}).$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 2
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Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
| Let the LHS be $A(n)$ and let the RHS be $2B(n)^4.$
Then $A(n+1)-A(n)=(n+1)^7+(n+1)^5=(n+1)^5(n^2+2n+2).$
$$\text {We have }\quad 2B(n+1)^4-2B(n)^4=$$ $$(*)\quad =2(B(n+1)^2+B(n)^2)\cdot (B(n+1)+B(n))\cdot (B(n+1)-B(n)).$$ Since $B(n)=n(n+1)/2$ we have $$B(n+1)^2+B(n)^2=(n+1)^2((n+2)^2+n^2)/4=(n+1)^2(n^2+2n+2)/2$$ $$\text { and}\quad B(n+1)+B(n)=(n+1)((n+2)+n)/2=(n+1)^2$$ $$\text { and }\quad B(n+1)-B(n)=n+1.$$ From these we compute $(*)$ (4th line from the top) and find it is equal to $(n+1)^5(n^2+2n+2)$, which is $A(n+1)-A(n).$ (2nd line from the top.).
So if $A(1)=B(1)$ then $A(n)=B(n)$ for all $n\in \Bbb N$ by induction on $n.$
Interesting identity. Another interesting one is $\sum_{x=1}^nx^3=(\sum_{x=1}^nx)^2=(n(n+1)/2)^2.$
| {
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"source": "stackexchange",
"question_score": "12",
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"answer_id": 4
} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| By power series we need the following two:
$$\cos x=1-\frac{x^2}{2}+o(x^2)$$
$$(1+x)^n=1+nx+o(x)$$
Thus
$$\sqrt {\cos 2x}=\left( 1-\frac{4x^2}{2}+o(x^2)\right)^\frac12=1-x^2+o(x^2)$$
$$\sqrt[3] {\cos 3x}=\left( 1-\frac{9x^2}{2}+o(x^2)\right)^\frac13=1-\frac{3x^2}{2}+o(x^2)$$
Thus
$$\sqrt {\cos 2x}×\sqrt[3] {\cos 3x}=\left(1-x^2+o(x^2)\right) \left(1-\frac{3x^2}{2}+o(x^2)\right)=1-\frac{5x^2}{2}+o(x^2)$$
And then:
$$\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}= \frac{1-\frac{x^2}{2}-1+\frac{5x^2}{2}+o(x^2)}{x^2}=\frac{2x^2+o(x^2)}{x^2}=2+o(1)\to2$$
| {
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derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I compared the limit of the function with $f(1)$:
$$\lim_{x\to 1}\space \frac{2}{x}\sin(x^3) = 2\sin(1) = f(1)$$
first question: was this step necessary?
next:
$$f'(1) =\lim_{h\to 0}\space \frac{ \frac2{h+1}\sin((h+1)^3)-2\sin(1)}h$$
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3)}{h+1}-2\sin(1)}h$$
and I know about $\lim_{x\to 0}\frac {\sin(x)}x = 1$, so:
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3) \cdot (h+1)^2}{(h+1)^3}-2\sin(1)}h$$
and then:
$$=\lim_{h\to 0}\space \frac{ 2(h+1)^2-2\sin(1)}h$$
and now I don't know how to get rid of the $h$ denominator
| I did not answer your exact question, I looked at the case of $x=0$ rather than $x=1$, but the argument is analogous.
Evaluate the derivative of $f : \mathbb{R} \to \mathbb{R}$, where $$f(x) : = \begin{cases}
\frac{2}{x} \sin(x^3), & x \in \mathbb{R} \backslash \{ 0 \}, \\
0, & \text{otherwise}.
\end{cases}$$
Let us first verify that $f$ is at least continuous at $x=0$, let alone differentiable. To this end, write $\sin(x)$ as its Taylor expansion $$\sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1}.$$ Then $$\frac{2}{x} \sin(x^3) = 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{3(2k+1) - 1} = 2\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{6k+2}. $$ We therefore see that $$\lim_{x \to 0} \frac{2}{x} \sin(x^3) =0,$$ verifying that $f$ is continuous at $0$. Now let us check whether $f$ is differentiable at $x=0$. Indeed, we observe that \begin{eqnarray*}
\lim_{\Delta x \to 0} \frac{f(0+\Delta x) - f(0)}{\Delta x} &=& \lim_{\Delta x \to 0} \frac{f(\Delta x) - f(0)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{2}{\Delta x \cdot \Delta x} \sin ( (\Delta x)^3 ).
\end{eqnarray*} For simplicity, let $z = \Delta x$ and note that we just need to determine whether the limit $$\lim_{z \to 0} \frac{2 \sin(z^3)}{z^2}$$ exists. Analogously to the above continuity argument, write $\sin(z^3)$ as its Taylor expansion, $$\sin(z^3) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{3(2k+1)}.$$ It then follows that \begin{eqnarray*}
\lim_{z \to 0} \frac{2}{z^2} \sin(z^3) &=& \lim_{z \to 0} 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{6k+2} \\
&=& 0,
\end{eqnarray*} so $f$ is differentiable with $f'(0)=0$.
| {
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"source": "stackexchange",
"question_score": "2",
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Steiner inellipse Hello it's related to my answer for Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq 3\frac{a+b+c}{ax+by+cz}$
My answer fails but I don't know why ... So I was thinking a generalization of the following formula:
$$\frac{IA^2}{CA\cdot AB}+\frac{IB^2}{BC\cdot AB}+\frac{IC^2}{CA\cdot BC}=1$$
I know that it's related to the Steiner inellipse and we have for a triangle ABC and the ellipse of foci $P$ and $Q$:
$$\frac{PA\cdot QA}{BA\cdot CA}+
\frac{PB\cdot QB}{CB\cdot AB}+
\frac{PC\cdot QC}{BC\cdot AC}=1$$
But in my proof I have also use the following formula:
\begin{align}
\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2} &=
\frac{1}{r^2}-\frac{1}{2rR} \\
IA^2+IB^2+IC^2 &= s^2+r^2+8rR \\
CA\cdot AB+BC\cdot AB+CA\cdot BC &= s^2+(4R+r)r \\
\frac{1}{CA\cdot AB}+\frac{1}{BC\cdot AB}+\frac{1}{CA\cdot BC} &=
\frac{1}{2rR}
\end{align}
So what's the new expression of:
\begin{align}
\frac{1}{BA\cdot CA}+\frac{1}{CB\cdot AB}+\frac{1}{BC\cdot AC} &= ? \\
\frac{1}{PA\cdot QA}+\frac{1}{PB\cdot QB}+\frac{1}{PC\cdot QC} &= ? \\
PA\cdot QA+PB\cdot QB+PC\cdot QC &= ? \\
BA\cdot CA+CB\cdot AB+BC\cdot AC &= ?
\end{align}
In function of the parameters of the inellipse and the triangle $ABC$ like the area and the side of the triangle or the semi major semi minor axes of the ellipse?
Edit: I have a good news
The centroid $M$ of the triangle $ABC$ correspond to the centre of the inellipse and we have the following relation for $P$ any interior point related to the triangle $ABC$:
$$PA^2+PB^2+PC^2=MA^2+MB^2+MC^2+3MP^2$$
Thanks a lot.
| A triangle $\triangle ABC$ with its centroid at the origin is the image of an origin-centered equilateral under a linear transformation. The transformation carries the equilateral's incircle to $\triangle ABC$'s Steiner inellipse. With an appropriate rotation, and by considering similar triangles (and/or ellipses) equivalent, we may assume that the transformation is a simple vertical scaling. This being so, ...
Consider the origin-centered equilateral $\triangle A_\star B_\star C_\star$ with vertex coordinates
$$A_\star := 2 r \operatorname{cis}\theta \qquad B_\star := 2 r\operatorname{cis}(\theta+120^\circ) \qquad C_\star := 2 r \operatorname{cis}(\theta-120^\circ)$$
where $\operatorname{cis}(\cdot) := (\cos(\cdot), \sin(\cdot))$. The (origin-centered) incircle of this triangle has radius $r$. For some non-negative $s<r$, apply the vertical-scaling transformation
$$(x,y) \to \left(\;x, \frac{ys}{r}\;\right)$$
chosen so as to turn the incircle of $\triangle A_\star B_\star C_\star$ into an origin-centered, axis-aligned ellipse with (horizontal) major radius $r$ and (vertical) minor radius $s$.
Let $A$, $B$, $C$ be the images of $A_\star$, $B_\star$, $C_\star$ under the transformation. One readily derives these relations:
$$\begin{align}
a^2 &= 12 \left( r^2 \sin^2\theta\phantom{\left(-60^\circ\right)}\; + s^2 \cos^2\theta \right) \\
b^2 &= 12 \left( r^2 \sin^2\left(\theta-60^\circ\right)+s^2 \cos^2\left(\theta-60^\circ\right) \right) \\
c^2 &= 12 \left( r^2 \sin^2\left(\theta+60^\circ\right)+s^2 \cos^2\left(\theta+60^\circ\right) \right) \\
\end{align}$$
Without too much trouble, one can derive
$$r^2 = \frac{1}{36}\left(a^2 + b^2 + c^2 + 2 t^2 \right) \qquad s^2 = \frac{1}{36}\left(a^2+b^2+c^2-2t^2\right)$$
where
$$t^4 = a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2$$
and also
$$
\tan 2\theta = \frac{\sqrt{3}(c^2-b^2)}{-2a^2+b^2+c^2} \quad\to\quad
\cos 2\theta = \frac{-2a^2+b^2+c^2}{2t^2}
$$
Now, the inellipse's foci are the points $P_{\pm} := (\pm\sqrt{r^2-s^2},0)$, and we have (after some manipulation)
$$\begin{align}
|\overline{AP_{\pm}}|^2 &= \frac{1}{9}\left(-a^2+2b^2+2c^2+t^2\mp 12 r t \cos\theta \right)
\end{align}$$
I haven't found a particularly-insightful way to write the right-hand side. However, the product of the "$+$" and "$-$" forms gives a convenient difference of squares that reduces quite nicely to this:
$$|\overline{AP_{+}}||\overline{AP_{-}}| = \frac{1}{3}bc \quad\text{thus, also}\quad |\overline{BP_{+}}||\overline{BP_{-}}| = \frac{1}{3}ca \quad |\overline{CP_{+}}||\overline{CP_{-}}| = \frac{1}{3}ab \tag{$\star$}$$
Consequently, it's no wonder that
$$\frac{|\overline{AP_{+}}||\overline{AP_{-}}|}{|\overline{AB}||\overline{AC}|} +
\frac{|\overline{BP_{+}}||\overline{BP_{-}}|}{|\overline{BC}||\overline{BA}|}+
\frac{|\overline{CP_{+}}||\overline{CP_{-}}|}{|\overline{CA}||\overline{CB}|} = 1$$
which is a somewhat weaker statement. In any case, $(\star)$ would seem to be the key to the relations you seek.
| {
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"source": "stackexchange",
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Why is $\int_0^{\pi/4} 5(1+\tan(x))^3\sec^2(x)\,dx$ equal to $18.75$ and not $3.75$? Why is $\int_0^{\pi/4} 5(1+\tan(x))^3\sec^2(x)\,dx$ equal to $18.75$ and not $3.75$?
I know the indefinite integral $\int 5(1+\tan(x))^3\sec^2(x)\,dx= \frac {5(1+\tan(x))^4} {4}+c$ by using $u$ substitution. Then shouldn't I evaluate that at $\frac{\pi}4$ minus that at $0$? Doing that gives $3.75$ but my texbook and wolfram alpha say the right answer is $18.75$
| Note that we have $\tan\left(\frac{\pi}{4} \right)=1$ and $\tan(0)=0$
Hence,
\begin{align}
\frac{5(1+\tan\left(\frac\pi4 \right))^4}{4}-\frac{5(1+0)^4}{4}&=\frac{5(2^4)}{4}-\frac{5}{4}\\&=\frac{5(15)}{4}\\&= \color{blue}5(3.75)\\&=18.75
\end{align}
A potential mistake is that you have forgotten to multiply by $5$. We have $\frac{15}4=3.75$ but I can't tell for sure unless more working is shown.
| {
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"answer_id": 0
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Proving that if $c^2 = a^2 + b^2$ then $c < a + b$ I am having some trouble in proving a conjecture that occurred to me some time ago, based on the Pythagorean theorem.
If, for a non-degenerate triangle, $$c^2 = a^2 + b^2$$
Then can the following be proven?
$$c < a + b$$
Is this statement always true?
| I guess your doubt is:
does a triple $(a,b,c)$ of positive numbers with $c^2=a^2+b^2$ also satisfy $c<a+b$, so that the three numbers are the sides of a triangle?
In particular, does a Pythagorean triple define a (right) triangle?
Indeed, the other two conditions, namely $a<b+c$ and $b<a+c$, are obviously satisfied because from $c^2=a^2+b^2$ we immediately get $c>a$ and $c>b$.
Also the third condition holds:
$$
c^2=a^2+b^2<a^2+2ab+b^2=(a+b)^2
$$
Note where the positivity assumption is applied.
More generally, suppose $c^2=a^2+b^2-2ab\tau$, where $-1<\tau<1$ (the cosine law should come to mind). Then
$$
c^2=a^2+2ab+b^2-2ab(1+\tau)=(a+b)^2-2ab(1+\tau)<(a+b)^2
$$
| {
"language": "en",
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"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours.
| Let us try modulo $\pmod3,\pmod8$ separately.
As $4\equiv1\pmod3,2\equiv-1\implies2^n\equiv(-1)^n$
$$3^n-4\cdot2^n+(-1)^n+6\equiv-(-1)^n+(-1)^n\equiv0\pmod3$$
For $n\ge1,8\mid4\cdot2^n$
$$3^n-4\cdot2^n+(-1)^n+6\equiv3^n+(-1)^n+6\pmod8$$
Now as $3^2\equiv1\pmod8,(3^2)^m\equiv1$
If $n$ is even, $$3^n+(-1)^n+6\equiv9^m+1+6\pmod8\equiv1+1+6$$
What if $n$ is odd $=2m+1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Invertible 4x4 matrix $$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
$$
Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, maybe the best idea is to use row operations and find an upper triangular of zeroes and then multiply the numbers on the diagonal to get the determinant. I have been doing some row operations and get this:
$$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
0 & -1 & -4 & 1 \\
0 & 0 & 2 & 6 \\
-1 & 0 & 0 & -12 \\
\end{pmatrix}
$$
I just need to get rid of the -1 on the last row. But I am stuck. Thank you for your assistance.
| Proof only by determinant
We have $$\begin{vmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{vmatrix}=A-B+C-D$$ where $$A=5\begin{vmatrix}
2 & 2 & 8 \\
6 & 2 & 8 \\
3 & 6 & 7 \\
\end{vmatrix}=5\left(2\begin{vmatrix}
2 & 8 \\
6 & 7 \\
\end{vmatrix}-2\begin{vmatrix}
6 & 8 \\
3 & 7 \\
\end{vmatrix}+8\begin{vmatrix}
6 & 2 \\
3 & 6 \\
\end{vmatrix}\right)=5(-68-36+240)=680,$$ $$B=6\begin{vmatrix}
2 & 2 & 8 \\
6 & 2 & 8 \\
2 & 6 & 7 \\
\end{vmatrix}=6\left(2\begin{vmatrix}
2 & 8 \\
6 & 7 \\
\end{vmatrix}-2\begin{vmatrix}
6 & 8 \\
2 & 7 \\
\end{vmatrix}+8\begin{vmatrix}
6 & 2 \\
2 & 6 \\
\end{vmatrix}\right)=6(-68-52+256)=816,$$ $$C=6\begin{vmatrix}
2 & 2 & 8 \\
6 & 6 & 8 \\
2 & 3 & 7 \\
\end{vmatrix}=6\left(2\begin{vmatrix}
6 & 8 \\
3 & 7 \\
\end{vmatrix}-2\begin{vmatrix}
6 & 8 \\
2 & 7 \\
\end{vmatrix}+8\begin{vmatrix}
6 & 6 \\
2 & 3 \\
\end{vmatrix}\right)=6(36-52+48)=192$$ and $$D=8\begin{vmatrix}
2 & 2 & 2 \\
6 & 6 & 2 \\
2 & 3 & 6 \\
\end{vmatrix}=8\left(2\begin{vmatrix}
6 & 2 \\
3 & 6 \\
\end{vmatrix}-2\begin{vmatrix}
6 & 2 \\
2 & 6 \\
\end{vmatrix}+2\begin{vmatrix}
6 & 6 \\
2 & 3 \\
\end{vmatrix}\right)=8(60-64+12)=64$$ Thus $$\begin{vmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{vmatrix}=680-816+192-64=-8\neq0$$ so your matrix has an inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574905",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$
Doing some manipulation we get:
$$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$
How can I prove this limit?
| HINT
$$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)=\frac{1}{n^2}\log\Big(1+\frac 1{n^2}\Big)^{n^2}+ \frac{2}{n^2}\log\Big(1+\frac 2{n^2}\Big)^\frac{n^2}{2}+...+ \frac{n}{n^2}\log\Big(1+\frac n{n^2}\Big)^\frac{n^2}{n}=\frac{1}{n^2}\sum_1^n k\log\left(1+\frac{k}{n^2}\right)^\frac{n^2}{k}\sim \frac{1}{n^2}\sum_1^n k=\frac{1}{n^2}\frac{n(n+1)}{2} \to \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
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"answer_count": 5,
"answer_id": 4
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Show that this polynomial has three different real roots There is polynomial $f(x)=x^3 +ax^2+bx+c$ and $ab=9c$ and $b<0$. Like in the title. I've only come up so far with $9x^3 + 9ax^2 + 9bx+ab=0$ and factoring here does nothing.
| Regarding the requirement that $a,c$ be real. Set $a = 3\mathrm{i}$, $b = -3$, $c = -\mathrm{i}$. Then $b < 0$, $ab = -9\mathrm{i} = 9c$, and $f(x) = x^3 + 3 \mathrm{i} x^2 - 3 x - \mathrm{i} = (x+\mathrm{i})^3$, so all three roots are the same and none are real. Consequently, for the problem to be correct, another contraint is required. That $a$, $b$, and $c$ must be real is a sufficient such constraint.
If $c = 0$, $ab=9c$ and $b<0$ force $a = 0$, so $f(x) = x^3 + bx = (x^2+b)x$, having the three real roots $\{-\sqrt{-b},0,\sqrt{-b}\}$, which are distinct.
From now on, we assume $c \neq 0$, so $0$ is not a root of $f$. First we address two cases where the roots are real but have duplicates, showing that if the roots are real, they are distinct. Then we show that having a conjugate pair of roots is impossible.
If the roots are real and not distinct, then either they are all the same or exactly two are the same.
Suppose (for purposes of contradiction) the roots are $u$, $u$, and $u$ (and $u \neq 0$). $f(x) = (x-u)^3= x^3 - 3u x^2 + 3u^2 x - u^3$, so \begin{align*}
a &= -3u \text{,} \\
b &= 3u^2 \text{, and} \\
c &= -u^3 \text{.}
\end{align*}
But this says $b > 0$. So the three roots cannot all be the same.
Now suppose (for purposes of contradiction) the roots are $u$, $u$, and $v$ (and $u \neq 0$, $v \neq 0$, and $v \neq u$). $f(x) = (x-u)^2(x-v) = x^3 -(2u+v)x^2 + (u^2 + 2uv)x - u^2 v$, so \begin{align*}
a &= -(2u+v) \text{,} \\
b &= u^2 + 2uv \text{, and} \\
c &= -u^2 v \text{.}
\end{align*}
From $ab=9c$, \begin{align*}
-(2u+v)(u^2 + 2uv) &= 9(-u^2 v) \\
-2u^3 -4 u^2 v - u^2 v - 2 uv^2 &= -9 u^2 v \\
-2u^3 - 2 uv^2 &= -4 u^2 v \\
-2u(u^2 - 2uv + v^2) &= 0 \\
-2u(u - v)^2 &= 0 \text{.}
\end{align*}
Since a product is zero only if one (or more) of the multiplicands is zero, either $-2u=0$ or $(u-v)^2 = 0$. The first says $u = 0$, a contradiction. The second says $v = u$, a contradiction. This shows not exactly two roots are the same.
Therefore, the if the roots are real, the three roots of $f$ are distinct.
The coefficients of $f$ are real. (Assumed additional constraint, see the aside before this solution.) Consequently, either all three roots are real, or one is real and two are a conjugate pair. Suppose two are a conjugate pair so the roots are $u$, $v + \mathrm{i}w$, $v - \mathrm{i}w$. By the above, we may take $u \neq 0$ and $w > 0$. Then
$f(x) = (x-u)(x-v - \mathrm{i}w)(x - v + \mathrm{i}w) = x^3 - (u+2v)x^2 + (2uv+v^2+w^2)x - (uv^2+uw^2)$, so \begin{align*}
a &= - (u+2v) \text{,} \\
b &= (2uv+v^2+w^2) \text{, and} \\
c &= - (uv^2+uw^2) \text{.}
\end{align*}
If $v = 0$, from $ab=9c$,$-(u)(w^2) = -9(uw^2)$, so $8 uw^2 = 0$ and either $u=0$ (contradiction) or $w = 0$ (contradiction). Therefore, $v \neq 0$.
From $ab=9c$, \begin{align*}
-(u+2v)(2uv+v^2+w^2) &= -9(uv^2+uw^2) \\
-2u^2v -u v^2 -u w^2 - 4uv^2-2v^3-2vw^2 &= -9uv^2 -9uw^2 \\
-2u^2v -2v^3-2vw^2 &= -4uv^2 -8uw^2 \\
(-2v)u^2 +(4v^2 +8w^2)u -2v(v^2+w^2) &= 0 \text{.}
\end{align*}
By the quadratic formula, $u = \frac{1}{v}\left( v^2 + 2w^2 \pm w\sqrt{3v^2+4w^2} \right)$. The only constraint we haven't used is $b < 0$, so we substitute these into the expression for $b$ and simplify to $b = 3v^2 + 5w^2 \pm 2w\sqrt{3v^2 + 4w^2}$. If we pick the positive root, each term is positive, so we must take the negative root and have $2w\sqrt{3v^2 + 4w^2} > 3v^2 + 5w^2$. Both sides are positive, so squaring only risks introducing negative solutions (of no interest to us), giving \begin{align*}
12v^2w^2 + 16 w^4 &> 9v^4 + 30 v^2w^2 + 25w^4 \\
0 &> 9v^4 + 18 v^2w^2 + 9w^4 \text{,}
\end{align*}
an impossibility. Therefore, it is not the case that $f$ has a conjugate pair of roots.
Therefore, there are three distinct roots, all of which are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578005",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the elliptic integral singular value.
It is also known or verified that
$\displaystyle K\left(\frac{1}{2}\right) =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.
Can one prove directly or analytically that
$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?
| We have
$$\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2\,\text{AGM}(a,b)}=\frac{\pi}{2\,\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)}\tag{1} $$
as a consequence of Lagrange's identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$.
On the other hand
$$ K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} = \frac{\pi}{2\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)}\tag{2} $$
and $$\text{AGM}\left(\sqrt{i},\sqrt{-i}\right)=\text{AGM}\left(1,\tfrac{1}{\sqrt{2}}\right) \tag{3}$$
so
$$K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} =\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+i)(x^2-i)}} = \int_{0}^{+\infty}\frac{dx}{\sqrt{x^4+1}}$$
as claimed. You can find an alternative, very short proof through Fourier-Legendre series expansions at page $29$ here.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?
| Since $|\sin t|\le |t|$ we have
$$|\sin(a)-\sin(b)|= 2\left|\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\right|\le |a-b|= \frac{1}{2n}-\frac{1}{2n+1}\sim \frac{1}{n^2}$$
with $a= \frac{1}{2n}$ and $b=\frac{1}{2n+1}$
Hence, the series is absolutely convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$z ≤ x + y$ implies $z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$
Suppose $x,y,z $ be nonnegative reals. Show that $z ≤ x + y\implies z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$.
My Proof: If $z=0$, then we are done. So, suppose $z>0$. Since $z ≤ x + y$, and $x$ and $y$ nonnegative, $z ≤ x + y+2xy+xyz$, which leads to $z(1+x+y+xy)\le (x+y+2xy)(1+z)$ and since $z\ne 0$, we get $z/(1 + z) ≤x/(1 + x) + y/(1 + y)$.
Is my proof correct?
| If
$z \le \sum_{i=1}^n x_i
$
then
we want to show that
$\dfrac{z}{1+z}
\le \sum_{i=1}^n \dfrac{x_i}{1+x_i}
$.
Let
$f(z)
=\dfrac{z}{1+z}
$.
We want to show that
$f(z)
\le \sum_{i=1}^n f(x_i)
$.
$f(z)
=\dfrac{z}{1+z}
=1-\dfrac{1}{1+z}
$
so
$f'(z)
=\dfrac1{(1+z)^2}
> 0
$.
$f$ is increasing so
$f(z)
\le f(\sum_{i=1}^n x_i)
$.
Therefore,
if we can show that
$f(\sum_{i=1}^n x_i)
\le \sum_{i=1}^n f(x_i)
$
we are done.
For $n=2$,
the difference is
$\begin{array}\\
f(x)+f(y)-f(x+y)
&=\dfrac{x}{1+x}+\dfrac{y}{1+y}-\dfrac{x+y}{1+x+y}\\
&=\dfrac{x}{1+x}+\dfrac{y}{1+y}-\dfrac{x}{1+x+y}-\dfrac{y}{1+x+y}\\
&=\dfrac{x}{1+x}-\dfrac{x}{1+x+y}+\dfrac{y}{1+y}-\dfrac{y}{1+x+y}\\
&=x(\dfrac{1}{1+x}-\dfrac{1}{1+x+y})+y(\dfrac{1}{1+y}-\dfrac{1}{1+x+y})\\
&=x\dfrac{(1+x+y)-(1+x)}{(1+x)(1+x+y)})+y\dfrac{(1+x+y)-(1+y)}{(1+y)(1+x+y)})\\
&=x\dfrac{y}{(1+x)(1+x+y)})+y\dfrac{x}{(1+y)(1+x+y)})\\
&=xy(\dfrac{1}{(1+x)(1+x+y)}+\dfrac{1}{(1+y)(1+x+y)})\\
&\ge 0\\
\end{array}
$
This easily generalizes.
$\begin{array}\\
\sum_{i=1}^n f(x_i)-f(\sum_{i=1}^n x_i)
&=\sum_{i=1}^n \dfrac{x_i}{1+x_i}-\dfrac{\sum_{i=1}^n x_i}{1+\sum_{i=1}^n x_i}\\
&=\sum_{i=1}^n \dfrac{x_i}{1+x_i}-\sum_{i=1}^n\dfrac{ x_i}{1+\sum_{j=1}^n x_j}\\
&=\sum_{i=1}^n (\dfrac{x_i}{1+x_i}-\dfrac{ x_i}{1+\sum_{j=1}^n x_j})\\
&=\sum_{i=1}^n x_i(\dfrac{1}{1+x_i}-\dfrac{ 1}{1+\sum_{j=1}^n x_j})\\
&=\sum_{i=1}^n x_i(\dfrac{(1+\sum_{j=1}^n x_j)-(1+x_i)}{(1+x_i)(1+\sum_{j=1}^n x_j)})\\
&=\sum_{i=1}^n x_i(\dfrac{\sum_{j=1,j\ne i}^n x_j}{(1+x_i)(1+\sum_{j=1}^n x_j)})\\
&\ge 0
\qquad\text{since each term is }\ge 0\\
\end{array}
$
and we are done.
Since
$f''(z)
=-\dfrac{2}{(1+z)^3}
< 0$,
by Jensen's inequality
we have
$f(\frac1{n}\sum x_i)
\ge \frac1{n}\sum f(x_i)
$
so that
$\sum f(x_i)
\le nf(\frac1{n}\sum x_i)
$.
| {
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"question_score": "2",
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Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we want to show all the solutions, we need to do something like $$\begin{cases}
\log x+C_1 & x>0\\
\log(-x)+C_2 & x<0
\end{cases}$$
Do we need to do change the constant every time there is a gap in the domain or is it just when the expression changes? For example, $$ \int \frac {x^5} {x^2-1} dx$$ which is $$ \frac {1} {2} \log |x^2-1| + \frac {x^4} {4} + \frac {x^2}{2} + C$$ and the domain is $$x \in \mathbb R \backslash \{-1,1\}$$ When we want to write all the solutions, is it something like $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x>1\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_3 & x<-1 \end{cases}$$ or is it that since the the first and last expression are the same they only have one constant associated? Meaning, the solutions are actually $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x \in \mathbb R \backslash [-1, 1]\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \end{cases}$$
| "Do we need to do change the constant every time there is a gap in the domain [...]?" The answer is Yes: the generic C stands for a locally constant function - a function constant on intervals. The values on disjoint intervals may be different.
| {
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"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Elementary Inequality from a local olympiad test The problem I'm stuck with is the following :
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$a+x \ge b+y \ge c+z \ge 0 $ and $ a+ b+c = x+y+z$
Prove that $ay+bx \ge ac+xz$
As easy as it sounds, I didn't achieve any significant progress after several tries.
Any advice, hint would be very appreciated. Thanks.
| Let us try to minimize
$$\varphi = a y + b x - ac - x z$$
We suppose that $a,b,c,x,y,z$ are given, satisfying the required conditions.
Without any computation one can decrease $\varphi$ by decreasing $b$ and $y$ by the same amount. Indeed, it doesn't change the condition
$a+b+c = x+y+z$ and it decreases $\varphi$ because $a+x\ge 0$. Hence we can suppose that $\boxed{b+y=c+z}$.
With the same argument, one can decrease $\varphi$ by decreasing $a$ and $x$ by the same amount, because $b+y\ge c+z$. We can do this until $\boxed{a+x=b+y}$.
Let $2 m:=a+x = b+y=c+z$. One has $6m = a+b+c+ x+y+z=2 (a+b+c)$
hence $\boxed{a+b+c = 3m}$ and $x + y+ z = 3m$. One has $\boxed{x = 2m-a}$, $\boxed{y = 2m-b}$, $\boxed{z = 2m-c}$
Replacing these values into $\varphi$, leads to
$$\boxed{\varphi = 2 (m-a)^2\ge 0}$$
Indeed,
$$ay+bx=a(2m-b)+b(2m-a)=2m(a+b) - 2ab=2m(3m-c)-2ab$$
and
$$-ac-xz = -ac-(2m-a)(2m-c) = -4m^2+2m(a+c)-2ac$$
hence
$$\varphi=2m^2+2ma-2a(b+c) = 2m^2+2ma-2a(3m-a)=2(m-a)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify this equation if $x$ is negative I am trying to simplify $7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}$, given that $x$ is negative. The answer is $-210x^3$, but I am getting $210x^3$. Below is my reasoning:
For a $ k > 0 $, let $ k = - x $. Then, $7\cdot\sqrt{5x}\cdot\sqrt{180x^5}$ = $7\cdot5\cdot6\cdot i\sqrt{k} \cdot i\sqrt{k^5}$ = $210\cdot(-1)\cdot k^3$ = $ 210 \cdot (-1) \cdot k^3 = 210\cdot (-1) \cdot (-x)^3 = 210x^3$.
What am I doing wrong?
Any help is greatly appreciated
| Since $k=-x$, we have $$7\cdot\sqrt{5x}\cdot\sqrt{180x^5}=7\cdot i\cdot\sqrt{5k}\cdot i\sqrt{180k^5}=-7\cdot\sqrt{900k^6}=-7\cdot\sqrt{900x^6}=-210x^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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