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How prove that $ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$? Let $ a,b,c>0$ be such that $ a+b+c=1$. How prove that
$ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$?
| There is no obvious symmetry in this inequality, so let's just try to simplify this with brute force. From $a + b + c = 1$:
*
*$1 - 3c = 3(1 - c) - 2 = 3(a + b) - 2$
*$1 - 4c = 4(1 - c) - 3 = 4(a + b) - 3$
*$c^2 = 1 + a^2 + b^2 -2a - 2b + 2ab$
After replacing these, our goal is now to show:
$3(a^3 + b^3 + a + b) \geq 8 - \frac{4 + 2ab}{a+b} \\
3(a+b)(a^3 + b^3 + a + b) + 2ab + 4 \geq 8(a+b)$
Let's not forget $a + b < 1$. Notice that if $a + b \leq \frac{1}{2}$, we have the result immediately.
Why?
Left hand side is obviously greater than $4 = \frac{8}{2} \geq 8(a+b)$, so from
now on we concentrate on $a + b \in (\frac{1}{2}, 1)$.
We can write $a^3 + b^3 = (a + b)^3 - 3ab(a+b)$, so let's substitute $u = a + b, v = ab$.
Now we have to prove:
$3u^2(u^2 - 3v + 1) + 2v + 4 - 8u \geq 0$
Further instructions: Use calculus. Partial differentiation shows that extrema of this function is located outside of the set we operate in, so any extrema is located on the border of our set.
Show that our function is greater than zero on this border, choose one point in the interior of our set and show that value of the function in it is greater than zero. Now you are done.
I'm curious: where did you find this inequality?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$ Question:
Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$
Attempt:
I have attempted several things with this problem. I will note what I feel is the most promising: note that $a(1-b)=\frac{1}{4}$ and $b(1-a)=\frac{1}{4}$ induces the following functions:
$$y=1-\frac{1}{4x} \ \ \ and \ \ y=\frac{1}{4(1-x)}$$
As a side note, these functions are inverses. I noticed that these functions intersect at $x=\frac{1}{2}$ and if we had chosen any value other than $\frac{1}{4}$ we would've ended up with more than one or no intersections. So I thought let $c$ be a real positive number and suppose $a(1-b),b(1-a)$ are both greater than $c$. Then the induced functions are: $$y=1-\frac{c}{x} \ \ \ and \ \ y=\frac{c}{1-x}.$$We have to show that the set $$\{x>0,y>0,y>1-\frac{c}{x},y>\frac{c}{1-x}\}$$ is empty when $c>\frac{1}{4}$. We note that the functions intersect at $$x=\frac{1}{2}(1\pm\sqrt{1-4c})$$ which is clearly not real when $c>\frac{1}{4}.$
| Suppose for some positive real numbers $a,b$ we have $a(1-b) > \dfrac{1}{4}$ and $b(1-a) > \dfrac{1}{4}$.
Then, $a(1-b) \cdot b(1-a) > \dfrac{1}{4} \cdot \dfrac{1}{4}$, which is equivalant to $a(1-a) \cdot b(1-b) > \dfrac{1}{4} \cdot \dfrac{1}{4}$.
Is this possible?
| {
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Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?
| Since $y(1)=1$ then by your answer we get
$$\frac A3+\frac B2+c+d=1\iff d=1-\frac A3-\frac B2-c$$
and then the solution is
$$y=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac A3-\frac B2-c\\=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proof based on Muirhead's inequality is pretty quick:
We multiply the left hand side with $\sqrt[3]{xyz} = 1$ and prove
$$x^{\frac{7}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{7}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{7}{3}} \leq x^3 + y^3 + z^3$$
with Muirhead ( $(3, 0, 0)$ majorizes $(\frac{7}{3}, \frac{1}{3}, \frac{1}{3})$).
I'm curious if there's a way to prove this without machinery of Muirhead's inequality and majorization. Also, this approach readily generalizes to proving
$$x^n + y^n + z^n \leq x^{n+1} + y^{n+1} + z^{n+1}$$
for non-negative $x, y, z$ such that $xyz = 1$.
Is there a way to prove this generalization without Muirhead?
| Another approach, not using the general power inequalities.
For convenience, we say $a=x^{\frac 13}$ and similar for $b$ and $c$.
Just apply the inequality arithmetic mean $>$ geometric mean in the following way (and cyclic permutations):
\begin{align}
\frac 79 a^9+\frac 19 b^9+\frac 19 c^9&=\frac{\underbrace{a^9+a^9+\cdots+a^9}_{\text{7 times}}+b^9+c^9}{9}\\
&\geq \sqrt[9]{\left(a^9\right)^7 b^9c^9}=a^7bc
\end{align}
Summing the cyclic permutations gives the required inequality.
In the general version, use the following: (where $a=x^{\frac 13}$ again)
\begin{align}
\frac{\underbrace{a^{3n+3}+a^{3n+3}+\cdots+a^{3n+3}}_{\text{3n+1 times}}+b^{3n+3}+c^{3n+3}}{3n+3}
&\geq \sqrt[b^{3n+3}]{\left(a^{3n+3}\right)^{3n+1} b^{3n+3}c^{3n+3}}=a^{3n+1}bc
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of the $11^\mathrm{th}$ power of the roots of the equation $x^5+5x+1=0$
Find the sum of the $11^\mathrm{th}$ power of the all roots of the equation
$$
x^5+5x+1=0
$$
My Attempt:
Let $R=\{\alpha,\beta,\gamma,\delta,\mu\}$ be the set of all roots of the equation ${x^5+5x+1=0}$, and let $x\in R$. Then we have
$$
\begin{align}
x^5 &= -\left(5x+1\right)\\
x^{10}&=25x^2+1+10x\\
\tag1 x^{11}&=25x^3+10x^2+x
\end{align}
$$
Taking the sum of $(1)$ on all elements $x\in R$ gives us
$$
\sum_{x\in R}x^{11}=25\cdot\sum_{x\in R}x^3+10\cdot\sum_{x\in R}x^2+\sum_{x\in R}x
$$
How can I solve the problem from this point?
| If $r$ is a root, then $r^5 = -5r - 1$, hence $r^{10} = (5r + 1)^2$ and $r^{11} = r(5r + 1)^2$. This greatly simplifies subsequent calculations. We can also note $r^3 = -5r^{-1} - r^{-2}$, so we get $$\begin{align*} r^{11} &= 25r^3 + 10r^2 + r \\ &= 25(-5r^{-1} - r^{-2}) + 10r^2 + r \\ &= r - 125r^{-1} + 10r^2 - 25r^{-2}\end{align*}.$$ So if $\{r_i\}_{i=1}^5$ are the five nonzero* complex roots of the given quintic $$f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5) = x^5 + 5x + 1,$$ then $\{r_i^{-1}\}_{i=1}^5$ are roots of the polynomial $$\begin{align*} f(1/x)x^5 &= x^5 \prod_{i=1}^5 (x^{-1} - r_i) = \prod_{i=1}^5 r_i (r_i^{-1} - x) \\ &= -r_1 r_2 r_3 r_4 r_5 \prod_{i=1}^5 (x - r_i^{-1}) \\ &= x^5 + 5x^4 + 1. \end{align*}$$ (* We know they are nonzero since $f(0) = 1$.)
Next, how do we find a polynomial whose roots are $\{r_i^2\}_{i=1}^5$? This is easily done by observing that $$\prod_{i=1}^5 (x - r_i^2) = \prod_{i=1}^5 (\sqrt{x} - r_i)(\sqrt{x} + r_i) = -f(\sqrt{x})f(-\sqrt{x}) = x^5 + 10x^3 + 25x - 1.$$ And a polynomial with roots $\{r_i^{-2}\}_{i=1}^5$ is, by a combination of the above reasoning, $$x^5 - 25x^4 - 10x^2 - 1.$$
Consequently, $$\begin{align*} \sum_{i=1}^5 r_i^{11} &= \sum_{i=1}^5 (r_i - 125r_i^{-1} + 10r^2 - 25r^{-2}) \\ &= 0 - 125 (-5) + 10(0) - 25(25) = 625 - 625 \\ &= 0. \end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of an unorthodox infinite series $ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $
This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.
| $$\begin{align}
\sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots
\\[1ex]
2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}}
\\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j}
\\ &= 1 + \sum_{j=1}^\infty \frac {2j+1}{2^j}
\\[1ex]
\sum_{k=1}^\infty \frac {2k-1}{2^k} & = 1 + \sum_{j=1}^\infty \frac {2j+1-(2j-1)}{2^j}
\\ & = 1 + \sum_{j=1}^\infty \frac{1}{2^{j-1}}
\\ & = \mathop{1 + \underbrace{\sum_{k=0}^\infty \frac{1}{2^{k}}}}_{\text{is this term familiar?}}
\\[2ex] & = 1 + 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots + \frac 1 {2^k} + \cdots
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.
| This is an alternate approach to derive the area of the circles
using Descartes four circle theorem.
WOLOG, assume $R = 1$. Let us call
*
*the outer circle (with radius $r_a = 1$) as $C_a$.
*the inner green circle (with radius $r_b = \frac12$) as $C_b$.
*the $1^{st}$ gray circle (the largest one at bottom) as $C_0$.
*the $2^{nd}$ gray circle (the one above $C_b$ and $C_0$) as $C_1$.
*in general, for any $k > 0$, we will call the gray circle above $C_b$ and $C_{k-1}$ as $C_k$.
*We can reflect the figure in question vertically. For $k < 0$, we will let $C_k$ be the mirror image of the circle $C_{-(1+k)}$.
The key is for any $k\in\mathbb{Z}$, the four circles $C_a, C_b, C_k$ and $C_{k\pm 1}$ are kissing each other. Let $r_k$ be the radius of $C_k$. We can apply Descartes four circle theorem to obtain:
$$\left(\frac{1}{r_{k\pm 1}} + \frac{1}{r_k} + \frac{1}{r_b} - \frac{1}{r_a} \right)^2
= 2 \left(\frac{1}{r_{k\pm 1}^2} + \frac{1}{r_k^2} + \frac{1}{r_b^2} + \frac{1}{r_a^2}\right)\tag{*1}$$
Since $r_a = 1$ and $r_b = \frac12$, this implies $\displaystyle\;\frac{1}{r_{k \pm 1}}\;$ are the two roots of quadratic equation
$$\left(\rho + \frac{1}{r_k} + 2 - 1 \right)^2
= 2 \left(\rho^2 + \frac{1}{r_k^2} + 2^2 + 1^2 \right)
$$
and hence
$$\frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} = 2\left(\frac{1}{r_k} + 1\right)
\quad\iff\quad
\frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} - \frac{2}{r_k} = 2$$
The RHS is an inhomogeneous linear recurrence relation on $\displaystyle\;\frac{1}{r_k}\;$
with constant term. Since the characteristic polynomial $(\lambda-1)^2$ has a double root at $1$, we know its solution must have the form $\displaystyle\;\frac{1}{r_k} = k^2 + \lambda k + \mu\;$. By symmetry,
$$r_k = r_{-(1+k)}\quad\implies\quad \lambda = 1 \quad\implies\quad
\frac{1}{r_k} = k(k+1) + \mu$$
To fix $\mu$, apply $(*1)$ to the case $C_a, C_b, C_0$ and $C_{-1}$, we get
$$\left(2\mu + 1\right)^2 = 2\left(2\mu^2 + 5\right)
\quad\implies\quad \mu = \frac{9}{4}
$$
From this, we get
$$r_k = \frac{4}{4k^2 + 4k + 9}
\quad\implies\quad
\text{Area} = \sum_{k=0}^\infty \frac{16\pi}{(4k^2+4k+9)^2}$$
reproducing what's in J.J's answer.
Update
To evaluate the sum, we will use following infinite product expansion of $\cosh(\pi x)$,
$$\cosh\pi x = \prod_{k=0}^\infty \left( 1 + \frac{x^2}{(k+\frac12)^2}\right)
$$
Taking logarithm, differentiate w.r.t $x$ and divide by $2x$, we get
$$\sum_{k=0}^\infty \frac{1}{(k+\frac12)^2 + x^2} = \frac{\pi}{2x} \tanh\pi x$$
Differentiate w.r.t $x$ and divide by $-2x$ once more, we get
$$\sum_{k=0}^\infty \frac{1}{((k+\frac12)^2 + x^2)^2}
= \frac{\pi}{4x^3}\tanh(\pi x) - \frac{\pi^2}{4x^2\cosh^2(\pi x)}
$$
With this, we find
$$\text{Area}
= \sum_{k=0}^\infty \frac{\pi}{((k+\frac12)^2 + 2)^2}
= \frac{\pi^2}{16}\left[\sqrt{2}\tanh(\pi\sqrt{2}) - \frac{2\pi}{\cosh^2(\pi\sqrt{2})}\right]
$$
Once again, this agrees with J.J's result.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
| Perhaps a more unorthodox approach: From the AM-GM inequality,
$$\sqrt[3]{\frac{1}{2}\sin^2x\cdot\frac{1}{2}\sin^2x\cdot\cos^2x} \le \frac{\frac{1}{2}\sin^2x\cdot + \frac{1}{2}\sin^2x\cdot + \cos^2x}{3}$$
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{\sin^2x + \cos^2x}{3}$$
But it is well known that $\sin^2x + \cos^2x = 1$. Hence,
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{1}{3}$$
$$\sin^4x\cos^2x\le\frac{4}{27}$$
Taking square roots of both sides,
$$-\frac{2}{\sqrt{27}} \le \sin^2x\cos x \le \frac{2}{\sqrt{27}}$$
Multiplying throughout by $2$,
$$-\frac{4}{\sqrt{27}} \le 2\sin^2x\cos x \le \frac{4}{\sqrt{27}}$$
But from the double angle formula, $2\sin^2x\cos x = \sin x \cdot2\sin x\cos x = \sin x \sin 2x$. Hence,
$$-\frac{4}{\sqrt{27}} \le \sin x \sin 2x \le \frac{4}{\sqrt{27}}$$
Equality for each bound occurs when $\frac{1}{2}\sin^2x = \cos^2x$, i.e. $\tan x = \pm \sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}+\dfrac{1}{x-z}$$
so
$$I=\dfrac{(x-z)^2}{16}+\dfrac{1}{x-z}=\dfrac{(x-z)^2}{16}+\dfrac{1}{2(x-z)}+\dfrac{1}{2(x-z)}\ge\dfrac{3}{4}$$
if and only if $(x-z)=2$,so $(x-y)(y-z)=8$
But we know $$(x-z)^2=[[(x-y)+(y-z)]^2\ge 4(x-y)(y-z)$$
so this is wrong,
Now I let $x-z=t$ it is clear $t\ge 4$,so
$$\dfrac{t^2}{16}+\frac{1}{t}=f(t)\Longrightarrow f'(t)\ge 0,t\ge 4$$
so
$$f(t)\ge f(4)=\dfrac{5}{4}$$
My Question: I fell my methods is ugly,I think this problem have other simple methods.Thank you
| You have reached
$$I = \frac{(x-z)^2}{16}+\frac1{x-z} = \frac{(x-z)^2}{16}+\frac4{x-z}+\frac4{x-z}-\frac7{x-z} \ge 3-\frac7{x-z} \ge \frac54$$
The last inequality is because $$x-z = \frac{16}{(x-y)(y-z)} \ge \frac{64}{(x-z)^2} \implies x-z \ge 4$$
P.S. Equality is when $x-y = y - z = 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$ How to find primitive of:
$$\frac{3x^4-1}{(x^4+x+1)^2}$$
I am having a faint idea of a type which may or maynot be in the primitve, i.e.:
$$\frac{p(x)}{x^4+x+1}$$
The problem is I am not getting an idea of a substitution to solve this problem.
I might show my work but it is totally useless, atleast in this case.
For reference:
| $\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\cdot \left(x^3+1+x^{-1}\right)^2}dx = \int\frac{(3x^2-x^{-2})}{(x^3+1+x^{-1})^2}dx$$
Now Let $$(x^3+1+x^{-1}) = t\;,$$ Then $$(3x^2-x^{-2})dx = dt$$
So $$\displaystyle I = \int\frac{1}{t^2}dt = -\frac{1}{t}+\mathbb{C} = -\frac{1}{x^3+1+x^{-1}}+\mathbb{C} = -\left(\frac{x}{x^4+x+1}\right)+\mathbb{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906320",
"timestamp": "2023-03-29T00:00:00",
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Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
| Your expression is
$$S = \sum_{k=1}^n k 10^k$$
We can pull out a factor of $10$ to get
$$S = 10 \sum_{k=1}^n k 10^{k-1}$$
Now, consider the function
$$f(x) = \sum_{k=1}^{n}kx^{k-1}$$
So $S = 10f(10)$. For $x \neq 1$, we can rewrite the sum as follows:
$$f(x) = \begin{align}
\frac{d}{dx}\sum_{k=1}^{n} x^k
&= \frac{d}{dx} \frac{x - x^{n+1}}{1-x}
\end{align}$$
where we have used the fact that $(x + x^2 + \cdots + x^n)(1-x) = x - x^{n+1}$ due to telescoping.
Therefore the desired sum is
$$S = 10f(10) = 10 \left.\left(\frac{d}{dx} \frac{x - x^{n+1}}{1-x}\right)\right|_{x=10}$$
I'll let you carry out the differentiation to finish the problem.
| {
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Simplify $x^3 - 4x^2 + 10x - 125$ I've been trying to simplify $x^3 - 4x^2 + 10x - 125$ for a while now, and I don't seem to progress.
I know that the factors of $125$ are $1$, $5$, $25$ and $125$, but none of these seem to help here. So far I have managed to get $x(x^2 - 4x + 10) - 125$.
Can I go any further than this? Thank you! By the way... it's my first post here, so if I didn't provide something necessary - excuse me.
| If you have access to any computer algebra system, you can solve for the roots of a cubic equation. (If you don't have access to a computer algebra system then writing everything out this fully is a waste of time!)
$$x^3 - 4x^2 + 10x - 125=(x-x_0)(x-x_1)(x-x_2)=\left(x-\frac{1}{3} 7^{2/3} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x-\frac{1}{3} 7^{2/3} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x+\frac{1}{3} \left(2\ 7^{2/3} \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}-\sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-4\right)\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Please help me to find a closed form for this integral:
$$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$
I suspect it might exist because there are similar integrals having closed forms:
$$\begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta(5)-\pi^2\zeta(3)\tag2\\
\int_0^1\frac{\ln^2(1+x)\ln x}x\mathrm dx&=\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)\tag3\\
\int_0^1\frac{\ln^3(1+x)\ln x}{x^2}\mathrm dx&=\frac34\zeta(3)-\frac{63}4\zeta(3)\ln2+\frac{23\pi^4}{120}\\&-\frac34\ln^42-2\ln^32+\frac{3\pi^2}4\ln^22-18\operatorname{Li}_4\!\left(\tfrac12\right).\tag4\end{align}$$
Thanks!
| An alternate form of the answers given by @Cleo and @Tunk-Fey as sum of $1$ and $1/2$ argumented polylogarithm-products with rational coefficients:
$$I = \frac{99}{16}\operatorname{Li}_5(1)-12\operatorname{Li}_5\left(\frac{1}{2}\right) + 15\operatorname{Li}_1\left( \frac{1}{2} \right)\operatorname{Li}_4(1) - 12\operatorname{Li}_1\left(\frac{1}{2}\right)\operatorname{Li}_4\left(\frac{1}{2}\right) - 15\operatorname{Li}_2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)-\frac{51}{4}\operatorname{Li}_1^2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)+12\operatorname{Li}_2(1)\operatorname{Li}_3\left( \frac{1}{2} \right) - \frac{2}{5}\operatorname{Li}_1^5\left(\frac{1}{2}\right),$$
where $\operatorname{Li}_n$ is the polylogarithm function, and specifically
$$\begin{align}
& \operatorname{Li}_5(1) \ \ \ = \zeta(5) \\
& \operatorname{Li}_5\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^5} \\
& \operatorname{Li}_4(1) \ \ \ = \zeta(4) = \frac{\pi^4}{90} \\
& \operatorname{Li}_4\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^4} \\
& \operatorname{Li}_3(1) \ \ \ = \zeta(3) \\
& \operatorname{Li}_3\left(\textstyle\frac{1}{2}\right) = \frac{7}{8} \zeta(3) - \frac{\pi^2}{12} \ln 2 + \frac{1}{6} \ln^3 2 \\
& \operatorname{Li}_2(1) \ \ \ = \zeta(2) = \frac{\pi^2}{6} \\
& \operatorname{Li}_2\left(\textstyle\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \ln^2 2 \\
& \operatorname{Li}_1\left(\textstyle\frac{1}{2}\right) = \ln2,
\end{align}$$
where $\zeta$ is the Riemann zeta function.
| {
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Prove $a^3+b^3+c^3\ge a^2+b^2+c^2$ if $ab+bc+ca\le 3abc$
if $a,b,c$ are positive real numbers and $ab+bc+ca\le 3abc$ Prove:$$a^3+b^3+c^3\ge a^2+b^2+c^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We are given $ab+bc+ca\le 3abc$ So$$2(ab+bc+ca)\le 6abc$$
so I can rewrite question as$$a^3+b^3+c^3+6abc \ge (a+b+c)^2$$
Also We can say$$9a^2b^2c^2 \ge (ab+bc+ca)^2 \ge 3abc(a+b+c)$$
My problem is is finding a way to use this facts and change the form of inequality to apply AM-GM.
| We only need to show that
$$(bc+ca+ab)(a^3+b^3+c^3)\ge3abc(a^2+b^2+c^2)$$
which is equivalent to
$$\sum_\text{cyc}a(b^4+c^4-b^3c-bc^3)\ge0$$
or
$$\sum_\text{cyc}a(b-c)^2(b^2+bc+c^2)\ge0$$
where $\sum_\text{cyc}$ is a summing operator $\sum_\text{cyc}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$
If $x$, $y$ and $z$ are positive numbers,prove:
$$(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We know that $$(x+y)(y+z)(z+x)\ge8xyz$$$$\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}\ge8xyz$$
So $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ is stronger inequality than $(x+y)(y+z)(z+x)\ge8xyz$. My Problem in Proving this inequality is that $\frac{1}{3}$ and $\sqrt[3]{}$ makes me think applying $3$ term AM-GM.but there is $8$ terms on Left hand side.Also I can't think of way to make $8$ on right hand side by $3$ term AM-GM.
By writing LHS in expanded form I was able to reach this$$LHS+xyz \ge 3\sum_{cyc}x\sqrt[3]{x^2y^2z^2}$$
| Because by AM-GM we obtain:
$$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\geq\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$
| {
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Finding the integral of $(x^2+4x)/\sqrt{x^2+2x+2}$ Can somebody explain me how to calculate this integral?
$$\int \frac{\left(x^2+4x\right)}{\sqrt{x^2+2x+2}}dx$$
| Let $x+1=\tan\theta$, so $x=\tan\theta-1$, $dx=\sec^{2}\theta d\theta$, and $\sqrt{x^2+2x+2}=\sqrt{(x+1)^2+1}=\sec\theta$.
Then $\displaystyle\int\frac{x^2+4x}{\sqrt{x^2+2x+2}}dx=\int\frac{(\tan\theta-1)^2+4(\tan\theta-1)}{\sec\theta}\sec^{2}\theta d\theta$
$\displaystyle=\int(\tan^{2}\theta+2\tan\theta-3)\sec\theta d\theta=\int(\sec^{2}\theta-1+2\tan\theta-3)\sec\theta d\theta$
$=\displaystyle\int(\sec^{3}\theta-4\sec\theta+2\sec\theta\tan\theta)\; d\theta$
$=\displaystyle\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)-4\ln|\sec\theta+\tan\theta|+2\sec\theta+C$
$=\displaystyle\frac{1}{2}(x+1)\sqrt{x^2+2x+2}-\frac{7}{2}\ln\left(\sqrt{x^2+2x+2}+x+1\right)+2\sqrt{x^2+2x+2}+C$.
| {
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Are there other methods to evaluate $\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$? Are there other methods to evaluate the following series?
$$\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$$
My attempt is as follows,
\begin{align}
\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\
\frac{1^{-4}+3^{-4}+\cdots+2^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\
\frac{y+2^{-4}(1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots)}{y} &=\frac{x}{y} \\
\frac{y+2^{-4}x}{y} &=\frac{x}{y} \\
\frac{x}{y} &= \frac{1}{1-2^{-4}}
\end{align}
| Your proof is fine.
The sigma notation can make some things cleaner. The below is the same argument, just written with different notation: $$x=\sum_{n=1}^\infty \frac{1}{n^4}\\y=\sum_{n=1}^\infty \frac{1}{(2n-1)^4}$$
Then $$x=\sum_{n=1}^{\infty} \left(\frac{1}{(2n-1)^4}+\frac{1}{(2n)^4}\right)=y+\frac{1}{2^4}x$$
and thus $$\frac{x}{y} = 1+\frac{1}{2^4}\frac{x}{y}\implies\\\frac{x}{y}=\frac{1}{1-\frac{1}{2^4}}$$
I suppose that the other approach is to note that every $n$ can be written uniquely as $2^k(2m+1)$. So:
$$\begin{align}x = \sum_{n=1}^\infty \frac{1}{n^4} &= \sum_{k=0}^\infty\sum_{m=1}^{\infty} \frac{1}{(2^k(2m+1))^4}\\
&=\left(\sum_{k=0}^\infty \frac{1}{2^{4k}}\right)\left(\sum_{m=0}^\infty \frac{1}{(2m+1)^4}\right)\\
&=\frac{1}{1-\frac{1}{2^4}}y
\end{align}$$
| {
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Proving Example 1.1.15 of secrets in inequalities
if $a,b,c,d$ are positive real numbers,Prove:$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2\ge\frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{9}{a^2+b^2+c^2}+\frac{16}{a^2+b^2+c^2+d^2}$$
I was reading the solution of it from book and something was not understandable for me.
I have problem in understanding when equality occurs in inequalities that I highlighted them.
for example the first one:$$a^2+c^2 \ge 2 ca$$ $$b^2+c^2\ge 2bc$$ $$a^2+b^2+2c^2 \ge 2(ac+bc)$$ $$4a^2+4b^2+8c^2\ge 8ac+ 8bc$$
but $8a^2+8b^2+8c^2 > 4a^2+4b^2+8c^2$ when $a,b,c$ are positive real numbers.So If i am wrong,where I made mistake?
Also about the third inequality I highlighted,I have problem at proving it.I would appreciate if someone helps me there.
| For the first:
*
*$\displaystyle \frac1{ac+bc}\ge\frac1{a^2+b^2+c^2}$ or $a^2+b^2+c^2\ge ac+bc$
*
*By AM_GM:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2}=ab,\frac{b^2+c^2}2\ge bc,\frac{c^2+a^2}2\ge ca$$
*Since $ab\ge0$, Adding all:
$$a^2+b^2+c^2\ge ab+bc+ca\ge ac+bc$$
For the second:
*
*$\displaystyle \frac{4}{b^2+c^2}\ge\frac1{a^2+b^2+c^2}$ or $4a^2+4b^2+4c^2\ge b^2+c^2$ or $4a^2+3b^2+3c^2\ge0$ which is true as $x^2\ge0,\forall x\in\mathbb R$
For the third:
*
*$\displaystyle \frac{18}{ab+bd+cd}\ge\frac{16}{a^2+b^2+c^2+d^2}$ or $\displaystyle a^2+b^2+c^2+d^2\ge\frac{8}{9}(ab+bd+cd)$
*
*Similiar to first we can prove $a^2+b^2+c^2+d^2\ge ab+bc+cd+da\ge ab+bd+cd\ge \frac{8}{9}(ab+bd+cd)$ since $\frac89<1$
| {
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Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or something like that..
Can someone explain to me simply how I would step by step factorize something like $4x^2 + 16x - 19$ ?
| If we multiply two monomials
$$(x-a)(x-b)=x^2-(a+b)x+ab$$
Think of this in terms of the arithmetic mean of the roots $\mu=(a+b)/2$ and the geometric mean $\gamma=\sqrt{ab}$:
$$x^2-2\mu+\gamma^2$$
Then
$$\begin{align}\mu^2-\gamma^2& =\left(\frac{a+b}{2}\right)^2-ab\\
& =\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}\\
& =\frac{a^2-2ab+b^2}{4}\\
& =\left(\frac{a-b}{2}\right)^2\\
& =\rho^2
\end{align}$$
Where $\rho=(a-b)/2$ is the "radius" from the mean $\mu$ to the roots $a,b$. We can recover the roots from
$$\{a,b\}=\mu\pm\rho=\mu\pm\sqrt{\mu^2-\gamma^2}$$
Taking your example, we see that $4x^2+16x−19=4(x^2+4x-19/4)=4(x^2-2(-2)x-19/4)$ so $\mu=-2$ and $\gamma=-19/4$, giving us the roots
$$\begin{align}
\{a,b\}& =-2\pm\sqrt{(-2)^2-(-19/4)}\\
& =-2\pm\sqrt{4+19/4}\\
& =-2\pm\sqrt{35/4}\\
& =-2\pm\sqrt{35}/2
\end{align}$$
If you need to, you can then rewrite the original expression as
$$\begin{align}
4x^2+16x−19& =4(x-(-2+\sqrt{35}/2))(x-(-2-\sqrt{35}/2))\\
\quad& =4(x+2-\sqrt{35}/2)(x+2+\sqrt{35}/2)\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a differential equation. Find a homogenous linear differential equation of order two of which $$y=e^x$$ and $$y=x+x^2$$ are two independent solutions. All I can think of is trial and error, how do I go backwards from these two equations to the DE?
| The equation is of the form $y''+ay'+by=0$.
Plugging $e^x$, $e^x+ae^x+be^x=0$, i.e. $1+a+b=0$.
Plugging $x+x^2$, $2+a(1+2x)+b(x+x^2)=0$.
Solving for $a$ and $b$,
$$a=\frac{-x^2-x+2}{x^2-x-1}\\b=\frac{2x-1}{x^2-x-1}.$$
Answer:
$$(x^2-x-1)y''+(-x^2-x+2)y'+(2x-1)y=0.$$
| {
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Selecting at least one ball of each color An urn contains five red, six white and seven blue balls. Five balls are selected without replacement. Find the probability that at least one ball of each color is selected.
Answer (attempt):
Getting red balls = $5 \choose 1$
Getting white balls = $6 \choose 1$
Getting blue balls = $7 \choose 1$
Remaining two balls = $15 \choose 2$
Total selections = $18 \choose 5$
P = $\frac{{5 \choose 1} {6 \choose 1}{7 \choose 1}{15 \choose 2}}{18 \choose 5}$
The answer seems to be wrong.
| The total number of combinations is:
$$\dbinom{5+6+7}{5}=\dfrac{18!}{5!\cdot13!}=8568$$
The number of combinations with no red balls is:
$$\dbinom{6+7}{5}=\dfrac{13!}{5!\cdot8!}=1287$$
The number of combinations with no white balls is:
$$\dbinom{5+7}{5}=\dfrac{12!}{5!\cdot7!}=792$$
The number of combinations with no blue balls is:
$$\dbinom{5+6}{5}=\dfrac{11!}{5!\cdot6!}=462$$
The number of combinations with no red balls and no white balls is:
$$\dbinom{7}{5}=\dfrac{7!}{5!\cdot2!}=21$$
The number of combinations with no red balls and no blue balls is:
$$\dbinom{6}{5}=\dfrac{6!}{5!\cdot1!}=6$$
The number of combinations with no white balls and no blue balls is:
$$\dbinom{5}{5}=\dfrac{5!}{5!\cdot0!}=1$$
So the probability of a combination without at least one ball of each color is:
$$\dfrac{1287+792+462-21-6-1}{8568}=\dfrac{2513}{8568}$$
And the probability of a combination with at least one ball of each color is:
$$1-\dfrac{2513}{8568}=\dfrac{6055}{8568}\approx0.7067$$
| {
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How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a general result of the form
$\frac{1}{n} = \frac{1}{a} + \frac{1}{b}$
and hence prove that any unit fraction can be expressed as the sum of two other distinct unit
fractions.
| $\frac{1}{n}$ - $\frac{1}{n + k}$ = $\frac{k}{n(n+k)}$
$\frac{k}{n(n+k)}$ = $\frac{1}{\frac{n(n+k)}{k}}$
So, as long as $\frac{n(n+k)}{k}$ is a whole number, we can say that $\frac{1}{\frac{n(n+k)}{k}}$ is a unit fraction and since
$\frac{1}{n}$ = $\frac{k}{n(n+k)}$ + $\frac{1}{n + k}$
$\frac{1}{n}$ can be expressed as a sum of two unit fractions as long as k has some value such that $\frac{n(n+k)}{k}$ is a whole number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Divergence of $u_{n+1}=1+\frac{n}{u_n}$ Let $u_n$ be defined by $u_0=1$ and $u_{n+1}=1+\frac{n}{u_n}$.
It can be shown easily that if it has a limit, then it must be $+\infty$.
Does $u_n$ diverge to $+\infty$ ?
What I have tried :
Let $x=u_n$
$U_{n+1}-u_n=1+\frac{n}x-x=f(x)$
$f'(x)=\frac{-n^2}{x^2}-1$
$f$ is sctrictly decreasing, $f\left(\dfrac{-1+\sqrt{1+4n}}2\right)=0$ which is its only $0$
But that diverges towards $\infty$ so I haven't been able to deduce anything from it.
| The first terms of the sequence are $u_0 = 1, u_1 = 1, u_2=2, u_3=2, \dots$
I prove by induction that
$$\forall n \geq 2,\qquad \ \sqrt{n} < u_n < \sqrt{n}+1.$$
From this follows that $\lim_n u_n = + \infty$.
For $n=2$ we are ok because $\sqrt{2} <2 < \sqrt{2} +1$.
For the inductive step we have
$$u_{n+1} = 1+ \frac{n}{u_n} < 1+ \frac{n}{\sqrt{n}} < 1+ \sqrt{n+1}$$
and
$$ u_{n+1} = 1+ \frac{n}{u_n} > 1+ \frac{n}{1+ \sqrt{n}} > \sqrt{n+1}$$
Where the last inequality holds because $\forall x >0$
$$ 1+ \frac{x}{1+ \sqrt{x}} - \sqrt{x+1} >0$$
holds.
| {
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How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seems to make it worse. Any help would be appreciated.
| A common trick: multiply by $1$.
That is: multiply $x(\sqrt{x^2+a^2}-\sqrt{x^2+b^2})$ by $\frac{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$
| {
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How to evaluate the integral $e^{-(c\ln(\frac{1}{x}))^s} dx$? Can anyone help me evaluate
$$\int_{\alpha}^1 \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} dx$$,
Where $0 \leq \alpha \leq 1$ and $s \in \mathbb{R}$.
I tried changing variables, integration by parts etc. and got nowhere. Any clues on how to handle this would be appreciated.
I tried integration by parts with $dv=1$ and $u = \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}}$. This gives us:
$$x \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} {\Huge|}_{\alpha}^1 - s c\int_{\alpha}^1 \left(c\ln\left(\frac{1}{x}\right)\right)^{s-1}\exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} dx .$$
Note that the second term reminds the incomplete Gamma function (with $ln(1/x)$ instead of $t$), however, i couldn't reach further than this.
| Consider:
$$
\begin{align*}
e^{-(c\ln\frac{1}{x})^s} &= e^{-(\ln(\frac{1}{x})^c)^s} \\
&= e^{-(\ln(\frac{1}{x^c}))^s} \\
&= \frac{1}{e^{(\ln(\frac{1}{x^c}))^s}} \\
&= \frac{1}{e^{\ln(\frac{1}{x^c})} \cdot e^{\ln(\frac{1}{x^c})} \ldots \text{s times} } \\
&= \frac{1}{\frac{1}{x^c} \cdot \frac{1}{x^c} \ldots \text{s times}} \,\,\,\,\, \text{[exponent and log cancel out] } \\
&= \frac{1}{(\frac{1}{x^c})^s} \\
&= \frac{1}{(\frac{1}{x^{cs}})} \\
&= x^{cs}
\end{align*}
$$
Now, your integral gets reduced to:
$$
\begin{align*}
\int_{\alpha}^{1}{x^{cs}}dx &= \left[\int{x^{cs}}dx\right]_{\alpha}^{1} \\
&= \left[\frac{x^{cs + 1}}{cs + 1} + C\right]_{\alpha}^{1} \\
&= \left[\frac{1^{cs + 1}}{cs + 1} + C - \frac{\alpha^{cs + 1}}{cs + 1} - C \right] \\
&= \left[\frac{1^{cs + 1}}{cs + 1} - \frac{\alpha^{cs + 1}}{cs + 1}\right] \\
&= \left[\frac{1^{cs + 1} - \alpha^{cs + 1}}{cs + 1}\right] \\
\end{align*}
$$
Since, $ s \in \mathbb{R}$ but you haven't given any information about $c$, we can simplify it up to:
$$
\begin{align*}
\int_{\alpha}^{1}{x^{cs}}dx &= \frac{1^{c} - \alpha^{cs + 1}}{cs + 1} \\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925394",
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Find the basis of the Ker(T) and Im(T) for $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$ For the linear application, $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$, find
*
*A matrix representation with respect to a basis of your choice
*A basis of Ker(T)
*A basis of Im(T)
What I have done so far...
Matrix representation: Taking a closer look at the map and the matrix operations, we can easily see that the matrix representation will be $\ldots$
Corrected Matrix:
$\begin{pmatrix} 0 & 0 & 0 & d \\ 0 & 0 & c & 0 \\ 0 & b & 0 & 0 \\ a & 0 & 0 & 0 \end{pmatrix} $
Because
$\begin{pmatrix} d & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & a \end{pmatrix} \begin{pmatrix} x^3 \\ x^2 \\ x \\ 1 \end{pmatrix} =T(ax^3+bx^2+cx+d)$
Basis of Ker(T): You have to find the set of all $ax^3+bx^2+cx+d \in P_3$ such that $$T(ax^3+bx^2+cx+d)=(0,0,0,0)^T$$
We consider
$\begin{pmatrix} d & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & a \end{pmatrix} \begin{pmatrix} x^3 \\ x^2 \\ x \\ 1 \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$
Which will result in the following equations:
\begin{cases}
dx^3=0 \\
cx^2=0 \\
bx=0 \\
a=0 \\
\end{cases}
Is this correct? I do not know how to proceed. I know we have to find $ax^3+bx^2+cx+d=0$ in $P_3$ such that T(ax^3+bx^2+cx+d)=0.
Basis of Im(T): If we just want to have a set of vectors which span the image, then it would be enough to take the column-vectors. But we want to have linear-independent vectors, which span our image.
What we have to do now: Consider $T^{T}$ matrix of T and apply the Gaussian-elimination-algorithm on it. The columns which you will obtain then form a basis of the image.
Need help
So the basis of the Im(T) is the vector columns of the matrix.
| You can try this way: choose basis $\displaystyle \{1,x,x^2,x^3\}$ of $P_3$. Then check:
$$T[x^3]=1$$
$$T[x^2]=x$$
$$T[x]=x^2$$
$$T[1]=x^3$$
So first element of basis is transofmed to fourth by $T$, second to third, third to second, fourth to first. So the matrix representation is:
$$T=\begin{bmatrix}0 && 0 && 0 && 1 \\ 0 && 0 && 1 && 0 \\ 0 && 1 && 0 && 0 \\ 1 && 0 && 0 && 0\end{bmatrix}$$.
Now you can check that $\det T \neq 0$, so Ker(T)=$\{0\}$, so basis of Ker(T) is $\{0\}$, Im(T)=$P_3$, so basis Im(T) is $P_3$, for example, $\displaystyle \{1,x,x^2,x^3\}$.
| {
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Let$\ n$ be an odd abundant number. What is the best upper bound to $\ \frac{\sigma(n)}{n}$? NOTE: Of course the bound of the product of primes below is incorrect. I knew very little of this stuff and it seems was too lazy and gaga (I still am gaga) to check that W|A was acting stupid.
In order to obtain my upper bound I use and extend a method I learnt from Greg Martin, who answered another question of mine.
Lemma: Let $i>1$ be an integer, and let $p$ be the largest prime dividing $i$. Then,
$$
\frac{\sigma(i)}i \le \bigg(1+\frac1p\bigg) \frac{\sigma(i/p)}{i/p}.
$$
Proof: Write $i=p^rm$ where $p\nmid m$. Then,
$$
\frac{\sigma(i)/i}{\sigma(i/p)/(i/p)} = \frac{\sigma(p^rm)}{p\sigma(p^{r-1}m)} = \frac{\sigma(p^r)\sigma(m)}{p\sigma(p^{r-1})\sigma(m)} = \frac{\sigma(p^r)}{p\sigma(p^{r-1})}.
$$
Since $\sigma(p^r) = \frac{p^{r+1}-1}{p-1}$, this gives
$$
\frac{\sigma(i)/i}{\sigma(i/p)/(i/p)} = \frac{p^{r+1}-1}{p(p^r-1)} \le \frac{p+1}p = 1+\frac1p.
$$
Dickson proved there are no odd abundant numbers with less than three prime factors, hence, if$\ P $ is the largest prime dividing$\ n$, we have$\ P\ge7$. So, let$\ n=k\times PA$, where$\ PA$ is any odd primitive abundant number and$\ k$ is an odd integer of the form$\ 3^a 5^b 7^c 11^d...$, where obviously all the exponents are non-negative. We shall use$\ 945$ as$\ PA$ simply because it is the first odd (and primitive) abundant number, but we'll show hereinafter that any$\ PA$ would do fine. Thus, we have $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(\frac{945k}{7}\right)}{\frac{945k}{7}}.$$ For a start, suppose that in the above factorization of$\ k$, the only positive exponent is$\ a$. Hence,$$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(5\times 3^{a+3}\right)}{5\times 3^{a+3}}=\frac{8\times 6\left(3^{a+4}-1\right)}{7\times 5\times 2\times 3^{a+3}}<\frac{72}{35}.$$ If only$\ a$ and$\ b$ are positive, $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(3^{a+3}5^{b+1}\right)}{3^{a+3}5^{b+1}}=\frac{8\left(3^{a+4}-1\right)\left(5^{b+2}-1\right)}{7\times 2\times 4\times 3^{a+3}5^{b+1}}<\frac{15}{7}.$$ If only$\ a, b$ and$\ c$ are positive, $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(3^{a+3}5^{b+1}7^c\right)}{3^{a+3}5^{b+1}7^c}=\frac{8\left(3^{a+4}-1\right)\left(5^{b+2}-1\right)\left(7^{c+1}-1\right)}{2\times 4\times 6\times 3^{a+3}5^{b+1}7^{c+1}}<\frac{5}{2}.$$ Having reached$\ P$, it is easy to see that from now on as the factorization gets larger by$\ p_n^\alpha$, we just multiply the last ratio by $\ \frac{\left(p_n^{\alpha+1}-1\right)}{p_n^\alpha(p_n-1)}<\frac{p_n}{p_n-1}$, leaving out, aside$\ 2$, only$\ 7$, and this is true for any $\ PA$.
As a consequence, $$\ \frac{\sigma\left(\frac{n}{7}\right)}{\frac{n}{7}}<\prod_{j=2}^\infty \frac{p_j}{p_j-1}<\frac{11.0454}{2}=5.5227.$$(if it necessary, for the last result see http://www.wolframalpha.com/input/?i=2*+3%2F2+*+5%2F4+*+7%2F6+*+... and click on >More terms in the approximation section) Therefore, we can conclude $$\ \frac{\sigma(n)}{n}<\frac{8}{7} 5.5227=6.3116\overline {571428}.$$
| Ummm...$\frac{\sigma(n)}{n}$ is unbounded above for odd $n$.
Lemma: For all $n\in\mathbb{N}$, we have $\frac{\sigma(n)}{n}=\sum_{d\mid n} \frac{1}{d}$.
Proof: Let $d_1<d_2<\dots<d_t$ be the divisors of $n$. Then $\frac{\sigma(n)}{n}=\frac{d_1}{n}+\cdots+\frac{d_t}{n}=\frac{1}{d_t}+\cdots+\frac{1}{d_1}$. QED.
Now if we choose an odd $n$ that is divisible by $3, 5, 7, \dots, 2k+1$, then $\frac{\sigma(n)}{n}\ge \frac{1}{3}+\frac{1}{5}+\frac17+\cdots+\frac{1}{2k+1}$. This is a partial sum of the odd harmonic series, which is divergent. So by choosing $k$ large enough, we can make $\frac{\sigma(n)}{n}$ as large as we please.
| {
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on a recursive sequence (exercise 8.14 Apostol). The exercise asks to prove convergence and find the limit of the sequence:$$a_n= \frac{b_{n+1}}{b_n},\text{ where } b_1=b_2 =1, b_{n+2} = b_{n} + b_{n+1}. $$
It also gives a hint: Show that $ \ b_{n+2}b_n - b_{n+1}^2 = (-1)^{n+1}$ by induction.
I am having problems proving $ \ b_{n+2}b_n - b_{n+1}^2 = (-1)^{n+1}$ by induction, could I have a helping hand?
| $$b_{n+3}b_{n+1}-b_{n+2}^2= (b_{n+2}+b_{n+1})b_{n+1}-b_{n+2}^2 = (b_{n+1}-b_{n+2})b_{n+2} + b^2_{n+1} = (b_{n+1}-b_{n+2})(b_{n+1} + b_n) + b^2_{n+1}= 2b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n -b_{n+2}b_n = (b^2_{n+1}-b_{n+2}b_n)+b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n$$
Now note that $$b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n = b_{n+1}(b_{n+1}+b_n-b_{n+2}) = 0$$
and therefore you get that $$b_{n+3}b_{n+1}-b_{n+2}^2= (b^2_{n+1}-b_{n+2}b_n)+b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n = -(b_{n+2}b_n-b^2_{n+1}) = -(-1)^{n+1} = (-1)^{n+2}$$
| {
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How to simplify this complex integral? How to approximate this integral as a function of a and b?
$$\int_0^\pi\int_0^{2\pi}\sqrt{(a-b\sin\varphi\cos\theta)^2+(b\cos\varphi)^2+(b\sin\varphi\sin\theta)^2}d\theta d\varphi$$
where a and b are two variables.
| As was mentioned: the integrand simplifies to: $\sqrt{a^2 + b^2 - 2 a b \cos\theta \sin\phi}$
Factor out $\sqrt{a^2 +b^2}$ and use the binomial theorem to expand, treating $\cos\theta \sin \phi$ as small. This gives:
$$\sqrt{a^2 +b^2} \left(1-\frac{a b \cos (\theta ) \sin (\phi )}{a^2+b^2}-\frac{a^2 b^2 \cos ^2(\theta ) \sin ^2(\phi )}{2 \left(a^2+b^2\right)^2}-\frac{a^3 b^3 \cos ^3(\theta ) \sin ^3(\phi )}{2 \left(a^2+b^2\right)^3}-\cdots\right)$$
Now these functions can be integrated directly. (The odd powers of $\cos$ integrate to $0$ over the range here.)
For any $a,b$ you get some power series in $\left(\frac{a b}{a^2+b^2}\right)$
vis: $$\frac{\pi ^2 \sqrt{a^2+b^2}\left(2097152-\frac{262144 a^2 b^2}{\left(a^2+b^2\right)^2}-\frac{184320 a^4 b^4}{\left(a^2+b^2\right)^4}-\frac{268800 a^6 b^6}{\left(a^2+b^2\right)^6}-\frac{525525 a^8 b^8}{\left(a^2+b^2\right)^8}\right)}{1048576}$$
| {
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PEMDAS question: $F(x) = 3x^2 - x+2$. Find $[f(a)]^2$ How should I go about doing this? $(3a^2-a+2)^2$?
Thus, $9a^4-a^2+4$
| $(3a^2-a+2)^2$ is correct but your expansion of this is incorrect. Remember that:$$(3a^2-a+2)^2=(3a^2-a+2)(3a^2-a+2)=3a^2(3a^2-a+2)-a(3a^2-a+2)+2(3a^2-a+2)$$
You might find this useful: expanding or removing brackets
Here is an example to illustrate the point:$$(3-1+2)^2=(4)^2=16$$If we did this using your result, we would get:$$(3-1+2)^2=3^2-1^2+2^2=9-1+4=12\ne16$$However:$$(3-1+2)^2=(3-1+2)(3-1+2)$$$$=3(3-1+2)-1(3-1+2)+2(3-1+2)$$$$=3(4)-1(4)+2(4)=12-4+8=16$$
More generally:$$(a+b+c)d=ad+bd+cd$$Now image $d=(e+f+g)$, then we would get:$$(a+b+c)d=(a+b+c)(e+f+g)$$$$=ad+bd+cd$$$$=a(e+f+g)+b(e+f+g)+c(e+f+g)$$
| {
"language": "en",
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"source": "stackexchange",
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Number theory problem from 11th Iberoamerican olympiads Given a number $n \in \mathbb{N}$, such that $n>1$, let us consider all the fractions of the form $1 \over{ab}$, where $a$ and $b$ are coprime natural numbers such that $0<a<b \leq n$ and $a+b>n$. Prove that the sum of all such fractions is equal to $1 \over{2}$.
| Induction on $n$. Let $S_n$ be your sum. The base case is easy:
$$S_2=\frac{1}{1\cdot 2}$$
In general,
$$S_{n+1} = S_n-
\left(
\sum
_{\stackrel{a=1}{(a,n+1-a)=1}}
^{\left\lfloor \frac n2 \right\rfloor}
\frac{1}{a(n+1-a)}\right)
+ \left(
\sum_{\stackrel{a=1}{(a,n+1)=1}}^{n}\frac{1}{a(n+1)}
\right)$$
That is, we remove the pairs $a<b$ with $a+b=n+1$, since they no longer belong in the sum, and we add in the pairs $a,b$ with $b=n+1$.
Now $$\frac{1}{a(n+1-a)} = \frac{1}{n+1}\left(\frac{1}{a} + \frac{1}{n+1-a}\right)\tag{1}$$
From that, it is not hard to show that:
$$\sum_{\stackrel{a=1}{(a,n+1-a)=1}}^{\left\lfloor\frac{n}2\right\rfloor} \frac{1}{a(n+1-a)} = \sum_{\stackrel{a=1}{(a,n+1)=1}}^{n}\frac{1}{a(n+1)}$$
Essentially, $(a,n+1-a)=1$ if and only if $(a,n+1)=1$, and then use $(1)$ on the left hand side.
I'd love to see a more geometric proof of this, or one using probability.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Finding determinant of a 3x3 matrix Assuming y is a nonzero real number, I need to find the determinant of this matrix:
$$ \left[
\begin{array}{cc}
1 & y & y^2 \\
y & y^2 & y^3 \\
y^2 & y^3 & y^4
\end{array}
\right] $$
Can anyone help me get started? I know the answer is 0 from using Mathematica. Thanks
Edit
If I pulled out y from a row and had
$$y*D \left[
\begin{array}{cc}
1 & y & y^2 \\
1 & y & y^2 \\
y^2 & y^3 & y^4
\end{array}
\right] $$
and $$ \left[
\begin{array}{cc}
1 & y & y^2 \\
1 & y & y^2 \\
y^2 & y^3 & y^4
\end{array}
\right] = 0$$
therefore,
$$ \left[
\begin{array}{cc}
1 & y & y^2 \\
y & y^2 & y^3 \\
y^2 & y^3 & y^4
\end{array}
\right] = 0$$
is this correct?
| Hint: Pick two rows (or two columns). Are they linearly independent?
| {
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Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|4m^2-8m-16$ or $d|4m^2+5m+4-(4m^2-8m-16)=13m+20$ but $d|15m+9$ thus $d|15m+9-(13m+20)=2m-11$ but $d|13m+20-(5m+3)=8m+17$ and $d|8m-88$ therefore $d|8m+17-(8m-88)=105=5.3.7$. So $4m^2+1=105k$ or $4m^2=104+105k'$ but $k'=4k''$ then $m^2=26+105k''$ or $4m^2+1=21k$ or $4m^2+1=35k$ or $4m^2+1=15k$ or $4m^2+1=5k$ or 3k or 7k. Are there solutions to these equations, please, or can I conclude that the two numbers are coprimes ?
| you could also follow this method :-
let $4m^2+1 $ and $4m^2+5m+4 $ are coprimes then $GCD(4m^2+1,4m^2+5m+4)=1$
which means there exist x,y such that ${\color{Magenta} x}(4m^2+1)+{\color{Magenta} y }(4m^2+5m+4)={\color{Red} 1}$ , so lets try to find x,y that might makes it true.
by division algorithm(assume $m=2s$ or $2s+1$ ) , $4m^2+1$ could be of the form $16n+1$ or $8n+1$ then $16n+1\times 8n+1$ gives the form of ${\color{Red} {24k+1}}$ and $4m^2+5m+4$ could be of the form $2n$ or $4n+1$ then $2n \times 4n+1$ gives the form of ${\color{Green} {2k}}$
now we need to check if :-
$GCD({\color{Red} {24k+1} }, {\color{Green} {2k}})=1$
so we need x,y such that ${\color{Magenta} x}({\color{Red} {24k+1} })+{\color{Magenta} y} {\color{Green} {2k}}=1$
let x=1 and y=-12 then
${\color{Magenta} 1}({\color{Red} {24k+1} })+{\color{Magenta} {-12}} {\color{Green} {(2k)}}=24k+1-24k=1$
| {
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Solve the PDE $xu_x-2yu_y+u=e^x,$ with the side condition $u(1,y)=y^2$ 1b. $xu_x-2yu_y+u=e^x,$ side condition $u(1,y)=y^2$
My attempt: This has been a super endurance and I hope I got the whole thing right. So anyway, here it goes ...oh and one more thing... can someone please show me how to solve the side condition step by step please. I know I'm kind of slow at the side condition part, but I really want to fully understand it...so far I can go up into have the $F(w)$ part isolated and then it has something to do with a dummy variable and afterwards substitute back, but where exactly? It would be greatly appreciated :).
$\frac{dy}{dx}=\frac{2y}{x}$
$\frac{-1dy}{2y}=\frac{1dx}{x}$
$\frac{-1}{2} \ln y = \ln x +C$
$\ln y^{\frac{-1}{2}} = \ln x +C$
$e^{\ln y^{\frac{-1}{2}}} = e^{\ln x +C}$
$y^{\frac{-1}{2}}=xe^C$
$\frac{y^{\frac{-1}{2}}}{x}=e^C$
$y^{\frac{-1}{2}}x^{-1}=e^C$
$y^{\frac{-1}{2}}x^{-1} = w \rightarrow x = \frac{w}{\sqrt{z}}$
$ z = y \rightarrow y =z$
$ W_x = -x^{-2}y^{\frac{-1}{2}}$
$W_y = \frac{-1}{2}x^{-1}y^{\frac{-3}{2}}$
$Z_y = 1$
$Z_x = 0$
$x[V_wW_x+V_zZ_x]-2y[V_wW_y+V_zZ_y]+v=e^{\frac{1}{w\sqrt{z}}}$
$-2yV_z+v=e^{\frac{1}{w\sqrt{z}}}$
$V_z+\frac{1}{-2y}v=\frac{1}{-2y}e^{\frac{1}{w\sqrt{z}}}$
$v(a) = e^{\int \frac{-1}{2y}} \rightarrow e^{-\frac{1}{2} \ln y} \rightarrow y^{\frac{-1}{2}}$
$y^{\frac{-1}{2}}V_z+y^{\frac{-1}{2}}\frac{1}{-2y}v=y^{\frac{-1}{2}}\frac{1}{-2y}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}V_z+y^{\frac{-3}{2}}\frac{1}{-2}v=y^{\frac{-3}{2}}\frac{1}{-2}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}V_z-y^{\frac{-3}{2}}\frac{1}{2}v=-y^{\frac{3}{2}}\frac{1}{2}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}v=\int \frac{-1}{2} y^{\frac{-3}{2}}e^{w^{-1}}e^{z^\frac{-1}{2}}$
$y^{\frac{-1}{2}}v=- y^{\frac{-3}{2}}e^{w^{-1}}e^{z^\frac{1}{2}} +F(w)$
$v=-y^{\frac{-1}{2}}e^{w^{-1}}e^{z^\frac{1}{2}}+F(w)y^{\frac{1}{2}}$
$u=-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}}x^{-1} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}}x^{-1} )y^{\frac{1}{2}}$
Now for the side condition
$u(1,y)=y^2$
Recall that our solution is
$u=-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}}x^{-1} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}}x^{-1} )y^{\frac{1}{2}}$
Under the side condition, $u(1,y)=y^2$, we have
$-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}} )y^{\frac{1}{2}}=y^2$
$F(y^{\frac{-1}{2}})=y^{\frac{3}{2}}+e^{{y^{(\frac{-1}{2}} )}^{-1}}e^{y^\frac{1}{2}}$
So if I let $ y^{\frac{-1}{2}} =r$ Would it be $y=r^2$ because I am multiplying by $-2$ to get rid of the negative.
Edit: maybe not.. more like $y = r^{-2}$
| Without looking at your attempt in details, a method of solving is shown below, so that you could compare with your calculus :
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"language": "en",
"url": "https://math.stackexchange.com/questions/937557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the first derivative $y=\sqrt\frac{1+\cosθ}{1-\cosθ}$ $$y=\sqrt\frac{1+\cosθ}{1-\cosθ}$$ my professor said that the answer is $$y'=\frac{1}{\cosθ-1}$$ she said use half angle formula but I just end up with $\frac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}$ I used the quotient rule. I also know that ${(1-\cosθ)^2}$ can be $\sinθ^2$ but i try to apply my identities but it's still wrong.
| $$ y^2=\frac {1+\cos\theta}{1-\cos\theta}.$$
Differentiating gives
$$\frac {d}{d\theta} y^2=\frac {d}{d\theta} \frac {1+\cos\theta}{1-\cos\theta} = -\frac{2\sin\theta}{(1-\cos\theta)^2}. $$
Using the chain rule gives for the left hand side
$$2y\frac {dy}{d\theta}. $$ Hence, $$\frac {dy}{d\theta}=-\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{(1-\cos\theta)^2},$$ a plot of which agrees with that of Jean-Claude Arbaut's answer.
Assuming this to be the case then retrospectively it would seem that
$$\text{sgn}(\sin\theta) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}.$$
Interesting.
Out of curiosity, let $\theta= \arcsin x$. Then
$$\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}=\frac{x}{\sqrt{1-\sqrt{1-x^2}} \sqrt{\sqrt{1-x^2}+1}}=\frac{x}{\sqrt{x^2}}=\frac{x}{\mid x\mid}=\text{sgn}(x) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Having problems proving $ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $ I'm trying to prove the following
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $$
using the identity $$ \sum_{r=0}^{n} \binom{n}{r} = 2^{n} $$
but I'm not able to. This is what I did so far,
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (1) \binom{n}{0} + (2) \binom{n}{1} + ... + (n+1) \binom{n}{n} $$
| Starting with the binomials expansion
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} t^{r} = (1+t)^{n}
\end{align}
it can be seen by differentiation that
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} r t^{r} = n t (1+t)^{n-1} = n \left[ (1+t)^{n} -
(1+t)^{n-1} \right].
\end{align}
Now by adding the original series to this result then
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} (r+1) t^{r} = (n+1) (1+t)^{n} - n (1+t)^{n-1}.
\end{align}
Now let $t=1$ to obtain
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} (r+1) &= 2^{n} (n+1) - 2^{n-1} n \\
&= 2^{n-1} (n+2)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition
When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$,
which WolframAlpha agrees to.
However when I try solving it using the derivative definition:
$$\lim_ {h\to 0} = \frac{f(x+h) - f(x)}{h} $$
I get:
$$2x \sin \left(\frac{1}{x+h} \right ) + h \sin \left(\frac{1}{x+h}\right)$$
which in return results in, as $h \to 0$:
$$2x \sin (1/x)$$
So what am I doing wrong when using the def of derivatives?
| Problem:
$\frac{d}{dx}\left(x^2\sin\left(\frac{1}{x}\right)\right)$
Use the product rule, $\frac{d}{dx}\left(uv\right)=v\frac{du}{dx}+u\frac{dv}{dx}$, where $u=x^2$ and $v=\sin\left(\frac{1}{x}\right)$:
$=x^2\left(\frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right)\right)+\left(\frac{d}{dx}\left(x^2\right)\right)\sin\left(\frac{1}{x}\right)$
Using the chain rule, $\frac{d}{dx}(\sin(\frac{1}{x}))=\frac{d\sin\left(u\right)}{du}\frac{du}{dx}$, where $u=\frac{1}{x}$ and $\frac{d}{du}(\sin(u))=\cos(u)$:
$=\left(\frac{d}{dx}\left(x^2\right)\right)\sin\left(\frac{1}{x}\right)+\boxed{\cos\left(\frac{1}{x}\right)\left(\frac{d}{dx}\left(\frac{1}{x}\right)\right)}x^2$
Use the power rule, $\frac{d}{dx}\left(x^n\right)=nx^{n-1}$, where $n=-1$: $\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}\left(x^{-1}\right)=-x^{-2}$:
$=\left(\frac{d}{dx}\left(x^2\right)\right)\sin\left(\frac{1}{x}\right)+\boxed{-\frac{1}{x^2}}x^2\cos\left(\frac{1}{x}\right)$
Simplify the expression:
$=-\cos\left(\frac{1}{x}\right)+\left(\frac{d}{dx}\left(x^2\right)\right)\sin\left(\frac{1}{x}\right)$
Use the power rule, $\frac{d}{dx}\left(x^n\right)=nx^{n-1}$, where $n=2$: $\frac{d}{dx}\left(x^2\right)=2x$:
Answer:
$$=-\cos\left(\frac{1}{x}\right)+\boxed{2x}\sin\left(\frac{1}{x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}dx$ How can I approach this integral? ($0<a \in \mathbb{R}$ and $n \in \mathbb{N}$)
$$\large\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx$$
Integration by parts doesn't seem to make it any simpler.
Hints please? :)
| Since the integrand is even then
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx
\end{align}
Using substitution $t=x^2$ we get
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx&=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx\\
&=\int_{0}^\infty t^{n-\frac{1}{2}}e^{- \frac{1}{2} at } \, dt
\end{align}
Then using substitution $u=\frac{1}{2} at$ we get
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx
&=\int_{0}^\infty t^{n-\frac{1}{2}}e^{- \frac{1}{2} at } \, dt\\
&=\int_{0}^\infty \left(\frac{2u}{a}\right)^{n-\frac{1}{2}}e^{- u } \, \frac{2du}{a}\\
&=\left(\frac{2}{a}\right)^{n+\frac{1}{2}}\int_{0}^\infty u^{n-\frac{1}{2}}e^{- u } \, du\\
&=\left(\frac{2}{a}\right)^{n+\frac{1}{2}}\Gamma\left(n+\frac{1}{2}\right)
\end{align}
where $\Gamma(z)$ is the gamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/941570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
| The denominators are fairly straightforward. I assume the next two are: (x^2 + 192)^9 and (x^2 + 243)^10? The recursion is: Given denominator (x^2 + k)^n, then the next denominator is (x^2 + k*(n/(n-1))^2)^(n+1). (Yes, this is same as the previous post, but I like to think in terms of recursions because it sometimes helps in decoding things.) As for the numerators I think have something based on the a common special function that's promising but it would help to know more specifics about the convolution. (@financequant)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Using a given identity to solve for the value of an expression This problem caught my eye in the book yesterday. Till now I still get stuck. Here it is:
If $$\frac{x}{x^2+1}=\frac{1}{3},$$ what is the value of $$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}?$$
The denominator is a cyclotomic polynomial which can be expressed as $$\frac{x^7-1}{x-1}$$ but I have no idea if this even helps.
| The first identity gives:
$$ x+\frac{1}{x}=3, $$
hence:
$$ x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 - 2 = 7, $$
$$ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)= 18 $$
and
$$ x^3+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3} = 18+7+3+1 = 29 $$
so the answer to your question is $\frac{1}{29}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Difficulty in "expressing radical as square" I have to get from this expression:
$(4+2\sqrt3)(\sqrt{2-\sqrt3})$
To this expression:
$\sqrt2+\sqrt6$
I tried to square $(4+2\sqrt3)$ and put it inside the radical, so:
$\sqrt{(16+12+16\sqrt3)(2-\sqrt3)}$
but eventually I get to: $\sqrt{52+32\sqrt3-28\sqrt3-48}$ and therefore $\sqrt{8+4\sqrt3}$ which will give me $2\sqrt{2+\sqrt3}$
From there I stopped because it didn't seem that I was getting anywhere.
I checked on wolfram alpha and using the step-by-step solution the first thing that it does is:
"Express $2-\sqrt3$ as a square using $2-\sqrt3 = \frac{1-2\sqrt3+(\sqrt3)^2}{2}$ which is clearly correct and from there it is relatively easy to proceed to the solution.
So my questions are: (1) what is that does not work in my method of putting the squared expression inside the radical (2) where can I fond more information on how to express a radical as a square, is there a procedure (something similar to completing the square for example)?
Thanks in advance.
| The idea is that you can compute square roots of numbers of the form $\sqrt{a+b\sqrt{d}}$. Indeed,
$$(x+y\sqrt{d})^2 = x^2 + dy^2 + 2xy\sqrt{d}. $$
We want $a = x^2 + dy^2$ and $b = 2xy$, so using $y = b/(2x)$, we get $a = x^2 + db^2/(4x^2)$, or $4x^4 - 4ax^2 + db^2 = 0$, which is a quadratic equation in $x^2$ whose solution is
$$ x^2 = \frac{4a\pm\sqrt{16a^2-16db^2}}{8} = \frac{a\pm\sqrt{a^2-db^2}}{2}. $$
For example, suppose we want to calculate $\sqrt{2-\sqrt{3}}$, so $a=2$ and $b=-1$. The formula gives $x^2 = (2\pm\sqrt{4-3})/2 = (2\pm 1)/2=3/2,1/2$. This gives the solutions
$$ x + y\sqrt{3} = \pm \left(\sqrt{\frac{3}{2}} - \sqrt{\frac{1}{6}} \sqrt{3}\right), \pm \left(\sqrt{\frac{1}{2}} - \sqrt{\frac{1}{2}} \sqrt{3}\right).$$
Wolfram alpha suggests using the second solution:
$$
\begin{align*}
(4+2\sqrt{3})\sqrt{2-\sqrt{3}} &= (4+2\sqrt{3})\left(-\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{2}} \sqrt{3}\right) \\ &= -4\sqrt{\frac{1}{2}} + 2\sqrt{\frac{1}{2}}\cdot 3 + \left(4\sqrt{\frac{1}{2}}-2\sqrt{\frac{1}{2}}\right)\sqrt{3} \\ &=
2\sqrt{\frac{1}{2}} + 2\sqrt{\frac{1}{2}} \sqrt{3} \\ &= \sqrt{2} + \sqrt{6}.
\end{align*}
$$
Of course, this is pretty ad hoc. However, this method will find an integral square root $\sqrt{a+b\sqrt{d}} = x+y\sqrt{d}$ if one exists, which could be helpful in other situations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/943491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How find this limit $\lim_{x\to 0}\frac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\frac{x^3}{3}+\frac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$ Find this limit
$$I=\lim_{x\to 0}\dfrac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$$
I think
$$I=6\lim_{x\to 0}\dfrac{\int_{0}^{x}(\sin{t}\ln{(1+t)}-t^2+\dfrac{t^3}{2})dt}{x^5}$$
Iuse $\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$
and
$$\ln{(1+x)}=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+o(x^3)$$
so
$$\sin{x}\ln{(1+x)}=x^2-\dfrac{x^3}{2}+\dfrac{1}{6}x^4+o(x^4)$$
so
$$I=\dfrac{1}{5}$$
Have other methods?
| The factors of the integrand and the denominator have power series expansions that converge near $0$. If you work out the first few terms, you get the following.
$\begin{align}
I& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\sin{t}\ln{(1+t)}\,dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}\\
\\
& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t-\dfrac{t^3}{6}+\dfrac{t^5}{120}+O(t^7)\right)\cdot\left(t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\
\\
& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t^2-\dfrac{t^3}{2}+\dfrac{t^4}{6}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\
\\
& =\lim_{x\to 0}\dfrac{\dfrac{x^3}{3}-\dfrac{x^4}{8}+\dfrac{x^5}{30}+O(x^6)-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\dfrac{x^5}{6}+O(x^7)}\\
\\
&= \lim_{x\to 0}\dfrac{\dfrac{x^5}{30}+O(x^6)}{\dfrac{x^5}{6}+O(x^7)}\\
&= \dfrac{1}{5}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving a recurrence equation that yields polynomials I am trying to solve the following recurrence equation:
$$
T(n) = kT(n - 1) + nd
$$
I have expanded the first 4 values ($n = 1$ was given):
$$\begin{align}
T(1) & = 1 \\
T(2) & = kT(2-1) + 2d = k + 2d \\
T(3) & = kT(3-1) + 3d = k(k + 2d) + 3d = k^2 + 2kd + 3d \\
T(4) & = kT(4-1) + 4d = k(k^2 + 2kd + 3d) + 4d = k^3 + 2k^2d + 3dk + 4d\\
\end{align}$$
I was able to convert the above into the following summation, in hopes of finding a closed-form solution:
$$
T(n) = k^{n - 1} + d \sum_{i=1}^{n-1} (i + 1) (k^{n-i-1})
$$
But after this point I am stuck. Was creating the summation the correct thing to do, or is there something better I could try?
| Using
\begin{align}
\sum_{j=0}^{m} t^{j} = \frac{1-t^{m+1}}{1-t}
\end{align}
then
\begin{align}
\sum_{j=0}^{m} j \, t^{j} = \frac{t - (m+1) t^{m+1} + m t^{m+2}}{(1-t)^{2}}
\end{align}
for which
\begin{align}
\sum_{j=0}^{m} (j+1) \, t^{j} &= \frac{1}{(1-t)^{2}} \left[ (1-t)(1-t^{m+1}) + t - (m+1) t^{m+1} + m t^{m+2} \right] \\
&= \frac{1}{(1-t)^{2}} \left[ 1 - (m+2) t^{m+1} + (m+1) t^{m+2} \right].
\end{align}
Now,
\begin{align}
\sum_{j=1}^{m} (j+1) t^{j} = \frac{1}{(1-t)^{2}} \left[ 1 - (m+2) t^{m+1} + (m+1) t^{m+2} \right] - 1.
\end{align}
This leads to
\begin{align}
\sum_{j=1}^{m} (j+1) t^{m-j} = \frac{1}{(1-t)^{2}} \left[ t^{m} - (m+2) t + (m+1) \right] - t^{m}.
\end{align}
Now for the difference equation
\begin{align}
T(n) = k T(n-1) + n d
\end{align}
where $T(1) = 1$ leads to the solution
\begin{align}
T(n) = k^{n-1} + d \sum_{j=1}^{n-1} (j+1) k^{n-j-1}
\end{align}
which can be seen in the form
\begin{align}
T(n) &= (1-d) k^{n-1} + \frac{d}{(1-k)^{2}} \left[ k^{n-1} - (n+1) k + n \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $({\sqrt{2}\!+\!1})^{1/n} \!+ ({\sqrt{2}\!-\!1})^{1/n}\!\not\in\mathbb Q$ How could we prove that for every positive integer $n$, the number
$$({\sqrt{2}+1})^{1/n} + ({\sqrt{2}-1})^{1/n}$$ is irrational?
I think it could be done inductively from a more general expression, but I don't know how.
I made an effort trying to solve it using many different methods.
| Use the fact that for every $n$ there exists a polynomial $P_n[x]$ monic of degree $n$ with integral coefficients so that we have the identity:
\begin{eqnarray*}
t^n + \frac{1}{t^n} = P_n(t+ \frac{1}{t})
\end{eqnarray*}
The $P_n$'s are connected with the Chebyshev polynomial of first kind since for $t = e^{i \theta}$ we get $2 \cos n \theta = P_n (2 \cos \theta)$
\begin{eqnarray*}
\cos n \theta &=& \frac{1}{2}P_n (2 \cos \theta) \\
\cos n \theta &=& T_n( \cos \theta)
\end{eqnarray*}
so
\begin{eqnarray*}
P_n(x) = 2 T_n(\frac{x}{2})
\end{eqnarray*}
For instance
\begin{eqnarray*}
T_3(x) = 4x^3 - 3 x \\
P_3(x) = 2 T_3(\frac{x}{2}) = x^3 - 3 x
\end{eqnarray*}
One checks easily the identity:
\begin{eqnarray*}
t^3 + \frac{1}{t^3} = P_3(t+\frac{1}{t})
\end{eqnarray*}
Back to our problem. We have $(\sqrt{2}-1)(\sqrt{2}+1) = 2 -1 = 1$ and so
$(\sqrt{2}-1) = \frac{1}{\sqrt{2} + 1}$ and similarly for any power
$$(\sqrt{2}-1)^{1/n} = \frac{1}{\sqrt{2} + 1)^{1/n}}$$.
Let $x =(\sqrt{2} + 1)^{1/n}$. Then
$$ x + \frac{1}{x} = (\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$$
Now if $x+\frac{1}{x}$ were rational then so would be $P_n(x+\frac{1}{x}) = x^n + \frac{1}{x^n} = \sqrt{2}+1 + \sqrt{2} - 1 = 2 \sqrt{2}$ which is not the case. We conclude that $x+\frac{1}{x} = (\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$ is irrational.
In fact $(\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$ is the unique positive root of the equation $(P_n[x])^2 - 8 =0$ or $ (T_n(\frac{x}{2}))^2 = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
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Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^2x$ we have:
$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$
But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$
Am I wrong somewhere?
| Multiplying both numerator and denominator by $\sec^4x$ yields
$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{d x}{1-3 \cos ^{2} x \sin ^{2} x} \\
=& \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{4} x}{\sec ^{4} x-3 \tan ^{2} x} d x \\
=& \int_{0}^{1} \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}-3 t^{2}} d t \quad \textrm{ where }t=\tan x \\
=& \int_{0}^{1} \frac{1+t^{2}}{t^{4}-t^{2}+1} d t \\
=& \int_{0}^{1} \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}-1} d t \\
=& \int_{0}^{1} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+1}\\=&\left[\tan ^{-1}\left(t-\frac{1}{t}\right)\right]_{0}^{1}\\=&\frac{\pi}{2}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Prove that $\pi > 24\small{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}}$ Prove that $\pi > 24\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$.
I tried using trig but I couldn't solve it. A hint I was given is to use half angle identities. This should be easy for someone who is experienced in inequalities, but I'm looking for a "low power" solution.
Thanks.
| $$\sqrt{3}=2\sin \frac{\pi}{3}$$
$$2+\sqrt{3}=2\left(1+\sin \frac{\pi}{3}\right)$$
$$2+\sqrt{3}=2\left(1+\cos \frac{\pi}{6}\right)$$
$$2+\sqrt{3}=2\left(2\cos^2 \frac{\pi}{12}\right)$$
$$\sqrt{2+\sqrt{3}}=2\cos \frac{\pi}{12}$$
$$2+\sqrt{2+\sqrt{3}}=2(1+\cos \frac{\pi}{12})$$
$$2+\sqrt{2+\sqrt{3}}=2\left(2\cos^2 \frac{\pi}{24}\right)$$
$$\sqrt{2+\sqrt{2+\sqrt{3}}}=2\cos \frac{\pi}{24}$$
$$2-\sqrt{2+\sqrt{2+\sqrt{3}}}=2\left(1-\cos \frac{\pi}{24}\right)$$
$$2-\sqrt{2+\sqrt{2+\sqrt{3}}}=2\left(2\sin^2 \frac{\pi}{48}\right)$$
$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=2\sin \frac{\pi}{48}$$
$$24\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=48\sin \frac{\pi}{48}$$
$$\pi > 24\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}} = 48\sin \frac{\pi}{48}$$
$$\pi > 48\sin \frac{\pi}{48}$$
$$2\pi > 96\sin \frac{\pi}{48}=48\left(2\sin \frac{\pi}{48}\right)=48\times \text{side length of regular 48-gon}$$
True because perimeter of circumscribed circle of regular 48-gon is > perimeter of polygon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$ I'm trying to prove $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$. Is the following a correct proof?
For all $n$ we have $0 \leq \left|\sqrt{n^2 + 1} - n\right| \leq \left|\sqrt{n^2+1} - 1 \right|$. For any $\epsilon > 0$ take $ N = \sqrt{(\epsilon+1)^2-1}$. Then for all $n > N$ we have $\left|\sqrt{n^2+1} - 1 \right| < \epsilon$ so $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - 1] = 0$ and by the squeeze theorem $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$.
| Another way:
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
$.
For $0 < x $,
$(1+x)^2
=1+2x+x^2
> 1+2x
$,
so
$\sqrt{1+2x} < 1+x$
or
$\sqrt{1+x} < 1+x/2$.
(You can directly prove this
by squaring
both sides.)
Putting
$x = 1/n^2$,
$\sqrt{1+1/n^2}
< 1+1/(2n^2)
$.
Therefore,
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
< n(1+1/(2n^2))
= n+1/(2n)
$,
so
$\sqrt{n^2+1}-n
< (n+1/(2n))-n
=1/(2n)
\to 0$
as $n \to \infty
$.
Note that can
be used to show that
$\sqrt{n^2+k}-n
\to 0$
for any fixed $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953214",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
| Let $y = f(x)$ be a function with domain $D$.
Let $\left[a,b\right] \le D$, then the global extrema are the extreme (largest and smallest) values in $\left[a,b\right]$.
There are two global extrema; global minima (least value) and global maxima (largest value)
Conceptually speaking,
$$f'(x) = 0\iff f(x) \text{ `makes a turning`}\iff f(x) \text{ is an extrema}$$
Let $c_1,\space c_2,\space \dots, c_n$ be the points at which $f'(x) = 0$ (ie, the extreme points)
Then,
$$\text{Global Minima }(m) = \min\{f(a), f(c_1),\dots, f(c_n), f(b)\}\\
\text{Global Maxima }(M) = \max\{f(a), f(c_1),\dots, f(c_n), f(b)\}$$
Then, the range of $f(x)$ in $\left[a,b\right]$ is simply the interval $(m,M)$
Given :: $f(x) = \sqrt{x-1} + 2\sqrt{3-x}$
Since $\sqrt{n}$ is only a real function when $n\ge 0$, then in $f(x)$,
$$x-1\ge 0 \quad \&\quad 3-x \ge 0\\
\Rightarrow 1 \le x \le 3\\
\Rightarrow x \in \left[1,3\right]$$
Now, to check for extrema,
$$\frac{d}{dx} f(x) = f'(x) = 0 \\
\Rightarrow \frac{1}{2\sqrt{x-1}} - \frac{1}{\sqrt{3-x}} = 0\\
\Rightarrow \frac1{2\sqrt{x-1}} = \frac1{\sqrt{3-x}}\\
\Rightarrow 4(x-1) = 3-x \\
\Rightarrow x = \frac{7}{5} = 1.4 \in \left[1,3\right]
$$
The suspects:
$$
f\left(\frac{7}{5}\right) = \sqrt{\frac{7 - 5}{5}} + 2\sqrt{\frac{15-7}{5}} = \sqrt{\frac{2}{5}} + 4\sqrt\frac{2}{5} = \sqrt5\cdot\sqrt2 =\sqrt{10} \approx 3.1622\\
f(1) = 0 + 2\sqrt{3-1} = 2\sqrt{2} \approx 2.8284\\
f(3) = \sqrt{3-1} + 0 = \sqrt{2}\approx 1.4142
$$
Clearly, $f\left(\frac{7}{5}\right) > f(1) > f(3)$
$\therefore $ The global Maxima is $\sqrt{10}$ and the global minima is $\sqrt{2}$
$$\boxed{\text{Range}(f) = \left[ \sqrt{2}, \sqrt{10} \right]}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$
It continues to infinity.I tried multiplying with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\frac{1}{8} -\frac{1}{4*6} -\frac{1}{4*6*8}-\frac{1}{4*6*8*10}$$.
Also if anybody is wondering the answer 0.5
| The $r$th term $\displaystyle\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$ where integer $r\ge1$
As the number of terms in the numerator $=$ the number of terms in the denominator $-1,$
If $\displaystyle S=\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)},$
Multiplying the numerator by the previous term of $1,3,5,\cdots$ (Arithmetic Series)
$\displaystyle -S=\sum_{n=1}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$
$\displaystyle=-\frac{-1}2+\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$
Now, $\displaystyle\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}=(1-1)^{\frac12}-1$
Now follow Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x}
=\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x
=\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}
$$
I noticed something interesting, namely that
$$
\begin{align*}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
& = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
& = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
= \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
\end{align*}
$$
The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$).
But I had problems proving the first equality. Does anyone have some quick hints?
| Subtract the two integrals in question and get
$$\int_0^{\pi} dx \frac{\cos{2 x}}{1+\frac12 \sin{2 x}} = \frac12 \int_0^{2 \pi} du \frac{\cos{u}}{1+\frac12 \sin{u}}$$
This may be shown to be equal to the complex integral
$$-\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} \frac{z+z^{-1}}{1+\frac{1}{4 i} (z-z^{-1})} = \oint_{|z|=1}\frac{dz}{z} \frac{z^2+1}{z^2+i 4 z-1}$$
The poles of the integrand within the unit circle are at $z=0$ and $z=-(2-\sqrt{3}) i$; their respective residues are $-1$ and $1$. By the residue theorem, therefore, the integral is zero and the two original integrals are equal.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Limit at negative infinity Evaluate $$\lim_{x\to -\infty}\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$$
I think the strategy is to divide the numerator and denominator by x^2. Help please. The textbook answer is $$\frac{1}{4+\sqrt{3}}$$
| Instead of dividing the numerator and denominator by $x^2$, it may be easier to factor out the appropriate factors from the radical expressions:
\begin{align}
\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}} & = \frac{\sqrt[3]{x^6\big(1+\frac{8}{x^6}\big)}}{4x^2+\sqrt{x^4\big(3+\frac{1}{x^4}\big)}} \\ & = \frac{\sqrt[3]{x^6}\sqrt[3]{1+\frac{8}{x^6}}}{4x^2+\sqrt{x^4}\sqrt{3+\frac{1}{x^4}}} \\ & = \frac{x^2\sqrt[3]{1+\frac{8}{x^6}}}{4x^2+x^2\sqrt{3+\frac{1}{x^4}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Zeros of a function of degree 4 I'm trying to show that the following function has no zeros
$$
60x^4-44x^3-25x^2-44x+60=0.
$$
I already tried using Eisenstein's criterium, but since the first and the last coefficient are both $60$, there is no prime which holds.
| In this answer I showed that the polynomial
$$
p(x) = x^4 + ax^3 + bx^2 + ax + 1
$$
has no real zeros if $|a| \leq 4$ and $b > 2 |a| - 2$.
We have
$$
60x^4-44x^3-25x^2-44x+60 = 60 \left(x^4 - \frac{44}{60}x^3 - \frac{25}{60}x^2 - \frac{44}{60}x+1\right),
$$
so in this case $a = -44/60$ and $b = -25/60$. Indeed, $|a| \leq 4$ and
$$
b = -\frac{25}{60} > -\frac{32}{60} = 2|a|-2,
$$
so the given polynomial has no real zeros.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove that $ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3) $ $$ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3) $$ where n is odd.
Plesase help mi with that equation.
| Let $n=2m+1\iff m=(n-1)/2$. Then, we have
$$\begin{align}\sum_{k=1}^{m}\binom{2k+1}{2}&=\sum_{k=1}^{m}\frac{(2k+1)(2k)}{2}\\&=\sum_{k=1}^{m}(2k^2+k)\\&=2\cdot\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}{2}\\&=\frac{m(m+1)(4m+5)}{6}\\&=\frac 16\times\frac{n-1}{2}\times \left(\frac{n-1}{2}+1\right)\times \left(4\cdot \frac{n-1}{2}+5\right)\\&=\frac{1}{24}(n-1)(n+1)(2n+3).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tricky sequences and series problem For a positive integer $n$, let $a_{n}=\sum\limits_{i=1}^{n}\frac{1}{2^{i}-1}$. Then are the following true:
$a_{100} > 200$ and
$a_{200} > 100$?
Any help would be thoroughly appreciated. This is a very difficult problem for me. :(
| $$\begin{gather}a_n = 1 + \underbrace {\frac{1}{2} + \frac{1}{3}}_{2\text{ terms}} + \underbrace {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}}_{4\text{ terms}} + \ldots + \underbrace { \frac{1}{2^{n -1}} + \frac{1}{2^{n -1}+1} + \ldots + \frac{1}{2^n -1}}_{2^{n-1}\text{ terms}}\\
> 1 + 2\cdot\frac{1}{4} + 4\cdot\frac{1}{8} + \ldots + 2^{n-1}\cdot\frac{1}{2^{n}}
\\ =1 + \underbrace { \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2}}_{(n-1) \text{ terms}} = 1 +\frac{n-1}{2} = \frac{n+1}{2}.
\end{gather}
$$
Thus, $a_n > \frac{n+1}{2}.$
On the other hand,
$$a_n < 1 + 2\cdot\frac{1}{2} + 4\cdot\frac{1}{4} + \ldots + 2^{n-1}\cdot\frac{1}{2^{n-1}} = 1+(n-1 )= n,$$
so the inequality $a_{100} > 200$ cannot be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge? Let $\alpha$ be such that $0\leq \alpha \leq 1$. Since $\sin n$ has no limit as $n$ tends to $\infty$, I'm having trouble with finding if the series $$\sum_{n=1}^{\infty}\sin \left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$$ is convergent? Thanks.
| This series is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin \left( 2\pi \sqrt{n^2+\alpha^2 \sin n+(-1)^n}\right)\\\\
&=\sin \left( 2\pi n \:\sqrt{1+\frac{\alpha^2\sin n}{n^2}+\frac{(-1)^n}{n^2}}\right)\\\\
&=\sin \left( 2\pi n \:\left(1+\frac{\alpha^2\sin n}{2n^2}+\frac{(-1)^n}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\\\
&=\sin \left( 2\pi n +\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\
&=\sin \left(\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\
&=\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)
\end{align}
$$
Now recall that $\displaystyle \sum_{n=1}^{+\infty}\frac{\sin n}{n}$ is convergent, moreover
$$
\sum_{n=1}^{+\infty}\frac{\sin n}{n}=\Im\sum_{n=1}^{+\infty}\frac{e^{in}}{n}=\Im\left(-\log(1-e^i)\right)=\frac{\pi-1}{2}.
$$
Then it is clear that your initial series $\displaystyle \sum_{n=1}^{+\infty} u_n $ is convergent, being the sum of convergent series.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$$
Finding $\cos\alpha$ and $\sin\beta$:
$$\begin{align}
\cos\alpha & = \frac{\sqrt{4-x^2}}{2}, \\[0.1in]
\sin\beta & = \sqrt{1-x^2}.
\end{align}$$
So plugging everything in:
$$\begin{align}
\sin(2\arcsin x + \arccos x) & = \frac{x}{2} \cdot x + \frac{\sqrt{4-x^2}}{2} \cdot \sqrt{1-x^2}\\[0.1in]
& = \frac{x^2 + \sqrt{4-x^2}\sqrt{1-x^2}}{2}.
\end{align}$$
But this doesn't seem to lead to the right side. Is my method incorrect?
| $\alpha=2\arcsin x$ means that $x=\sin(\frac\alpha2)$, not that $\frac x2=\sin\alpha$
$\frac\alpha2=\arcsin x$
$x=\sin(\frac\alpha2)$
an alternate and faster way is to use
$\arcsin x+\arccos x=\frac\pi2$
| {
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"url": "https://math.stackexchange.com/questions/967041",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 1
} |
Modular Arithmetic: using congruence to find remainder How do I use the fact that if $a = b \pmod n$ and $c = d\pmod n$ then $ac = bd\pmod n$ to find the remainder when $3^{11}$ is divided by $7$?
| $3^{11} \equiv (3^2)^5\cdot 3 \equiv 2^5 \cdot 3 \equiv 96 \equiv 5 \pmod 7$
Of course, it's actually easier to employ Fermat's Little Theorem (if you're permitted) like so:
$3^{11} \equiv 3^6 \cdot 3^5 \equiv 1 \cdot (3^2)^2 \cdot 3 \equiv 2^2 \cdot 3 \equiv 12 \equiv 5 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How calculate the shaded area in this picture? Let the centers of four circles with the radius $R=a$ be on 4 vertexs a square with edge size $a$. How calculate the shaded area in this picture?
| Let $(APD)$ be the area of the figure $APD$.
And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.
First, we have
$$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$
Second, we have
$$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$
Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triangle KAD$ is a equilateral triangle.
So, since we have
$$\begin{align}(K(E)AD(M))&=(\text{sector}\ AKD)+(\text{sector}\ DKA)-(\triangle KAD)\\&=\frac{\pi}{6}a^2+\frac{\pi}{6}a^2-\frac{\sqrt 3}{4}a^2\\&=\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2,\end{align}$$
we have
$$\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2=x+y+2z.\tag3$$
Solving $(1),(2),(3)$ gives us
$$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$
| {
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"url": "https://math.stackexchange.com/questions/977998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question on simplification of $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$? I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
| Since $2 = (2n+3) - (2n+1)$ then
\begin{align}
\frac{2}{(2n+1)(2n+3)} &= \frac{(2n+3) - (2n+1)}{(2n+1)(2n+3)} = \frac{1}{2n+1} - \frac{1}{2n+3}.
\end{align}
The series then becomes
\begin{align}
\sum_{n=1}^{\infty} \frac{2}{(2n+1)(2n+3)} &= \sum_{n=1}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) \\
&= \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \\
&= \frac{1}{3}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Limit of a Sequence involving 3rd root I'm not finding any way to simplify and solve the following limit:
$$
\lim_{n \to \infty} \sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}
$$
I've tried multiplying by the conjugate, but this give a more complex limit.
| $$\begin{align*}&\sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}=\\\\
&=\sqrt{n^2+n+1}-n +\left(n-\sqrt[3]{n^3+n^2+n+1}\right)=\\ \\
&=\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n } +\dfrac{n^3-n^3-n^2-n-1}{n^2+n(n^3+n^2+n+1)^{\frac{1}{3}}+(n^3+n^2+n+1)^{\frac{2}{3}}}= \\\\
&=\dfrac{n+1}{\sqrt{n^2+n+1}+n } +\dfrac{-(n^2+n+1)}{n^2+n(n^3+n^2+n+1)^{\frac{1}{3}}+(n^3+n^2+n+1)^{\frac{2}{3}}} \underset{n\to\infty}\to \dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}
\end{align*}$$
| {
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"url": "https://math.stackexchange.com/questions/980006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6})= -\pi \sin(\pi \cdot \frac{1}{6})= \frac{-\pi}{2}$
And I put the $x$-value into $f(x)$ to find the $y$-value
$f(\displaystyle\frac{1}{6})=\cos(\pi \cdot \frac{1}{6})=\frac{\sqrt{3}}{2}$
Now I use the formula $y=m(x-x_0)+y_0$
$y= \displaystyle\frac{-\pi}{2}(x-\frac{1}{6})+\frac{\sqrt{3}}{2}$ = $\displaystyle\frac{-\pi \cdot x}{2} + \frac{\sqrt{3}}{2} + \frac{\pi}{12}$
Am I correct? Because my book got an different answer like $6 \pi x +12y =6\sqrt{3}+ \pi$
| multiplying your equation
$y=-\frac{\pi}{2}x+\frac{\sqrt{3}}{2}+\frac{\pi}{12}$ by $12$ you will get
$12y+6\pi x=6\sqrt{3}+\pi$
| {
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Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd)
+ (ad+bc)\sqrt{5}$
My work:
Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.
Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$
The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$
The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$
The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$
The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.
If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.
| You can use a similar trick as in complex numbers. Note that $(a+b\sqrt5)(a-b\sqrt5)=a^2-5b^2$, so $\frac{1}{a+b\sqrt5}=\frac{a-b\sqrt5}{a^2-5b^2}$.
| {
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Interesting functional equations problem? $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$
Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$.
How would we solve this? I noticed that if you plug in $\frac{x-1}{x}$ in for $x$, and then again, we can solve $f(x)$ after bashing it out with systems of equations. However, the resulting function has a few values for which it is undefined. I tried finding those values by going back to the original equation, but it doesn't seem like a "good" answer. Since I then wanted to verify that I had the correct function first, I plugged the actual function back into the original equation. However, the LHS and RHS didn't match up. Is there a better way to solve this other than just bashing?
Thanks in advance!
| Sorry, this wouldn't fit into a comment, so I had to post this in as an answer, but when I plugged $f$ back into the original equation, I got:
\begin{align*}
f(x) + 3 f\left(\frac{x-1}{x}\right) &= \frac{1}{28}\left(x - 3 - \frac{3}{x} - \frac{9}{x-1}\right) + 3\left(\frac{1}{28}\right)\left(\frac{x-1}{x} - 3 - \frac{3}{\frac{x-1}{x}} - \frac{9}{\frac{x-1}{x}-1}\right)\\
&= \left(\frac{1}{28}\right)\left(x - 3 - \frac{3}{x}-\frac{9}{x-1}+3\left(\frac{x-1}{x}-3-\frac{3x}{x-1}+9x\right)\right)\\
&= \left(\frac{1}{28}\right)\left(x - 3 - \frac{3}{x}-\frac{9}{x-1}+\frac{3x-3}{x} - 3 - \frac{9x}{x-1} + 27x\right)\\
&= \left(\frac{1}{28}\right)\left(28x - 6 + \frac{3x-6}{x}-\frac{9+9x}{x-1}\right)\\
&= \left(\frac{1}{28}\right)\left(28x^3-28x^2-6x^2+6x-3x^2+6x+3x+6-9x+9x^2\right)\\
&= \left(\frac{1}{28}\right)\left(28x^3-28x^2+6x+6\right)\\
&= x^3-x^2+\frac{3x}{14} + \frac{3}{14},
\end{align*}
which doesn't match the given $7x$. If you could tell me where I made my mistake, it would be greatly appreciated. Thanks in advance!
| {
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Area between curves $y=x^3$ and $y=x$ I've tried to done one of my homework problems for several times, but the answer doesn't make sense to me. The question asks to find the area between $y=x^3$ and $y=x$. Those are odd functions, and I'm pretty sure that the area between their graphs should be 0. However, I keep getting 1/2 as my answer. Can anyone please check my work to see if I did something wrong?
$$
\begin{align}
A &= \int_0^1 \left(x-x^3\right) \,\operatorname{d}\!x + \int_{-1}^0 \left(x^3-x\right) \,\operatorname{d}\!x \\
&= \left. \left( \frac{x^2}{2} - \frac{x^4}{4} \right) \right\vert_0^1 + \left. \left( \frac{x^4}{4} - \frac{x^2}{2} \right) \right\vert_{-1}^0 \\
&= \left( \frac{1}{2} - \frac{1}{4} - 0 \right) + \left[ 0 - \left( \frac{1}{4} - \frac{1}{2} \right) \right] \\
&= \left(\frac{1}{4}\right)+\left[-\left(\frac{-1}{4}\right)\right] \\
&= \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\end{align}
$$
Thanks ahead!
| The area between two curves is always positive. See the below graph.
The area in green and orange is the area you are finding. It is always going to be positive because it exists. When you subtract the two curves, you are finding the area between the curves, regardless of their position relative to the x axis.
Also, you could have just multiplied one of your integrals by 2 to get the answer because they are odd functions :P
| {
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How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$? From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$
For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-integer $a$?
Integrating by parts,
$$\begin{align}
J_n &= \left.-\frac{1}{n+1} \frac{1+x^n}{(1+x)^{n+1}}\right\lvert_0^1+ \frac{n}{n+1} \int_0^1 dx \frac{x^{n-1}}{(1+x)^{n+1}}
\\&=\frac{1}{n+1}\left(1-2^{-n} \right) + \frac{n}{n+1} \int_0^1 dx \left[ \frac{1+x^{n-1}}{(1+x)^{n+1}} - \frac{1}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1} + \frac{1}{n+1} \left[\left(1-2^{-n} \right) -\int_0^1 dx \frac{n}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1}.
\end{align}.
$$
Because $J_0 = 1$, we therefore have $$J_n = \frac{1}{n+1} \mathrm{for\,} n \in \mathbb{N}.$$
| $$\begin{align*}
\int_{0}^{1}\frac{1+x^{a}}{\left(1+x\right)^{a+2}}dx
&=\int_{0}^{1}\frac{1}{\left(1+x\right)^{a+2}}dx+\int_{0}^{1}\frac{x^{a}}{\left(1+x\right)^{a+2}}dx \\
&=\left.-\frac{1+x}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1}+\left.\frac{\left(1+x\right)x^{a+1}}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1} \\
&=\frac{1}{a+1}.
\end{align*}$$
| {
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Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$
2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$
I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???
| Here is a simple way. Let $z_{1}=cos\theta_{1}+isin\theta_{1}$,$z_{2}=cos\theta_{2}+isin\theta_{2}$, $z_{3}=cos\theta_{3}+isin\theta_{3}.$
By $z_{1}+z_{2}+z_{3}=0$ we get the sum of the cosines are zero and the sum of the sines is zero.
Squaring the equations
$cos\theta_{1}+cos\theta_{2}=-cos\theta_{3}$ and the
$sin\theta_{1}+sin\theta_{2}=-sin\theta_{3}$ we get
$cos(\theta_{1}-\theta_{2})=-\dfrac{1}{2}$.
Therefore $\theta_{1}-\theta_{2}=\dfrac{2\pi}{3}$. Likewise we get similar equations.
Since the triangle with sides $|z_{1}|$ and $|z_{2}|$ is isosceles, and
the angle $(z_{1},z_{2})$ is $120^{o}$ then the other two angles are $30^{o}$.
We do the same for $z_{2},z_{3}$ and we get the same result.
So the angles of the triangle are all $30^{0}+30^{0}=60^{0}$ and hence the triangle is equilateral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area bounded by the graphs $y = x^2 – 2x – 6$ and $y = 6 – x^2$? Find the area bounded by the graphs of the following functions
$$y = x^2 – 2x – 6$$ and $$y = 6 – x^2$$
Are the points $-3$ and $2$? Where to go form there?
| The points of intersection can be found by solving
$$x^2-2x-6=6-x^2 \Longleftrightarrow x^2-x-6=0 \Longleftrightarrow (x-3)(x+2)=0$$
Thus $x=\color{red}-2$ and $x=\color{red}+3$ are the points of intersection.
In the interval $[-2,3]$, $6-x^2>x^2-2x-6$, therefore the area of the bounded region is
$$\int_{-2}^36-x^2-(x^2-2x-6)\,dx$$
$$=\displaystyle \int_{-2}^3 -2x^2+2x+12 \,dx$$
$$=\displaystyle \left. \left(-\frac{2}{3}x^3+x^2+12x\right)\right|_{-2}^3$$
$$=\boxed{41\frac{2}{3}}$$
| {
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"url": "https://math.stackexchange.com/questions/993778",
"timestamp": "2023-03-29T00:00:00",
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How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$
I did the first parts :
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$
$\omega=-1$ is a root so :
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5})$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (-\cos \frac{\pi}{5} + i \sin \frac{\pi}{5})$$
$$\omega-\omega^4= 2 \cos \frac{\pi}{5}$$
Did similar method for $\omega^3-\omega^2$ and got the desired result.
For this part I only know the first part which is:
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}$$
$$2\cos {\pi}{5} + 2\cos \frac{3\pi}{5}=\omega-\omega^4+\omega^3-\omega^2=1$$
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}=\frac{1}{2}$$
$$\cos \frac{\pi}{5} \cos \frac{3\pi}{5} = \frac{1}{4} (\omega-\omega^4)(\omega^3-\omega^2)$$
$$\frac{1}{4} (\omega^4-\omega^3-\omega^7+\omega^6)$$
How do I find $\omega^7$ and $\omega^6$ ???? please help.
Also I would love if I receive support for the last part too, I don't know how to make a quadratic equation
| Since $\omega^5+1=0$, you have that $\omega^5=-1$, so $\omega^6=\omega^5\cdot\omega=-\omega$ etc.
For the quadratic equation, use that $(x-a)(x-b)=x^2-(a+b)x+ab$, let $a=\frac{1}{2}(\omega-\omega^4)=\cos(\pi/5)$ and $b=\frac{1}{2}(\omega^3-\omega^2)$. You calculated values for $a+b$ and $ab$ just beforehand, so you have the quadratic equation (namely $x^2-x-1$).
| {
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find critical point $f(x,y) = 8 + 2y^3+2x^3 - 3xy$ $f(x,y) = 8 + 2y^3+2x^3 - 3xy$
$f_x = 6x^2 - 3y$
$f_{xx} = 12x$, $x= 0$, $y=0$
$f_y = 6y^2 -3x$
$f_{yy}=12y$
$f_{xy}=-3$
at $(0,0)$, $D = -9 < 0$
Is it one saddle point $(0,0)$?
| You need to first find ALL the points where $f_x = f_y = 0$.
$f_x = 6x^2 - 3y = 0$ gives us $y = 2x^2$ (1)
$f_y = 6y^2 - 3x = 0$ gives us $x = 2y^2$ (2)
Substitute (1) into (2) to get $x = 2(2x^2) = 8x^4$, i.e. $8x^4-x = 0$.
This has another real root besides $x = 0$, which will give you another critical point.
| {
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The dilogarithm function. Question on an identity of it Upon reading a journal article about manipulating series using the dilogarithm function, I have a few questions. But before I ask them, let me give the information the article provides.
Consider the series $\displaystyle\sum \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$
The related power series is $f(z) = \displaystyle\sum \frac{1}{k^2}z^k = z + \frac{z^2}{4} + \frac{z^3}{9} + ...$
The derivative of this is $f'(z) = \displaystyle\sum \frac{1}{k}z^{k-1} = 1 + \frac{z}{2} + \frac{z^2}{3} + ... = -\frac{\ln(1-z)}{z}$
The series we have provided is just the derivative of the dilogarithm function, and so
$Li'_{2}(z) = -\displaystyle\frac{\ln(1-z)}{z}$.
Now there is an identity that says $Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C$ where $C$ is a constant.
From here, taking $z = 1$ gives us $C = 2Li_{2}(-1) = 2(-1 + \displaystyle\frac{1}{4} - \displaystyle\frac{1}{9} + \displaystyle\frac{1}{16} - ...)$.
This can be related to $Li_{2}(1)$ by using the fact that the even terms of $\displaystyle\sum \frac{1}{k^2} = \sum \frac{1}{(2k)^2} = \frac{1}{4}Li_{2}(1)$. And so the odd terms must be $\displaystyle\frac{3}{4}Li_{2}(1)$ and so the alternating sum I gave earlier is the difference of the even terms and the odd terms which is $-\displaystyle\frac{1}{2}Li_{2}(1)$. This shows that $C = -Li_{2}(1)$.
Here are my questions now.
Where on earth does $Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C$ come from? I do not see that at all.
I don't see how $-(\displaystyle\frac{1}{2})Li_{2}(1)$ is the alternating series. Theres no way to generate $-1$ from it which is the first term of the alternating series.
| Extending the hint in the comment section, one has
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)=\frac{\ln\left(1+\frac{1}{x}\right)}{x}=\frac{\ln(1+x)-\ln(x)}{x},$$
and
$$\frac{d}{dx} Li_2\left(-x\right)= -\frac{\ln\left(1+x\right)}{x}. $$
This implies
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)+ \frac{d}{dx} Li_2\left(-x\right) =
-\frac{\ln(x)}{x}. $$
Integrating the above expression one arrives at
$$ Li_2\left(-\frac{1}{x}\right)+ Li_2\left(-x\right) =
-\frac{1}{2}\ln^2 x + K. $$
To determine the constant of integration $K$ we select $x=1$ in the above equality:
$$2Li_2(-1)= 0+K= K $$
In summary,
$$Li_2\left(-\frac{1}{x}\right)+Li_2(-x)=2 Li_2(-1)-\frac{1}{2}\ln^2(x), $$
as claimed.
| {
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
| Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$
and
$$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$
| {
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Factoring $x^4 + 4x^2 + 16$ I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.
I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for factoring $x^4 + 4x^2 + 16$?It factors into two irreducible quadratic trinomials, which is where I think the problem is stemming from.
Thanks in advance.
| Here is another method that might generalize to more situations.
Using DeMoivre's Theorem, the roots of $x^6-64 = 0$ are $x = 2e^{ik\pi/3}$ for $k = 0,1,2,3,4,5$.
So, the polynomial is the product of the following complex linear factors:
$(x-2)(x+2)(x-2e^{i\pi/3})(x-2e^{i5\pi/3})(x-2e^{i2\pi/3})(x-2e^{i4\pi/3})$
Now pair up factors that are complex conjugates, and simplify to get:
$(x-2)(x+2)(x^2-(4\cos\frac{\pi}{3})x+4)(x^2-(4\cos\frac{2\pi}{3})x+4)$
$= (x-2)(x+2)(x^2-2x+4)(x^2+2x+4)$
Alternatively, you can first use the difference of squares factorization, and then use the sum of cubes and difference of cubes factorizations:
$x^6-64 = (x^3+8)(x^3-8) = (x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$
| {
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How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
| By Hölder's inequality
$$\left(\frac a{\sqrt{1+a}}+\frac b{\sqrt{1+b}}+\frac c{\sqrt{1+c}}\right)^2\Big(a(1+a)+b(1+b)+c(1+c)\Big)\ge(a+b+c)^3,$$
so we only need to prove
\begin{align*}(a+b+c)^3&\ge\frac92\Big(a(1+a)+b(1+b)+c(1+c)\Big)\\(a+b+c)(2(a+b+c)^2-9)&\ge9(a^2+b^2+c^2)\\\end{align*}
AM-GM tells us that $(a+b+c)^2=(a+b+c)^2+9-9\ge6(a+b+c)-9$, so it suffices to prove
\begin{align*}(a+b+c)(12(a+b+c)^2-27)&\ge9(a^2+b^2+c^2)\\4(a+b+c)^2&\ge3(a^2+b^2+c^2)+9(a+b+c)\\a^2+b^2+c^2+8(ab+bc+ca)&\ge9(a+b+c)\end{align*}
And I'll leave this last inequality up to you. Use the fact that $abc=1$, which haven't been used so far.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$ Prove that for any positive integer n,$$ 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 $$
Could anyone give me a hint on this question? Does it have something to do with Riemann Sums?
| Hints:
$$2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k}+\sqrt{k+1}} < \frac{1}{\sqrt{k}}< \frac{2}{\sqrt{k-1}+\sqrt{k}}=2(\sqrt{k}-\sqrt{k-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$ by induction Prove that $$\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$$ for all $n\in \mathbb{N}$ where $n\geq2$. I've already proven the base case for $n=2$, but I don't know how to make the next step.
Is the base case right?
for $n=2$
$$\sum^{2}_{k=1} \frac{1}{\sqrt{k}}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+2}{2}>\sqrt{2}$$
| Assume $\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$.
Then
$$\sum^{n}_{k=1} \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n+1}}>\sqrt{n} + \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}}.$$
Now since $n\ge 2$, we have $$\frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}} > \frac{\sqrt{n}\sqrt{n}+1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}.$$
So we have
$$\sum^{n+1}_{k=1} \frac{1}{\sqrt{k}} > \sqrt{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why is $x(x+2)(x-3)$ not $x^2+2x(x-3)$? How would you explain the principle why $x(x+2)(x-3)$ is not $x^2+2x(x-3)$ but $(x^2+2x)(x-3)$? This may involve the fundamentals of eliminating parenthesis.
| This is the distributive property for multiplication: $(a+b)(c+d)=a(c+d)+b(c+d)$. In your case:
$$k(a+b)(c+d)=(ka+kb)(c+d)=ka(c+d)+kb(c+d).$$
This comes from:
$$k(a+b) = \overbrace{(a+b)+(a+b)+\dotsc+(a+b)}^{k\text{-times}} = a+b+a+b+\dotsc+a+b=\overbrace{a+a+\dotsc+a}^{k\text{-times}}+\overbrace{b+b+\dotsc+b}^{k\text{-times}}=ka+kb.$$
Generalizing a bit more:
$$(k+p)(a+b) = \overbrace{(a+b)+(a+b)+\dotsc+(a+b)}^{(k+p)\text{-times}} = a+b+a+b+\dotsc+a+b=\overbrace{a+a+\dotsc+a}^{(k+p)\text{-times}}+\overbrace{b+b+\dotsc+b}^{(k+p)\text{-times}}=(k+p)a+(k+p)b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to write a fourier series using periodic boundary conditions Would writing
$$
f(x) = x^2
$$
as a Fourier series using periodic boundary conditions on $-L < x < L$ with a basis of
$$
e^{\frac{i\pi nx}{L}}
$$
be just
\begin{align}\bigl\langle e^{\frac{i\pi nx}{L}},x^{2}\bigr\rangle & = \Bigl\langle e^{\frac{i\pi nx}{L}},\sum_ma_mx^{2}\Bigr\rangle\\
&=\int_{-L}^Lx^{2}e^{\frac{i\pi nx}{L}}dx\\
&=\biggl[\frac{x^{2}}{\frac{in\pi}{L}}-\frac{2x}{(\frac{in\pi}{L})^{2}}+\frac{2}{(\frac{in\pi}{L})^{3}}\biggr]e^{\frac{i\pi nx}{L}}\Biggr|_{-L}^{L}
\end{align}
| You should write
\begin{align}
c_n &= \langle x^2, e^{-in\pi x/L}\rangle\\
&= \frac{1}{2\pi}\int_{-L}^Lx^2\bigg(\cos\Big(\frac{n\pi x}{L}\Big) - \sin\Big(\frac{n\pi x}{L}\Big)\bigg)dx
\end{align}
Since $x^2\sin$ is an odd term, the sine terms integrate to zero so we are left with
$$
c_n = \frac{1}{\pi}\int_0^Lx^2\cos\Big(\frac{n\pi x}{L}\Big)dx = \frac{2\cos(n\pi)}{n^2} = \frac{2(-1)^n}{n^2}
$$
for $n\neq 0$. When $n = 0$, we have $c_0 = \frac{\pi^2}{3}$. By the convergence theorem, $c_n = c_{-n}$ so
$$
x^2 = \frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos\Big(\frac{n\pi x}{L}\Big)
$$
An interesting part to this Fourier series is that by using Parseval's identity for $x\in(-\pi, \pi)$ for $L$, we have
\begin{align}
\frac{\pi^4}{9}+2\sum_{n=1}^{\infty}\frac{4}{n^4} &= \frac{1}{\pi}\int_0^{\pi}x^4dx\\
&= \frac{\pi^4}{5}\\
\sum_{n=1}^{\infty}\frac{1}{n^4} &= \frac{\pi^4}{90}\\
\zeta(4) &= \frac{\pi^4}{90}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$
Then how I proceed.
| As above write $x=a^{-1}$, $y=b^{-1}$, $c=z^{-1}$, so we are required to prove
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2}, \qquad xyz=1.$$
Consider now that $\frac{x}{y+z}$, $\frac{y}{z+x}$, $\frac{z}{x+y}$ are in the same order as $x,y,z$. Then by the Rearrangement Inequality
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{xy}{y+z} + \frac{yz}{z+x} + \frac{zx}{x+y} $$
and also
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{xz}{y+z} + \frac{yz}{z+x} + \frac{zy}{x+y}; $$
averaging the two, we get
$$ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{x+y+z}{2} \ge \frac{3}{2}$$
the last step forllowing from AM-GM.
| {
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"url": "https://math.stackexchange.com/questions/1009022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$ Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$
No idea how to even start.
| Using the suggested factorizations,
and using
$\begin{array}\\
k^2-k+1
&=k(k-1)+1\\
&=(k-1+1)(k-1)+1\\
&=(k-1)^2+(k-1)+1\\
\end{array}
$
(this is really the key),
$\begin{array}\\
\prod_{k=2}^n \dfrac{k^3-1}{k^3+1}
&=\prod_{k=2}^n \dfrac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\\
&=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n (k+1)}\dfrac{\prod_{k=2}^n (k^2+k+1)}{\prod_{k=2}^n (k^2-k+1)}\\
&=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n (k+1)}\dfrac{\prod_{k=2}^n (k^2+k+1)}{\prod_{k=2}^n ((k-1)^2+(k-1)+1)}\\
&=\dfrac{\prod_{k=1}^{n-1} k}{\prod_{k=3}^{n+1}k}\dfrac{\prod_{k=2}^n (k^2+k+1)}{ \prod_{k=1}^{n-1} (k^2+k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{n^2+n+1}{3}\\
&=\dfrac23\dfrac{ n^2+n+1}{n^2+n}\\
&=\dfrac23(1+\dfrac{ 1}{n^2+n})\\
& \to \dfrac23
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the value of this function If $f:\mathbb{R} \rightarrow\mathbb{R}$ is a function which satisfies $$f(x)+f(y)=\frac{f(x-y)}{2}\cdot \frac{f(x+y)}{2}$$ for all $x\in\mathbb{R}$ and $f(1)=3$, what is $f(6)$?
| The function you described does not exist because no value of $f(0)$ makes sense:
*
*$f(0) + f(0) = \frac{f(0 - 0)}{2} \cdot \frac{f(0+0)}{2} = \frac{f(0)^2}4$
implies that $8f(0) = f(0)^2$ which means $f(0)$ can either be $0$ or $8$.
*$3+f(0) = f(1) + f(0) = \frac{f(1-0)}{2}\cdot \frac{f(1+0)}{2} = \frac{f(1)^2}{4} = \frac{9}{4}$ means that $f(0) = \frac{9}{4} - 3 = -\frac34$, which leads to a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$
$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.
$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$
So Solutions of the above equation are $x=6$ and $x=8$
Can we solve it using Derivative or Algebraic way
Thanks
| HINT:
Write $8-x = 2 a$, $\ x-6=2b$. Then $\ a+b=1$ and $a^4+b^4=1$. Draw some level curves.
$\bf{added:}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.
I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
| If you are allowed to use Taylor series, you could start with $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ Now, use $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replacing $y$ by the previous developments, you then have $$\sqrt{1+\tan(x)}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{11 x^3}{48}-\frac{47 x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1+\sin(x)}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{48}+\frac{x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}=\frac{x^3}{4}-\frac{x^4}{8}+O\left(x^5\right)$$ and finally $$\frac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3}=\frac{1}{4}-\frac{x}{8}+O\left(x^2\right)$$ which shows the limit and how it is approached when $x$ goes to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Triangle in Triangle I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.
| I start with coordinates, choosing one edge length equal to $1$ w.l.o.g.:
$$B=(0,0) \qquad A=(x,y) \qquad C=(1,0) \qquad P=\lambda A+(1-\lambda)B$$
with $y\neq 0$ and some $\lambda\in[0,1]$ we'll need to determine.
Then using some computation (which I did using projective geometry, but there are other ways as well) you get
$$
R=(x^3\lambda + xy^2\lambda - x^2\lambda - y^2\lambda + y^2,
y (x^2\lambda + y^2\lambda - x))/(x^2 + y^2 - x)
\\
Q=(x^4\lambda + 2x^2y^2\lambda + y^4\lambda - 2x^3\lambda - 2xy^2\lambda + x^2\lambda + y^2\lambda - y^2,0)/((x - 1)(x^2 + y^2 - x))
$$
The line $QP$ is then orthogonal to $BC$ iff they have the same $x$ coordinate, i.e. if
$$x^4\lambda + 2x^2y^2\lambda + y^4\lambda - 2x^3\lambda - 2xy^2\lambda + x^2\lambda + y^2\lambda - y^2 = \lambda x(x - 1)(x^2 + y^2 - x)$$
Subtract the right hand side, cancel out a common factor of $y^2$ and you end up with
$$\lambda=\frac1{x^2 + y^2 - x + 1}$$
Now let's turn that back into triangle lengths. I had edge lengths
$$a^2 = BC^2 = 1 \qquad c^2 = AB^2 = x^2+y^2 \qquad b^2 = AC^2 = (1-x)^2+y^2$$
so I had
$$a^2+b^2+c^2 = 2x^2+2y^2-2x+2$$
and therefore
$$\lambda=\frac{2a^2}{a^2+b^2+c^2}$$
Note that this formulation of the result is invariant under rescaling: if you scale all the coordinates by the same amount, then that amount cancels out of the above expression. Which fits in nicely with the fact that the convex combination of $P$ describes the same point no matter the scale. So at this point we are justified to drop the assumption of $a=1$ and are therefore back in the general case.
But you were not asking about this factor $\lambda$ but instead about the length $PB$. For this you have to multiply by $AB=c$, so the final result is
$$PB=\frac{2a^2c}{a^2+b^2+c^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Triple integration question involving sketching
Sketch the region $R=\{(x,y,z):0\le z\le 1-|x|-|y|\}$ and evaluate the integral
$$\iiint\limits_R(xy+z^2)dV$$
I REALLY need someone to confirm this!!!!!! This is what I did:
I used wolfram alpha to calculate the triple integrals, http://www.thestudentroom.co.uk/attachment.php?attachmentid=337725&d=1415577300 so there is no mistakes unless I got my bounds wrong which I don't think I did. but someone PLEASE CONFIRM THIS ANSWER!
| Take a region of the 3D graph, and plot points on it. I would recommend the range $x \in \mathbb{R}:[-2,2]$ and $ y \in \mathbb{R}:[-2,2]$. This gives you some positive, and some negative. When you do this, you make a note of the symmetry of the graph, as well as the points of intersection with the xy-plane, since that gives you the bonds of integration.
Per request, here is the solution:
In the 1st quadrant of the XY-plane, the graph appears to be a triangle from $(0,0) \rightarrow (1,0) \rightarrow (0,1)$. Thus, the line of the constraint is $z=1-x-y$. Notice that the aboslute value signs are dropped, because we are dealing with the first quadrant only. This is due to the symmetry of the structure. In general, you would have to separate the integration into four pieces, and solve each separately.
$$\iiint\limits_R (xy+z^2)dV=\iint\limits_A\int_0^{1-x-y} (xy+z^2)dzdA$$
The z-constraint is performed first, which is why it is explicitly stipulated with its constraint equation.
$$\iint\limits_A \int_0^{1-x-y} (xy+z^2)dzdA=\iint\limits_A \left [ xyz+\frac{z^3}{3} \right ]_0^{1-x-y} dA$$
Algebra follows to finish substituting for $z$ and get the integrand solely interms of $x$ and $y$.
$$=\iint\limits_A \left [ xy(1-x-y)+\frac{(1-x-y)^3}{3}-0-0\right ]dA$$
Wolfram Alpha for $(1-x-y)^3$.
$$=\iint\limits_A \left [xy-x^2-xy^2+\frac{1}{3}*(-x^3-3x^2y+3x^2-3xy^2+6xy-3x-y^3+3y^2-3y+1)\right ]dA$$
Now we can apply the constraint to $y$. It is the line of intersection of the $z$ surface with the XY-plane: $y=1-x$
$$=\int \int_0^{1-x} \left [xy-x^2-xy^2+\frac{1}{3}*(-x^3-3x^2y+3x^2-3xy^2+6xy-3x-y^3+3y^2-3y+1)\right ]dydx$$
$$=\int \left [ \frac{x(1-x)^2}{2}-(1-x)x^2-\frac{x(1-x)^3}{3}+\frac{1}{3}*(-(1-x)x^3-\frac{3x^2(1-x)^2}{2}+3x^2(1-x)-x(1-x)^3+3x(1-x)^2-3x(1-x)-\frac{(1-x)^4}{4}+(1-x)^3-\frac{3(1-x)^2}{2}+(1-x)) \right ]dx$$
Now we can apply the true constraints on $x$, which are $\left [0,1\right]$.
$$=\int_0^1 \left [ \frac{x(1-x)^2}{2}-(1-x)x^2-\frac{x(1-x)^3}{3}+\frac{1}{3}*(-(1-x)x^3-\frac{3x^2(1-x)^2}{2}+3x^2(1-x)-x(1-x)^3+3x(1-x)^2-3x(1-x)-\frac{(1-x)^4}{4}+(1-x)^3-\frac{3(1-x)^2}{2}+(1-x)) \right ]dx$$
At this point, the algebra is just tedious. Just integrate after expanding all powers, and substitute.
| {
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"url": "https://math.stackexchange.com/questions/1013127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
| If $k=z$, there are no solution by Fermat's Last Theorem.
When $k=1$ a solution can be $x=1$, $y=2$, $z = 3$, because $1^{3} + 2^{3} = 3^{2}$
Another $k=38$ have a solution can be $x =3$, $y=5$, $z=2$ and there are lots of $k$ can be
solution of that equation, but now I don't know any technique for count how many or what $k$ give a solution.
EDIT:
I found the following solution, I think it isn't the best but:
\begin{equation}
x^{3} + y^{3} = (x+y)[(x-y)^{2}+xy]
\end{equation}
we can think that $(x+y) = z^{2}$ and $k =[(x-y)^{2}+xy]$.
now using $xy = \frac{1}{4}[(x+y)^2 - (x-y)^{2}]$
we can found $3$ solutions,
$1) k =[(x-y)^{2}+xy] \iff x+y$ is a square;
$2) k =[\frac{3}{2}(x-y)^{2}+\frac{(x+y)^2}{2}] \iff \frac{x+y}{2}$ is a square;
$3) k =[3(x-y)^{2}+(x+y)^2] \iff \frac{x+y}{4}$ is a square;
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as a derivative of a function $f$ at some value, let's call that value $c$, as follows :$$f'(c) = \lim_{x\to 0}\frac{f(c + x) - f(c)}{x} = \frac{\sqrt[3]{ax + b} - 2}{x} = \frac{5}{12}$$
*Now I deduce the following: $$f(c + x) = \sqrt[3]{ax + b}$$ and $$f(c) = 2$$
*Use limits as follows: $$\lim_{x\to 0} f(c + x) = f(c) = 2$$ that is, $$\lim_{x\to 0} f(c + x) = \lim_{x\to 0} \sqrt[3]{ax + b} = \sqrt[3]{b} = 2$$ now solve for $b$, $$\sqrt[3]{b} = 2 \Leftrightarrow b = 8$$
*Since I know that $$f(c + x) = \sqrt[3]{ax + 8}$$, I can solve for $a$, which is $$a = \frac{[f(c+x)]^3 - 8}{x} = \frac{[f(c+x)]^3 - [f(c)]^3}{x}$$
*Let $g(x) = [f(x)]^3$, such that $$g'(x) = 3\cdot [f(x)]^2 \cdot f'(x)$$
so $$g'(c) = 3\cdot [f(c)]^2 \cdot f'(c) = 3 \cdot 4 \cdot \frac{5}{12} = 5$$
*Rephrase $g'(c)$ using first principles such that $$g'(c) = \lim_{x \to 0}\frac{g(c + x)- g(c)}{x}= \lim_{x \to 0}\frac{[f(c + x)]^3 - [f(c)]^3}{x} = \lim_{x \to 0} a = 5$$
*Since $a$ is a constant, $\lim_{x \to 0} a = a$, that is, $$a = 5$$
*My solution: $b = 8, a = 5$.
Please have a look at my approach and give me any hints/suggestions regarding the solution and/or steps taken.
| It is fine. However, there is a short proof if you know this
$(1+x)^\frac1n=1+\frac1nx+o(x)$, where $\lim_{x\to 0}\frac{o(x)}{x}=0.$
So $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{b^\frac13 \sqrt[3]{\frac a bx + 1}-2}{x}=\lim_{x \to 0}\frac{(b^\frac13-2)+ \frac13 b^\frac13 \frac abx+b^\frac13o(x)}{x}=\frac{5}{12}.$$
So, $b^\frac13-2=0$ and $\frac13 b^\frac13 \frac ab=\frac{5}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
Calculating the value of $\frac{a-d}{b-c}$ If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of:
$$\frac{a-d}{b-c}$$
Where $a,b,c,d$ are real numbers.
Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this value. Please help.
| This might not be the most mathletic solution, but here it is anyway.
Note that for the given information to make sense, we must have $c\neq d$ and $b\neq d$. From these two observations, we also see that $b\neq c$ since otherwise the two given equations are equal, i.e., $2=3$ which is nonsense. Let $k=\frac{a-d}{b-c}$.
So we can then form a system of three equations in four unknowns: $$a-b-2c+2d=0, \\ a-3b-c+3d=0,\\ a-kb+kc-d=0.$$
So the values of $a,b,c,d$ lie in the nullspace of the matrix $$\begin{bmatrix} 1 & -1 & -2 & 2\\ 1 & -3 & -1 & 3\\1 & -k & k & -1\end{bmatrix}.$$
We can apply the usual row reduction procedure to see that the nullspace of that matrix is equal to the nullspace of $$\begin{bmatrix} 1 & -1 & -2 & 2\\0 & 1 & -\frac{1}{2} & -\frac{1}{2} \\0 & 0 & \frac{k}{2}+\frac{5}{2} & -\frac{k}{2}-\frac{5}{2}\end{bmatrix}.$$
Now if $\frac{k}{2}+\frac{5}{2}\neq 0$, then we can continue the row reduction to obtain $$\begin{bmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1\\0 & 0 & 1 & -1\end{bmatrix}.$$ But this means that $a=b=c=d$, which we know is not the case. Thus $\frac{k}{2}+\frac{5}{2}=0$ must be true, i.e., $k=-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
| Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have
$$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$
Hence,
$$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$
\begin{align}
\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx & = \int_0^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx + \int_{1/2}^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2} (1-x)\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = 2 \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2}\sin(\pi x) \cos^{2k-1}(\pi x)dx\\
& = \frac2{\pi^2}\int_0^1 t^{2k-1} \arccos(t) dt - \frac1{\pi}\int_0^1 t^{2k-1} dt\\
& = \frac1{2\pi^{3/2}} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} - \frac1{2 \pi k}
\end{align}
Use this in $\spadesuit$ and the approrpriate Taylor series will give you the answer. I am away now since I have a class to teach. Will add more details if needed sometime later.
Updated
We have
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}2 \dfrac1{(1-x)^{3/2}}$$
Integrate the above once and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} + \frac{c_1}x$$
To obtain $c_1$, take $\lim_{x \to 0}$, the left hand side is $\Gamma(1/2)$. For the right hand side limit to even exist, we need $c_1 = -\sqrt{\pi}$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} - \frac{\sqrt{\pi}}x$$
Integrate again and divide by $x$ to obtain
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \sqrt{\pi} \left(\dfrac{\log\left(1-\sqrt{1-x} \right) - \log\left(1+\sqrt{1-x} \right)}x \right) - \sqrt{\pi}\frac{\log(x)}x + \frac{c_2}x$$
To obtain $c_2$, take limit $x \to 0$. For the right hand side limit to even exist, we need $c_2 = 2\sqrt{\pi} \log(2)$. Hence,
$$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \dfrac{2\sqrt{\pi}\log2-2\sqrt{\pi} \log\left(1+\sqrt{1-x}\right)}{x}$$
Integrate again to obtain
\begin{align}
\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^3\Gamma(k)} x^{k} = \frac{\sqrt{\pi}}{2} \left(2\text{Li}_2(1/2(\sqrt{1-x}-1)) - 2\text{Li}_2(1/2(\sqrt{1-x}+1))\\
+ \left(\log(1-\sqrt{1-x}) - \log(1+\sqrt{1-x}) \right)\left(\log(1-\sqrt{1-x}) + \log(1+\sqrt{1-x}) -2\log2\right)\right) - \sqrt{\pi} \frac{\log^2(x)}2 + 2\sqrt{\pi} \log2 \log(x) + c_3
\end{align}
Now plug in $x=1$ in the above to get the sum of the first series and the sum of the second series is $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\pi^2}6$$
I shall add more details if needed.
| {
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"url": "https://math.stackexchange.com/questions/1021647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 0
} |
Proving this inequality $\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< (x+\frac{1}{n-1})$ How can I prove this inequality $$\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< \left(x+\frac{1}{n-1}\right)$$ if $n$ and $x$ are positive integer number $$x>=1$$ $$n>1$$
| As A.D. noted, the inequality is wrong. So what is the correct upper bound?
Let
$$ u_{km} = \dfrac{1}{kx}\sqrt[n]{(k x)^n + \sqrt[n]{((k+1) x)^n + \sqrt[n]{\ldots + \sqrt[n]{(mx)^n}}}}$$
with $x \ge 0$, $m \ge k \ge 1$, $n \ge 1 $, so the left side of your inequality is $\lim_{m \to \infty} x u_{1m}$.
We have
$$ u_{km} = \sqrt[n]{1 + \dfrac{k+1}{k^n x^{n-1}} u_{k+1,m}} \le 1 + \dfrac{k+1}{n k^n x^{n-1}} u_{k+1,m}$$
with $u_{mm} = 1$.
We get
$$ u_{1m} \le \sum_{j = 0}^{m-1} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} < \sum_{j = 0}^{\infty} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} $$
Since $n \ge 2$, $(j+1)!/(j!)^n \le (j+1)!/(j!)^2 = (j+1)/j!$ and $x^{n-1} \ge x$ so
$$ u_{1m} < \sum_{j=0}^\infty \dfrac{j+1}{(n x)^j j!} = \left(1 + \dfrac{1}{n x}\right) e^{1/(n x)}$$
Thus a correct inequality is
$$ \sqrt[n]{x^n + \sqrt[n]{(2 x)^n + \sqrt[n]{\ldots }}} < \left(x + \dfrac{1}{n}\right) e^{1/(nx)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
| You can continue, using $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 1-2\sin^2(\frac{x}{2}) \iff \cos x -1 = -2\sin^2(\frac{x}{2})$ after your last step so that $$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2} = \lim_{x\to0} \frac{1+2\Big(\frac{\cos x - 1}{x^2}\Big)}{6} = \lim_{x\to0} \frac{1+2\Big(\frac{-2\sin^2(\frac{x}{2})}{x^2}\Big)}{6} = \lim_{t\to0} \frac{1-\Big(\frac{\sin t}{t}\Big)^2}{6} = \lim_{t\to 0} \frac{1-1}{6} = \color{red}{\fbox{0}}$$
where $t=\frac{x}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the least square solution and optimal least square solution of the linear system Find the least square solution and optimal least square solution of the linear system:
$$x_1 + 2x_3 = 1$$
$$x_2 +3x_3 = 0$$
$$-x_1 + x_2 + x_3 =0$$
$$-x_2 -3x_3 =1$$
Letting $A$ be the matrix of coefficents, I get that $A^*A$ is
$$A^*A =\left(\begin{array}{rrr} 2 & -1 & 1 \\
-1 & 3 & 7 \\
1 & 7 & 23 \end{array}\right)$$
which is not invertible. How can I get the least square solution then? Can someone help me with this please? Thanks.
| You want the least-squares solution of
$$
\left[\begin{array}{rrr}1 & 0 & 2 \\
0 & 1 & 3 \\
-1 & 1 & 1 \\
0 & -1 & -3 \end{array}\right]x
=\left[\begin{array}{r}1 \\ 0 \\ 0 \\ 1\end{array}\right]
$$
What this means is that you want to minimize the distance between the two sides. Let $c_{1}$, $c_{2}$, $c_{3}$ denote the column vectors of the coefficient matrix. The following is orthogonal to $c_1$:
$$
c_{2}' = c_{2}-\frac{(c_2,c_1)}{(c_1,c_1)}c_1=c_2+\frac{1}{2}c_1
=\left[\begin{array}{c} 1/2 \\ 1 \\ 1/2 \\ -1 \end{array}\right]
$$
And the following is orthogonal to $c_1$, $c_2$:
$$
c_{3}'=c_3-\frac{(c_3,c_1)}{(c_1,c_1)}c_1
-\frac{(c_3,c_2')}{(c_2',c_2')}(c_1+\frac{1}{2}c_2) \\
= c_{3}-\frac{1}{2}c_1-\frac{15/2}{5/2}(c_2+\frac{1}{2}c_1) \\\
= c_{3}-2c_1 -3c_2 = \left[\begin{array}{c} 0\\0\\0\\0\end{array}\right].
$$
Now we know that $x_1=-2, x_2=-3, c_3=1$ is in the null space of the original problem, which makes the least squares solution non-unique. So we'll choose a particular least squares solution to minimize $x_1^{2}+x_2^{2}+x_3^{2}$. In other words, there are two minimization problems required to solve this problem.
Let $c_{4}$ denote the column vector on the right. A best solution $x$ is where
$$
c_{4} - \frac{(c_4,c_1)}{(c_1,c_1)}c_1-\frac{(c_4,c_2')}{(c_2',c_2')}(c_2+\frac{1}{2}c_1) \\
= c_4 -\frac{1}{2}c_1-\frac{-1/2}{5/2}(c_2+\frac{1}{2}c_1) \\
= c_4 -\frac{2}{5}c_1+\frac{1}{5}c_2.
$$
Therefore, a least squares solution is
$$
x = \left[\begin{array}{c}2/5 \\ -1/5 \\ 0\end{array}\right]
$$
But we may vary this over the null space of the matrix
$$
x=\left[\begin{array}{c}2/5 \\ -1/5 \\ 0\end{array}\right]
+\alpha\left[\begin{array}{c}2 \\ 3 \\ -1\end{array}\right]
$$
The solution vector of smallest length is where $\alpha$ is chosen so that the above is orthogonal to the second vector on the right:
$$
x=\left[\begin{array}{c}2/5 \\ -1/5 \\ 0\end{array}\right]-\frac{1/5}{14}\left[\begin{array}{c}2 \\ 3 \\ -1\end{array}\right]
= \left[\begin{array}{r}13/35 \\ -17/70 \\1/70\end{array}\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solutions does this have, modulo $p$? $p$ is a prime number, and $p<14$. Furthermore, $x,y\in\mathbb{Z}$ and $0\leq x<13$ and $0\leq y<13$. Let $a,b,c$ denote unknown variables. How many solutions does the following system of equations have, modulo $p$? (that is, two solutions are the same if they are congruent):
\begin{cases}
a+b+c=1\\
a+2b+(x+1)c=y+1\\
a+3b+(x^2+2x+2)c=3y+1
\end{cases}
| You have to compute the determinant:
$$\left|\begin{array}{ccc}
1&1&1\\
1&2&x+1\\
1&3&x^2+2x+2
\end{array}
\right|$$
It is
$$2x^2+4x+4-3x-3-x^2-2x-2+x+1+3-2=x^2+1$$
which is not $0$ for $p\in\{3,7,11\}$
For $p=2$, $p=5$ and $p=13$ the determinant can be $0$. Namely:
*
*If $p=2$, when $x$ is odd
*If $p=5$, when $x\in\{2,3,7,8,12,13\}$
*If $p=13$, when $x=5$ or $x=8$.
In all other cases, the determinant is not $0$ and there is only one solution. To consider the other possibilities, let's apply Gauss-Jordan:
$$\left(\begin{array}{ccc|c}
1&1&1&1\\
1&2&x+1&y+1\\
1&3&x^2+2x+2&3y+1
\end{array}
\right)$$
$$\left(\begin{array}{ccc|c}
1&1&1&1\\
0&1&x&y\\
0&2&x^2+2x+1&3y
\end{array}
\right)$$
$$\left(\begin{array}{ccc|c}
1&1&1&1\\
0&1&x&y\\
0&0&x^2+1&y
\end{array}
\right)$$
We see that the system has no solution when $p\mid x^2+1$ but $p\nmid y$. When $p\mid x^2+1$ and $p\mid y$, the matrix has rank $2$. Therefore, the set of solutions has dimension $3-2=1$. Since the filed is $\Bbb F_p$ which has $p$ elements, there are $p$ solutions in this case.
| {
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"url": "https://math.stackexchange.com/questions/1024004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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First Order Ordinary Differential Equation by Any Method (1R-24) I just need for someone to check my work and suggest a better way to solve this if one exists. I can use any method but not numerical or any other iterative series approximation. The following documents my process of attempting to solve this problem. It does not appear to be separable, reducible to separable such as with an appropriate change of variable like the u-substitution $u=\tfrac{y}{x}$, or linear, or reducible to linear? But is it exact or can it be made to be exact?
$$y' = \frac{y}{x + (y + 1)^2}$$
Rearrange terms and expand then test for total differential (exactness):
$$Mdx + Ndy = 0$$
$$-ydx + (x + y^2 + 2y + 1)dy = 0$$
$$ \frac {\partial M}{\partial y} = -1, \frac {\partial N}{\partial x} = 1$$
It is not exact but can it be made to be exact with an integrating factor in x? Here $P$ is $M$ and $Q$ is $N$. $P$ and $Q$ are used to imply that ODE is not exact but is being worked on to find an integrating factor.
$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{Q} \left[ \frac {\partial P}{\partial y} - \frac {\partial Q}{\partial x} \right]$$
$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{x + y^2 + 2y + 1} \left[ -1 - 1 \right] = \frac {-2}{x + y^2 + 2y + 1}$$
This will not work because of the y-terms. Plus it would be far too complex to use. We need to find a simpler integrating factor. How about deriving one in y? Like before $P$ is $M$ and $Q$ is $N$.
$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{P} \left[ \frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right]$$
$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{-y} \left[ 1 - (-1) \right] = - \frac {2}{y}$$
It appears we may have made a breakthrough in finding an integrating factor in y.
$$ \int \frac {dF}{F} = -2 \int \frac {dy}{y}$$
$$ln|F| = -2 \ln|y|$$
$$F = y^{-2}$$
So far so good. Let us now multiply the rearranged ODE by $F$ and test once more for exactness. As before $P$ is $M$ and $Q$ is $N$.
$$FPdx + FQdy = 0$$
$$y^{-2}(-y)dx + y^{-2}(x + y^2 + 2y + 1)dy = 0$$
$$-\frac {1}{y}dx + \left(\frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}\right)dy = 0$$
$$ \frac {\partial M}{\partial y} = \frac {1}{y^2}, \frac {\partial N}{\partial x} = \frac {1}{y^2}$$
The integrating factor worked. We now have exactness and can finally proceed in solving this ODE. We may begin with either term $M$ or $N$.
$$u(x,y) = \int Mdx = - \int \frac {1}{y}dx = - \frac {x}{y} + k(y)$$
$$ \frac {\partial u}{\partial y} = \frac {\partial \left[-\frac{x}{y} + k(y)\right]}{\partial y} = \frac {x}{y^2} + k(y)' = N = \frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}$$
We are now at the home stretch. We just need to isolate and solve for $k(y)$ and insert it back into our premature solution $u(x,y)$.
$$ k(y)' = 1 + \frac {2}{y} + \frac {1}{y^2}$$
$$ k(y) = \int \left[1 + \frac {2}{y} + \frac {1}{y^2}\right]dy = y + 2\ln|y| - \frac {1}{y} + c$$
$$u(x,y) = - \frac {x}{y} + k(y) = - \frac {x}{y} + y + 2\ln|y| - \frac {1}{y} + c$$
$$y + 2\ln|y| - \frac {1}{y}(x + 1) = c$$
I believe I may have solved it but I just need someone to check my work.
| I think you could go faster to the solution rewriting $$\frac{dy}{dx} = \frac{y}{x + (y + 1)^2}$$ as $$\frac{dx}{dy} = \frac{x + (y + 1)^2}{y}$$ that is to say $$\frac{dx}{dy}-\frac{x}{y} =\frac{ (y + 1)^2}{y}=y+2+\frac{1}{y}$$ which gives for the homogeneous equation $x=Cy$. So, the equation write now $$y \frac{dC}{dy}=y+2+\frac{1}{y}$$ that is to say $$\frac{dC}{dy}=1+\frac{2}{y}+\frac{1}{y^2}$$ and then $$C=-\frac{1}{y}+y+2 \log (y)+K$$ So, finally $$x=K y+y^2+2 y \log (y)-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in excel the software cannot handle it (because of truncation of significant figures) and the results are nonsense.
I thought to approximate the result using the first few terms of a series in increasing powers of $k$. I can multiply out the first few terms and examine the patterns in the following pyramid...
$$(1+k)^1 = k +1 $$
$$(1+k)^2 = k^2 +2k +1 $$
$$(1+k)^3 = k^3 +3k^2 +3k +1 $$
$$(1+k)^4 =k^4+4 k^3+6 k^2+4 k+1$$
$$(1+k)^5 =k^5+5 k^4+10 k^3+10 k^2+5 k+1$$
$$(1+k)^6 =k^6+6 k^5+15 k^4+20 k^3+15 k^2+6 k+1$$
So for example, for $N=3$ we would obtain the sum
$$
S= k^3 +4k^2 +6k +1
$$
The results suggest a solution with a pattern of the form
$$
S = a + bk^1 +ck^2+dk^3...
$$
I can see that $a=N$. The other coefficients increase monotonously and it might be possible to determine a formula for the coefficients from the pattern. Although the general pattern is not convergent, it is possible that for certain restricted ranges of $N$ and $k$ a convergent formula could be obtained. If so then it is possible that a useful approximation of S can be obtained just by evaluating the first few terms in the series.
But is there a well-known general formula for the terms in this series or can one be derived algebraically from the original formula?
UPDATE
following on from the answer by User73985...
$$ S=\sum_{i=1}^N (1+k)^i = N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$
So
$$ S= N + \sum_{j=2}^{N+1}\frac{(N+1)!}{(N+1-j)!j!}k^{j-1}$$
then
$$ S= N
+ \frac{(N+1)!}{(N-1)!2!} k^{1}
+ \frac{(N+1)!}{(N-2)!3!} k^{2}
+ \frac{(N+1)!}{(N-3)!4!} k^{3} +...
$$
giving
$$ S= N
+ \frac{(N+1)(N)}{2!} k^{1}
+ \frac{(N+1)(N)(N-1)}{3!} k^{2}
+ \frac{(N+1)(N)(N-1)(N-2)}{4!} k^{3} +...
$$
thus
$$ S= N
+ \frac{N^2+N}{2} k^{1}
+ \frac{N^3-N}{6} k^{2}
+ \frac{N^4-2 N^3-N^2+2 N}{24} k^{3} +...
$$
For $N=1 to 10,000$ and $k= 2.40242 * 10^{-12}$ this formula can be truncated to
$$ S = N + \frac{N^2+N}{2} k^{1}
$$
and then gives results very close to those expected. Because $k$ is so small relative to $n$ the terms in higher powers of $k$ can be ignored. Note that the coefficient of $k^1$ is consistent with that found by examination of the coefficients in the "pyramid" presented above.
| Using
$$ \sum_1^N (1+k)^i = \frac{(k+1)\,((k+1)^N-1)}{k}$$
(which since this is a geometric series is not hard to prove)
we get
$$ \sum_1^N (1+k)^i = \frac{\left(\sum_{j=0}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$
$$ = \frac{1 + (N+1)k + \left(\sum_{j=2}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$
$$ = N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges or diverges? We will denote the set of prime numbers with $\mathcal P$.
We know that the sum $\sum_{n=1}^{\infty}\frac1n$ and $\sum_{n=2}^{\infty}\frac1{n\ln n}$ diverges. It is also known that $\sum_{p \in \mathcal P} \frac1p$ is also diverges, where the sum runs over the $p$ primes.
How could we decide that $\sum_{p \in \mathcal P} \frac1{p\ln p}$ is converges or not?
| We begin by proving a weaker version of Mertens' first theorem:
$$\left\lvert\sum_{p\leqslant n} \frac{\ln p}{p} - \ln n\right\rvert \leqslant 2\tag{1}$$
for all $n\geqslant 2$.
Although Mertens' first theorem isn't too hard to prove, a complete proof would be too long for this answer, so we only prove
$$\sum_{p\leqslant n} \frac{\ln p}{p} \leqslant 2\ln n\tag{2}$$
for $n\in \mathbb{N}\setminus\{0\}$, which suffices.
For each prime $p\leqslant n$, there are $k(p,n) := \left\lfloor\frac{n}{p}\right\rfloor$ multiples of $p$ that are $\leqslant n$, and hence
$$\prod_{p\leqslant n} p^{k(p,n)} \mid n!$$
and
$$\sum_{p\leqslant n} \left\lfloor \frac{n}{p}\right\rfloor\ln p \leqslant \ln n!$$
for all $n\geqslant 1$. Thus we have
\begin{align}
\sum_{p\leqslant n} \frac{\ln p}{p} &= \frac{1}{n}\sum_{p\leqslant n} \frac{n}{p}\ln p\\
&< \frac{1}{n}\sum_{p\leqslant n} \left(\left\lfloor \frac{n}{p}\right\rfloor + 1\right)\ln p\\
&\leqslant \frac{1}{n}\ln n! + \frac{1}{n}\sum_{p\leqslant n}\ln p\\
&\leqslant \frac{1}{n}\ln (n^n) + \ln n\\
&= 2\ln n.
\end{align}
Using $(2)$, we obtain the estimate
$$\sum_{n < p \leqslant n^2} \frac{1}{p\ln p} = \sum_{n < p \leqslant n^2} \frac{\ln p}{p(\ln p)^2} < \frac{1}{(\ln n)^2}\sum_{n < p \leqslant n^2}\frac{\ln p}{p} \leqslant \frac{2\ln (n^2)}{(\ln n)^2} = \frac{4}{\ln n}$$
for every $n \geqslant 2$. Then
\begin{align}
\sum_{p} \frac{1}{p\ln p} &= \frac{1}{2\ln 2} + \sum_{k=0}^\infty \sum_{2^{2^k} < p \leqslant 2^{2^{k+1}}} \frac{1}{p\ln p}\\
&\leqslant \frac{1}{2\ln 2} + \sum_{k=0}^\infty \frac{4}{\ln 2^{2^k}}\\
&= \frac{1}{2\ln 2} + \frac{4}{\ln 2} \sum_{k=0}^\infty \frac{1}{2^k}\\
&= \frac{1}{2\ln 2} + \frac{8}{\ln 2}\\
&< +\infty.
\end{align}
The obtained bound for the sum is of course ridiculously large, but we were only interested in proving convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 2
} |
Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$ I'm doing some questions from Anton, 8th edition, page 543, question 13. I've found a answer but it does not match with the answer given at the last pages.
Questions asks to solve $\int{\frac{2x^2-9x-9}{x^3-9x}}$
So:
$$\frac{2x^2-9x-9}{x(x^2-9)}$$
then:
$$\frac{2x^2-9x-9}{x(x-3)(x+3)}$$
(using $a(x-x')(x-x'')$)
Continuing:
$$2x^2-9x-9 = \frac{A}{x} + \frac{B}{(x-3)} + \frac{C}{(x+3)}$$
Solving the product:
$$A(x^2-9)+Bx(x+3)+Cx(x-3)$$
Ended up with a system, because: $(A+B+C)x^2+(3B-3C)x+(-9A)$
\begin{cases} A+B+C =2 \\ 3B-3C = -9 \\ -9A = -9\end{cases}
then:
$A = 1$
$B = -1$
$C = 2$
The answer I found:
$$ ln |x| - ln|x-3| + 2 ln|x+3| + C $$
answer given by author:
$$ ln |\frac{x(x+3)^2}{x-3}| + C$$
What am I missing?
| Using properties of logarithm ($\ln a + \ln b = \ln (ab)$) you have
$$\ln |x| - \ln |x-3| + 2 \ln |x+3| = \ln \frac{|x|(x+3)^2}{|x-3|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many zeros does $f(x)= 3x^4 + x + 2 $ have?
How many zeros does this function have?
$$f(x)= 3x^4 + x + 2 $$
| $f(x)$ has $4$ zeros by the fundamental theorem of algebra.
Here is an alternative way to find that none of them is real:
$$ \ \ \ \ \ \ \ \ 3x^4+x+2=0 \ \ \ \ \ (1)\\ 3x^4=-x-2,$$ which, as $0\ne-2$, is equivalent to $$3x^3=-1-\frac{2}{x}.$$
Letting $g(x)=3x^3$ and $\displaystyle h(x)=-1-\frac{2}{x}$ we have $$
\left\{
\begin{array}{c}
x>0 \\
g(x)=h(x)\\
g(x)>0 \\
h(x)<-1 \\
\end{array}
\right.
\text{and} \left\{
\begin{array}{c}
x<0 \\
g(x)=h(x)\\g(x)<0 \\
h(x)>-1 \\
\end{array}
\right. .$$
The first has no solutions whereas the second implies $$\left\{
\begin{array}{c}
-1<g(x)<0 \\
-1<h(x)<0 \\
\end{array}
\right. \leftrightarrow \left\{
\begin{array}{c}
-\frac{1}{\sqrt[3]{3}}<x<0 \\
x<-2 \\
\end{array}
\right. ,$$ which is an inconsistent system. Thus, $(1)$ is impossible in $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
| $14=2^{2^2}-2$and$18=2^{2^2}+2$$13=\frac{22}2+2$$24=\frac{(2^2)!}{\frac22}$$20=\sqrt{{22}^2}-2=(2^2)!-2^2$$22=\sqrt{\left[(2^2)!-2\right]^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 0
} |
The convergence of a recurrcively defined sequence. Let $a_1=\sqrt{2}$ and $a_n=\sqrt{2+a_{n-1}}$ determine the convergence of the sequence and find its limit.
I know the sequence converges to $2$ and i can show this informally. But I don't know how to prove that formally.
| Once proved the convergence by induction as suggested by Henno, you might be interested in seeing directly that the limit is $2$. So let's find $a_n$ (without the dependence from $a_{n-1}$) and then take the limit as $n\to \infty$. Here we have: $$ a_1=\sqrt{2}\\ a_2=\sqrt{2+\sqrt{2}}=\sqrt{\sqrt{2}(\sqrt{2}+1)}=\sqrt[4]{2}\sqrt{\sqrt{2}+1}\\ a_3=\sqrt{2+\sqrt[4]{2}\sqrt{\sqrt{2}+1}}=\sqrt{2^{1/4}\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}=2^{1/8}\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)} \\ a_4=\sqrt{2+2^{1/8}\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}=2^{1/16}\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}} \\ a_5=\sqrt{2+2^{1/16}\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}}= \\2^{1/32}\sqrt{2^{15/16}+\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}} \\ \vdots \\ a_n=2^{1/2^n}\sqrt{2^{1-1/2^{n-1}}+\sqrt{2^{1-1/2^{n-2}}+...+\sqrt{\sqrt{2}+1}}}.$$
The limit is $$\lim_{n\to \infty} a_n = \lim_{n\to \infty} 2^{1/2^n} \times \lim_{n\to \infty} \sqrt{2^{1-1/2^{n-1}}+\sqrt{2^{1-1/2^{n-2}}+...+\sqrt{\sqrt{2}+1}}}= \\ \sqrt{2+\sqrt{2+\sqrt{2+...}}}=L \\ 2+\sqrt{2+\sqrt{2+...}}=2+L=L^2 \\ L^2-L-2=0.$$
Finally, using the quadratic formula we find $L=-1 \vee L=2$. Since $L>0$, it is $L=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a recurrence with diagonalization? Considering the recurrence $F_n=F_{n-1}+3F_{n-2}-3F_{n-3}$ where $F_0=0$, $F_1=1$ and $F_2=2$. Use diagonalization to find a closed form expression for $F_n$.
So I first continued the recurrence to find $F_3=5$, $F_4=8$, $F_5=17$ ... etc
From this is I got a vector with three consecutive terms in the recurrence to be
$u_k = \begin{pmatrix} -3F_{n-3} \\ 3F_{n-2} \\ F_{n-1}\end{pmatrix}$
From here get a matrix $A$ from the terms of the recurrence:
$A = \begin{pmatrix} F_4 & F_3 & F_2 \\ F_3 & F_2 & F_1 \\ F_2 & F_1 & F_0 \end{pmatrix} =
\begin{pmatrix} 8 & 5 & 2 \\ 5 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix}$
Then the action of $A$ on $u_k$ would produce the $u_{k+1}$ term. [Correct me if I am wrong.]
So, $Au_k = \begin{pmatrix} 8 & 5 & 2 \\ 5 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} -3F_{n-3} \\ 3F_{n-2} \\ F_{n-1}\end{pmatrix} =
\begin{pmatrix} -24F_{n-3} + 15F_{n-2} + 2F_{n-1} \\ -15F_{n-3} + 6F_{n-2} + F_{n-1} \\ -6F_{n-3} + 3F_{n-2} + 0 \end{pmatrix}$
Then $u_0 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$
From this point on I am unsure how to follow through with completing the problem. I have followed this Fibonacci Recurrence Problem - Most Helpful to figure out what to do next but have fallen short.
I see that the next step should be to find $\det(A - \lambda I)$ :
$0 = \det(A - \lambda I) = \begin{vmatrix} 8-\lambda & 5 & 2 \\ 5 & 2-\lambda & 1 \\ 2 & 1 & -\lambda \end{vmatrix}$
$=(8-\lambda)\begin{vmatrix} 2-\lambda & 1 \\ 1 & -\lambda\end{vmatrix} - 5\begin{vmatrix} 5 & 1 \\ 2 & -\lambda\end{vmatrix} + 2\begin{vmatrix} 5 & 2-\lambda \\ 2 & 1\end{vmatrix}$
$= (8-\lambda)(-2\lambda^2-1)-5(-5\lambda-2)+2(1+2\lambda)$
$= -16\lambda^{2}-8 + 2\lambda^3 + \lambda + 29\lambda + 12$
$= 2\lambda^3-16\lambda^2+30\lambda+4$
$= 2(\lambda^3-8\lambda^2+15\lambda+2)$
$= 2((\lambda-5)(\lambda-3)\lambda+2)$
From the source above, I can't figure out what the next steps should be? Can anyone help? Thanks!
| Your matrix $A$ is not correct. It should represent the recurrence. You define $u_k = \begin{pmatrix} F_{k+2} \\ F_{k+1} \\ F_{k}\end{pmatrix}$ and the matrix $A$ is such that $u_{k+1}=Au_k$, so $A=\begin{pmatrix} 1 & 3 & -3 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ where the first line is the recurrence and the next two lower the past values. This is the matrix you want to diagonalize, and it has nice eigenvalues. You need to find a matrix $P$ such that $A=P^{-1}DP$ with $D$ diagonal. The diagonal entries of $D$ will be the eigenvalues of $A$ and the nice thing is that $A^n=P^{-1}D^nP$. Powers of a diagonal matrix are rather easier than a general matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
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