Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this?
| Let's start here and simplify the problem.
$$\frac{(x+3)}{(x-1)(x+1)} - \frac{(x-2)}{(x+1)(x+1)}$$
I'm going to call $(x+1)=a$ and $(x-1)=b$
$$\frac{(x+3)}{(b)(a)} - \frac{(x-2)}{(a)(a)}$$
So, on the right side we have (a)(a) in the denominator, and only one in the denominator on the right. So we have to multiply the term on the right by $\frac{a}{a}$.
The term on the left has (b) in the denominator, while the one on the right doesn't. So we're going to to multiply the one on the right by $\frac{b}{b}$.
Let's take those steps, and we get:
$$\frac{(x+3)(a)}{(b)(a)(a)} - \frac{(x-2)(b)}{(a)(a)(b)}$$
See how both denominators are identical!? Great! Now we can put both of the numerators on top of the denominator we worked so hard to get.
$$\frac{(x+3)(a)-(x-2)(b)}{(b)(a)(a)}$$
Now all you have to do is plug back in $a=(x+1)$ and $b=(x-1)$, open up the brackets and collect like terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a simpler proof?
| The cube of a binomial is
$$
(u+v)^3=u^3+v^3+3uv(u+v)
$$
Let us observe that in this case
\begin{align}
&u^3+v^3&&=2a\\
&uv&&=[a^2-(a^2+b^3)]^{1/3}=-b\\
&u+v&&=x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What is the expected value of the product of randomly selected balls
An urn contains four balls numbered 1, 2, 5, and 7. If a person
selects a set of two balls at random, what is the expected value of
the product of the numbers on the balls?
My thoughts:
$E(X)=\sum_{k=1}^{n}a_{k}b_{k}$
$E(X) = E_{1}(X)*E_{2}(X)$
The probability of choosing the first ball is $\frac{1}{4}$, so the expected value of the first ball is $E_{1}(X)=\tfrac{1}{4}\sum(1+2+5+7) = 3.75$
The probability of choosing the second ball is $\frac{1}{3}$(because 3 balls left after the first move). But what would be the $E_{2}(X)$, since we don't know which ball was chosen first and can't say $E_{2}(X)=\tfrac{1}{4}\sum(2+5+7)$
| There's only 6 different ways to draw two balls, and each of those are equally likely, so I would just calculate it as:
$$
\frac{1\cdot 2+1\cdot 5+1\cdot 7+2\cdot 5+2\cdot 7+5\cdot 7}{6}
=\frac{2+5+7+10+14+35}{6}
=\frac{73}{6}
=12\frac{1}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the sum of the $81$ products in the $9 \times 9$ multiplication grid? What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
| Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:
First, show that this is true for $n=1$:
$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$
Third, prove that this is true for $n+1$:
$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$
$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+\frac{n^3+2n^2+n}{2}+\frac{n^3+2n^2+n}{2}+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$
$\frac{n^4+6n^3+13n^2+12n+4}{4}=$
$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$
Please note that the assumption is used only in the part marked red.
Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $z^5 +32 =0$
Solve $z^5 +32 =0$
My attempt : $$z^5 = -32$$
Multiply the powers on both sides by $\frac{1}{5}$
we get $$z = 2 * (-1)^\frac{1}{5}$$
Now I'm stuck at this step I don't know how to proceed. Kindly help.
| We have $$\left(-\dfrac z2\right)^5=1$$
Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
If $x^5-1=0,$
either $x-1=0\iff x=1$
or $x^4+x^3+x^2+x+1=0$
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
as $x\ne0,$ divide both sides by $x^2$ to find $$0=x^2+\dfrac1{x^2}+x+\dfrac1x+1=\left(x+\dfrac1x\right)^2+\left(x+\dfrac1x\right)-1$$
Solve for $x+\dfrac1x$ then solve for $x$
| {
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"url": "https://math.stackexchange.com/questions/1823051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \implies\ ... (x-y)(13x+13y-12)=0\implies x=y\ or\ 13x+13y-12=0$
The first equation has infinite answers and the second has none(since $gcd(13,13)$ does not divide $12$), am I right??
| The solution to this system of equations there. http://www.artofproblemsolving.com/community/c3046h1046718__4
$$\left\{\begin{aligned}&x^2+qy=z^2\\&y^2+qx=v^2\end{aligned}\right.$$
Use this decision.
$$x=2psb^2-a^2p^2$$
$$y=2abp^2-b^2s^2$$
$$q=as(4bp-as)$$
$$z=a^2p^2+2psb^2-abs^2$$
$$v=2abp^2-psa^2+b^2s^2$$
In our case, it is necessary to $b=1$ ; $p=1$ and $(a,s) - (\pm3;\pm1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many words of length $n$ can we make from $0, 1, 2$ if $2$'s cannot be consecutive? How many words we can make from $0,1,2$? The restriction is we can't put the digit $2$ after the digit $2$.
My solution:
I tried to solve it with Inclusion-Exclusion Principle. Count the number of the words without restriction and then I decreased the options that I don't want. The number of ways to make words without restriction $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot \ldots \cdot 3$, so this is $3^n$. And then the number that I don't want to count is:
$$\binom {n-1}{2} - \binom {n-2}{3} + \cdots + (-1)^n \cdot \frac{ (n-(n+1)}{(i+1)!}$$
so my solution is $$3^n - \sum_{i=1}^n-1 \frac{(-1)^n \cdot (n-i)}{(i+1)!}$$
This is not the answer. Can someone help me? What did I do wrong?
Does the problem have any solution with a recurrence relation??
Thanks for your help!
| This answer is based upon the Goulden-Jackson Cluster Method which is a convenient method to derive a generating function for problems of this kind.
We consider words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1,2\}$$ and the set $\mathcal{B}=\{22\}$ of bad words, which are not allowed to be part of the words we are looking for.
We derive a function $F(x)$ with the coefficient of $x^n$ being the number of wanted words of length $n$.
According to the paper (p.7) the generating function $F(x)$ is
\begin{align*}
F(x)=\frac{1}{1-dx-\text{weight}(\mathcal{C})}
\end{align*}
with $d=|\mathcal{V}|=3$, the size of the alphabet and with the weight-numerator $\mathcal{C}$ with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[22])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[22])&=-x^2-\text{weight}(\mathcal{C}[22])x
\end{align*}
and get
\begin{align*}
\text{weight}(\mathcal{C}[22])=-\frac{x^2}{1+x}
\end{align*}
It follows:
A generating function $F(x)$ for the number of words built from $\{0,1,2\}$ which do not contain the subword $22$ is
\begin{align*}
F(x)&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\
&=\left(1-3x+\frac{x^2}{1+x}\right)^{-1}\\
&=\frac{1+x}{1-2x-2x^2}\\
&=1+3x+8x^2+\color{blue}{22}x^\color{blue}{3}+60x^4+164x^5+448x^6\\
&\qquad+1224x^7+3344x^8+9136x^9+\cdots\tag{1}
\end{align*}
We see e.g. the term $22x^3$ indicates there are $\color{blue}{22}$ different words of length $\color{blue}{\text{three}}$ built from $\{0,1,2\}$ which do not contain $22$. Out of the $3^3=27$ different words of length three built from $\{0,1,2\}$ there are precisely five words
\begin{align*}
022,122,222,221,220
\end{align*}
which are to exclude, resulting in a total of $27-5=22$ words.
Using partial fraction decomposition we can derive a representation of $F(x)$ as geometric series expansion. From this representation we can easily obtain the coeffcients of $x^n$ which are the number of wanted words, i.e. the words which do not contain $22$.
We obtain
\begin{align*}
F(x)&=\frac{1+x}{1-2x-2x^2}\\
&=\frac{2+\sqrt{3}}{2\sqrt{3}}\cdot\frac{1}{1-(1+\sqrt{3}x)}-\frac{2-\sqrt{3}}{2\sqrt{3}}\cdot\frac{1}{1-(1-\sqrt{3}x)}\\
&=\frac{2+\sqrt{3}}{2\sqrt{3}}\sum_{n=0}^\infty(1+\sqrt{3})^nx^n-\frac{2-\sqrt{3}}{2\sqrt{3}}\sum_{n=0}^\infty(1-\sqrt{3})^nx^n\\
\end{align*}
and conclude:
The number of words of length $n$ built from $\{0,1,2\}$ which do not contain $22$ is
\begin{align*}
\frac{2+\sqrt{3}}{2\sqrt{3}}(1+\sqrt{3})^n-\frac{2-\sqrt{3}}{2\sqrt{3}}(1-\sqrt{3})^n\qquad\qquad n\geq 0
\end{align*}
which is a sequence starting with
\begin{align*}
1,3,8,22,60,164,448,1224,3344,9136,\ldots
\end{align*}
in accordance with (1).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A variation of Ahmed's integral $\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $ Given that the closed form exist, evaluate the following Integral:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
| Let
$J(a)=\int_0^\infty\frac1{1+t^2} \ln \frac{t^2+2t\sin 2a+1}{t^2-2t\sin 2a+1}dt
$
and establish $J’(a)=8a\csc(2a)$. Note\begin{align}
&I=\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx
=\frac14\int_{0}^{1}\sin^{-1}x \ d\left(\ln \frac{4+2x-x^2}{4-2x-x^2} \right)
\end{align}
Integrate by parts and then substitute $x=\frac{2t}{1+t^2}$ to obtain
\begin{align}
I= &\ \frac\pi8\ln5-\frac14(J(\frac{3\pi}{20})-J(\frac{\pi}{20}))
=\frac\pi8\ln5 -\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}2a\csc(2a)da\\
\overset{ibp}
=&\ \frac\pi8\ln5 - a \ln\tan a\bigg|_{\frac{\pi}{20}}^{\frac{3\pi}{20}}
+\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da\\
=&\ \frac\pi8\ln5 - \frac{3\pi}{20}\ln \tan \frac{3\pi}{20}
+\frac{\pi}{20}\ln \tan \frac{\pi}{20}-\frac25 G\\
\end{align}
where$\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da =-\frac25G$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic manipulation of a limit. What are the algebraic manipulations and steps that makes the limit
\begin{equation}
\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
\end{equation}
equal to
\begin{equation}
\lim_{x\to2}(x^2+2x+4)
\end{equation}
It's probably trivial, I just don't seem to be able to see it.
| Based on @Kushal Bhuyan hint
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ formula
In our case
$$\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
=\lim_{x\to2}\left(\frac{x^3-2^3}{x-2}\right)
$$
Then $x^3-2^3=(x-2)(x^2+2x+2^2)$, therefore:
$$\require{cancel}\lim_{x\to2}\left(\frac{\cancel{(x-2)}(x^2+2x+2^2)}{\cancel{x-2}}\right)
=\lim_{x\to2}(x^2+2x+4)=2^2+2\cdot 2+4=\color{blue}{12}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}$ I have a problem with finding the following limit. I suspect that it should be easy, but really I don't have a clue.
$$
\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}
$$
| Note that if $(n-1)x \leqslant u \leqslant nx \leqslant v \leqslant (n+1)x,$ then
$$\frac{1}{4 + v^2} \leqslant \frac{1}{4 +n^2x^2} \leqslant \frac{1}{4 + u^2}, $$
and
$$\int_{nx}^{(n+1)x}\frac{dv}{4 + v^2} \leqslant \frac{x}{4 +n^2x^2} \leqslant \int_{(n-1)x}^{nx}\frac{du}{4 + u^2}, $$
Summing we get
$$\int_x^\infty \frac{dv}{4 + v^2} \leqslant \sum_{n=1}^\infty \frac{x}{4 +n^2x^2} \leqslant \int_0^\infty \frac{du}{4 + u^2} .$$
Using the squeeze principle,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{x}{4 +n^2x^2} = \int_0^\infty \frac{du}{4 + u^2}.$$
Hence,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{\sin x}{4 +n^2x^2} = \lim_{x \to 0} \frac{\sin x}{x} \sum_{n=1}^\infty \frac {x}{4 +n^2x^2} \\ = \int_0^\infty \frac{du}{4 + u^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck
$$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$
Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$
Edit: x tends to infinity, not to 0. I transcribed wrongly.
| This is simply false.
As $\lim_{x \to 0} x + n = n$ for all $n \in \mathbb{N}$ and the square root is continuous, we have $\lim_{x \to 0} \sqrt{x+n} = \sqrt{n}$.
Then, you must show
\begin{align*}
a + \sqrt{2} b + \sqrt{3}c = 0 \iff a+b+c =0.
\end{align*}
Take $a = 2$, $b = -\sqrt{2}$ and $c=0$, then the LHS is satisfied, but the RHS is not.
With $a=1$, $b=-1$ and $c=0$, the LHS is not satisfied, but the RHS is.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, which will obviously obey the equation $\left(a,\frac{a^2}{a-1}\right)$. The derivative is calculated as
\begin{align*}
f(x) = \frac{x^2}{x-1} = x^2(x-1)^{-1} \Rightarrow f'(x) = 2x(x-1)^{-1} -x^2(x-1)^{-2} = \frac{x^2-2}{(x-1)^2} \Rightarrow f'(a) = \frac{a^2-2}{(a-1)^2}.
\end{align*}
Filling this into the equation we get
\begin{align*}
y - \left(\frac{a^2}{a-1}\right) = \frac{a^2-2}{(a-1)^2}\left(x-a\right).
\end{align*}
Obviously this equation will contain the point $(2,0)$, so we fill it in and obtain the answer:
\begin{align*}
\frac{a^2}{a-1} = \frac{a^2-2}{(a-1)^2}\left(a-2\right) \Rightarrow a^2 +2a-4=0 \Rightarrow (a+1)^2=5 \Rightarrow a = -1\pm\sqrt 5.
\end{align*}
The equations are getting so complicated I'm pretty sure I'm doing it wrong, but I'm not quite sure what I'm doing wrong so I'm wondering if StackExchange could offer some input on my method?
| Your method is correct, but note that, as noted in the comment, your derivative is wrong.
In general: if a line passes thorough the points $P=(x,f(x))$ and $Q=(x_Q,y_Q)$ than its slope is :
$$
m=\frac{f(x)-y_Q}{x-x_Q}
$$
and if the line is tangent to the function $y=f(x)$ at $P$ than this slope is the value of the derivative at this point:
$m=f'(x)$, so we have the equation:
$$
\frac{f(x)-y_Q}{x-x_Q}=f'(x)
$$
where $x$ is the abscissa of the point of tangency.
| {
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"timestamp": "2023-03-29T00:00:00",
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evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1)} + \frac{C}{(x-1)} = \frac{-1}{x}+\frac{1/2}{(x+1)} + \frac{1/2}{(x-1)} $$
Back to the integral:
$$2(-\int_2^\infty\frac{1}{x}+\frac{1}{2}\int_2^\infty\frac{1}{(x+1)} + \frac{1}{2}\int_2^\infty\frac{1}{(x-1)})dx$$
$$\lim_{a\to2..{b\to\infty}}2[(-\ln(x)+\frac{\ln(x+1)}{2} + \frac{\ln(x-1)}{2}]_a^b$$
$$2[(-\ln(b)+\frac{\ln(b+1)}{2} + \frac{\ln(b-1)}{2}]-2[(-\ln(2)+\frac{\ln(3)}{2} + \frac{\ln(1)}{2}] $$
$$(-\infty+\infty+\infty) + (\ln(3/4)) $$
So, the integral does not converge? I am unsure about the last part, if anyone can confirm me the answer or explain why it's wrong... thank you guys!
| Short answer: yes, it does: the denominator doesn't cancel and the asymptotic behavior is $\sim x^{-3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Question regarding roots of a cubic polynomial If $\alpha$, $\beta$ and $\gamma$ are the roots of a cubic equation with
$$\alpha + \beta + \gamma = 1$$
$$\alpha^2 + \beta^2 + \gamma^2 = 2$$
$$\alpha^3 + \beta^3 + \gamma^3 = 3$$
Then find the value of
$$\alpha^4 + \beta^4 + \gamma^4$$
| I will use the notation of this question of mine. The $\text{EXP}$ map gives:
$$ \exp\left(-\sum_{m\geq 1}\frac{p_m}{m}x^m\right) = \sum_{r=0}^{3}(-1)^r e_r x^r \tag{1}$$
hence a polynomial having $\alpha,\beta,\gamma$ as roots is:
$$ p(x)=x^3-x^2-\frac{x}{2}-\frac{1}{6} \tag{2} $$
and the $\text{LOG}$ map gives:
$$ p_4 = -4 [x^4]\log\left(1 - x - x^2/2 - x^3/6\right)=\color{red}{\frac{25}{6}}.\tag{3}$$
Once we have $(2)$, we may also go this way: any root of $p(x)$ fulfils $x^4=x^3+\frac{x^2}{2}+\frac{x}{6}$, hence:
$$ p_4 = p_3+\frac{p_2}{2}+\frac{p_1}{6} = 3+1+\frac{1}{6}.\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac {\sin x}{1-\sin x}-\frac {\sin x}{1+\sin x}\equiv 2\tan^2 x$ I need assistance with proving the following identity:
$$\frac {\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x} \equiv 2\tan^2 x$$
What I have done so far is expanded them:
$$\frac {\sin x\;(1+\sin x)}{(1-\sin x)(1+\sin x)}-\frac {\sin x\;(1-\sin x)}{(1+\sin x)(1-\sin x)}$$
So therefore:
$$\frac {\sin x+\sin^2 x}{1-\sin^2x}-\frac {\sin x-\sin^2 x}{1-\sin^2 x}$$
I'm completely stuck on what to do next. Any pointers would be appreciated. Thanks for your time!
| As $\cos^2x=(1-\sin x)(1+\sin x)$
$$\dfrac{\sin x}{1-\sin x}=\dfrac{\sin x-1+1}{1-\sin x}=-1+\dfrac{1+\sin x}{\cos^2x}=\tan^2x+\sec x\tan x$$
Similarly, $$\dfrac{\sin x}{1+\sin x}=\sec x\tan x-\tan^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
evaluate if integral converge & determine antiderivative The problem is i need to
study the convergence of A and B
and find the antiderivative of C
$$A=\int_0^\infty \frac{\sin(x) +x}{\sqrt x + x^3}dx$$ $$B=\int_0^\infty \frac{1}{\sqrt {e^x-1}(x^2+x^{1/3} )}dx$$
$$C=\int\frac{1}{x}\sqrt\frac{2-x}{2+x}dx$$
in C i try substitution $$U =\sqrt\frac{2-x}{2+x}$$
| hi sami i look if the problem C u need to use substitution 5 time
$$u=\dfrac{1}{x+2}$$
& then after that
$$X=\sqrt{4u-1}$$
&
$$w=X+1$$
&
....
but the final answer is $$\ln\left(\left|\sqrt{-\dfrac{x-2}{x+2}}-1\right|\right)-\ln\left(\sqrt{-\dfrac{x-2}{x+2}}+1\right)+2\arctan\left(\sqrt{-\dfrac{x-2}{x+2}}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$
Could someone please explain to my why this is the case?
Thanks a lot.
| Another approach based on Taylor expansions.
Consider $$y=x(\sqrt{x^2+1} - x)=x^2\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ and remember that, for small $y$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$. Replace $y$ by $\frac{1}{x^2}$ to get $$\sqrt{1+\frac{1}{x^2}}-1=\frac{1}{2 x^2}-\frac{1}{8 x^4}+O\left(\frac{1}{x^5}\right)$$ which makes $$y=\frac{1}{2}-\frac{1}{8 x^2}+O\left(\frac{1}{x^3}\right)$$ showing the limit and also how it is approched
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
\end{bmatrix} $
is at the same time $3D$ rotation matrix and for it the sum of entries in every column (row) is constant (here $-\dfrac{1}{3}+ \dfrac{2}{3} + \dfrac{2}{3} = 1)$.
The same is true if we change the order of columns in it.
For example:
$A_1=\begin{bmatrix}
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\end{bmatrix} $ $A_2=\begin{bmatrix}
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\end{bmatrix} $
Questions:
Is it any systematic way to find other non-trivial (without $0$ and $1$) rotation matrices with this property?
Especially it is interesting whether the above rotation matrices are the only ones with rational entries ?- maybe someone knows other rotation matrices exist where the sum of entries is constant..
and...
Can it be proved in some way that if the sum of entries in columns for a rotation matrix is constant then it should be equal to the length of column vectors?
| This answer is not an independent one, it's just a supplement to Yves'
answer to have made his formula for generating rational values of required matrix more explicit (what is important for me and maybe useful for others)
So we have Yves' equation $ p^2+3g^2=1$ what can be written as $\left({\dfrac{a}{d}}\right)^2+3{\left({\dfrac{b}{d}}\right)^2} =1 $
then $ a^2+3b^2=d^2 $
and $(d-a)(d+a)=3b^2$ so we have $ d-a=3 $ and $d+a=b^2$.
Solution for $a$, $d$ as a $f(b)$: $d=\dfrac{b^2+3 }{2}$
$a=\dfrac{b^2-3 }{2}$.
Now we have generator for numbers $a$ and $d$ assuming $b$ is odd.
Examples (when $b$ is coprime with 3, although it seems not to be neccesary at general case):
$b=5$ : $d=\dfrac{5^2+3 }{2}=14$ $a=\dfrac{5^2-3 }{2}=11$.
$b=7$ : $d=\dfrac{7^2+3 } {2}=26$ $a=\dfrac{7^2-3}{2}=23$.
$b=11$ : $d=\dfrac{11^2+3 }{2}=62 $ $a=\dfrac{11^2-3}{2}=59$.
$b=13$ : $d=\dfrac{13^2+3 }{2}=86 $ $a=\dfrac{13^2-3}{2}=83$. ...
...infinite number of solutions..
Although there is an infinite number of matrices with properties given in the question the matrix listed as first seems still to be unique: it is the sole non-trivial descendant of $4$ (!) families of matrices:
*
*matrices with rational entries,
*rotation matrices,
*matrices with constant row (column) sum of entries,
*and symmetrical ones.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Antiderrivative of ${d^2 y \over dx^2} = 1-x^2$
At any point $(x,y)$ on a curve, ${d^2 y \over dx^2} = 1-x^2$, and an equation of the tangent line to the curve at the point $(1,1)$ is $y=2-x$. Find an equation of the curve.
This is what I've done
$${d^2 y \over dx^2} = 1-x^2
\\ \int dy' = \int (1-x^2)dx
\\y' = x- {x^3 \over 3} +C
\\ {dy\over dx}=x- {x^3 \over 3} +C
\\\int dy =\int (x- {x^3 \over 3} +C)dx
\\y= {x^2 \over 2}-{1\over 3} \cdot {x^4 \over 4}+c_1x + c_2
\\ y= {x^2 \over 2}-{x^4 \over 12}+c_1x +c_2$$
Do I now substitute (1,1) in this? I don't this is right. Someone help me. Thank you!
| Hint:
Since $y=2-x$ is the tangent to the curve at $(1,1)$ we have $y'(1)=-1$, so
\begin{align*}
1-\frac13+c_1&=-1\\
\frac12-\frac1{12}+c_1+c_2&=1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I'm looking for a matrix $M$ with $\det(M)=a^2+b^2+c^2+d^2$ In order to show that $(a^2+b^2+c^2+d^2)(A^2+B^2+C^2+D^2)= \alpha^2+\beta^2+\gamma^2+\delta^2$ with $a,b,c,d,A,B,C,D,\alpha,\beta,\gamma,\delta \in \mathbb Z$.
I would like to find a matrix with determinant $a^2+b^2+c^2+d^2$ in order to multiply the matrix with another with determinant $A^2+B^+C^2+D^2$.
| The matrix, where $i = \sqrt{-1}$,
\begin{pmatrix}
a+ib & -c+id\\
c+id & a-ib
\end{pmatrix}
works. When we multiply two matrices of the above form, we get a matrix of the same form.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove the sum of squares of 3 rationals cannot be 7
Prove there isn't $r_1, r_2,r_3 \in \mathbb{Q}$ so that ${r_1}^2 +
{r_2}^2 + {r_3}^2=7 \tag1$
From (1) we get $a^2 + b^2 + c^2=7n^2 \tag2$ where $a,b,c,n \in \mathbb{N}$. I have tried playing with parity of these numbers, without success.
UPDATE
Suppose $n$ is even. Then either $a, b, c$ are all even or only one of them, let's say $a$ is even. The latest case is not possible because applying modulo 4 to (2) we get $2=0$. So $a, b, c$ are all even. Repeatedly simplifying by 4, we can reduce this case to $n$ odd.
Now suppose $n$ is odd. Then either $a, b, c$ are all odd or two of them, let's say $a,b$ are even. The latest case is not possible because applying modulo 4 to (2) we get $1=3$.
The only case I cannot cover is $a,b,c,n$ all odd.
| This is about 2-adic restrictions. First, odd squares of integers are $1 \pmod 8.$ Integer squares can only be $0,1,4 \pmod 8$ in any case. Therefore the sum of three integer squares cannot be $7 \pmod 8.$
Next, if the sum of three squares is divisible by $4,$ so $x^2 + y^2 + z^2 = k$ with $k \equiv 0 \pmod 4,$ then $x,y,z$ must be even so we can divide through and get integers $\left( \frac{x}{2} \right)^2 + \left( \frac{y}{2} \right)^2 +\left( \frac{z}{2} \right)^2 = \frac{k}{4}.$ This is all you need to deal with $x^2 + y^2 + z^2 = 7 n^2$ in integers.
Also worth mentioning Aubry-Davenport-Cassels, there is a geometric proof that, if a number is the sum of three rational squares, it is also the sum of three integer squares. This is presented in Serre's little book.
About $7$ itself, if we have $u^2 + v^2 + w^2 = k$ with $u,v,w$ not all divisible by $7,$ then we can solve $x^2 + y^2 + z^2 = 49k$ with $x,y,z$ all nonzero $\pmod 7.$ That is, we choose one of $(u,v,w)$ or $(-u,v,w)$ or $(u,-v,w)$ or $(u,v,-w)$ (and rename as $(u,v,w)$ again) so that $u + 2 v + 4 w \neq 0 \pmod 7.$ Then we take
$$ x = 3u+6v - 2w, \; \; \; y = -2u+3v +6w, \; \; \; z = 6u -2v +3 w. $$
All are nonzero $\pmod 7.$ This is the rational orthogonal matrix
$$
\frac{1}{7}
\left(
\begin{array}{rrr}
3 & 6 & -2 \\
-2 & 3 & 6 \\
6 & -2 & 3
\end{array}
\right)
$$
as in PALL 1940
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$5$ numbers add up to 3231.What is the $6$th number? This is Q27 from Australian Maths 2013.
$3$ different non-zero digits are used to form $6$ different $3$-digit numbers.The sum of $5$ of them is $3231$.What is the $6$ th number?
What I tried:
Let $a,b,c$ be the different digits.
$(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b) =3231 $
From there,I can see that
$a+2b+2c =10x +1 $,where $x$ is some integer.
$2a+b+2c =10j+(3-x)$,where $j$ is some integer.
$2a+2b+c =32-j$
Using substitution to sub in the values of $j$ and $x$,
$221a +212b+122c=3231$,which leads me back to where I started from...
| With three digits $a,b,c$, You should be able to get at most six different $3$-digits numbers, and they are: $abc,acb,bac,bca,cab,cba$So when you add them up, the equation should be$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)=3231+n$$where $n$ is the unknown $6$th number.
By quick estimation you can find that,
When $a+b+c=15, n=99$, does not qualify;
When $a+b+c=16, n=321$, does not qualify;
When $a+b+c=17, n=543$, does not qualify;
When $a+b+c=18, n=765$, qualified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$
Prove that
$$
\begin{vmatrix}
1 & a^2 + bc & a^3 \\
1 & b^2 + ac & b^3 \\
1 & c^2 + ab & c^3 \\
\end{vmatrix}
=(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$
My work
I am getting - sign extra , please tell me why I am getting this
| Alternatively, it is easily seen that the determinant of the given matrix is a homogeneous polynomial $P(a,b,c)$ in $a$, $b$, and $c$ of degree $5$. Furthermore, $P(a,b,c)=0$ if any two of the three inputs are equal. Therefore, $P(a,b,c)$ is divisible by $(b-c)(c-a)(a-b)$. Hence, $$P(a,b,c)=(b-c)(c-a)(a-b)\,Q(a,b,c)$$ for some homogeneous quadratic polynomial $Q(a,b,c)$. It is evident that $Q(a,b,c)$ is symmetric in $a,b,c$, whence $Q(a,b,c)=p\,\left(a^2+b^2+c^2\right)+q\,\left(bc+ca+ab\right)$ for some constants $p,q$. As $$-4=P(-1,0,1)=(-1)(2)(-1)\,(2p-q)$$ and $$-30=P(-1,0,2)=(-2)(3)(-1)\,(5p-2q)\,,$$
we get $p=-1$ and $q=0$. That is, $P(a,b,c)=-(b-c)(c-a)(a-b)\left(a^2+b^2+c^2\right)$.
Edit: Just noticed that Jack D'Aurizio had mentioned this polynomial technique.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inverse of matrix with very structured submatrix Does this matrix admit an easy analytic expression for its inverse?
$$\begin{bmatrix}
a_1 & 0 & 0 & 0 & 0 &0&\dots&0 \\
a_2 & 1 & -b & 0 & 0&0&\dots&0 \\
a_3 & 0 & 1 & -b & 0 & 0&\dots&0\\
a_4 & 0 & 0 & 1 & -b & 0 &\dots&0\\
\vdots \\
a_{N-1} & 0 & 0 & 0 & 0 & 0 & \dots & -b \\
a_N & 0 & 0 & 0 & 0 & 0 & \dots & 1
\end{bmatrix}$$
The special thing is that it has a very easy structure, apart from row one/column one.
| If we call your matrix $A$, then
$$
A^{-1} =
\begin{pmatrix}
\frac{1}{a_1} & 0 & 0 & 0 & 0 & \dots \\
-\frac{1}{a_1b^2} \sum_{k=2}^Na_kb^k & 1 & b & b^2 & b^3 & \dots \\
-\frac{1}{a_1b^3} \sum_{k=3}^Na_kb^k & 0 & 1 & b & b^2 & \dots \\
-\frac{1}{a_1b^4} \sum_{k=4}^Na_kb^k & 0 & 0 & 1 & b & \dots \\
\vdots & \vdots & & \ddots & \ddots & \\
-\frac{a_N}{a_1} & 0 & 0 & 0 & \dots & 1
\end{pmatrix},
$$
which you can verify by a direct calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit involving the Sine integral function $$
\mbox{Prove that}\qquad
\lim_{x \to \infty}\left[\vphantom{\large A}%
x\,\mathrm{si}\left(x\right)+ \cos\left(x\right)\right]
= 0
$$
where we define
$$\mathrm{si}\left(x\right) =
- \int^{\infty}_{x}\frac{\sin\left(t\right)}{t}\,\mathrm{d}t
$$
I have no clue how to start. I have verified the result using wolframalpha.
| Define $$Si(x) = \int^x_0 \frac{sin t}{t} dt$$
By the relation $Si(x) = \frac{\pi}{2} + si(x) $
And the asymptotic series expansion of $si(x)$:
$$si(x) = -\frac{\cos x}{x} (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\frac{\sin x}{x} (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7)) $$
Therefore
$$xsi(x)+\cos x = -\cos x (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\sin x (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7))+\cos x \\
= -\cos x (- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\sin x (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7))$$
When x goes up to infinity, $\cos x$ falter between $[-1,1]$ but $x^n$ rise to infinity so $1/x^n$ goes to 0 where $n \geq 0$ is a natural number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| We notice this looks a bit like $(x^2-y^2)^2$, so we write
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= (x^4 - 2x^2y^2+y^4) - 9x^2y^2 \\
&= (x^2-y^2)^2 - (3xy)^2 \\
&= (x^2 + 3xy - y^2)(x^2 - 3xy - y^2).
\end{align*}
It factors further, but no longer over the integers (there will be square roots involved).
We also could have used $(x^2+y^2)^2$ instead of $(x^2-y^2)^2$, to get
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= (x^4 + 2x^2y^2+y^4) - 13x^2y^2 \\
&= (x^2+y^2)^2 - (\sqrt{13}\;xy)^2 \\
&= (x^2 + \sqrt{13}\;xy + y^2)(x^2 - \sqrt{13}\; xy + y^2).
\end{align*}
The two factorizations may look different, but they are the same four linear factors combined in two different ways.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Points on the elliptic curve for Ramanujan-type cubic identities Given the rational Diophantine equation,
$$t^3 - t^2 - \tfrac{1}{3}(n^2 + n)t - \tfrac{1}{27}n^3=w^3\tag1$$
Two points are,
$$t_0 = 0\tag2$$
$$t_2 = \frac{-(1 + 2 n) (1 + 11 n + 42 n^2 + 14 n^3 + 13 n^4)}{9 (7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4)}\tag3$$
Question: How do we find an intermediate point $\color{blue}{t_1}$ between $t_0$ and $t_2$; that is, one with numerator and denominator less than a quartic?
It's the missing piece in this family,
$$\sqrt[3]{t_0+x_1}+\sqrt[3]{t_0+x_2}+\sqrt[3]{t_0+x_3}= \sqrt[3]{z_0}\tag4$$
$$\sqrt[3]{\color{blue}{t_1}+x_1}+\sqrt[3]{\color{blue}{t_1}+x_2}+\sqrt[3]{\color{blue}{t_1}+x_3}= \sqrt[3]{z_1}\tag5$$
$$\sqrt[3]{t_2+x_1}+\sqrt[3]{t_2+x_2}+\sqrt[3]{t_2+x_3}= \sqrt[3]{z_2}\tag6$$
where,
$$z_0=-(2n+1)+3\sqrt[3]{\tfrac{n(n^2+n+1)}{3}}\tag7$$
$$z_2=\frac{-3(2 + n)^3 (1 + n + n^2)}{7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4}\tag8$$
and the $x_i$ are the three roots of,
$$x^3 + x^2 - \tfrac{1}{3}(n^2 + n)x + \tfrac{1}{27}n^3=0\tag9$$
P.S. Once $\color{blue}{t_1}$ is found, and assuming it also has additional properties, it is easy to find $z_1$ as a rational root of a nonic. (The relations found by davidoff303 was just the special case $n=-3$.) There are infinitely many rational points $t_i$ but I am interested in those with the smallest height.
| The cubic
\begin{equation*}
u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0
\end{equation*}
can be shown to be equivalent to the elliptic curve
\begin{equation}
y^2=x^3+1296n^2(n^2+n+1)^2
\end{equation}
by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient MS-DOS version of Derive. The curve has torsion points of order $3$ when $x=0$.
The transformations are straightforward but lengthy, so are not given.
Tests suggest that the rank is greater than $0$ unless $n=1$. They suggest the rank is often greater than $1$.
Using simple experimental results it is possible to show that
\begin{equation}
(-4(n-1)(2n+1), \, 4(n+2)(7n^2+n+1))
\end{equation}
gives a simple non-torsion point on the curve, unless $n=1$.
Applying the transformations to this point and adding the torsion points just gives the values of $t$ not wanted, the quartic one given in the original question and two very similar points
\begin{equation*}
t=\frac{n^2(19n^4+17n^3+24n^2+14n+7)}{(1-n)(31n^4+41n^3+15n^2-7n+1)}
\end{equation*}
and
\begin{equation*}
t=\frac{n(n-1)(31n^4+41n^3+15n^2-7n+1)}{3(2n+1)(13n^4+14n^3+42n^2+11n+1)}
\end{equation*}
Doubling the point above gives a much more complex form. There might be massive cancellations reducing the corresponding value of $t$ but I did not try it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is a better way of calculating this, but my knowledge is not very good (the questions are at the end of the explanation):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor = \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
by Bertrand's postulate $\lfloor \sqrt{p_{i+1}} \rfloor \lt \lfloor \sqrt{2 \cdot p_i} \rfloor$ so it is possible to replace in a first step $\lfloor \sqrt{3} \rfloor$ by $\lfloor \sqrt{2 \cdot 2} \rfloor$:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2 \cdot 2} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
then recursively doing the same with the next prime $\lfloor \sqrt{5} \rfloor \lt \lfloor \sqrt{2\cdot 3} \rfloor \lt \lfloor \sqrt{2\cdot (2 \cdot 2)} \rfloor$, etc. the result is:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$
$\sqrt{2} \lt 2$:
$\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}} \lt 2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} $
Finally, $\sum_{i=1}^n 2^i = (2^{i+1}-1)-2^0 = 2^{i+1}-2$, thus:
$2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} \lt 2 \cdot (2^{(\frac{n}{2}+1)}-2)=2^{(\frac{n}{2}+2)}-4$
So finally the expression would be:
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P \lt 2^{(\frac{n}{2}+2)}-4$
I would like to ask the following questions:
*
*Are the calculations correct or there are gaps / errors?
*Is there a better approach? Thank you!
UPDATE 2016/7/17: Thank you very much for the answers, for the sake of completeness I have sketched the solutions everybody kindly provided, compared with $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$. The reason is that this sequence is expected to grow slower than the sequence of the sum of the first $n$ elements, so it is a good expression to compare with. The only solution that was too big was mine, kindly improved by mathlove, that is why I wanted to ask for better values really under the value of the sum of the first $n$ natural numbers.
Blue: $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$
Red (mathlove) = $\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$
Yellow (Marco Cantarini) = $n\sqrt{n\log\left(n\right)+n\log\left(\log\left(n\right)\right)}-\frac{2\left(n-1\right)\sqrt{n-1}}{3\sqrt{\log\left(2\right)}}+1$
Green (LeGrandDODOM) (leading term only): $\displaystyle\frac 23 n^{\frac 32}\sqrt{\ln n}$
| We can obtain a better bound. We need first some preliminary result:
Proposition 1:$$\sum_{k=1}^{N}\sqrt{k}\in\left(\frac{2}{3}N\sqrt{N},\frac{4N+3}{6}N\sqrt{N}\right).
$$ Proof: Let $$S=\sum_{k=1}^{N}\sqrt{k}
$$ using Abel's summation and the bound $t-1\leq\left\lfloor t\right\rfloor \leq t
$ we have $$S=N\sqrt{N}-\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t}dt>\frac{2}{3}N\sqrt{N}$$ $$S<\frac{4N+3}{6}N\sqrt{N}.\square$$ The second thing we need is the Rosser's theorem and the last thing is the bound $p_{n}<n\log\left(n\right)+n\log\left(\log\left(n\right)\right),n>5
$. Now we can study our sum. We have, using partial summation, $$\sum_{i\leq n}\left\lfloor \sqrt{p_{i}}\right\rfloor \leq\sum_{i\leq n}\sqrt{p_{i}}=n\sqrt{p_{n}}-\sum_{i=1}^{n-1}i\left(\sqrt{p_{i+1}}-\sqrt{p_{i}}\right)
$$ $$=n\sqrt{p_{n}}-\sum_{i=1}^{n-1}i\left(\frac{p_{i+1}-p_{i}}{\sqrt{p_{i+1}}+\sqrt{p_{i}}}\right)\leq n\sqrt{p_{n}}-\sum_{i=1}^{n-1}\frac{i}{\sqrt{p_{i}}}
$$ $$<n\sqrt{p_{n}}-\sum_{i=2}^{n-1}\sqrt{\frac{i}{\log\left(i\right)}}\leq n\sqrt{p_{n}}-\frac{1}{\sqrt{\log\left(2\right)}}\frac{2}{3}\left(n-1\right)\sqrt{n-1}+1
$$ $$\color{red}{<n\sqrt{n\log\left(n\right)+n\log\left(\log\left(n\right)\right)}-\frac{2\left(n-1\right)\sqrt{n-1}}{3\sqrt{\log\left(2\right)}}+1},\, n>5.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
where $\lfloor\,\cdot\,\rfloor$ represents the floor function.
My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?
| There are no solutions. For $x\to196^-$ the function value is $12342$, and for $x=196$ it is $12348$, with no values in between.
| {
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"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $7$?
I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7} + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$.
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?
| By Fermat's little theorem, $3^6\equiv 1 \bmod 7$, so that $3^{35}\equiv 3^5\equiv 5\bmod 7$. Hence $7\mid (3^{35}-5)$, and explicitly, $3^{35}=7\cdot 7147363585571386.$
Edit: "shifting interest to the factorisation": $$3^{35}-5=2\cdot 7\cdot 1729363\cdot 2066472911,$$ using integer factorisation algorithms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| $2xyz$ even $\rightarrow$ $x$ even and $y$,$z$$\in$ odd OR $x$,$y$,$z$$\in$ even
*
*$x=2k$ $y$,$z$$\in$odd:
$4k^2 + y^2 + z^2 = 4kyz$
$4k(k-yz) +y^2 + z^2 = 0$
We have this problem: $4|y^2 + z^2$ so $y$ or $z$ must be even but are odd.
*$x=2k, y=2l, z=2m$:
$4k^2 + 4l^2 + 4m^2 = 16klm$
$k^2 + l^2 + m^2 = 4klm$
And the problem is infinite (problem is the same like at the beginning - 4 instead of 2 nothing change) but $x,y,z$ should be finite.
There is no answer over $\Bbb{N}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Regarding irrationality of $\sqrt{5}$ In the proof of irrationality of $\sqrt{5}$, Hardy and Wright define $x=\frac{\sqrt{5}-1}{2}$. From that I know $1-x=x^2$. But then the authors say that when $1$ is divided by $x$ the remainder is $1-x=x^2$. But I am unable to follow how ?
| When $1$ is divided by $x$, you get $$\frac1x = \frac{x+1-x}{x} = \
\frac xx + \frac{1-x}{x} = 1 + \frac{1-x}{x}$$
The quotient of the division is $1$ and the remainder is the numerator $1-x$ of the "fractional part". Note that $0\leq \frac{1-x}{x} < 1$ here, so this makes sense.
This is analogous to $$5\div 3 = \frac{3+2}3 = \frac33 +\frac23=1 +\frac23$$
so the quotient is $1$ and the remainder is $2$ (the numerator of the "fractional part") since $0\leq\frac23<1$.
| {
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$3\times 3$ matrix with eigenvalues are given
If $\displaystyle P=\begin{bmatrix}
0 & -2 & 3\\
-1& 1& -1\\
a & 2 & b
\end{bmatrix}$ for some $a,b\in \mathbb{R},$ suppose $1$ and $2$ are eigen values of $P$
and $\displaystyle P\begin{bmatrix}
1\\
0\\
-3
\end{bmatrix} =\begin{bmatrix}
3\\
0\\
-3
\end{bmatrix},$ Then $\displaystyle P^4\begin{bmatrix}
1\\
-1\\
0
\end{bmatrix} =$
$\bf{My\; Try::}$ Here $$\displaystyle \begin{bmatrix}
0 & -2 & 3\\
-1& 1& -1\\
a & 2 & b
\end{bmatrix}\begin{bmatrix}
1\\
0\\
-3
\end{bmatrix} =\begin{bmatrix}
3\\
0\\
-3
\end{bmatrix}$$
After multiplication we get $$\begin{bmatrix}
3\\
1\\
a-b
\end{bmatrix}=\begin{bmatrix}
3\\
1\\
-3
\end{bmatrix}$$
So we get $a-3b=-3$
Now how can i solve it after that, Help required, Thanks
| $$tr(P)=\lambda_1+\lambda_2+\lambda_3$$
$$b+1=1+2+\lambda_3\implies\lambda_3=b-2$$
$$det(P)=\lambda_1\lambda_2\lambda_3$$
$$-a-2b-6=2(b-2)\implies a+4b=-2$$
on the other hand
$$a-3b=-3$$
if
$$\displaystyle \begin{bmatrix}
0 & -2 & 3\\
-1& 1& -1\\
a & 2 & b
\end{bmatrix}\begin{bmatrix}
1\\
0\\
-3
\end{bmatrix} =\begin{bmatrix}
-9\\
2\\
-3
\end{bmatrix}$$
| {
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"url": "https://math.stackexchange.com/questions/1852824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)$ is the equations and $Q(x)$ is the quotient and $D(x)$ is the divisor and $R(x)$ is the remainder.
My thought process
so I had set $$ax^2+bx+c=Q(x)\times(x-6)\times(x-P)+0$$
and in the next equation:
$$cx^2+bx+a=Q(x)\times(x-P)(x-Q)+0$$
I really don't know where to go from here
| It is true that $x=0$ is not a root of $ax^2+bx+c$, since if so then $c$ would be zero, and the second polynomial $cy^2+by+a$ is linear with at most one root. Similarly, $y=0$ is not a root of $cy^2+by+a$.
Consider
$$cy^2+by+a = y^2 \left(c+\frac by + \frac a{y^2}\right)$$
If $x=0$ is not a root of $ax^2+bx+c$, i.e. $p\ne 0$, then $y=\dfrac16$ and $y=\dfrac1p$ are the two roots of $cy^2+by+a$. Then
$$pqr = p\cdot \frac 16\cdot \frac1p$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Question on the coefficients of $(1+x+x^2+x^3+x^4)^{496}$
Consider the expansion $$(1+x+x^2+x^3+x^4)^{496} = a_0+a_1x+\cdots+a_{1984}x^{1984}.$$
$\quad$ (a) Determine the greatest common divisor of the coefficients $a_3,a_8,a_{13},\ldots,a_{1983}$.
$\quad$ (b) Prove that $10^{340} < a_{992} < 10^{347}$.
Is there an easier way to solve this and is there a formula for the multisection for just the sum of the coefficients and not including $x^k$?
I thought of using the Multisection formula to prove (a). That is, $$\sum_{k \equiv r \pmod{m}}a_kx^k = \dfrac{1}{m} \sum_{s=0}^{m-1} \epsilon^{-rs} f(\epsilon^s x)$$ where $\epsilon$ is a primitive $m$th root of unity and $\displaystyle f(x) = (1+x+x^2+x^3+x^4)^{496}$.
Thus we have $r = 3, m = 5$ and so \begin{align*}\sum_{k \equiv 3 \pmod{5}}a_kx^k &= \frac{1}{5} \sum_{s=0}^{4} w^{-3s} (1+w^s x+(w^{s}x)^2+(w^{s} x)^3+(w^{s}x)^4)^{496}\\&=\dfrac{1}{5}\sum_{s=0}^4 w^{-3s}\left(\dfrac{(w^s x)^{5}-1}{w^s x-1}\right)^{496}\\&= \dfrac{1}{5}\sum_{s=0}^4 w^{-3s} \left(\dfrac{x^{5}-1}{w^s x-1}\right)^{496}.\end{align*}
| About point $(b)$, we may notice that $992=\frac{1984}{2}$, hence $a_{992}$ is the largest coefficient (we are dealing with palyndromic polynomials) and
$$ a_{992} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(e^{-2iz}+e^{-iz}+1+e^{iz}+e^{2iz}\right)^{496}\,dz $$
is a real integral not so difficult to approximate:
$$ a_{992} = \frac{5^{496}}{2\pi}\int_{-\pi}^{\pi}\left(\frac{1+2\cos(z)+2\cos(2z)}{5}\right)^{496}\,dz $$
gives:
$$ a_{992} \approx \frac{5^{496}}{2\pi}\int_{-\infty}^{+\infty}e^{-496z^2}\,dz = \frac{5^{496}}{8\sqrt{31\pi}}$$
so $a_{992}$ is between $10^{\color{red}{344}}$ and $10^{\color{red}{345}}$.
About point $(a)$, given $f(x)=\left(\frac{1-x^5}{1-x}\right)^{496}=\sum_{n\geq 0}a_n x^n$, by the discrete Fourier transform:
$$ \sum_{k\equiv 3\!\pmod{5}}\!\!\!a_k x^k = \frac{1}{5}\sum_{s=0}^{4}w^{-3s}\left(\frac{x^5-1}{w^s x-1}\right)^{496}=\frac{(x^5-1)^{496}}{5}\sum_{s=0}^{4}\frac{w^{2s}}{(w^s x-1)^{496}} $$
so:
$$ \sum_{k\equiv 3\!\pmod{5}}\!\!\!a_k x^k =(1-x^5)^{496}\sum_{j\geq 0}\binom{495+5j+3}{5j+3}x^{5j+3}.$$
If we have $g(x)=\sum_{n\geq 0}g_n x^n$, then $\frac{g(x)}{1-x}=\sum_{n\geq 0}G_n x^n$, with $G_n=\sum_{k=0}^{n}g_k$. The sequences $\{g_n\}_{n\geq 0}$ and $\{G_n\}_{n\geq 0}$ have the same greatest common divisor, hence in our case it is enough to find the greatest common divisor of the sequence
$$ \left\{\binom{495+5j+3}{495}\right\}_{j\geq 0}. $$
The $\gcd$ is preserved by the forward difference operator, and the $j$-th term of our sequence is a polynomial in $j$ with degree $495$, with leading term $C_{495}\cdot j^{495}$. The $\gcd$ of the whole sequence is so
$$ \gcd\left(C_{495},\binom{495+3}{495}\right) = \color{red}{2\cdot 31\cdot 71\cdot 83}=365366.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A couple of series questions that I just can't figure out (Calc 2)
Show that
$$
\begin{align}
\left(\frac{\pi}{2}\right)^2\left[\int_0^{\pi/2}\cos^{2n}t\ dt-\int_0^{\pi/2}\cos^{2n+2}t\ dt\right]&=\frac{\pi^3}{8}\left[\frac{(2n-1)!!}{(2n)!!}-\frac{(2n+1)!!}{(2n+2)!!}\right]\\[10pt]
&=\frac{\pi^3}{8}\frac{(2n-1)!!}{(2n+2)!!}
\end{align}
$$
Deduce that
$$
0<\frac{(2n)!!}{(2n-1)!!}I_{2n}\le\frac{\pi^3}{8}\frac{1}{2n+2}
$$
I am working on an assignment which requires me to show that
$$
\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}
$$
These are the last two parts that I cannot for the life of me figure out. If someone could expand it so I can understand it better that would be greatly appreciated. Thank you!
Photo of problems
| I'm assuming that you already show that the closed form of this Wallis' integral is $$\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\frac{\pi}{2}\tag{1}
$$ hence $$\frac{\pi^{2}}{4}\left(\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx-\int_{0}^{\pi/2}\cos^{2n+2}\left(x\right)dx\right)
$$ $$=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}-\frac{\left(2n+1\right)!!}{\left(2n+2\right)!!}\right)=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!\left(2n+2\right)-\left(2n+1\right)!!}{\left(2n+2\right)!!}\right)
$$ $$=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!\left(2n+2-2n-1\right)}{\left(2n+2\right)!!}\right)=\color{red}{\frac{\pi^{3}}{8}\frac{\left(2n-1\right)!!}{\left(2n+2\right)!!}}\tag{2}
$$ by the definition of the double factorial. For the second part note that, if $\left|x\right|\leq\frac{\pi}{2}
$, we have $$x^{2}\leq\frac{\pi^{2}}{4}\sin^{2}\left(x\right)
$$ so $$\frac{\pi^{2}}{4}\left(\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx-\int_{0}^{\pi/2}\cos^{2n+2}\left(x\right)dx\right)=\int_{0}^{\pi/2}\cos^{2n}\left(x\right)\frac{\pi^{2}}{4}\sin^{2}\left(x\right)dx\geq I_{2n}
$$ hence from $(2)$
$$I_{2n}\frac{\left(2n\right)!!}{\left(2n-1\right)!!}\leq\frac{\pi^{3}}{8}\frac{\left(2n-1\right)!!}{\left(2n+2\right)!!}\frac{\left(2n\right)!!}{\left(2n-1\right)!!}=\color{red}{\frac{\pi^{3}}{8}\frac{1}{2n+2}}$$
as wanted.
| {
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$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| $123+4+5+6+7+8+9=162>144$. Since any other way of having a three-digit number used in the summation will be strictly larger, we know none using a three-digit number exist. Similarly, we know no arrangement with a number greater than three digits will yield a sum of $144$.
Using only one two-digit number, the largest it could be is $1+2+3+4+5+6+7+89 = 117<144$, so no arrangements with only one two-digit number exist. Similarly, this also implies that no arrangements with all one-digit numbers exist since it will be smaller than even this.
Using two two-digit numbers, we already know that $1+2+3+45+6+78+9=144$ is a solution.
By changing it from $78+9$ to $7+89$ we have effectively increased the sum by $9$, similarly moving where the two-digit number is to the left will decrease the total by $9$, implying that $1+2+34+5+6+7+89=144$ is also a solution.
One can reason that any other pair of locations for the two-digit numbers will be either larger than or less than $144$
Using three two-digit numbers: through trial and error, we search for the one using the smallest available: $12+34+5+6+78+9=144$.
Using the same observation as before, we can shift one of the two digit numbers up and the other down to balance eachother out: $12+3+45+67+8+9=144$
One can reason that no others exist using three two-digit numbers.
We look for four two-digit numbers: the smallest possible is $12+34+56+78+9=189>144$ so no more exist.
The complete list then:
$$1+2+3+45+6+78+9\\
1+2+34+5+6+7+89\\
12+34+5+6+78+9\\
12+3+45+67+8+9$$
for a total of four arrangements
| {
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"url": "https://math.stackexchange.com/questions/1855576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.
Find the smallest $\alpha\in\mathbb{R}$ such that, for all $x,y,z\in\mathbb{R}$, the following inequality holds $$\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1\,.$$
This question is inspired by this thread. According to this related link, it suffices to assume that $x$, $y$, and $z$ are positive. Mathematica indicates that $\alpha=3$ is the lowest possible value, where the sole equality case is $x=y=z=1$. I am looking for a nice solution.
| If $x=y=z=1$ we get $\alpha\geq3$.
But for $\alpha=3$ it's enough to prove our inequality for non-negatives $x$, $y$ and $z$.
Since $3(x^2-x+1)^3-(x^6+x^3+1)=(x-1)^4(2x^2-x+2)\geq0$,
our inequality follows from Holder:
$$3\prod_{cyc}(x^2-x+1)\geq\sqrt[3]{\prod\limits_{cyc}(x^6+x^3+1)}\geq x^2y^2z^2+xyz+1$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/1856820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers!
I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$...
| Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1\implies b ={a^2+1\over a+2}$$
Write $c=a+2$ and then $$b = {(c-2)^2+1\over c} = {c^2-4c+5\over c}= c-4 +{5\over c}$$
so $c\mid 5\implies a+2\in \{-5,-1,1,5\} \implies a\in \{-7,-3,-1,3\}$ and respectively $b\in \{-10,-10,2,2\}$. This should not be difficult to finsih.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and get the right side
My Proof:
\begin{align*}
(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\left(\dfrac{1}{\sqrt{n+1}+\sqrt{n} }\right)\\
&=(-1)^{n}\left(\sqrt{n+1}+\sqrt{n} \right)^{-1}\\
&=(-1)^{n}\left(\sqrt{n}\left(1+\sqrt{1+\dfrac{1}{n}}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\dfrac{1}{n}\right)^{\dfrac{1}{2}} \right)^{-1}\\
&=\mbox{Note that :}\left(1+x\right)^{\alpha}=1+\mathcal{O}\left( x\right) \\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\mathcal{O}\left(\dfrac{1}{n} \right)\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(2+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\mathcal{O}\left(\dfrac{1}{n}\right)\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)\\
\end{align*}
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
*
*Am i right ?
| It is correct to me. In short,
$$
\begin{align}
(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\frac1{\sqrt{n+1}+\sqrt{n}}
\\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+\sqrt{1+1/n}}
\\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+1+O(1/n)}
\\\\&=\frac{(-1)^n}{2\sqrt{n}}\left(1-O(1/n) \right)
\\\\&=\frac{(-1)^n}{2\sqrt{n}}+O\left(\frac1{n^{3/2}} \right).
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solution of differential equation $\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$ Solve the given differential equation.
$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$
I have been trying to solve given differential equation using elementary approaches but no manipulation is a leading to a solvable form. Could someone help me with this?
| $$y'(x)=\frac{1}{xy(x)\left(x^2\sin(y(x)^2)+1\right)}\Longleftrightarrow$$
Write the differential equation in terms of $x$.
Notice: $\frac{\text{d}y(x)}{\text{d}x}\cdot\frac{\text{d}x(y)}{\text{d}y}=1$
$$\frac{1}{x'(y)}=\frac{1}{yx(y)\left(x(y)^2\sin(y^2)+1\right)}\Longleftrightarrow$$
$$x'(y)=y\sin(y^2)x(y)^3+yx(y)\Longleftrightarrow$$
$$x'(y)-yx(y)=y\sin(y^2)x(y)^3\Longleftrightarrow$$
$$\frac{2y}{x(y)^2}-\frac{2x'(y)}{x(y)^3}=-2y\sin(y^2)\Longleftrightarrow$$
Let $q(y)=\frac{1}{x(y)^2}$, which gives $v'(y)=-\frac{2x'(y)}{x(y)^3}$:
$$q'(y)+2yq(y)=-2y\sin(y^2)\Longleftrightarrow$$
Let $r(y)=\exp\left[\int2y\space\text{d}y\right]=e^{y^2}$.
Multiply both sides by $r(y)$:
$$e^{y^2}q'(y)+2yq(y)e^{y^2}=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$
Substitute $2ye^{y^2}=\frac{\text{d}}{\text{d}y}\left(e^{y^2}\right)$:
$$e^{y^2}q'(y)+\frac{\text{d}}{\text{d}y}\left(e^{y^2}\right)q(y)=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$
Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}y}\left(q(y)e^{y^2}\right)=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}y}\left(q(y)e^{y^2}\right)\space\text{d}y=\int-2y\sin(y^2)e^{y^2}\space\text{d}y\Longleftrightarrow$$
$$q(y)e^{y^2}=-\frac{e^{y^2}\left(\sin(y^2)-\cos(y^2)\right)}{2}+\text{C}\Longleftrightarrow$$
$$q(y)=\frac{2\text{C}e^{-y^2}+\cos(y^2)-\sin(y^2)}{2}\Longleftrightarrow$$
$$\frac{1}{x(y)^2}=\frac{2\text{C}e^{-y^2}+\cos(y^2)-\sin(y^2)}{2}\Longleftrightarrow$$
$$x(y)=\pm\frac{e^{\frac{y^2}{2}}}{\sqrt{\frac{e^{y^2}\left(\cos(y^2)-\sin(y^2)\right)}{2}+\text{C}}}$$
So, we get the following solution:
$$x=\pm\frac{e^{\frac{y(x)^2}{2}}}{\sqrt{\frac{e^{y(x)^2}\left(\cos(y(x)^2)-\sin(y(x)^2)\right)}{2}+\text{C}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$? The primary question was:
What is the minimum value for $(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
$\color{red}{\text{But sorry guys! I messed it up! my question is:}}$ (Edited)
What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
I feel like this needs to deal with AM-GM inequality but I don't know how to apply that onto this case. It is easy but like I still don't know how to do it! Can someone help me with it?
| Given $$ \frac{(1-a)(1-b)(1-c)}{abc}$$
Now Using $a+b+c=1\;,$ We get
$$\frac{(b+c)(c+a)(a+b)}{abc} = \left[\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \left(\frac{c}{b}+\frac{a}{b}\right)\cdot \left(\frac{b}{c}+\frac{a}{c}\right) \right]$$
Now Using $\bf{A.M\geq G.M},$ We get $$\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \left(\frac{c}{b}+\frac{a}{b}\right)\cdot \left(\frac{b}{c}+\frac{a}{c}\right)\geq 2\sqrt{\frac{bc}{a^2}}\cdot 2\sqrt{\frac{ca}{b^2}} \cdot 2\sqrt{\frac{ab}{c^2}} = 8$$
and equality hold when $\displaystyle a=b=c = \frac{1}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
How to decompose $x^3-1$ I need to decompose $x^3-1$, I know the Binomial theorem, and finding roots of a polynomial, how should I approach this?
| You have $(x-1)(x^2+x+1)$ now we need a root for the second polynomial.
We have $x^2+x+1=0\iff x^2+x+\frac{1}{4}=\frac{-3}{4}\iff(x+\frac{1}{2})^2=\frac{-3}{4}\iff x+\frac{1}{2}=\frac{\pm\sqrt{-3}}{2}$
$\iff x=\frac{\pm\sqrt{-3}-1}{2}$
So the polynomial is equal to:
$(x-1)(x+\frac{1+\sqrt{3}i}{2})(x+\frac{1-\sqrt{3}i}{2})$
| {
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"url": "https://math.stackexchange.com/questions/1863240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equation with respect to $x$,
$2b^2x+2a^2yy'=0\\
y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
Differentiating again with respect to x,
$y''=-\frac{b^2}{a^2}{\cdot}\frac{y-xy'}{y^2}$
a) At $(0,b)$
$y'=0\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{b-(0)(0)}{b^2}=-\frac{b}{a^2}$
$\displaystyle{\kappa=\frac{|-\frac{b}{a^2}|}{\left[1+\left(0\right)^2\right]^\frac{3}{2}}=\frac{b}{a^2}}$
b) At $(a,0)$
$y'=\infty\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{\frac{y}{y'}-x}{\frac{y^2}{y'}}=\infty$
I am not sure, if my solution to part (b) is correct. The book says, the answer must be $\frac{a}{b^2}$.
| Alternatively, observe that the given equation can be parametrized as follows: $$(x(t), y(t)) = (a \cos t, b \sin t).$$ Then the curvature is given by $$\kappa(t) = \frac{|x'(t) y''(t) - x''(t) y'(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}.$$ We compute $$(x'(t), y'(t)) = (-a \sin t, b \cos t), \\ (x''(t), y''(t)) = (-a \cos t, -b \sin t);$$ hence $$\kappa(t) = \frac{|ab|}{(a^2 \sin^2 t + b^2 \cos^2 t)^{3/2}}.$$ Since the choice $t = \pi/2$ corresponds to $(0,b)$, and the choice $t = 0$ corresponds to $(a,0)$, it is now trivial to obtain $\kappa(0) = |ab|/b^3 = a/b^2$, and $\kappa(\pi/2) = |ab|/a^3 = b/a^2$, for $a, b > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using CS inequality to find maximum of a function I am trying to us Cauchy-Schwarz inequality to find the maximum of:
$$|(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)|$$
Where $a$, $b$, and $c$ are real numbers, and $a+b+c=0$ and $a^2+b^2+c^2=2$.
What I've tried:
$$a_1=a,\ a_2=b,\ a_3=c$$
$$b_1=b\sqrt{a-b}, \ b_2=c\sqrt{b-c}, \ b_3=a\sqrt{c-a}$$
Which gives:
$$(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)\leq(a^2+b^2+c^2)(a-b^3+b-c^3+c-a^3)$$
And I can rearrange and apply the constraints to get
$$(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)\leq2(-b^3-c^3-a^3)$$
But I'm not sure where to go from here. How can I use CS to get to an answer here?
| Let $a=1$, $b=-1$ and $c=0$. Hence, we get a value $2$.
We'll prove that it's a maximal value.
Indeed, since the condition gives $ab+ac+bc=-1$, we obtain:
$|\sum\limits_{cyc}(a^3b^2-a^3c^2)|=|(ab+ac+bc)(a-b)(a-c)(b-c)|=|(a-b)(2a+b)(a+2b)|$.
Since the condition gives $a^2+ab+b^2=1$, it remains to prove that
$$4(a^2+ab+b^2)^3\geq(a-b)^2(2a^2+5ab+2b^2)^2$$
which is $(a+b)^2a^2b^2\geq0$. Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the Inequality $\frac{1}{1-x}-\frac{x(3-x)(2-x)(13x^4-50x^3+89x^2-84x+36)}{4(1-x)(2x(1-x))^2}<1$ Can anyone suggest any hints to prove the following inequality:
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
for all $x \in (0,1)$?
| $$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{1}{1-x}-1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{x}{1-x}.$$
For $x\in(0,1)$ we can prove that
$$\frac{(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(2x(1-x))^2} > 1,$$
or
$$(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36) - 4(2x(1-x))^2 >0.$$
Taking in account that for $x\in(0,1)$
$$x(1-x)\leq \dfrac14,\quad 3-x>2,\quad 2-x>1,$$
$$13x^4 - 50x^3 + 89x^2 - 84x + 36 = 0.5x^4 + 12.5(x-1)^4 +\dfrac1{14}(14x-17)^2+\dfrac{20}7\geq \dfrac{20}7,$$
easy to prove the required inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated
| $$\tan ^{ -1 } \left( \frac { \sqrt { 4+x } +\sqrt { 4-x } }{ \sqrt { 4+x } -\sqrt { 4-x } } \right) =\tan ^{ -1 } \left( \frac { { \left( \sqrt { 4+x } +\sqrt { 4-x } \right) }^{ 2 } }{ \left( \sqrt { 4+x } -\sqrt { 4-x } \right) \left( \sqrt { 4+x } +\sqrt { 4-x } \right) } \right) =\\ =\tan ^{ -1 }{ \left( \frac { 4+x-2\sqrt { 16-{ x }^{ 2 } } +4-x }{ 2x } \right) = } \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) } \\ \frac { d }{ dx } \left( \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) } \right) =\frac { 1 }{ 1+{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) }^{ 2 } } \frac { \left( \frac { 2x }{ \sqrt { 16-{ x }^{ 2 } } } \right) 2x-2\left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }{ 4{ x }^{ 2 } } =\\ =\frac { 4{ x }^{ 2 } }{ 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }^{ 2 } } \frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right) }{ 4{ x }^{ 2 }\left( \sqrt { 16-{ x }^{ 2 } } \right) } =\frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right) }{ \left( \sqrt { 16-{ x }^{ 2 } } \right) \left( 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }^{ 2 } \right) } $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics problem involving binomial coefficient I found this interesting problem in a Romanian mathematical magazine while preparing for the USAMO. Let $k$ be a non-zero natural number. Determine $x,y,z \in \Bbb N$ such that
$$\binom {z+k}{x+y} - \binom {z}{x} \le k \space and \space 2x+y \le z.$$
| I suspect you are looking for less of a brute force argument than those above.
Firstly, note that $(x,y,z)=(0,0,n)$ is a solution for $n \geq 0$ (for any appropriate convention for $0$ choose $0$), since the LHS of the first inequality is just $0$. Therefore, assume $x+y \geq 1$.
A useful identity is obtained as follows: consider partitioning a set of size $z+k$ into two subsets $A$ and $B$ of sizes $z$ and $k$ respectively. To choose $x+y$ objects from the $z+k$, we can independently choose $n$ objects from $A$ and $m$ objects from $B$ such that $n+m=x+y$, then sum over the possible values of $n$:
$$\binom{z+k}{x+y}=\sum_{j=0}^N\binom{k}{j}\binom{z}{x+y-j},$$
where $N = \min\{x+y, k\}\geq 1$. Note that the second given inequality gives $z\geq x+y$ so these are the correct limits. Applying this inequality further, we obtain:
$$\binom{z}{x}\leq \binom{z}{x+y}; \text{and}$$
$$\binom{z}{r}\geq \binom{2x+y}{r}.$$
Consequently,
$$\binom{z+k}{x+y}-\binom{z}{x}\geq \sum_{j=1}^N\binom{k}{j}\binom{2x+y}{x+j}\geq k.$$
Therefore for the first given inequality to hold, we must have $N=1$ and
$$\binom{2x+y}{x+1}=1;$$
that is, $x+1=2x+y$ and so $x+y=1$, which automatically ensures $N=1$ for every $k$.
It is then straightforward to check the two cases:
$x=1$, $y=0 \implies$
$$\binom{z+k}{x+y}-\binom{z}{x}=\binom{z+k}{1}-\binom{z}{1}=k.$$
$x=0$, $y=1 \implies$
$$\binom{z+k}{x+y}-\binom{z}{x}=\binom{z+k}{1}-\binom{z}{0}=z+k-1.$$
In summary the solutions are:
$$\color{red}{(0,0,n); (1,0,n+2) \text{ and } (0,1,1) \text{ for } n\geq 0}$$
as seen in the other answers.
| {
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power series representation of a function by differentiation Find power series representation for $\frac{1}{(7+x)^2}$
What I tried...
$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
$$\frac{1}{7(1-(-{x \over7}))}=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{7^{n+1}}$$
$${d\over dx}(\frac{1}{7+x})={d\over dx}(\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{7^{n+1}}$$
$$\frac{1}{(7+x)^2}=\sum_{n=1}^{\infty}\frac{(-1)^nnx^{n-1}}{7^{n+1}}$$
$$\frac{1}{(7+x)^2}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(n+1)x^n}{7^{n+2}}$$
The correct answer:$$\sum_{n=0}^{\infty}\frac{(-1)^{n}(n+1)x^n}{7^{n+2}}$$
I don't understand why we are not adding +1 to the $(-1)^n$ term when we reduce n from 1 to 0.
| Hint: Instead of differentiation you could also apply the formula for the binomial series expansion with $\alpha=-2$
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1\tag{1}
\end{align*}
We obtain
\begin{align*}
\frac{1}{(7+x)^2}&=\frac{1}{7^2}\cdot\frac{1}{(1+\frac{x}{7})^2}\\
&=\frac{1}{7^2}\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{x}{7}\right)^n\qquad\qquad |x|<7\tag{2}\\
&=\frac{1}{7^2}\sum_{n=0}^\infty\binom{n+1}{n}(-1)^n\left(\frac{x}{7}\right)^n\tag{3}\\
&=\sum_{n=0}^\infty\frac{(-1)^n(n+1)}{7^{n+2}}x^n\tag{4}\\
\end{align*}
and the claim follows.
Comment:
*
*In (2) we apply the binomial expansion series with $\alpha=-2$ valid for $\left|\frac{x}{7}\right|<1$
*In (3) we use the binomial identity
$
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
$
*In (4) we use $\binom{n+1}{n}=\binom{n+1}{1}=n+1$
| {
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"url": "https://math.stackexchange.com/questions/1869005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| using the formula $$a^2-b^2=(a-b)(a+b)$$ we obtain
$$(x-1)^2-(x-5)^2=(x-1-x+5)(x-1+x-5)=4(2x-6)=8(x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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APICS Mathematics Contest 1999: Prove $\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$ is a constant function of $x$ This is question 3 from the APICS Mathematics Competition paper of 1999:
Prove that $$\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ is a constant function of $x$.
Expanding it seems rather daunting, in particular the last term, and nothing I've tried has been useful towards cancelling terms out.
It was assigned in a pre-calculus course, so it should be possible to solve without using derivatives. However, showing that $f'(x)=0$ would obviously be a valid solution.
Any ideas are greatly welcome.
| Let
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$
Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$.
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(-\beta -x\right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)+\sin\left(x+\beta \right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)-\color {red}{\sin\left(\alpha-\beta +x+\alpha\right)\sin(x+\beta)}$$
Now use $2\sin A \sin B = \cos (A-B) - \cos (A+B)$.
$$E=\sin^2(x+\alpha)-\frac{1}{2}\left[\cos(2\alpha-2\beta)-\cos(2x+2\alpha)\right]$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\color{green}{\cos(2x+2\alpha)}$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\left[1-2\sin^2(x+\alpha)\right]$$
$$E=\frac{1}{2}-\frac{1}{2}\cos(2\alpha-2\beta)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding true roots of $x^2 + 10^9x +1$ using the Quadratic Formula I've been given a hint to rearrange the quadratic formula to:
$$b\sqrt{1-\frac{4ac}{b^2}}$$
but am still dealing with a numerical error at $4/10^{18}$
Can I solve this numerically with the Taylor Series?
| Almost as H. H. Rugh answered,
consider the case of $$x^2+a x+1=0$$ where $a$ is a huge number.
The roots are given by $$x_{1,2}=-\frac{1}{2} \left(a\pm\sqrt{a^2-4}\right)=-\frac{a}{2} \left(1\pm\sqrt{1-\frac 4{a^2}}\right)$$ Now, using Taylor series for small $y$ (or the generalized binomial expansion)$$\sqrt{1-y}=1-\frac{y}{2}-\frac{y^2}{8}-\frac{y^3}{16}-\frac{5 y^4}{128}+O\left(y^5\right)$$ replace $y$ by $\frac 4{a^2}$ to get $$x_1=-a+\frac{1}{a}+\frac{1}{a^3}+\frac{2}{a^5}+O\left(\frac{1}{a^6}\right)$$ $$x_2=-\frac{1}{a}-\frac{1}{a^3}-\frac{2}{a^5}+O\left(\frac{1}{a^6}\right)$$
| {
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Find all functions that satisfy $f(\frac{x+4}{1-x}) + f(x) = x$ I found the following task in a book and I would be interested if someone has an idea to solve it:
Find all the functions $f$ that satisfy $f(\frac{x+4}{1-x}) + f(x) = x$.
My ideas:
Assuming that $f$ is a power series or making a substitution.
I tried several approaches, but without success.
| Hint:
Consider $T(x+1)=\dfrac{T(x)+4}{1-T(x)}$ ,
Let $T(x)=U(x)+1$ ,
Then $U(x+1)+1=\dfrac{U(x)+5}{-U(x)}$
$U(x+1)+1=-1-\dfrac{5}{U(x)}$
$U(x+1)=-2-\dfrac{5}{U(x)}$
Let $U(x)=\dfrac{V(x+1)}{V(x)}$ ,
Then $\dfrac{V(x+2)}{V(x+1)}=-2-\dfrac{5V(x)}{V(x+1)}$
$\dfrac{V(x+2)}{V(x+1)}=-\dfrac{2V(x+1)+5V(x)}{V(x+1)}$
$V(x+2)+2V(x+1)+5V(x)=0$
$V(x)=\theta_1(x)(-1+2i)^x+\theta_2(x)(-1-2i)^x$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$V(x)=\theta_1(x)e^{x\ln(-1+2i)}+\theta_2(x)e^{x\ln(-1-2i)}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$V(x)=\theta_1(x)e^{\frac{x\ln5}{2}+(\pi-\tan^{-1}2)ix}+\theta_2(x)e^{\frac{x\ln5}{2}-(\pi-\tan^{-1}2)ix}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$V(x)=\Theta_1(x)5^\frac{x}{2}\sin((\tan^{-1}2)x)+\Theta_2(x)5^\frac{x}{2}\cos((\tan^{-1}2)x)$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$\therefore T(x)=\dfrac{\Theta_1(x+1)5^\frac{x+1}{2}\sin((\tan^{-1}2)(x+1))+\Theta_2(x+1)5^\frac{x+1}{2}\cos((\tan^{-1}2)(x+1))}{\Theta_1(x)5^\frac{x}{2}\sin((\tan^{-1}2)x)+\Theta_2(x)5^\frac{x}{2}\cos((\tan^{-1}2)x)}+1$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$T(x)=\dfrac{\Theta(x)\sqrt5\sin((\tan^{-1}2)(x+1))+\sqrt5\cos((\tan^{-1}2)(x+1))}{\Theta(x)\sin((\tan^{-1}2)x)+\cos((\tan^{-1}2)x)}+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
Hence $f(2-2\tan((\tan^{-1}2)(x+1)))+f(2-2\tan((\tan^{-1}2)x))=2-2\tan((\tan^{-1}2)x)$
| {
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How to evaluate $x^6+x^4+x^3+x^2+1=0$? There is a hint in the question, use the factorization of $x^5+x+1$.
| How about first multiplying by $(x-1)$ ?
$$(x-1)(x^6+x^4+x^3+x^2+1)= x^7-x^6+x^5 - x^2+x-1= (x^2-x+1)(x^5-1)$$
and carry on from there?
Added: Not sure if worth mentioning, but doing the same on $x^5+x+1$ requires an extra little trick:
$$ (x-1)(x^5+x+1)= x^6-x^5 (+x^3 - x^3) +x^2-1=(x^3-x^2+1)(x^3-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\lim_{(x,y) \to (0,0)} (x^2 +y^2 -x^3 y^3)/(x^2 +y^2) =1$ How can I go about proving that $$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1 ?$$
I checked some lines along $x, y$ and $x=y$ and it all gave $1$
| The first step,
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$
Now study this and find $a$
$1-a=1-\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2}$
$a=\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2}$
$\frac {1}{a}=\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2}{x^3y^3} $
$=\lim_{(x,y) \to (0,0)} \frac{x^2}{x^3 y^3}+\frac{y^2}{x^3 y^3}=\lim_{(x,y) \to (0,0)} \frac{1}{xy^3}+\frac {1}{x^3y}=\infty$
Then $a$ must be $0$
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1-a = 1$
| {
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Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that $(217)^\frac{1}{3} \simeq 6.0092$
My attempt :
$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$
$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$
$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$
$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$
How can I proceed after this ? Is there another method ? Is my method correct ?
| If we MUST use the binomial coefficients, this becomes rather tedious:
$$\frac{1+x^2+x}{(1-x)^2}=\frac{(1-x)^2+3x}{(1-x)^2}=1+\frac {3x}{(1-x)^2}$$$$=1-3x(1-x)^{-2}=1-3(1-x)^{-1}+3(1-x)^{-2}$$
$$=1-3\sum\binom{-1}{r}(-1)^rx^r+3\sum\binom{-2}{r}(-1)^rx^r$$
$$=1+3\sum \left(-\binom{-1}{r}+\binom{-2}{r}\right)(-1)^rx^r$$
Let's try to prove that the binomial difference if equal to $(-1)^r\times r$. For $r=0$ the sum is $0$, for $k=1$ the sum is $-1$, assuming this is true for $r=m$, let's check for $r=m+1$: $$-\binom{-1}{m+1}+\binom{-2}{m+1}=-\frac{-1-m}{1+m}\binom{-1}{m}+\frac{-2-m}{1+m}\binom{-2}{m}$$ $$=(-1)m(-1)^m-\frac 1{m+1}\binom{-2}{m}=m(-1)^{m+1}+(-1)^{m+1}$$
So the expansion becomes $$1+\sum_r3rx^r$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.
My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int \frac{\ln ({t})}{(1-t^2)^{3/2}} dt.$$
After this what to do?
| Using the substitution $x=\sec(\theta)$, we get
$$
\begin{align}
\int\frac{x\log(x)}{\left(x^2-1\right)^{3/2}}\,\mathrm{d}x
&=-\int\log(x)\,\mathrm{d}\left(x^2-1\right)^{-1/2}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\frac{\mathrm{d}x}{x\sqrt{x^2-1}}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\mathrm{d}\theta\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\sec^{-1}(x)+C
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Real function with complex antiderivative $\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}$? Consider this indefinite integral (I'm interested in the interval $x>0$):
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx$$
By substitution:
$$u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}, \qquad x=\frac{1}{2u \sqrt{u^2-1}}$$
We get a closed form antiderivative:
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \tanh^{-1} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}} +C$$
Now the inverse hyperbolic tangent is real only for the argument in $(-1,1)$. But in our case the argument for all real $x$ is $ > 1$.
How can the function with real values on $x>0$ have a complex antiderivative?
Edit
To be clear, this is the correct antiderivative. I.e. by differentiating it we get the function under the integral.
How would you explain this without appealing to complex analysis, i.e. branches? This is a real valued function for positive $x$, so this can be given as an assignment to a first year calculus student for example
Edit 2
The derivation of the antiderivative after the substitution:
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \int \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}} \frac{dx}{\sqrt{x^2+1}}=\sqrt{2} \int u \frac{dx}{\sqrt{x^2+1}}$$
$$\sqrt{x^2+1}=x(2u^2-1)$$
$$dx=-x\frac{2u^2-1}{u(u^2-1)}du$$
$$\frac{dx}{\sqrt{1+x^2}}=-\frac{du}{u(u^2-1)}=\frac{du}{u(1-u^2)}$$
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \int \frac{du}{1-u^2}=\sqrt{2} \tanh^{-1} u$$
| Consider
$$
u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}>
\sqrt{\frac{1}{2}+\frac{1}{2}}=1
$$
Assuming your substitutions are correct, the integral becomes
$$
\sqrt{2}\int\frac{1}{1-u^2}\,du=
\frac{\sqrt{2}}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\,du=
\frac{\sqrt{2}}{2}\log\left|\frac{1+u}{1-u}\right|+c
$$
You cannot apply the substitution $u=\tanh v$, as $u>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| For $n=3$ reminder is $5$.
If $n$ is even, then $2^n+n^2$ is also even, hence not prime.
Therefore, we should only consider odd numbers $n$.
In fact, we should consider only cases $n=6k\pm 1$, $n=6k\pm 3.$
Since $$2^n+n^2=2^n+1+(6k\pm1)^2-1=3(2^{n-1}-2^{n-2}+\dots-1)+6T=0\mod 3,$$ this number is not prime for $n=6k\pm 1$, so it only remains to check case $n=6k\pm 3$:
$$2^n+n^2=2^n+(6k\pm 3)^2=2^n+3\mod 6=2^n+1+2\mod 6,$$
$$2^n+n^2=(2+1)(2^{n-1}-2^{n-2}+\dots+2-1)+2\mod 6=6C-3+2\mod 6,$$
$$2^n+n^2=5\mod 6.$$
Finally, we can conclude that desired reminder is $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The indefinite integral $ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$ $$ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\frac{\sin x + \cos x}{\sqrt{\cos 2x}}}$$
$$ \sqrt 2 \int \frac{\sin \left(2x +\alpha+ \frac{\pi}{4}\right) + \sin \left(\alpha -\frac{\pi}{4}\right)}{\cos^3 x\cdot\sqrt{\cos 2x}}$$
How I can do it after this and get rid of square root?
| $$I=\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha}\int \frac{\sin x}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx+\sin{\alpha}\int \frac{1}{\cos^2 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha} I_1+\sin{\alpha}I_2,$$ where
$$I_1=\int \tan{x}\sqrt{\frac{1+\tan{x}}{1-\tan{x}}} d(\tan{x}),$$
$$I_2=\int \sqrt{\frac{1+\tan{x}}{1-\tan{x}}}d(\tan{x}).$$
Substitute $t=\tan{x}$ in both integrals:
$$I_1=\int t\sqrt{\frac{1+t}{1-t}}dt,$$
$$I_2=\int \sqrt{\frac{1+t}{1-t}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct? Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.
Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$
Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$
For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$
Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.
So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.
Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$
So, we get:
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$
$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$
$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$
$\Rightarrow G_n=3^n-2^n$
The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.
In addition, please feel free to offer advice and/or constructive criticism about my proof.
| Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).
| {
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Does the limit exist for $\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$? $$\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\le \lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3}}=\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2}x^{\frac{3}{2}}}=\frac{\pi}{\sqrt{2}\cdot2}$$
So the limit of $\displaystyle\frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$ is finite?
| Use equivalents:
*
*$\arctan x\sim_{+\infty}\dfrac\pi 2$,
*$\sqrt{2x^3+4x^2+}x\sim_{+\infty}\sqrt{2x^3}$,
hence $\; \dfrac{x^{\tfrac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\sim_{+\infty}\dfrac{\pi x^{\tfrac{3}{2}}}{2\sqrt{\mathstrut 2} x^{\tfrac{3}{2}}}=\dfrac{\pi\sqrt{\mathstrut2}}4$.
| {
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Find $\int_0^1\frac{\ln^2(1-x)}{x}\ dx$ In solving $\displaystyle\int_0^\frac{\pi}{4}\dfrac{\ln(\sin x)\ln(\cos x)}{\sin x\cos x}\ dx,$ I have found that this is equal to $\dfrac{1}{16}\displaystyle\int_0^1\dfrac{\ln^2(1-x)}{x}\ dx.$ WolframAlpha says that the desired value is $\dfrac{\zeta(3)}{8},$ so I suspect a conversion to a series is necessary.
How do I prove $\displaystyle\int_0^1\dfrac{\ln^2(1-x)}{x}\ dx=\displaystyle\sum_{n=1}^\infty\dfrac{2}{n^3}$?
Note that the above integral can also be given as $\displaystyle\int_0^1\dfrac{\ln^2x}{1-x}\ dx$, which I know is equal to $\displaystyle\sum_{n=0}^\infty x^n\ln^2x.$
Also for reference, here is a picture of my original work to get to this point.
| Here is an approach that makes use of an Euler sum.
We will first find a Maclaurin series expansion for $\ln^2 (1 - x)$. As
$$\ln (1 - x) = - \sum_{n = 1}^\infty \frac{x^n}{n},$$
we have
\begin{align*}
\ln^2 (1 - x) &= \left (- \sum_{n = 1}^\infty \frac{x^n}{n} \right ) \cdot \left (- \sum_{n = 1}^\infty \frac{x^n}{n} \right ).
\end{align*}
Shifting the summation index $n \mapsto n + 1$ gives
\begin{align*}
\ln^2 (1 - x) &= x^2 \left (- \sum_{n = 0}^\infty \frac{x^n}{n + 1} \right ) \cdot \left (- \sum_{n = 0}^\infty \frac{x^n}{n + 1} \right )\\
&= \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{x^{n + 2}}{(k + 1)(n - k + 1)},
\end{align*}
where the last line has been obtained by applying the Cauchy product.
Shifting the summation indices as follows: $n \mapsto n - 2, k \mapsto k - 1$ gives
\begin{align*}
\ln^2 (1 - x) &= \sum_{n = 2}^\infty \sum_{k = 1}^{n - 1} \frac{x^n}{k(n - k)}\\
&= \sum_{n = 2}^\infty \sum_{k = 1}^{n - 1} \left (\frac{1}{nk} + \frac{1}{n(n - k)} \right ) x^n\\
&= 2 \sum_{n = 2}^\infty \frac{x^n}{n} \sum_{k = 1}^{n - 1} \frac{1}{k}\\
&= 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},
\end{align*}
where $H_n$ is the $n$th harmonic number.
Now evaluating the integral. From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as
\begin{align*}
\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx &= 2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^2}.
\end{align*}
From properties of the harmonic numbers we have
$$H_n = H_{n - 1} + \frac{1}{n},$$
thus
\begin{align*}
\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx &= 2 \sum_{n = 2}^\infty \frac{H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{1}{n^3} = 2 \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{1}{n^3}.
\end{align*}
Each sum can be readily found. They are:
$$\sum_{n = 1}^\infty \frac{1}{n^3} = \zeta (3) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3).$$
A proof of the result for the second sum containing the harmonic number can, for example, be found here. Thus
$$\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx = 4 \zeta (3) - 2 \zeta (3) = 2 \zeta (3),$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$
$\bf{My\; Try::}$ Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$
Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$
So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$
Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$
So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$
Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$
Now How can I solve after that, Help Required, Thanks
| Here is a 3rd solution (inspired by yours, @Christian Blatter, but using different variables).
Let $$Z_1=\frac{z^2_{1}}{z_{2}z_{3}}, \ Z_2=\frac{z^2_{2}}{z_{3}z_{1}}, Z_3=\frac{z^2_{3}}{z_{1}z_{2}} \ \ \ (*)$$
Let us remark that all $Z_k \in \mathbb{U}$ (unit circle).
Thus the initial condition is replaced by $Z_1+Z_2+Z_3=-1 \ \ \ (1)$
Another immediate relationship is $Z_1Z_2Z_3=1 \ \ \ (2)$
At last, is we set $w=Z_1Z_2+Z_2Z_3+Z_3Z_1$, we have
$$w=\frac{z_{1}z_{2}}{z^2_{3}}+\frac{z_{3}z_{1}}{z^2_{2}}+\frac{z_{2}z_{3}}{z^2_{1}}$$
Using the rule $u \in\mathbb{U} \Longrightarrow \bar{w}=\dfrac{1}{w}$ :
$\bar{w}=Z_1+Z_2+Z_3$ thus $w=-1 \ \ \ (3) \ $ (by using (1))
Taking into account (1), (2), (3), by Vieta's formulas, $\{Z_1,Z_2,Z_3\}$ is the set of roots of
$$Z^3+Z^2-Z-1=0$$
i.e., $1$ (multiplicity 1) and $-1$ (multiplicity 2).
One can assume, using definitions (*), that $Z_1=1,Z_2=Z_3=-1$ ; thus
$$\begin{cases}z_1^2=z_2z_3\\ z_2^2=-z_3z_1\\ z_3^2=-z_1z_2\end{cases} \ \ \ (4)$$
or, equivalently:
$$\dfrac{z_1}{z_2}=\dfrac{z_3}{z_1}=-\dfrac{z_2}{z_3}=:\zeta \ \ \ (5)$$
(i.e., we have given the name $\zeta$ to this common ratio).
Thus, by elementary operations on (5), one finds
$$\zeta^3=-1 \ \ \ \Longrightarrow \ \ \zeta=-1, e^{i \pi/3}, e^{-i \pi/3} \ \ \ (6)$$
Thus, using (5), the set of solutions of the initial equation is
$$\begin{cases}z_1=z_1\\ z_2=\zeta^2 z_1\\ z_3=\zeta z_1\end{cases} \ \ \ (7)$$
where $z_1$ is any element of $\mathbb{U}$.
Thus $|z_1+z_2+z_3|=|z_1||1+\zeta+\zeta^2|=2|z_1|=2$ whatever the value of $\zeta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The significance of the number in front of a series? I believe I have solved the series to $-4/9$ however I'm not sure of the significance of the 9 in front of sigma. Is this simply multiplying the final result or is it something different?
The series in question is
$$9\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n$$
| Your solution of the series is correct. According to the formula for geometric power series
$$\sum_{n=1}^\infty q^n=\frac{q}{1-q}\qquad\qquad |q|<1$$
we obtain
\begin{align*}
\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n=\frac{-\frac{4}{5}}{1+\frac{4}{5}}=-\frac{4}{9}
\end{align*}
We do not explicitely need to write a multiplication sign between a factor and a series to denote the multiplication. We can simply write $9\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n$
and the meaning is
\begin{align*}
9\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n=9\cdot\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n=9\cdot\left(-\frac{4}{9}\right)=-4
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to derive $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ Looking at an proof of $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ that argues that $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ but I can't see how this is accomplished since the integral is generalized (it is a limit it the endpoint) and also if using an geometric sum to derive this it is divergent in this endpoint (value $=1$). How is this solved in a rigorous proof?
| So there is a post in @Robert Z's link in the comments that proves
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$
So I won't restate what has already been done, but will just show that
$$ \int_0^{1} \int_0^1 \frac{1}{1 - xy} dx dy = \sum_{n=1}^{\infty} \frac{1}{n^2} $$
Observe that
$$ 1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
Thus
$$ \int_0^x \frac{1}{1-x'} dx' = x + \frac{1}{2} x^2 + \frac{1}{3}x^3 ... $$
Thus:
$$ - \ln(1 - x) = x +\frac{1}{2}x^2 + \frac{1}{3}x^3 ... $$
Moving another step, observe that if we divide by $x$ and integrate AGAIN,
Our terms of the form $$\frac{1}{n} x^n \rightarrow_{divide} \frac{1}{n} x^{n-1} \rightarrow_{integrate} \frac{1}{n} \frac{1}{n} x^n = \frac{1}{n^2} x^n $$
So it then follows
$$ \int_{0}^{x} - \frac{\ln(1-x')}{x'} dx' = x + \frac{1}{4}x^2 + \frac{1}{9} x^3 + ... = \sum_{n=1}^{\infty} \frac{x^n}{n^2} $$
So now suppose we integrate to $x=1$ we then yield:
$$ \int_{0}^{1} - \frac{\ln(1 - x')}{x'} dx' = 1 + \frac{1}{4} + \frac{1}{9} ... = \sum_{n=1}^{\infty} \frac{1}{n^2} $$
So now the final punch. Observe (treating $x$ constant)
$$ \int_0^1 \int_0^1 \frac{1}{1 - xy} dx dy = -\int_0^1 \frac{\ln(1-x)}{x} dx $$
So we now have the desired equivalence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the expression has infinitely many solutions
Does $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$ have infinitely many solutions in positive integers $(m,r)$?
Does $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$ have infinitely many solutions in positive integers $(m,r)$ where $m \equiv 3 \pmod{4}$?
I wasn't sure how to prove that there are infinitely many solutions, but for both of them we need $48r^2$ to be one less than a perfect square. Thus we need $48r^2 = x^2-1$ and so $x^2-48r^2 = 1$, which has infinitely many solutions, but we also need $m$ to be an integer. How do we ensure that in both cases?
| The first has no solutions with positive integer $m$. However, the second case,
$$\begin{aligned}
m&=\dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)\\
&=\dfrac{1}{18}\left(x-1\right)\\
&\equiv {3\pmod{4}}
\end{aligned}$$
has infinitely many given by,
$$x= \frac{ (7+\sqrt{48})^{6k+3}+(7-\sqrt{48})^{6k+3}}{2}=1351,\; 9863382151,\; 72010600134783751\dots$$
$$r = \frac{ (7+\sqrt{48})^{6k+3}-(7-\sqrt{48})^{6k+3}}{2\sqrt{48}}=195,\; 1423656585,\; 10393834843080975\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why in the binomial expansion are the powers of x positive? In Binomial expansion with $n$ as a rational, positive number then,
$$(x+y)^n =\binom{n}{0}y^n + \binom{n}{1} y^{n-1} x +\binom{n}{1} y^{n-2}x^2 +\cdots+\binom{n}{n} x^{n}$$
While if $\alpha$ is a rational, non-positive number and $\lvert x \rvert< 1$ then,
$$(1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2} x^2 +\cdots$$
Imagine if $\alpha$ is $-3$ then how can $x$ be going up in positive powers?
| Here is one way to see this - from the theory of GP, it may be familiar to write for a common ratio $|x| < 1$,
$$\frac1{1-x} = 1+x+x^2+x^3 + \cdots$$
Using $x \to -x$, we get
$$\frac1{1+x} = 1-x+x^2-x^3 + \cdots$$
If you differentiate that, you get
$$-\frac1{(1+x)^2} = -1+2x-3x^2+4x^3 -\cdots$$
$$\implies \frac1{(1+x)^2} = 1-2x+3x^2-4x^3 +\cdots$$
I hope you can extend easily to the case $\alpha = -3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885815",
"timestamp": "2023-03-29T00:00:00",
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Expand$\frac{ x^{k+1}}{k+1} -\frac{ (x-1)^{k+1}}{k+1} $using the binomial theorem The required is to expand
$$\frac{x^{k+1}}{(k+1)} - \frac{(x-1)^{k+1}}{(k+1)}$$
using the binomial theorem.
Here is my solution, is it correct?
$$x^k -\frac{kx^{k-1}}{2} + \frac{\binom{k}{2}x^{k-2}}{3} - \frac{\binom{k}{3}x^{k-3}}{4} + \frac{\binom{k}{4}x^{k-4}}{5} + .... - \frac{\binom{k}{k - 3}x^{4}(-1)^{k-3}}{4} - \frac{\binom{k}{k - 2}x^{3}(-1)^{k-2}}{3} - \frac{kx^{2}(-1)^{k-1}}{2} - x(-1)^{k} - \frac{-1^{k+1}}{k+1}$$
| The binomial theorem states:
$$
\left(a + b\right)^n = \sum_{m=0}^n \binom{n}{m} \cdot a^m \cdot b^{n-m}
$$
Applying it for $(x-1)^{k+1}$, with $a = x$, $b=-1$ and $n=k+1$ we have:
$$\begin{eqnarray}
\left(x-1\right)^{k+1} &=& \sum_{m=0}^{k+1} \binom{k+1}{m} \cdot x^m \cdot \left(-1\right)^{k+1-m} \cr &=& x^{k+1} + \sum_{m=0}^{k} \binom{k+1}{m} \cdot x^m \cdot \left(-1\right)^{k+1-m} \cr
&=& x^{k+1} - \cdot \sum_{m=0}^{k} \binom{k+1}{m} \cdot x^{m} \cdot \left(-1\right)^{k-,}
\end{eqnarray}
$$
Now, observe that
$$
\binom{k+1}{m} = \frac{\left(k+1\right)!}{\left(k+1-m\right)! \cdot m!} = \frac{k+1}{k+1-m} \cdot \frac{k!}{\left(k-m\right)! \cdot m!} = \frac{k+1}{k+1-m} \cdot \binom{k}{m}
$$
Combining these:
$$\begin{eqnarray}
\frac{x^{k+1}-\left(x-1\right)^{k+1}}{k+1} &=& \cdot \sum_{m=0}^{k} \frac{1}{k+1-m} \binom{k}{m} \cdot x^{m} \cdot \left(-1\right)^{k-m} \cr
&\stackrel{m \to k - p }{=}& \cdot \sum_{p=0}^{k} \frac{1}{p+1} \binom{k}{k-p} \cdot x^{k-p} \cdot \left(-1\right)^{p} \cr &=& \cdot \sum_{p=0}^{k} \frac{1}{p+1} \binom{k}{p} \cdot x^{k-p} \cdot \left(-1\right)^{p}
\end{eqnarray}$$
Added:
Answering the OP's questions posed in comments, here is the verification of the result in Mathematica:
orig[k_Integer,x_]:=(x^(k+1)-(x-1)^(k+1))/(k+1)
expanded[k_Integer,x_]:=Sum[Binomial[k,p]/(p+1)(-1)^p x^(k-p),{p,0,k}]
In[3]:= Table[orig[k,x]==expanded[k,x]//Simplify,{k,0,6}]
Out[3]= {True,True,True,True,True,True,True}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| Since the OP admitted that they can't continue from Dr.MV's hint, I'll do it myself:
$$\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}=\frac{1}{1-b} \left(\sum_{n=0}^\infty a^n-b \sum_{n=0}^\infty (ab)^n \right)=$$
Here we use geometric series, so we need to have $|a|<1$ and $|ab|<1$: *
$$=\frac{1}{1-b} \left(\frac{1}{1-a}-\frac{b}{1-ab} \right)=\frac{1}{(1-a)(1-ab)}$$
* Correction due to Batominovski's comment
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can I integrate $\int \arctan(\sec x + \tan x) dx$ I got this problem in my homework exercise:
$$\int \arctan(\sec x + \tan x) dx$$
I simplified it to
$$\int \arctan\left(\dfrac{1+\sin x}{\cos x}\right) dx$$
$$=\int \arctan\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right) dx$$
now, I tried putting $\sqrt{\dfrac{1+\sin x}{1-\sin x}} = \tan t$
Then $x = \arcsin(\sin^2 t-\cos^2 t)$ but it becomes complete mess after that!
| By integration by parts we obtain
$$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C.$$
Since we have the identity (see Qwerty's answer):
$$\sec x + \tan x={1+\sin(x)\over \cos(x)}=\tan(\pi/4+x/2),$$
the primitive can be simplified (!?) to
$$x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C=x\arctan(\tan(\pi/4+x/2))-\dfrac{x^2}{4}+C\\=\frac{x^2+\pi x}{4}-\pi x
\left\lfloor \frac{x}{2\pi}+\frac{3}{4}\right\rfloor+C.$$
P.S. Note that if we erroneously write that $\arctan(\sec x + \tan x)=\pi/4+x/2$, then we get a different answer
$$\int \arctan(\sec x + \tan x) dx=\int (\pi/4+x/2) dx=\frac{x^2+\pi x}{4}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| There is another inductive way, I don't know if better or worse:
Our goal is to prove that $2^{n+1}>(n+1)^4$, assuming that $2^n>n^4$ and, as noted, $n\ge 17$. Let's estimate $2^{n+1}-(n+1)^4$:
$$2^{n+1}-(n+1)^4=\big(2^n-[(n+1)^4-n^4]\big)+(2^n-n^4)>(2^n-n^4)+(2^n-n^4)>0$$
To show the first inequality we have to check:
$$(n+1)^4-n^4<n^4$$
but
$$\left(\frac{n+1}n\right)^4\le\left(\frac{18}{17}\right)^4<2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that this identity holds? By comparing the first terms of the Taylor expansion at $0$, it seems that
for $|x|<4/27$ the following identity holds:
$$\ln\left(\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^{n-1}}{2n+1}\right)=
\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^n}{n}.$$
I tried by differentiating both sides, but I am completely lost in a terrible mess. I wonder if there is a better strategy to handle it. Any idea?
| Let's try to use your idea of looking at derivatives: For $x=0$ both sides result in $0$, so indeed, if the derivatives are the same then we have the equality.
Let $g(x)$ be the left-hand side and $f(x)$ the right-hand side. The derivative of $g(x)$ is
$$g'(x)=\frac{1}{\sum_{n=1}^\infty\binom{3n}{n}\frac{x^{n-1}}{2n+1}}\sum_{n=2}^\infty\binom{3n}{n}\frac{(n-1)}{2n+1}x^{n-2}$$
and the derivative of the right-hand side is
$$f'(x)=\sum_{n=1}^\infty\binom{3n}{n}x^{n-1}$$
Just to simplify things, let's rewrite all these sums starting at $0$:
$$g'(x)=\frac{1}{\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}}\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n$$
$$f'(x)=\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n$$
So $g'(x)=f'(x)$ is equivalent to
$$\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n=\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)$$
Now let's use the fact that $(\sum_{n=0}^\infty a_n)(\sum_{n=0}^\infty b_n)=\sum_{n=0}^\infty\left(\sum_{j=0}^n a_jb_{n-j}\right)$, so
\begin{align*}
\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)\hspace{-80pt}&\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\binom{3j+3}{j+1}\frac{x^j}{2j+3}\binom{3n-3j+3}{n-j+1}x^{n-j}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}\right)x^n.
\end{align*}
Finally, the problem is equivalent to showing that for all $n$,
$$\binom{3n+6}{n+2}\frac{1}{2n+5}=\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}$$
For $n=0$ both sides yield $3$, so perhaps the rest will follow by induction.
EDIT: For $n=1$, the left-hand above side yields 12 but the right-hand side yields 24, so apparently this is not true (if there is no mistake in my arguments).
| {
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"url": "https://math.stackexchange.com/questions/1897288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\binom{x+2}{5} = 126$ for $x$ How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one solution exists?
| Use brute force approach, We can easily find x=7 is the solution.
And x=7 is the only one solution. Because:
There's no solution for x<3, because $\binom{x+2}{5}=0$ for x<3
for x>=3:
$\binom{x0+2}{5} < \binom{x+2}{5} < \binom{x1+2}{5}$ for every x1 > x > x0 >=3; and x,x1,x0 is real number
so
$\binom{x0+2}{5} < \binom{7+2}{5} = 126 < \binom{x1+2}{5}$ for every x1 > 7 > x0 >=3; and x1,x0 is real number
Conclusion: x=7 is the unique answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
I determined that since $2a^2-1 \equiv 1,7 \pmod{8}$ we must have $b^2+2 \equiv 3 \pmod{8}$ in order for the fraction to be an integer. I didn't see how to find a contradiction from here.
| Hint $\ $ If a prime $\,p\,$ divides numerator and denominator then $\,{\rm mod}\ p\!:\ b^2\equiv -2,\,$ and $\, (2a)^2\equiv 2.\,$ Since both $\,2\,$ and $\,-2\,$ are squares mod $\,p\,$ it follows by reciprocity that $\,p\equiv 1\pmod 8.\,$ But $\,b^2+2\,$ must be divisible by a prime $\,q\not\equiv 1\pmod 8,\,$ else $\,{\rm mod}\ 8\!:\ b\equiv 1\,\Rightarrow\,b^2+2\equiv 3,\,$ contradiction. Since $\,q\,$ divides the denominator but not the numerator, the fraction is not an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it with Mathematica, but I can't figure out how to prove it.
The integral seems quite problematic to me. If the limits were finite, I would do this:
$$\frac{1}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{1}{a^4-b^4}(\sqrt{x^4+a^4}-\sqrt{x^4+b^4})$$
Then, for one of the integrals we will have:
$$\int_A^B \sqrt{x^4+a^4} dx=a^3 \int_{A/a}^{B/a} \sqrt{1+t^4} dt$$
This integral is complicated, but quite well known.
On the other hand $\int_0^\infty \sqrt{1+t^4}dt$ diverges, so I can't consider the two terms separately.
But the integral behaves like I can! If we look at the final expression, it seems like $\int_0^\infty \sqrt{1+t^4}dt=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}}$ even though it can't be correct.
I have to somehow arrive at Beta function, since we have a squared Gamma as an answer.
I'm interested in this integral, since it represents another kind of mean for two numbers $a$ and $b$. If we scale it appropriately:
$$I(a,b)=\frac{8 \sqrt{\pi}}{\Gamma(1/4)^2 } \int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}= \frac{4}{3} \frac{a^2+ab+b^2}{a^3+ab(a+b)+b^3}$$
So, $1/I(a,b)$ is a mean for the two numbers.
| $$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$
Because the $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx $ is convergent so the integration can be linearly seperated.
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx - \int_0^\infty(\sqrt{x^4+b^4}-x^2) dx$$
Out of symmetry, the question becomes how to solve integration $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx$ in terms of $a$.
Let $a^2 \sqrt{t} = \sqrt{x^4+a^4}-x^2 \implies x^2 = a^2 \frac{1-t}{2\sqrt{t}}$ which brought us to an old problem: see the second answer in this post by votes sort.
After simplification the integration becomes $\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} a^3$.
Therefore, the original formula should be $$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx = \frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} (a^3-b^3)$$.
Hence the question get solved.
| {
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"url": "https://math.stackexchange.com/questions/1898568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Finding a set of 3 positive prime numbers that satisfy a polynomial equation What is the set of positive primes p,q,r that satisfy the equation: $p^{4}+2p+q^{4}+q^{2}=r^{2}+4q^{3}+1$. It can be easily shown (using parity concepts) that either p or r must be 2, but it is not clear to me how to advance any further since after proper substitution for p or r did not find an appropriate factorization.
| We have
$$ r^2=p^4+2p+q^4-4q^3+q^2-1$$
*
*If $q\ge 5$, then $q^4-4q^3+q^2-1\ge q^3+q^2-1>0$ and hence $r^2>p^4\ge 16$, so $r\ge 5$.
*If $q=3$, then $r^2=p^4+2p-19$, so $r=2$ leads to $(p^3+2)p=23$, contradiction
*If $q=2$, then $r^2=p^4+2p-13$, so $r=2$ leads to $(p^3+2)p=17$, contradiction.
We conclude that $r\ne 2$, hence by your parity argument
$$ p=2.$$
The equation becomes
$$r^2=q^4-4q^3+q^2+19 $$
Again we check small values of $q$ separately: $q=2$ leads to $r^2=7$; $q=3$ leads to $r^2=1$,so in both cases no prime $r$.
We conclude $$q\ge 5.$$
Now
$$ r^2=q^4-4q^3+q^2+19<q^4-4q^3+2q^2+4q+1=(q^2-2q-1)^2.$$
On the other hand,
$$ (q^2-2q-2)^2=q^4-4q^3+8q+4=r^2-q^2+8q-15=r^2-(q-4)^2+1\le r^2$$
with equality only for $q=5$, which makes $r=13$. Thus the only valid solution is
$$p=2,\qquad q=5,\qquad r=13. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor Expansion of a Trig Function Did a problem in a lecture today with Taylor Expansions.
The question is:
Find the Taylor Expansion of
$
z=\sin(x^2+y^2)
$
To the second order at the point (1, -1)
I am happy with the solution until she gets to
$
\frac{d^2(f)}{d(y)^2}(1,-1)=-4\sin(2)=-3,637
$
According to my own calculations,
$
\frac{d^2(f)}{d(y)^2}(1,-1)=2\cos(2)-4\sin(2)=-4,469
$
She checked her answers and insists that she is correct. I am confused as to where that $2\cos(2)$ went. If someone could explain it to me that would be awesome.
$
\frac {df}{dy}=2y\cos (x^2+y^2)
$
And
$
\frac {d^2(f)}{d(y)^2}=2\cos(x^2+y^2)-4(y)^2 \sin(x^2+y^2)
$
(Using product rule)
| Your computations are fine. We have
$$
f(x,y):=\sin(x^2+y^2)
$$ giving by the chain rule
$$
\frac {df}{dy}(x,y)=2y\cos (x^2+y^2), \qquad \frac {d^2f}{d^2y}(x,y)=2\cos(x^2+y^2)-4y^2 \sin(x^2+y^2)
$$ and
$$
\left.\frac {d^2f}{d^2y}(x,y)\right|_{(1,-1)}=2\cos(2)-4 \sin(2)=-4.469483380\cdots.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| Let $a_n=x^n+y^n$.
Then
$$3a_n=(x+y)a_n=x^{n+1}+xy^n+yx^n+y^{n+1}=a_{n+1}+xya_{n-1}=a_{n+1}+3a_{n-1}$$
and so
$a_{n+1}=3a_n-3a_{n-1}$.
Also, $a_0=x^0+y^0=2$, $a_1=x+y=3$.
So $a_2=3a_1-3a_0=3$, $a_3=3a_2-3a_1=0$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $x+y+z$ From Equations Including Radicals Let $(x,y,z)$ be the real solution to the system of equations
\begin{align*}
x+y &= \sqrt{4z-1} ,\\
y+z &= \sqrt{4x- 1} , \\
z+x &= \sqrt{4y-1} .
\end{align*}
Find $x+y+z.$
I could add all the equations up but that doesn't do any good. Thanks in advance!
| To initiate Arthur's hints in the comments:
$$x^2 + 2xy + y^2 = (x + y)^2 = 4z - 1$$
$$y^2 + 2yz + z^2 = (y + z)^2 = 4x - 1$$
$$z^2 + 2zx + x^2 = (z + x)^2 = 4y - 1$$
Rather than adding all three equations, we subtract them pairwise. Then assuming $x \neq y \neq z$, we have:
$$2y(x - z) + (x + z)(x - z) = 4(z - x) \implies 2y + x + z = -4$$
$$2z(y - x) + (y + x)(y - x) = 4(x - y) \implies 2z + y + x = -4$$
$$2x(y - z) + (y + z)(y - z) = 4(z - y) \implies 2x + y + z = -4$$
Equating, we get
$$2y + x + z = 2z + y + x = 2x + y + z = -4.$$
Therefore, we obtain
$$x + y + z = -4 - y = -4 - z = -4 - x.$$
This implies that
$$x = y = z,$$
which contradicts our earlier assumption.
Therefore, the original system of equations has no real solution.
Added August 27 2016
Therefore, for the original system of equations to have a real solution, we must have $x = y = z$.
Consequently, the system is reduced to solving the lone equation
$$x + x = \sqrt{4x - 1}$$
$$(2x)^2 = 4x - 1$$
$$4x^2 - 4x + 1 = 0$$
$$x = \frac{1}{2}.$$
Thus,
$$x = y = z = \frac{1}{2}$$
is the only real solution.
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of the difference between two exponential functions
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$
I did nothing by now.
| We have that
$$\lim_{x \to +\infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right)=
\lim_{x \to +\infty}\left((x+3)\exp\left[\frac{\ln(x+3)}{x}\right]-x\exp\left[\frac{\ln(x)}{x+3}\right]\right).
$$
Moreover
$$(x+3)\exp\left[\frac{\ln(x+3)}{x}\right]=(x+3)\cdot\left(1+\frac{\ln(x+3)}{x}+O(\ln^2(x)/x^2)\right)\\
=x+3+(x+3)\frac{\ln(x+3)}{x}+o(1)$$
and
$$x\exp\left[\frac{\ln(x)}{x+3}\right]=x\cdot\left(1+\frac{\ln(x)}{x+3}+O(\ln^2(x)/x^2)\right)\\
=x+\frac{x\ln(x)}{x+3}+o(1).
$$
Hence we obtain
$$\lim_{x \to +\infty}\left(x+3+(x+3)\frac{\ln(x+3)}{x}-x-\frac{x\ln(x)}{x+3}\right)
=\lim_{x \to +\infty}\left(3+(x+3)\frac{\ln(x)+\ln(1+3/x)}{x}-\frac{x\ln(x)}{x+3}\right)\\
=\lim_{x \to +\infty}\left(3+\ln(x)\cdot \left(\frac{x+3}{x}-\frac{x}{x+3}\right)\right)\\
=\lim_{x \to +\infty}\left(3+\ln(x)\cdot \left(\frac{6x+9}{x(x+3)}\right)\right)=3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the number of 3-element subsets from the set {1,2,3,...,11,12,13} such that the sum of the 3 elements is divisible by 3 I found this problem from an old math questionnaire.
How many 3-element subsets of {1,2,3,...,11,12,13} are there for which the sum of the 3 elements is divisible by 3?
At first, I tried listing down all of them. But then, I realized they were too many and so, I just stopped. Next, I tried using stars and bars technique. I represented the 3 elements as a,b,c and their sum as a+b+c=m where m is the set of positive integers divisible by 3 from 3 to 39. From there, I tried out each equation with stars and bars from a+b+c=3 to a+b+c=15, and found a total of 185 subsets satisfying the given conditions. I stopped at here since I noticed that if I continued doing this from a+b+c=18 to a+b+c=39, I would have an error since I was not able to limit the values of a,b,c at most 13.
I do not know how to proceed right now. Can anyone help?
| A more general approach is given by generating functions.
We may consider the bivariate polynomial
$$ q(x,y)=(1+yx)(1+yx^2)\cdot\ldots\cdot(1+yx^{12})(1+yx^{13}) $$
and the coefficient of $y^3$ in $q(x,y)$, that is
$$ [y^3]\,q(x,y) = r(x) = x^6+x^7+2 x^8+3 x^9+4 x^{10}+5 x^{11}+7 x^{12}+8 x^{13}+10 x^{14}+12 x^{15}+14 x^{16}+15 x^{17}+17 x^{18}+17 x^{19}+18 x^{20}+18 x^{21}+18 x^{22}+17 x^{23}+17 x^{24}+15 x^{25}+14 x^{26}+12 x^{27}+10 x^{28}+8 x^{29}+7 x^{30}+5 x^{31}+4 x^{32}+3 x^{33}+2 x^{34}+x^{35}+x^{36} $$
then sum the coefficients of the monomials of the form $x^{3k}$. If $\omega=\exp\left(\frac{2\pi i}{3}\right)$, that sum is just
$$ \frac{1}{3}\left( r(1)+r(\omega)+r(\omega^2) \right) $$
or the coefficient of $y^3$ in the polynomial $\frac{1}{3}\left(q(1,y)+q(\omega,y)+q(\omega^2, y)\right)$, where:
$$\frac{1}{3}\left(q(1,y)+q(\omega,y)+q(\omega^2, y)\right) = (1+y)^4\cdot\left(1+20 y^2+14 y^3+\ldots\right)$$
gives us the answer:
$$ 14+4\cdot 20+4 = \color{red}{98}. $$
Not by chance, this number is close to one third of the total number of $3$-subsets, given by $\binom{13}{3}=286$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A beautiful triangle inequality Prove that in any triangle with side lengths $a, b, c$ the inequality:
$$(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\geq \sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
For demonstration, I tried to use substitutions Ravi or algebraic calculation but without success. Maybe someone has an idea of saving. Thank you!
| Since $\sqrt{b}+\sqrt{c}-\sqrt{a}=\sqrt{b+c+2\sqrt{bc}}-\sqrt{a}>\sqrt{a}-\sqrt{a}=0$, we need to prove that
$$(a+b-c)^2(a+c-b)^2(b+c-a)^2\geq(a^2+b^2-c^2)(a^2+c^2-b^2)(b^2+c^2-a^2)$$
We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$ and from here we can assume that
$a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $b^2+c^2-a^2\geq0$.
Indeed, if $a^2+b^2-c^2<0$ and $a^2+c^2-b^2<0$ so $a^2<0$, which is contradiction.
Now our inequality follows from
$$(a+b-c)^2(a+c-b)^2-(a^2+b^2-c^2)(a^2+c^2-b^2)=2(b-c)^2(b^2+c^2-a^2)$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| Let
\begin{align}
y&=\sin ^{-1}\left(\frac{2^{x+1}3^x}{1+36^x}\right)\\
&=\sin ^{-1}\left(\frac{2^x\cdot2\cdot3^x}{1+36^x}\right)\\
&=\sin ^{-1}\left(\frac{2\cdot6^x}{1+6^{2x}}\right)\\
&=2\tan ^{-1}6^x
\end{align}
Thus
\begin{align}
\frac{dy}{dx}&=\frac{2}{1+6^{2x}}\cdot6^x\log 6\\
\implies \frac{dy}{dx}&=\frac{2\cdot6^x\log 6}{1+36^x}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute this limit $\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}$ $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$
such that $2\alpha n\le n$
I tried to use Stirling formula and we get
$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\lim_{n\to ∞}\frac{1}{n}\log\frac{n^{\frac{3n}{2}}}{2\pi(n-2\alpha n)^{\frac{3((n-2\alpha n)}{2}{(2\alpha n)}^{3\alpha n}}}=$$
$$=\lim_{n\to ∞}\log{\frac{n^{\frac{3}{2}}}{2\pi(n-2\alpha n)^{\frac{3((1-2\alpha )}{2}{(2\alpha n)}^{3\alpha }}}}$$
but I couldn't continue
| Assuming that $$\binom{m}{n} = \frac{\Gamma(m + 1)}{\Gamma(n + 1)\Gamma(m - n + 1)}\tag{1}$$ whenever the RHS is defined we can see that the sequence $$a_{n} = \binom{n}{2\alpha n}$$ satisfies $$\frac{a_{n + 1}}{a_{n}} = \frac{\Gamma(n + 2)}{\Gamma(2\alpha n + 2\alpha + 1)\Gamma(n + 2 - 2\alpha n - 2\alpha)}\cdot\frac{\Gamma(2\alpha n + 1)\Gamma(n - 2\alpha n + 1)}{\Gamma(n + 1)}$$ so that $$\frac{a_{n + 1}}{a_{n}} = \frac{n\Gamma(2\alpha n)(n - 2\alpha n)\Gamma(n - 2\alpha n)}{\Gamma(2\alpha n + 2\alpha)(n + 1 - 2\alpha n - 2\alpha)(n - 2\alpha n - 2\alpha)\Gamma(n - 2\alpha n - 2\alpha)}$$ and hence
\begin{align}
L &= \lim_{n \to \infty}\frac{a_{n + 1}}{a_{n}}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{\Gamma(2\alpha n)\Gamma(n - 2\alpha n)}{\Gamma(2\alpha n + 2\alpha)\Gamma(n - 2\alpha n - 2\alpha)}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{(2\alpha n/e)^{2\alpha n}((n - 2\alpha n)/e)^{n - 2\alpha n}}{((2\alpha n + 2\alpha)/e)^{2\alpha n + 2\alpha}((n - 2\alpha n - 2\alpha)/e)^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{(2\alpha n)^{2\alpha n}(n - 2\alpha n)^{n - 2\alpha n}}{(2\alpha n + 2\alpha)^{2\alpha n + 2\alpha}(n - 2\alpha n - 2\alpha))^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{(2\alpha)^{-2\alpha}}{1 - 2\alpha}\lim_{n \to \infty}\left(\frac{n}{n + 1}\right)^{2\alpha n}\frac{1}{(n + 1)^{2\alpha}}\frac{(n - 2\alpha n)^{n - 2\alpha n}}{(n - 2\alpha n - 2\alpha))^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{(2\alpha e)^{-2\alpha}}{1 - 2\alpha}\lim_{n \to \infty}\left(\frac{n - 2\alpha n - 2\alpha}{n + 1}\right)^{2\alpha}\left(\frac{n - 2\alpha n}{n - 2\alpha n - 2\alpha}\right)^{n(1 - 2\alpha)}\notag\\
&= (2\alpha e)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}\lim_{n \to \infty}\left(1 - \frac{2\alpha}{n(1 - 2\alpha)}\right)^{n(2\alpha - 1)}\notag\\
&= (2\alpha)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}\notag
\end{align}
and hence $a_{n}^{1/n}$ tends to the same limit $L = (2\alpha)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}$ and therefore $$\frac{1}{n}\log a_{n} \to \log L = -\{(1 - 2\alpha)\log(1 - 2\alpha) + 2\alpha \log 2\alpha\}$$ We have used the Stirling's approximation $$\lim_{x \to \infty}\frac{\Gamma(x)}{(x/e)^{x}\sqrt{2\pi/x}} = 1$$ and the result makes sense only when $0 < \alpha < 1/2$. The terms with $\sqrt{2\pi/\cdots}$ are not seen in above limit evaluation because after some cancellation in numerator and denominator they tend to $1$ as $n \to \infty$.
Silly Note: A lot of typing in $\mathrm\LaTeX$ could be simplified by replacing $2\alpha$ with $a$ but alas! I chose to be in sync with OP's notation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Are there other good ways to look at this differential equation or is this it? \begin{align}
\tan \frac \alpha 2 & = u \\[4pt]
\alpha & = 2\arctan u \\[4pt]
d\alpha & = \frac{2\,du}{1+u^2} \\[4pt]
\sin\alpha & = \sin(2\arctan u) = 2(\sin\arctan u) (\cos \arctan u) \\[4pt]
& = 2\,\frac u {\sqrt{1+u^2}} \cdot \frac 1 {\sqrt{1+u^2}} = \frac{2u}{1+u^2}
\end{align}
$$ \frac{\hspace{6cm}}{} \qquad \S \qquad \frac{\hspace{6cm}}{} $$
The well known substitution above is for the moment my preferred approach to solving the differential equation
$$
\frac{d\alpha}{\sin\alpha} = \frac{d\beta}{\sin\beta}. \tag{differential equation}
$$
One gets
$$
\frac{d\alpha}{\sin\alpha} = \frac{\left( \dfrac{2\,du}{1+u^2} \right)}{\left( \dfrac{2u}{1+u^2} \right)} = \frac{du} u
$$
and writing $v = \tan \dfrac\beta 2$, we then have
$$
\frac{du} u = \frac {dv} v
$$
so that $\log u = \log v + \text{constant}$ and so $u = v\times \text{constant}$, and finally
$$
\tan \frac\alpha 2 = \text{constant} \times \tan \frac\beta 2. \tag{solution}
$$
So my question is whether there are other ways to approach this that are either better or otherwise worth some attention.
| From the first time I heard about Weierstrss substitution, it also became my favored way for solving things such as $$\int \frac{dx}{a+b \sin(x)+c\cos(x)}=-\frac{2 }{\sqrt{b^2+c^2-a^2}}\tanh ^{-1}\left(\frac{(a-c) \tan
\left(\frac{x}{2}\right)+b}{\sqrt{b^2+c^2-a^2}}\right)$$
For things such as $$I=\int \frac{dx}{\sin(x)}=2i\int \frac{dx}{e^{ix}-e^{-ix}}$$ the change of variable $$e^{ix}=t \implies x=-i \log(t)\implies dx=-i\frac{dt} t$$ is also interesting since it makes $$I=2\int \frac {dt}{t^2-1}=\log\left(\frac{1-t}{1+t} \right)= -2 \tanh ^{-1}(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A question related to compute infinite summation How to solve this: $$ 1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\cdots $$ So, the $n^{th}$ term of the sum can be written as $$a_n=\frac{2\cdot5\cdot8\cdots (2+3(n-1))}{6\cdot12\cdot18\cdots (6n)} = \frac{2\cdot5\cdot8 \cdots (2+3(n-1))}{6^n (n!)}$$ So the above sum can be written as $$ 1+\sum_{n=1}^\infty a_n .$$ Now how should I proceed to solve this. Please help me.
| Consider
$$
1+\sum^\infty_{n=1}\prod^n_{i=1}\frac{2+3(i-1)}{6i}=1+\sum^\infty_{n=1}\prod^n_{i=1}\left(\frac{1}{2}-\frac{1}{6i}\right)=1+\sum^\infty_{n=1}2^{-n}\prod^n_{i=1}\left(1-\frac{1}{3i}\right)
$$
Product above can be expressed in terms of Gamma-functions:
$$
\prod^n_{i=1}\left(1-\frac{1}{3i}\right)=\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}
$$
Thus, we have
$$
1+\sum^\infty_{n=1}2^{-n}\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}=\sum^\infty_{n=0}2^{-n}\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}
$$
By definition
$$
\left(\!
\begin{array}{c}
x \\
y
\end{array}
\!\right) = \frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}
$$
In our case $x=n-\frac{1}{3},\quad y=-\frac{1}{3}$. Which leads us to:
$$
\sum^\infty_{n=0}\left(\!
\begin{array}{c}
n-\frac{1}{3} \\
-\frac{1}{3}
\end{array}
\!\right)2^{-n}
$$
Recall generating function:
$$
\frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\left(\!
\begin{array}{c}
m+n \\
m
\end{array}
\!\right)z^n
$$
Setting $z=\frac{1}{2}$ and $m=-\frac{1}{3}$ gives us:
$$
\sum^\infty_{n=0}\left(\!
\begin{array}{c}
n-\frac{1}{3} \\
-\frac{1}{3}
\end{array}
\!\right)2^{-n}=\left(1-\frac{1}{2}\right)^{-\frac{2}{3}}=2^{\frac{2}{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Combinatorial proof of a binomial identity: $\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$ I am to find a combinatorial argument for the following identity:
$$\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$$
For the right hand side, I was think that would just be number of ways to choose at least $s-1$ elements out of a $[2r-s]$ set. However, for the left hand side, I don't really know what it is representing.
Any help would be greatly appreciated!
| Suppose we seek to verify that
$$\sum_{k=1}^r {2r\choose 2k-1} {k-1\choose s-1}
= 2^{2r-2s+1} {2r-s\choose s-1}$$
where presumably $s\ge 1$. The lower limit is set to $k=1$ as the
first binomial coefficient is zero when $k=0.$
Introduce
$${2r\choose 2k-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2r-2k+2}} \frac{1}{(1-z)^{2k}}
\; dz.$$
This provides range control and vanishes when $k\gt r$
so we may extend the range to infinity, obtaining
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}}
\sum_{k\ge 1} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz.$$
This yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}}
\sum_{k\ge s} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}}
\sum_{k\ge 0} {k+s-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}}
\frac{1}{(1-z^2/(1-z)^2)^s}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}}
\frac{1}{((1-z)^2-z^2)^s}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}}
\frac{1}{(1-2z)^s}
\; dz.$$
This is
$$[z^{2r-2s+1}] \frac{1}{(1-2z)^s}
= 2^{2r-2s+1} {2r-2s+1+s-1\choose s-1}
\\ = 2^{2r-2s+1} {2r-s\choose s-1}$$
and we have the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| Suppose $a,b,c,d\geq 0$ and set $x=a^4,\,y=b^4,\,z=c^4,\,w=d^4$. Then
$$f(x,y,z,w)=x^{5/4}+y^{5/4}+z^{5/4}+w^{5/4}$$
is strictly convex and thus restricted to the domain
$$\{x,y,z,w\geq 0: x+y+z+w=16\}$$
it attains the maximum only in the extremal points, which are the four vertices $(16,0,0,0)$ and cyclicals.
If one number is less then zero than taking the moduli the value of $a^5+b^5+c^5+d^5$ strictly increases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2}$ I want to evaluate the sum
$ \displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2} $
My approach so far has been the possible use of shifted Legendre polynomial. We know
$Q_n(x)=\displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}x^j$
But I am not being able to relate these two. I see that
$ \displaystyle \int_0^1Q_n(x)dx=\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{j+1}$
but how do I get $\dfrac{1}{(j+1)^2}$ ? I need some hint for this.
| Suppose we seek a closed form of the sum
$$\sum_{q=0}^n (-1)^{n+q} {n\choose q} {n+q\choose q} \frac{1}{(q+1)^2}.$$
This is
$$\sum_{q=0}^n (-1)^{n+q} \frac{q+1}{n+1}
{n+1\choose q+1} {n+q\choose q} \frac{1}{(q+1)^2}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^{n+q}
{n+1\choose q+1} {n+q\choose q} \frac{1}{q+1}.$$
Observe that
$$[z^q] \frac{1}{(1-z)^{n+1}} = {n+q\choose q}$$ and hence
$$\frac{1}{n} [z^{q+1}] \frac{1}{(1-z)^{n}} = \frac{1}{q+1} {n+q\choose q}.$$
We introduce
$$\frac{1}{q+1} {n+q\choose q} =
\frac{1}{n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+2}}
\frac{1}{(1-z)^n} \; dz$$
and obtain for the sum
$$\frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\frac{1}{(1-z)^n}
\sum_{q=0}^n (-1)^q {n+1\choose q+1} \frac{1}{z^{q+1}}
\; dz
\\ = - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\frac{1}{(1-z)^n}
\sum_{q=1}^{n+1} (-1)^q {n+1\choose q} \frac{1}{z^q}
\; dz
\\ = - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\frac{1}{(1-z)^n}
\left(-1 + \left(1-\frac{1}{z}\right)^{n+1}\right)
\; dz.$$
Now the first piece here yields
$$- \frac{(-1)^n}{n(n+1)} \times -1 \times
[z^0] \frac{1}{(1-z)^n} = \frac{(-1)^n}{n(n+1)}.$$
The second piece is
$$ - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\frac{1}{(1-z)^n} \frac{(z-1)^{n+1}}{z^{n+1}}\; dz
\\= \frac{1}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\frac{1}{(1-z)^n} \frac{(1-z)^{n+1}}{z^{n+1}}\; dz
\\= \frac{1}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1-z) \; dz.$$
This vanishes when $n\ge 1,$ which we have assumed anyway.
It follows that the desired answer is
$$\frac{(-1)^n}{n(n+1)}.$$
This matches the result by @JackD'Aurizio.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to solve $\int\tfrac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx$? I am trying to find a solution for the following integration for a physics problem. Could anyone give hint on how to do this?
$$
\int\frac{\sqrt{x^2+1}}{\left(x^2+a\right)^\frac{3}{2}}dx
$$
Thanks for your time.
| Allow us to use generalized binomial expansion theorem on the $\sqrt{x^2+1}$ and the $u$ substitution $x^2+a=u$.
$$\sqrt{x^2+1}=\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}=x+\frac12x^{-1}-\frac18x^{-2}+\dots$$
$$\begin{align}
\int\frac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx & = \int\frac{\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}}{(x^2+a)^{3/2}}dx \\
& = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int\frac{x^{-2n}}{(x^2+a)^{3/2}}(2xdx) \\
& = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int u^{-3/2}(u-a)^{-n}du \\
\end{align}$$
Again, binomial expand the $(u-a)^{-n}$
$$\begin{align}
& = \frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\int u^{-3/2}u^{-1-k}du \\
& = C+\frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\frac{(x^2+a)^{-(3/2)-k}}{-(3/2)-k} \\
\end{align}$$
I guess you could say its no longer an integral problem and more of a summation problem.
Also, don't expect anything simple, because according to Wolframalpha, the integral involves elliptic integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$ Let $a, b, c, d \in \mathbb{R}_{>0}$, then prove that
$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$
Can this be done without using AM-GM inequality, or without using any identity/theorem of inequality? I don't want it to be concise or elegant, I just want rigorous steps that show explicitly how we achieve that result.
Thanks in advance
| Let $\frac{a}{b}=x^4$, $\frac{b}{c}=y^4$, $\frac{c}{d}=z^4$ and $\frac{d}{a}=t^4$ for positives $x$, $y$, $z$ and $t$.
Hence, we need to prove that $x^4+y^4+z^4+t^4\geq4xyzt$, which is true because
$$x^4+y^4+z^4+t^4-4xyzt=(x^2-y^2)^2+(z^2-t^2)^2+2(xy-zt)^2\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number?
| There is no need to find an explicit formula for $\sum_{n=1}^{600}F_n^2$ (but this is doable and I'll show it in the second part of this answer). The Fibonacci sequence obeys the relation $F_{n+2}=F_{n+1}+F_{n}$ and $8$ is a Fibonacci number, hence the sequence $\{F_n\pmod{8}\}_{n\geq 1}$ has a short period and the same applies to $\{F_n^2\pmod{8}\}_{n\geq 1}$:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9 & 10 & 11 & 12\\ \hline F_n\pmod{8} & 1 & 1 & 2 & 3 & 5 & 0 & 5 & 5 & 2 & 7 & 1 & 0 \\ \hline F_n^2\pmod{8} & 1 & 1 & 4 & 1 & 1 & 0 & 1 & 1 & 4 & 1 & 1 & 0\\ \hline\end{array}$$
By direct inspection, we have that $F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2$ is always a multiple of $8$, hence $6\mid 600$ implies $8\mid\sum_{n=1}^{600}F_n^2$.
Now the brute-force approach. Let $\sigma=\frac{1+\sqrt{5}}{2}$ and $\bar{\sigma}=\frac{1-\sqrt{5}}{2}$. We have $\sigma+\bar{\sigma}=1$, $\sigma\bar{\sigma}=-1$ and
$$ F_n = \frac{\sigma^n-\bar{\sigma}^n}{\sqrt{5}},\qquad F_n^2 = \frac{\sigma^{2n}+\bar{\sigma}^{2n}-2(-1)^n}{5} $$
hence it is enough to show that $8$ is a divisor of $L_2+L_4+\ldots+L_{1200}$, where $L_n=\sigma^{n}+\bar{\sigma}^n$ is a Lucas number. Lucas numbers obey the same recurence relation as Fibonacci numbers, hence
$$ L_2+L_4+\ldots+L_{1200} = L_{1201}-L_{1} $$
can be easily proved by induction and it is enough to show that $L_{1201}\equiv L_1\pmod{8}$.
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline L_n\pmod{8} & 2 & 1 & 3 & 4 & 7 & 3 & 2 & 5 & 7 & 4 & 3 & 7 & 2 & 1 \\ \hline \end{array}$$
By direct inspection, the period of the sequence $\{L_n\pmod{8}\}_{n\geq 1}$ is $12$, hence $L_{1201}\equiv L_1\pmod{8}$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| Your first way is wrong and, unfortunately, a common mistake. With the same argument you would conclude that
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=0
$$
which is a big error (see below).
You can change $\sin x$ into $x$ when it's a factor, not a summand. For instance, if you have
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}
$$
you can as well compute
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}=
\lim_{x\to0}\frac{1-x-1}{x(\sqrt{1-x}+1)}=-\frac{1}{2}
$$
because then
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}=
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\frac{x}{\sin x}=
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\lim_{x\to0}\frac{x}{\sin x}=-\frac{1}{2}\cdot 1
$$
For the second way you have used Taylor expansions and, indeed, the first limit above can be computed correctly with them:
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=
\lim_{x\to0}\frac{x-\left(x-\dfrac{x^3}{6}+o(x^3)\right)}{x^3}=\frac{1}{6}
$$
Do you see? The quotient $\dfrac{\sin x}{x}$ is “like $1$”, but the difference $x-\sin x$ is “like $\dfrac{x^3}{6}$”.
You can tackle the computation of the limit by observing that
\begin{align}
(\sin x)^2-(x\cos x)^2
&=(\sin x-x\cos x)(\sin x+x\cos x) \\
&=\left(x-\frac{x^3}{6}+o(x^3)-x\left(1-\frac{x^2}{2}+o(x^2)\right)\right)
(x+o(x)+x(1+o(1)) \\
&=\left(\frac{x^3}{3}+o(x^3)\right)(2x+o(x)) \\
&=\frac{2}{3}x^4+o(x^4)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
square root in algebra Have a question about roots.
$$ \sqrt{9} = 3^2 = 3 \times 3 $$
$$ \sqrt{9a^6} = 3a^3 $$
But why does $ \sqrt{a^6} = a^3 $?
Is it correct vision?
$$ \sqrt{a^6} = \sqrt{a\times a \times a \times a \times a \times a} = \sqrt{2(a \times a \times a)} $$
Can anybody add more details about my question?
Thanks!
| -----$\sqrt{9} = 3^2 = 3 * 3$
Incorrect.
$\sqrt{9} = \sqrt{3^2} = 3$
====
----but why $\sqrt{a^6}=a^3$
Because $\sqrt{a^6} = \sqrt{(a^3)^2} = a^3$.
[Assuming $a \ge 0$. If $a < 0$ then $a^6 = |a|^6 =|a^6| > 0$ and $\sqrt{a^6} = |a|^3$. For the rest of the post I'm assuming $a \ge 0$.]
=====
----Is it correct vision ?
$\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{2(a∗a∗a)}$
Not quite. $\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{(a∗a∗a)*(a*a*a)}=\sqrt{a^3*a^3} = \sqrt{(a^3)^2} = a^3$
Note:
$(a*a*a*a*a*a) \ne 2(a*a*a)$
$(a*a*a*a*a*a) = (a*a*a)*(a*a*a) = (a*a*a)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Determinant of a symmetric $7\times 7$-matrix Is there any simple method to calculate the determinant of the following symmetric $7\times 7$-matrix
$$M:=\begin{pmatrix}
1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\
0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\
0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\
0 & 0 & 0 & 1 & -a_{3} & a_{2} & -a_{1}\\
a_{2} & -a_{1} & a_{4} & -a_{3} & a_0 & 0 & 0\\
a_{3} & -a_{4} & -a_{1} & a_{2} & 0 & a_0 & 0\\
a_{4} & a_{3} & -a_{2} & -a_{1} & 0 & 0 & a_0\\
\end{pmatrix}$$
where $a_i$ are real numbers.
| The determinant of $M$ is given by
$$
\det(M)=(a_0 - a_1^2 - a_2^2 + a_3^2 - a_4^2)(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^2.
$$
Note that the matrix is not symmetric.
For the symmetric variant we have
$$
\det(M)=(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^3,
$$
where we could use rules of the determinant for block matrices (the upper left corner being the identity).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.
| When we do the Tayor series expansion of the numerator, we don't care about the $x^3$ and higher powered terms. They are all going to 0. I am going to drop them as I go.
$\lim_\limits{x\to 0}\frac {1 - (1 - \frac {x^2}{2})(1-2x^2)^\frac12(1-\frac {9x^2}{2})^\frac 13}{x^2}$
Now do the binomial expansion and continue to drop the $x^3$ and higher powered terms.
$\lim_\limits{x\to 0}\frac {1 - (1 - \frac {x^2}{2})(1-x^2)(1-\frac {3x^2}{2})}{x^2}$
Multiply it out, (still dropping high powered terms)
$\lim_\limits{x\to 0}\frac {1 - (1 - (\frac 12+1+\frac 32) x^2)}{x^2}$
And evaluate.
$ 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square of both sides and I tried to factor and failed.
| The following is a shortcut that works for this particular equation.
Multiplying both sides by the sum of the radicals (which is always $\ne 0$) gives:
$$(3.5^2+x^2)-(3.0^2+x^2)=0.25(\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2})$$
$$\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2} = 13$$
Adding the latter to the original equation gives:
$$2 \sqrt{3.5^2+x^2} = 13.25$$
From there $x^2 = (\frac{13.25}{2})^2 - 3.5^2 = 43.890625 - 12.25 = 31.640625$, thus $x = \pm 5.625$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$ Let $x,y,z \in [0;1]$. Prove the inequality
$$(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$$
My work so far:
Let $f(x)=(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 -6$, where $x\in[0;1]$
$f(0)=(y^4+z^4)+(y^5+z^5)+(y)^6+(y-z)^6+(z)^6 -6$
| @Alex Silva starts off with the reduction
$$\begin{align*} & (x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \leq \\& \leq 2(x^2+y^2+z^2)+(x-y)^2+(y-z)^2+(z-x)^2\end{align*}$$
Now note that
$$(x - y)^2 \leq (1 - x)^2 + (1 - y)^2$$
which is true because it is equivalent to
$$-2xy \leq -2x + 1 - 2y + 1 \implies x + y \leq 1 + xy \implies y(1 - x) \leq 1 - x$$
The last inequality is certainly true because $(1 - x) \geq 0$ and $ y \in [0,1]$. From here, note that
$$(1 - x)^2 \leq 1 - x^2 \qquad x \in [0,1]$$
So we get
$$(x - y)^2 \leq 2 - x^2 - y^2$$
We can repeat this argument for the other terms to get
$$(x - y)^2 + (x - z)^2 + (y - z)^2 \leq 6 - 2(x^2 + y^2 + z^2)$$
$$\implies 2(x^2 + y^2 + z^2) + (x - y)^2 + (x - z)^2 + (y - x)^2 \leq 6$$
Again, credits to Alex Silva for giving the reduction. Let me know if there are any algebra mistakes.
EDIT:
In general, we have for $x,y \in [0,1]$ the following fact
$$|x - y|^k \leq (1 - x)^k + (1 - y)^k \leq (1 - x^k) + (1 - y^k)$$
where the first inequality follows from the fact that given $x$ and $y$, the distance between them is always less than the distance between $x$ and $1$ and/or the distance between $y$ and $1$. The second inequality follows from the fact that
$$(1 - x^k) = (1 - x)(1 + x + \dots + x^{k-1}) \geq (1 - x)$$
$$\implies (1 - x^k) \geq (1 - x) \geq (1 - x)^k$$
so really the proof is unnecessary because of this general case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\left(-4+3 \sin^2 \theta\right)$ from the right side, but I don't know where to go from there.
| We need to establish $$\sin2\theta\ge-\dfrac{1+3\cos^2\theta}2=-\dfrac{2+3(1+\cos2\theta)}4$$
$$\iff4\sin2\theta+3\cos2\theta\ge-5$$
Now
$4\sin2\theta+3\cos2\theta=5\sin\left(2\theta+\arcsin\dfrac45\right)$
Finally for real $\theta,$ $$-1\le\sin\left(2\theta+\arcsin\dfrac45\right)\le1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
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