Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| We have
$\displaystyle 2 + \frac{10}{9} \sqrt{3} = 1 + \sqrt{3} + 1 + \frac{1}{9} \sqrt{3}$
$\displaystyle = 1 + \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}^2} + \frac{1}{\sqrt{3}^3}= \bigl(1 + \frac{1}{\sqrt{3}}\bigr)^3$
Similarily, we have $2-\frac{10}{9} \sqrt{3} = \left(1-\frac{1}{\sqrt{3}}\right)^3$. Hence
$\displaystyle \sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} ~ = ~ \bigl(1 + \frac{1}{\sqrt{3}}\bigr) +\bigl(1-\frac{1}{\sqrt{3}}\bigr) = 2$.
PS: I have to admit that the first line is a little bit unmotivated. You need some experience in order to see these transformations (whereas verification is trivial). Alternatively, you can hope that both summands of this complicated sum actually lie in $\mathbb{Q}(\sqrt{3})=\{p+q \sqrt{3} : p,q \in \mathbb{Q}\}$, make the Ansatz $2+\frac{10}{9} \sqrt{3} = (p+q \sqrt{3})^3=\dotsc$ and solve for $p,q$, which gives $p=1$ and $q=\frac{1}{3}$ as a possible solution. The answer by lab bhattacharjee explains how to calculate the sum more "mechanically".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 2
} |
Partial fraction $\frac{x}{(1+x)^2}$ How do you compute $\dfrac{x}{(1+x)^2}$ using partial fractions? The reason I ask is because when I try to solve it I keep getting an impossible $A, B$.
$A(1+x) + B(1+x) = x$
$A + Ax + B + Bx = x$
$(A+B)x = 1$
$(A+B) = 0$
However a practice problem I'm working on implies this can be solved by partial fractions, but doesn't go into detail.
| $$\frac{x}{(1 + x)^2} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2}$$
$$x=A(1+x)+B $$
put $x=-1$ gives $B=-1$
and compare coefficient of x from both side gives $A=1$
so answer is $$\frac{x}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proving the inequality $\tan(1)\le\sum_{k=1}^{\infty} \frac{\sin(1/k^2)}{\cos^2 (1/(k+1))}$ How am I supposed to prove this inequality?
$$\tan(1)\le\sum_{k=1}^{\infty} \frac{\sin\left(\frac{1}{k^2}\right)}{\cos^2 \left(\frac{1}{k+1}\right)}$$
Jordan inequality might be an option but led me nowhere. What else to try?
| If $k \in \mathbb{N}$, $k \geq 2$ and $x \in [0, \sqrt{\frac{1}{2}}]$ then
$$\frac{\partial}{\partial x}\sin(\frac{x}{k}) \cos(x) = \frac{1}{k} \cos(\frac{x}{k}) \cos(x) - \sin(\frac{x}{k}) \sin(x) \geq \frac{1}{2 k} - \frac{x^2}{k} \geq 0
$$
so $\sin(\frac{x}{k})\cos(x)$ is non-decreasing on this interval. In particular
$$
\frac{\sin(\frac{1}{k(k+1)})}{\cos(\frac{1}{k})} \leq \frac{\sin(\frac{1}{k^2})}{\cos(\frac{1}{k+1})}
$$
for all $k\geq 2$. This inequality also holds in fact for $k=1$ since $\sin(1) < \sin(\pi - 2) = \sin(2)$.
Now we get the following inequalities for all $k \geq 1$:
$$\begin{eqnarray}
\tan(\frac{1}{k}) - \tan(\frac{1}{k+1}) &=& \frac{\sin(\frac{1}{k}) \cos(\frac{1}{k+1}) - \cos(\frac{1}{k}) \sin(\frac{1}{k+1})}{\cos(\frac{1}{k}) \cos(\frac{1}{k+1})}\\
& = &\frac{\sin(\frac{1}{k(k+1)})}{\cos(\frac{1}{k}) \cos(\frac{1}{k+1})} \leq \frac{\sin(\frac{1}{k^2})}{\cos^2(\frac{1}{k+1})}
\end{eqnarray}
$$
and summing from $1$ to $\infty$ gives the required result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Can an odd perfect number be divisible by $825$? I know that an odd perfect number cannot be divisible by $105$. I wonder if that's also the case for $825$.
| No it cannot.
Let $n=\prod p_i^{\alpha_i}=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots$ where $p_i$ is the $i$th prime number and let $S(n)$ be the sum of the divisors of $n$.
Suppose $n$ is an odd perfect number divisible by $825=3\cdot 5^2\cdot 11$, then $S(n)=2n$ and $\alpha_1=0,\alpha_2\ge 1,\alpha_3\ge 2,\alpha_5 \ge 1$.
Since
$$
\begin{align}
S(n)& = n\left(1+\frac{1}{3}+\cdots+\frac{1}{3^{\alpha_2}}\right)
\left(1+\frac{1}{5}+\cdots+\frac{1}{5^{\alpha_3}}\right)\cdots
\\
& = n \prod_i \left(\sum_{j=0}^{\alpha_i} p_i^{-j}\right)
\end{align}$$
then we must have $\alpha_2\ge 2$, since $1+1/3 = 4/3$ but $S(n)=2n$ is not divisible by 4. Likewise $\alpha_4 \ge 2$ since $1+1/11=12/11$.
Then either $\alpha_2=2$, and since $1+1/3+1/9=13/9$ it must be that $13\mid n$. Or $\alpha_2>2$. Both cases lead to contradictions so there cannot be such an odd perfect number.
If $\alpha_2=2$ then
$$
2 = \frac{S(n)}{n} \ge \frac{13}{9} \left(1+\frac{1}{5}+\frac{1}{25}\right)
\left(1+\frac{1}{11}+\frac{1}{121}\right)\left(1+\frac{1}{13}\right) > 2
$$
which cannot be, or if $\alpha_2>2$
$$
2 = \frac{S(n)}{n} \ge \left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)
\left(1+\frac{1}{5}+\frac{1}{25}\right)
\left(1+\frac{1}{11}+\frac{1}{121}\right) > 2
$$
which also cannot be.
This is adapted from this solution mentioned in the comments to the linked question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How to prove $(\frac{n+1}{e})^nHow to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.
| Induct - the base case $n = 1$ is trivial. We will show both sides of the equality by looking at ratios; indeed, $\displaystyle\frac{(n+2)^{n+1} e^n}{e^{n+1} (n+1)^n} = \frac{1}{e}\cdot\left(1+\frac{1}{n+1}\right)^{n} \cdot(n+2) < \frac{1}{1+1/(n+1)} \cdot (n+2) = n+1$ since $\left(1 + \frac{1}{n+1}\right)^{n+1} < e.$
Hence, $\left(\frac{n+2}{e}\right)^{n+1} < \left(\frac{n+1}{e}\right)^{n}\cdot (n+1) < n!\cdot (n+1) = (n+1)!$ by hypothesis. For the other side, the ratio is $\displaystyle\frac{1}{e}\cdot \left(1 + \frac{1}{n+1}\right)^{n+1}\cdot(n+2)$; it suffices to show that this is greater than $n+1.$ That is, $\displaystyle \frac{1}{e}\cdot\left(\frac{n+2}{n+1}\right)^{n+1} > \frac{n+1}{n+2},$ which is equivalent to $\left(1+ \frac{1}{n+1}\right)^{n+2} > e.$ To prove this, we use the fact that $\ln (1+h) > \frac{h}{1+h}.$ Then $(n+2)\ln\left(1+\frac{1}{n+1}\right) > (n+2)\frac{1/(n+1)}{(n+2)/(n+1)} = 1,$ whence the desired inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Expand: $e^{\sin z}$ at $z = 0$ I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.
ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)!}} = \sum_{n = 0}^\infty \frac{\small{B(1, -1/3!, \dots , (-1)^n/(2n+1)!)} z^n}{n!}$$
| Exponential formula.
Supose
$$ f(x)=a_1 x+{a_2 \over 2}x^2+{a_3 \over 6}x^3+\cdots+{a_n \over n!}x^n+\cdots.$$
Then
$$\exp f(x)=e^{f(x)}=\sum_{n=0}^\infty {b_n \over n!}x^n,$$
where
$$b_n=\sum_{\pi=\left\{\,S_1,\,\dots,\,S_k\,\right\}} a_{\left|S_1\right|}\cdots a_{\left|S_k\right|},$$
and the index $\pi$ runs through the list of all partitions $\{S_1,\ldots,S_k\}$ of the set $\{1,\ldots,n\}$. (When $k=0$ the product is empty and so equals $1$.)
We have $a_1=-a_3=a_5=-a_7=\cdots=1$ and $a_2=a_4=a_6=\cdots=0$.
So suppose we want $b_4$. There are $15$ partitions of $\{1,2,3,4\}$: one corresponding to $1+1+1+1$, six corresponding to $2+1+1$, three corresponding to $2+2$, four corresponding to $3+1$, and one corresponding to $4$. In this case we can discard the ones containing even numbers, so we need only consider $1+1+1+1$ and $3+1$. We get
$$
b_4 = a_1^4 + 4a_3 a_1 = 1 - 4 = -3.
$$
Similarly,
$$
\begin{align}
b_0 & & & =1 \\[6pt]
b_1 & = a_1 & & = 1 \\[6pt]
b_2 & = a_1^2 & & = 1 \\[6pt]
b_3 & = a_1^3 + a_3 & & = 1-1 = 0 \\[6pt]
b_4 & = a_1^4 + 4a_3 a_1 & & = 1 - 4 = -3 \\[6pt]
b_5 & = a_1^5 + 10a_3 a_1^2 + a_5 & & = 1 - 10 + 1 = -8 \\[6pt]
b_6 & = a_1^6 + 20 a_3 a_1^3 + 10 a_3^2 + 6a_5 a_1 & & = 6 - 20 + 10 + 1 = -3 \\[6pt]
b_7 & = a_1^7 + 35 a_3 a_1^4 + 70 a_3^2 a_1 + 21a_5 a_1^2 + a_7 & & = 1 - 35 + 70 + 21 - 1 = 56
\end{align}
$$
And so on.
The radius of convergence is $\infty$, because this is a composition of two entire functions, and is therefore an entire function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$? Numerically, it looks that the limit is
$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,
but I have not been able to demonstrate it analytically. Does anyone have a idea on how to deal with this limit?
| Here is how you advance, compute the Taylor series of $\frac{1}{\sqrt{x-y}}$ at the point $x=1$
$$ \frac{1}{\sqrt{x-y}} = {\frac {1}{\sqrt {1-y}}}-\frac{1}{2}\, \left( 1-y \right) ^{-3/2} \left( x-1
\right) +O \left( \left( x-1 \right) ^{2} \right) $$
$$ \implies \frac{1}{\sqrt{x-y}} \sim {\frac {1}{\sqrt {1-y}}}. $$
So, when $x$ close to $1$, our integral behaves as
$$\int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {x-y}}dy \sim \int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {1-y}}dy$$
$$=\int _{0}^{1}\!{\frac{ \left( -u+1 \right)^{3/2}}{\sqrt {-u+2}u}}{du}\sim \int _{0}^{1}\!{\frac {1}{\sqrt {2}u}}{du}.$$
The last integral does not converge. Note that, we used the change of variables $1-y=u$ and
$$ \frac{ ( -u+1 )^{3/2}}{\sqrt{-u+2}u} \sim \frac{1}{\sqrt{2}\,u} \quad \mathrm{when}\quad u\sim 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point.
| Alt. hint: following up on OP's $^{\tiny\text{(?)}}\,$ approach, with the missing $\,=3\,$ appended...
I tried to expand the left hand side: $$\;xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)} = 3 \tag{1}$$
Before that, multiplying the original equation by $\,\left(x\color{red}{-}\sqrt{x^2+3}\right)\left(y\color{red}{-}\sqrt{y^2+3}\right)\,$ gives:
$$
(-3)\,(-3) = 3 (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) \;\;\iff\;\; (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) = 3 \tag{2}
$$
Expanding $(2)$ and subtracting from $(1)$ then gives:
$$\require{cancel}
\begin{align}
y\sqrt{x^2+3} + x\sqrt{y^2+3} = 0 \;\;&\iff\;\; y\sqrt{x^2+3} = -x\sqrt{y^2+3} \\
&\implies\;\;y^2(\cancel{x^2}+3)=x^2(\cancel{y^2}+3) \\
&\iff\;\; x^2 = y^2 \\
&\iff\;\; (x-y)(x+y) = 0
\end{align}
$$
The case $\,x-y = 0\,$ has the only solution $\,x=y=0\,$, which leaves $\,x+y=0\,$ either way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$? We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$
What is the limit of this function as $n \rightarrow \infty$?
My idea:
$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$
Is this correct?
| Your method is wrong. Consider the first values of the expression:
*
*for $n=1$ you get $1$;
*for $n=2$ you get $1/4+2/4=3/4$;
*for $n=3$ yoy get $1/9+2/9+3/9=6/9=2/3$.
One could make the conjecture that these terms are always bigger than $1/2$, which can be proved by induction. Let
$$
f(n)=\frac{1}{n^2}\sum_{i=1}^n i
$$
We have $f(1)=1>1/2$. Suppose the assertion holds for $f(n)$ and consider
\begin{align}
f(n+1)&=\frac{1}{(n+1)^2}\sum_{i=1}^{n+1} i\\
&=\biggl(\frac{1}{(n+1)^2}\sum_{i=1}^{n} i\biggr)+\frac{n+1}{(n+1)^2}\\
&=\frac{n^2}{(n+1)^2}\biggl(\frac{1}{n^2}\sum_{i=1}^n i\biggr)+\frac{1}{n+1}\\
\text{(by induction hypothesis)}\qquad&>\frac{1}{2}\frac{n^2}{(n+1)^2}+\frac{1}{n+1}\\
&=\frac{1}{2}\frac{n^2+2n+2}{(n+1)^2}\\
&=\frac{1}{2}\biggl(\frac{n^2+2n+1}{(n+1)^2}+\frac{1}{(n+1)^2}\biggr)\\
&=\frac{1}{2}\biggl(1+\frac{1}{(n+1)^2}\biggr)\\
&>\frac{1}{2}
\end{align}
Therefore your conclusion that the limit is $0$ cannot be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Integration of a rational function from +/- infinity I am trying to calculate the integral
$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$
where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?
| Hint: Use partial fraction to write the integral as
$$ {\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}} = \frac{A+Bx}{b^2 + (a+x)^2} + \frac{C+D x}{1+c(a-x)^2}, $$
then you can write it as
$$ \frac{A}{b^2 + (a+x)^2} + \frac{C}{1+c(a-x)^2} + \frac{Bx}{b^2 + (a+x)^2} + \frac{D x}{1+c(a-x)^2}, $$
and finally make the change of variables $x-a=t$ and $x+a=t$ and things will be easy. See here for similar techniques and read the comments under my answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the maximal $t$ satisfying a family of inequalities Given $c \in (0,1)$, find the maximal positive $t$ satisfying the following:
$$\forall n \in \{1,2,\ldots \}: 1+\frac{c}{n+(1-c)} \le \left(1+\frac{1}{n+t}\right)^{c}$$
My progress thus far:
*
*A special case is $c = \frac{1}{2}$: the inequality simplifies to $3t-1 \le n(1-4t)$, which implies $t \le \frac{1}{4}$. $t=\frac{1}{4}$ actually works.
*The inequality $\left(1+\frac{1}{n+t}\right)^{c} \le \frac{c}{n+t}$ shows $t \le 1-c$.
| The correct $t$ is $\frac{1-c}{2}$, as I was able to guess in the comments using vadim123's calculations.
Proof: define $f(x)=\ln(1+\frac{1}{\frac{1}{x}+t})^c - \ln(1 + \frac{c}{\frac{1}{x}+1-c})$. We are given that $f(\frac{1}{n}) \ge 0$ for any $n \ge 1$.
Simplifying $f$, we find $f(x)=c \ln(1+\frac{x}{1+tx})-\ln(1+\frac{cx}{1+x(1-c)})$.
The first derivative is $f'(x)=c (\frac{1}{(1+tx)(1+(t+1)x)}-\frac{1}{(1+x)(1+(1-c)x)})$
If $t=\frac{1-c}{2}$, the first derivative is positive for positive $x$, since it amounts to $(1+x(1-c))(1+x) \ge (1+x\frac{1-c}{2})(1+x\frac{3-c}{2})$, which upon expanding it becomes $x^2(1-c^2) \ge 0$. This shows that $f$ is increasing, hence $f(\frac{1}{n}) \ge f(0) = 0$.
In general, $f'(x) \ge 0$ iff $x(1-c-t(t+1))\ge 2t+c-1$ (assuming $x \ge 0$). Assume $1-c\ge t > \frac{1-c}{2}$ also works. Since $ 2t+c-1 > 0$, $f$ is necessarily decreasing in a small right neighborhood of $0$, so $f(\frac{1}{n})$ is negative for big enough $n$ since $f(0)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find eigenvectors of $A$ Let $A=\begin{pmatrix} 0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2 \end{pmatrix}$
I find eignevalues $-1,0,1,2$ and $0$ has multiplicity of $2$.
But I only find one eigenvector of $0$ and all the other eigenvalues have eigenvectors, so anyway $A$ has four eigenvectors and diagonalizable. But I think $0$ must have $2$ eigenvectors and one of eigenvalue can't have eigenvector. What's wrong with me?
I corrected matrix!!!!!
Then the C.P is $x(x^3-2x^2-x+2)=x(x+1)(x-1)(x-2)$ and the eigenvalues that I got is right.
| $$\begin{align} \det(A-xI) &= \det(\begin{pmatrix}-x&0&0&0\\1&-x&0&-2\\0&1&-x&1\\0&0&1&2-x\end{pmatrix})\\ &=-x(-x(-x(2-x)-1)-2) \\&=-x(2x^2-x^3+x-2)\\&=-x(-x^2(x-2)+x-2) \\&=-x(x-2)(-x^2+1)\end{align}$$
so $x=0,x=2,x=1,x=-1$ are eigenvalues
then we must find X such that :
$(\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}-\lambda_iI)X=0$
$\to$
$\lambda$=0$ :\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}X=0 \to x_1=2x_4$ and $x_2=-x_4$and $x_3=-2x_4$ our vector can be $(-2x_4,x_4,2x_4,x_4)$as(2,-1,-2,1)
$x=2:$
$\begin{pmatrix}-2&0&0&0\\1&-2&0&-2\\0&1&-2&1\\0&0&1&2-2\end{pmatrix}X=0$$\to $$x_1=0,x_2=-x_4,x_3=0\,\,$as $\,\,(0,-1,0,1)$
$x=1:$
$\begin{pmatrix}-1&0&0&0\\1&-1&0&-2\\0&1&-1&1\\0&0&1&2-1\end{pmatrix})X=0$ $\to $
$x_1=0,x_2=-2x_4,x_3=-x_4$
as $\,\,(0,-2,-1,1)$
$x=-1:$
$\begin{pmatrix}1&0&0&0\\1&1&0&-2\\0&1&1&1\\0&0&1&2+1\end{pmatrix}X=0$$\to$$x_1=0,x_2=2x_4,x_3=-3x_4\,\,$as $\,\,(0,2,-3,1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Definite integration of a trigonometric function
How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$
my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$
I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.
| $$\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{2a \sin^2 x}{a^2\sin^2 x +b^2 cos^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{a^2\csc^2 x +b^2 \cot^2 x \csc^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{(\cot^2 x+1)(a^2+b^2\cot^2 x)} dx$$
Let $u=\cot x$
$$\displaystyle 2a\int_0^{\infty} \dfrac{du }{(u^2 +1)(a^2+b^2u^2)} $$
$$\displaystyle 2a\int_0^{\infty} \left(\dfrac{1}{(u^2+1)(a^2-b^2)}-\dfrac{b^2 }{(a^2 -b^2)(a^2+b^2u^2)}\right) du $$
$$\displaystyle \frac{2a}{a^2-b^2}\int_0^{\infty} \left(\dfrac{1}{(u^2+1)}-\dfrac{b^2 }{(a^2+b^2u^2)}\right) du $$
And I hope you'll take it from here...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the value of $x^3-x^{-3}$ given that $x^2+x^{-2} = 83$
If $x>1$ and $x^2+\dfrac {1}{x^2}=83$, find the value of the expression$$x^3-\dfrac {1}{x^3}$$
a) $764$
b) $750$
c) $756$
d) $760$
In this question from given I tried to approximate the value of $x$ which should just above to 9 then I tried to calculate the value of cubic expression but all options are close enough to guess. Any idea to solve it?
| First, notice that
$$\left(x-\frac{1}{x}\right)^2=x^2-2+\frac{1}{x^2}=83-2=81$$
Then, use the difference of two cubes formula:
$$x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)=9\cdot(83+1)=756$$
We take the positive root of $81$ because $x>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$? I suspect that
$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
{{\pi}\over{2}}$$
for $C>0$.
I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?
| So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:
$ \frac{y^2}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{A}{y^2-\frac{2-C-\sqrt{C(C-4)}}{2}}+\frac{B}{y^2-\frac{2-C+\sqrt{C(C-4)}}{2}}=\frac{A(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})+B(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{y^2(A+B)-1(A(\frac{2-C-\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}$.
Therefore,we want $A+B=1 \wedge A(\frac{2-C+\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4}}{2})=0$. This is a linear system, that you can solve to find A and B, and then calculate the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| $$(5 + 2\sqrt{3}) = (7 + \sqrt{3})(a - \sqrt{3b})$$
$$=7a-7\sqrt b\sqrt 3 +a\sqrt 3-3\sqrt b$$
compare the coefficients of $\sqrt 3$ of both sides
we get$$7a-3\sqrt b=5$$ and $$a-7\sqrt b=2$$
solve for a and b
$a=29/46$ and $\sqrt b=-9/46$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$ $\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$
*
*$x=2\tan^{-1}t$
*$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$
*$dx=\dfrac{2}{1+t^2}\,dt$
*$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$
Is this the right idea? If so what do I do next?
$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$
$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$
$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$
| If $x=2\arctan t$ then $\cos x = \cos\left(2\arctan t\right)$. Use the fact that $\cos(2u)=\cos^2u-\sin^2u$. So you get
$$
\cos x = \cos^2\left(\arctan t \right) -\sin^2\left(\arctan t \right)
$$
If $\varphi=\arctan t$ then $\dfrac t1=t = \tan\varphi=\dfrac{\text{opposite}}{\text{adjacent}}$, so $\cos\varphi=\dfrac{\text{adjacent}}{\text{hypotenuse}}$. So $\text{opposite}=t$, $\text{adjacent}=1$, and by the Pythagorean theorem $\text{hypotenuse}=\sqrt{1+t^2}$.
Then $\cos=\dfrac{\text{adjacent}}{\text{hypotenuse}}$ and so
$$
\cos^2(2\arctan t) = \cos^2 (\cdot)-\sin^2(\cdot) = \left(\frac{1}{\sqrt{1+t^2}}\right)^2 - \left(\text{something similar}\right)^2 = \cdots\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$ I have this integral to evaluate:
$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$
I have tried substitution, trig identity and integration by parts but I'm just going round in circles.
Can anyone explain the method I need to work this out?
| Let's try some imaginary stuff.
$\dfrac{x^4}{x^2+1}=\dfrac{Ax^3}{x+i}+\dfrac{Bx^3}{x-i} \implies A=B= \dfrac{1}{2}$
$\dfrac{x^4+1}{x^2+1}= \dfrac{1x^3}{2(x+i)}+\dfrac{1x^3}{2(x-i)}+\dfrac{1}{x^2+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got $$s + 13 = 4 + 4\sqrt{7-s}+ 7 – s$$
I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
| You're almost at the end result:
After achieving $ 2s + 2 = 4 \sqrt{7-s} $, square both sides again to generate a quadratic equation.
$$ 4 ( s + 1 )^2 = 16 ( 7 - s ) = 4 (s^2 + 2s+ 1)$$
Solving which is quite easy.
$$\begin{align}
s^2 + 2s + 1 &= 28 - 4s \\
s^2 + 6s - 27 &= 0 \\
(s + 9) (s - 3) &= 0
\end{align}$$
NOTE
You have to neglect $ s = -9 $ from the solutions as it'll give you $ \sqrt{-9 + 13} - \sqrt{ 7 - (-9)} = \sqrt{16} = +2 - (+4) \neq 2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Help with school solid geometry In the triangular pyramid $MABC$ all side edges equals $1$, $\angle AMB = \angle BMC = 60 ^\circ$, $\angle AMC = 45 ^\circ$.
Find:
1) square of the $\triangle ABC$;
2) dihedral angle on the $AB$ edge;
3) $\rho(M, ABC)$.
Help me, please.
P.S. sorry for my English, I am not a native speaker.
| Wow the question recalls my old memory about high school.
Actually you know
$$AM = MC = 1$$
and because of the law of cosines
$$cos(\angle AMC) = \frac{\sqrt{2}}{2} = \frac{1^2+1^2-AC^2}{2\cdot 1 \cdot 1}\Rightarrow AC = \sqrt{2-\sqrt{2}}.$$
The area can be computed by Heron's formula.
Notice $\triangle AMC = \triangle ABC$, so $\angle ABC = \pi/4.$
Draw a line(in plane $ABC$) through the middle point $D$ of $AB$ and it intersects $BC$ at $E$. We know
$$AD = BD = DE = AB/2 = \frac{1}{2}\text{ and }BE = \frac{\sqrt{2}}{2}$$
And you can show $AB\perp \triangle MDE$ by $AB\perp MD, DE$. So $\angle MDE$ is what you want to solve.
Again invoking the law of cosine for $\angle MBE$, $ME^2 = 1+\frac{1}{2}-\frac{\sqrt{2}}{2} = \frac{3-\sqrt{2}}{2}$. Now use the same law to $\angle MDE:$
$$\cos\angle MDE = \frac{\frac{3}{4}+\frac{1}{4}-ME^2}{2\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2}} = \frac{3-\sqrt{3}}{3}.$$
Immediately from this result, draw a line $MF$ in plane $MDE$ intersecting $CD$ at $F$. Since $AB\perp MDE\Rightarrow AB\perp MF$ and $MF\perp CD$, we have $MF\perp ABC$, and
$$\rho(M,ABC) = MD\cdot\sin\angle MDE.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need help with integral and change the variable I have
$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$
and I need to make change of the variable $u = \sin x$.
| Hint: Try this if you want:
$$x\sqrt { x } =u\\ x={ u }^{ \frac { 2 }{ 3 } }\\ \sqrt { x } ={ u }^{ \frac { 1 }{ 3 } }\\ dx=\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 } }du\\ \int _{ 0 }^{ \pi } \sqrt { x } \cos x~ dx=\int _{ 0 }^{ \pi }{ u } ^{ \frac { 1 }{ 3 } }\cos { u } ^{ \frac { 2 }{ 3 } }\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 } }du=\frac { 2 }{ 3 } \int _{ 0 }^{ \pi\sqrt\pi }{ \cos { u } ^{ \frac { 2 }{ 3 } }du } $$
$\int { \cos { u } ^{ \frac { 2 }{ 3 } }du } $ does not have an analytical expression. You can try on wolframalpha.com .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$n^5-n$ is divisible by $10$? I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$. However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!
| Clearly, $n$ and $n^5$ are of the same parity. Hence, $2 \vert (n^5-n)$.
To check for divisibility by $5$, note that
\begin{align}
n^5 - n & = (n^2+1)n(n+1)(n-1)\\
& = (n^2-4+5)n(n+1)(n-1)\\
& = (n^2-4)n(n+1)(n-1) + 5n(n+1)(n-1)\\
& = (n-2)(n-1)n (n+1)(n+2) + 5n(n+1)(n-1)\\
\end{align}
Clearly, $5 \vert 5n(n+1)(n-1)$. Also, $(n-2)(n-1)n (n+1)(n+2)$ is a product of $5$ consecutive numbers and hence $5$ divides it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 0
} |
How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
| From the formulas: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\\\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
we have: $\cos(a)\cos(b)=\frac{1}{2}\big(\cos(a+b)+\cos(a-b)\big)$
Apply this formula:
$\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{11\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \\=\frac{1}{2}\big(\cos(\pi)+\cos(\frac{10\pi}{12})\big)\frac{1}{2}\big(\cos(\pi)+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(-1-\cos(\frac{2\pi}{12})\big)\big(-1+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(1-\cos(\frac{2\pi}{12})^2\big)$
You certainly know: $\cos(\frac{2\pi}{12})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$
Result is: $\frac{1}{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Limit of a Sequence involving cubic root I succeed in finding the following limit applying binomials and squeeze theorem:
$$\lim(\sqrt{n+1} - \sqrt{n}) = \lim\frac{1}{\sqrt{n+1} + \sqrt{n}} = 0$$ because $0 \leq \frac{1}{\sqrt{n+1} + \sqrt{n}} \leq \frac{1}{\sqrt{n}}$
But I need help because I'm not finding any way to simplify and solve the following limit:
$$\lim(\sqrt[3]{1-n^3} + n)$$
| The point is that $\sqrt[3]{n^3-1}$ is awfully close to $n$ when $n$ is large. For a crude estimate, note that
$$\left(n-\frac{1}{n}\right)^3=n^3-3n+\frac{3}{n}-\frac{1}{n^3}\lt n^3 -1$$ if $n\ge 2$. It follows that if $n\ge 2$ then
$$ 0\lt n-\sqrt[3]{n^3-1} \lt n-\left(n-\frac{1}{n}\right)=\frac{1}{n},$$
and the rest follows by Squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true? $a_1a_2+a_2a_3+...+a_na_1\leq n$
Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true?
$$a_1a_2+a_2a_3+...+a_na_1\leq n$$
By observing:
When $n=1$, this is trivial;
When $n=2$, $ab\leq(\frac {a+b} 2)^2=1\leq2$;
When $n=3$, $ab+bc+ca\leq(\frac {a+b} 2)^2+(\frac {b+c} 2)^2+(\frac {c+a} 2)^2=\frac 1 2(a+b+c)^2-\frac 1 2(ab+bc+ca)$
$\Rightarrow ab+bc+ca\leq3$;
When $n=4$, $ab+bc+cd+da=(a+c)(b+d)\leq(\frac{a+b+c+d} 2)^2=4$.
But I can't find a more general way to prove these at once. Please help.
Thank you.
| The inequality does not hold in general. Indeed, take $a_1=a_2=\frac{n}{2}-1$ and
$a_3=..=a_n=\frac{2}{n-2}.$ Then the left hand side of our inequality is greater that $n^2/4-n+1$ which can be made greater than $n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
can the following sum be simplified For $n \ge 3$, define
$$f(n) = \sum_{k=3}^n {n \choose k}{k-1 \choose 2}.$$
Is there a closed form expression for $f$?
| We will use the identities
$$
\begin{align}
\sum_{k=m}^n\binom{n}{k}\binom{k}{m}
&=\sum_{k=m}^n\binom{n}{m}\binom{n-m}{k-m}\\
&=\binom{n}{m}2^{n-m}\tag{1}
\end{align}
$$
and
$$
\binom{k-1}{2}=\binom{k}{2}-\binom{k}{1}+\binom{k}{0}\tag{2}
$$
noting that
$$
\binom{k-1}{2}=\left\{\begin{array}{}
0&\text{for }k\in\{1,2\}\\
1&\text{for }k=0
\end{array}\right.\tag{3}
$$
Applying $(1)$, $(2)$, and $(3)$, we get
$$
\begin{align}
\sum_{k=3}^n\binom{n}{k}\binom{k-1}{2}
&=-\binom{n}{0}\binom{-1}{2}+\sum_{k=0}^n\binom{n}{k}\binom{k-1}{2}\\
&=-1+\sum_{k=0}^n\binom{n}{k}\left(\binom{k}{2}-\binom{k}{1}+\binom{k}{0}\right)\\
&=-1+\binom{n}{2}2^{n-2}-\binom{n}{1}2^{n-1}+\binom{n}{0}2^{n-0}\\[9pt]
&=(n^2-5n+8)2^{n-3}-1\tag{4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
| WOLOG, suppose $c$ is the largest. If $a>b$, then $(a-b)(b-c)(c-a)\le 0$. So we may suppose $a<b<c$. Let $b-a=x\ge 0, c-b=y\ge 0$, then $3a+2x+y=1$, thus $2x+y\le 1$. The problem becomes find maximum value of
$$
xy(x+y), \quad x, y \ge0.
$$
So we may suppose $2x+y=1$, then
$$
xy(x+y)=x(1-2x)(1-x)=:f(x), \quad 0\le x\le \frac{1}{2}.
$$
Put $f'(x)=1-6x+6x^2=0$, when $x=\frac{3-\sqrt{3}}{6}$, $f(x)$ attains the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(x+1)^{10}.(x^2 -1)^8$$
| From the binomial theorem you know that
$$(x+1)^{10}=\sum_{k=0}^{10}\binom{10}kx^k$$
and
$$(x^2-1)^8=\sum_{k=0}^8\binom8k(-1)^{8-k}x^{2k}=\sum_{k=0}^8\binom8k(-1)^kx^{2k}\;,$$
so
$$\begin{align*}
(x+1)^{10}(x^2-1)^8&=\left(\sum_{k=0}^{10}\binom{10}kx^k\right)\left(\sum_{k=0}^8\binom8k(-1)^kx^{2k}\right)\\\\
&=\left(1+10x+\ldots+10x^9+x^{10}\right)\left(1-8x^2+\ldots-8x^{14}+x^{16}\right)\;.
\end{align*}$$
The $x^{20}$ term in the product will be the sum of all terms of the form $(ax^k)(bx^\ell)$ such that $k+\ell=20$. Find all of those combinations of $k$ and $\ell$, and you’re nearly there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $a$ and $b$ are non-negative real numbers, prove that $ab(a+b) \leq a^2+b^2$.
If $a$ and $b$ are non-negative real numbers, prove that $ab(a+b) \leq a^2+b^2$.
Is is a geometric mean? How to prove it?
| Note that if $a+b\le 2$, then $ab(a+b)\le 2ab$. And we have $2ab\le a^2+b^2$, since $(a-b)^2\ge 0$. So the desired inequality does hold if $a+b\le 2$.
If the condition $a+b\le 2$ is violated, then the inequality can fail. For let $a=b=k$. Then $ab(a+b)=2k^3$ and $a^2+b^2=2k^2$. If $k\gt 1$ then $2k^3\gt 2k^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$
Without loss of generality we can let $x=\max{\{x,y,z\}}$
Proof: case 1
$x\ge y\ge z$
we can easily prove
$$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$
and
$$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$
so we have
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$
Then use
$$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$
$$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$
$$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$
to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.
Now,I have another idea: using Holder inequality
we have
$$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$
$$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$
and the following link has some discussion about this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230
and
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872
and Vasc gave the hint:
$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$
How prove this hint?Thank you everyone.
and my other idea is as follows:
let $a=\min(a,b,c)$
we can prove
$$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$
\begin{align*}
&\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2}
\end{align*}
Summing up
\begin{align*}
&\sum_{cyc}{\sqrt{a^2+bc+2}}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\
=&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8}
\end{align*}
By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!
| Let's solve
$$ \min \Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg]$$
subject to
$$
x+y+z=3 \\
x\ge0 \\
y\ge 0 \\
z\ge 0
$$
Formally, this is done with Kuhn-Tucker condition. Write those down, exploit symmetries, note that non-negativity constraints do not bind.
Solution: $x=y=z=1$. Hence, the minimum value of the objective is $6$.
The other inequality admits the same proof.
details: see wiki or original paper for details of KKT conditions.
$$
L(x,y,z,\lambda,\mu_1,\mu_2,\mu_3) = -\Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg] + \lambda[x+y+z=3] + \mu_1[x-0] +\mu_2[y-0] +\mu_3[z-0]
$$
KKT conditions:
\begin{align}
[x]\qquad
-\frac{1}{2}\eta_1 2x -\frac{1}{2}\eta_2 z-\frac{1}{2}\eta_3 y &= \lambda - \mu_1 \\
[y]\qquad
-\frac{1}{2}\eta_1 z -\frac{1}{2}\eta_2 2y-\frac{1}{2}\eta_3 x &= \lambda - \mu_2 \\
[z]\qquad
-\frac{1}{2}\eta_1 y -\frac{1}{2}\eta_2 x-\frac{1}{2}\eta_3 2z &= \lambda - \mu_3 \\
[\lambda]\qquad\qquad\qquad\qquad\quad
x+y+z&=3\\
[\mu_1]\qquad
\mu_1\ge 0\qquad\text{and}\qquad \mu_1x&=0 \\
[\mu_2]\qquad
\mu_2\ge 0\qquad\text{and}\qquad \mu_2y&=0 \\
[\mu_3]\qquad
\mu_3\ge 0\qquad\text{and}\qquad \mu_3z&=0 \\
\end{align}
where $\eta_1 = (x^2+yz+2)^{-1/2},\eta_2 = (y^2+zx+2)^{-1/2}$ and $\eta_1 = (z^2+xy+2)^{-1/2}$
KKT methods states that the minimum of the objective should be a solution (for some $\lambda,\mu_1,\mu_2,\mu_3$) to the above system.
Wlog, let $x=\max\{x,y,z\}$ so that $x>0$ and $\mu_1=0$.
Case 1: $y=z=0$. Solve for $\lambda$ from $[x]$ and note that $\mu_2 <0$, therefore this case is empty.
Case 2: $y=0$ and $z=3-x\ne0$. Then $\mu_3=0$. Solve for $\lambda$ in $[x],[z]$, equate the two expressions and deduce that $x=z$. Observe that $\mu_2<0$ (at $x=z$), therefore this case is empty.
Case 3: $\mu_1=\mu_2=\mu_3=0$ so that non-negativity constraints are slack. Equate LHS of conditions $[x],[y],[z]$, combine with condition $[\lambda]$ to solve for $x,y,z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 0
} |
Probability of adjacent seating A homework question states:
A room holds two rows of six seats each. Two friends are assigned
randomly to the 12 seats. What is the probability that the 2 friends
sit in adjacent seats?
Note: Friends sitting behind friends don't count. Friends sitting
diagonally adjacent to each other don't count. Only friends setting
beside each other (left/right) in the same row count.
$$
\cdot~~~~~= Empty~seat
$$
$$
\times = Occupied~seat
$$
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
By drawing out the favorable possibilities:
$
\begin{bmatrix}
\times & \times & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \times & \times & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \times & \times & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \times & \times & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \times & \times \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$
$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\times & \times & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \times & \times & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \times & \times & \cdot & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \times & \times & \cdot \\
\end{bmatrix}
$$
\begin{bmatrix}
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \times & \times \\
\end{bmatrix}
$
It seems like there are a total of 10 favorable situations.
I hope I'm right in saying there are a total of ${12 \choose 2}$ total possible situations (friends can sit in any two seats)?
So is the probability that 2 friends sit adjacent to each other in this room of 12 seats:
$$ \frac{10}{{12 \choose 2}} = \frac{10}{66} = 0.1515152$$
Whether that's right or wrong, I guess, what's the better mathematical approach (using the whole ${X \choose Y}$ thing to think about this problem?
| The calculation is correct, and efficiently done. There are $\binom{12}{2}$ equally likely ways to select $2$ seats from $12$.
You then counted the "favourables" well, though it was quite unnecessary to list separately the favourables in the first row and the favourables in the second.
Since you did the problem in a nice way, let me do it an uglier way. Call the people Alicia and Bob, and seat Alicia first. We could (i) put her in an end seat or (ii) not in an end seat.
The probability we put Alicia in an end seat is $\frac{4}{12}$, Given she is in an end seat, the probability Bob ends up beside her is $\frac{1}{11}$.
The probability Alicia is not in an end seat is $\frac{8}{12}$, and then Bob has probability $\frac{2}{11}$ of ending up beside her. Thus our probability is
$$\frac{4}{12}\cdot\frac{1}{11}+\frac{8}{12}\cdot\frac{2}{11}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/419053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Create Fourier-Series of f(x) = x if 0 < x < Pi and 0 if Pi < x < 2*Pi I tried the following to create the Fourier-series of the function:
$$ f(x) = \begin{cases} x & 0<x<\pi \\ 0 & \pi < x < 2 \pi \end{cases}$$
This is what I tried:
$$a_0 = \frac{1}{\pi}\int\limits_0^\pi x dx = \frac{1}{\pi}\cdot\left(\frac{x^2}{2}|_0^\pi\right) = \frac{1}{\pi}\cdot\frac{\pi^2}{2} = \frac{\pi^2}{2\cdot\pi} = \frac{\pi}{2}$$
Is that right?
$$a_n = \frac{1}{\pi}\int\limits_0^\pi x \cdot \cos(nx) dx = \frac{1}{\pi}\cdot\left(\frac{x \cdot \sin(nx)}{n} + \frac{\cos(nx)}{n^2}\right)|_0^\pi$$
Assuming that $\sin(nx)$ is always $0$ and $\cos(nx)$ is $(-1)^n$. Is that ok??
$$ = \frac{1}{\pi}\cdot\left(\frac{0}{n} + \frac{(-1)^n}{n^2}\right) = \frac{1}{\pi}\cdot\left(\frac{(-1)^n}{n^2}\right) = \frac{(-1)^n}{\pi n^2}$$
That's already wrong I'm sure!
$$b_n = \frac{1}{\pi}\int\limits_0^\pi x \sin(nx) dx = \frac{1}{\pi}\cdot\left(\frac{x \cdot \cos(nx)}{n} + \frac{\sin(nx)}{n^2}\right)|_0^\pi$$
Assuming that $\sin(nx)$ is always $0$ and $\cos(nx)$ is $(-1)^n$. Is that ok??
$$ = \frac{1}{\pi}\cdot\left(\frac{\pi \cdot (-1)^n}{n}\right) = \frac{\pi\cdot(-1)^n}{\pi n} = \frac{(-1)^n}{n}$$ voila, thats wrong again :-)
A solution would be great, but I also need to understand what was my mistake.
Thank you in advance!
| Plugging in $x=0$ and $x=\pi$ in $\sin (nx)$ indeed gives you $0$.
Plugging in $x=\pi$ in $\cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $\cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/422646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solutions for $p$ where $2 p^2 - 1 = q^2$ Consider this equation:
$$2 p^2 - 1 = q^2$$
where $p$ and $q$ are prime.
After vigorous checking, I couldn't find any solutions $p>29.$
Is it so that $p=5,\;p = 29$ are the only solutions?
Edit:
Found solutions $p>29$ with the help this.
The solution for $p,q$ can be expressed as:
$$\begin{align}p &=a_n= {1\over4} \left(2 (3-2 \sqrt{2})^n+\sqrt{2} (3-2 \sqrt{2})^n+2 (3+2 \sqrt{2})^n-\sqrt{2} (3+2 \sqrt{2})^n\right)&\\
q &=b_n=- {1\over2} \left((3-2 \sqrt{2})^n+\sqrt{2} (3-2 \sqrt{2})^n+ (3+2 \sqrt{2})^n-\sqrt{2} (3+2 \sqrt{2})^n\right)&\end{align} $$
thus:
$$\begin{align}
2 p^2 - 1 = q^2\rightarrow\text{True}&\implies\\
p = a_n,\;q=b_n,1<n\\
\end{align}$$
So when both $a_n$ and $b_n$ are both primes, then $2 a_n^2 - 1 = b_n^2$.
Solutions for $n$ I've found so far are:
$$\begin{align}
n=2&&a_n = 5&&b_n = 7\\
n=3&&a_n = 29&&b_n = 49\\
n=15&&a_n = 44560482149&&b_n = 63018038201\\
n=30&&a_n = 19175002942688032928599&&b_n = 19175002942688032928599\\
\end{align}$$
| There are at least 2 more solutions, you just haven't searched far enough
OEIS A163742 gives solutions $(63018038201, 44560482149)$ and $(19175002942688032928599, 13558774610046711780701)$. It is conjectured that these are the only 4 prime pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
failed application of magicry in Taylor expansion of $1/x^2$ near $x=2$ It's straightforward to find the Taylor expansion for $\frac{1}{x^2}$ near $x=2$ using the the Taylor series definition.
This is turns out to be $\frac{1}{4} - \frac{1}{4} (x-2) + \frac{3}{16}(x-2)^2 + \cdots$
I was trying to be cute by finding the expansion using the geometric series:
$$
\begin{align}
\frac{1}{x^2} &= \frac{1}{1 -(1 - x^2)} = \frac{1}{1-z} &\text{ where } z = 1-x^2\\
\end{align}
$$
This is where I run into trouble. My initial guess was to expand around $x=0$ and then shift over by 2.
$$
\begin{align}
&1 + z + z^2+ \cdots\\
&1 + (1-x^2) + (1-x^2) ^2 + \cdots \\
&1 + (\ 1-(x-2)^2\ ) + (\ 1-(x-2)^2\ )^2+\cdots
\end{align}
$$
I looked at the Wikipedia page on the geometric series and see that the formula is different when the common ratio is not 1.
$$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$$
This helped me to see that I had missed the $a$ term. I'm taking it on faith that the common ratio is less than one. The article points out that when $|r|<1$, the series will be $\frac{a}{1-r}$, which is the familiar geometric series.
Somehow, my substitution seems to be wrong. Any help will be appreciated.
| Your idea is sound... iff
$$|1-x^2|<1\implies 0<x^2<2\implies |x|<\sqrt2$$
Now, you want to develop around $\,x=2\,$...not good, uh?
My idea:
$$\frac1x=\frac1{2+(x-2)}=\frac12\frac1{1+\left(\frac{x-2}2\right)}=\frac12\left(1-\frac{x-2}2+\frac{(x-2)^2}4-\ldots\right)\implies$$
$$\frac1{x^2}=\frac1x\frac1x=\frac14\left(1-\frac{x-2}2+\frac{(x-2)^2}4-\ldots\right)^2=$$
$$=\frac14\left(1-(x-2)+\frac34(x-2)^2-\frac38(x-2)^3+\ldots\right)$$
The only thing you need now to justify the above is to prove that
$$\left|\frac{x-2}2\right|<1\iff |x-2|<2\iff\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Product and Sum of Polynomial Roots The ratio of the sum of the roots of the equation, $8x^3+px^2-2x+1=0 $ to the product of the roots of the equation $5x^3+7x^3-3x+q=0 $ is $3:2$. What is the value $\frac{p-q}{p+q}$?
Well I found out the sum of the roots of 1st equation is $-\frac{p}{8}$ and product of roots of second equation is $-\frac{d}{5}$.
Now what to do further? How to get in terms of $\frac{p-q}{p+q} $ ?
| Using Vieta's Formulas,
the sum of the roots of $\ 8x^3+px^2-2x+1=0$ is $\frac{-p}8$
and the product of the roots of $\ 5x^3+7x^3-3x+q=0$ is $\frac {-q}5$
So, $$\frac {\frac{-p}8}{\frac {-q}5}=\frac32\implies \frac pq=\frac{3\cdot 8}{2\cdot 5}=\frac{12}5$$
Now, use Componendo and dividendo
or use $$\frac{p-q}{p+q}=\frac{\frac pq-1}{\frac pq+1}=\frac{\frac {12}5-1}{\frac {12}5+1}=\frac{12-5}{12+5}$$
or use $$\frac p{12}=\frac q5=r(\ne0)\text{(say)}\implies p=12r,q=5r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given that $xyz=1$ , find $\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}$? I think I solved this problem, but I don't feel $100$ percent sure of my solution. We have:
$xy=\large {\frac 1z}$
$xz=\large \frac 1y$
$yz=\large \frac 1x$
So let's substitute these into our sum:
$\large \frac{1}{1+x+\frac 1z}+\frac{1}{1+y+\frac 1x}+\frac{1}{1+z+\frac 1y}$
If we rewrite with a common denominator we get
$\large \frac{z}{1+z+xz}+\frac {x}{1+x+xy}+\frac{y}{1+y+zy}$
If $\large \frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac {x}{1+x+xy} +\frac{y}{1+y+zy}+\frac{z}{1+z+xz}$ , $;$ then $(x, y, z)=1$ and we can compute that the sum is equal to $1$.
The problem I have with this is that I found what $(x, y, z)$ $had$ to be, but I but I only had one weak restriction on the variables. Secondly, what are other way of doing this that I can learn from? Thanks.
| \begin{align*}
\frac{1}{1+x+xy}+\frac{1}{1+z+xz}+\frac{1}{1+y+yz}&=\frac{1}{1+x+xy}+\frac{xy}{xy(1+z+xz)}+\frac{x}{x(1+y+yz)}\\
&=\frac{1}{1+x+xy}+\frac{xy}{xy+1+x}+\frac{x}{x+xy+1}\\
&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
If $ 5x+12y=60$ , what is the minimum of $\sqrt{x^2+y^2}$? I know this can be easily done by solving for $y$ and substituting, so that you only have to find the minimum value of the parabola $\large x^2 + \left ( \frac {60-5x}{12} \right)^2$ using standard techniques, but is there a less messy way to do this using inequalities? I tried various things such as $AM-GM$, but I don't get anywhere. One thing which I did notice about the restriction is that it is of the form $ax+by=ab$, but I don't know how to make any use of that. Thanks.
| You need to find the foot of the perpendicular to the line $5x + 12y = 60$ that passes through the origin. The line has slope $-\frac{5}{12}$, so any perpendicular has slope $\frac{12}{5}$. The perpendicular passes through the origin, so its equation is $y=\frac{12}{5}x$. Find the intersection of the two lines:
\begin{aligned}
5x + 12y &= 60\\
y &= \frac{12}{5}x\\
5 x + \frac{12 \cdot 12}{5} x &= 60\\
25x + 144 x &= 300\\
169x &= 300\\
x &= \frac{300}{169}\\
y &= \frac{12}{5} \cdot \frac{300}{169}\\
y &= \frac{720}{169}.
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 6
} |
Prove or disprove ${{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$ Prove or disprove $\displaystyle{{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$
This is not homework. I'm trying to prove something related to Catalan numbers, and I'm stuck here.
I tried this: this is the coefficient of $x^a$ in $\displaystyle{(1+x)^{2a-1}+x(1+x)^{2a-3}+\dots x^{a-1}(1+x)}$
If we take $(1+x)^{2a-1}$ as the first term, then this is a geometric series of $a$ terms with $\displaystyle{\frac{x}{(1+x)^2}}$ as the common factor. Applying the formula for the summation of a geometric series, we get $$\displaystyle{\frac{(1+x)^{2a-1}[\displaystyle{\frac{x^a}{(1+x)^{2a}}}-1]}{\displaystyle{\frac{x}{(1+x)^2}-1}}}$$
On solving this, we get $$\displaystyle{[(1+x)^{2a+1}-x^a(1+x)](1+x+x^2)^{-1}}$$ $$\displaystyle{=[(1+x)^{2a+1}-x^a(1+x)][1-x(1+x)+x^2 (1+x)^2 -x^3 (1+x)^3\dots]}$$
Finding the coefficient of $x^a$ in this expression seems to be the sum of mutiple expressions again!
EDIT: Potato has shown that this is fase by substituting $a=3$. Could someone then give the general expression of the sum?
Thanks in advance!
| This is false. You can see this by testing $a=3$.
$${ 5 \choose 3} + { 3 \choose 2 } + 1 = {6 \choose 4}-1\neq {6 \choose 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How many real roots for $ax^2 + 12x + c = 0$? If $a$ and $c$ are integers and $2 < a < 8$ and $-1 < c$, how many equations of the form $$ax^2+12x+c=0$$ have real roots?
| As @DonAntonio give hint a quadratic eqn have real roots iff
$$b^2-4ac\ge0$$
$$144-4ac\ge0\implies ac\le36$$
since $2<a<8$ and $c>-1$ so $a$ will be in set {$3,4,5,6,7$} and $c$ {$0,1,2,3,\cdots$}
from $$ac\le36$$
firstly we will check combination for equality so factors of $36$ in given range
$36=3\times 12,4\times 9,6\times 6$
so possible combination of a and c will :
for $a=3,$ $c$ will {$0,1,2,3,4,5,6,7,8,9,10,11,12$}
for $a=4,$ $c$ will {$0,1,2,3,4,5,6,7,8,9$}
for $a=5,$ $c$ will {$0,1,2,3,4,5,6,7$} (this will hold only "<" relation since $5$ is not divider of $36$)
for $a=6,$ $c$ will {$0,1,2,3,4,5,6$}
for $a=7,$ $c$ will {$0,1,2,3,4,5$} (same case as $a=5$)
so possible eqn are $13+10+8+7+6=44$
$44$ equation having a and c in given range have real roots
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$.
Let $a,b,c>0 $ where $abc=1$ and $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$. Prove that $$ (a+b+c)^3(a^2+b^2)(b^2+c^2)(c^2+a^2) \ge \big((a^2b+b^2a)(b^2c+c^2b)(c^2a+a^2c)\big) ^3.$$
Thanks
| By AM-GM, $\left(a^2+b^2\right)\geq 2ab$ and similarly for two other pairs. Hence, $$8=\prod_{\text{cyc}}\,\left(a^2+b^2\right) \geq \prod_{\text{cyc}}\,(2ab)=8\,(abc)^2=8\,.$$
Therefore, the equality holds, implying that $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| You made a mistake when you applied the limit $n \to \infty$ to the $\left(1 + \frac 1n\right)$ terms but not other terms outside. This is not a valid limit application because the whole thing is still in an indeterminate $\infty - \infty$ form.
I can offer you a tedious method that does not require you to estimate the growth of the two terms in the difference. (In fact, if you expand out everything in $O(\ldots)$, you only need basic algebra and some fundamental limit theorems.)
Define
\begin{align*}
A_n & = \sqrt{n^2 + n} = \sqrt[6]{(n^2 + n)^3} = \sqrt[6]{n^6 + 3n^5 + 3n^4 + n^3}
= n\sqrt[6]{1 + O(1/n)}\\
B_n & = \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^3 + n^2)^2} = \sqrt[6]{n^6 + 2n^5 + n^4}
= n\sqrt[6]{1 + O(1/n)}.
\end{align*}
We want to find $\lim_{n\to\infty} A_n - B_n$.
Let us factor $A_n^6 - B_n^6$ as follows:
\begin{align*}
A_n^6 - B_n^6 & =
(A_n - B_n)(A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5) \\
\therefore A_n - B_n & =
\frac{A_n^6 - B_n^6}{A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5}
\end{align*}
We know that $A_n^6 - B_n^6 = n^5 + 2n^4 + n^3 = n^5(1 + O(1/n))$. All terms in the denominator are of the form $A^i_nB^{5-i}_n$, so
$$
A^i_nB^{5-i}_n = \sqrt[6]{(n^6 + 3n^5 + 3n^4 + n^3)^i(n^6 + 2n^5 + n^4)^{5-i}}
= n^5\sqrt[6]{1 + O(1/n)}.
$$
Therefore,
\begin{align*}
\lim_{n\to\infty} A_n - B_n
& = \lim_{n\to\infty}\frac{n^5(1 + O(1/n))}{6n^5\sqrt[6]{1 + O(1/n)}} \\
& = \frac 16
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
Find the value of $x+y$ if $x^2+y^2+10 = 2\sqrt2x+4\sqrt2y$
If $$x^2+y^2+10 = 2\sqrt{2}x+4\sqrt{2}y$$ then the value of $(x+y)$ is:
a) $4\sqrt{2}$
b) $3\sqrt{2}$
c) $6\sqrt{2}$
d) $9\sqrt{2}$
Please teach me its basics and how to solve it?
| HINT:
We have $$x^2-2\cdot x\cdot \sqrt2 +(\sqrt2)^2+y^2-2\cdot y\cdot 2\sqrt2+(2\sqrt2)^2=10-2-8$$
$$\implies(x-\sqrt2)^2+(y-2\sqrt2)^2=0$$
We know, if $z$ is real , $z^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Obviously, AM-GM seems to be suitable for LHS.
For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to do.
Can someone please teach me? Thank you.
p.s. That $\sqrt [27]{}$ is really terrible...
| I found a solution that uses a bit of calculus.
First using the identity you have already found, we figure the inequality is equivalent to
$$\sqrt[3]{\frac{a^3+b^3+c^3+24}{27}} = \frac{a+b+c}{3} \le \sqrt[27]{\frac{a^3+b^3+c^3}{27}}.$$
Setting $x = \frac{a^3+b^3+c^3}{27}$ we find it equivalent to saying
$$(x+\frac89)^9-x \ge 0.$$
This is where the calculus enters the equation. It is a standard method to prove that all the zeros of $p(x) = (x+\frac89)^9-x$ are in fact negative. Since $x\ge0$ by definition this proves the inequality.
[The calculus: Take the derivative $p'(x) = 9(x+\frac89)^8-1$ which has two zeros $x_1,x_2$ smaller than zero. Show that $p(x_1),p(x_2)>0$.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
How prove this inequality for $a,b,c,d$ are real numbers let $a,b,c,d$ are real numbers,show that
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}\ge6\sqrt{|ad-bc|}$$
This problem is creat by China's famous mathematician hua luogeng,http://en.wikipedia.org/wiki/Hua_Luogeng
when he is child,and Now this problem has some ugly methods,I think this inequality has nice methods,Thank you
| HINT:
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}$$ $$=2\sqrt{a^2+c^2}+\sqrt{(a-\sqrt{3}b)^2+(c-\sqrt{3}d)^2}+\sqrt{(a+\sqrt{3}b)^2+(c+\sqrt{3}d)^2}$$ $$\geq 2\sqrt{2|ac|}+\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}$$
Now,
$$\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}\geq\sqrt{2(|(a-\sqrt{3}b)(c-\sqrt{3}d)|+|(a+\sqrt{3}b)(c+\sqrt{3}d)|)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b^2+c^2 &\geq ab+bc+ca\\
2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\
\left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0
\end{align*}
$$
Which is obviously true.
However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:
$$
\begin{align*}
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\
a+b+c &\geq a+b+c
\end{align*}
$$
Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?
Edit: Please don't help me with the original inequality, I'll figure it out.
| Just assume, wlog $a\leq b\leq c$. Then this equation is all you need:
$$a^3+b^3+c^3=a^2b+b^2c+c^2a+\underset{\geq 0}{\underbrace{(c^2-a^2)(b-a)}}+\underset{\geq 0}{\underbrace{(c^2-b^2)(c-b)}}\geq a^2b+b^2c+c^2a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Discriminant of a trinomial
Let $b,c \in \mathbb{Z} $ and let $n \in \mathbb{N} $, $n \ge 2. $ Let $f(x) = x^{n} -bx+c$. Prove that $$\hbox{disc} (f(x)) = n^{n }c^{ n-1}-(n-1)^{n-1 }b^{n }.$$
Here $\hbox{disc} (f(x)) = \prod_{i} f'(\alpha_{i} )$ where $\alpha_{1}, \dots, \alpha_{n}$ are the roots of $f(x)$.
After some calculations I obtained $\hbox{disc} (f(x)) = \frac{\prod_{i} \alpha_{i}(n-1)b \ - \ nc }{\prod_{i} \alpha_{i}} $, but I'm afraid this is the wrong way.
| Here is a brute force approach:
$f'(x) = nx^{n-1} - b$, and we want to compute $\prod_i f'(\alpha_i)$. We do this by looking for the minimal polynomial with roots $\alpha_i^{n-1}$.
Note that
$$\begin{array}%
x^n - bx+c = 0 &\Leftrightarrow x(x^{n-1} - b) = -c \\
&\Leftrightarrow x^{n-1} (x^{n-1} - b)^{n-1} = (-1)^{n-1}c^{n-1}
\end{array}$$
let $y_i = \alpha_i^{n-1}$, and $z_i = f'(\alpha_i) = ny_i - b$. We have found the minimal polynomial for $y_i$:
$$y(y-b)^{n-1} = (-1)^{n-1}c^{n-1}$$
and we want the product $\prod_i z_i$. Consider the change of variable $z = ny-b$, i.e. $y = \frac{z+b}{n}$. Substitute into the above equation, we get
$$\frac{z+b}{n} \left(\frac{z - (n-1)b}{n}\right)^{n-1} = (-1)^{n-1}c^{n-1} \\
\Leftrightarrow (z+b)(z-(n-1)b)^{n-1} - (-1)^{n-1} n^n c^{n-1} = 0$$
The constant term of this polynomial in $z$ is $(-1)^n \prod_i z_i$, therefore
$$\prod_i z_i = (-1)^n \left((-1)^{n-1}b^n(n-1)^{n-1} - (-1)^{n-1} n^n c^{n-1}\right) = n^n c^{n-1} - (n-1)^{n-1}b^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/440452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Maxima/Minima Problem I am working on this Maxima and Minima Problem :
Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$
I tried it solving using the lagrange's method of multipliers. I get the following equations
$x+3x\theta+2y\theta=0$
$y+4y\theta+2x\theta=0$
$2z=0$
$3x^2+4xy+6y^2=140$
So i get $z=0$
Please suggest how to go about solving for $x$ and $y$ ?
| Those called homogeneous problems can be easily solved
Calling $d^2 = x^2+y^2$ and making $y = \lambda x$ we have
$$
d^2 = x^2(1+\lambda^2)
$$
but from $3x^2+4 x y +6 y^2 = 140$ we obtain
$$
x^2(3+4\lambda+6\lambda^2) = 140
$$
then
$$
d^2 = \frac{140(1+\lambda^2)}{3+4\lambda+6\lambda^2}
$$
now deriving
$$
\frac{d}{d\lambda}d^2 = \frac{200(2\lambda^2-3\lambda-2)}{(3+4\lambda+6\lambda^2)^2}
$$
and the condition for local minimum/maximum is
$$
2\lambda^2-3\lambda-2 = 0
$$
obtaining $\lambda = \{-\frac 12, 2\}$ and
$$
\min d^2 = 20\Rightarrow d = \sqrt{20}\\
\max d^2 = 70\Rightarrow d = \sqrt{70}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/441137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove more general identity:
$$
\begin{array}{|c|}
\hline \\
\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\
\hline
\end{array}
, \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
| In equation $(7)$ of this answer, I compute that
$$
\sum_{k=1}^n\tan^2\left(\frac{k\pi}{2n+1}\right)=n(2n+1)
$$
Thus, with $n=2m+1$,
$$
\begin{align}
\sum_{k=0}^{n-1}\frac1{\cos^2\left(\frac{k\pi}{n}\right)}
&=\sum_{k=0}^{2m}\sec^2\left(\frac{k\pi}{2m+1}\right)\\
&=2\sum_{k=0}^m\tan^2\left(\frac{k\pi}{2m+1}\right)+2(m+1)\\[6pt]
&=2m(2m+1)+2(m+1)\\[12pt]
&=(2m+1)^2\\[12pt]
&=n^2
\end{align}
$$
Residue Theory Answer
For odd $n$, consider the function
$$
f(z)=\frac{n/z}{z^n-1}\left(\frac{z-1}{z+1}\right)^2\tag{1}
$$
All the singularities are simple, except the singularity at $-1$.
$$
\left(\frac{z-1}{z+1}\right)^2=1-\frac4{z+1}+\frac4{(z+1)^2}\tag{2}
$$
Furthermore,
$$
\frac{\mathrm{d}}{\mathrm{d}z}\frac{n/z}{z^{\raise{2pt}n}-1}=\frac{n-n(n+1)z^n}{z^2(z^{\raise{2pt}n}-1)^2}\tag{3}
$$
Using $(2)$ and $(3)$, we get the residue of $f$ at $-1$ to be
$$
-4\cdot\frac{-n}{-1-1}+4\cdot\frac{n+n(n+1)}{(-1-1)^2}=n^2\tag{4}
$$
The residue of $f$ at $0$ is $-n$ and the residue at each $n^{\text{th}}$ root of unity is $-\tan^2(\theta/2)$.
Since the integral of $f$ over an increasing circle vanishes, the sum of the residues must be $0$. Therefore,
$$
\sum_{k=0}^{n-1}\tan^2\left(\frac{k\pi}{n}\right)=n^2-n\tag{5}
$$
and therefore,
$$
\sum_{k=0}^{n-1}\sec^2\left(\frac{k\pi}{n}\right)=n^2\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 2
} |
How to show that $f(x,y)$ is continuous. How to show that $f(x,y)$ is continuous.
$$f(x,y)=\frac{4y^3(x^2+y^2)-(x^4+y^4)2x\alpha}{(x^2+y^2)^{\alpha +1}}$$ for $\alpha <3/2$.
Please show me Thanks :)
| The only point where we can have trouble is $(x,y)=0$. To figure out what happens there, use the polar coordinates, as suggested in the comment of davin. We have $x=r\cos\theta$, $y=r\sin\theta$ and therefore
$$x^2+y^2=r^2,\qquad x^4+y^4=r^4\left(\cos^4\theta+\sin^4\theta\right)$$
so that
$$f=r^{3-2\alpha}\left(4\sin^3\theta-2\alpha\cos^5\theta-2\alpha\cos\theta\sin^4\theta\right).\tag{1}$$
Obviously, as $\alpha<3/2$ and $(x,y)\rightarrow(0,0)$, the limit of $f$ in (1) exists and is equal to the value of $f$ at $(0,0)$ (equal to zero).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/444913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Suppose $xyz=8$, try to prove that $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2$ Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always less than 1 for $t>0$.
I'm just wondering if we can have some elegant simple prove. I guess we should use Jensen's inequality. Thanks.
Suppose $x,y,z\in R^+$ and $xyz=8$, try to prove that $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2.$$
Please note that the usual AM-GM inequality may not do its trick here as the equality is rather hold on the boundary.
Thanks.
| It can be proved by contradiction. Set $a=\sqrt{1/(1+x)}$, $b=\sqrt{1/(1+y)}$ and $c=\sqrt{1/(1+z)}$. We have $$8a^2b^2c^2=(1-a^2)(1-b^2)(1-c^2).$$ We need to prove $a+b+c<2$. Assume that $a+b+c\geq 2$, we have
$$1-a^2=(1+a)(1-a)< 2(b+c-1)$$
Since $(1-b)(1-c)>0$, we have $$1-a^2<2bc$$
Then, we have
$$(1-a^2)(1-b^2)(1-c^2)<8a^2b^2c^2.$$
It is a contraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/446863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ Please help to prove this inequality.
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.
Thanks.
| By symmetry, the minimum value of the expression
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
under the constraint $(x^2-1)(y^2-1)(z^2-1)=8^3$ will occur when $x=y=z$. From this, it follows that the minimum occurs when $x=y=z=3$, and the minimum is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\leq1$ Let $a$, $b$, $c$ and $d$ are non-negative numbers such that $abc+abd+acd+bcd=4.$ Prove that:
$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}\leq1$
I simplified it and it turns out that it suffices to show
$19\le3(ab+ac+ad+bc+bd+cd)+abcd$
subject to $abc+abd+acd+bcd=4.$
I then tried Lagrange multiplier but it didn't help too much...
| Let $a+b+c+d=3u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Hence, we need to prove that
$$\prod_{cyc}(a+3)\geq\sum_{cyc}(a+3)(b+3)(c+3)$$ or
$$3(ab+ac+bc+ad+bd+cd)+2(abc+abd+acd+bcd)+abcd\geq27$$ or
$$18v^2w^2+8w^4+t^4\geq27w^4$$ or
$$19w^4-18v^2w^2-t^4\leq0$$ or
$$w^2\leq\frac{9v^2+\sqrt{81v^4+19t^4}}{19}.$$
But $$24(3v^4-4uw^3+t^4)=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0,$$
which gives $t^4\geq4uw^3-3v^4$.
Hence, it remains to prove that
$$w^2\leq\frac{9v^2+\sqrt{81v^4+19(4uw^3-3v^4)}}{19}$$ or
$$9v^2+\sqrt{24v^6+76uw^3}\geq19w^2,$$
which is obvious by Maclaurin: $u\geq w$ and $v^2\geq w^2$.
A proof of Maclaurin see here: How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
show that $\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}(1-\frac{e^{-ab}(ab+2)}{2})$ show that
$$\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}\left(1-\frac{e^{-ab}(ab+2)}{2}\right)$$
for $a,b> 0$
I would like someone solve it using contour but also I would to see different solution using different way to solve it
is there any help
thanks for all
| You can consider employing the residue theorem. The trick is to use a simple contour despite the pole at $z=0$. A way to attack this is to deform the usual semicircular contour at the origin so as not to include the pole.
Consider first the case $a>0$. Rewrite the integral as
$$\frac12 \int_{-\infty}^{\infty} dx \frac{\sin{a x}}{x (x^2+b^2)^2} = \frac{1}{4 i} \int_{-\infty}^{\infty} dx \frac{e^{i a x} - e^{-i a x}}{x (x^2+b^2)^2}$$
Therefore consider the contour integral
$$\frac{1}{4 i} \oint_C dz \frac{e^{i a z}}{z (z^2+b^2)^2}$$
where $C$ is as described above, a semicircle of radius $R$ in the upper half plane, with a semicircle of radius $\epsilon$ about the origin jutting into the upper half plane. Note then that the pole at the origin is not within $C$.
Then the above contour integral is equal to
$$\frac{1}{4 i} \int_{-R}^{-\epsilon} dx \frac{e^{i a x}}{x (x^2+b^2)^2} + \frac{\epsilon}{4} \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i a \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi}+b^2)^2} + \frac{1}{4 i} \int_{\epsilon}^{R} dx \frac{e^{i a x}}{x (x^2+b^2)^2} + \\ \frac{R}{4} \int_0^{\pi} d\theta\, e^{i \theta} \frac{e^{i a R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta}+b^2)^2} $$
We take the limit as $R \to \infty$ and $\epsilon \to 0$ and the contour integral then becomes
$$\frac{1}{4 i} PV \int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x (x^2+b^2)^2} -\frac{\pi}{4 b^4} $$
where $PV$ denotes the Cauchy principal value. Note that the fourth integral vanishes as
$$\frac{1}{R^4} \int_0^{\pi} d\theta \, e^{-a R \sin{\theta}} \le \frac{2}{R^4} \int_0^{\pi/2} d\theta \, e^{-2 a R \theta/\pi} = \frac{\pi}{R^5}\frac{1-e^{-a R}}{a} $$
The contour integral is also equal to $(1/(4 i)) i 2 \pi$ times the residue of the integrand at the pole $z=i b$. Because we have a double pole, the residue has value
$$\begin{align}\frac{d}{dz} \left [ \frac{e^{i a z}}{z (z+i b)^2}\right ]_{z=i b} &= -\left (\frac{a b+2}{4 b^4} \right ) e^{-a b}\end{align}$$
Therefore
$$\frac{1}{4 i} PV \int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x (x^2+b^2)^2} = \frac{\pi}{4 b^4} \left [1-\frac12 (2+a b) e^{-a b} \right ]$$
when $a>0$. Similarly, one may find that, for $a>0$:
$$\frac{1}{4 i} PV \int_{-\infty}^{\infty} dx \frac{e^{-i a x}}{x (x^2+b^2)^2} = -\frac{\pi}{4 b^4} \left [1-\frac12 (2+a b) e^{-a b} \right ]$$
(Note that the contour integral in this case is $(1/(4 i) (-i 2 \pi)$ times the residue at $z=-i b$ because the contour in the lower half plane is taken clockwise rather than counterclockwise.) Therefore, for $a>0$,
$$\int_0^{\infty} dx \frac{\sin{a x}}{x (x^2+b^2)^2} = \frac{\pi}{2 b^4} \left [1-\frac12 (2+a b) e^{-a b} \right ]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
How does $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$? I'm studying trigonometry and one of the key triangles (60 30 90) has a side of $\frac{\sqrt{3}}{2}$. I understand that it is derived from an equilateral triangle of sides 1, however I can't see how $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$ and why $\frac{\sqrt{3}}{2}$ is simpler to use than $\sqrt{\frac{3}{4}}$. I obtained $\sqrt{\frac{3}{4}}$ from $\sqrt{1^2 - (\frac{1}{2})^2}$
| $$\sqrt{\frac{3}{4}}=\sqrt{3\cdot4^{-1}}=\sqrt{3\cdot2^{-2}}=\sqrt{3}\cdot2^{-1}=\frac{\sqrt{3}}{2}$$
As to which to use, normally people tend to like to have as little surds as possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$ Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$.
I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but it just makes it more complex:
$$\frac{\tan(42^{\circ}) - \tan(6^{\circ})}{1+\tan(42^{\circ})\tan(6^{\circ})}= \frac{\tan(24^{\circ}) + \tan(12^{\circ})}{1 - \tan(24^{\circ})\tan(12^{\circ})}.$$
Can anyone here please provide a hint?
| Here is a proof that I got after back calculation.
Let $\theta=36^{\circ}$ Then, one can show that $\theta$ satisfies the following equation $$(\cos \theta +1)(4\cos^2\theta -2\cos\theta-1)=0$$since $\cos 36^{\circ}>0$, $\theta $ satisfies $$4\cos^2\theta-2\cos \theta-1=0\\ \Rightarrow 2\cos 2\theta-2\cos \theta=-1\ \\ \Rightarrow \cos 2\theta-\cos \theta=-\frac{1}{2}$$ Let $\phi=12^{\circ}$. Then $\theta=3\phi$ and $$\cos 6\phi-\cos 3\phi=-\frac{1}{2}\\ \Rightarrow \cos 6\phi+1=\cos 3\phi+\frac{1}{2}=\cos 3\phi+\cos 5\phi\quad (\because 5\phi=60^{\circ})\\ \Rightarrow 2\cos^2 3\phi=2\cos 4\phi \cos \phi\\ \Rightarrow \frac{\cos 36^{\circ}}{\cos 48^{\circ}}=\frac{\cos 12^{\circ}}{\cos 36^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}+\sin 6^{\circ}\sin 42^{\circ}}{\cos 6^{\circ}\cos 42^{\circ}-\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}+\sin 24^{\circ}\sin 12^{\circ}}{\cos 24^{\circ}\cos 12^{\circ}-\sin 24^{\circ}\sin 12^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}}{\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}}{\sin 24^{\circ}\sin 12^{\circ}}\quad (\mbox{by Componendo-Dividendo})\\ \Rightarrow \tan 6^{\circ}\tan 42^{\circ}=\tan 24^{\circ}\tan 12^{\circ}\hspace{6cm}\Box$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
| In the first line you have $4-4.5=\sqrt{(4-4.5)^2}$, which isn't true, because $-0.5\neq 0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 1
} |
Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cdot 2-1)^2=-2\cdot 4$
lets take $n=4$ so $(2\cdot 4-3)^2-(2\cdot 4-1)^2\neq-2\cdot 16$
I did something wrong?
Thanks!
| You have proven that the base case works.
Induction Step: Assume that the claim holds true for $n=k$.
It remains to prove the claim true for $n=k+2$. Observe that:
$$ \begin{align*}
&1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2+(2(k+2)-3)^2-(2(k+2)-1)^2 \\
&= [1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2]+[(2k+1)^2-(2k+3)^2] \\
&= [-2k^2]+[(2k+1)^2-(2k+3)^2] \qquad \text{by the induction hypothesis}\\
&= [-2k^2]+[((2k+1)+(2k+3))((2k+1)-(2k+3))] \qquad \text{difference of squares}\\
&= [-2k^2]+[(4k+4)(-2)] \\
&= -2(k^2)-2(4k+4) \\
&= -2(k^2+4k+4) \\
&= -2(k+2)^2 \\
\end{align*} $$
as desired. This completes the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to verify these series expansions? We know that $$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots (-1)^n \frac{x^{n+1}}{n+1} + \cdots = \log (1+x) $$ whenever $-1<x<1$. What can we say if $x=1$?
That is, does the series $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots (-1)^n \frac{1}{n+1} + \cdots $$ converge to $\log 2$? If so, how to prove this fact rigorously?
And what about the behavior of the series $$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots (-1)^n \frac{x^{2n+1}}{2n+1} + \cdots, $$
which converges to $\arctan x$ for $-1<x<1$, at the points $x = \pm 1$?
| Note that for $x\lt1$
$$
\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\gt0
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
=x^{2k-2}-x^{2k-1}\gt0
$$
Thus, as $x\to1^-$,
$$
f(x)=\sum_{k=1}^\infty\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
$$
is a positive series, increasing termwise to
$$
f(1)=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)
$$
which converges. Thus, by dominated convergence,
$$
\lim_{x\to1^-}f(x)=f(1)
$$
The same argument works for
$$
g(x)=\sum_{k=1}^\infty\left(\frac{x^{4k-3}}{4k-3}-\frac{x^{4k-1}}{4k-1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ p \equiv 1 \pmod{4}$, prove $((\frac{p-1}{2})!)^2 \equiv -1 \pmod {p}$ where p is prime. Characteristics: The fields where $ p \equiv 1 \pmod{4}$ has half the number from 1 to $\frac{p-1}{2}$ both in positive and the negative. There can be paired up such that when multiplied together, they equal $ -(p-1) \equiv +1 \pmod p$. Only one pair equals -1 and that is $\pm 1$.
Why does that happen?
| If we start from Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$ for a prime $p$, then we can write
$$\begin{align}
-1 &\equiv (p-1)! = \prod_{k=1}^{\frac{p-1}{2}} k \cdot \prod_{m = 1}^{\frac{p-1}{2}} (p-m)\\
&\equiv \left(\frac{p-1}{2}\right)! \cdot (-1)^{\frac{p-1}{2}}\prod_{m = 1}^{\frac{p-1}{2}} (m-p)\\
&\equiv (-1)^{\frac{p-1}{2}} \left(\left(\frac{p-1}{2}\right)!\right)^2
\end{align}$$
for odd primes $p$. Now if $p \equiv 1 \pmod{4}$, the factor $(-1)^{\frac{p-1}{2}}$ is $1$, hence
$$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -1 \pmod{p}$$
then.
It is Wilson's theorem that follows from pairing up each $1 < k < p-1$ with its inverse, leaving $(p-1)! \equiv 1\cdot(p-1) \equiv -1 \pmod{p}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/461288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need proofread for deriving quadratic equation formula How to Solve quadratic equation $$ax^{2}+bx+c=0$$ such as $$a \neq 0$$
Divide by a both side from the equation
such as
$$\frac{a}{a}x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2}+\frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2} + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a}=0$$
$$\Rightarrow (x + \frac{b}{2a})^2 - (\frac{b}{2a})^{2} + \frac{c}{a} = 0$$
$$\Rightarrow (x + \frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^2}{4a^{2}} - \frac{c}{a}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2}}{4a^{2}} - \frac{4ac}{4a^{2}}$$
$$\Rightarrow (x + \frac{b}{2a})^2 = \frac{b^{2} - 4ac}{4a^{2}}$$
$$\Rightarrow (x + \frac{b}{2a}) =\pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$
$$\Rightarrow x = -\frac{b}{2a} \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}$$
$$\Rightarrow x = \frac{-b\pm\sqrt{b^{2} - 4ac}}{2a}$$
| You have an error on your second-to-last line.
$$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
is correct, but when you move the $\frac{b}{2a}$ over, you forgot to change its sign. Correctly, this becomes
$$x + \frac{b}{2a} - \frac{b}{2a} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$ \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/461735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following :
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
Please suggest as getting no clue on this... thanks..
| In addition to AWertheim's answer, in case you haven't studied telescopic series yet, this is how you can calculate $\displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n}$:
First you see that you can write:
$ \displaystyle \frac{1}{n^{2}+n} = \frac{1}{n} - \frac{1}{n+1} $
This technique is called partial fraction decomposition if you don't know it already.
Now see what happens:
$\displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n} = \displaystyle \sum_{n=1}^{1999}(\frac{1}{n} - \frac{1}{n+1}) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \cdot \cdot \cdot +(1/1999 - 1/2000) = 1 + (-1/2+1/2) + (-1/3 +1/3) +\cdot \cdot \cdot + (-1/1999+1/1999)-1/2000= 1 - 1/2000 = \frac{1999}{2000}$
So, each term cancels the previous term, except the first and last terms. This kind of series is called telescopic series.
Also note that: $\displaystyle \sum_{n=1}^{1999} 1 = 1999$
So, your sum is:
$\displaystyle \sum_{n=1}^{1999} (1 + \frac{1}{n^{2}+n}) = \displaystyle \sum_{n=1}^{1999} 1 + \displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n} = 1999+ \frac{1999}{2000} = \frac{3999999}{2000} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ Both x and y are real numbers and x > y .
Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ .
I tried solving the second equation to get $y=(13-x)/(x+1)$ and substituted that in equation one to get a quartic equation in x : $x^4+3 x^3-25 x^2-69 x+154 = 0$ .
I tried a few rational root test guesses to no avail. Then I checked wolfram alpha to find out that it has no rational roots. So I think there must be a better way to compute $x^2+6y$ if x > y.
| HINT:
We have , $$x^2-xy+y^2+3(xy+x+y)=15+3\cdot13$$
$$\implies (x+y)^2+3(x+y)-54=0$$
Solve for $x+y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \implies a < \frac{2}{a}\\
\text{Let}\hspace{1cm} B=\frac{a+\frac{2}{a}}{2}=\frac{a^2+2}{2a}
$$
$B$ is greater than $a$ because:
$$ \begin{array} {aa} B>a & \implies \frac{a^2+2}{2a}>a \\
& \implies a^2 + 2 > 2a^2 \\
& \implies 2 > a^2 \\
& \implies a^2 < 2
\end{array}$$
If $B^2$ is less than $2$, then $B^2-2<0$, but:
$$\begin{array} {aa} B^2-2 < 0 & \implies \left( \frac{a^2+2}{2a} \right)^2 - 2 < 0 \\
& \implies \frac{a^4+4a^2+4}{4a} - \frac{8a^2}{4a^2} < 0 \\
& \implies \frac{(a^2-2)^2}{(2a)^2} < 0
\end{array}$$
Which is a contradiction since the left hand side of the inequality will be positive for all values of $a$.
But I think we must be able to find such a $B$ since based on my understanding of this answer, we can find a another rational number whose distance from $a$ is less than the distance between $a$ and $\sqrt{2}$
Therefore, I have 2 questions to ask:
*
*Why does this approach work in the case of $a^2>2$ but not when $a^2<2$?
*How should I approach these kind of questions since it seems that there are a few ways to construct a $B$ that satisfies a given set of restrictions? For example, see here (the proof is immediately before the section "13. The Completeness Axiom".
| The substitution $$a \leftarrow \frac{a^2+2}{2a} $$ is exactly what you get from Newton's method for the function $f(a)=a^2-2$. This will result in iterations that are greater than $\sqrt{2}$ since $f$ is an increasing convex function on $(0, \infty)$. I will sketch two alternatives, both based on Newton's method, to get iterations less than $\sqrt{2}$.
First idea. Use some other function than $f(a) = a^2-2$ with Newton's method. For example, if $0 \leq p < \sqrt{2}$ then
$$
\frac{1}{\sqrt{2}+p} = \frac{\sqrt{2}-p}{2-p^2}
$$
and so $\sqrt{2}$ is a root of the function $$f(a) = \frac{2-p^2}{a+p}-a+p = \frac{2-a^2}{a+p}.$$
This function is convex and decreasing on $(0, \infty)$. Applying Newton's method to this function results in the iteration
$$
a \leftarrow \frac{p a^2+4 a+2 p}{a^2+2 p a+2}
$$
which produces an increasing sequence that converges to $\sqrt{2}$ from below as required. For example for $p=0$ you get
$$
a \leftarrow \frac{4a}{a^2 + 2}
$$
which is exactly the harmonic mean of $a$ and $\frac{2}{a}$ as suggested by haruspex. For $p=1$ and $p=\frac{7}{5}$ you get
$$
a \leftarrow \frac{a^2+4a+2}{a^2+2a+2} \textrm{ and } a \leftarrow \frac{7 a^2+20 a+14}{5 a^2+14 a+10}
$$
respectively, to give just two other examples. Substituting $a=\frac{7}{5}$ in the latter results in $\frac{1393}{985}$ which is less than $4\times 10^{-7}$ below $\sqrt{2}$.
Second idea. For this idea I will assume that $a\geq 1$. As you already found out we have $$a < \sqrt{2} < \frac{a^2+2}{2a}.$$
Therefore we could try to take some weighted average of $a$ and $\frac{a^2+2}{2a}$ that ends up below $\sqrt{2}$. So we're looking for some factor $\lambda \in (0,1)$ such that
$$
\lambda a + (1-\lambda)\frac{a^2+2}{2a} < \sqrt{2}
$$
for all $a \in [1, \sqrt{2})$. It is not difficult to show that this is the case exactly if $\lambda \in (3 - 2\sqrt{2}, 1)$. We can take $\lambda = \frac{1}{5}$ or $\lambda = \frac{5}{29}$ for example to get the iterations
$$
a \leftarrow \frac{3 a^2+4}{5a} \textrm{ and } a \leftarrow \frac{17 a^2 + 24}{29a}
$$
respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Finding unit's digit in exponentiation Could someone please explain to me how to find the unit's digit in the following expression:
$$7^{95} - 3^{58}$$
| Unit Digit means $\pmod{10}$
Observe that $7\equiv7\pmod{10},7^2=49\equiv9,7^3\equiv9\cdot7\equiv3,7^4\equiv3\cdot7\equiv1$
$3\equiv3\pmod{10},3^2=9\equiv9,3^3=27\equiv7,3^4=81\equiv1$
So, both $3,7$ have a cycle with period $=4$ which can be also confirmed using Euler's Totient Theorem with $\phi(10)=\phi(2)\phi(5)=4$ and $(3,10)=(7,10)=1$
As $95=23\cdot4+3$ and $58=4\cdot14+2,$
$7^{95}=(7^4)^{23}\cdot7^3\equiv1^{23}\cdot3\pmod{10}$
and $3^{58}=(3^4)^{14}\cdot3^2\equiv1^{14}\cdot9\pmod{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$ Find $$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$$
My approach :
$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} =\lim_{n \to \infty} \frac{1}{n^2}\sum^{2n}_{r =1} \frac{\frac{r}{n}}{\sqrt{1+\frac{r^2}{n^2}}} $$
If I put $\frac{r}{n} =t $ then we can write it
$$\lim_{n \to \infty} \frac{1}{n^2}\sum^{2n}_{r =1} \frac{t}{\sqrt{1+t^2}} $$
Will it help some how here.. and how can we change the limits then.. please suggest thanks.
| HINT:
Putting $2n=m,$
$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} $$
$$=4\lim_{n \to \infty} \frac{1}{2n}\sum^{2n}_{r =1} \frac{r}{\sqrt{(2n)^2+4r^2}} $$
$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac r{\sqrt{m^2+4r^2}}$$
$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac {\frac rm}{\sqrt{1+4\left(\frac rm\right)^2}}$$
$$=4\int_0^1\frac{xdx}{\sqrt{1+4x^2}}$$
as $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
what will be the probability? Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
my approach was,the number of likely combinations are (1,6),(2,6),(3,6),(2,5),(1,5),(3,5),(2,4),(3,4),(3,3) and total combinations will be 3(for 1,2,3 in first roll)*6( in second roll).so probability should be 9/18.but answer was not accepted.please help me understanding the correct approach.
| Alicia is playing the game. She wins if the sum total of values is at least $6$.
If the first toss results in a $4$ or a $5$, she won't get a chance to retoss, and she won't win.
Instant win: If she gets a $6$ on the first toss, then she has won quickly. This has probability $\dfrac{1}{6}$.
We now examine what happens if the first toss is a $1$, $2$, or $3$. Examine the cases one at a time.
First toss is a 1: If at first she gets a $1$, (probability $\frac{1}{6}$), then she gets to retoss. To win, she needs a $5$ or a $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{2}{6}$.
First toss is a 2: In this case, to win she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{3}{6}$.
First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{4}{6}$.
Thus the probabilility Alicia wins is $\dfrac{1}{6}+ \dfrac{1}{6}\cdot \dfrac{2}{6}+ \dfrac{1}{6}\cdot \dfrac{3}{6}+ \dfrac{1}{6}\cdot \dfrac{4}{6}$. Simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$.
Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$.
Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$.
Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.
Is my proof correct? Is there a shorter way to prove this?
| The case $\ x=0,\ y=0\ $ is of the form $\ x=y$.
Now without loss of generality set $\ y=c\cdot x$, which implies $\ c^n=1$, which implies $\ y=\pm\ x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Determining the general form of $10^x \bmod 210$ While solving a problem I came across solving $10^x\bmod 210$ for various values of $x$. It seems that the values repeat after an interval of 6 for $x\geq4$. Can any one explain how can solve this equation for some value of $x$ and show that it repeats at an interval of 6.
Note that $210 = 2\times3\times5\times7$ is a product of the first four primes.
Thanks
| $10^x \equiv 0 \mod 2$ and $\mod 5$ for $x \ge 1$.
$10 \equiv 1 \mod 3$ so $10^x \equiv 1 \mod 3$ for all $x \ge 0$.
$10 \equiv 3 \mod 7$ and $3^6 \equiv 1 \mod 7$ so $10^{6k} \equiv 1 \mod 7$ for all $k$. Moreover since $3^2$ and $3^3$ don't work, $10^x \equiv 1 \mod 7$ only for multiples of $6$. If $x, y \ge 1$ and $x = y + 6 k$,
$10^x - 10^y = 10^y (10^{6k} - 1) \equiv 0 \mod 2, 5, 3$ and $7$, and therefore $\mod 210$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The Diophantine equation $x^2 + 2 = y^3$ How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)
| Although I prefer an answer without ring theory here is a solution by using the extension $\mathbb{Z}[\sqrt{-2}]$.
All variables in this proof are integers.
$x^2+2=y^3$ factors as $(x+\sqrt{-2})(x-\sqrt{-2})=y^3$.
Since
$1)$ $\mathbb{Z}[\sqrt{-2}]$ is a UFD.
$2)$ the LHS factors into $2$ conjugates which implies that the LHS must have $2n$ primefactors. (A conjugate factors analogue to its Original in a UFD , this is easy to show when using the norm)
3) the RHS must have $3m$ prime factors and the smallest common multiple of $2$ and $3$ is $6$.
We can conclude that :
$1)$ both LHS and RHS has $6A$ prime factors.
$2)$ Since we have two conjugates on the LHS we can conclude that $(x+\sqrt{-2})$ is a cube in $\mathbb{Z}[\sqrt{-2}]$.
Hence we get the equation $(x+\sqrt{-2})=(a+b\sqrt{-2})^3$
We proceed by expanding the cube :
$x+\sqrt{-2} = a^3 - 6 a b^2 + (3 a^2 b-2 b^3)\sqrt{-2}$
We can solve the sqrt part $3 a^2 b-2 b^3 = 1$ because $b^2$ must be $1$ because $b$ is a factor on the LHS !!
Let $b=1$ then we get $3a^2 - 2 = 1$ hence $a=1$.
It follows $x=a^3 - 6a b^2 = 5$.
If we took $b=-1$ or factored $x-\sqrt{-2}$ we get the same or a negative solution for $x$ hence $x=5$ is the only positive solution.
We thus get $5^2 + 2 = 3^3$
Q.E.D.
mick
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the given answer of $-n(2n+1)$
| Note that
$$(2n-1)^2-(2n)^2=1-4n,$$
and therefore your sum is equal to
$$\underbrace{(1+1+\ldots+1)}_{n\;\mathrm{times}}-4\left(1+2+\ldots+n\right)=n-4\cdot\frac{n(n+1)}{2}=-n(2n+1).$$
The only thing one needs to know is the arithmetic progression sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| HINT
$A+B+C = 180$
$A+B = 180 - C$
We'll apply tangent function:
$\tan (A+B) = \tan (180 - C)$
We'll consider the identity:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x\tan y}$
$\frac{\tan A + \tan B}{1-\tan A\tan B} = \frac{\tan 180 - \tan C}{1+\tan 180\tan C}$
But $\tan 180 = 0$, therefore, we'll get:
$\frac{\tan A + \tan }{1-\tan A\tan B}$ = $\frac{0 - \tan C}{1+0}$
$\frac{\tan A + \tan B}{1-\tan A\tan B} = -\tan C$
We'll multiply by $(1-\tan A\tan B)$:
$\tan A + \tan B = -\tan C +\tan A\tan B\tan C$
Hence
$\tan A + \tan B+ \tan C = \tan A\tan B\tan C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 5
} |
How to solve $f_{n}(x)=xf_{n-1}(x)-f_{n-2}(x)$? How to solve the following recurrence relation
$$f_{n}(x)=xf_{n-1}(x)-f_{n-2}(x)$$
with the initial conditions $f_{1}(x)=x, f_{2}(x)=x^2-1$?
The answer is that $f(x+\frac{1}{x})=\frac{1-x^{2n+2}}{x^{n}(1-x^2)}$ (it may be wrong).
| You can use a brute force method. Possibly the least elegant solution, but here goes:
Compute $f_n(x)$ for a few $n$:
$$\begin{aligned}
&f_1(x) = x\\
&f_2(x) = x^2-1\\
&f_3(x) = x^3-2x\\
&f_4(x) = x^4-3x^2+1\\
&f_5(x) = x^5-4x^3+3x\\
&f_6(x) = x^6-5x^4+6x^2-1\\
&f_7(x) = x^7-6x^5+10x^3-4x\\
&f_8(x) = x^8-7x^6+15x^4-10x^2+1
\end{aligned}$$
Do you recognize the binomial coefficients? Ask yourself if this can be generalized to $$f_n(x) = x^n-\binom{n}{2}x^{n-2}+\binom{n-1}{3}x^{n-4}-\binom{n-2}{4}x^{n-6}+\dots$$
Assume it can and prove it by induction, as follows:
Doublecheck that the initial conditions are met by plugging in $n=1$ and $n=2$. Then calculate $f_{n+1}$ under the above assumption.
$$f_{n+1}(x) = xf_n(x)-f_{n-1}(x)=\\
x\left[ x^n-\binom{n}{2}x^{n-2}+\binom{n-1}{3}x^{n-4}-\binom{n-2}{4}x^{n-6}+\dots \right]-\\\left[ x^{n-1}-\binom{n-1}{2}x^{n-3}+\binom{n-2}{3}x^{n-5}-\binom{n-3}{4}x^{n-7}+\dots \right]=\\
x^{n+1} - \left[\binom{n}{2}+1\right]x^{n-1} + \left[\binom{n-1}{3}+\binom{n-1}{2}\right]x^{n-3}-\left[\binom{n-2}{3}+\binom{n-2}{4}\right]x^{n-5}+\dots\\ =
\{\text{Well-known binomial identity, cf. Pascal's triangle}\} = \\
x^{n+1} - \binom{n+1}{2}x^{n-1} + \binom{n}{3}x^{n-3} - \binom{n-1}{4}x^{n-5}+\dots$$
which completes the proof.
Comments:
*
*Again, brute force-ish, and not very elegant. Also, the expression for $f_n(x)$ might be possible to simplify significantly.
*The $\dots$ here and there might need to be taken care of more cautiously.
*The binomial coefficients outside Pascal's triangle are assumed to be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute the integral $\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$
Compute the indefinite integral
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}
$$
My Attempt:
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx
$$
Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}$ to get
$$
\begin{align}
2t\,dt &= \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\,dx\\
2tdt &= \frac{-4x^2+2x+2}{(x^2-x+1)^2}dx
\end{align}
$$
I don't know how to proceed from here.
| According to http://en.wikipedia.org/wiki/Euler_substitution, this integral can have these four approaches to solve:
Approach $1$:
Let $u=x+\sqrt{x^2+x+1}$ ,
Then $x=\dfrac{u^2-1}{2u+1}$
$dx=\dfrac{2u(2u+1)-(u^2-1)2}{(2u+1)^2}du=\dfrac{2u^2+2u+2}{(2u+1)^2}du$
$\therefore\int\dfrac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$
$=\int\dfrac{\dfrac{2u^2+2u+2}{(2u+1)^2}}{\left(\left(\dfrac{u^2-1}{2u+1}\right)^2-\dfrac{u^2-1}{2u+1}+1\right)\left(u-\dfrac{u^2-1}{2u+1}\right)}du$
$=\int\dfrac{\dfrac{2u^2+2u+2}{(2u+1)^2}}{\dfrac{(u^2-1)^2-(u^2-1)(2u+1)+(2u+1)^2}{(2u+1)^2}\times\dfrac{u^2+u+1}{2u+1}}du$
$=2\int\dfrac{2u+1}{u^4-2u^3+u^2+6u+3}du$
Other approaches are similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Real solutions of the equation $x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x}$ Solve the equation
$$x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x},$$
where $x \in \mathbb{R}$.
| I'm assuming we are solving over the reals.
Set $a=\sqrt{3-x}$ and likewise for $b$ and $c$.
Then $x=ab+bc+ca$ and we see $$a^2+x = 3,\qquad b^2+x=4,\qquad c^2+x=5$$
Substituting for $x$, we find $$(a+b)(a+c) = 3,\qquad (b+c)(b+a)=4,\qquad (c+a)(c+b)=5$$
Thus $$(a+b)(a+c)(b+c)=\sqrt{60}$$
This leads to $$a+b=\frac{\sqrt{60}}{5},\qquad b+c=\frac{\sqrt{60}}{3},\qquad a+c=\frac{\sqrt{60}}{4}$$
So $$a = a+b+c - (b+c) = \frac{47}{2\sqrt{60}} - \frac{20}{\sqrt{60}} = \frac{7}{2\sqrt{60}}$$
Thus $$x = 3-a^2 = 3 - \frac{7^2}{240} = \boxed{\frac{671}{240}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Proving $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$, given $abc=1$ Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$
The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequality.
| $$\Longleftrightarrow (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 6$$
since by $AM-GM$ inequality
$$ (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}$$
$$\Longleftrightarrow2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}\ge 6$$
$$\Longleftrightarrow (ab+bc+ac)^2\ge 3(a+b+c)$$
since by $AM-GM$
$$(ab+bc+ac)^2\ge 3abc(a+b+c)=3(a+b+c)$$
By Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Generating functions and finding coefficient of $x^{3n}$ So I have this homework question I've been trying all day to answer, any help would be appreciated:
Determine using generating functions in how many ways can you choose $3n$ letters out of $\{a, b, c\}$ so that each letter is chosen at most $2n$ times?
| As @user84413's answer tells you, you are after:
$\begin{align*}
[x^{3 n}] (1 + x + x^2 + \dotsb + x^{2 n})^3
&= [x^{3 n}] \left( \frac{1 - x^{2 n + 1}}{1 - x} \right)^3 \\
&= [x^{3 n}] (1 - 3 x^{2 n + 1} + 3 x^{4 n + 2} - x^{6 n + 3}) \cdot (1 - x)^{-3} \\
&= ([x^{3 n}] - 3 [x^{n - 1}]) \sum_{k \ge 0} (-1)^k \binom{3}{k} x^k \\
&= ([x^{3 n}] - 3 [x^{n - 1}]) \sum_{k \ge 0} \binom{k + 3 - 1}{3 - 1} x^k \\
&= \binom{3 n + 2}{2} - 3 \binom{n + 1}{2} \\
&= \frac{(3 n + 2) (3 n + 1)}{2} - 3 \frac{(n + 1) n}{2} \\
&= 3 n^2 + 3 n + 1 \\
&= (n + 1)^3 - n^3
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How find this $\sum_{n=1}^{\infty}a_{n}$ let $$a_{1}=1,a_{n+1}=\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}}-\sqrt{2}$$
find the value
$$\sum_{n=1}^{\infty}a_{n}$$
my try:let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$,then we have
$$S_{n+1}=S_{n}+a_{n+1}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}$$
and I find $S_{1}=1,S_{2}=2-\sqrt{2},S_{3}=3-\dfrac{3}{2}\sqrt{2}$
I guess
$$S_{1}>S_{3}>\cdots>S_{2n-1},S_{2}>S_{4}>\cdots>S_{2n}$$
so I try prove
$$S_{2n+1}-S_{2n-1}<0$$
if $n$ is odd. then
$$S_{n+2}=S_{n+1}+\dfrac{1}{S_{n+1}}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}+\dfrac{1}{S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}}-\sqrt{2}$$
let $$f(x)=x+\dfrac{1}{x}-\sqrt{2}\Longleftrightarrow S_{n+2}=f(S_{n+1})=f(f(S_{n}))$$
where $x=S_{n}$
and $$S_{n+2}-S_{n}=\dfrac{1}{x}-\sqrt{2}+\dfrac{1}{x+\dfrac{1}{x}-\sqrt{2}}-\sqrt{2}=\dfrac{(1-\sqrt{2}x)^3}{x(x^2-\sqrt{2}x+1)}$$
so if $x\in \left(\dfrac{1}{\sqrt{2}},1\right)$, then we have $$S_{n+2}<S_{n}$$
where $n$ is odd numbers.
my question: How can I determine
$x=S_{n}$(n is odd) in $(\dfrac{1}{\sqrt{2}},1)$?
and I think this problem have other nice methods.Thank you
| Very nice problem. Here is my approach:
If the series converges, it converges to $\frac{1}{\sqrt{2}}$ since $a_n\rightarrow 0$. Taken in consideration this, we define $$b_n=a_1+\cdots+a_n-\frac{1}{\sqrt{2}}$$
We want to prove that $b_n$ converges (to zero).
We have
$$
b_{n+1}=b_n+a_{n+1}=b_n-\sqrt{2}+\frac{1}{b_n+\frac{1}{\sqrt{2}}}=b_n-\frac{\sqrt{2}b_n}{b_n+\frac{1}{\sqrt{2}}}
$$
Note that if $b>0$ then
$$
0<\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}<2b
$$
Note also that
$$
b-\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}>\frac{-1}{\sqrt{2}}
$$
For the other side, if $-\frac{1}{\sqrt{2}}<b<0$ then
$$
0>\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}>2b
$$
It implies that if $b_n>0$ then $b_n>b_{n+1}>-b_n$. In same way, if $-\frac{1}{\sqrt{2}}<b_n<0$, then $b_n<b_{n+1}<-b_n$. These inequalities can be written in compact form as:
$$
|b_{n+1}|<|b_n|
$$
and therefore $\{|b_n|\}$ converges to $L$. It only remains to prove that $L=0$.
For this, note that if $\frac{1}{\sqrt{2}}>L>0$, then
$$
1=\limsup\frac{|b_{n+1}|}{|b_n|}=\limsup|1-\frac{\sqrt{2}}{b_n+\frac{1}{\sqrt{2}}}|<1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve for the two variables. I used the method of substitution to finally obtain $x(x+1)=6(21-2y)$. Then I figured that both sides must be non-negative, and hence the value of $y$ has to be less than $11$. Then we test for values less than $11$ which makes $6(21-2y)$ a product of two consecutive numbers. WE get $3$ as a value of $y$, and also $7$. But putting $7$ in (1) does not work.
But there is a flaw in my reasoning. When I get to the 'consecutive' part, I start assuming $x$ and $y$ are integers. Since the person who asked me is not familiar with complex numbers, WE can assume that the solutions are real. However, I am interested in extending the values of $x$ and $y$ beyond the reals, if there are any.
NOTE: From (2), we get the value of $y^2$ in terms of $x$. Then we figure out the value of $x$ in (1) and then substitute $y$ for $90-x^2$, which gives us $x(x+1)$.
| $$\sqrt x + y = 6\tag{1}$$
$$x^2 + y^2 = 90\tag{2}$$
I'll pursue a "different take" and solve for $y$ in terms of $x$:
First, we can consider the task as one of finding the points of intersection of the curve $\sqrt x + y = 6$, and the circle $x^2 + y^2 = 90$. Plotting both equations in Wolfram Alpha gives us:
It is quite evident that there is one and only one point of intersection: $(x, y) = (9, 3)$.
Second approach, more algebraic:
From $(1)$, $y = 6-\sqrt x$.
Substituting $y = 6 -\sqrt x$ into $(2)$ gives us
$$\begin{align} x^2 +(6 - \sqrt x)^2 & = 90 \\ \\
x^2 + 36 - 12\sqrt x + x & = 90 \\ \\
x^2 + x - 12\sqrt x = 54\tag{3}\end{align}$$
Now, putting $t = \sqrt x$ or $t^2 = x$ gives us the equation:
$$t^4 + t^2 - 12 t - 54 = 0$$
$t = 3$ is one root (and this corresponds to $x = t^2 = 9$. Using polynomial division to factor gives us $$t^4 + t^2 - 12 t - 54 = (t - 3)(t^3 + 3t^2 +10t + 18) = 0$$
Only three other possible roots: one is real, and two complex. The approximate value of the second real root $t$ is $-2.1887$. But this implies $t = \sqrt x \approx -2.1887$ which is impossible, since the square root function returns the positive root. So in the end, our only fruitful solution remains $x = 9,\;y = 3$.
(If you're curious, It's a very elaborate process to find the exact form for the second real solution $t$: From Wolfram, here's a step by step:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim\limits_{x\to-3}\frac{x^2-9}{2x^2+7x+3}$ I'm going through my textbook, solving problems and am confused because the textbook seems to be giving me the incorrect answer. The question is:
$$ \lim_{x \to -3} \frac{x^2-9}{2x^2+7x+3}$$
The way that I solved this was,
$$\frac{9-9}{...} = \frac{0}{...}$$
Which would end up being $0$, however the book gives me an answer of $6/5$
| $\require{cancel}$
Note that both the numerator and denominator evaluate to $0$ at $x = -3$. That means we have an indeterminate limit, which simply means more work needs to be done.
Factor numerator and denominator:
$$\frac{x^2-9}{2x^2+7x+3} = \dfrac{(x + 3)(x - 3)}{(x + 3)(2x + 1)}$$ Cancel the line terms.
$$\frac{(\cancel{x + 3})(x - 3)}{(\cancel{x + 3})(2x + 1)} = \frac{(x - 3)}{(2x +1)}$$
Now evaluate the limit of the right-hand side as $x\to -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real .
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ but that is actually $x^6-8 x^3+27$ , close but not exact )
I'm run out of ideas , so any help is appreciated .
Note : I know that the answer is $$(x^2+2 x+3) (x^4-2 x^3+x^2-6 x+9)$$ , but I want to know how !
| In such cases, we can start with
$$(x^2)^3+(ax)^3+b^3-3\cdot x^2\cdot ax\cdot b=x^6+x^3(a^3-3ab)+b^3$$
Comparing with the given expression, $b^3=27, a^3-3ab=-10$
We know, the only real value of $b$ is $3\implies a^3-9a+10=0$
Clearly, $a=2$ satisfies the last equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Question on definite integral How to compute the following:
$$ \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x(1+x^2)}dx $$
If you could, please, solve it completely, for, I am afraid, a hint won't do for me.
| Let $\displaystyle I(a) = \int_{0}^{\infty}\frac{\tan^{-1}(ax)}{x.(1+x^2)}dx$
Now Diff. both side w.r.to $a$ , We Get
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+a^2x^2).x.(1+x^2)}.x.dx$
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+x^2).(1+a^2x^2)}dx$
$\displaystyle \frac{dI(a)}{da} = \frac{1}{(a^2-1)}\int_{0}^{\infty}\left(\frac{a^2}{1+a^2x^2}-\frac{1}{1+x^2}\right)dx$
$\displaystyle \frac{dI(a)}{da}= \frac{a}{(a^2-1)}.\frac{\pi}{2}-\frac{1}{(a^2-1)}.\frac{\pi}{2}$
$\displaystyle \frac{dI(a)}{da} = \frac{\pi}{2}.\frac{1}{a+1}$
Now $\displaystyle \int \frac{dI(a)}{da}da = \frac{\pi}{2}\int\frac{1}{a+1}da$
$\displaystyle I(a) =\frac{\pi}{2}.\ln |a+1|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving trigonometric inequality with elementary means In a set of olympiad problems, the last,hardest one was:
We know that $\sin(x)<x<\tan(x)$ for every real $x\in (0,\pi/2)$. What is the maximum integer value of $n$ such that holds for every $x\in(0,\pi/2)$
$$\sin(2x)+\tan(2x)>nx$$
By taylor expansion and term by term sum the answer is immediatly $4$.
But what was the intended, elementary way of solving it(using the information they gave)?
| I'm assuming that the maximal (integer) $n$ such that
$$\sin (2x) + \tan (2x) > nx$$
holds for $0 < x < \frac{\pi}{4}$ is sought, and not for $\frac{\pi}{2}$. Otherwise no such $n$ exists.
So looking at
$$\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} = \frac{\sin (2x)}{2x}\left(1 + \frac{1}{\cos(2x)}\right) < 1 + \frac{1}{\cos (2x)},$$
letting $x \searrow 0$ immediately shows $n \leqslant 4$. On the other hand,
$$\begin{align}
\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} &= \frac{\tan (2x)}{2x}\left(\cos (2x) + 1\right)\\
&= \frac{\tan (2x)}{2x}2\cos^2 x\\
&= \frac{\cos^2 x}{x}\cdot \frac{2\tan x}{1 - \tan^2 x}\\
&= 2\frac{\cos^2 x}{1-\tan^2 x}\cdot \frac{\tan x}{x}\\
&> 2 \frac{\cos^2 x}{1 - \tan^2 x}.
\end{align}$$
But we know that $0 < \sin x < \tan x$ for $0 < x < \pi/2$, so we have
$$\sin^2 x < \tan^2 x \iff 1 - \cos^2x < \tan^2 x \iff 1-\tan^2 x < \cos^2 x,$$
and for $0 < x < \pi/4$, we have $0 < 1 - \tan^2 x$, so
$$\frac{\cos^2 x}{1-\tan^2 x} > 1,$$
whence, for $0 < x < \pi/4$, we have
$$\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} > 2,$$
or
$$\sin (2x) + \tan (2x) > 4x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get maximum value of a function with three variables satisfying a condition? The maximum value of the function
$$f(x, y, z) = \left(x -\frac{1}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{1}{3}\right)^2$$
subject to the constraints-
$x + y + z = 1; x \ge 0; y \ge 0; z \ge 0$
what is the actual method to find that ? it may be simple but i dont know , please help me to understand .
| This is a smooth function defined in a closed interval and as such would achieve its maximum (and minimum) on the boundary or where the gradient is $0$. Calculus (particularly Lagrangian formulation) is an effective way to handle this. Other ways include using a suitable substitution (if you can find one), using suitable inequalities etc.
Here I will demonstrate the inequality way. WLOG, let $x \ge y \ge z$. We can show that $f(1, 0, 0) - f(x, y, z) \ge 0$.
$$f(1, 0, 0) - f(x, y, z) = \frac{2}{3} - \left(x -\frac{1}{3}\right)^2 - \left(y - \frac{1}{3}\right)^2 - \left(z - \frac{1}{3}\right)^2 \\
= \frac{1}{3} + \frac{2}{3}(x+y+z) - (x^2+y^2 + z^2) = 1 - (x^2+y^2 + z^2)$$
Now $x^2 + y^2 + z^2 \le (x+y+z)^2 = 1$ completes this proof. The maximum is $f(1, 0, 0) =\frac{2}{3}$ which is achieved on the boundary at $(x, y, z) = (1, 0, 0)$ or any permutation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| Since $$\frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1}-\sqrt{k}) \ge \frac{1}{\sqrt{k+1}}$$
We find.
$$\begin{align}
& \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge 2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k}) = 2(\sqrt{n+1}-1) \ge 2\sqrt{n} - 2\\
\text{and}\quad & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k=1}^{n-1} \frac{1}{\sqrt{k+1}} \le 1 + 2\sum_{k=1}^{n-1}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{n} - 1
\end{align}
$$
As a result,
$$2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} S_n = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $\int_0^1 \exp\left(\frac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}\right)\sqrt{\frac{1-8x^2+16x^4}{1+7x^2-8x^4}}dx=e-1$ show that
$$\int\limits_0^1 {\exp \left( {\frac{{4x\sqrt {1 - {x^2}} }}{{\sqrt {8{x^2} + 1} }}} \right)} \sqrt {\frac{{1 - 8{x^2} + 16{x^4}}}{{1 + 7{x^2} - 8{x^4}}}} dx = e - 1$$
I think this is nice integral,This problem is my china frend give me do it at yesterday, But I can't prove it. Thank you
my try:let
$$u=\dfrac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}$$
then $$du=-\dfrac{4(8x^4+2x^2-1)}{\sqrt{1-x^2}(8x^2+1)^{\frac{3}{2}}}$$
| Let us note that under the change of variables $y=\sqrt{\frac{1-x^2}{1+8x^2}}$ we have $x=\sqrt{\frac{1-y^2}{1+8y^2}}$ and the interval $(0,\frac12)$ is mapped to $(\frac12,1)$ and vice versa. Also,
$$\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx=\frac{3(4y^2-1)dy}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}$$
Now our integral can be written as
\begin{align}
&\int_0^1e^{4xy}\frac{|1-4x^2|}{\sqrt{(1-x^2)(1+8x^2)}}dx=\\
=&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_{\frac12}^1e^{4xy}\frac{4x^2-1}{\sqrt{(1-x^2)(1+8x^2)}}dx=\\
=&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_0^{\frac12}e^{4xy}\frac{3(1-4y^2)}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}dy=\\
=&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_0^{\frac12}e^{4xy}\frac{3(1-4x^2)}{(1+8x^2)\sqrt{(1-x^2)(1+8x^2)}}dx=\\
=&\int_0^{\frac12}e^{4xy}\frac{4(1-4x^2)(1+2x^2)}{(1+8x^2)\sqrt{(1-x^2)(1+8x^2)}}dx=\\
=&\int_0^{\frac12}e^{4xy}(4xy)'_xdx=\\
=&\left[\exp 4x\sqrt{\frac{1-x^2}{1+8x^2}}\,\right]_{0}^{\frac12}=\\
=&e-1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| Note that
$$\sin{\frac{3 \pi}{5}} = \sin{\frac{2 \pi}{5}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$
$$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=\pi/5$. Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$
which means that
$$\cos{x} = \frac{1 + \sqrt{5}}{4}$$
$$\sin{x} = \frac{\sqrt{10-2\sqrt{5}}}{4}$$
Then, using $\sin{2 x} = \sin{3 x}$ as noted above, we have
$$4 \sin{x} \sin{2 x} = 8 \sin^2{x} \cos{x} = 8 \frac{10-2 \sqrt{5}}{16} \frac{1+\sqrt{5}}{4} = \frac{10-10-2 \sqrt{5}+10\sqrt{5}}{8} = \sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Product of Permutations Can someone explain how to multiply two permutations? I cannot get myself to understand. Please be very simple and explain each step.
| If you have two permutations displayed in this form:
$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 2 & 5 & 1\end{pmatrix}$$
and
$$\tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 5 & 4 & 3 & 2 & 1\end{pmatrix}$$
you can reorder the columns of $\tau$ so that the first line of $\tau$ matches the second line of $\sigma$, and you obtain $\tau = \begin{pmatrix}3 & 4 & 2 & 5 & 1\\ 3 & 2 & 4 & 1 & 5 \end{pmatrix}$, and $\tau\sigma$ now is described by the first line of $\sigma$ and the last line of $\tau$, i.e
$$\tau\sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\3 & 2 & 4 & 1 & 5\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I show that this sequence converges to $0$? For $x \in [0,1]$, let $a_k = k^2x(1-x^2)^k$. How can I show that $\lim_{k \to \infty}a_k = 0$? I don't know what to do because $k^2 \to \infty$ and $(1-x^2)^k \to 0$, so I would have to prove that $\infty \cdot 0 = 0$.
| We see that $\frac{a_{k+1}}{a_k} = \frac{(k+1)^2x(1-x^2)^{k+1}}{k^2x(1-x^2)^{k}} = (1 + \frac{2}{k} + \frac{1}{k^2})(1-x^2)$.
Since $x < 1$, $1 - x^2 < 1$, say equal to $1 - \epsilon$, with $\epsilon > 0$.
Since $\lim_{k \to \infty} (1 + \frac{2}{k} + \frac{1}{k^2}) = 1$, we can take $k$ large enough so that $(1 + \frac{2}{k} + \frac{1}{k^2}) < 1 + \epsilon$.
Hence $\frac{a_{k+1}}{a_k} \leq (1+ \epsilon)(1 - \epsilon) = 1 - \epsilon^2 < 1$ for large enough $k$. Say for $k > N$.
Then we have, for $k > N$, $a_k = a_N\frac{a_{N+1}}{a_N} \cdots \frac{a_k}{a_{k-1}} \leq a_N(1- \epsilon^2)^{k - N}$
This tends to zero clearly as $k \to \infty$, since $(1 - \epsilon^2) < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $ we got this integration problem
$$\int_{}^{}\frac{1}{x(1+x)^3}dx $$
it seems a fairly simple problem but what i am struggling with it is doing its partital fractions
$\int_{}^{}\frac{1}{x(1+x)^3}dx $ =$\int_{}^{}\frac{a}{x}dx$ +$\int_{}^{}\frac{b}{(1+x)^3}dx$+
$\int_{}^{}\frac{c}{(1+x)^2}dx$ + $\int_{}^{}\frac{d}{1+x}dx$ now how to get values of $a,b,c,d$ ? it seems confusion by doing it by $a(1+x)^3+$ $bx+$ $cx(1+x)$ $+d(x(1+x)^2) =1$ is there any other way to break it into partial fractions and solve it ?
| Ok, your equation is right:
$$a(1 + x)^3 + bx + cx(1 + x) + d(1 + x)^2 = 1$$
Rewriting this:
$$\begin{align}
a(1 + 3x + 3x^2 + x^3) + bx + cx + cx^2 + dx(1 + 2x + x^2) & = 1 \\
a + 3ax + 3ax^2 + ax^3 + bx + cx+cx^2+dx+2dx^2+dx^3& = 1 \\
\end{align}$$
Grouping the terms by the powers of x:
$$\begin{align}
(a+d)x^3 + (3a+2d+c)x^2 + (3a+b+c+d)x + a = 1
\end{align}$$
Now as you can see, there are no $x^3, x^2, $ or $x$ terms on the right side of the equation. Therefore because the RHS (right hand side) has to equal the LHS (left hand side), we can make a system of equations for $a, b, c$ and $d$
$$\begin{align}
x^3: a +d= 0 \\
x^2: 3a + 2d+c = 0 \\
x : 3a+b+c+2d = 0 \\
Constants: a = 1
\end{align}$$
Solving this we get $a = 1,$ $b = -1,$ $c = -1,$ $d = -1$
So the partial fraction decomposition is equal to:
$$\frac{-1}{(1+x)^3} + \frac{-1}{(1+x)^2} + \frac{1}{x} + \frac{-1}{(1+x)}$$
The rest you can integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
| Since
$$1 = \sin^2(\theta) + \cos^2(\theta)$$
dividing by $\sin^2(\theta) \cos^2(\theta)$, we get that
$$4\csc^2(2 \theta) = \csc^2(\theta) + \sec^2(\theta)$$
Replace $\theta$ as $\dfrac{x}{2^n}$, we get that
$$\sec^2\left(\dfrac{x}{2^{n+1}}\right) = 4\csc^2\left(\dfrac{x}{2^n}\right) - \csc^2\left(\dfrac{x}{2^{n+1}}\right)$$
Divding by $4^{n+1}$, we get that
$$\dfrac{\sec^2\left(\dfrac{x}{2^{n+1}}\right)}{4^{n+1}} = \dfrac{\csc^2\left(\dfrac{x}{2^n}\right)}{4^n} - \dfrac{\csc^2\left(\dfrac{x}{2^{n+1}}\right)}{4^{n+1}}$$
Now telescopic summation gives us
$$\sum_{n=0}^{N} \dfrac{\sec^2\left(\dfrac{x}{2^{n+1}}\right)}{4^{n+1}} = \csc^2\left(x\right) - \dfrac{\csc^2\left(\dfrac{x}{2^{N+1}}\right)}{4^{N+1}}$$
Now letting $N \to \infty$, we get what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
how to solve this differential equation $\frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ by substitution? I've tried with this differential equation $\displaystyle \frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ , put $u=xy$ then $\displaystyle\frac{du}{dx}=x\frac{dy}{dx}+y$ So,
It will be after editing
$\displaystyle \frac{du}{dx} = \frac{1+u}{(1-u)} + \frac{u}{x}$
I am stuck here, keeping in mind only these three techniques are allowed: substitution, separating variable, and converting it to homogeneous equation
| Approach $1$:
$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$
$(x-x^2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{1}{x}-y$ ,
Then $y=\dfrac{1}{x}-u$
$\dfrac{dy}{dx}=-\dfrac{1}{x^2}-\dfrac{du}{dx}$
$\therefore x^2u\left(-\dfrac{1}{x^2}-\dfrac{du}{dx}\right)=x\left(\dfrac{1}{x}-u\right)+1$
$-u-x^2u\dfrac{du}{dx}=2-xu$
$x^2u\dfrac{du}{dx}=(x-1)u-2$
$u\dfrac{du}{dx}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$
This belongs to an Abel equation of the second kind.
Let $v=\dfrac{1}{x}$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{1}{x^2}\dfrac{du}{dv}$
$\therefore-\dfrac{u}{x^2}\dfrac{du}{dv}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$
$u\dfrac{du}{dv}=(1-x)u+2$
$u\dfrac{du}{dv}=\left(1-\dfrac{1}{v}\right)u+2$
Let $s=u-v$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}-1$
$\therefore(s+v)\left(\dfrac{ds}{dv}+1\right)=\left(1-\dfrac{1}{v}\right)(s+v)+2$
$(s+v)\left(\dfrac{ds}{dv}+\dfrac{1}{v}\right)=2$
$(s+v)\dfrac{ds}{dv}+\dfrac{s}{v}+1=2$
$(s+v)\dfrac{ds}{dv}=1-\dfrac{s}{v}$
$(v-s)\dfrac{dv}{ds}=v(s+v)$
Approach $2$:
$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$
$(yx+1)\dfrac{dx}{dy}=x-yx^2$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-y\left(u-\dfrac{1}{y}\right)^2$
$yu\dfrac{du}{dy}+\dfrac{u}{y}=-yu^2+3u-\dfrac{2}{y}$
$yu\dfrac{du}{dy}=-yu^2+\dfrac{(3y-1)u}{y}-\dfrac{2}{y}$
$u\dfrac{du}{dy}=-u^2+\dfrac{(3y-1)u}{y^2}-\dfrac{2}{y^2}$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $u=\dfrac{1}{v}$ ,
Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=-\dfrac{1}{v^2}+\dfrac{3y-1}{y^2v}-\dfrac{2}{y^2}$
$\dfrac{dv}{dy}=\dfrac{v^3}{y^2}-\dfrac{(3y-1)v^2}{y^2}+v$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:
$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$
is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:
$$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} = \lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} \cdot \frac{1 \over x}{1 \over x} = \lim_{x\to -\infty}\frac{\frac{x}{x}\cdot(2+\frac7x)}{\sqrt{\frac{x^2}{x^2}\cdot(1+\frac{2}{x}-\frac{1}{x^2})}} = \frac{1\cdot(2+0)}{\sqrt{1\cdot(1+0-0)}} =\frac{2}{\sqrt{1}}=2$$
Where did I make a mistake?
| We have that : $\frac{2x+7}{\sqrt{x^2+2x-1}}=\frac{x(2+\frac{7}{x})}{\sqrt{x^2(1+\frac{2}{x}-\frac{1}{x^2})}}=\frac{x(2+\frac{7}{x})}{(\left | x \right |)\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}}$. Now use the fact that $\left | x \right |=-x$ for negative $x$ and youre done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Binomial Coefficient Identities I am trying to simplify the following fraction, which I think is equal to 1 but I am not sure.
$$\frac{\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}}{\frac{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}}$$
I tried to use the identity
$$\left(\begin{array}{c}
n\\
r
\end{array}\right)=\left(\begin{array}{c}
n-1\\
r-1
\end{array}\right)+\left(\begin{array}{c}
n-1\\
r
\end{array}\right)
$$
I have done the following
Step 1:
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}$$
Step 2:
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}$$
From there I get stuck here
$$\frac{\left[\left(\begin{array}{c}
b\\
k
\end{array}\right)-\left(\begin{array}{c}
b-1\\
k
\end{array}\right)\right]}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left[\left(\begin{array}{c}
r+b\\
n
\end{array}\right)-\left(\begin{array}{c}
r+b-1\\
n
\end{array}\right)\right]}$$
Is there any other identity that would be more useful for this problem? If not does anyone have a useful hint for where to proceed from here?
| If we use the identity $\displaystyle\binom{m}{n}=\frac{m}{n}\binom{m-1}{n-1}$, we obtain
$$\displaystyle\frac{\binom{b-1}{k-1}\binom{r+b}{n}}{\binom{b}{k}\binom{r+b-1}{n-1}}=\frac{\binom{b-1}{k-1}\frac{r+b}{n}\binom{r+b-1}{n-1}}{\frac{b}{k}\binom{b-1}{k-1}\binom{r+b-1}{n-1}}=\frac{k}{b}\cdot\frac{r+b}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Divisors of $9!$ which are perfect square, cube or power $4$ [1] Total no. of positive divisors of $9!$ which are perfect square.
[2] Total no. of positive divisors of $9!$ which are perfect cube.
[3] Total no. of positive divisors of $9!$ which are perfect power 4.
$\underline{\bf{My\; Try}}$:: prime factors of $9! = 2^7 \times 3^4 \times 5 \times 7$
[1] Total no. of Divisors which are perfect square $9! = (2^3)^2\cdot (3^2)^2$ is $ = (2+1)\cdot (3+1) = 12$
[2] Total no. of Divisors which are perfect cube $9! = (2^2)^3\cdot (3^3)^1$ is $ =(3+1)\cdot (1+1) = 8$
[3] Total no. of Divisors which are perfect cube $9! = (2^4)^1\cdot (3^4)^1$ is $ =(1+1)\cdot (1+1) = 4$
Now My Question is that is it Right or not. If not please explain me
Thanks
| We use your prime factorization. Note that the perfect squares will have to be of shape $2^{2m} 3^{2n}$, where $m$ and $n$ are non-negative integers, and $2m\le 7$, $2n\le 4$.
For $m$ we can choose any of $0$, $1$, $2$, or $3$ ($4$ choices). For each of these choices, we can let $n=0$, $1$, or $2$. Thus the number of square divisors is $(4)(3)$, your answer.
For perfect cubes, our divisor will have shape $2^{3m}3^{3n}$. There are $3$ choices for $m$, and $2$ choices for $n$, giving a total of $(3)(2)$.
Fourth power is similar. We get $(2)(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3 \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be very nice to see a different solution from mine.
| Since $\sin(x)-a\cos(x)\le0$,
$$
\sin(x)-a^3\cos(x)\le(a-a^3)\cos(x)\tag{1}
$$
Since $a\in\mathbb{N}$, $a-a^3\le0$. If $\cos(x)\ge0$, then $(1)$ implies that $\sin(x)-a^3\cos(x)\le0$. So assume wlog that $\cos(x)\le0$. Therefore, $\sin(x)\le a\cos(x)\le0$, and thus,
$$
a\le\tan(x)\tag{2}
$$
Since $x$ is in the third quadrant, $(2)$ implies that
$$
\sin(x)-a^3\cos(x)\le\frac{a^3-a}{\sqrt{1+a^2}}\tag{3}
$$
Maximizing
$$
f(a)=\dfrac{a^3-a}{\sqrt{1+a^2}}\dfrac1{\sqrt{1+a^6}}\tag{4}
$$
with
$$
f'(a)=-\dfrac{a^4+1}{\sqrt{(a^6+1)^3(a^2+1)}}(a^4-4a^2+1)\tag{5}
$$
gives $f'(a)=0$ at $a=\sqrt{2+\sqrt3}\implies f(a)\le\frac13$.
Therefore, putting $(3)$ and $(4)$ together yields
$$
\sin(x)-a^3\cos(x)\le\frac13\sqrt{1+a^6}\tag{6}
$$
However, since $a\in\mathbb{N}$ and $f(2)=\dfrac{6}{5\sqrt{13}}$, we can improve the $\dfrac13$ to $\dfrac{6}{5\sqrt{13}}$. That is, with the restriction that $a\in\mathbb{N}$, we have a slightly stronger inequality:
$$
\sin(x)-a^3\cos(x)\le\frac{6}{5\sqrt{13}}\sqrt{1+a^6}\tag{7}
$$
For $x\le1000000$, the closest we get to equality in $(7)$ is $x=11270$ and $a=2$. For those values,
$$
\sin(x)-a\cos(x)=-0.00000638535498889561
$$
and
$$
\frac{\sin(x)-a^3\cos(x)}{\sqrt{1+a^6}}=0.33281742491402203493
$$
where
$$
\frac{6}{5\sqrt{13}}=0.33282011773513747321
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
show that the straight lines $(a^2-3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ form with the lie $ax+by+c=o$ an equilateral triangle show that the straight lines
$(a^2-3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ form with the lie ax+by+c=o
an equilateral triangle whose area is $\frac{c^2}{\sqrt{3}(a^2+b^2)}$
is there any other way to solve without intersection points. Also the Angle between the first two lines is 60 degrees.any another way to solve this.
| As I was saying, if we have three constants $F,G,H$ and $$ F x^2 + G x y + H y^2 = 0, $$ we divide through by $x^2$ to arrive at $$ F + G \frac{y}{x} + H \frac{y^2}{x^2} = 0. $$ Or, as I said, using the traditional $m = \frac{y}{x},$ we get
$$ F + G m + H m^2 = 0,$$ or in order
$$ H m^2 + G m + F = 0.$$
We suspect we are describing two lines passing through the origin. If so, the slope(s) $m$ satisfy
$$ m = \frac{-G \pm \sqrt {G^2 - 4 HF}}{2H}. $$
So, if $G^2 - 4 F H > 0,$ we do have two different slopes and two lines. For your original problem $$ F = a^2 - 3 b^2, G = 8 a b, H = b^2 - 3 a^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} .... + \frac {1}{\sqrt{n}} < 2\sqrt{n}$ Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} ... + \frac {1}{\sqrt{n}} < 2\sqrt{n} $
My attempt was as follows:
Lets assume the inequality is true for n = k
$S_k = 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} ... + \frac {1}{\sqrt{k}} $
$ => S_k < 2\sqrt{k} $
$ => S_k < 2\sqrt{k + 1} $
We need to prove that
$ => S_k + \frac {1}{\sqrt{k+1}} < 2\sqrt{k + 1} $
$ => \frac {1}{\sqrt{k+1}} < 2\sqrt{k + 1} - S_k $
Now I don't know where to go from here please help
| $$2\sqrt{k} + \frac {1}{\sqrt{k+1}} = \frac {2\sqrt{k^2+k}+1}{\sqrt{k+1}} < \frac {2\sqrt{k^2+k+\dfrac14}+1}{\sqrt{k+1}} = \frac{2\left(k+\dfrac12\right)+1}{\sqrt{k + 1}}= 2\sqrt{k + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$$
Can we find a known value for $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(k)}\right\rfloor$ for any $k\in\mathbb{N}?$
| First show that the sum converges by grouping summands in pairs, so it is enough to check the approximating sums with $n$ running from $1$ to $2^k-1$, say. For these, split the sum into the pieces where $\left\lfloor\frac{\log n}{\log 2}\right\rfloor$ is constant: $$\sum_{n=1}^{2^k-1} \frac{(-1)^n}{n} \left\lfloor \frac{\log n}{\log 2} \right\rfloor = \sum_{r=0}^{k-1} \sum_{n=2^r}^{2^{r+1}-1} r\frac{(-1)^n}{n} = \sum_{r=0}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n} = \sum_{r=1}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n}$$
Split $\frac{1}{n}$ for the even n's into $\frac{1}{n/2}-\frac{1}{n}$ and rearrange the inner sums:
$$\begin{align}\sum_{r=1}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n}&= \sum_{r=1}^{k-1} r \left( \sum_{n=2^r}^{2^{r+1}-1} \frac{-1}{n} + \sum_{n=2^{r-1}}^{2^r-1} \frac{1}{n}\right)\\&
=1+\sum_{r=1}^{k-2}(r+1-r) \sum_{n=2^r}^{2^{r+1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n}\\&= \sum_{n=1}^{2^{k-1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n}.\end{align}$$
The first sum is $\log 2^{k-1} + \gamma + o(1)$, the second is $(k-1) \log 2 + o(1)$.
For $k$ instead of $2$, you can do the same rearranging, but the sums no longer overlap, so it's trickier to handle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Maximize the volume of a rectangular box in the first octant with one vertex in the plane $x+2y+3z=3$ Find the volume of the largest rectangular box in the first octant with the three faces in the coordinate planes and one vertex in the plane $x+2y+3z=3$.
Now I know that $V=xyz$ and I have set $$V=xy\left(1-\frac{x}{3}-\frac{2y}{3}\right).$$
After I take the partial derivatives with respect to $x$ and $y$, I am not sure what to do. Any help would be great! Thanks.
| Since $V=xy-\frac{1}{3}x^2y-\frac{2}{3}xy^2$,
$V_{x}=y-\frac{2}{3}xy-\frac{2}{3}y^2=0$ and $V_{y}=x-\frac{1}{3}x^2-\frac{4}{3}xy=0$.
Multiplying by 3 in each equation and dividing by $y$ in the first equation and by $x$ in the second equation (since $x>0$ and $y>0$) gives
$2x+2y=3$ and $x+4y=3$. (Now solve for x and y.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$
(A) $a+b \over a-b$ (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None
EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (a-b)^2=0$$
but this is getting me nowhere.
| Hint :
$$\sqrt{1+\frac{4}{x}+\frac{4}{x^2}}=\sqrt{\frac{1}{x^2}(x+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.