Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find following If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find the value of $f(\alpha,\beta)=\dfrac{\beta^3}{2}\mathrm{cosec}^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)+\dfr... | $$f(\alpha,\beta)=\dfrac{\beta^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}+\alpha\right)}{\alpha^2}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}-\beta\right)}{\alpha^2}$$
$$f(\alpha,\beta)=\beta\left(\alpha^2+\beta^2+\alpha\sqrt{\alpha^2+\beta^2}\right)+\alpha\left(\alpha^2+\beta^2-\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $1 - 3xy + x^3 + y^3 = 0$ I am trying to solve the equation
$$
1 - 3xy + x^3 + y^3 = 0
$$
in the real numbers. I can see several (infinitely many) solutions:
*
*$x=y = 1$
*$x+ y = -1$
I am trying to show that these are all of them.
For example, if $x = y$, then $x^2(-3+2x) = -1$ forcing $x = y = 1$
I can't... | If you substitute $y$ with $-x-1$ you get the identity
$$x^3+(-x-1)^3-3x(-x-1)+1=0$$
This means that the polynomial $x^3+y^3-3xy+1$ is divisible by $x+y+1$. Performing long division you get the following factorization:
$$x^3+y^3-3xy+1= (x+y+1)(x^2-xy+y^2-x-y+1)$$
so the problem becomes proving that $(x;y)=(1;1)$ is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove by induction that $f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}$ is a multiple of 3 if it is an integer I'm trying to solve the following problem by induction but I'm getting stuck.
For positive integers $a$ and $b$, define $$f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}.$$ If $f(a,b)$ is an integer, p... | Note that $$f(a,b)=\frac{a^2+b^2+1}{ab}$$ So, if $ab\mid a^2+b^2+1$, then $f(a,b)$ is an integer.
Note that $a^2\equiv0,1\pmod3$.
If $3\mid a$, then $$a^2\equiv0\pmod3\to a^2+b^2+1\equiv1,2\pmod3$$so $3$ doesn't divide it, but $3\mid ab$. This implies that $f(a,b)$ is not an integer. The same analysis holds when $3\mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the $n^{th}$ derivative of $y=\frac {x}{x^2+a^2}$ Find the $n^{th}$ derivative of $y=\dfrac {x}{x^2+a^2}$.
My Attempt:
$$y=\dfrac {x}{x^2+a^2}$$
$$y=x.(x^2+a^2)^{-1}$$
Differentiating both sides with respect to $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$... | Hint:
Rewrite the definition of $y$ as $(x^2+a^2)y=x$, and apply Leibniz' formula for the $n$-th derivative of a product of functions: for $n>1$, we have
$$(x^2+a^2)y^{(n)}+2nx y^{(n-1)}=0,$$
whence the recurrence relation
$$y^{(n)}=-\frac{2nx}{x^2+a^2}y^{(n-1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power.
My try
Since $x, y, z$ are positive integers it follows that $x+2y, y+2z, z+2x \geq 3$. We have following system of equations:
$$\begin{ali... | Assume the prime is $p \neq 3.$ Note that we may keep (simultaneously) dividing out $p$ from $x,y,z$ until at least one of them is not divisible by $p\; , \;$ with the result that $$ \gcd(x,y,z) = 1 $$
We do have such primitive solutions when
$$ (x,y,z) = (1,1,1) \; \; , \; \; (x,y,z) = (13,7,1) $$
when both primes ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is... | Note
$$x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}=x(1+\frac1{2x}+O(\frac1{x^2}))=x+\frac12+O(\frac1x)$$
$$x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}=x+1+O(\frac1x)$$
Therefore,
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )$$
$$=\lim\limits_{x \to \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate $\sin\alpha$
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate
$\sin\alpha$
My proof:
$\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\\
\frac{\sin\left(\alpha-\frac{\pi}{4}\right)}{\cos\left(\alpha-\frac{\pi}{4}\right)}... | $\tan (\alpha -\frac {\pi} 4)=\frac {\tan (\alpha)-1} {1+\tan \alpha}=\frac 1 3$ and this gives $\tan \alpha =2$. Can you find $\sin \alpha$ from this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving HM-GM inequality using AM-GM I want to prove
$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} ≤ \sqrt[n]{a_1\cdots a_n}$$
Using AM-GM inequality.
My attempt:
Consider arbitrary $x_1,\cdots,x_n$ We know that
$$\sqrt[n]{x_1\cdots x_n} ≤ \frac{x_1 + \cdots + x_n}{n}$$
Replacing all $x$ with $\frac{1}{x}$
$$... | That's pretty much the standard proof that AM-GM implies GM-HM. The only variation among specific presentations of them is how much detail to go into. For example, you could argue you can just write$$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}}\le\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\implies\sqrt[n]{x_1\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the sum of $k/2^k, k=1$ to $n$ Let $S=1/2+2/2^2+3/2^3+...+n/2^n$
I try searching on the internet and see only the version of $k=1$ to infinity. I put this equation on Wolfram Alpha and get $(2^{n+1}-n-2)/2^n$ but I dunno how to do that. Please help
| This is an example of what is called an arithmetico-goemetric series. We can write it more compactly as $$S_n = \displaystyle\sum\limits_{k=1}^n \frac{k}{2^k}$$
The common ratio for the denominators is $2$, so we will multiply the entire series by $2$:
\begin{align}
S_n &= \,\qquad\frac{1}{2} +\frac{2}{4} +\frac{3}{8} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$ I am attempting the following trigonometry problem. Given that in an acute angled triangle $\Delta \text{ABC}$, the following equalities hold true
$$\cos^2(A)+\cos^2(B)+2\sin(A)\sin(B)\cos(C)=\dfrac{15}{8}\\
\cos^2(B)+\cos^2(C)+2\sin(B)\sin(C)... | Hint: Note that $$2\sin A\sin B\cos C-2\cos A\cos B\cos C=2\cos(180^\circ-A-B)\cdot\cos C=2\cos^2 C$$
So, the first equation becomes $$\cos^2 A+\cos^2 B+\cos^2 C+2\cos A\cos B\cos C=\frac{15}8$$This is symmetric, which makes it quite useful.
Now, we subtract the second equation from this to get $$\cos^2 A+2\cos A\cos B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is t... | We have indeed that
$$5+2\sqrt{6} =(\sqrt 3 + \sqrt 2)^2$$
$$5-2\sqrt{6} =(\sqrt 3 - \sqrt 2)^2$$
and therefore
$$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=-\sqrt{3}(\sqrt 3 + \sqrt 2+\sqrt 3 - \sqrt 2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Does $\sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ converge uniformly on $(-\infty,\infty)$?
Does $\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
converge uniformly on $(-\infty,\infty)$?
My attempt:
No. Consider the case where $x=0$, then
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = 0$.
For ... | For (1) you got the idea but the argument is not exactly right.
Indeed the series does converges for all $x$, as you show
$$ \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = \begin{cases} 1+ x^2 & \text{if } x\neq 0,\\ 0 & \text{if }x=0.\end{cases}.$$
However it does not converges to a continuous function. Thus the conver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Computing $\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$ We fix $y \ge 0$. What is the following limit?
$$\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$$
My attempt so far is as follows:
By Stirling's fo... | I've chosen to ignore the floor function; you can probably modify this argument to include the floor function if you wish.
The logarithm is
\begin{align}
&-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(\frac{n}{n-y\sqrt{n}})
\\
&=-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(1 + \frac{y}{\sqrt{n}-y})
\\
&= -y\sqrt{n} +(n - y\sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Paradox in indefinite integral Today,I met a problem about calculating such an indefinite integral$$ I=\int \frac{1}{(1+x^2)(1+x^{2019})}$$ My thought is as follows $$
\begin{aligned} I & = \int \frac { 1 } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = \int \frac { 1 } { \left( 1 + x ^ { - ... | In going from the first line to the second line, you make a substitution. Thus, if
$$
I =\int \frac{1}{(1+x^2)(1+x^{2019})}dx = \int f(x)\,dx,
$$
Then
$$
\int \frac { 1 } { \left( 1 + x ^ { - 2 } \right) \left( 1 + x ^ { - 2019} \right) } \left( - \frac { 1 } { x ^ { 2 } } \right) d x = \int f(1/x)\,dx.
$$
We have no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find: $\lim\limits_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$ Calculate the following limit:
$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$$
I need find this limit without l'Hospital or Taylor series.
Wolfram alpha gives $\frac{1}{2}$
My try is:
Let: $a=\sqrt{... | Note
$$\sqrt{x^{2}+1}-x=x \left(\sqrt{1+\frac1{x^2}}-1\right)=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}$$
$$\sqrt[3]{x^{3}+1}-x=x \left(\sqrt[3]{1+\frac1{x^3}}-1\right)=\frac{\frac1{x^2}}{\sqrt[3]{\left(1+\frac1{x^2}\right)^2}+\sqrt[3]{1+\frac1{x^2}}+1}$$
Thus,
$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How far can I rewrite an equation before the derivative is not equivalent? I've run into a problem that I can't explain to my class.
We are looking at the derivative for the equation $\frac{x}{y}+\frac{y}{x}=3y$. We calculated it to be $\frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}$ and we also verified it with Wolfram Alpha.
A... | Along a curve $f(x,y) = 0$ the derivative $y'$ is given by
$$
y_f'(x,y)
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
.
$$
Given some smooth function $\varphi$ which doesn't vanish on the curve, we can create a new function $g(x,y) = \varphi(x,y) \, f(x,y)$ that also vanishes along the curve. For this we get
$$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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General solutions of $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3$ Note: this actually has nothing to do with probability theory. Fell free to skip ahead and solve the equations below.
Since I am unable to solve this continuous version of a problem about moments, I am now trying to do something simple but related. Let's say $X$ is ... | Let the values be equal to $t$.
Then $xy+yz+zx = \frac{ (x+y+z)^2 - (x^2+y^2+z^2) }{2} = \frac{t^2-t}{2}$
and $xyz = \frac{ (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z)}{-3} = \frac{t^3-t - 3(t^2-t)/2 \times t}{-3} = \frac{-t^3+3t^2 - 2t } { 6}$.
So, these are roots to the cubic
$$X^3 - 6tX^2 + 3(t^2-t)X+(t^3-3t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ Tf(x)= \frac{1}{\sqrt{\pi}}\int_0^x\frac{f(x)}{\sqrt{x-t}}dt$ show that , $Tf(x)\in L^1$ If $Tf(x)= \frac{1}{\sqrt{\pi}}\int_0^x\frac{f(t)}{\sqrt{x-t}}dt$. Show that $T:L^1[0,1]\to L^1[0,1]$ and $TTf(x) = \int_0^1 f(t)dt$
$\textbf{My attempt:}$
Suppose that $\frac{1}{\sqrt{\pi}}\chi_{[0,x]}(t)\frac{f(t)}{\sqrt{... | Part (a) looks fine to me (I would argue you're using Tonelli's theorem instead of Fubini's, but people often combine them).
Part (b) has many errors. For one, you end up with an expression that has $x,y,t$ as free variables instead of just $x$. Here is one solution:
$$\begin{align*}TTf(x) &= \frac{1}{\sqrt{\pi}} \in... | {
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"url": "https://math.stackexchange.com/questions/3464909",
"timestamp": "2023-03-29T00:00:00",
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Present $\alpha^{-1}\beta$ as a product of disjoint cycles, where $\alpha=(1\ 6)(3\ 2\ 8)$ and $\beta=(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$. Let $\alpha=(1\ 6)(3\ 2\ 8)$ and $\beta=(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$
Present $\alpha^{-1}\beta$ as a product of disjoint cycles
$$\alpha^{-1} = (8\ 2\ 3)(1\ 6)$$
$$\alpha^{-1}\beta... | We wish to calculate the following in disjoint cycle notation
\begin{eqnarray*}
\alpha^{-1} \beta = (8\ 2\ 3)(1\ 6)(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10).
\end{eqnarray*}
Start with the element $1$ and reading left to right, $1$ is mapped to $6$ in the second cycle & the $6$ is mapped to $3$ in the fourth cycle, so $1$ maps t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer
| If $m/5$ is odd, $11|2^5+3^5|2^m+3^m$ while $181|3^5+4^5|3^m+4^m$, so$$\operatorname{lcm}(11,\,181)=1991|(2^m+3^m)(3^m+4^m)=6^m+8^m+9^m+12^m.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$? I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive rational numbers.
... | Another way for positive variables.
We need to prove that
$$\sum_{cyc}2\sqrt{(1-a)(1-b)}\geq\sum_{cyc}(2a+2\sqrt{ab})$$ or
$$\sum_{cyc}(1-a+1-b-2a-2\sqrt{ab})\geq\sum_{cyc}(1-a-2\sqrt{(1-a)(1-b)}+1-b)$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\geq\sum_{cyc}(\sqrt{1-a}-\sqrt{1-b})^2$$ or
$$\sum_{cyc}\frac{(a-b)^2}{(\sqrt{a... | {
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"url": "https://math.stackexchange.com/questions/3474228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equa... | It is $$\frac{x}{y+1}+\frac{y}{x+1}\le 1$$ since
$$x(x+1)+y(y+1)\le (x+1)(y+1)$$ This is equivalent to
$$0\le x(1-x)+(1-y)(1-x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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What is the center of mass of the liquid with a density of $10 -x^{2}$ in a cone? (drawing included) Problem:
A straight circular cone with the height $2$ m and the radius $1$ m has the pointy end downward and is filled with a liquid. In a flat plane parallel with the bottom surface of the cone and at a distance of $x$... | Your work looks good but there is a mistake in the integrand:$$\int_0^2 dm = \frac{\pi}{4} \int_0^2 (10-x^2) \color{red}{x^2}dx = \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{0}^{\infty}\frac{x^{s}}{ax^{3}+bx^{2}+cx+d}dx$ I know that the following formulas are true:
$$\int_{0}^{\infty}\frac{x^{s}}{ax+b}dx=\frac{-\pi}{a\sin\left(\pi s\right)}\left(\frac{b}{a}\right)^{s}$$
and
$$\int_{0}^{\infty}\frac{x^{s}}{ax^{2}+bx+c}dx=\frac{\pi\sin\left(s\cos^{-1}\left(\frac{b}{2\sqrt{ac}}\right)\... | Write $1/(ax^3 + b x^2 + c x + d)$ in partial fractions:
$$ \frac{1}{a x^3 + b x^2 + c x + d} = \sum_\alpha \frac{1}{(3 \alpha^2 a + 2\alpha b + c)(x - \alpha)} $$
where the sum is over the roots of $a x^3 + b x^2 + c x + d$ (assuming these are distinct). Of course you'll also want to assume none of these roots are po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= ... |
Goal: Establish the fact that in order to solve $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$
we only need to solve
$$
x=\sqrt[3]{6+x}
$$
Define $f(x)$:
$$
f(x) = \sqrt[3]{x+6}$$
From line 3, we are looking to solve:
$$
x = f(f(f(x)))$$
Notice that $f(x)$ is strictly increasing. In other words, for all $a$ and $b$ in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
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$\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$ If P is a point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Whose foci are S and S’.Let angle PSS’=$\alpha$ and $PS’S=\beta$ then prove that $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$
I know that S(ae,0) and S’(-ae,0) and $b^2=a^2(1-... | Write the coefficients of $P$ as $x=a\cos t,\,y=b\sin t$ so$$\begin{align}PS^2&=a^2(\cos t-e)^2+b^2\sin^2 t\\&=a^2(1-e\cos t)^2,\\PS^{\prime2}&=a^2(1+e\cos t)^2,\\SS^{\prime2}&=4a^2e^2,\end{align}$$and$$\begin{align}\cos\alpha&=\frac{PS^2+SS^{\prime2}-PS^{\prime2}}{2PS\cdot SS^\prime}\\&=\frac{e-\cos t}{1-e\cos t},\\\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Given the matrices $A\in M_{m\times n}, B\in M_{n\times p}: \ B\neq 0$, suppose $AB=0$. Prove: there exists a non-trivial solution for $Ax=0$ Given two matrices: $A_{m\times n}$ and $B_{n\times p}$, such that $B\neq 0$,
suppose $AB=0$.
Prove: there exists a non-trivial solution for $Ax=0$.
My Attempt:
suppose by cont... | If $B = [b_1,...,b_p] \neq 0$, there is some $i \in \{1,...,p\}$ such that $b_i \neq 0$. Then, by definition of the matrix product, $Ab_i = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^... | Another way :
$$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$
$$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$
$$a_{n+1}=a_{n+1-2r}+\underbrace{2n-1+2n-3+\cdots}_{r\text{ terms}}$$
Here $n+1=101,n+1-2r=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem :
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me :
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$
But i can't get by my... | Hint:
$$\dfrac{\cos4x}{\cos3x}=\dfrac{2\cos^22x-1}{\cos x(4\cos^2x-3)}=\dfrac{2(1-2\sin^2x)^2-1}{\cos^2x(4\cos^2x-3)}\cdot\cos x$$
Set $\sin x=u$
$$I=\int_0^{\sqrt3/2}\dfrac{8u^4-8u^2+1}{(1-u^2)(1-4u^2)}du$$
Writing $u^2=v$ and use Partial Fraction Decomposition
$$\dfrac{8v^2-8v+1}{(1-v)(1-4v)}=2+\dfrac a{1-v}+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int \frac{x^2}{1+x^5}dx$ I tried factoring the denominator and performing partial fraction decomposition, but the algebra becomes too cumbersome for me...
Is there a cleaner approach for this integral?
| There is a clean way to integrate. Note that $x^5+1$ factorizes as
$$x^5+1= (1+x)(x^2-2\phi_+x+1)(x^2-2\phi_-x+1)$$
with $\phi_{\pm} = \frac{1\pm\sqrt5}{4}$. So, decompose the integrand,
$$\frac{5x^2}{1+x^5}=\frac1{x+1}-\phi_- \frac{2x+2}{x^2-2\phi_+x+1}
-\phi_+ \frac{2x+2}{x^2-2\phi_-x+1}$$
The integral for the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod ... | Another way:
$(\pm1)^2\equiv1,(\pm2)^2\equiv4,(\pm3)^2\equiv9,(\pm4)^2\equiv6\pmod{10}$
The last digit of $n^2,n$ will be same if $n=10m,10m+1,10m+5,10m+6$
$(10m+6)^2\equiv36+20m\pmod{100}$
We need $10m+6\equiv36+20m\pmod{100}\iff10m\equiv-30\pmod{100}\iff m\equiv-3\pmod{10}$
WLOG $m=10r-3$
$10m+6=10(10r-3)+6=100r-24$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Find minimum value of $P = \frac{y + z}{x}$
Let $x,y,z > 0$ such that $3x(x+y+z)=yz$ . Find the minimum value of $$P=\frac{y+z}{x}$$
$$3x(x+y+z)=yz\Leftrightarrow 3(x+y)(x+z)=4yz$$
Or $$3(\frac{x}{y}+1)(\frac{x}{z}+1)=4$$
Let $a=x/y, b=x/z ( a,b>0)$
We have: $$4=3(a+1)(b+1)\ge3 \cdot 2 \sqrt a \cdot 2 \sqrt b=12 \... | Let $y+z=2u$ and $yz=v^2$, where $v>0$.
Thus, since by AM-GM $$\frac{y+z}{2}\geq\sqrt{yz},$$ we obtain $u\geq v$,
$$3x^2+6ux-v^2=0$$ or $$x=-u+\sqrt{u^2+\frac{v^2}{3}}$$ and
$$\frac{y+z}{x}=\frac{2u\left(\sqrt{u^2+\frac{v^2}{3}}+u\right)}{\frac{v^2}{3}}\geq\frac{2v\left(\sqrt{v^2+\frac{v^2}{3}}+v\right)}{\frac{v^2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that $r^{2} = 3$ from a geometric series
The sum of the first two terms of geometric sequence $a_n=a_0r^n$ is $7$, and the sum of the first six terms is $91$.
Show that $r^{2} = 3$.
How do we go about this?
Here's what I have tried -
$\displaystyle \Rightarrow \frac{\dfrac{a(r^{6}-1)}{r-1}}{\dfrac{a(r^{2}-1)}{r-... | Use this identity $a^3- b^3 = (a-b)(a^2 +ab+b^2)$, and then solve a quadratic equation.
$$\Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = 13$$
$$\Rightarrow \frac{(r^{2}-1)(r^4+r^2+1)}{(r^{2}-1)} = 13$$
$$\Rightarrow r^4+r^2+ 1= 13$$
$$\Rightarrow r^4+r^2-12=0$$
You'll find something like $(r^2 -3)(r^2 +4)=0$
BINGO!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$
Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$
How to solve this problem without using derivatives?
Michael Rozenberg's edit:
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1... | $$ \forall_{y\ge -1}\quad\sqrt{1+y}\ \le\ 1+\frac y2 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ =\ \sqrt{1+(x-3)}\ \le\ 1 +\frac{x-3}2 $$
and
$$ \forall_{x\ge -1}\quad\sqrt{x+1}\ =\ 2\cdot\sqrt{1+\frac{x-3}4}\ \le
\ 2+\frac{x-3}4 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ \le\ 2+x $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx$ I want to show that
$$\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx=\frac{\psi^{(1)}(\frac{t}{2\pi})-4\psi^{(1)}(\frac{t}{\pi})}{8\pi}t+\frac14
$$
where $\psi^{(1)}(z)$ is the ... | We first modifiy the integral to be calculated, by remarking that, changing $x=-u$,
\begin{align}
I&=\int^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)}\,dx\\
&=-\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{-tu} +1\right)}\,du\\
&=-\int_{-\infty }^{0}\frac{ue^{tu} }{\left( u^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to evaluate $\iint_A \frac{1}{x^2+y^2}\,\mathrm dx\,\mathrm dy,$ where $A=[\frac{1}{a},a]\times[0,1]$ I want to compute
$$\iint_A \frac{1}{x^2+y^2}\,\mathrm dx\,\mathrm dy$$ where $A:=\left[\frac{1}{a},a\right]\times[0,1]$.
I got that this is equal to $\int_{1/a} ^a \frac{1}{x}\arctan \Big(\frac{1}{x}\Big) \mathrm ... | Let's catch up with you:
$$\begin{align}
\iint\limits_A \frac{1}{x^2+y^2}dx\,dy &= \int\limits_{\frac{1}{a}}^a\int\limits_0^1\frac{1}{x^2+y^2}dy\,dx\\
&=\int\limits_{\frac{1}{a}}^a\int\limits_0^1\frac{1}{x^2(1 + (y/x)^2)}dy\,dx.
\end{align}$$
Taking $u = y/x, \frac{\partial u}{\partial y} = 1/x$ so the last integral be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int \frac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}dx$ $$\int \dfrac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \dfrac{\sqrt{1-2\sin^2x\cos^2x}\cdot\sin x}{\sin^4x\cos x}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$\int -\dfrac{\sqrt{1-2t^2(1-t^2)}}{t(1-... | Continue the second approach below with integration by parts
$$I=\int \dfrac{\sqrt{t^4+1}}{t^3}d
=-\frac12 \frac{1+t^4}{t^2} +\int \frac{tdt}{\sqrt{1+t^4}}dt$$
where
$$\int \frac{tdt}{\sqrt{1+t^4}}=\frac12 \int \frac{d(t^2)}{\sqrt{1+(t^2)^2}}=\frac12 \sinh^{-1}t^2
$$
Thus,
$$I =- \frac{1+t^4}{2t^2} +\frac12 \sinh^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the positive integers $a,b,$ and $c$ s.t. $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}.$
Let $a,b,c$ be positive integers such that $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}$. Determine the values of $a, b,$ and $c$.
To solve this simplific... | Let $x,y,z\in\mathbb{R}$ such that $x^3 = 5, y^3 = 4$, and $z^3 = 2$. Then $y=z^2, y^2 = 2z$, and $yz = 2.$ We want to simplify $3\sqrt{x-y}$. Note that
$$\begin{align}
(x-y)(x+y)^2&=(x-y)(x^2+2xy+y^2)\\
&=x^3-y^3+x^2y-xy^2\\
&=1+(xz)^2-2xz\\
&=(xz-1)^2.
\end{align}$$
Hence
$$\begin{align}
3\sqrt{x-y} &= 3\cdot\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$ I came across the following:
$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 1... | If I consider the equations you provide with your "ideas so far":
\begin{align}
1 \times 9 + 1 &= 10 \\
12 \times 9 + 2 & = 110 \\
123 \times 9 + 3 & = 1110 \\
\vdots\\
123456789 \times 9 + 9 & = 1111111110, \\
\end{align}
The first equation being true, this system is equivalent to the system composed of their succes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
How to factorize $X^5-1$ into irreductible factors? I have to factorize the polynomial $P(X)=X^5-1$ into irreducible factors in $\mathbb{C}$ and in $\mathbb{R}$, this factorisation happens with the $5$th roots of the unity.
In $\mathbb{C}[X]$ we have $P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$.
In $\mathbb{R}[X]$ the... | A way to obtain an explicit factorisation using Chebyshev polynomials:
Using the recurrence relation:
$$P_{n+1}(t)=2tP_n(t)-P_{n-1}(t),\qquad P_0(t)=1,\;P_1(t)=t,$$
we readily obtain
$$P_5(t)=2t\, P_4(t)-P_3(t)=16t^5-20t^3+5t,$$
so that, for $t=\cos\frac\pi{10}$, we obtain the equation
$$\cos5\Bigl(\frac{\pi}{10}\Bigr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
... | Using @Andrew Chin's comments, we face for the second series
$$S_\infty=\sum_{n=0}^\infty\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}$$
Let $$a_n=\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}=\frac{\Gamma (n+1)}{\left(1+\frac{1}{x}\right)_n}$$ where appear Pochhammer symbols. So
$$S_p=\sum_{n=0}^p a_n=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system :
$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$
My try but I think not complete :
The system equivalent :
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end... | The procedure you followed was correct, including getting the right values of $y$. However, you made mistakes for one value of $x$ (i.e., $\frac{17}{3}$ is not correct) and for both your values of $z$.
However, here's a slightly easier way to deal with the equations instead. You can use the symmetry to multiply your $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Nature of the critical point $(0,0)$ of the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $(0,0)$ is a critical point. Observe,
\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{x... | You have that
$$
g_a(x)=f(x,ax^2 ) = 2\left( {a^2 - a} \right)x^4 + \left( {1 - a} \right)x^6
$$
With $0<a<1$ the function $g_a(x)$ has a local maximum. With $a>1$ the function has a local minimum. This means that $(0,0)$ is a saddle point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$ for non-negative real $a_i$ For any integer $n$ and any nonnegative real numbers $a_1,\ldots,a_n$ we have
$$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$$
It seems to use Holder inequality, but I can't get the $\frac... | Remark1: @fedja gave an excellent solution by proving a continuous version of the OP which implies the OP (see the link @Martin R pointed out in comment for the OP, which was deleted now). I rewrote fedja's solution for the OP (discrete version).
Remark2: Actually, although @fedja proved a continuous version, his proof... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Distance from point $D$ to $BI$
I found that : $BI = 2\sqrt 2, DI = \sqrt 5, DB = 2\sqrt 2$. It means $\triangle BDI$ is isosceles triangle. Lets say $B$ to $DI$ is $a = \sqrt {BI^2 - ( \frac 12 DI)^2} = \sqrt{\frac{27}{4}}$. Lets say $t$ is distance from D to BI.
$t. BI = a. DI$
$t = \frac{\sqrt{27}}{4}$.
But why my ... | Note that $BI = \sqrt5$. So, △BDI is isosceles with DI = BI instead. Then, I to DB is
$a = \sqrt {BI^2 - ( \frac 12 DB)^2} = \sqrt{3}$ and
$$t = \frac{DB}{BI}a = \frac{2\sqrt2}{\sqrt5}\cdot\sqrt3=\frac{\sqrt{24}}5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Closed form of recurrence relation $a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}$ Given the following sequence:
$$\begin{cases}a_{0}=2\\a_{1}=\frac{5}{2}\\a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}\end{cases}$$
Prove that
$$a_{n}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}$$
For all natural numbers $n$.
My try:... | You want to proof that $a_n = f(n)$ for any $n\inℕ$ with
$$ f(n) := 2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}.$$
Check of two (!) initial conditions:
*
*$f(0) = 2 = a_0 \checkmark$
*$f(1) = \frac52 = a_1 \checkmark$
Inductive step:
$$\begin{aligned}
a_{n+1} &= a_n (a_{n-1}^2 - a_0) - a_1\\
&= f(n) (... | {
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Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that
$$x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
x_{n+2}-2 x_{n+1} &= 2 x_{n... | $\begin{align}x_{n+2}=2x_{n+1}+2^n(x_2-2x_1)&&2x_{n+1}=2[2x_n+2^{n-1}(x_2-2x_1)]=4x_{n}+2^n(x_2-2x_1)\\
x_{n+2}=4x_n+2^n(x_2-2x_1)2&&4x_n=4[2x_{n-1}+2^{n-2}(x_2-2x_1)]=8x_{n-1}+2^n(x_2-2x_1)\\
x_{n+2}=8x_{n-1}+2^n(x_2-2x_1)3&&8x_{n-1}=8[2x_{n-2}+2^{n-3}(x_2-2x_1)]=16x_{n-2}+2^n(x_2-x_1)\end{align}$
$\dots$
$x_{n+2}=2^n... | {
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Help proving this inequality Give $x$ , $y$ , with $x,y \in \Re$ , and $x^2 + y^2 = 1$ , show that .
$$ \left(\frac{1}{1+x^2}\right) + \left(\frac{1}{1+y^2}\right) + \left(\frac{1}{1+xy}\right) \ge \left(\frac{3}{1+((x+y)/2))^2}\right)$$
This is my approach:
from $x^2 + y^2 = 1$ , we can see that $xy \le \frac{1}{2}$ ,... | Let $xy=t$.
Thus, by AM-GM $\frac{1}{2}\geq t\geq-\frac{1}{2}$ and we need to prove that $$\frac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}+\frac{1}{1+xy}\geq\frac{3}{1+\frac{x^2+y^2+2xy}{4}}$$ or
$$\frac{3}{2+t^2}+\frac{1}{1+t}\geq\frac{3}{1+\frac{1+2t}{4}}$$ or
$$(1-2t)(5t^2+3t+1)\geq0$$ and we are done!
| {
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"source": "stackexchange",
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show that $x_1^2+x_2^2+\cdots+x_{2020}^2\geqslant x_1x_2+x_2x_3+\cdots+x_{2020}x_1$ how can i use AM-GM inequality to show that $$x_1^2+x_2^2+\cdots+x_{2020}^2\geqslant x_1x_2+x_2x_3+\cdots+x_{2020}x_1 ?$$
Where $x_1,x_2,\ldots,x_n$
are any real numbers.
to get an idea I tried to do for particular cases with $n=3,4.$
... | The idea is the same:
$$
x_1x_2 + x_2x_3 + \cdots + x_{2020}x_1 \leq \frac{x_1^2 + x_2^2}{2} + \frac{x_2^2 + x_3^2}{2} + \cdots + \frac{x_{2020}^2 + x_1^2}{2} = x_1^2 + x_2^2 + \cdots + x_{2020}^2
$$
| {
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"source": "stackexchange",
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Different result of counting number between 1 and 2000 which the sum of all the digits is 25. How many number between 1 and 2000 which the sum of all the digits is 25?
I have tried to answer in two ways.
First way
Let the digits is $A$, $B$, $C$, and $D$.
$A+B+C+D=25$ with $A\leq 1$, $B\leq 9$, $C\leq 9$, and $D\leq 9$... | The first way is incorrect.
Every combination of $A,B,C,D$ maps to a valid combination of $E,F,G,H$, such as $A=0, E=1$ and $A=1, E=0$. However, every combination of $E,F,G,H$ does not map to a combination of $A,B,C,D$: for example, $(E,F,G,H)=(3,0,0,0)$ leads to $A = -2$.
This means that you are counting solutions tha... | {
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Does the series converge, problem with $(-1)^n$, but Leibnitz impossible. Hi my problem is connected with this series
$$
\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt[3]n + (-1)^{n(n+1)/2}}
$$
I was trying to sum it by $4$ elements, but it didn't end up well.
I don't know how to deal with this problem, Leibnitz, Dirichlet, Abe... | We consider the series
$$
\sum\limits_{n = 4}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}
{{\sqrt[3]{n} + \left( { - 1} \right)^{\frac{{n\left( {n + 1} \right)}}
{2}} }}}
$$
instead your series. If we prove that this series is convergent then your series is convergent also. We can rewrite the series as
$$
\sum\lim... | {
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"source": "stackexchange",
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Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$.
|
We are looking for a generating function $G(z)$ where the coefficient of $z^n$ gives the wanted number of solutions. We want to find a proper representation for
\begin{align*}
G(z)=\sum_{{{x_1,x_2,x_3\geq 1}\atop{x_1\geq x_2\geq x_3}}\atop{x_1<x_2+x_3}}z^{x_1+x_2+x_3}\tag{1}
\end{align*}
In order to simpli... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$ If $ a $ and $ b $ are real numbers that satisfy the equation $ 17 (a ^ 2 + b ^ 2) -30ab-16 = 0 $, the maximum value of the expression$$\sqrt{16a^2+4b^2-16ab-12a+6b+9}$$goes:
a) $1$
b) $3$
c) $5$
d) $7$... | The hint.
Let $2a-b=k.$
Thus, we need to find a maximal value of $\sqrt{4k^2-6k+9}.$
Now, find all values of $k$, for which the equation
$$k^2(17(a^2+b^2)-30ab)=16(2a-b)^2$$ has solutions and choose from these values such that $\sqrt{4k^2-6k+9}$ will get a maximal value.
I got that it happens for $k=-\frac{5}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11$ Hello everyone: I want to prove that the following holds for all $a,b>0$: $$\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11.$$
My attempt: We notice by AM-GM that $\frac1a+2a\geq 2\sqrt 2$ for all $a>0$. So we only need to prove that $$\frac3b+2\sqrt{b^2+8}>11-2\sqrt{2}.$$
... | Firstly, it should be $$\frac{3}{b}+4\sqrt{b^2+8}\geq11-2\sqrt2,$$ which is true by C-S and AM-GM:
$$\frac{3}{b}+4\sqrt{b^2+8}=\frac{3}{b}+\frac{4}{3}\sqrt{(1+8)(b^2+8)}\geq$$
$$\geq\frac{3}{b}+\frac{4(b+8)}{3}\geq4+\frac{32}{3}>11-2\sqrt2.$$
Secondly, we can prove your second inequality by the similar way:
$$\frac{3}{... | {
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Solving differential equation by separating variables Solve the following differential equation:
$$\frac yx \frac {dy}{dx}=\sqrt {1+x^2+y^2+x^2y^2}$$
I have tried like this:
$$\frac yx \frac{dy}{dx}=\sqrt {1+x^2+y^2(1+x^2)}\\
\implies \frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}\\
\implies \int \frac {ydy}{\sqrt {1+y... | $$\frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}$$
$$\implies \int \frac {ydy}{\sqrt {1+y^2}}=\int {x}{\sqrt{1+x^2}}~dx$$
$$\implies \sqrt {1+y^2}=\dfrac{\left(x^2+1\right)^\frac{3}{2}}{3}+c$$where $c$ is integrating constant.
| {
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"source": "stackexchange",
"question_score": "2",
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Check Proof by Contradiction Please check the following proof.
Result to Prove: If $a$ and $b$ are odd integers, then $4\not| (a^2+b^2)$
Assume, to the contrary, that there exists two odd integers $a$ and $b$ such that $4|(a^2+b^2)$.
Then, $(a^2+b^2)=4n$ for some $n\in\Bbb Z$.
Let $a=2x+1$ and $b=2y+1$ for some $x,y\... | $2(2x^2+2y^2+2x+2y+1)=4n$ does not imply $2x^2+2y^2+2x+2y+1=4n$, but $2x^2+2y^2+2x+2y+1=2n$. Other than that, it's fine.
| {
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Find a, b, c and d of a function $f(x) = ax^4 + bx^3 + cx^2 + dx + 1$, given values f(1), f(2), f(3), ... I have a (very-likely) 4-grade function, and I know all the values, but I am having problems to find the formula of the function itself.
The number without the x is a one for sure.
The functions points are:
(0 | ... | Given $7$ distinct data points , we can use polynomial interpolation in the Lagrange from such that:
$\color{red}{\left(0,1\right)},\color{green}{\left(1,9\right)},\color{pink}{\left(2,41\right)},\color{blue}{\left(3,129\right)},\left(4,321\right),\color{olive}{\left(5,681\right)},\color{orange}{\left(6,1289\right)}$
... | {
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"source": "stackexchange",
"question_score": "1",
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Trouble finding limits at $-\infty$. I'm being asked to evaluate $\lim\limits_{x\rightarrow\infty}f(x)$ and $\lim\limits_{x\rightarrow -\infty}f(x)$ for the following funtion: $f(x)=\frac{4x^3+2}{x^3+\sqrt{4x^6+3}}$.
For the first limit I proceed as follows:
$\begin{align*}
\lim\limits_{x\rightarrow\infty}f(x)&=\lim\li... | When $x<0$ we have $x^{-3} \sqrt{4x^6+3} = -\sqrt{x^{-6}(4x^6+3)} = - \sqrt{4 + 3x^{-6}}$.
| {
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"source": "stackexchange",
"question_score": "1",
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Divide $ yy' = (y + 1)^2 $ by $ (y + 1)^2 $ I was reading from Ordinary Differential Equations (Lesson 4C page 36) and came across this question:
Find a 1-parameter family of solutions of the differential equation
$$ (a) \quad yy' = (y + 1)^2 $$
and the particular solution for which $\quad y(2) = 0$
The first step was ... | In differential equations and calculus it is often assumed that
$$y'=\frac{dy}{dx}$$
where $y$ is the dependent variable and $x$ is the independent variable. In your case, given $y \neq 1$
$$\frac{yy'}{(y+1)^2} = \frac{(y+1)^2}{(y+1)^2}=1 \implies \frac{y}{(y+1)^2}\frac{dy}{dx}=1 \implies \frac{y}{(y+1)^2}\,dy=dx$$
by ... | {
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If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \leq A \leq B$, then det $A \leq$ det $B$. Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.
If $A$ and $B$ are linear transformations on a finite-dimensional inner prod... | A verbose ("for dummies") version of Omnomnomnom's and Andreo's proof is as follows.
We have two cases: (a) $B$ is non-invertible, and (b) $B$ is invertible.
Case (a): since $B$ is not invertible, det $B = 0$. Also, the null-space of $B$ has some non-zero vector $x$. It follows from the positivity of $(B-A)$ that $\big... | {
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Find all positive integers $n $ such that $n^2+n+7$ devided by 13. An idea please Find all positive integers $n $ such that $n^2+n+7$ devided by 13.
An idea please
| I'm assuming you want the value to be divisible by 13.
Since the values mod 13 (remainder when divided by 13) repeat every 13, we just need to test 1, 2, 3, 4, 5, 6, 7, 8, 9, ..., 13.
1: 1 + 1 + 7 = 9
2: 4 + 2 + 7 = 13 works
3: 9 + 3 + 7 = 19
4: 16 + 4 + 7 = 27
5: 25 + 5 + 7 = 37
6: 36 + 6 + 7 = 49
7: 49 + 7 + 7 = 63
... | {
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Condition for $\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6}+ . . . =\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .$ I saw this problem in a 'classical algebra' textbook:
If $y=2x^2-1$ prove that under certain condition we have:
$\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6}+ . . . =\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5... | By What is the correct radius of convergence for $\ln(1+x)$?, we need $$\dfrac1{x^2},\dfrac1{y^2}<1$$
Now,
$$2\sum_{r=0}^\infty\dfrac{(1/y)^{2r+1}}{2r+1}=\ln\dfrac{1+1/y}{1-1/y}=\ln\dfrac{y+1}{y-1}$$
$$\implies-\ln\left(1-\dfrac1{x^2}\right)=\ln\left(\dfrac{y+1}{y-1}\right)=-\ln\left(\dfrac{y-1}{y+1}\right)$$
$$\implie... | {
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"source": "stackexchange",
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Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:
If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.
Initial attempts at AM-GM are not very helpful for me, I don't think.... | COMMENT.-I find this inequality quite difficult to prove by ordinary means. However it seems to me we could do something valuable as follows: The maximum of $f(x)=\dfrac{(x^2+1)^2}{(x^8+1)(x^4+1)}$ is equal to $1.695$ and it is taken at $x=0.717$ Putting $f(x)$ and $f(y)$ maximum we must have $z=\dfrac{3-xy}{x+y}=1.733... | {
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Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdo... | Use Binary AM-GM as
$$3x+\frac{1}{15x}\ge \frac{1}{\sqrt{5}}, ~\text{equality when}~ x=\frac{1}{\sqrt{45}}\tag1$$
$$5y+\frac{25}{y} \ge 2\sqrt{125}, ~\text{equality when}~ y=\sqrt{5} \tag2$$
$$z+\frac{1}{36z} \ge \frac{1}{3} ~ \text{equality when} z=\frac{1}{6} \tag3$$
Adding them, we get
$$3x+\frac{1}{15x}+5y+\frac{2... | {
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"timestamp": "2023-03-29T00:00:00",
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Trisected sides of a scalene triangle Scalene triangle $\bigtriangleup ABC$ has area 45. Points $P_1$ and $P_2$ are located on side $AB$ such that $AP_1 = P_1P_2 = BP_2$. Additionally, the points $Q_1$ and $Q_2$ are located on side $AC$ such that $AQ_1 = Q_1Q_2 = CQ_2$. The area of the intersection of triangles $BQ_1Q_... | I'm not sure I understand what you mean by "No answer". If every scalene triangle gives the same fraction (strongly implied by the problem), then by taking a limit (scalene triangles that approximate an equilater one better and better), we would expect an equilateral triangle to give the same fraction as well, so that ... | {
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Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$. This is the final step of one old competition problem, and in textbook it simply says "reader can prove this", but I found this a bit nontrivial.
Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$.
I tried ... | Trigonometric Proof
This follows almost immediately from $$\sin 20^\circ \sin 40^\circ \sin 80^\circ =\frac{\sqrt 3}{8},\tag{1}$$
which is a corollary of
$$\sin x \sin (60^\circ-x) \sin (60^\circ+x)=\frac{\sin(3x)}{4}.$$
My hint is to write one of $\sin 20^\circ$ from $\sin^2 20^\circ$ as $2\sin 10^\circ \cos 10^\circ$... | {
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If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(... | Let $(x-3)=y$ Then we get
$$\sum_{r=0}^{2m} A_r(1+y)^r=\sum_{r-0}^{2n} B_r~ y^r$$
Compare the coefficient of $y^n$ both sides, then
$${n \choose n}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}=B_n$$ or
$${n+1 \choose n+1}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}$$
Next use $${n \choose r}+{n \choose r... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 \ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 $ Let real $a,b,c,d > 0$. Show that
\begin{align}
\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 &\ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 \\
&= (a b^2 c^4 d^8 )^4 +(b c^2 d^4 a^8 )^4 +(c d^2 a^4 b^8 )^4 +(d a^2... | Let $a b^2 c^4 d^8=t,$ $b c^2 d^4 a^8=z,$ $c d^2 a^4 b^8=y$ and $d a^2 b^4 c^8=x$.
Thus, $$\sum_{cyc}\frac{x^2}{y}=\sum_{cyc}\frac{d^2a^4b^8c^{16}}{cd^2a^4b^8}=\sum_{cyc}c^{15}$$ and we need to prove that
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{t}+\frac{t^2}{x}\geq\sqrt[4]{64(x^4+y^4+z^4+t^4)},$$ which is a known prob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Which one of the following is true?
The image of the half plane $\text {Re} (z) + \text {Im} (z) >0$ under the map $w = \frac {z-1} {z+i}$ is given by
$(\text A)$ $|w| > 1$
$(\text B)$ $|w| < 1$
Let $z = x + iy$ with $x + y >0.$ Then I found that $$|w| = \frac {\sqrt {\left (x^2 + y^2 - x + y \right )^2 + \left (y -... | To continue from where you got up to, you could have tried
$$\begin{align}&\sqrt{\left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}}<x^{2}+\left(y+1\right)^{2}
\\
&\iff \left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}-\left(x^{2}+\left(y+1\right)^{2}\right)^{2}<0
\\
&\iff \left(x^{2}+\left(y+1\right)^{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expectation and variance of the squared distance between $X$ and $Y$ Given that $X$ and $Y$ are two independent univariate random variables sampled uniformly from the unit interval [0,1]. I am trying to find the expected value and the variance of the random variable $Z = \mid X - Y \mid ^2 $ (the squared distance betw... | If I substitute 0 and 1 for a and b respectively to:
\begin{align}
&= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3) \\
&=\frac{1}{6}
\end{align}
Would the answer (the expected value of Z) be $\frac{1}{6}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?
Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}... | Your derivative of the logarithm is wrong. It is not $\frac 2{(1-x)^2}$. It is $\frac{2}{1-x^2}$.
$$\left[\ln\frac{1+x}{1-x}\right]'=\frac{\left[\frac{1+x}{1-x}\right]'}{\frac{1+x}{1-x}}=\frac{\frac 2{(1-x)^2}}{\frac{1+x}{1-x}}=\frac{2}{(1+x)(1-x)}=\frac{2}{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$ $$\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$$\lim_{x\to 0}\frac{xe^x+x-2e^x+2}{x^3}$$
Limit is in the form "$\frac{0}{0}$", using L'hopital once:
$$\lim_{x\to 0}\frac{e^x+xe^x+1-2e^x}{3x^2}=\lim_{x\to 0}\frac{-e^x+xe^x+1}{3x^2}$$
Again the limit is of the form "... | series expansion of $\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $
around $x=0$
$$\lim_{x\rightarrow 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$\displaystyle \lim_{x\rightarrow 0}\frac{x\bigg(\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\bigg)+1\bigg)-2\bigg(1+\frac{x}{1!}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induct... | For each $n = 1, 2, 3, \ldots$, we have
$$
\begin{align}
a_{n+1} &= \frac{3a_n -1 }{3 - a_n } \\
&= \frac{ 3a_n - 9 + 8 }{ 3 - a_n } \\
&= -3 + \frac{8}{ 3 - a_n }.
\end{align}
$$
Thus we obtain
$$ a_2 = - 3 + \frac{8}{3-a_1} = \frac{3a_1 - 1}{3 - a_1}. $$
$$ a_3 = -3 + \frac{ 8 }{ 3 - a_2 } = -3 + \frac{8}{3 - \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Asymptotic expression to $e^{-k^2/2n}$ I want to show that $$\Bigl(1-{k\over n}\Bigr)^{-(1+n-k)/2}\Bigl(1+{k\over n}\Bigr)^{-(1+n+k)/2}$$ is asymptotic to $e^{-k^2/2n}$.
I, in my earlier post, was satisfied with the given answer. But only now I noticed that the answer provided has a flaw since though it claims to produ... | You have
\begin{align}
\Bigl(1-{\tfrac kn}\Bigr)^{-\frac{1+n-k}2}\Bigl(1+{\tfrac kn}\Bigr)^{-\frac{1+n+k}2}
&=
\exp\left[-\tfrac{1+n-k}2\,\log\left(1-\tfrac kn\right)-\tfrac{1+n+k}2\,\log\left(1+\tfrac kn\right)\right]\\ \ \\
&=\exp\left[-\tfrac{1+n-k}2\,\left(-\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)-\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help... | Disclaimer. In all general cases below, you need to choose your (real) constants such that all terms are well-defined.
First note $2x^2-3=\frac12(2x-\sqrt6)(2x+\sqrt6)$ so we can decompose the integrand as $$\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}=\frac52\cdot\left(\frac{1}{\left(2 x-\sqrt{6}\right) \sqrt{3 x^2-2 x+1}}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
2018 MathCounts: Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If $D(m) + D(n) = 125$, what is the value of $m+n$? The question was asked in the 2018 Raytheon MATHCOUNTS National Competition. It appears in the video around the 35:34 mark:
https://www.youtube.com/watch?v=dSnOLW_W6og&t=2134
Let $D(... | We want to solve $n^2-3n+m^2-3m = 250$ for $n,m\ge 3$.
Multiplying by $8$ and symmetrizing the polynomials, we get that the equation is equivalent to solving
$$j^2-9 + k^2-9 = 1000,$$
with $j=2n-3$, $k=2m-3$, or
$$j^2+k^2=1018.$$
Mod $3$ we have $$j^2+k^2\equiv 1\pmod{3},$$ so one of $j$ or $k$ is divisible by $3$. W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have th... | For $x=y=z=\sqrt3$ we'll get a value $\sqrt3-1$.
But, $$\sum_{cyc}\frac{x - 1}{y^2}-\sqrt3+1=\sum_{cyc}\left(\frac{x}{y^2}-\frac{2}{y}+\frac{1}{x}\right)-\sum_{cyc}\left(\frac{1}{x^2}-\frac{1}{xy}\right)+\frac{xy+xz+yz}{xyz}-\sqrt3=$$
$$=\sum_{cyc}\frac{(x-y)^2}{y^2x}-\sum_{cyc}\frac{(x-y)^2}{2x^2y^2}+\frac{xy+xz+yz}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $ I'm trying to solve $$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $$
The value should be $\displaystyle \frac{\... | Suppose $\displaystyle\int\limits_0^x\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt$. It's well known that $\arctan z=\dfrac{\pi}{2}-\text{arccotan} z$, so our integral is $$\dfrac{\pi}{2}\int\limits_0^x\frac{dt}{(t^2+1)\sqrt{t^2+2}}\, dt-\int\limits_0^x\frac{\text{arccotan}\left(\frac{x}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$ I want to prove that for $a,b,c>0$ we have
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=
\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\... | We need to prove that:
$$\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\leq\frac{3}{\sqrt7},$$ where $a$, $b$ and $c$ are positive numbers.
Indeed, by C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\right)^2\leq\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\sum_{cyc}(17a+2b+2c).$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{a}{(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum of $a + b + c$ Let $a, b, c$ be non-negative real numbers such that
$$abc[(a − b)(b − c)(c − a)]^2 = 1$$
Find the minimum of $a + b + c$.
Source: https://cms.math.ca/crux/v43/n5/public_Chow_et_al_43_5.pdf
My Attempt:
$a \mapsto \frac{x+y}{2}$
$b \mapsto \frac{y+z}{2}$
$c \mapsto \frac{x+z}{2}$ $\;$ w... | Remarks: @youthdoo said there is a AM-GM solution. I found it.
Problem: Let $a, b, c\ge 0$ with $abc[(a-b)(b-c)(c-a)]^2 = 1$. Prove that
$$a + b + c \ge \sqrt[9]{\frac{531441}{16}}.$$
Sketch of a proof:
WLOG, assume that $a > b > c$.
Let $0 < x_1 < x_2 < x_3$ be the three real roots of $2x^3 - 54x^2 + 243x - 243 = 0$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $m_a\geq \dfrac{b^2+c^2}{4R}$ Let triangle ABC, $m_a$ the lenght of the median from A, $b,c$ the lenghts of the segments AC and AB respectively and R the circumradius. Prove that:
$m_a\geq \dfrac{b^2+c^2}{4R}$.
I found this in a book and the hint was to denote M the midpoint of BC and $A_2$ the second inters... | As your hint says, from the power of $M$ to $(O)$ we get
$$AM \cdot MA_2 = BM \cdot MC = \frac{1}{4}a^2$$
$$\Longrightarrow AA_2 = AM + MA_2 = \frac {AM^2 + \frac {1}{4}a^2}{AM} = \frac {\frac {1}{2}(b^2 + c^2) - \frac {1}{4}a^2 + \frac {1}{4}a^2}{m_a} = \frac {b^2 + c^2}{2m_a}$$
Since a diameter is greater or equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding constants $a$ and $b$ using derivatives 'Suppose $f(x)= 3(ax-b/x)^3$. Given that $f(3/2)= 3$ and $f'(3/2)= 30$, find $a$ and $b$.'
I've tried chain rule and getting $a$ or $b$ on its own and substituting back into the function, but I feel like I'm overcomplicating it.
Thanks for any help.
| We have $f'(x)=9(a+\frac{b}{x^2})(ax-\frac{b}{x})^3$
$f(\frac{3}{2})=3$ so $3(a.\frac{3}{2}-b\frac{2}{3})^3=3$; $f'(\frac{3}{2})=30$ so $9(a+b\frac{4}{9})(a\frac{3}{2}-b\frac{2}{3})^3$
Implies $9a-4b=6$ and $9a+4b=30$, so $a=2$ and $b=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find expression for following summation
Find expression of:
$$
\sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ???
$$
|
$$X=\sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ?$$
Consider
$$\begin{align}
S:=\sum_{r=1}^n \dbinom{n}{r} e^{ir\theta}
&= \sum_{r=1}^n \dbinom{n}{r} (\underbrace{e^{i\theta}}_{:=a})^r \\
&= \sum_{r=1}^n \dbinom{n}{r} a^r \\
&= -1+\sum_{r=0}^n \dbinom{n}{r} a^r1^{n-r} \\
&= -1+(1+a)^n \\
&= (1+e^{i\theta})^n-1\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If x, y and z are rational and strictly positive and if $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$, show that $\sqrt{x^2+y^2+z^2}$ is rational This is a question from "Pathfinder to Mathematical Olympiads", I know that we need to prove that $x^2+y^2+z^2$ is a perfect square of a rational number. I did prove it but I'm w... | Replace $z$ by $-z$, we get $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ and do not change $x^2 + y^2 + z^2$, then we can calculate that $xy + yz + zx = 0$, so $x^2 + y^2 + z^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) = (x + y + z)^2$, and $\sqrt{x^2 + y^2 + z^2} = x + y + z$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \qua... | By your work $$\theta=\pm40^{\circ}+120^{\circ}k,$$ where $k\in\mathbb Z,$ which gives $$\{\pm40^{\circ},\pm80^{\circ},\pm160^{\circ}\}.$$
We see these roots in your picture.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If I take a line, take the cube, add a constant, over what range is the cube root roughly linear? If a line is linear
$$y= mx+b$$
If I cube the expression
$$y^{3} = (mx +b)^{3}$$
And now add a constant
$$y^{3} = (mx +b)^{3}+C$$
Roughly where is the line still linear if I take the cube root again?
$$y = ((mx +b)^{3}... | We have that
$$\begin{align*}
y &= \sqrt[3]{(mx+b)^3+C} \\
&= \sqrt[3]{(mx+b)^3\cdot\left(1+\dfrac{C}{(mx+b)^3}\right)} \\
&= (mx+b)\cdot\sqrt[3]{1+\dfrac{C}{(mx+b)^3}} \\
\end{align*}$$
As $x\to\infty$, or as $x\to-\infty$, we have that $\dfrac{C}{(mx+b)^3}\to0$ and $\sqrt[3]{1+\dfrac{C}{(mx+b)^3}}\to1$.
So $\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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why is $e^{\frac{\pi i}{3}}$ a root of $f(x) = x^{2}-x+1$? I am not sure how this comes about since, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$
| $x=e^{i\pi/3}=1/2+\sqrt{3}i/2.$
Then $$f(e^{i\pi/3})=[1/2+i\sqrt{3}/2]^2-[1/2+i\sqrt{3}/2]+1=1/4-3/4+i\sqrt{3}/2-1/2-i\sqrt{3}/2+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int\limits_{-\infty}^{\infty}\frac{\sin x}{x}\cdot\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}dx$ I would greatly appreciate it if someone could find a mistake in my solution to the following problem.
Evaluate the integral
$$
I=\int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}... | What’s wrong is that you replaced $\hat g(y)$ in the integrand by $\int_{-\infty}^\infty\hat g(y)\mathrm dy$, and there’s no reason why you should be able to do that. What you can do, though, is to argue that the convolution of two rectangular pulses of widths $\frac13$ and $\frac15$ has width less than $1$, so you can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i ... | We have, for instance,
$$
169 + 49 + 25 + 9 + 4=256
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$
I tried to prove it by ind... | First note that
\begin{align}x_k&\geq 1+\frac1{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{k\cdot(k+1)}\\&=1+\frac{1}{2}-\frac{1}{k+1}=\frac{3}{2}-\frac{1}{k+1}=\frac{3k+1}{2(k+1)}.\end{align}
Therefore,
$$k^2x_k^2\geq \frac{k^2(3k+1)^2}{4(k+1)^2}.$$
For $k\ge 2$,
$$\frac{1}{k^2x_k^2}\leq \frac{4(k+1)^2}{k^2(3k+1)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Limit with or without lhopital I am having trouble seeing what is the limits for the following expression. I tried L'hopital without any major success.
$$
\lim_{x \to 0} (1+\frac{1}{x^\frac{1-a}{a}})^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right)
$$
where $0 < a <1$.
Any views? Thanks
| As Brian Moehring's question comment indicates, it's reasonable to assume that $x \to 0+$ as you will get undefined behavior if you allow $x$ to be negative for fractional powers. Thus, this is what this answer will use.
You have
$$\begin{equation}\begin{aligned}
\lim_{x \to 0^{+}} \left(1+\frac{1}{x^\frac{1-a}{a}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that
$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
find the value of rationals $a,b,c.$
Solution I tried: I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \fra... | $\alpha=1+\sqrt[3]{2}$ is a root of $(x-1)^3=2$, and so
$$
3 = \alpha^3 - 3 \alpha^2 + 3 \alpha = \alpha(\alpha^2 - 3 \alpha + 3)
$$
Therefore,
$$
\alpha^{-1}=\frac13(\alpha^2 - 3 \alpha + 3)
$$
Now use that
$$
\alpha^2 = 1+2\sqrt[3]2+\sqrt[3]4
$$
This solution works in general because every nonzero algebraic number is... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i... | Taking a hint from @ChiefVS, one solution is $\alpha\approx 0.785398-i\, 0.346574$
All solutions are then:
$\{\alpha+2k\pi, -\alpha+2k\pi\}, \quad k\in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $2^n = 10^6$ for $n$ I am trying to solve $2^n = 10^6$ for $n$. Is there some way that I can solve this analytically (say, using base $2$ logarithms), or must I solve it numerically? Thank you.
| Well, sure. $\log_2 2^n = \log_2 10^6$
$n = 6\log_2 10$ or
$\log_{10} 2^n = \log_{10} 10^6$
$n \log_{10} = 6$
$n = \frac {6}{\log_{10}2}$ or
$\log_b 2^n = \log_b 10^6$
$n\log_b 2^n = 6\log_b 10$
$n = 6\frac {\log_b 10}{\log_b 2}$ for any $b > 0; b\ne 1$ are all acceptable answers.
But it's worth noting that
1) $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If the roots of $x^2-7x+1=0$ are $p$ and $q$, find a quadratic equation with roots $p^2+q^2$ and $\sqrt{p}+\sqrt{q}$
Let $p$ and $q$ be the roots of $x^2-7x+1=0$. Find a quadratic equation with roots of $p^2+q^2$ and $\sqrt{p} + \sqrt{q}$.
Attempt to solve:
What I got from the previous quadratic equation
$$p+q=-\frac... | Let $x^2=y$ and $x$ be a root of $$ax^2+bx+c=0$$
$$(ax^2+c)^2=(-bx)^2$$
$$\implies(ay+c)^2=b^2y\iff a^2y^2+y(2ca-b^2)y+c^2=0$$
So, in the first case, $$a=c=1,b=-7$$
For the second, $$\dfrac{a^2}1=\dfrac{b^2-2ca}7=\dfrac{c^2}1$$
$\implies a^2=c^2\iff c=\pm a$
But $\dfrac ca=\sqrt p\sqrt q\ge0$ for real $p,q$
So, $c=a,$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Two projectiles at an initial height with a missing angle of projection
David and Goliath are testing what they learned in school, that the
distance travelled by launching a projectile is dependent on a range
of variables; the launch angle, height and velocity of projection.
As Goliath is significantly taller, Da... | From
\begin{equation*}
\cos \theta = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) },
\end{equation*}
multiplying both numerator and denominator by $v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h}$ gives
\begin{align*}
\cos \theta &= \frac{(\sqrt{3}v + \sqrt{3v^2 + 80h})(v\sin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something ... | Use the relationship,
$$ x^3+y^3+z^3 = (x+y+z)^3 - 3(x+y)(y+z)(z+x)$$
to evaluate,
$$a+b+c= (\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c})^3
-3(\sqrt[3]{a} + \sqrt[3]{b})(\sqrt[3]{b} + \sqrt[3]{c})(\sqrt[3]{c} + \sqrt[3]{a} )$$
$$= 0^3 -3(0- \sqrt[3]{c})(0- \sqrt[3]{a})(0- \sqrt[3]{b} )=3\sqrt[3]{abc}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Differentiating secant inverse I want to show that $\cfrac{d}{dx}[\operatorname{arcsec} x]=\cfrac{1}{\vert x \vert\sqrt{x^2-1}}$. Here is my attempt:
\begin{align*} y=\sec^{-1}x &\iff x=\sec y \\ \frac{d}{dx}[x]&=\frac{d}{dx}[\sec y] \\ 1 &=\frac{\cos^2y}{\sin y} \frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{\sin y}{\cos^2y}... | You have
$$x = \sec y = \frac{1}{\cos y} \implies \cos y = \frac{1}{x} \tag{1}\label{eq1A}$$
Also, you have, assuming that $\sin y \ge 0$, that
$$\sin^2 y + \cos^2 y = 1 \implies \sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|} \tag{2}\label{eq2A}$$
Note you have a mistake in your di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}$ calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals
Let
$$
a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}
$$
My goal is to calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals.
What I have tried so far and got ... | if we have $f(x) > 0$ and $f'(x) > 0,$ then
$$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$
Taking $a=n+1,$ $ b = 2n,$
$$ \int_{n}^{2n} \; \sqrt x \; dx \; < \; \sum_{j=n+1}^{2n} \; \sqrt j \; < \; \int_{n+1}^{2n+1} \; \sqrt x \; dx $$
and a suitable ant... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
My Attempt:
$c$ cannot be $5$ since $70 \times 5 > 343$.
It can't be $4$ either since if we put $c = 4$, we get... | You are correct in saying that $c=1,2,3$ but this is where you then made the mistake. We need to consider each of these cases separately
When $c=3$, we have $30a+50b<133$. Here, $b$ cannot be greater than $3$ as $50\times3=150>133$. So we have $b=1,2$.
When $b=1$, then we have $30a<83$ meaning $a<2.766\ldots$ so $a=1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
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