Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.
| $x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} $ or $\dfrac{-b- \sqrt{b^2-4ac}}{2a}$
If $x=\dfrac{p}{q} \implies b^2-4ac=k^2$, and $k$ is odd. ($odd-even=odd$).
Considering $a,b,c$ odd.
$k^2 \equiv 1 \mod 8$
$b^2 \equiv 1 \mod 8$
$4ac \equiv 4 \mod 8$
$b^2-4ac=-3 \mod 8 \implies 5 \mod 8$, a contradiction.(Either of $a$ or $c$ ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 11,
"answer_id": 7
} |
Stirling's Approximation and Ramanujan's Proof of Bertrand's Postulate Perhaps, the most astounding step in Ramanujan's proof of Betrand's postulate is his application of Stirling's approximation.
He starts with the following inequality:
$\log\Gamma(x) - 2\log\Gamma(\frac{1}{2}x + \frac{1}{2}) \le \log[x]! - 2\log[\fra... | To make the bounds explicit we can start with the earlier form of Stirling's Approximation
$$
\log \Gamma(z) = \left(z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log (2\pi)
+ \sum_{n=1}^{k-1} \frac{B_{2n}}{2n(2n-1)z^{2n-1}}+R_k(z)
$$
If we take $k-1$ terms in the sum then $|R_k|$ is bounded by the $k$th term (see the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/342520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$ I came across the following problem :
Let $a,b,c$ are non-zero real numbers .Then the minimum value of $a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}?$ This is a multiple choice question and the options are $0,6,3^2,6^2.$
I do not ... | Did you try $AM-GM$?
$\dfrac{(a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})}{6} \ge 1$
$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge 6$
OR
$\sum (a_i-\dfrac{1}{a_i})^2 \ge 0$
$\sum (a_i)^2+\dfrac{1}{(a_i)^2} \ge 2 \cdot i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding determinant of a simple matrix Can someone please explain how to compute the determinant of $J_n - I_n$ where $j_n$ it a matrix of ones?
E.g. for $n=5$ we get the following matrix
$$\left(\begin{array}{ccccc}
0 & 1 & 1 &1 &1 \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0 \end{array}... | Let us put
$$
D =\left|\begin{array}{ccccc}
0 & 1 & 1 &1 &1 \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0
\end{array} \right|
$$
Performing the operation $L_1 \gets \sum_{k=1} L_k$ and putting $m=n-1$, we obtain
$$
D =\left|\begin{array}{ccccc}
m & m & m &m &m \\
1 & 0 & 1 &1 &1 \\
1 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Max and Min of $f(x,y)$ Let $f(x,y)=x(y \log y-y)-y \log x$. Find $\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})$.
| Note that $$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)}) \leq \smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}f(x,\frac{1}{2})=\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})$$
Since $-\frac{x\log 2+x+\log x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate rational function $\frac{x^2}{1+x^4}$ Integrate $$\int\frac{x^2dx}{1+x^4}$$
I've factored the denominator to $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and got stuck.
| Or, use $$\frac{x^2}{1+x^4}=\frac{1}{2} \frac{2x^2}{1+x^4}=\frac{1}{2}\left(\frac{x^2-1}{1+x^4}+\frac{x^2+1}{1+x^4}\right)$$ and this idea.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is ..... I came across the following problem that says:
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
1.Always positive
2.Always negative
3.zero
4.Sometimes positive and sometimes negative.
I have to determine which of the aforementioned options is right.
Now ... | I am posting 2 ways of solving this:
$(1)$ If $x,y$ belong to positive real numbers only:
We know that $x^2 + y^2 \geq 2xy$.
Hence we can say that $x^2 + y^2 > xy$
Hence even if $xy$ is negative $x^2 + y^2$,which is positive, is always greater than $xy$ making the sum $x^2 + y^2 + xy$ always positive.
$(2)$ If $x,y$ be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 2
} |
Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
| \begin{align}\sum_{k=0}^n(k+1) \binom{n}{k}&=\sum_{k=0}^nk \binom{n}{k}+\sum_{k=0}^n \binom{n}{k}=n\sum_{k=1}^n \binom{n-1}{k-1}+2^n\\&=n\sum_{k=0}^{n-1} \binom{n-1}{k}+2^n=n2^{n-1}+2^n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
arithmetic progression of triangle sides Let $gcd(a,b,c)=1$ such that $a^2, b^2, c^2$ are in arithmetic progression. Show they can be written in the form
$a=-p^2+2pq+q^2$
$b=p^2+q^2$
$c=p^2+2pq-q^2$
for relatively prime integers $p,q$ of different parities.
so if $a^2, b^2, c^2$ are in arithmetic progression, then
$a... | Since $a^2, b^2, c^2$ are in arithmetic progression, $a^2+c^2=2b^2$.
If $a$ is even, then so is $c$, so $4 \mid a^2+c^2=2b^2$, so $b$ is also even, giving a contradiction.
Thus $a$ is odd. Similarly $c$ is odd, so $b$ is also odd.
$$(a-b)(a+b)=a^2-b^2=b^2-c^2=(b-c)(b+c)$$
$$\frac{a-b}{2}\frac{a+b}{2}=\frac{b-c}{2}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A question in Complex Analysis $\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$ My problem is to integrate this expression:
$$\int_0^{2\pi}\log(1-2r\cos x +r^2)dx.$$
where $r$ is any constant in $[0,1]$.
I know the answer is zero. Can you explain you idea to me or just prove that?
Maybe you will use the "Cauchy integral theor... | Copying from my blog.
Let $$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ Some preliminary results on $I(a)$. Note that we have $$I(a) = \underbrace{\displaystyle \int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Linear Programming: Modifying Coefficients of the Objective Function Consider a final tableau with entries:
\begin{array} {|c|c|}\hline BV & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & RHS \\ \hline x_3 & 0 & -\frac{1}{2} & 1 & 1 & 2 & 0 & -1 & 4 \\ \ x_1 & 1 & \frac{1}{2} & 0 & 2 & -1 & 0 & -2 & 2 \\ x_6 & 0 & 2 & 0 & ... | After having calculating all the missing $C^\pi_{NBV}$ of the given tableau, we should have following optimal tableau:
\begin{array} {|c|c|}\hline BV & z & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & RHS \\ \hline z & 1 & 0 & \frac{5}{2} & 0 & 0 & \frac{1}{2} & 0 & 3 & 13 \\ \hline x_3 & 0 & 0 & -\frac{1}{2} & 1 & 1 & 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra mapping question Does there exist a matrix $A$ such that
$$
A\left( \begin{array}{ccc}
1 \\
0 \\
0 \\
0 \\
\end{array} \right) = \left( \begin{array}{ccc}
1 \\
0 \\
0 \\
0 \\
\end{array} \right)$$
$$
A\left( \begin{array}{ccc}
0 \\
1 \\
0 \\
0 \\
\end{array} \right) = \left( \begin{array}{ccc}
0 \\
1... | If $A$ exists, $A$ satisfies
$$A\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}=
\begin{pmatrix}
1&0&0&0\\
0&1&1&0\\
0&0&0&1\\
0&0&0&0\end{pmatrix}$$
So
$$A=
\begin{pmatrix}
1&0&0&0\\
0&1&1&0\\
0&0&0&1\\
0&0&0&0
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding all $\alpha$ such that a matrix is positive definite I have
$A = $
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
and I want to find all $\alpha$ such that $A$ is positive definite.
I tried
$ x^tAx = $
$
\left[\begin{array}{r}
x & y & z
\... | The matrix is positive definite if and only if their eigenvalues are positive, so we calculate the characteristic polynomial $\chi_A(x)=\det(xI-A)$ and we solve for $x$ and we find:
$$\lambda_1=\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}-\frac{1}{2}\alpha+3,\quad\lambda_2=3-\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}- \frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
An integral inequality related to Taylor expansion Problem. Let $f:[a,b]\to\mathbb{R}$ be a function such that $ f\in C^3([a,b])$ and $f(a)=f(b)$. Prove that $$ \left|\int\limits_{a}^{\frac{a+b}{2}}f(x)dx-\int\limits_{\frac{a+b}{2}}^{b}f(x)dx\right|\leq\frac{(b-a)^4}{192}\max_{x\in [a,b]}|f'''(x)|.$$
Any idea are welco... | Idea : Making use of successive Integration by parts ,
$\int P(x)f^{(3)}(x)\,dx = P(x)f^{(2)}(x)-P^{(1)}(x)f^{(1)}(x)+P^{(2)}(x)f(x)-\int P^{(3)}(x)f(x)\,dx$
Consider the two third degree monic polynomials, $P_1(x)$ and $P_2(x)$.
Now, we compute the difference of the definite integrals:
$\int\limits_{\frac{a+b}{2}}^b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving $x^2+19=y^5$ I was given several exercises and there is a particular one, I am not able to solve.
Let it be given that $Pic(\mathbb{Z}[\sqrt{−19}])$ is a finite group of order $3$. Use this to find all integral solutions of the equation $x^2 + 19 = (x + \sqrt{−19})(x-\sqrt{−19})= y^5$.
I have absolutely no ide... | Let $x$ and $y$ be integers which satisfy $x^2 + 19 = y^5$. Then $(x + \sqrt{-19})(x - \sqrt{-19}) = (y)^5$.
If you can find that $(x + \sqrt{-19})$ and $(x - \sqrt{-19})$ are coprime ideals, (if $19 \mid x$ would hold, then $19 \mid y$ and $19 = y^5 - x^2$ where the RHS is divisible by $19^2$ - contradiction) then $(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
If $z$ is a complex number of unit modulus and argument theta If $z$ is a complex number such that $|z|=1$ and $\text{arg} z=\theta$, then what is $$\text{arg}\frac{1 + z}{1+ \overline{z}}?$$
| $z$ can be written as $\cos\theta+i\sin\theta\implies \bar z=\cos\theta-i\sin\theta$
So, $$\frac{1+z}{1+\bar z}=\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}$$
$$=\frac{2\cos^2\frac \theta2+2i\cos\frac\theta2\sin\frac\theta2}{2\cos^2\frac \theta2-2i\cos\frac\theta2\sin\frac\theta2}$$
$$=\frac{\cos\frac\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end... | There is a mechnical procedure, as follows.
Any polynomial function of $r = \sqrt 2 + \sqrt 3$ must have the form $a + b\sqrt 2 + c\sqrt 3 + d\sqrt 6$ for rational $a,b,c,d$. Consider the set of numbers of that form as a vector space $V$ over the rationals. It has dimension 4.
Now calculate $r^0, r^1, r^2, r^3, r^4$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Is there a formula for $\sum_{n=1}^{k} \frac1{n^3}$? I am searching for the value of $$\sum_{n=k+1}^{\infty} \frac1{n^3} \stackrel{?}{=} \sum_{n = 1}^{\infty} \frac1{n^3} - \sum_{n=1}^{k} \frac1{n^3} = \zeta(3) - \sum_{n=1}^{k} \frac1{n^3}$$
For which I think I need to know the value of $$\sum_{n=1}^{k} \frac1{n^3}$$
D... | We have
$$S_n = \sum_{k=n+1}^{\infty} \dfrac1{k^3} = \int_{n^+}^{\infty} \dfrac{d \lfloor t \rfloor}{t^3} = \left. \dfrac{\lfloor t \rfloor}{t^3} \right \vert_{t=n^+}^{\infty} + 3 \int_{n^+}^{\infty} \dfrac{\lfloor t\rfloor dt}{t^4} = -\dfrac1{n^2}+3 \int_{n^{+}}^{\infty} \dfrac{t-\{t\}}{t^4} dt$$
Hence, we get that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 2
} |
Show that $ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$ Verify the following identity:
$$ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$$
| Method $1:$
As
$\displaystyle\tan^2x=\sec^2x-1=(\sec x-1)(\sec x+1),\frac{\tan x}{\sec x-1}=\frac{\sec x+1}{\tan x}$
$$\implies\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}=\frac{\sec x-1}{\tan x}+\frac{1+\sec x}{\tan x}=\frac{2\sec x}{\tan x}=\frac{\dfrac2{\cos x}}{\dfrac {\sin x}{\cos x}}=\frac2{\sin x}$$
Method ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then w... | By Riemann sum we have
$$\frac{1}{n^3}\sum_{k=1}^nk(k+1)=\frac{1}{n}\sum_{k=1}^n\frac{k}{n}(\frac{k+1}{n})\sim_\infty \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^2\to\int_0^1x^2dx=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Evaluating the integral $ \int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \ dx$ I'm curious about the proper way to evaluate $$ \int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \, dx = \text{Re} \int_{-\infty}^{\infty} \frac{e^{i(x- \frac{1}{x})}}{1+x^{2}} \, dx$$ using... | Let $x-1/x = t$. Now note that $t^2 = x^2 + \dfrac1{x^2} -2 \implies \left(x+\dfrac1x \right)^2 = t^2+4$. We then get that
$$\left(1 + \dfrac1{x^2} \right)dx = dt \implies \dfrac{dx}{1+x^2} = \dfrac{x^2}{(1+x^2)^2} dt = \dfrac{dt}{\left(x+1/x \right)^2} = \dfrac{dt}{t^2+4}$$
Hence, the integral becomes
$$\int_{-\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Determine all $2\times2$ matrices $A$ such that $A^2=0$. This is from Lang's introduction to Linear Algebra page no 61.
Determine all $2\times 2$ matrices $A$ such that $A^2 = 0$.
Let $A=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$
$A^2=\begin{pmatrix}
a^2+bc & ab+bd \\
ac+cd & d^2+cb
\end{pmatrix}... | Well you have your 4 equations:
$$\cases{a^2+bc=0 \\ ab+bd=0 \\ ac+cd=0 \\ d^2+bc=0}$$
And you know that if $A^2=0$ then $det(A^2)=det(A)^2=0$ and thus $det(A)=0$. So you have a fifth equation:
$$ad-bc=0 \Rightarrow ad=bc$$
Now replace this in your system and factorize it:
$$\cases{a^2+ad=a(a+d)=0 \\ ab+bd=b(a+d)=0 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
| HINT
$$\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}} = \dfrac{x^2 \sqrt{1+1/x^4}}{x^2 \sqrt[3]{1+1/x^6}} = \dfrac{\sqrt{1+1/x^4}}{\sqrt[3]{1+1/x^6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Derivation of the general forms of partial fractions I'm learning about partial fractions, and I've been told of 3 types or "forms" that they can take
(1) If the denominator of the fraction has linear factors:
$${5 \over {(x - 2)(x + 3)}} \equiv {A \over {x - 2}} + {B \over {x + 3}}$$
(2) If the denominator of the frac... | It comes from two facts for polynomials $f(x)$ and $g(x)$:
*
*We can find $a(x)$ and $b(x)$ such that $\gcd(f(x),g(x))=a(x)f(x)+b(x)g(x)$.
*We can write $f(x)=q(x)g(x)+r(x)$, where $\deg(r)<\deg(g)$.
For your second example, we can use fact 1 to write $1=a(x)(x-1)+b(x)(x^2+4)$. Multiplying by $2x+3$ yields:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 2
} |
Proving the inequality $\ln(\cos x)\ge \frac{-x^2}{\cos^2(x)}$ for $x\in [0,\frac{\pi}{2}]$ I want to prove the following inequality: $\ln(\cos x)\ge \dfrac{-x^2}{\cos^2(x)}$
Please I'm stuck with this problem, maybe considering the equivalent inequality $\ln(\sec x)\ge (x \sec x)^2$ would help, but I'm not sure.
I wan... | $$
\ln (1+x) = x - \frac{x^2}2 + \frac{x^3}3 - \cdots.
$$
$$
\ln\cos x = \ln\left(1 -\frac{x^2}{2} + \frac{x^4}{24}-\cdots\right) \ge\ln\left(1 -\frac{x^2}{2}\right)
$$
$$
=\frac{-x^2}{2} -\frac{x^4}{8} + \cdots \ge -\frac{x^2}{1.5} \ge-\frac{x^2}{\cos^2 x}\text{ for $x$ close enough to $0$.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find $\gcd(a^{2^m}+1,a^{2^n}+1)$ when $m \neq n$? How to prove the following equality?
For $m\neq n$,
$\gcd(a^{2^m}+1,a^{2^n}+1) = 1 $ if $a$ is an even number
$\gcd(a^{2^m}+1,a^{2^n}+1) = 2 $ if $a$ is an odd number
Thanks in advance.
| Let $p$ divides $a^{2^n}+1\implies a^{2^n}\equiv-1\pmod p$
If $m>n,$ taking $2^{m-n}$th power, $(a^{2^n})^{2^{m-n}}\equiv(-1)^{2^{m-n}}\pmod p\implies a^{2^m}\equiv1\pmod p$
So, $p$ divides $ a^{2^m}-1$
If $p$ divides $ a^{2^m}+1,p$ will divide $(a^{2^m}+1)-(a^{2^m}-1)=2$
So, if $a$ is even, $a^{2^n}+1$ will be odd $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the points of discontinuity: $f(x) = (x^4+x^3+2x^2)/\tan^{-1}(x)$ if $x\ne0$ and $f(0)=10$ Here is the question:
Find the points of discontinuity:
$$f(x) =
\begin{cases}
\frac{x^4+x^3+2x^2}{\tan^{-1}x} & \text{if} \ x\ne0 \\ \\
10 & \text{if} \ x=0 \\
\end{cases}.$$
$\frac{x^4+x^3+2x^2}{\tan^{-1}x}$ being th... | Applying L'Hospital:
$$\lim_{x\to 0}\frac{x^4+x^3+2x^2}{\tan^{-1}x}=\lim_{x\to 0}\frac{4x^3+3x^2+4x}{\frac{1}{1+x^2}}=0\ne 10$$
Therefore, $f$ is not continuous at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| By Cauchy-Schwarz
$$(a\sqrt{1-b^2}+b\sqrt{1-a^2})^2 \leq (a^2+b^2)(1-b^2+1-a^2)=2(a^2+b^2)-(a^2+b^2)^2$$
$$=1-(1-a^2-b^2)^2\leq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to find the solutions $x$ of $ 2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$ Let
$$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$$
where $90^{\circ}<x<180^{\circ}$.
My idea: $$2\sin{11^{\circ}}\sin{71^{\circ}}\... | $$2\sin 11^\circ*\sin 71^\circ*\sin (x+30)^\circ=\sin 2013^\circ*\sin 210^\circ$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=\sin 213^\circ*-\sin 30^\circ$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*(\sin 213^\circ-\sin 71^\circ+\sin 71^\circ)$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to compute $\lim\limits_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$ without using L'hopital? I can compute
$$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$$ using L'hopital and the limit equals $\frac{\sqrt{3}}{12}$, but is there another way to compute this limit without using L'hopi... | Let's replace $x$ by $x+\pi/3$ and let $x \to 0$.
I will be as simple-minded as I can.
We use $\sin(\pi/3) = \sqrt{3}/2$ and $\cos(\pi/3) = 1/2$.
$\cos(x+\pi/3) = \cos(x) \cos(\pi/3) - \sin(x)\sin(\pi/3)
= \cos(x)/2 - \sin(x)\sqrt{3}/2
$.
As $x \to 0$,
$\cos(x+\pi/3) \to 1/2$.
$\sin(3(x+\pi/3))
= \sin(3x+3\pi/3)
= \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
finding Moment generating function and CDF with pmf The random variable X has the pmf f(-1)=1/4, f(0)=1/8, f(1)=1/4, f(2)=3/8
a) How would you draw the c.d.f with points (-2,F(-2)), (-1,F(-1)), (0,F(0)), (1,F(1)), (2, F(2)), (3,F(3))
b)Write the MGF of this distribution
c) Range of and p.m.f of X^(2)
For the cdf i tri... | (1) I do not think you are expected to give single "equation."
We have $F(-2)=\Pr(X\le -2)=0$.
Similarly, $F(-1)=\Pr(X\le -1)=\frac{1}{4}$.
Also, $F(0)=\Pr(X\le 0)=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$.
Also, $F(1)=\Pr(X\le 1)=\frac{5}{8}$.
Also, $F(2)=1$. And $F(3)=1$.
To understand what we did, remember we are calcu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/378922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This impl... | As in general we have
$$
\mathbf{A}^2 - \big( \lambda_1 + \lambda_2 \big) \mathbf{A}
+ \lambda_1 \lambda_2 \mathbf{I} = 0, \tag 1
$$
we see that in this case
$$
\lambda_1 = 2, \quad \lambda_2 = 3. \tag 2
$$
So the general solution is given by
$$
\bbox[16px,border:2px solid #800000]
{ \mathbf{B} \pmatrix{ 3 & 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Is $B = A^2 + A - 6I$ invertible when $A^2 + 2A = 3I$? Given:
$$A \in M_{nxn} (\mathbb C), \; A \neq \lambda I, \; A^2 + 2A = 3I$$
Now we define:
$$B = A^2 + A - 6I$$
The question:
Is $B$ inversable?
Now, what I did is this:
$A^2 + 2A = 3I \rightarrow \lambda^2v + 2\lambda v = 3v \rightarrow \lambda_1 = 1, \lambda_2... | Note that
$$
\begin{align}
(A-I)B
&=(A-I)(A^2+A-6I)\\
&=(A-2I)(A^2+2A-3I)\\[4pt]
&=0
\end{align}
$$
Thus, unless $A=I$, and therefore, $B=-4I$, $(A-I)B=0$ implies that $B$ is not invertible.
Clarification: Suppose that $B^{-1}$ exists, then
$$
\begin{align}
A-I
&=(A-I)BB^{-1}\\
&=0B^{-1}\\[6pt]
&=0
\end{align}
$$
Thus,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-... | $$\text{So, }(2^x)^2- 2^x\cdot2\cos\left(\frac{x^2+x}6\right)+1=0$$ which is a quadratic equation in $2^x$
As $x$ is real, so would be $2^x$
so,the discriminant must be $\ge 0$
i.e., $$\{2 \cos\left(\frac{x^2+x}6\right)\} ^2-4\cdot1\cdot1\ge0$$
$$\implies4\left(\cos^2\left(\frac{x^2+x}6\right)-1\right)\ge0$$
$$\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})... | Hint:
$$
\left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right)=\frac{9-17}4=-2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividi... | $$
\begin{array}{rcl}
6(8^x + 27^x) &=& 7(12^x + 18^x) \\
6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\
\end{array}
$$
Substitute $a\!=\!2^x$ and $b\!=\!3^x$ for simplicity:
$$
\begin{array}{rcl}
6(a^3 + b^3) &=& 7(a^2b + ab^2) \\
6(a+b)(a^2 - ab + b^2) &=& 7ab(a+b) \\
6(a^2 - ab + b^2) &=& 7ab \\
6a^2 -13ab + 6b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 3,
"answer_id": 0
} |
Evaluting the limit $\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$ I'm attempting to evaluate the limit
$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$
I got it reduced to the following
$\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{1}{\left(x-2\right)^{2}}-\frac{3}{\left(x-2\right)^{4}}}+1}{1-\frac... | \begin{align*}
&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}\\
=&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{x-2\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{z\rightarrow\infty}\frac{1}{\sqrt{z^2-3}-z}\\
=&\lim_{z\rightarrow\infty}\frac{\frac{1}{z}}{\sqrt{1-\frac{3}{z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (s... | Hint of 1. Substitute $x=\frac{3}{2}y$
But I have no idea about 2...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
Compute the remainder when $67!$ is divided by $71$. This is how far I've been able to get.
By using Wilson's Theorem:
$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$
EDIT: Here is how I proceede... | Hint: $$69 \equiv -2 \pmod{71}, \qquad 68 \equiv -3\pmod{71}.$$
Further hint: $72 = 6 \cdot 12$, so the inverse you need to compute should not be too difficult...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $ Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$
Thanks!
| $$\dfrac{x^4}{x^2+1}=\dfrac{(x^4+2x^2+1)-2x^2-1}{x^2+1}=\\
=\dfrac{(x^2+1)^2-2x^2-2+1}{x^2+1}=x^2+1-2+\dfrac{1}{x^2+1}=x^2-1+\dfrac{1}{x^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Init... | $$ \begin{align*} \int \frac{\cos^2 x}{\left(1 + \sin^2 x\right)^2} \ dx &= \int \frac{\sec^2 x}{\left(\sec^2 x + \tan^2 x\right)^2} \ dx \\ &= \int \frac{1}{\left(2\tan^2x + 1\right)^2}\ d(\tan x) \\ \left(\tan x \mapsto \frac{\tan u}{\sqrt2} \right) &= \frac{1}{\sqrt{2}}\int \frac{\sec^2 u}{\sec^4 u} \ du \\&= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First,... | $$x^4 + y^4 \ge \frac{(x^2+y^2)^2}{2} \\
2x^4+2y^4\ge x^4+2x^2y^2+y^4 \\
x^4-2x^2y^2+y^4 \ge0 \\ (x^2-y^2)^2\ge0$$
Notice that $x^2-y^2\in R$, so all real number with an even exponent are always greater or iqual to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Relation between chords length and radius of circle Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show... | Let $P$ be the point where the two chords (and a diameter) meet. Let $h$ (and $k$) be the distance from $P$ to the midpoint of the $2a$ chord (respectively, the $2b$ chord); that is, say $P$ divides the chord into sub-segments of length $a+h$ and $a-h$ (respectively, $b+k$ and $b-k$). Note that $P$ divides a diameter i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$ How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?
| When $2n+1$ is prime, $4n+2$ will only have $2$ and $2n+1$ as its prime factor.
From the question, $4n+3$ is also a prime.
As $4n+1,\,4n+2,\,4n+3$ are three consecutive numbers, one of them must be divisible by $3$ and we know that $4n+2$ and $4n+3$ are not, thus we have the conclusion that $4n+1$ is divisible by $3$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Is it possible to use this fact to prove something like:
$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}... | Well, assuming that $a,b,c\neq 0$, we can see that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&=\frac{1}{a+b+c}
\\ \\
\frac{ab+bc+ac}{abc}&=\frac{1}{a+b+c}
\\ \\
a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2&=abc
\\ \\
(ab+b^2+ac+bc)c+(ab+b^2+ac+bc)a&=0
\\ \\
(ab+b^2+ac+bc)(c+a)&=0
\\ \\
(a+b)(b+c)(c+a)&=0
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $y$ value of canonical ellipse. I have an ellipse:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;
Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?
Hope someone can help me - thanks in advance :... | Given:
$\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$
We can subtract $\dfrac{x^2}{a^2}$ from both sides:
$\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}$
Next multiplying both sides by $b^2$:
$\left(\dfrac{y^2}{b^2}\right)b^2=\left(1-\dfrac{x^2}{a^2}\right)b^2$
Which becomes:
$\dfrac{b^2y^2}{b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$ Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?
I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.
| Hint: Use $(x)(x+1)(x+2)(x+3)+1 = (x)(x+3)(x+2)(x+1)+1 =(x^2+3x)(x^2+3x+2)+1= (x^2+3x)^2+2(x^2+3x)+1=(x^2+3x+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ... | I came across very similar problems while doing Maths Olympiad in high school and had only one approach to it then.
Observe the property for $a,b,c \in \Bbb K$ $$ \text {if } a+b+c=0,\ \text{ then }\ a^3+b^3+c^3=3abc \quad (\ast)$$
This can easily be proved by expanding $(a+b+c)^3$
$$\begin{array}{rll}(a+b+c)^3&=&a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
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Solution to second order nonlinear ODE I need to find and exact solution for the following ODEs $$y''=-3y'+2y+2x+3,\qquad y(0)=2$$ $$y(1)=-4+5\exp\left(-3/2+\left(\sqrt{17}\right)/2\right)$$ and $$y''=2y^3-6y-2x^3;$$ $$1\leq x\leq2;$$ $$y(1)=2;$$ $$y(2)=5/2$$
| First equation $y''+3y'-2y=2x+3$ has a solution which is a sum of solution to the homogeneous ODE $y''+3y'-2y=0$ and a particular solution of inhomogeneous problem. By solving the characteristic equation $a^2+3a-2=0$, we can easily check that roots are $\frac{-3\pm\sqrt{17}}{2}$. Therefore, solution to this problem is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$
| In order to solve this, it would be best to convert the $\sec x$ into $\frac{1}{\cos x}$. and convert $\tan x$ to $\frac{\sin x}{\cos x}$ So we will have
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$$
Now we have to get rid of the fraction in the denominator. So multiply by $\cos x$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$ How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?
It is part of a question about finding limits at mooculus.
| $$\begin{array}{r|rrrr|r}
& 1 & 0 & 0 & 0 & -1296 \\
-6 & & -6 & 36 & -216 & 1296 \\
\hline
& 1 & -6 & 36 & -216 & 0
\end{array}
$$
This shows that
$$
x^4-1296=(x+6)(x^3-6x^2+36x-216)
$$
Of course you can also use
$$
a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)
$$
with $a=x$ and $b=-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Probability of no matching pairs of shoes There are 3 different pairs ( i.e. 6 units say aa, bb, cc) of shoes in a lot. Now three person come and pick the shoes randomly (each get 2 units). Let p be the probability that no one is able to wear shoes (i.e. no one gets a correct pair) , then the value of $\frac{13p}{4-p}$... | We use a technique that is definitely overkill for this problem, the Principle of Inclusion/Exclusion. But the idea might be useful for a larger problem, say $5$ people and $5$ pairs of shoes.
Let our people be P, Q, and R. There are $\binom{6}{2}$ ways for P to pick two shoes. And there are $\binom{3}{1}$ ways to pic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Algebraic expression in its most simplified form I am trying to simplify the algebraic expression:
$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$
I am having trouble though. My current thoughts are:
$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$... | Comment: After the calculation very well described by AWertheim and amWhy, we end up wit $\dfrac{x^2-3x-4}{x^2+3x+2}$.
Note that the calculation is not over. The top and bottom each factor nicely, and there will be some cancellation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to solve the irrational system of equations? Solve the system of equations $$\begin{cases} \sqrt{x+2y+3}+\sqrt{9 x+10y+11}=10,&\\[10pt] \sqrt{12 x+13y+14}+\sqrt{28 x+29y+30}=20. \end{cases} $$
| Hint: Try $t=x+y+1$ to get
\begin{cases} \sqrt{t+y+2}+\sqrt{9t +y+2}=10,&\\[10pt] \sqrt{12t+y+2}+\sqrt{28t+y+2}=20. \end{cases}
And solve by squaring.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determine the general solution for $2\cos 2x−5\cos x+2=0$ Determine the general solution for $2\cos 2x−5\cos x +2=0$
my answer I got was : $1.05+n\pi, 4.19+n\pi$
| $$\implies 2(2\cos^2x-1)-5\cos x+2=0\iff \cos x(4\cos x-5) =0$$
If $\cos x=0,x =\frac{(2m+1)\pi}2$ where $m$ is any integer
If $4\cos x-5=0, \cos x=\frac54>1$ but for real $x, -1\le \cos x\le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help finding eigenvectors? The given matrix is:
$$
\begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad
$$
I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$
which factors down to $$(x+1)^2 * (x-3) = 0$$
I see that it has eigenvalues of -1 and 3.
I know I'm almost there, I plugged in ... | I am going to use the approach you are using so you can see your issues.
We are given: $A = \begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}$
We set up and and solve: $|A - \lambda I| = 0$, which yields:
$$\left|\begin{matrix}3-\lambda & 1 & 6 \\ 2 & 1-\lambda & 0 \\ -1 & 0 & -3-\lambda\end{matrix}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Surface integration over section of paraboloid below a plane Let S be the finite portion of the surface $z = 4x^2 + y^2$ cut off by the plane
$z = 8x + 4y - 4$.
Evaluate the surface integral $(x, y, 3z)\,dS$ over the region $S$ where the normal to $S$ points upwards.
I can do this using the divergence theorem, but I d... | The plane equation is not right. The points satisfying parametrization you gave live on the paraboloid $z = 4x^2 +y^2$, they only intersect the plane at some points.
Well if you wanna use divergence theorem:
The integration region is as follows:
The bottom is the surface $z=4x^2+y^2$, while it has a cap formed by the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality Of Four Variables Suppose $a,b,c,d$ are real numbers greater than $1$. Prove that
$$abc+abd+acd+bcd-3abcd<1.$$
| The inequality holds because
\begin{align*}
&4\times\left[\,3abcd-(abc+abd+acd+bcd)+1\,\right]\\
\\
=&2(a-1)(b-1)(c-1)(d-1)\\
\\
&+(a+1)(b+1)(c-1)(d-1)+(a+1)(b-1)(c+1)(d-1)\\
&+(a+1)(b-1)(c-1)(d+1)+(a-1)(b+1)(c+1)(d-1)\\
&+(a-1)(b+1)(c-1)(d+1)+(a-1)(b-1)(c+1)(d+1)\\
\\
&+(a+1)(b-1)(c-1)(d-1)\\
&+(a-1)(b+1)(c-1)(d-1)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Solve the following in non-negative integers: $3^x-y^3=1$. Solve the following in non-negative integers: $$3^x-y^3=1$$
Of course $(x,y)=(0,0)$ is a trivial solution. After seeing that I proceeded like this:
$$3^x-y^3=1$$$$\implies3^x-1=y^3$$$$\implies2(3^{x-1}+3^{x-2}+ \cdots +3^1+1)=y^3$$$$\therefore2|y$$
So let $y=2k... | If $y+1=3^a$ with $a\ge1\iff y>1$
$y^2-y+1=3(3^{2a-1}-3^a+1)$ which can not be power of $3$ unless $2a-1=a\iff a=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
The final answer is supposed to be:
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I walked through every possible expansion; I com... | A good idea for this sort of thing is to use Wolfram Alpha to ensure that the two things are, indeed, equal. In this case, they are, so we can spend some time looking to factor.
$$\begin{align}
\frac{k(k+1)(2k+1)}{6} + (k + 1)^2 &= \frac{k(k+1)(2k+1) + 6(k + 1)^2}{6}\\
&= \frac{(k+1)(k(2k+1) + 6(k + 1))}{6}\\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac... | The 2nd evaluation can be obtained from the residue theorem:
*
*Using parity, write the integral as $\displaystyle \frac12\int_{-\infty}^{\infty}\frac{\ln(1+x^4)\,dx}{(1+x^2)^2}$.
*Interpret this as a complex integral and pull the integration contour to, say, $i\infty$. The result will be given by the residue at 2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 3
} |
The number of distinct real roots of the folllowing determinant
The number of distinct real roots of determinant $$
\begin{vmatrix}
\csc x & \sec x & \sec x \\
\sec x & \csc x & \sec x \\
\sec x & \sec x & \csc x \\
\end{vmatrix}
=0$$ lies in the interval $\frac{-\pi}{4} \le x \... | We need to solve $\cos(x) + 2\sin(x) = 0$ and $\cos(x) - \sin(x) = 0$. The first equation is solved by setting $u = \cos(x)$ and solving for $u$, i.e.
\begin{align*}
u+2\sqrt{1-u^2} &= 0\\
u &= -2\sqrt{1-u^2} \\
u^2 &= 4(1-u^2) \\
5u^2 &= 4
\end{align*}
and checking which values of $\cos^{-1}(\pm 2/\sqrt{5})$ lie in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't h... | $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
(take L.C.M. in numerator and denominator)
$$\dfrac{\dfrac{{(1-2x)}-{(1-x)}}{1-2x}}{\dfrac{(1-2x)-2{(1-x)}}{1-2x}}$$
$$\dfrac{\dfrac{1-2x-1+x}{1-2x}}{\dfrac{1-2x-2+2x}{1-2x}}$$
$$\dfrac{\dfrac{-x}{1-2x}}{\dfrac{-1}{1-2x}}$$
since $$\dfrac{\dfrac ab}{\dfrac cd}=\dfrac a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seem... | A. $x_n>0$ , $n\in \mathbb{N}$.
B.
We will show that $\{x_2, x_3, \ldots, x_n, \ldots \}$ is monotonically non-increasing.
$$
x_n-x_{n+1} = x_n - \frac{x_n}{2} - \frac{1}{x_n} = \frac{x_n}{2} - \frac{1}{x_n} =
\frac{x^2_n-2}{2x_n}, \qquad n>1.
$$
But $x^2_n-2>0$, when $n>1$, because
$$
x^2_n-2 = \left( \dfrac{x_{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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derive the Maclaurin series by using partial fractions. derive the Maclaurin series for the function
$(x^3+x^2+2x-2)/(x^2-1)$ by using partial fractions and a known Maclaurin series.
question.
how can I use partial fractions in this case?
Is this case the special one?
plz provide me a hint. thanks.
| Using polynomial long division:
$$\frac{(x^3+x^2+2x-2)}{(x^2-1)}=x + 1 + \frac{3x-1}{x^2-1}$$
For partial fraction decomposition, think of it as the inverse of finding a common denominator for two fractions: first factor the denominator:
$$x + 1 + \frac{3x-1}{x^2-1} = x + 1 + \frac{3x-1}{(x - 1)(x+ 1)}$$
Then we split... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Upper bound for expression involving logarithms Let $N = 2^p$ for some $p \in \mathbb{N}$. Find the smallest upper bound for $\frac{N}{2}\log\left(\frac{N}{2}\right) + \frac{N}{4}\log\left(\frac{N}{4}\right) + \ldots + 1$
I guess I could first rewrite this to $\frac{2^p}{2}\log\left(\frac{2^p}{2}\right) + \frac{2^p}{4}... | Fix $k = \log 2 = 1$. Then expression yields
$$\begin {eqnarray}
S & = & \frac {N} {2} (\log N - k) + \frac {N} {4} (\log N - 2k) + \cdots + \frac {N} {2^{p - 1}} (\log N - (p - 1)k) + 1 \nonumber \\
& = & N \log N (\frac {1} {2} + \frac {1} {4} + \cdots + \frac {1} {2^{p - 1}}) - N (\frac {1} {2} + \frac {2} {4} + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the values of the constants in the following identity $2x^3+3x^2-14x-5=(ax+b)(x+3)(x+1)+C$ I'm working through identities but I can't figure out how to get further than multiplying out the above to get :
$$2x^3+3x^2-14x-5=2ax^3+3ax^2+3ax+bx^2+3bx+bx+3b+C$$
can someone give me a hint on what to do next?
| By expanding and grouping like terms together, we obtain:
$$ \begin{align*}
2x^3+3x^2-14x-5&=(ax+b)(x+3)(x+1) +C\\
2x^3+3x^2-14x-5&=(ax+b)(x^2+4x+3) +C\\
2x^3+3x^2-14x-5&=(ax)(x^2+4x+3)+b(x^2+4x+3) +C\\
2x^3+3x^2-14x-5&=(ax^3+4ax^2+3ax)+(bx^2+4bx+3b) +C\\
2x^3+3x^2-14x-5&=(a)x^3+(4a+b)x^2+(3a+4b)x+(3b+C)
\end{align*} $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
Find $x,y,z \in \mathbb Q$ such that: $$x + \frac 1y, y + \frac 1z, z+
\frac 1x \in \mathbb Z$$
Here is my thinking:
$$x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z\\ \implies \left ( x + \frac 1y\right ) \left ( y +... | Let's start with your observation, that $|xyz| = 1$, and therefore $xyz = \pm 1$. If we take any solution $(x, y, z)$ with $xyz = -1$ and change it to $(-x, -y, -z)$, it is still a solution and we have $(-x)(-y)(-z) = 1$, so it's enough to consider the solutions with $xyz = 1$ and obtain the rest by flipping all the si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
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How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \fr... | Answering only "How can a sum of only positive numbers equal 0?" : look at $1^2 + 2^2 + 3^2 + 4^2 + 5^2+ ... = 0$ which is a meaningful result if the series is understood as zeta-series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Why is $\frac{1}{\frac{1}{X}}=X$? Can someone help me understand in basic terms why $$\frac{1}{\frac{1}{X}} = X$$
And my book says that "to simplify the reciprocal of a fraction, invert the fraction"...I don't get this because isn't reciprocal by definition the invert of the fraction?
| $$y=\frac1{\frac 1 x} $$
$$y'(x)=\left(\frac1{\frac 1 x}\right)' = -\left(\frac 1 {\left(\frac 1x\right)^2}\right)\left({\frac 1 x}\right)' =
\frac 1 {\left(\frac 1x\right)^2} \cdot {\frac 1 {x^2}} = \frac {y^2(x)}{x^2}$$
So we have that $$x^2dy = y^2dx\\
\int \frac{dy}{y^2} = \int \frac{dx}{x^2}\\
-\frac{1}{y} = -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/421620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 12,
"answer_id": 5
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Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$
If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$
Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.
| Adding the triangle inequalities:$$a^{2}>(b-c)^{2}\\b^{2}>(a-c)^{2}\\c^{2}>(b-a)^{2}$$
We get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$
Now divide by $a^{2}+b^{2}+c^{2}$ to get,$$1>2-2\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\\\rightarrow \frac{ab+bc+ca}{a^2+b^2+c^2}>\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Explanation of this step in a modular arithmetic problem
The multiplicative inverse of $5$ is $7$, when using mod $34$.
$$\begin{align*}
5\cdot x&=3\\[0.1in]
7\cdot 5\cdot x &=7\cdot 3\\[0.1in]
1\cdot x &=7\cdot 3\\[0.1in]
x&=21
\end{align*}$$
I don't understand this part:
$$\begin{align*}
7\cdot 5\cdot x &=7\cdot ... | Since $5x = 3 \pmod{34}$, you can multiply both sides of this equation by $7$ (the inverse of $5$ to obtain $7 \cdot 5x = 7 \cdot 3 \pmod{34}$. (The reason why this works is because $a=b \pmod{n}$ implies $ca = cb \pmod{n}$, and you should check this yourself if you haven't.)
Therefore since
$$
7 \cdot 5 = 35 = 34 + 1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recursion problem help The following are the teachers example problems. The issue is that I don't understand the exact steps they took to go from $f(0)$ to $f(1)$ to $f(2)$ to $f(3)$. What I'm asking here is if someone could be so kind to show me how the answers for $f(0)\ldots f(3)$ are derived for each problem.
Let $... | You are given $f(0)=3$ and $$\tag1f(n)=(-1)^nf(n)+4n\quad\text {for }n\ge 1.$$
To compute $f(1)$, you let $n=1$ in equation $(1)$, i.e. $f(1)=(-1)^1f(1-1)+4\cdot 1$. Since you knwo $f(09=3$, this gives you $f(1)=(-1)^1\cdot 3+4=1$. To compute $f(2)$, you let $n=2$ in $(1)$, which gives you $f(2)=(-1)^2f(1)+4\cdot 2 = 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find area of the surface obtained by rotating curve around x-axis? I got a curve $y=a \cdot cosh \frac{x}{a}$ where $|x|\le b$. The task is to find area of the surface obtained by rotating curve around x-axis.
Here is my solution. Unfortunately the result is not identical with the result of the textbook. Would you pl... | Hint: Use whe well-known hyperbolic identity
$$\cosh^2 x-\sinh^2x=1\iff 1+\sinh^2x=\cosh^2x$$
Use the above and you'll save about 80% of the work you did...:)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is e... | It boils down to showing
$$\frac{2k+1}{2k+2} \leqslant \frac{\sqrt{3k+1}}{\sqrt{3k+4}}$$
for $k \geqslant 2$. Squaring that, we need to show
$$\left(1-\frac{1}{2k+2}\right)^2 = 1 - \frac{1}{k+1} + \frac{1}{4(k+1)^2} \leqslant 1 - \frac{3}{3k+4}.$$
From then on, it should not be difficult.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Upper Bound of Logarithm For $1\leq x < \infty$, we know $\ln x$ can be bounded as following:
$\ln x \leq \frac{x-1}{\sqrt{x}}$.
Then what is the upper bound of $\ln x$ for following condition?
$2\leq x <\infty$
| To get your first upper bound we may start with the expansion :
$$\ln(1+x)=x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 + O(x^5)$$
with the upper bound :
$$\frac x{\sqrt{1+x}}=x - \frac{x^2}2 + \frac{3x^3}8 - \frac{5x^4}{16} + O(x^5)$$
To stay in the same spirit you may use the expansion :
$$\ln\left(1+\frac x2\right)=\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
Wi... | What about by integration by parts? It looks pretty simple:
$$u=x^2\;,\;\;u'=2x\\v'=x\sqrt{1-x^2}\;,\;\;v=-\frac13(1-x^2)^{3/2}$$
and thus
$$\int x^2\cdot x\sqrt{1-x^2}\,dx=-\frac13x^2(1-x^2)^{3/2}+\frac23\int x(1-x^2)^{3/2}dx=$$
$$-\frac13x^2(1-x^2)^{3/2}-\frac2{15}(1-x^2)^{5/2}+C$$
Note: We used above the following:
... | {
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"url": "https://math.stackexchange.com/questions/434765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
the value of $\lim\limits_{n\rightarrow\infty}n^2\left(\int_0^1\left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$ This is exercise from my lecturer, for IMC preparation. I haven't found any idea.
Find the value of
$$\lim_{n\rightarrow\infty}n^2\left(\int_0^1 \left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$$
Thank you
| By integration by parts,
\begin{align*}
\int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx
&= \left[ -(1-x)(1+x^{n})^{\frac{1}{n}} \right]_{0}^{1} + \int_{0}^{1} (1-x)(1 + x^{n})^{\frac{1}{n}-1}x^{n-1} \, dx \\
&= 1 + \int_{0}^{1} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx
\end{align*}
so that we have
\begin{align*}
n^{... | {
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"source": "stackexchange",
"question_score": "10",
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Differentiate the following w.r.t. $\tan^{-1} \left(\frac{2x}{1-x^2}\right)$ Differentiate : $$ \tan^{-1} \left(\frac {\sqrt {1+x^2}-1}x\right) \quad w.r.t.\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
| Putting $x=\tan\theta,$
$$\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
$$=\quad \tan^{-1} \left(\frac{2\tan\theta}{1-\tan^2\theta}\right) $$
$$=\quad \tan^{-1}(\tan2\theta)=n\pi+2\theta=n\pi+2\tan^{-1}x $$ where $n$ is any integer
$$\frac {\sqrt {1+x^2}-1}x=\frac {\sqrt {1+\tan^2\theta}-1}{\tan\theta}=\frac {\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier metho... | Multiply with $\sqrt{x-1/x}-\sqrt{1-1/x}$ to get
$$x-1=(x-\frac1x)-(1-\frac1x) =x\left(\sqrt{x-1/x}-\sqrt{1-1/x}\right)$$
so $$1-\frac1x = \sqrt{x-1/x}-\sqrt{1-1/x}$$
and by adding
$$ x+1-\frac1x =2\sqrt{x-\frac1x}$$
Now with $z=x-\frac1x$, this is simply
$$z+1=2\sqrt z,$$
a quadratic in $\sqrt z$ with (double) solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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For a convex function, the average value lies between $f((a+b)/2)$ and $(f(a) + f(b))/2$ Suppose that $f\in C^2$, $f''(x)\geq 0$ $\,\,\,\forall x \in [a,b]$. I want to show that $$\frac{1}{2}(b-a)(f(a)+f(b))\leq \int_a^bf(t)\,dt\leq (b-a)f\left(\frac{a+b}{2}\right).$$If we divide by $b-a$, we see that the left term is ... | Here is an answer that doesn't assume $f$ is differentiable.
$f$ is convex iff $\frac{f(x)-f(y)}{x-y}$ is non-decreasing in both $x$ and $y$. If $f$ is convex, $f$ is continuous.
Assume $a\lt b$ and let $c=\frac{a+b}{2}$.
Since $\dfrac{f(x)-f(c)}{x-c}$ is non-decreasing,
$$
D^-=\sup_{x\lt c}\frac{f(x)-f(c)}{x-c}
\le\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
system of equations solving for positive $a,b,c$ i need help
i need to find positive number $a,b,c$ solving this system of equations?
$$(1-a)(1-b)(1-c)=abc$$
$$a+b+c=1$$
I found that $0<a,b,c<1$ and I try to solve it by try $(1-a)=a$, $(1-b)=b$ and $(1-c)=c$ and got that
$a=0.5,b=0.5,c=0.5$ but it contradicts the se... | by expanding the left hand side you will have
$$1-(a+b+c)+ab+bc+ac-abc=abc
$$
using the second equality we will have:
$$2abc=ab+bc+ca
$$
And using the substitution method we can replace $a$ by $1-(b+c)$ then we have:
$$2bc(1-(b+c))=(1-(b+c))(b+c)+bc
$$
$$\Rightarrow 2bc-2b^2c-2bc^2=b+c-b^2-c^2-2bc+bc
$$
$$\Rightarrow 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there... | With $s=\sin 10^\circ$ and $c=\cos 10^\circ$, we have $\sin 20^\circ =\sin(2\cdot 10^\circ)=2sc$ and $\sin100^\circ=c$ and $\sin 30^\circ =\frac12$, hence
$$ \sin10^\circ\sin20^\circ\sin30^\circ = s\cdot2sc\cdot \frac12=ssc=\sin10^\circ\sin10^\circ\sin100^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Indefinite integral $\int{\frac{dx}{x^2+2}}$ I cannot manage to solve this integral:
$$\int{\frac{dx}{x^2+2}}$$
The problem is the $2$ at denominator, I am trying to decompose it in something like $\int{\frac{dt}{t^2+1}}$:
$$t^2+1 = x^2 +2$$
$$\int{\frac{dt}{2 \cdot \sqrt{t^2-1} \cdot (t^2+1)}}$$
But it's even hard... | $$
\frac{dx}{x^2+2} = \frac{dx}{2\left(\frac{x^2}{2} + 1\right)} =\frac{dx}{2\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{dx/\sqrt{2}}{\sqrt{2}\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{1}{\sqrt{2}}\cdot\frac{du}{u^2+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Expressing a combination of sine and cosine as a single cosine Prove that:
$$\dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t = \cos \left(\omega t + \dfrac\pi4\right)$$
Obviously, if we are evaluating the right side of the equation, it would a easier. It would only take the following steps:
$cos \left(\om... | If you want to prove it as if you didn't know what it actually should be equal to, then here what people usually do. Assume, you have the expression $$a \cos x + b \sin x$$ and you want to find what it is equal to. The very first step is to notice the if you pull some factor outside of the brackets, so the whole expres... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The value of $w$ also has a max error of $p\%$ Suppose $\frac{1}{w}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ where each variable $x,y,z$ can be measured with a max error of $p\%$
Prove that the calculated value of $w$ also has a max error of $p\%$
I guess I need to take its derivative i.e $-\frac{1}{w^2}dw=-\frac{1}{x^2}... | From this line
$$-\frac{1}{w^2}dw=-\frac{1}{x^2}dx-\frac{1}{y^2}dy-\frac{1}{z^2}dz$$
The percentage error of $w$ is
$$p_w=100\frac{dw}{w}\%$$
and likewise for the other variables, therefore
$$\frac{1}{w}p_w=\frac{1}{x}p_x+\frac{1}{y}p_y+\frac{1}{z}p_z$$
So assuming a maximum error of $p\%$ on the independent variables... | {
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"source": "stackexchange",
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Domain of $\sqrt{1+\frac1x}$ I solved this as follows:
$1+\frac1x \ge 0$
$\frac1x \ge -1$ (subtract $1$ from both sides)
$1 \ge -x$ (multiply $x$ to both sides, cancel out $x$ from bottom of left side)
$-1 \le x$ (multiply both sides by $-1$, change sign)
$x \ge -1$
Another way to solve is:
$1+\frac1x \ge 0$
$\frac{(x+... | Here's a better way at looking at this:
$$\begin{align*}
1+\frac{1}{x} &\ge 0 \\
x+1 &\ge 0 \tag{multiply both sides by $x$ when $x >0$} \\
&\textrm{or} \\
x+1 &\le 0 \tag{multiply both sides by $x$ when $x < 0$}
\end{align*}$$
In the first case, we restrict $x>0$, so every permissible value of $x$ satisfies the inequa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is perfect square is: I wanted to know, how can i determine the no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is a perfect square.($x \in R$)
I have tried, since $x$ is real the discriminant must be $\geq 0$.
$D = 4(6-k)^2 -4(16+9k)(1... | HINT:
Let $(16x^2+12x+39) + k(9x^2 -2x +11)=(ax+b)^2$
$\implies (16+9k)x^2+(12-2k)x+3+11k=a^2x^2+2abx+b^2$
Comparing the coefficients of $x^2,x,x^0$
we have $a^2=16+9k,2ab=12-2k\implies ab=6-k, b^2=3+11k$
$\implies (6-k)^2=(ab)^2=(16+9k)(3+11k)$ which is a Quadratic Equation in $k$ on re-arrangement
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Number Theory divisibilty How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$?
Also, how can I format my questions to allow for squares instead of doing the ^ symbol.
| First we prove two theorems that we will use
If $a,\;b,\;c\;\in \mathbb{N}$ with $a\neq0$ and $x,y\in\mathbb{N} $such that $a\mid b$ and $a\mid c$, then, $a\mid bx+cy$;
Show:$a\mid b$ and $a\mid c$ implies that there $m,n\in\mathbb{N}$ such that $b=a\cdot m$ and $c=a\cdot n$;$$bx+cy=am\cdot x+an\cdot y=a(mx+ny)\Longr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integration $\int \frac{\sqrt{x^2-4}}{x^4}$ Problem :
Integrate $\int \frac{\sqrt{x^2-4}}{x^4}$
I tried : Let $x^2-4 =t^2 \Rightarrow 2xdx = 2tdt$
$\int \frac{\sqrt{x^2-4}}{x^4} \Rightarrow \frac{t^3 dt}{\sqrt{t^2+4}(t^4-8t+16)}$
But I think this made the integral too complicated... please suggest how to proceed.. Tha... | HINT:
As the radical contains $x^2-4,$
put $x=2\sec\theta$
$$\implies \int\frac{\sqrt{x^2-4}}{x^4}dx=\int \frac{2|\tan\theta|}{\sec^4\theta} 2\sec\theta\tan\theta d\theta=4\cdot \text{sign}(\tan\theta)\int \sin^2\theta \cos \theta d\theta$$
Putting $\sin\theta=u,$
$$4\text{sign}(\tan\theta)\int \sin^2\theta \cos \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The pr... | One thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.
$$\begin{align}
\frac{3}{\frac{5}{\sqrt{5}}}&=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}\cdot\frac{\frac{\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work:
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use
$$\frac... | This would not be the best answer but may help if you are interested in computing the first terms of the Taylor expansion.
Just use the fact that $\frac{1}{1-z}= \sum_{n=0}^\infty z^n $ for $|z|<1$ and the binomial formula, i.e.
\begin{align}
\frac{1}{(1+x^3)^4} &= \frac{1}{1+4x^3+6x^6+4x^9+x^{12}} \\
&= \sum_{n=0}^\in... | {
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"url": "https://math.stackexchange.com/questions/450900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$ without using trigonometry? $$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
Is there any way to find the answer without using trigonometry, like this?
Hint by Parth Thakkat:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$... | $$\int \sqrt{x+1}/x^{7/2} dx=\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx$$
$1+\frac{1}{x}=t$ then $\frac{1}{x^2}dx=-dt $ and $\frac{1}{x}={t-1}$
$$\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx=\int\frac{1}{x} \sqrt{1+\frac{1}{x}} \frac{dx}{x^2}=-\int(t-1)\sqrt {t}dt=\int (t^{1/2}-t^{3/2})dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Partial Fractions I here would like to clear my doubt on the question below:
$$\frac{1}{x(x-1)(x-2)}\;,$$
that is, we want to bring it into the form:
$$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$
in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynom... | No, your values are not correct: they do not satisfy the equation $3A+2B+C=0$. The equation $-3A+B+C=0$ has nothing to do with the problem, so numbers obtained by using it simply aren’t relevant. I suspect that you got the equation $-3A+B+C=0$ by making a sign error in collecting the coefficients of $x$ in $A(x-1)(x-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximum and minimum function on area Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $.
I will grateful for hints and yours help.
| What could be said about this "skewed superellipse" $ \ 3x^4 \ + \ xy \ + \ y^4 \ = \ 6 \ $ is that since the curve has symmetry about the origin [if a point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x,-y) \ $ ], we should expect that extrema are located at corresponding points in opposing quadrants. user64494 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find a solution of 9x = 24 (mod 21) I need help finding a solution of $9x\equiv {24}\pmod {21}$.
Here is what I tried, but it's wrong.
mod x is the positive value of x. mod $21 = 21.$
$9x\equiv {24}\pmod {21}$.
$9x = 24*21$
$x = 24*21/9 = 56$
| Since $21=3\cdot 7$ and $(3,7)=1$, we have:
$$9x\equiv 24\pmod{21}\Leftrightarrow 9x\equiv 24\pmod{3}\wedge9x\equiv 24\pmod{7}$$
Now, $9x\equiv 24\pmod{3}\Leftrightarrow 0x\equiv 0\pmod{3}$, which is always true and
$$9x\equiv 24\pmod{7}\Leftrightarrow 2x\equiv 3\pmod{7}\Leftrightarrow 4\cdot2x\equiv 4\cdot 3\pmod{7}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
} |
Error in approximation of nonzero root to $x^2=\sin x$ using Taylor's cubic polynomial. I have successfully obtained the root's approximation $r=\sqrt{15}-3$ as I'm supposed to as following:$$\begin{align}
\displaystyle f(x)=\sin x &= x - \frac{x^3}{6} + E\\
x^2 &\approx x - \frac{x^3}{6}\\
r^2 &= r - \frac{r^3}{6}\\
r... | The series for $\sin x$ is an alternating series. So for numbers $x$ near $r$, we have $\sin x\lt x-\frac{x^3}{3!}+\frac{x^5}{5!}$.
Since $r^2=r-\frac{r^3}{3!}$, the absolute value of the difference is $\lt \frac{(0.9)^5}{5!}$. This brings is a little below $\frac{1}{200}$. Using $\sqrt{15}-3$ instead of $0.9$ brings ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you do partial fraction $\frac2{x^2 - 8}$? How do you do partial fraction on $\frac2{x^2 - 8}$ ? Or are there other method of doing? I tried trig substitution but could not get the answer.
| It's very simple:
$$\frac{2}{x^2-8} = \frac{2}{(x+2\sqrt 2)(x-2\sqrt 2)} = \frac{\frac{1}{4} \sqrt 2}{x-2\sqrt 2}-\frac{\frac{1}{4} \sqrt 2}{x+2\sqrt 2}$$
Thus, for example, $$\int \frac{2}{x^2-8} dx = \frac{1}{4} \sqrt 2 \int \frac{dx}{x-2\sqrt 2} - \frac{1}{4} \sqrt 2 \int \frac{dx}{x+2\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Average of all 6 digit numbers that contain only digits $1,2,3,4,5$ How do I find the average of all $6$ digit numbers which consist of only digits $1,2,3,4$ and $5$?
Do I have to list all the possible numbers and then divide the sum by the count? There has to be a more efficient way, right?
Thank you!
| The generating function of these numbers is
$$f(x) = \prod_{q=0}^5
\left(x^{10^q} + x^{2 \times 10^q}+ \cdots + x^{5 \times 10^q}\right).$$
Hence the number of values is given by
$$f(1) = \prod_{q=0}^5 5 = 5^6,$$
(this part is trivial but it confirms that we have the right generating function).
Similarly, the sum of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
} |
Sum of integer parts of different numbers I have the sum of all these integer parts of different numbers
$$
\lfloor 1\rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{15} \rfloor
$$
I don't have any idea to solve this exercise.
| More generally, for nonnegative integer $n$, $\lfloor \sqrt{x} \rfloor = n$ if
$n \le \sqrt{x} < n+1$, i.e. $n^2 \le x < (n+1)^2$. Since $(n+1)^2 - n^2 = 2n+1$,
there are $2n+1$ integers $x$ for which this is true. Thus
$$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2}\rfloor + \ldots + \lfloor \sqrt{(n+1)^2 - 1}\rfloor =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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.