Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show that $(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$ for some integers $A,B,C$ How do you solve this question?:$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ is for all $x$. Here $A$, $B$ and $C$ is constants. I need to know $A$, $B$ and $C$ to pass this. They are integers.
I got this far:
(LS = left side)
$$LS = ... | Anything forall x implies an identity so you can put any * values of x to get C. for e.g.:
$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$
$$
\left\{
\begin{array}{c}
49=A+C\quad x=0 \\
1=-A+C\quad x=\pi/2\\
18=B+C\\
\end{array}
\right.
$$
Solving these we get:
$$
\left\{
\begin{array}{c}
A=24 \\
C=25\\
B=-7\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How is $ i^{-1} = -i$ and $i^{-3} = i$? Now I know that with positive powers of $i$ the cycle is: $i , -1 , -i , 1\ldots$
The negative power cycle is: $-i , -1 , i , 1 \ldots$
Can someone explain to me how $\frac 1 {\sqrt{-1}}$ is equal to $-i$ and $\frac 1 {-\sqrt{-1}}$ is equal to $i$?
| $\frac{1}i=i^{-1}$ is true by definition and $n+\frac{1}n=\frac{n^2+1}n$ (since $n=\frac{n^2}n$ and $\frac{n^2}n+\frac{1}n=\frac{n^2+1}n$) is also true. If you plug $i$ in $\frac{i^2+1}i=\frac{-1+1}i=\frac{0}i=0$.
Therefore $i+\frac{1}i=0$ and $\frac{1}i=-i$. By our original statement (namely $i^{-1}=\frac{1}i$) we can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$
Let $x,y,z>0$. Prove that
$$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
| Alternative proof without full expanding or uvw:
Let
$$u = \frac{1}{x + 2y}, \quad
v = \frac{1}{y + 2z}, \quad w = \frac{1}{z + 2x}.$$
We have
\begin{align*}
&\mathrm{LHS} - \mathrm{RHS} \\
=\,& u^2 + v^2 + w^2 - \frac{1}{xy + yz + zx}\\
=\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + uv + vw + wu - \frac{1}{xy + yz ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Use the range for $\theta$ to determine the indicated function value Problem:
$$\sin\theta = \frac {1}{2}; \frac {\pi}{2} \leq \theta \leq \pi, \text{find} \cos\theta$$
So I have that $\sin\theta = \frac {1}{2}$ and that the range for $\theta$ is $\frac {\pi}{2} \leq \theta \leq \pi$.
I'm assuming I need to find the ... | Solving $\sin\theta = \frac {1}{2}$, we get $\theta = \frac {\pi}{3}$ or $\frac {2\pi}{3}$.
Edit: my mistake and should be $\theta = \frac {\pi}{6}$ or $\frac {5\pi}{6}$
Note that only the latter fits the given range.
Thus, $\cos\theta = \cos \frac {5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = … = - \cos \frac {\pi}{6} = -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5... | Hint: $\displaystyle\int\frac{dy}{y^3} = \int y^{-3} dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a rational function Calculate the limit
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}$$
I divided by the highest degree of x, which is $x^{4}$, further it gave
$$ \frac{-\frac{1}{6}}{\frac{64}{3}} = \frac{-1}{128}$$
which is wrong... what is my error?
| At $0$ we have
$$x^4=o(x^2)\quad\text{and}\quad x^3=o(x^2)$$
so
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}=\lim_{x \to 0}\frac{3x^2}{4x^2}=\frac34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Simplifying a product written in Capital Pi Notation I'm having some trouble figuring out how to simplify Capital Pi Notation.
What I tried was to expand the multiplication with various n and tried to find a pattern. Could someone point me in the right direction on how to approach these problems?
$$\prod_{k=2}^{n} \l... | To get this question off the unanswered list, I'll convert my comment into an answer.
Notice that $1 - \dfrac{1}{k^2} = \dfrac{k^2-1}{k^2} = \dfrac{(k-1) \cdot (k+1)}{k \cdot k}$. Therefore, we have:
$\displaystyle\prod_{k = 2}^{n}\left(1-\dfrac{1}{k^2}\right) = \prod_{k = 2}^{n}\dfrac{(k-1) \cdot (k+1)}{k \cdot k} =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite geometric progression involving square terms The sum of an infinite geometric progression is 15 and the sum the squares of these terms is 45. Find the series.
The formula for sum of infinite GP is $\frac{a }{1-r} $ and I got two equations $15=\frac {a}{1-r} $ snd $45=\frac {a^2}{1-r^2} $. Solving I got $3=\fr... | Clearly, $a\ne0, r^2\ne1$
We have $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{1-r}{1+r}$
But, $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{45}{15^2}=\frac15$
Now apply Componendo and dividendo
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to divide two inequalities I would like to know if somebody knows how to properly divide one inequality by another, as a resolution method similar to when we divide one equality by another.
Take this as an example: $x^2 - y^2 \lt 8$ and $x + y \gt 3$.
I want to know if it is possible to factor the left hand side of... | Suppose $x^2-y^2 < 8$ and $x+y > 3$. If $y - x \le -\dfrac{8}{3}$, then $x-y \ge \dfrac{8}{3}$, and hence $x^2-y^2 = (x-y)(x+y) > \dfrac{8}{3} \cdot 3 = 8$, a contradiction since $x^2-y^2 < 8$.
Thus, any $(x,y)$ which satisfy both $x^2-y^2 < 8$ and $x+y > 3$ will also satisfy $y - x < -\dfrac{8}{3}$.
However, the con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Inverse Trigonometric functions - Boyce & Diprima 2.2.19 The problem asks for a solution to the initial value problem:
\begin{align}
&\sin(2x)dx+\cos(3y)dy=0\\
&y\left(\frac{\pi}{2}\right)=\frac{\pi}{3}
\end{align}
The problem is separable and I arrive at the following implicit solution:
\begin{align}
\sin(3y)-3\cos^2(... | After integrating you have:
$$ - \frac{1}{2} \cos{2x} + \frac{1}{3} \sin{3y} = K, \quad K \in \mathbb{R}, $$
which is as nice as an explicit solution. Substitute the initial data to find the value of the constant of integration:
$$ - \frac{1}{2} \cos{\pi} + \frac{1}{3} \sin{\pi} = \frac{1}{2} = K, $$
and you are do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calcula... | Seeing the column of $1$'s, a first thought would be to subtract the first row from the other two. This turns out to give you a factor $b-a$ in the second row and a factor $c-a$ in the third.
$$
\begin{align}
\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
&=
\begin{vmatrix}
1 &... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$. Trigonometry
$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$
then the value of
$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?
NOTE: can somebody help me
$\cos^2\alpha \... | Let $t=\sin^2(\alpha)$ and $s=\sin^2(\beta)$. Then, multiplying both sides of the given identity by $s(1-s)$ gives:
$$
(1-t)^2s+t^2(1-s)=s(1-s).
$$
Bringing the RHS over to the LHS simplifies to $(s-t)^2=0$ so $s=t$. Now, the expression you want to evaluate is just
$$
\frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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An improper integral involving nested square roots I have:
$$I_{a,b}= \int_b^{ + \infty } \left( \sqrt {\sqrt {x + a} - \sqrt x } - \sqrt {\sqrt x - \sqrt {x - b} } \right)dx$$
with $a>0$ and $b>0$.
I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.... | We have:
$$\sqrt{x+a}-\sqrt{x}=\frac{a}{\sqrt{x}+\sqrt{x+a}},\qquad \sqrt{x}-\sqrt{x-b}=\frac{b}{\sqrt{x}+\sqrt{x-b}}$$
and:
$$f_{a,b}(x)=\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}=\frac{\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}}{\frac{\sqrt{a}}{\sqrt{\sqrt{x}+\sqrt{x+a}}}+\frac{\sqrt{b}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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The sequence $(a_0,a_1,a_2,\cdots,)$ satisfies $ a_{n+1}=a_n+2a_{n−1}$. What is $a_5$? Assume that the sequence $(a_0,a_1,a_2,\cdots,)$ satisfies the recurrence $\displaystyle a_{n+1}=a_n+2a_{n−1}$. We know that $a_0=4$ and $a_2=13$. What is $a_5$?
I got $a_1=5, a_3=23, a_4=49, a_5=95$
| For just the 5th term this is overkill, but anyway.
Define the generating function:
$\begin{align*}
A(z)
&= \sum_{n \ge 0} a_n z^n
\end{align*}$
Write the recurrence shifted (subtraction in indices gets nasty), multiply by $z^n$, sum over $n \ge 0$, recognise the sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
... | {
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"url": "https://math.stackexchange.com/questions/916379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve such a nonlinear ODE, the analytical solution of which is known! I have the following ODE with initial/boundary value conditions:
$$\left. \begin{aligned}
\left(x^2-10 x-y^2\right)y\, y'(x)+(x-5) y^2 y'(x)^2-(x-5) y^2=0
;\qquad (\text{ODE})\\
y(0)^2=25;\qquad y'(0)^2=\frac{3-\sqrt{5}}{2} \qquad\qquad\qquad... | Let $U = x^2 + y^2$ and $V = (x-10)^2 + y^2$.
Multiply the equation by $20$ and look at individual terms, we have:
$$\begin{align}
20(x-5) (yy')^2
&= \left[(x^2+y^2) - ((x-10)^2+y^2)\right] (yy')^2\\
&= (U-V)(yy')^2\\
\\
20(x^2 - 10x - y^2) yy'
&= 2yy'\left[(x-10)\left(x^2+y^2\right) - x\left((x-10)^2+y^2\right)\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
| Hint:
Try this:
$$\frac{1}{2-x} = \frac{\frac{1}{2}}{1-\frac{x}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty 2^{-(n+1)}x^n$$
Your method gives the Taylor expansion around $x=1$ with radius of convergence of $|x-1| < 1$ so for $x = 0$ you have $|x-1| = |-1| = 1 \not < 1$ you are on t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | The question posed was whether the OP's proof is correct. Mark Bennet pointed out one flaw in comments, namely that you can't divide by $x-y$ if $x=y$. It's worth noting another problem with the proof: You can indeed suppose wlog that $x\ge y$, from which $x^2\cdot x\ge x^2\cdot y$ follows (because $x^2\ge0$), but t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Two variables Diophantine equation $x^3 - 2y^2 =54$ How can we solve $x^3 - 2y^2 =54$ Diophantine equation in natural numbers? Modular arithmetic method is not useful maybe?
| Since $x$ must be even, $x=2z$, we have:
$$ 4z^3- y^2 = 27 \tag{1}$$
or:
$$z^3 = \frac{y+3\sqrt{-3}}{2}\cdot\frac{y-3\sqrt{-3}}{2}=w\cdot\overline{w},\tag{2}$$
where $w$ lies in the ring of integers of $\mathbb{Z}[\sqrt{-3}]$, that is a UFD.
This implies that, if $w$ and $\overline{w}$ are coprime, $w=\frac{y-1}{2}+3\... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Properties of a recursively-defined sequence using induction This is a homework problem. Not expecting the solution, just a nudge in the right direction!
$N$ is a function defined inductively as follows:
$$N(1) = N(2) = N(3) = 1$$
$$N(n) = N(n−1) + N(n−3) \quad \text{for } n > 3.$$
a. Prove by induction on $a$, $$N(n)... | (a.1) Basis ($a=1$): you assume $n>4$ and:
$$
N(3)N(n-2) + N(1)N(n-3) + N(2)N(n-4) = N(n-2) + N(n-3) + N(n-4)
=N((n-1) - 1) + N((n-1)-3) + N(n-3) = N(n-1) + N(n-3) = N(n)
$$
Obs: you can check manually the case $n=4$ where the induction formula is not guaranteed to work using $N(0)=0$.
(a.2) Inductive step: assume the ... | {
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"url": "https://math.stackexchange.com/questions/920866",
"timestamp": "2023-03-29T00:00:00",
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Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\... | Here is another way to prove such identities:
Get the groebner basis of all the relevant polynomial equalities for what you are dealing with. In this case it's very simple since we only have $\sin\theta=s$ and $\cos\theta=c$ and the relevant polynomial equality is $s^2+c^2-1=0$. This is your groebner basis - no extra w... | {
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"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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Find $\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ Find $\displaystyle \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ without a calculator.
L'Hopital's rule only gives me the limit for one side. Not two. How would I solve this problem?
| One definition of the absolute value function is $|x| = \sqrt{x^2}$. For negative values of $x$, observe that
$$\frac{x}{|x|} = \frac{x}{-x} = -1$$
Thus,
\begin{align*}
\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 + 2}} & = \lim_{x \to -\infty} \frac{x}{\sqrt{x^2\left(1 + \dfrac{2}{x^2}\right)}}\\
& = \lim_{x \to -\infty} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{... | You look like you are on the right path, and have the triangle drawn out correctly. I recommend using the substitution $\frac{x}{\sqrt{1-x^2}}=\tan(\theta)$. This should convert your integral into $$7\int \! \tan(\theta)\sin(\theta)\cos(\theta) \, \mathrm{d}\theta=7\int \!\sin^2(\theta) \, \mathrm{d}\theta$$ Now using ... | {
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"timestamp": "2023-03-29T00:00:00",
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Testing convergence of the series of $n^p((n-1)^{-1/2}-n^{-1/2})$ Exercise 8.15 (l) of Analysis by Apostol states:
Test for convergence: $$\sum\limits_{n = 2}^\infty n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \right)$$
The solution I have states as the first step: "Note that
$$ n^p\left(\frac{1}{\sqrt{n-1}}-\frac... | You start with (ignoring the $n^p$ for the time being)
$$\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}},$$
then you rationalise the numerator, using $\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}$, getting
$$\frac{n-(n-1)}{\sqrt{n-1}\sqrt{n}(\sqrt{n}+\sqrt{n-1})}.$$
Now t... | {
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Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2}... | Your inequality holds good if $x>-2$, not $2$.
| {
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"url": "https://math.stackexchange.com/questions/923478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Show that $n^3+2n$ is divisible by 3 for all $n\ge 1$ i want to prove it with mathematical induction :
first i am tried with n=0
then it is divisible by zero then i move to next step change all n with K then i get this product :
$$P(K)=K^3+2K = 3m$$
Note: $3k$ because we multiply any no. with $3$ is divisible by $3$
no... | I know you want to prove by induction, but anothter proof would be to observe that every integer is congruent to either 0, 1, or 2 (mod 3). Then we want to see if for every integer n:
$n^3 + 2n \equiv 0\pmod{3}$
Well, take all integers n such that $n \equiv 0\pmod{3}$. Then $n^3 \equiv 0\pmod{3}$ and $2n \equiv 0\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
Show that $ \sum_{k=0}^{r} \binom{r-k}{m} \binom{s+k}{n} = \binom{r+s+1}{m+n+1} $? I can't resolve this exercise and I need a tip.
$$ \sum_{k=0}^{r} \binom{r-k}{m} \binom{s+k}{n} = \binom{r+s+1}{m+n+1} $$
where $ n \geq s $.
| Suppose we seek to evaluate
$$\sum_{k=0}^r {r-k\choose m} {s+k\choose n}$$
where $n\ge s$ and $m\le r.$
Introduce
$${r-k\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-k-m+1}} \frac{1}{(1-z)^{m+1}} \; dz.$$
Note that this is zero when $k\gt r-m$ so we may extend the sum in $k$
to $k=\infty.$
Introduce... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Closed-forms for several tough integrals These integrals came up in the process of finding solution to Vladimir Reshetnikov's problem. I wonder if there are closed-forms for the following integrals:
\begin{array}{1,1}
&[\text{1}] &\quad\int_0^1\frac{\operatorname{Li}_3(ax)}{1+2x}\ dx\\[12pt]
&[\text{2}] &\quad\int_0^1\... | $\def\Li{{\rm Li}}$I think I know how Mr. Tunk-Fey's approach to find solution to Mr. Vladimir Reshetnikov's problem:
\begin{equation}
I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx
\end{equation}
He used integration by parts method. Here might be the way he evaluated $I$:
Let
\begin{equation}
u=\ln(1+2x)\quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 0
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How to prove the sequence $x_{n+1} = \frac{x_n}2 + \frac 1{x_n}$ is a Cauchy sequence Here is a question I do not know how to prove. Thanks for your helping！
Prove that the sequence $$x_{n+1} = \dfrac{x_n}2 + \dfrac 1{x_n}, x_0 = 1$$is a Cauchy sequence.
| Another method:
Firstly a bit of pedantry. The sequence is decreasing only after the second element. The first three elements of the sequence are $ 1, \frac{3}{2}, \frac{17}{12} $.
Now consider the function $f$ which takes $ x_n \to x_{n + 1} $.
$$ f(x) = \frac x 2 + \frac 1 x \implies f'(x) = \frac{x^2 - 2}{ 2x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Multiplying both sides of an equation when there's a limit on one side?
Determine the value of $a\in\mathbb{R}$, such that $\displaystyle\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$
My attempt:
\begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
&\implies\lim_{x\to 1} x^2+(3-a)x+3a=7x-7\\
&\implies\lim_{x\to ... | There is a safer way to think. If you try to plug $1$ into the limit, you'll get a division by zero. So, your only hope for this limit to make sense, is if $x^2 + (3 - a)x + 3a$ is divisible by $x - 1$, so you can simplify and eliminate the division by zero. This means that $1$ must be a root of $x^2 + (3-a)x + 3a$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 11,
"answer_id": 0
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Lucas Numbers Proof $L_n = \alpha^n + \beta^n$ Proof by Induction:
Lucas numbers are recursively defined as:
$L_n = L_{n-1} + L_{n-2}$ where $L_1 = 1$ and $ L_2 = 3 $for $n \ge 3$
Show that: $L_n = \alpha^n + \beta^n$ for $\alpha = {1+\sqrt{5}\over2}$ and $\beta = {1-\sqrt{5}\over2}$
Is this the correct approach?
bas... | Our induction assumption is that the formula holds for all $i\le n+1$, and we show that the formula holds for $i=n+2$. So we are using what is called strong induction.
Actually, it is enough to have as induction assumption that the formula holds at $n=k$ and $n=k+1$, and show it holds at $n=k+2$.
So we know that for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solving for $x$ given $y = 2x - 9$ and $y = 5$. When does the 5 come into play? I am doing some problems. It states "Isolate the left side for each of the following equations. Then solve for x, assuming the value of y is 5 in a all cases." I noticed that I can get two different answers, depending on when I decide to pl... | You have a mistake in the first. The second has no mistake.
For the first, note that
$$\begin{align}9+y=2x&\iff \frac{1}{2}\left(9+y\right)=\frac 12\times 2x\\&\iff \frac{9+5}{2}=x\\&\iff x=7.\end{align}$$
Note that you need to multiply $9+y$ by $\frac 12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$ I decided to follow a recent trend and ask a question about logarithmic integrals :)
Is there a closed form for this integral?
$$\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$$
| This integral is equal to
$$ -4\big( \zeta(-3,-1,-1,-1) +\zeta(-3,-1,1,-1) +\zeta(-3,1,-1,1) +\zeta(3,-1,-1,-1) +\zeta(3,-1,1,-1) +\zeta(3,1,-1,1) \big) $$
in terms of the multiple zeta function, which can also be simplified to
$$ 2\zeta(-5,-1)-2\zeta(-5,1)+2\zeta(5,-1)+{\textstyle\frac32}\zeta(5,1)+4\zeta(-3,1,1,1), $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 1,
"answer_id": 0
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If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ $x,y,z$ are positive real numbers such that $$xyz=1$$ Prove that $\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$. I have no idea how to solve this problem.
I've tried Engel form of Cauchy inequali... | $$\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$$
Since $xyz=1$ we substitute, $x=\frac{a}{b} , y= \frac{b}{c} , z= \frac{c}{a} $
Now, by Engel form(also called Titu lemma)
$$ \sum_{cyc}\frac {a^2}{c^2 + ab }\ge\frac {(a^2 + b^2 +c^2)^2}{a^2b^2 + b^2c^2 + c^2a^2+a^3b + b^3c + c^3a} $$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\pi$ By Hand? All over the internet the only hand equation i found was
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
| One easy-to-understand improvement to your method, which I I don't see used much, is:
$$\pi/6 = \arctan \left ( \frac{\sqrt{3}}{3} \right ) \\
= \int_0^{\frac{\sqrt{3}}{3}} \frac{1}{1+x^2} dx \\
= \sum_{n=0}^\infty \frac{(-1)^{n} \left ( \frac{\sqrt{3}}{3} \right )^{2n+1}}{2n+1} \\
= 3^{-1/2} \sum_{n=0}^\infty \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 3
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Do we simplify the Proof by Contradiction? Prove the following by contradiction:
Suppose $a,b\in\mathbb{Z}$. If $4|\left(a^2+b^2\right)$, then $a$ and $b$ are not both odd (in other words, $a$ and $b$ are even)
So, I did this:
Assume $a$ and $b$ are odd
Let $a={2k+1}$
Let $b={2l+1}$
$4|\left((2k+1\rig... | $$
(2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2
$$
This means that $\left(a^2 + b^2\right) = 4 \lambda + 2$, meaning that $4$ could not possibly divide $\left(a^2 + b^2\right)$, since there's always a remainder of $2$.
...by the way, this means that for $4$ to divide $a^2 + b^2$, both $a$ and $b$ must ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve for X in Cubic poynomial I've been given a Polynomial (Cubic)
$$k=\frac16x\cdot(x+1)\cdot(2x+1)$$
If $k$ is given, is there any way to solve for $x$?
| We solve the problem in a very special case only, where $x$ happens to be a positive integer.
Since $\frac{x(x+1)(2x+1)}{6}\gt \frac{x^3}{3}$, we have $x^3\gt 3k$, and therefore $x\gt (3k)^{1/3}$.
Note that $2(x+1)^3=2x^3+6x^2+6x+2$ while $x(x+1)(2x+1)=2x^3+3x^2+x$. It follows that if $x$ is a positive integer then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| $5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)=-x^3(x-5)(x+5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with implicit differentiation simplifications STEP 1: $$ (x+y)^{3} = x^3 + y^3 $$
STEP 2: $$ 3(x+y)^2 (1+ dy/dx) = 3x^2 + 3y^2(dy/dx) $$
STEP 3: $$ 3(x+y)^2 + \frac{dy}{dx}\cdot3(x+y)^2 = 3x^2 + 3y^2(dy/dx)$$
STEP 4: $$\frac{dy}{dx} \cdot 3(x+y)^2 = 3x^2 + 3y^2(dy/dx) - 3(x+y)^2$$
STEP 5: $$\frac{dy}{dx} = (3x^2 ... | Hint $x^3 + y^3 = (x+y)(x^2-xy+y^2)$
so we have the expression
$$(x+y)^3=(x+y)(x^2-xy+y^2)$$
$$(x+y)[(x+y)^2-(x^2-xy+y^2)]=0$$
$$(x+y)(3xy)=0$$
$$x^2y+y^2x=0$$
which when differentiated becomes
$$x^2\displaystyle\frac{dy}{dx}+y(1)+y^2(1)+x\displaystyle\frac{dy}{dx}=0$$
$$\displaystyle\frac{dy}{dx}(x^2+x)=-(y^2+y)$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $$9^x=5^x+2\sqrt{20^x}+4^x$$
I'm not really sure where to start. I tried simplifying with logarithms and factoring out the x but it ended up looking just as complicated..
| \begin{align*}
9^x&=5^x+2\sqrt{20^x}+4^x\\
9^x-5^x-4^x&=2\sqrt{20^x}\\
\text{At this point, one solution jumps out}&\\
9^2-5^2-4^2\space
=\space 81-25-16\space
=\space 40&=2\sqrt{20^2}\\
\implies x=2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Not a perfect square of the form for any integer x. Now a days, I become good fan of this site, as this site making me to learn more math..hahaha.
Okay!
Can we prove that $x^3 + 7$ cannot be perfect square for any positive/negative or odd/even integer of $x$.
I checked with number up to x = 1,...1000. I realized that,... | Rewrite our Diophantine equation as $y^2+1=x^3+8$. Note that $x$ cannot be even, for $y^2+1$ cannot be divisible by $4$.
So $x$ must be odd. We have
$$y^2+1=x^3+8=(x+2)(x^2-2x+4).$$
Suppose $x\equiv 1\pmod{4}$. Then $x+2\equiv 3\pmod{4}$, which is impossible, since a positive number of the form $4k+3$ cannot divide $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall ... | Let $x=q^e-1$. We have $q^{ke}-1 = \exp(k\cdot\ln (1 + x))-1$. Because $\nu_p (x)\geq 1$, we have $\nu_p (\ln (1 + x)) = \nu_p(x)\geq 1$ (look at power series), so that $k\cdot \ln(1+x)$ is in the radius of convergence of the $p$-adic exponential (here we use the fact that $p$ is odd).
Looking again at the power seri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to find closed form of summation of Fibonacci Sequence? I created two formulas to prove a binary theory involving the Fibonacci sequence.
(1) $\sum_{i=0}^n F_{2i+1} $
Equation (1) is the sum of all Fibonacci numbers up to $F_n$ where every $i$ in $F_i$ is an odd number.
I had come up with a closed form where (1) ... | The Fibonacci numbers have the form $(\alpha - \beta) F_{n} = \alpha^{n} - \beta^{n}$, where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now
\begin{align}
\sum_{k=0}^{n} (-1)^{k} F_{k} &= \frac{1}{\alpha - \beta} \sum_{k=0}^{n} \left( (- \alpha)^{k} - (- \beta)^{k} \right) \\
&= \frac{1}{\alpha - \beta} \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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A Polynomial that Passes through the following four points? I'm trying to do this for practice but I'm just going nowhere with it, I'd love to see some work and answers on it.
Thanks :)
Find a polynomial that passes through the points (-2,-1), (-1,7), (2,-5), (3,-1). Present the answer in standard form.
What I've tried... | $$f(x) = a x^3 + b x^2 + c x + d$$
\begin{eqnarray*}
-1 &=& a (-2)^3 + b(-2)^2 + c(-2) + d \\
7 &=& a (-1)^3 + b(-1)^2 + c(-1) + d \\
-5 &=& a(2)^3 + b (2)^2 + c (2) + d \\
-1 &=& a(3)^3 + b(3)^2 + c (3) + d
\end{eqnarray*}
or
\begin{eqnarray*}
-8 a + 4b -2c + d &=& -1\\
-{\ }a{\ } +{\ } b -{\ } c + d &=&{\ } 7 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the order of the elements in the given groups I have to find the order of the following elements in the given groups:
*
*$(1 \ \ 2 \ \ 3) \ (1 \ \ 2\ \ 4) \text{ in } S_5$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 1 & 2 & 5
\end{pmatrix}=\begin{pmatrix}
1 & 3
\end{pmatrix}\begin{pmatrix}
2 & 4
\end{pmatri... | The first two computations are correct.
The matrix $A$ is invertible (because $\det A=1$), so $A^3$ can't be the zero matrix. Once you found that $A^2=-I_2$ ($I_2$ the identity matrix), you know that $A^4=(A^2)^2=I_2^2=\dots$
Thus you know the order of $A$, because…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For all real $\theta$ prove that $ \cos(\sin\theta) \gt \sin(\cos\theta)$ How do I prove this? Im not able to even start it. Help please!
| Try graphing both of the functions.
Also, this answer may help:
$$
\begin{align}
&\cos (\sin x) - \sin (\cos x) > 0\\
\implies &\cos (\sin x) - \cos ( π/2 - \cos x ) > 0\\
\implies &2 \sin \left[ \frac{\pi}{4} + \frac{1}{2}(\sin x - \cos x) \right]\cdot \sin \left[ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
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Does a maximum value exist for this expression? Let $x$, $y$, $z$ be positive real numbers and $x + y + z =3$. Does a maximum value exist for this expression?
$$\displaystyle E = \frac{x}{2 y+3 z}+\frac{4 y}{5 z + 6 x}+\frac{7 z}{8x+9 y}.$$
I tried
Put
$$a=2 y+3 z,\quad b=6 x+5 z,\quad c=8 x+9y.$$
Then
$$x = -\dfrac... | The expression $E$ is not bounded above. Let $x$ be very large, and $y$ and $z$ small.
Edit: The condition $x+y+z=3$ has been added. The expression $E$ is still not bounded above. Let $x$ be very close to $3$, making $y$ and $z$ close to $0$. The term $\frac{x}{2y+3z}$ can be made arbitrarily large.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use this result to solve Ax=x for x. Given matrix A = \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix}
and x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}
Answer: Given $Ax = x$. Subtract from x from both sides to get $Ax - x = 0$. We kn... | First it is important, that $\textbf A \textbf x - \textbf x = \textbf 0 \Rightarrow (\textbf A - \textbf I)\textbf x=0$. The vector $\textbf x$ has to be on the right side of $(\textbf A-\textbf I)$. Otherwise it wouldn´t work. And you have to subtract $\textbf I$ from $\textbf A$.
$$ \left( \begin{bmatrix}2 & 1 & 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What are the ranges of triangle angles? Lets say, that $\alpha \le \beta \le \gamma$. As shown here, $60 \le \gamma \lt 180$. What are the minimum and maximum values of $\alpha$ and $\beta$?
The answer:
$$0\lt \alpha \le 60 \\ 0 \lt \beta \lt 90\\ 60 \le \gamma \lt 180$$
| If $\alpha \le \beta \le \gamma$ are triangle angles, then their ranges are:
$$
0^\circ \lt \alpha \le 60^\circ\\
0^\circ \lt \beta \lt 90^\circ\\
60^\circ \le \gamma \lt 180^\circ$$
Proof
The lower $\alpha$, $\beta$ and the upper $\gamma$ bounds can be determined from the case, when $\gamma \rightarrow 180^\circ$, $\a... | {
"language": "en",
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\p... | Hints:
*
*Note $z_1z_2=1$
*You could write your (2) as $z_1=e^{i2\pi/3}$ and $z_2=e^{-i2\pi/3}$.
*Taking the $10$th power of an exponential is easy
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve this differential equation: $x^2dy-y^2dx+xy^2(x-y)dy=0$
$$x^2dy-y^2dx+xy^2(x-y)dy=0$$
What I tried:
$$\frac{x^2}{y^2} \frac{dy}{dx}+x(x-y)\frac{dy}{dx}=1\\$$
Let $h=-1/x, \; k=-1/y,\; dh=1/x^2 \, dx, \; dk=1/y^2 \,dy$
$$\frac{dk}{dh}+\frac{(k-h)}{k^2} \frac{dk}{dh}=1\\
\frac{dk}{dh}+\frac hk=1+\frac1k\... | The equation is $\displaystyle \frac{dy(x)}{dx}\cdot{\frac{x^2}{y(x)^2}}+\frac{dy(x)}
{dx}\cdot{x(x-y(x))}=1$.
Multiplying by $y(x)^2$ we get $$\displaystyle \frac{dy(x)}{dx}\left[x^2+x(x-y(x))y(x)^2\right]=y(x)^2 \rightarrow \frac{dy(x)}{dx}=\frac{y(x)^2}{x^2+x^2y(x)^2-xy(x)^3}$$
Now, $\displaystyle \frac{dy(x)}{dx}=\... | {
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"url": "https://math.stackexchange.com/questions/953648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty... | it is $\frac{|x| \sqrt{6-\frac{2}{x^2}}}{x(1+\frac{2}{x})}=-\frac{\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}$ and the searched limit is $-\sqrt{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Different methods to multiply The sum and the multiplication in $\mathbb{Z}_p$ correspond to the the sum and the multiplication of powerseries.
Example for sum:
$$(3 \cdot 7^0+4 \cdot 7^1+2 \cdot 7^2)+(5 \cdot 7^0+3 \cdot 7^1)= \\ (3+5) \cdot 7^0+(4+3) \cdot7^1+2 \cdot 7^2= \\ (7+1) \cdot 7^0+7 \cdot 7^1+2 \cdot 7^2= \... | The other method for multiplication is a reflected version of "long multiplication".
You'll forgive me if I start on the left with the highest power of 7 (in your example) and align on the right, because what we're looking at here is working arithmetic in base 7. But that just means you need to read the other way acro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Residue Calculus (Computing an Improper Integral)
Use residue calculus to compute the integral $\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz$
My solution
If we add to the interval $I_{R}=[-R,R]$ add the semicircle $\gamma_{R}$ in the upper half plane with centre at the origin and radius $R$, then we obtain... | Yes. Simply rewrite $$\int \frac{1}{(z^2+16)(z^2+25)} \mathrm{d}z= -\frac{1}{225} \int \frac{1}{\frac{z^2}{25}+1} \mathrm{d}z +\frac{1}{144} \int \frac{1}{\frac{z^2}{16}} \mathrm{d}z$$ and substitute $u=\frac{z}{5}$ for the first integral and $v=\frac{z}{4}$ for the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:
$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.
So far I have done:
$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equ... | In general, we have distinct primes $p,q$; we have integers $a \in [0,p)$ and $b \in [0,q)$; and we want $x$ such that $x \equiv a$ mod $p$ and $x \equiv b$ mod $q$.
To solve this equation, the first step is to calculate $r = p^{-1}$ mod $q$. Assuming we have done that, the rest is easy: just set $x = (b-a)rp + a$. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An equality about sums of Bernstein polynomials: $\sum_{k=0}^n \left(x - \frac kn\right)^2 p_{nk}(x) = \frac{x(1-x)}{n}$ We defined the Bernstein polynomials as following
$$
p_{nk}
\ = \
\frac{n!}{k!(n-k)!} x^k(1-x)^{n-k}$$
I have to show this:
$$
\sum_{k=0}^n \left(x - \frac kn\right)^2 p_{nk}(x) \ = \ \frac{x(1-x)}{... | It is simpler to start from $\sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)$, using the fact that $\sum_{k=0}^nk^2p_{n,k}(x)=n(n-1)x^2+nx$. This gives
\begin{align}
\sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)&=x^2-\frac{2x}n\sum_{k=0}^nkp_{n,k}(x)+\frac 1{n^2}(n(n-1)x^2+nx)\\
&=x^2-\frac {2x}n\cdot nx+\frac 1{n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical Induction Question, Proof Help Prove using Mathematical Induction that for all natural numbers ($n>0$):
$$
\frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}.
$$
Proof by Induction:
Let P(n) denote 1/ √1 + 1/ √2 + … + 1/ √n ≥ √n
Base Case: n = 1, P(1) = 1/... | Assuming $\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$, we have
\begin{align*}
\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} &= \sum_{i=1}^n \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} \\
&\geq \sqrt{n} + \frac{1}{\sqrt{n+1}} \\
&= \frac{\sqrt{n^2+n} + 1}{\sqrt{n+1}} \\
&\geq \frac{\sqrt{n^2} + 1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | $$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-7x})(\sqrt{x^2+2x}+\sqrt{x^2-7x})}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}$$
$$=\lim_{x\to \infty}\frac{9x}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}=\lim_{x\to \infty}\frac{\frac{9x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-7x}}{x}}$$$$=\lim_{x\to \infty}\frac{9}{\sqrt{\frac{x^2+2x}{x^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+... | $$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2=(5^n+2)(5^{2n}-1)=(5^n+2)(5^n-1)(5^n+1)$$
Do you see middle term is divisible by $4$?
| {
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"source": "stackexchange",
"question_score": "1",
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Proving a Binomial Identity Problem $\boldsymbol{25}$ [$\boldsymbol{5}$ Points]: Show that
$$
\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}=2^n
$$
Hint: Denote the left hand side by $f(n)$ and prove that $f(n+1)=2f(n)$.
Original Image
Can you please help me with problem 25. I need to prove that $f(n+1)=2 f(n)$, where $f(n)$ i... | Showing the hint is just a straightforward calculation. If $f(n) = \sum \binom{n+k}{k} \frac{1}{2^k}$, then
\begin{align*}
f(n+1) &= \sum_{k=0}^{n+1} \binom{n+k+1}{k} \frac{1}{2^k} \\
&= \sum_{k=1}^{n+1} \binom{n+k}{k-1} \frac{1}{2^k} + \sum_{k=0}^{n+1} \binom{n+k}{k} \frac{1}{2^k} \\
&= \frac{1}{2} \sum_{k=0}^n \binom... | {
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"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluate a linear system of three equations Solve for $x, y\ \text{and}\ z\ $:
$x-3z=10\\
-x+y+2z=7\\
2x+2y-5z=-8$
My working:
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
-1 & 1 & 2 & 7 \\
2 & 2 & -5 & -8
\end{array}\right) =
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 3 & 0 & -11
\end{a... | You seem to have 62/3 = 24. That gives you errors in both R2+R3/3 and just R3/3 in the last line. The rest seems to be correct. Fix that and you end up with (-52, -11/3, -62/3) which you can check does fit your original system.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic equation $4x^2+4x=7$ using quadratic formula Solve using quadratic formula. $4x^2+4x=7$
So $4x^2+4x-7=0$
$A=4$ $b=4$ $c=-7$
$$x=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}=\frac{-4\pm\sqrt{16+112}}{8}=\frac{-4\pm\sqrt{128}}{8}$$
What's next?
| Next is simplifying:
$$x=\frac{-4\pm\sqrt{4^2-4\cdot4\cdot-7}}{2\cdot4}=\frac{-4\pm\sqrt{16+112}}{8}=-\frac12\pm\frac18\cdot8\sqrt{2}=-\frac12\pm\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$
| Hint: $$x^2 + 1 = x^2 + xy + yz + zx = (x+y)(x+z)$$
Also, you will need the inequality
$$
\sqrt{AB}\le \frac12(A+B)
$$
Solution:
$$ \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}
\\= \frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(x+z)}}
+\frac{z}{\sqrt{(x+y)(x+z)}}
\\
\le \frac12\left[
\frac ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the derivatives to $f(x)=4/x^2$ and $g(t)=(t-5)/(1+\sqrt{t}\,)$ I have these two assignments:
Find the derivatives to (a) $f(x)=4/x^2$ and b) $g(t)=(t-5)/(1+\sqrt{t}\,)$ by using the definition $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$$
a) This is my attempt at (a); am I correct?
$$\lim_{h \to 0} \frac{\lef... | *
*First function
$$f'(x)=\lim_{h \to 0} \frac{\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)=\lim_{h \to 0}\frac{1}{h}\frac{-4h^2-8xh}{(x+h)^2x^2}=\lim_{h \to 0}\frac{1}{h}\frac{-4h(h+2x)}{(x+h)^2x^2}= \lim_{h \to 0}\frac{-4(h+2x)}{(x+h)^2x^2}= \frac{-8x}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the element $z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to $U_{12}$ Show that the element $\displaystyle z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to U12
What I don't understand:
*
*In what way ... | I might be misunderstanding your question, but here goes nothing!
It seems like we want to prove that $z^{12} = 1$. Now
$$ z = i(\cos\frac{\pi}{3} - i\sin\frac{\pi}{3}) = i(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = ie^{-i\frac{\pi}{3}} = ie^{-i\frac{2\pi}{6}} $$
$$z^6 = i^6(e^{-i\frac{2\pi}{6}})^6 = -1$$
$$z^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$
*
*The below problem is taken from Joseph Edwards book Integral Calculus for beginners.
How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2... | First of all, we use De Moivres’ Theorem to express $\sin 5x $ in terms of $\cos x$ and $\sin x$.
$$
\begin{aligned}
&\cos 5 x+i \sin 5 x\\=&(\cos x+i \sin x)^{5} \\
=& \cos^{5} x+5 i \cos ^{4} x \sin x-10 \cos^{3} x \sin ^{2} x-10 i \cos ^{2} x \sin ^{3} x+5 \cos x \sin ^{4} x+i \sin ^{5} x
\end{aligned}
$$
Comparing... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 3
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $... | I do not follow "So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are..." and I think that is the source of trouble.
If $9$ divides $30+x+y$, then either $30+x+y=36$ or $30+x+y=45$.
If $11$ divides $12+x-y$, then for sure $12+x-y=11$.
Add one of the first two equations to the second. Either $42+2x=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Finding an angle $\theta$ in a complex number If we know that $z = \frac{1}{\sqrt2}(\cos\theta+i\cdot\sin\theta)$ and also that $z = \frac{(\sqrt3-1)+i(\sqrt3+1)}{4}$
How can I find $\cos\theta$ and $\sin\theta$? Using a calculator it gives me the angle $\frac{5\pi}{12}$. I tried the following:
$$\cos\theta = \frac{\Re... | $$
\cos(\frac{5\pi}{12}) = \cos(\frac{3\pi}{4}-\frac{\pi}{3})=\cos(\frac{3\pi}{4})\cos(\frac{\pi}{3})+\sin(\frac{3\pi}{4})\sin(\frac{\pi}{3})=\frac{-\sqrt{2}}{2}\cdot\frac{1}{2}+\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = $$ $$\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{3}-1}{2\sqrt{2}}
$$
Now work backwards, amigo.
| {
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"url": "https://math.stackexchange.com/questions/984547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I u... | To continue the work of Anastasiya-Romanova but not using complex analysis
For $I_1$:
Notice:
$p(z)=1+2z^2+2z^4=2\Big(z^2+az+b\Big)\Big(z^2-az+b\Big)$
where $a=\sqrt{\sqrt{2}-1}$ and $b=\dfrac{\sqrt{2}}{2}$
Therefore:
$f(z)=\displaystyle \dfrac{2z^2}{1+2z^2+2z^4}=\dfrac{z}{2a\Big(z^2-az+b\Big)}-\dfrac{z}{2a\Big(z^2+az... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989021",
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"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 1
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Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice spec... | We can use a contour integral in the complex plane, as I showed here for the case $n=3$. Now, however, we use
$$\oint_C dz \frac{\log^2{z}}{z^n-1}$$
where $C$ is the modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.
Let's evaluate this integral over the contours. ... | {
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"source": "stackexchange",
"question_score": "13",
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Prove the $n$th Fibonacci number is the integer closest to $\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n$
Prove that the $n$th Fibonacci number $f_n$ is the integer that is closest to the number $$\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n.$$
Hi everyone, I don't really understand the prob... | We prove the hint by induction on $n$. For $n = 1$, $LHS < \dfrac{1}{\sqrt{5}} < \dfrac{1}{2}$, so the statement is true for $n = 1$. Assume it is true for $n = k$, that is: $|R_k|=\left|\dfrac{1}{\sqrt{5}}\cdot \left(\dfrac{1-\sqrt{5}}{2}\right)^k\right| < \dfrac{1}{2}$, we have:
$|R_{k+1}| = \dfrac{1}{\sqrt{5}}\cdot... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove this formula for $\pi$ I have to use a certain approximation for $\pi$ for my computer science class, but I don't really understand what's going on, other than that this is related to the Taylor polynomial for arctangent. Could someone please prove that this is equal to $\pi$?
It's on the first formula under the... | To give an explanation:
It is well known that $\tan'(x)=1+\tan^2(x)$. By deriving $\tan(\arctan(x))=x$ we get: $$\arctan'(x)\cdot\tan'(\arctan(x))=\arctan'(x)\cdot\left(1+(\tan(\arctan(x)))^2\right)=\arctan'(x)\cdot\left(1+x^2\right)=1\iff\arctan'(x)=\frac{1}{1+x^2}$$ and by the standard formula of geometric series we ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ Problem :
The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$
Can I use this in numerator and denominator :
The minimum value of $a +\frac{1}{a}$
Using A... | hint: let
$$f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+x^{-3}}=3\left(x+\dfrac{1}{x}\right)\ge 6$$
because
$$\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2=\left(x+\dfrac{1}{x}\right)^6-\left(x^3+\dfrac{1}{x^3}\right)^2$$
so
$$f(x)=\l... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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On the determinant of a matrix The matrix
$$\left[\begin{array}{ccc}
30&20&30\\
40&50&20\\
30&30&20
\end{array}\right]$$
I tried solving it for myself and got $12000$, but math way tells me its $-1000$.
I dont understand how you get a negative, Where did I mess up my calculations?
I did
30 x ((50x20)-(30x20)) = 12000
... | We want to find the determinant of
$$\begin{pmatrix} 30 & 20 & 30 \\
40 &50 &20 \\
30 &30 &20 \end{pmatrix}.$$
There are many ways of writing this down, but let's do it in the expand-by-minors way. First, since everything is divisible by $10$, let's factor out a $10$. Since it's a $3 \times 3$ matrix, this affects the ... | {
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$ 1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})... $ Converges By comparison test? For what value of real constant $a$ does the following series converge?
$$
1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqr... | Okay, I think I can guess at the pattern with the most recent addition
$$
\begin{align}
&\sum_{n=0}^\infty\left(\frac1{\sqrt{4n+1}}+\frac1{\sqrt{4n+3}}-\frac{a}{\sqrt{2n+2}}\right)\\
&=\sum_{n=0}^\infty\frac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-a\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}\\
&=\sum_{n=0}^\infty\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Equivalence of summations Show that $$\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2}\equiv \frac{1}{n}\sum^{n}_{i=1}x_{i}^{2} - \bar{x}^{2}.$$ Note that $\bar{x} = \frac{1}{n}\sum^{n}_{i=1} x_{i}$. So I have started by:
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{... | \begin{align}
\sum_{i=1}^n (x_i-\bar x)^2 & = \sum_{i=1}^n (x_i^2 -2\bar x x_i + \bar x^2) \\[10pt]
& = \left(\sum_{i=1}^n x_i^2\right) -2\bar x \left( \sum_{i=1}^n x_i\right) + n\bar x^2 \\[10pt]
& = \left(\sum_{i=1}^n x_i^2\right) -2\bar x \left( n\bar x\right) + n\bar x^2 = \cdots\cdots\cdots
\end{align}
| {
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Evaluate $ \lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n+r}$ by expressing as Riemann sum $$ \lim_{n\to \infty} \left(\frac{1}{n+1} + \frac{1}{n+2} + ..+ \frac{1}{2n}\right)$$
How do i find the limit by expressing it as a definite integral of an appropriate function via Riemann Sums?
I do know that for riemann sums
$\Del... | Notice That : $$\frac{1}{n+r}=\frac{1}{n}\frac{1}{1+\frac{r}n}$$
So our some can be written as
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r} =
\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}$$
Now by replacing $\frac{1}{n}$ with $dx$,$\frac{r}{n}$ with $x$ and $\sum$ with $\int$
$$\lim_{n\to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Construct a triangle given an angle and two medians Construct, with ruler and compass, a triangle $ABC$ knowing the angle $\widehat{A}$ and $m_a$ and $m_b$, where $m_a$ and $m_b$ are the medians relative to the vertices $A$ and $B$, respectively.
| If we set $m_a=3x,m_b=3y$ and consider the following construction:
We have that:
$$GU = 2x\sin\varphi,\quad GV=x\sin\varphi,\quad UV=M_BW=3x\sin\varphi,$$
$$M_B V^2 = y^2-x^2\sin^2\varphi,\quad AU=2x\cos\varphi,\quad AW=AU-M_BV, $$
hence:
$$\cot A = \frac{AW}{M_B W}=\frac{2x\cos\varphi-\sqrt{y^2-x^2\sin^2\varphi}}{3x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002418",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we... | Hint: this is also
$$
\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}
=\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\sqrt{\dfrac{c}{1-c}}
$$And then you can for instance use the Lagrange multiplicators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| Since $14+4\sqrt{10}\gt14-4\sqrt{10}\gt0$, the difference of the square roots is equal to something positive, i.e.,
$$\sqrt{14+4\sqrt{10}}-\sqrt{14-4\sqrt{10}}=u\gt0$$
If we square both sides, we have
$$\begin{align}
u^2&=(14+4\sqrt{10})-2\sqrt{14+4\sqrt{10}}\sqrt{14-4\sqrt{10}}+(14-4\sqrt{10})\\
&=28-2\sqrt{14^2-4^2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^... | For $\boldsymbol{r\in(-1,1)}$
$$
\begin{align}
1
&=\frac1{2\pi i}\oint\frac1{z-r}\mathrm{d}z\tag{1}\\
&=\frac1{2\pi i}\int_{-\pi}^\pi\frac1{e^{i\theta}-r}\mathrm{d}e^{i\theta}\tag{2}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-re^{i\theta}}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{3}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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A few inequality problems I can't seem to get; Cauchy and the Mean Inequality Chain
*
*Prove that $ \sqrt{\frac{2x^2-2x+1}{2}}\geq\frac{1}{x+\frac{1}{x}} $ for $ 0 < x < 1. $
This one seems reminiscent of the quadratic mean on the left, maybe $\sqrt{\frac{(x-1)^2+x^2}{2}}$, but I can't find a way to compare it to t... | For the first question, you can try as follows.
Since
\begin{align*}
\sqrt{\frac{2x^2-2x+1}{2}}-\frac{x}{x^2+1}=\frac{1}{2}\cdot \frac{\sqrt{4x^2-4x+2}\cdot (x^2+1)-2x}{x^2+1}=:f(x),
\end{align*}
if we set
\begin{gather*}
a(x):=\sqrt{4x^2-4x+2}=\sqrt{4\left(x-\frac{1}{2}\right)^2+1},
\end{gather*}
the numerator of $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| This is the correct formula if you want to square: $ (\sqrt{x} +x)^2 = x^2 + 2x\sqrt{x} +x$.
| {
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"source": "stackexchange",
"question_score": "1",
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Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$ \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$... | The integral is undefined at 3. We must use an improper integral.
$$
\int_0^3 \frac{1}{\sqrt{3-x}} dx = \lim_{x \to 3^-}\int_0^b \frac{1}{\sqrt{3-x}} dx
$$
using a u substitution where u = 3 - x, then du = -dx which gives the following:
$$
\int \frac{-1}{\sqrt{u}}du = -2\sqrt{u}
$$
Putting this back in terms of x giv... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Here is a "by hand" method.
Note that $\sqrt [3] 4 =(\sqrt [3] 2)^2$ and $(\sqrt [3] 4)^2=2\sqrt[3] 2$.
This means that the powers of $y=\sqrt [3] 2-\sqrt [3] 4$ can all be written as polynomials in $x=\sqrt [3] 2$.
But since $x^3=2$ we can always reduce the polynomial to at most quadratic.
$$y=x-x^2$$
$$y^2=x^2-2x^3+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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Solving Trigonometric Limits compute the following limit:
$$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\ \mbox{as}\ x\to0.$$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
but without ... | Note that the numerator of the expression under limit operation can be expressed as $$\begin{aligned}f(a, x) &= \cos (a + 2x) - 2\cos(a + x) + \cos a\\
&= \cos (a + 2x) + \cos a - 2\cos(a + x)\\
&= 2\cos(a + x)\cos x - 2\cos(a + x)\\
&= 2\cos(a + x)(\cos x - 1)\end{aligned}$$ and hence the desired limit $L$ can be eval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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telescopic series & sum of sequences - how to? The entire web shows the same example basicly(or something extremely similar): $$\sum_{n=2}^\infty \frac{-2}{(n+1)(n+2)}= \frac{-2}{3}$$
Is there a way to find this final answer without doing the pretty long procedure?
In addition to it I want to apply the telescopic s... | By partial fractions,
$$\frac{-2}{(n+1)(n+2)} \equiv \frac{A}{n+1} + \frac{B}{n+2}$$
For some $A$ and $B$.
Thus,
$$-2 \equiv A(n+2)+B(n+1)$$
By letting $n=-1$ and $n=-2$ in turn, we can see that $A=-2$ and $B=2$. So the sum is now:
$$\sum_{n=2}^{n=k}{\left(\frac{2}{n+2} - \frac{2}{n+1}\right)}=2\sum_{n=2}^{n=k}{\left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers n such that $\;\frac{n^2-9}{n^2-5n+4}$ is an integer.
Find all integers such that $\;\dfrac{n^2-9}{n^2-5n+4}\;$ is an integer.
I am really struggling to figure this out. I can tell that -3,3, and 5 are solutions but I don't know how to show that these are the only solutions or if they even are the o... | The denominator is $(n-1)(n-4)$. If $n^2\ne 9$, we must have that $n-4$ divides $n^2-9$.
Note that $\frac{n^2-9}{n-4}=n+4+\frac{7}{n-4}$. So the only candidates are those $n$ such that $n-4$ divides $7$. That gives the short list $n=5$, $n=3$, $n=11$, and $n=-3$. Test them all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n ... | A slightly better bound would be
$$\frac{2^n+3}{3^n-1}=\frac{2^n(1+3\cdot 2^{-n})}{3^n(1-3^{-n})}\leq \frac{5/2}{2/3}\frac{2^n}{3^n}$$
| {
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"url": "https://math.stackexchange.com/questions/1020306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove a limit for $g(x)$ with the definition of limit only We have the function $g(x) = x^3+1$
Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.
I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) ... | $$|x^2+2x+4|=|x^2-4x+4+6x-12+12|=|(x-2)^2+6(x-2)+12| \leq |\delta^2+6\delta+12|$$
So for small $\delta$ you have (for example) $|x^2+2x+4|<13$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 +... | Another way: Sum the obvious AM-GMs:
$$\tfrac12 x^6 + \tfrac12 x^4 \ge x^5, \quad \tfrac12 x^4 + \tfrac12 x^2 \ge x^3$$
$$\tfrac12 x^2 + \tfrac34 \ge \sqrt{\tfrac32}|x| \ge x \implies \tfrac12 x^2 + \tfrac34 > x$$
with $\frac12x^6 \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Remainder of a combination Problem from a contest:
What is the remainder when $\binom{169}{13}$ is divided by $13^5$?
I thought that Wolstenholme's/Babbage's would help, but not entirely sure how.
| $$\binom{169}{13}=13\cdot\frac{168\cdot \ldots \cdot 157}{12!}=13\cdot\frac{(13^2-1)\cdot\ldots\cdot(13^2-12)}{12!}$$
$$\frac{1}{12!}\prod_{k=1}^{12}(13^2-k)=\prod_{k=1}^{12}\left(1-\frac{13^2}{k}\right) = 1-13^2 H_{12}+C\cdot13^4$$
and since $H_{12}\equiv 0\pmod{13^2}$ by Wolstenholme's theorem, it follows that:
$$\fr... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$8x +9y = 5$ where $x,y \in \mathbb{Z}$
Solve the following Diophantine equation algebaically: $$8x+9y=5$$
Give 3 possible solutions for the equation
I have the following:
The Diophantine equation has solutions $x,y \iff 8x=5\mod{9}$ has a solution $x \equiv\mod{9}$
Since $\gcd(8,9)=1$, by Bezout's Lemma, for $r,t ... | Choose $x$ or $y$ ? Let's solve for $x$: $x = \dfrac{5-9y}{8} = \dfrac{5-y}{8} - y \Rightarrow 5-y = 8k \Rightarrow y = 5 - 8k \Rightarrow x = k - (5-8k) = 9k-5$. Thus:
$(x,y) = (9k-5,5-8k), k \in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Indefinite integral evaluation of fraction Here's the indefinite integral to evaluate:
$F(x) = \int \left(\frac{3x + 4}{x + 3}\right)dx$
Here are my steps (verbose):
$$\begin{align}
u & = x + 3 \\
(x + 3)dx & = du \\
\left(\frac{d}{dx}(x + 3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + \frac{d}{dx}(3)\right)dx & = du \\... | You don't lose any information when you split up the equation; they are equal quantities. The fact that one method nets you a "$9+C$" while another method only nets you another constant $"D"$ is perfectly valid. If you had initial conditions to work with, you would simply find that $9+C = D$, so your integral results ... | {
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"url": "https://math.stackexchange.com/questions/1026112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I hav... | If you consider these imaginary parts of elliptic integrals as closed-forms, then
$$\frac{1}{2}-\frac{\sqrt 2}{3}-\frac{\sqrt{2} \log
\left(\sqrt{2}-1\right)}{8}+\frac{\sqrt{2} \log
\left(\sqrt{2}+1\right)}{8}-\frac{\sqrt 2}{8}\Im\left(F\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)+\frac{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
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Prove that $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$ How can I prove $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$?
We have
$(a+b-c)^2+=a^2+b^2+b^2+2ab-2bc-2ca$,
$(b+c-a)^2+=b^2+c^2+a^2+2bc-2ca-2ab$
$(c+a-b)^2+=c^2+a^2+b^2+2ca-2ab-2bc$
But I don't know how to show the required result?
Please give hint. Thank you
| We have
$$(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 = 3a^2+3b^2+3c^2-2ab-2bc-2ca$$
From AM-GM, we have $a^2+b^2 \geq 2ab$, $b^2+c^2 \geq 2bc$ and $c^2+a^2 \geq 2ca$. Hence,
$$2(a^2+b^2+c^2) \geq 2(ab+bc+ca) \implies 3(a^2+b^2+c^2) \geq 3ab+3bc+3ca$$
Hence,
\begin{align}
(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 & = 3a^2+3b^2+3c^2-2ab-2bc-2ca\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find this limit $\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$
Give real numbers $a,b$ such that $0<a<b$ and $m=\dfrac{a+b}{2}<\dfrac{\pi}{4}$,Evaluate
$$\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$$
where
$$f(x)=\dfrac{(1+\cos{(2m-2x)})\cos{(a-x)}\cos{(... | I think your first identity is wrong:
$$\frac{1+\cos(2m-2x)}{\sin(2m-2x)}=\tan\left(x-m+\frac{\pi}{2}\right)$$
So your function simplifies to:
$$ f(x)=\tan\left(x-m+\frac{\pi}{2}\right)\tan\left(\frac{\pi}{4}+\frac{a-x}{2}\right)\tan\left(\frac{\pi}{4}+\frac{b-x}{2}\right)$$
You can use the triple tangent identity: if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$ \int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right)$ How to show the following identity from Wolfram?
$$
\int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right) =-2\pi \sqrt 2\left[\frac{5 \pi^2}{3} + 16 \left(3 \ln 2 + G - 4 \right) \right].
$$
Here $G$ is Ca... | We start from the integral
$$
\int_{0}^{\infty} \frac{x^u }{(1+s x)(1+t x) }dx= \frac{\pi}{\sin \pi u} \cdot \frac{s^u - t^u}{s^u t^u (s-t)}
$$
for $s$, $t>0$ and $0 < u < 1$.
Integrating with respect to $t$ we get
\begin{eqnarray}
\int_{0}^{\infty} \frac{x^u \log ( 1 + \frac{1}{t x})}{(1+s x)}= - \frac{\pi}{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
For what $a,b$ such that $ax^2+by^2 = z^2$? This post made me think about this question. What is the criterion on positive integer $a,b$ such that,
$$ax^2+by^2 = z^2$$
can be solved in positive integers $x,y,z$?
(Three broad classes are: 1) $a =\square$; 2) $a+b = \square$; and 3) if the Pell-like eqn $z^2-by^2 = a$ i... | I'm probably the only proponent of this approach for solving Diophantine equations.
Prefer to understand why the solution is, instead of using modular arithmetic. Especially for complex equations modular arithmetic to use makes no sense. And this approach allows us immediately to find out whether there is a solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}... | The final quadratic is an upward drawn parabola. You can solve for $4sin^2 t -2 sin t -1 =0 $ and take the values for t such that sint >0 in [-$\frac{\pi}{2}$,$\frac{\pi}{2}$].
The roots are $\frac{1-\sqrt{5}}{4}$ and $\frac{1+\sqrt{5}}{4}$. So $\frac{1+\sqrt{5}}{4}$ =< sint and sin t <= $\frac{1-\sqrt{5}}{4}$ in [... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\qu... | $$\underset{n\rightarrow\infty}{\lim}\frac{v_{n+1}}{v_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1\right)\cdots\left(n+1+n-1\right)\left(n+1+n\right)\left(n+1+n+1\right)}{n\cdots\left(n+n\right)}=$$
$$=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1+n\right)\left(n+1+n+1\right)}{n}=\infty$$ because you h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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.