Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expressio... | Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can... | By C-S
$$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$
Thus, it's enough to prove that
$$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0,$ where $f$ decreases, which says that it... | {
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How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| $x^3+1=0\implies (x+1)(x^2-x+1)=0$
If $x+1=0,x=-1$
else $x^2-x+1=0$ then $x=\frac{1±\sqrt{1-4}}{2\cdot1}=\frac{1±\sqrt3i}{2}$ using the well known Quadratic Formula.
Alternatively, using Euler's identity and Euler's formula like other two solutions,
for $x^n=-1=e^{i(2m+1)\pi}$ as $e^{i\pi}=-1$ where $m$ any any in... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| It's a convexity inequality: the function $f(x) = \frac{1}{x}$ is convex on $\mathbb{R}^{+*}$ (its second derivative is positive), so for any $a,b,c\in\mathbb{R}^{+*}, \frac{1}{3}(f(a)+f(b)+f(c))\geq f(\frac{a+b+c}{3})$, which is $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq\frac{3}{a+b+c}$, or finally $\frac{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Solving $\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$ What is the answer of this:
$\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$
| $\textbf{Hint}$ : First, try to relate this to the pythagorean theorem. Also note.
$\sqrt{(x - 2)^2 + (y - 1)^2} = \sqrt{2}$
$(x - 2)^2 + (y - 1)^2 = 2$
Think about circles.
| {
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Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
| Note that
$$ \tan \left(\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) \right)=z
$$
so that $$\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) = \arctan(z) + n \pi $$
for the appropriate integer... | {
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"timestamp": "2023-03-29T00:00:00",
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| Let $x=\sqrt{1+i\sqrt3}+\sqrt{1-i\sqrt3}$. Then
$$x^2=(1+i\sqrt3)+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}+(1-i\sqrt3)=2+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}$$
so
$$(x^2-2)^2=4(1+i\sqrt3)(1-i\sqrt3)=4(1+3)=16$$
so
$$x^4-4x^2-12=(x^2-6)(x^2+2)=0$$
Thus $x\in\{\sqrt6,-\sqrt6,i\sqrt2,-i\sqrt2\}$, so it remains to determine which ... | {
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Solving for x with exponents (algebra) So I am trying to help a friend do her homework and I am a bit stuck.
$$8x+3 = 3x^2$$
I can look at this and see that the answer is $3$, but I am having a hard time remembering how to solve for $x$ in this situation.
Could someone be so kind as to break down the steps in solving... | Here's a way without the formula. Using "completing the square"
\begin{align*}
3x^2-8x-3&=0\\
x^2-\frac{8}{3}x-1&=0\\
\left(x-\frac{4}{3}\right)^2-\frac{16}{9}-1&=0\\
x-\frac{4}{3}&=\pm \sqrt{\frac{25}{9}}\\
x &= \frac{4}{3}\pm \frac{5}{3}\\
x = \frac{9}{3} = 3\ \ &\mathrm{or}\ \ x = -\frac{1}{3}.
\end{align*}
| {
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Identity proof $(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$ In a proof from a textbook they use the following identity (without proof):
$(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$
Is there an easy way to prove the above? I suppose maybe an induction proof will be appropriate, but I would really lik... | Yes there is. First let's look at the series:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} \, . $$
This is a geometric series with $n$-terms, whose first term is $x^{n-1}$ and whose common ratio is $y/x$. Applying the standard formula we obtain:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum\limits_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$ by induction.
Theorem (Principle of mathematical induction):
Let $G\subseteq \mathbb{N}$, suppose that
a. $1\in G$
b. if $n\in \mathbb{N}$ and $\{1,...,n\}\subseteq G$, then $n+1\in G$
Then $G=\mathbb{N}$
Proof by Induction
Prove that
... | You can also notice that $$\sum\limits_{i=1}^n \frac{i}{2^i} = \sum\limits_{i=1}^n \frac{1}{2^i}+ \sum\limits_{i=2}^{n} \frac{1}{2^i} + \dots + \sum\limits_{i=n}^n \frac{1}{2^i}= \sum\limits_{j=1}^n \sum\limits_{i=j}^n \frac{1}{2^i}$$ So you can find the result without knowing it in advance; this is only a geometric se... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation wi... | Using complex numbers, and setting $z=e^{i\theta}$,
$$a\frac{z-z^{-1}}{2i}+b\frac{z+z^{-1}}2=c,$$
or
$$(b-ia)z^2-2cz+(b+ia)=0.$$
The discriminant is $c^2-b^2-a^2:=-d^2$, assumed negative, then the solution
$$z=\frac{c\pm id}{b-ia}.$$
Taking the logarithm, the real part $$\ln\left(\dfrac{\sqrt{c^2+d^2}}{\sqrt{a^2+b^2}}\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The intersection of a line with a circle Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$
What I did:
$y^2=10-x^2$
$y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$
$ x+ 2 = y=\sqrt{10-x^2}$
If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$
But then, and here is where the problems... | If you do everything correct, the solution of $x+2=-\sqrt{10-x^2}$ would be almost the same - $(1,-3)$, $(-3,1)$. It's refer to intersaction of $y = -(x+2)$ and original circle.
By the way, your answer is correct.
I have no idea how you'd get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2})$.
| {
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Basic Integral question Prove that $$ \int_0^\pi \frac{dx}{\alpha - \cos(x)}\ = \frac{\pi}{\sqrt{\alpha^2 - 1}}$$ for $\alpha > 1$
| Put $z = e^{ix}$ so that $dz = i e^{ix} dx$ and $dx = 1/z/i \,dz$. The integral becomes
$$\frac{1}{2} \int_{|z|=1} \frac{2}{2\alpha - z - 1/z} \frac{1}{iz} dz $$
Simplifying this yields
$$\frac{1}{2i} \int_{|z|=1} \frac{2}{2\alpha z - z^2 - 1} dz$$
The poles are at
$$z = \alpha \pm \sqrt{\alpha^2-1},$$
Of these, only t... | {
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Show $x^2 +xy-y^2 = 0$ is only true when $x$ & $y$ are zero. Show that it is impossible to find non-zero integers $x$ and $y$ satisfying $x^2 +xy-y^2 = 0$.
| Let $g=\gcd(x,y)$. Let $x=rg$ and $y=sg$. Then
$$g^2r^2+g^2rs-g^2s^2=0$$
$$r^2+rs-s^2=0$$
$$s^2-r^2=rs$$
$$(s+r)(s-r)=rs$$
We've divided out all common factors, so $\gcd(r,s)=1$. For any prime $p$, if $p|r$, $p$ does not divide $s$ and therefore divides neither $s+r$ nor $s-r$. The same argument applies if $p|s$. ... | {
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Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.
The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
| Let $y=x+\frac{1}{x}$, try now to express $x$ as a function of $y$.
We have
$$x^2-xy+1=0$$
$$x=\frac{y \pm \sqrt{y^2 -4}}{2}$$
Substitute this value for x in your expression for $f$.
$$f(y)=\left(\frac{y \pm \sqrt{y^2 -4}}{2}\right)^2+\left(\frac{2}{y \pm \sqrt{y^2 -4}}\right)^2$$
$$f(y)=y^2-2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Implicit solution to an ODE I have come accross a problem which leads to the following ODE:
$\frac{dy}{dx}= \frac{y}{x}-\frac{1}{h}\frac{\sqrt{x^2+y^2}}{x}$,
where $h>0$ is a parameter. I am not able to solve it, however maple gives an implicit
solution:
$\frac{x^{1/h}y}{x}+\frac{x^{1/h}\sqrt{x^2+y^2}}{x}=$constant.
I... | The important observation is that th equation is homogeneous. Indeed
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\frac{{\sqrt {{x^2} + {y^2}} }}{x} \cr
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \cr
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sq... | {
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"timestamp": "2023-03-29T00:00:00",
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Why $\cos^2 (2x) = \frac{1}{2}(1+\cos (4x))$? Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$
I don't understand this, how I must to multiply two trigonometric functions?
Thanks a lot.
| Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want.
EDIT
The identity
$$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$
Setting $A = B = \the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Derivative of a split function We have the function:
$$f(x) = \frac{x^2\sqrt[4]{x^3}}{x^3+2}.$$
I rewrote it as
$$f(x) = \frac{x^2{x^{3/4}}}{x^3+2}.$$
After a while of differentiating I get the final answer:
$$f(x)= \frac{- {\sqrt[4]{\left(\frac{1}{4}\right)^{19}} + \sqrt[4]{5.5^7}}}{(x^3+2)^2}$$(The minus isn't behind... | Although collecting powers is a good idea as suggested, things are made clearer after taking formal logarithm before differentiating:
$$\ln f=2\ln x+\frac{3}{4}\ln x -\ln(x^3+2)$$
$$\frac{f'}{f}=\frac{2}{x}+\frac{3}{4x}-\frac{3x^2}{x^3+2}=\frac{11}{4x}-\frac{3x^2}{x^3+2}=-\frac{x^3-22}{4x(x^3+3)}$$
$$f'=-\frac{x^\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
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Adding sine waves of different phase, sin(π/2) + sin(3π/2)? Adding sine waves of different phase, what is $\sin(\pi/2) + \sin(3\pi/2)$?
Please could someone explain this.
Thanks.
| $\sin(\pi+x)=\sin \pi\cos x+\cos\pi\sin x=-\sin x$ as $\sin \pi=0,\cos\pi=-1$
This can be achieved directly using "All-Sin-Tan-Cos" formula or using $\sin 2A+\sin 2B=2\sin(A+B)\cos(A-B)$, $\sin(\pi+x)+\sin x=2\sin\left(x+\frac \pi 2\right)cos\left(\frac \pi 2\right)=0$
So, $\sin(x)+\sin(\pi+x)=0$
More generally, $\sin(... | {
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Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?
| Since
$$
\left(x-\frac{7+\sqrt{37}}2\right)\left(x-\frac{7-\sqrt{37}}2\right)=\left(x-\frac72\right)^2-\frac{37}4=x^2-7x+3\;,
$$
the recurrence relation
$$x_n=7x_{n-1}-3x_{n-2}$$
has solutions
$$
x_n=c_1\left(\frac{7+\sqrt{37}}2\right)^n+c_2\left(\frac{7-\sqrt{37}}2\right)^n\;.
$$
With $c_1=c_2=1$ we have $x_0=2$ and $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of squares in a rectangle Given a rectangle of length a and width b (as shown in the figure), how many different squares of edge greater than 1 can be formed using the cells inside.
For example, if a = 2, b = 2, then the number of such squares is just 1.
| Without loss of generality we may assume that that the width $b$ is $\ge$ to the height $a$. Note that there are $b+1$ vertical lines and $a+1$ horizontal lines in the picture.
If $a\ge 2$, the number of $2\times 2$ squares is $(a-1)(b-1)$. This is because the top left corner of a $2\times 2$ square can be any corner ... | {
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Solve $2a + 5b = 20$ Is this equation solvable? It seems like you should be able to get a right number!
If this is solvable can you tell me step by step on how you solved it.
$$\begin{align}
{2a + 5b} & = {20}
\end{align}$$
My thinking process:
$$\begin{align}
{2a + 5b} & = {20} & {2a + 5b} & = {20} \\
{0a + 5b} & =... | Note that $2a$ must have as its unit digit $0, 2, 4, 6,$ or $8$ (because it's even!).
Similarly, note that $5b$ must have as its unit digit $0$ or $5$.
With a bit of thinking, you can see that for $2a + 5b$ to be $20$, we need to ensure that $2a$ has a unit digit of $0$ (hence $a$ is a multiple of $5$).
So let $a$ be a... | {
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Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \fr... | The mixing ratios give
$$
\begin{bmatrix}
\frac15&\frac45\\
\frac47&\frac37
\end{bmatrix}
\begin{bmatrix}A\\B\end{bmatrix}=\begin{bmatrix}C\\D\end{bmatrix}
$$
That is, $1$ can of $C$ is $\frac15$ can of $A$ and $\frac45$ can of $B$, $1$ can of $D$ is $\frac47$ can of $A$ and $\frac37$ can of $B$.
Because
$$
\begin{alig... | {
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"timestamp": "2023-03-29T00:00:00",
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convergence tests for series $p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$ If the sequence:
$p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$
Prove that the sequence
$((n+1/2)p_n^2)^{n=\infty}_{1}$ is decreasing.
and that the series $(np_n^2)^{n=\infty}_{1}$ is convergent.
Any hints/ answe... | Firstly, it should be noted that $\dfrac{p_{n+1}}{p_n}=\dfrac{2n+1}{2n+2}=\dfrac{n+\frac{1}{2}}{n+1},$ therefore,
$$\frac{a_{n+1}}{a_n}=\frac{\left(n+1+\frac{1}{2} \right)p_{n+1}^2}{\left(n+\frac{1}{2} \right)p_{n}^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)^2}{\left(n+\frac{1}{2} \right)\left(n+1 ... | {
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"answer_id": 0
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Trigonometric Identity - $\tan(\pi/2 -x) - \cot(3\pi/2 -x) + \tan(2\pi-x) - \cot(\pi-x) ...$ $$\tan\left(\frac{\pi}{2} -x\right) - \cot\left(\frac{3\pi}{2} -x\right) + \tan(2\pi-x) - \cot(\pi-x) = \frac{4-2\sec^{2}x}{\tan{x}}$$
L.S.
$= \cot{x} - \tan{x} - \tan{x} + \cot{x}$
$= 2\cot{x} - 2\tan{x}$
$= 2\left(\frac{\co... | You have simplified your initial expression to $2 \cot(x) - 2 \tan(x)$. Make use of the following trigonometric identities: $$\cot(x) = \dfrac1{\tan(x)}$$ $$\sec^2(x) - \tan^2(x) = 1$$ and simplify to get what you want.
Move your mouse over the gray area below for a complete solution.
$$2 \cot(x) - 2 \tan(x) = 2 \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement hold... | If we define
$$
f_n(t)=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{1}
$$
then we have
$$
\begin{align}
f_n'(t)
&=\sum_{k=0}^\infty\binom{2k+n}{k}kt^{k-1}\\
&=\sum_{k=0}^\infty\binom{2k+n-1}{k-1}(2k+n)t^{k-1}\\
&=\sum_{k=0}^\infty\binom{2k+n+1}{k}(2k+n+2)t^k\\[6pt]
&=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)\tag{2}
\end{align}
$$
If w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/237810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 0
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numbers' pattern It is known that
$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\
9+10+11+12 &=& 13+14+15 \\\
16+17+18+19+20 &=& 21+22+23+24 \\\
25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots
\end{array}$$
There is something similar for square numbers:
$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13... | There is the famous $3^3+4^3+5^3=6^3$. But I don't know if there are others.
At least not where the whole sequence is consecutive...
One gets the problem of three triangular numbers T1 T2 T3 for which one wants to solve $T1^2+T2^2=2 \cdot T3^2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{la... | $$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$
Take sum from k=1 to n
$$\sum_{k=1}^n(k+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
$$\sum_{k=2}^{n+1}k^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
Add 1 on both sides
$$\sum_{k=1}^{n}k^4 + (n+1)^4 = ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Finding eigenvalues and eigenvectors, $2\times2$ matrices
Find all eigenvalues and eigenvectors:
a.) $\pmatrix{i&1\\0&-1+i}$
b.) $\pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}$
For a I got:
$$\operatorname{det} \pmatrix{i-\lambda&1\\0&-1+i-\lambda}= \lambda^{2} - 2\lambda i + \lambda - i - 1
$$
For b... | Basic tools
For $2 \times 2$ matrices the characteristic polynomial is:
$$
p(\lambda) = \lambda^{2} - \lambda\, \text{tr }\mathbf{A} + \det \mathbf{A}
$$
The roots of this function are the eigenvalues, $\lambda_{k}$, k=1,2$.
The eigenvectors solve the eigenvalue equation
$$
\mathbf{A} u_{k} = \lambda_{k} u_{k}
$$
Ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Value of $\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ I remember that a couple of years ago a friend showed me and some other people the following expression:
$$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$
As shown below, I can prove that this limit exists by the monotone convergence theorem. I als... | We can prove this by simple squeezing.
Using the fact that $A>B>0$ implies $A^n-B^n \leq n(A-B)A^{n-1}$, and the facts that $e^{-x}\leq\frac{1}{1+x},e^x\geq 1+x+\frac{x^2}{2}$, we have:
$$ e^{-k} - \left(1-\frac{k}{n}\right)^n \leq n\left(e^{-k/n}-1+\frac{k}{n}\right)e^{-k\,\frac{n-1}{n}}\leq e\cdot\frac{k^2 e^{-k}}{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/244657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?
| $\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
As $\angle DAB=18^\circ, \angle DAC=(42-18)^\circ=24^\circ, $
Similarly, $\angle CBD=18^\circ$
Let $\angle ACD=x,$ so, $\angle DCB=96^\circ-x$
So, in $\triangle ADC, \angle ADC=180^\circ-(24^\circ+x)=156^\circ-x$
Similarly, from $\triangle BCD, \angle BDC=72^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Arc Length Problem: Assimilation of A Constant The problem I am working on is,
I understand the motivation of the step, but I am not quite sure how the author the solution manual absorbs the constant into the other term. What is the process?
EDIT:
Here is another one:
Here is my attempt to try and figure out what the... | $\quad\quad\quad$First question: $$1 + \frac14\left(x^8 - 2 +\frac{1}{x^8}\right)\tag{1.1}$$
$$=\frac14\cdot 4 + \frac14\left(x^8 - 2 + +\frac{1}{x^8}\right)\quad=\quad\frac14\left(4 + x^8 -2 +\frac{1}{x^8}\right)\tag{1.2}$$
$$=\frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$
$$=\frac14\left(x^4 + \frac{1}{x^4}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An upper bound for $\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$? Does anyone know of a reasonable upper bound for the following:
$$\sum_{i = 1}^m \frac{\binom{i}{k}}{2^i},$$
where we $k$ and $m$ are fixed positive integers, and we assume that $\binom{i}{k} = 0$ whenever $k > i$.
One trivial upper bound uses the identity ... | I like Pot's argument better, but here's another approach for those who may be interested.
I get an upper bound of $2$, independent of $k$ and $m$.
Applying summation by parts, we have, for $k \geq 2$,
$$\begin{align*}
\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} &= \binom{m}{k} \sum_{i=1}^m \frac{1}{2^i} - \sum_{i=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/247261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Rationalizing the denominator of $\frac {\sqrt{10}}{\sqrt{5} -2}$ I have the expression
$$\frac {\sqrt{10}}{\sqrt{5} -2}$$
I can't figure out what to do from here, I can't seem to pull any numbers out of either of the square roots so it appears that it must remain as is.
| To rationalize the denominator i. e. turn the denominator rational multiply both numerator and denominator by
the conjugate of the denominator $-\sqrt{5}-2$ or its symmetric $\sqrt{5}+2$, expand both and
simplify
$$\begin{eqnarray*}
\frac{\sqrt{10}}{\sqrt{5}-2} &=&\frac{\sqrt{10}\left( \sqrt{5}+2\right) }{
\left( \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to simplify polynomials I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
I tried combining like terms
$$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
$$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$
$$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$
It says the answer is $$3x^4-7x^3+x^2+8x+17$... | Observe the magical power of color:
$$\color{blue}{5}x^\color{blue}{2}+3x^4-7x^3+\color{green}{5}x+\color{orange}{8}+\color{blue}{2}x^\color{blue}{2}+(\color{green}{-4})x+\color{orange}{9}+(\color{blue}{-6})x^\color{blue}{2}+\color{green}{7}x.$$
Instead of Color-Me-Elmo, we have Color-Me-Like-Terms-And-Combine (not as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Sum of infinite series with arctan: $\sum_{n=1}^{\infty}\left(\arctan\left(\frac{1}{4}-n\right)-\arctan\left(-\frac{1}{4}-n\right)\right)$ I'm trying to find the value of the following sum (if exist):
$$\sum_{n=1}^{\infty}\left(\arctan\left(\frac{1}{4}-n\right)-\arctan\left(-\frac{1}{4}-n\right)\right)$$
where, $\arcta... | Consider
$$\tag{1}f(x):=\sum_{n=1}^\infty \arctan\left(\left(\frac 14-n\right)x\right)-\arctan\left(\left(-\frac 14-n\right)x\right)$$
and let's rewrite the derivative of $f$ :
\begin{align}
\tag{2}f'(x)&=\frac 4{x^2}\sum_{n=1}^\infty\frac{1-4n}{(4n-1)^2+\bigl(\frac 4x\bigr)^2}-\frac{-1-4n}{(4n+1)^2+\bigl(\frac 4x\bigr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Show that the expression, $\frac{(n)(n+1)}{2}$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$. $F(n) =\frac{(n)(n+1)}{2}$
Show that $F(n)$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$.
Let, $n = 2$
$F(2) =\frac{2(2+1)}{2}$
$F(2) = 3$
Let, $n = 4$
$F(4) =\frac{4(... | $n$ and $n+1$ are relatively prime, or perhaps more simply, if $n$ is even, then $n+1$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Evaluate
$$
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
$$
The answer is $\frac{1}{2}$, have no idea how to arrive at that.
| Alternatively:
$$ \sqrt{\frac{x^3}{x-1}} - x = x\sqrt{\frac{x}{x-1}} - x = x\left(\sqrt{1+\frac{1}{x-1}}-1\right) $$
For small $\alpha$ we have $\sqrt{1+\alpha}\approx 1+\alpha/2$, so for large $x$ we get
$$\cdots \approx x\left(1+\frac{1}{2x-2}-1\right) = \frac{x}{2x-2} \to \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
How to find $(-64\mathrm{i}) ^{1/3}$? How to find $$(-64\mathrm{i})^{\frac{1}{3}}$$
This is a complex variables question.
I need help by show step by step.
Thanks a lot.
| If you transform $-64i$ to polar form, you get $r=\sqrt{0+(-64)^2}=64$ and $\theta=-\pi/2$.
Then you have $$(-64i)^{1/3} = r^{1/3}\cdot (\cos(\theta*\frac{1}{3})+i\sin(\theta*\frac{1}{3})) = 64^{1/3}\cdot (\cos((-\pi/2)*\frac{1}{3})+i\sin((-\pi/2)*\frac{1}{3})$$
$$= 4\cdot (\cos(-\pi/6)+i\sin(-\pi/6))$$
Given that
$$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to integrate this $\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}} $ without trigonometric substitution? I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$$
| $$\int\frac{dx}{(a^2+x^2)^{\frac n2}}=\int1\cdot \frac1{(a^2+x^2)^{\frac n2}}dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}-\int\left(\frac{-n}2\frac{2x\cdot x}{(a^2+x^2)^{\frac n2+1}} \right) dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}+n\int \left(\frac{(a^2+x^2-a^2)}{(a^2+x^2)^{\frac n2+1}}\right)dx $$
$$=\frac x{(a^2+x^2)^{\frac n... | {
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"url": "https://math.stackexchange.com/questions/260831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Using product and chain rule to find derivative. Find the derivative of
$$y =(1+x^2)^4 (2-x^3)^5$$
To solve this I used the product rule and the chain rule.
$$u = (1+x^2)^4$$
$$u' = 4 (1+x^2)^3(2x)$$
$$v= (2-x^3)^5$$
$$v' = 5(2-x^3)^4(3x^2)$$
$$uv'+vu'$$
$$((1+x^2)^4)(5(2-x^3)^4(3x^2)) + ((2-x^3)^5 )(4 (1+x^2)^3(2x... | The problem is in your differentiation of $$v= (2-x^3)^5$$
You have: $$v'= 5(2-x^3)^4(3x^2)$$
However, the derivative of $2-x^3$ is $-3x^2$. Thus, $$v' = 5(2-x^3)^4(-3x^2)=-15x^2(2-x^3)^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Working out digits of Pi. I have always wondered how the digits of π are calculated. How do they do it?
Thanks.
| From one of my favorite mathematicians,
$$
\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k \ge 0} \frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}.
$$
Looking here, we see some many formulae. Some of particular interest are:
$$
\pi=\sum_{k \ge 0}\frac{3^k-1}{4^k}\zeta(k+1),\quad \zeta(s)=\sum_{k\ge 0}k^{-s}.
$$
$$
\frac{1}{6}\pi^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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What are the A,B,C parameters of this ellipse formula? I am looking at
$$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$$
This is a rotating ellipse formula, where $h,k$ are the centroid of the ellipse. I have tried looking around for $A,B,C$ parameters, and I see that they are from Quadratic formula. But to be frank,... | I would like to explain the transforms on a picture.
$$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1$$
$$x=x'.\cos \alpha - y'. \sin \alpha $$
$$y=x'.\sin \alpha + y'. \cos \alpha $$
We can get easily the result above from the picture
$$\cos \alpha. x=\cos \alpha (x'.\cos \alpha - y'. \sin \alpha) $$
$$\sin \alpha. y=\sin \al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)... | Cauchy-Swartz:
$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+ \cdots +\frac{a_n^2}{b_n} \geq \frac{(a_1+a_2+ \cdots +a_n)^2}{b_1+b_2+ \cdot +b_n}$
So,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$
| {
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"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Simple integral help How do I integrate $$\int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx$$
Where $\lceil x \rceil $ is the ceiling function, and $\left\{x\right\}$ is the fractional part function
| The main idea is to divide $(0,1)$ into "good" intervals. I'lll give only the main steps of computation
$$
\int\limits_{(0,1)} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx
=\sum\limits_{n=1}^\infty\int\limits_{n\leq \frac{1}{x}<n+1} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
| Exercises:
Try dividing $420 - 1$ by integers $2, 3, ..., 7$ and note the respective remainders are $1, 2, 3, 4, 5, 6$.
Then note that $420 = \textrm{lcm}\,(2, 3, 4, 5, 6, 7)$
Any $n = k\cdot 420 - 1$, $k\in \mathbb{Z}$, will yield the same remainders when divided by $2, 3, ..., 7$.
Try dividing $840 - 1 = 839\,$ by... | {
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"url": "https://math.stackexchange.com/questions/268619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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How can I find two independent solution for this ODE? Please help me find two independent solutions for $$3x(x+1)y''+2(x+1)y'+4y=0$$ Thanks from a new beginner into ODE's.
| We suppose that the $\displaystyle y=\sum_{n=0}^{\infty}a_{n}x^n$ is a solution.
Then
$$y=a_{0}+a_{1}x+...+a_{n}x^n+... $$
$$y'=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+...+(n+1)a_{n+1}x^n+...$$
$$ y''=2a_{2}+3.2a_{3}x+4.3a_{4}x^2+...+(n+2)(n+1)a_{n+2}x^n+... $$
replacing this in the DE,
$3x^2y''+3xy''+2xy'+2y... | {
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$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem
I tried this approach:
$$
\frac{1}{\sqrt{n^3+1}}\le\frac... | Hint:
Use the fact that:
$$\sum_{i=1}^n{i}=\frac{n^2 +n}{2}$$
And:
$$\frac{1}{\sqrt{n^3+1}}\le\frac{i}{\sqrt{n^3+i}}\le\frac{i}{\sqrt{n^3+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/270216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.
Things should be known:
*
*Roots of a Quadratic Equations can be identified by:
The roots can be figured out by:
$$\frac{-b \pm \sqrt{d}}{2a},$$
whe... | The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$.
The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.
Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it me... | {
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Recurrence relation by substitution I have an exercise where I need to prove by using the substitution method the following
$$T(n) = 4T(n/3)+n = \Theta(n^{\log_3 4})$$
using as guess like the one below will fail, I cannot see why, though, even if I developed the substitution
$$T(n) ≤ cn^{\log_3 4}$$
finally, they ask m... | This recurrence has the nice property that we can give an explicit value for all $n$, not just powers of three. Let $$ n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k$$ be the representation of $n$ in base three. With $T(0) = 0$, we have by inspection
$$ T(n) = \sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j
\sum_{k=j}^{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove by induction that, for $n≥1, \sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}$? Hi can you help me solve this:
I have proved that $p(1)$ is true and am now assuming that $p(k)$ is true. I just don't know how to show $p(k+1)$ for both sides?
| If you want to do it by a conventional "blind" induction, suppose that for a certain $k$ we have
$$\sum_1^k \frac{1}{r(r+1)}=\frac{k}{k+1}.\tag{$1$}$$
We want to prove that
$$\sum_1^{k+1} \frac{1}{r(r+1)}=\frac{k+1}{k+2}.\tag{$2$}$$
Note that the left-hand side of $(2)$ is the left-hand side of $(1)$, plus $\dfrac{1}{(... | {
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"timestamp": "2023-03-29T00:00:00",
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Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$ Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$
| Hint: Consider the Taylor expansion of the summand about $1/n$ for $n$ large:
$$\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}} \approx \left (1 + \frac{7}{2 n^2} \right ) - \left (1 - \frac{8}{3 n^2} \right ) = \frac{37}{16 n^2}$$
Use the comparison test.
| {
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If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| The function $t\mapsto t^3$ is convex on ${\mathbb R}_{>0}$. Given that $${1\over x}+{1\over y}+{1\over z}=1\qquad(*)$$ we therefore have by Jensen's inequality
$$x^2+8y^2+27 z^2={1\over x} x^3 + {1\over y}(2y)^3+{1\over z}(3z)^3 \geq\left({1\over x} x+ {1\over y}2y +{1\over z} 3z\right)^3 =216\ ,$$
where equality hold... | {
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Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$ Please help me calculate this:
$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method.
Thanks in advance for solution / hints us... | One thing you should learn, is that analysts like to think of functions as power series (or at worst Laurent series) In this sense, L'Hopital's rules is essentially saying that "When we have a function $ \frac {(x-7)f(x)}{(x-7)g(x)}$, then we can 'fill in' the hole and carry along our own merry way".
So, if we don't ha... | {
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Is my limit evaluation correct? I'm studying for my calculus exam and I have the following limit:
$$\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )$$
My solution is:
$$\begin{align*}
&\lim\limits_{n\ \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}... | As $n$ grows, the number of terms increases. So you need extra care.
The sum is equal to $\sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}$.
We have
$\left | \sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}\right| \leq n \cdot \frac{1}{\sqrt{n^3}}=\frac{1}{\sqrt{n}}$.
Now you can take a limit and the limit is $0$.
| {
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How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| We have that
$$
x^2=(x-1+1)^2=(x-1)^2+1+2(x-1)
$$
and so for $x\neq 1$:
$$
\frac{x^2}{x-1}=\frac{(x-1)^2+1+2(x-1)}{x-1}=x-1+\frac{1}{x-1}+2=x+1+\frac{1}{x-1}.
$$
| {
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"source": "stackexchange",
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Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}$ How to prove the following inequalities without using Bernoulli's inequality?
*
*$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1},$$
*$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}.$$
My proof:
*
... | The inequality to be shown is
$$(n+1)^{n+1}\geqslant(n+3)^n,
$$
for every positive integer $n$. Introduce the function $u$ defined by
$$
u(x)=(x+1)\log(x+1)-x\log(x+3),
$$
then standard computations yield
$$
u'(x)=\frac3{3+x}+\log\left(\frac{1+x}{3+x}\right),\qquad
u''(x)=\frac{3-x}{(1+x)(3+x)^2},
$$
hence $u'$ incr... | {
"language": "en",
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"source": "stackexchange",
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Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
| A slight variant of lab bhattacharjee's method provides a simpler solution:
Let
$$ I(a) = \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = 2\int_{0}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx. $$
Then by a simple application of multivariable calculus,
\begin{align*}
I'(a)
&= 2 \left. \frac{\sqrt{a^2-x^2}}{1+x^2} \right|_... | {
"language": "en",
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"source": "stackexchange",
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Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer w... | \begin{align*}
\begin{vmatrix}b^2c^2&bc&b+c\\c^2a^2&ca&c+a\\a^2b^2&ab&a+b\end{vmatrix}
\stackrel{\large R_1-R_3\,,\,R_2-R_3}{\longrightarrow}
&(c-a)(c-b)
\,\begin{vmatrix}b^2(c+a)&b&1\\a^2(c+b)&a&1\\a^2b^2&ab&a+b\end{vmatrix}\\
\stackrel{\large C_1\,-\,ab\,C_2}{\longrightarrow}
&(c-a)(c-b)c
\,\begin{vmatrix}b^2&b&1\\a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How many binomial coefficients are equal to a specific integer ($\binom{n}{r} = 2013$ or $\binom{n}{r} = 2014$)?
*
*Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2013$.
*Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2014$.
My Attempt for $(1)$:
By simple g... | $$\binom n {r+1}\ge \binom nr \iff \frac{\binom n {r+1}}{\binom nr}\ge 1\iff\frac{n-r}{r+1}\ge1\iff r\le \frac{n-1}2$$
So, $$\binom n r\le \binom n {r+1}\iff r\le \frac{n-1}2$$ and $$\binom n r\ge\binom n {r+1}\iff r\ge\frac{n-1}2$$
For any integer $u,\binom n1=u\implies n=u$ will always have a solution in integers.
... | {
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"timestamp": "2023-03-29T00:00:00",
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Question about theta of $T(n)=4T(n/5)+n$ I have this recurrence relation $T(n)=4T(\frac{n}{5})+n$ with the base case $T(x)=1$ when $x\leq5$. I want to solve it and find it's $\theta$. I think i have solved it correctly but I can't get the theta because of this term $\frac{5}{5^{log_{4}n}}$ . Any help?
$T(n)=4(4T(\fra... | Hint :$\frac {\log{n}}{\log4}=\log_4{n}$
So $5^{\log_4{n}}=n^{\frac {1}{\log_5{4}}}$
Use this.
| {
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Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{2... | $f_1(x)=x+1$
$f_1(x^5)=x^5+1$
$f_1(x^5)$, when divided by $f_1(x)$ leaves a remainder 0.
$f_2(x)=x^2+x+1$
$f_2(x^5)=x^{10}+x^5+1$
$f_2(x^5)$, when divided by $f_2(x)$ leaves a remainder 0.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ How can I find the formula for the following equation?
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$
More importantly, how would you approach finding the formula? I have fo... | You can use these formulas for Problems like this , $$\frac{1}{a.b}=\frac{1}{(b-a)}\left(\frac{1}{a}-\frac{1}{b}\right)$$ $$\frac{1}{a.b.c}=\frac{1}{(c-a)}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$
Hope it'll help you .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate integral with quadratic expression without root in the denominator $$\int \frac{1}{x(x^2+1)}dx = ? $$
How to solve it? Expanding to $\frac {A}{x}+ \frac{Bx +C}{x^2+1}$ would be wearisome.
| It is not hard to express it as partial fraction, if you write $1$ in the numerator as $x^2+1 - x^2$.
Hence, $$\dfrac1{x(x^2+1)} = \dfrac{x^2+1-x^2}{x(x^2+1)} = \dfrac{x^2+1}{x(x^2+1)} - \dfrac{x^2}{x(x^2+1)} = \dfrac1x - \dfrac{x}{x^2+1}$$
Hence,
$$\int \dfrac{dx}{x(x^2+1)} = \int \dfrac{dx}x - \int\dfrac{x}{x^2+1}dx ... | {
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Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that:
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$
$(a,b) = (a,a+1)$ are solutions.
My motivation is for th... | If you want the ratio $(a^2+b^2-1)/(ab)$ to be equal to $a$, choose $b=a^2-1$.
Then the numerator is $$a^2+b^2-1=a^2+(a^2-1)^2-1=a^4-a^2=a^2(a^2-1),$$
which divided by the denominator $ab=a(a^2-1)$ gives the result $a$.
This answers in the affirmative the question of the "update" of the OP, since here $a$ can be any $m... | {
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Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$.
My reasoning:
$$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$
$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + ... | $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab+ac+bc) - 3abc$ is the right factorisation
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$ What is the limit of the following function which consists of an exponential and algebraic expression? $$ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}\;\;?$$
| $$\begin{align}\lim\limits_{x \to 0}\ (e^x + x)^ \frac {1} {x} &=\lim\limits_{x \to 0}\ (1+x+\frac{x^2}{2!}+\dots + x)^ \frac {1} {x}\\ &=\lim\limits_{x \to 0}\ (1+2x+\frac{x^2}{2!}+\dots )^ \frac {1} {x}=e^2\end{align}$$Note that since $x \to 0$ we can assume $\frac{x^2}{2!}+ \frac{x^3}{3!}+\cdots =0$ and $$\lim_{x \t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$.
Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$
And I appreciate hint rather than explicit solution.
Thank You.
| Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] <\frac{2}{9}\left(m!\right)^3.$$
So you have to go at least far enough so that
$$
\frac{2}{9}\left(m!\right)^3>10^{6},
$$
or $m! > \sqrt[3]{4500000} > 150$. So ... | {
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half-angle trig identity clarification I am working on the following trig half angle problem. I seem to be on the the right track except that my book answer shows -1/2 and I didn't get that in my answer. Where did I go wrong?
$$\sin{15^{\circ}} = $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \cos{30^{\c... | In the $4^{th}$ equation, you should have $$\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2} \cdot \frac{2}{2}}=\pm\sqrt{\frac{2-\sqrt{3}}{4}}=\pm\frac{1}{2}\sqrt{2-\sqrt{3}}.$$ Since $30/2$ is in the first quadrant, the answer should be the positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$ I came across these two series $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$$ and $$\sum_{n=1}^\infty \frac{(-1)^n(n^2-1)}{n^2+(n^2-1)^2}.$$ Does anyone recognize the sums of these series?
| Using the Residue Theorem, we can show that, for sufficiently well-behaved functions $f(z)$:
$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \sum_{k=1}^m \mathrm{Res}_{z=z_k} \pi \csc{(\pi z)} f(z)$$
where the $z_k$ are the poles of $f$. I will not show this for now, only that I note that $f(z) = (z^2 + (z^2-1)^2)^{-1}$ i... | {
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Two Problem: find $\max, \min$; number theory: find $x, y$
*
*Find $x, y \in \mathbb{N}$ such that $$\left.\frac{x^2+y^2}{x-y}~\right|~ 2010$$
*Find max and min of $\sqrt{x+1}+\sqrt{5-4x}$ (I know $\max = \frac{3\sqrt{5}}2,\, \min = \frac 3 2$)
| The second problem can be solved using routine calculus. There are also various non-calculus solutions. These in principle involve only basic machinery from algebra or trigonometry, but are in practice more complicated than the calculus solution.
The derivative of our function is
$$\frac{1}{2}\left(\frac{1}{\sqrt{x+1}... | {
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Product of numbers Pair of numbers whose product is $+7$ and whose sum is $-8$.
Factorise $x^{2} - 8x + 7$.
I can factorise but it's just I can't find any products of $+7$ and that is a sum of
$-8$. Any idea? Thanks guys!
Thanks.
| $(x-a)(x-b)=x^2-(a+b)x+ab$
So, you are looking for the roots of $x^2-(+2)x+(-3)$, that is, $x^2-2x-3$, which is $=x^2-2x+1-4=(x-1)^2-4$, and this is $0$ iff $(x-1)^2=4$, that is, $x-1=\pm 2$, i.e., $-1,3$ is your solution.
Similarly for the other one: $x^2-8x+7=(x-4)^2-16+7=\dots$
Or, the other way: $7+1=8$ and $7\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that for any odd integer:$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$ I'm trying to figure out how to prove that for any odd integer, the floor of:
$$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$$
Any help is appreciated to construct this proof!
Thanks guys.
| Let $n$ be an odd integer.
Then there exists an integer $k$, such that:
$$n=2k+1$$
It follows that:
$$\begin{align}
\left\lfloor\frac{n^2}{4}\right\rfloor &= \left\lfloor\frac{(2k+1)^2}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k+1)}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k)}{4}+\frac{1}{4}\right\rfloor\\
&=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence.
$$\begin{gather}
T(n) = T(n − 3) + 1/2\\
T(0) = T(1) = T(2) = 1.
\end{gather}$$
I tried solving it using the forward iteration.
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) ... | I believe this is the right answer:
$$
\\
T(n) = T(n - 3) + \frac{1}{2}
\\
T(n) = T(n - 6) + \frac{1}{2} + \frac{1}{2} = T(n - 6) + \frac{2}{2}
\\
T(n) = T(n - 9) + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = T(n - 9) + \frac{3}{2}
\\
T(n) = T(n - 12) + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = T(n - 12) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$ using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du$ evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du = \dfrac{\pi}{\sin( \pi p)}$. I am having trouble finding what is $p$. I set $u = x^4$, I figure $du = 4x^3 dx$, I am unsure though how to find ... | If $u=x^4$ and $du=4x^3 dx$ then:
$$dx = \frac{1}{4x^3} du = \frac{1}{4{(x^4)}^{3/4}} du =
\frac{1}{4{u}^{3/4}} du$$
So your integral is:
$$\int_0^\infty \dfrac{dx}{1+x^4} = \frac{1}{4}\int_0^\infty \frac{u^{-3/4}du}{1+u} = \frac{\pi}{4\sin(\pi/4)}=\frac{\pi }{2 \sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0Evaluate by complex methods
$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$
Sis.
| This integral is $1/2$ the integral over $[0,2 \pi)$. Let $z=e^{i \theta}$, $d\theta = dz/(i z)$; the result is
$$\frac{1}{2 i}\oint_{|z|=1} \frac{dz}{z} \frac{-\frac{1}{4} (z^2-1)^2}{(a z^2-(1+a^2)z+a)(b z^2-(1+b^2)z+b)}$$
which can be rewritten as
$$\frac{i}{8} \oint_{|z|=1} \frac{dz}{z} \frac{(z^2-1)^2}{(a z-1)(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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how to solve, $(x^2-2x+2y^2)dx+2xydy=0$? solve the following differential equation.
$\tag 1(x^2-2x+2y^2)\,dx+2xy\,dy=0$
$\frac{dy}{dx}=\frac{2x-x^2-2y^2}{2xy}$
dividing (1) throughout by $y^2$ we have,
$\tag 2 \left(\frac{x^2}{y^2}+2-2\frac{x}{y^2}\right)dx+2(\frac{x}{y})dy=0$
| Illustrating @DavidMitra's idea: multiply through by $x$:
$$(x^3-2x^2+2xy^2)\,dx+2x^2y\,dy=0$$
We want to find a function $F(x,y)$ such that
$$\frac{\partial F}{\partial x} = x^3-2x^2+2xy^2$$
$$\frac{\partial F}{\partial y} = 2x^2y$$
From the former equation, we integrate with respect to $x$ and get
$$F(x,y) = \frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| Starting with pipi's simplification of the mess:
$$
\frac{b - c}{x + a^2} + \frac{c - a}{x + b^2} + \frac{a - b}{x + c^2} = 0 \\
(b - c) (x + b^2) (x + c^2) + (c - a) (x + a^2) (x + c^2) + (a - b) (x + a^2) (x + b^2) = 0
$$
Suprisingly, this turns out a linear equation for $x$, with solution:
$$
x = a b + a c + b c
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
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Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + ... | You don't know that $\gcd(m^2-n^2, m^2+n^2)=1$, you need to prove it.
Now if $p|m^2-n^2$ and $p|m^2+n^2$ then $p$ divides their sum and their difference. Use this.
And to proceed further you need extra information, which you probably left out but it is the key: $m,n$ are relatively prime and of opposite parity....
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
| I will be very non-rigorous here.
Letting $m = j+k$,
$\begin{align}
\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}
&\approx \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{m}{j^3+(m-j)^3}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{(j+(m-j))(j^2-j(m-j)+(m-j)^2)}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Solving for x with radicals and negative exponents How do I go about solving for $x$ in this equation?
$$\displaystyle -x^{-\large\frac{3}{4}} + \frac{15^{\large\frac{1}{4}}}{15} = 0$$
| Hint: $$-x^{-3/4} + \frac{15^{1/4}}{15} = 0 \iff -\frac {1}{x^{3/4}} + \frac{15^{1/4}}{15^{4/4}} = 0$$
$$\iff -\frac {1}{x^{3/4}} + \frac{1}{15^{3/4}} = 0$$
$$\iff \dfrac{x^{3/4}}{15^{3/4}} = 1 \iff \left(\frac{x}{15}\right)^{3/4} = 1$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of analyticity for $\operatorname{Log}(z^4 - 1)^{1/2}$ How do I solve for the analyticity of $\operatorname{Log}(z^4 - 1)^{1/2}$?
So I factored out $$z^4 - 1 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4 - 1$$
Now the $$\operatorname{Re}(z^4 - 1) = x^4 - 6x^2y^2 + y^4 - 1 < 0 \implies x^4 - 6x^2y^2 + y^4 < 1$$ but I'm... | You're doing fairly well. As for the inequality, find that polynomial's roots first:
$$
x^4-6x^2y^2+y^4-1 = 0 \\
x^2 = 3y^2 \pm \sqrt{9y^4-y^4+1} = 3y^2 \pm \sqrt{8y^4+1} \\
x_{1,2,3,4} = \pm \sqrt { 3y^2 \pm \sqrt{8y^4+1}}
$$
So inequality can be rearranged as
$$
(x-x_1)(x-x_2)(x-x_3)(x-x_4) > 0
$$
As for the second r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Probability of choosing two equal bits from three random bits Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal?
I would like to know if the following analysis is correct and/or if there is a better way to think about it.
$$\Pr[\text{choose two equal bits}]... | Clearly the intuition tells us that the answer should be $\frac{1}{2}$, because choosing 2 random bits at random without replacing is just like simply choosing two random bits which are equal with probability $\frac{1}{2}$.
There are 3 way to choose two bits out of 3, i.e. choosing $B_1$ and $B_2$ or $B_2$ and $B_3$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310508",
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"source": "stackexchange",
"question_score": "1",
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Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
I e... | I know it is a very late to answer this, but i came across a very nice method to prove this.
$$\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$\displaystyle \iff (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313605",
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"source": "stackexchange",
"question_score": "10",
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Statistical Probability We make sweets of $5$ different flavors. The sweets are made in batches of $1.8$ million sweets per batch and then combined/mixed together into a total random batch of $9$ million sweets.
We then pack into tubes of $14$ sweets and we would like to know the probability of getting all $5$ flavors... | The details depend on whether you want a theoretically exact answer, or an excellent approximation. Since this is packaged as an applied problem, we work with the approximation. So we will assume that each sweet has one of $5$ values, independently of the values of the previously chosen sweets. The number $9,000,000$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the inverse of a polynomial in Laurent series division ring? In the Laurent series division ring, how can we find the inverse of a given polynomial? For example how can you find the inverse of $3x^{-2} +x^{-1}+5x+7x^4$. Is there a certain formula to find the inverse?
I've tried to find the inverse for t... | $$ \frac{1}{7 x^4 + 5 x + x^{-1} + 3 x^{-2}} = \frac{x^2}{7 x^6 + 5 x^3 + x + 3}$$
The polynomial $P(x) = 7 x^6 + 5 x^3 + x + 3$ is irreducible over the rationals, with six
distinct complex roots $r_1$ to $r_6$ (none of them real, by the way). We have
a partial fraction decomposition
$$ \frac{x^2}{7 x^6 + 5 x^3 + x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding the asymptote of a function Find the asymptote of $f(x)$ as $x \rightarrow \infty$ where
$$f(x) = \frac{x^3 -x^2 +2}{x-1}$$
The answer key mentioned that the answer is $h(x)=x^2$ where $h(x)$ is the equation of the asymptote of $f(x)$. I beg to differ with this solution:
$$
\begin{align*}
f(x) = \frac{x^3 -x^2 ... | With a rearrangement,
$$
\begin{align*}
f(x) -x^2
&=\frac{2}{x-1}\\
f(x) -h(x)
&=\frac{2}{x-1}\\
\lim_{x\rightarrow\infty}f(x) -h(x)
&=\lim_{x\rightarrow\infty}\frac{2}{x-1}\\
&=0
\end{align*}
$$
If you think about it, the idea is that as $x \rightarrow \infty$, the difference between $f(x)$ and $h(x)$ tends to $0$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$
My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv... | Why not use Hensel's Lemma ?
First, let us find a solution modulo $\,7\,$ of $\,f(x)=x^{11}+x^8+5\pmod 7\,$.
Clearly, $\,r=1\,$ is a root. Now, define
$$t:=-\frac{f(1)}{7}f'(1)^{-1}=-(11+8)^{-1}=(-19)^{-1}=30^{-1}=18\pmod{49}\Longrightarrow $$
$$s:=1+18\cdot 7=127=29\pmod{49}\,\,\,\text{is a root of}\,\,\,f(x)\pmod{49}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Conditional Probability with balls in an urn
Two balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that
(a) b... | The probability that the contents of the urn are two red is indeed $\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\frac{1}{2}$. The derivation could have been done more quickly.
Question (a) asks for the probability both are red given that the two drawn balls are red. Let $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
| a triffle answer;we have $2005=401.5$ so $p=5,401$ also: $$a^5+b^5=(a+b)(a^4+b^4-ab(a^2-ab+b^2)) \tag I$$.also we have
$$a^3+b^3=(a+b)(a^2-ab+b^2) \tag {II} $$ from the $p|a^3+b^3$ we have 2 cases : 1) if $p|a+b$ then from $(I)$ we have $p|a^5+b^5$
2)if $p|(a^2-ab+b^2)$ by attention to $(II)$,$p|(-ab)(a^2-ab+b^2)$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers. I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers."
I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found.
I used Wolfram Alpha to get... | (Note that this method is pretty explanatory and slow, you can do it faster).
Let $u = 2\theta$, then we have:
$$ \sin 4\theta = \sin 2u $$
We know that:
$$ \sin 2u = 2\sin u\cos u$$
Now put $u = 2\theta$ back in:
$$ \sin (2 \cdot 2\theta) = 2\sin 2\theta \cos 2\theta $$
$$ \sin (4\theta) = 2\sin 2\theta \cos 2\theta $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral?
$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$
| $$\quad\int \frac{1}{x+\large\frac{1}{x^2}}\,dx \quad = \quad\int \frac{1}{\Large\frac {x^3+ 1}{x^2}}\,dx\quad= \quad\int \frac {\color{blue}{\bf x^2\,dx}}{\color{red}{\bf x^3 + 1}}\quad(x \neq -1)$$
Let us substitute $\;\;\color{red}{\bf u} = \color{red}{\bf x^3 + 1} \;\implies\; du = 3x^2\,dx \;\implies \;\color{blue... | {
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"timestamp": "2023-03-29T00:00:00",
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Common basis for subspace intersection
Let $ W_1 = \textrm{span}\left\{\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\1\\1\end{pmatrix}\right\}$, and $ W_2 = \textrm{span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}3\\0\\-1\end{pmatrix}\right\}$. Find a basis for $W_1 \cap W_2$
I first though... | Let's try to make it easier: if you take a minute to read carefully what is $\,W_2\,$ , you'll see that
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in W_2\iff y=0$$
Now, since an element of $\,w_1\in W_1\,$ has the general form
$$w_1=a\begin{pmatrix}1\\2\\3\end{pmatrix}+b\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}a+2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle?
Triangle
$$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$
Circle
$$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)... | You want to find the intersection points of the lines and the cycle. You can to this by substituting.
For instance: line $a$.
$$9 = (27-1.8\lambda_1 -25)^2 + (-2.25 + 0.25\lambda_1+2)^2 + (4+0.7\lambda_1)^2$$
Now you can calculate the value of $\lambda_1$, using the Quadratic formula.
You can ran into 3 cases: the di... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$ Find $x$ such that $$\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$$
I knew the answer was $3$.
| This is asking for the last digit of that sum. So we may ignore everything but the last digit of each number. Now eliminate the few last terms for a minute and consider:
$$(1^1+\cdots +9^9)+(1^{11}+\cdots +9^{19})+\cdots+(1^{2001}+\cdots +9^{2009})\equiv \,?\pmod{10}$$
Notice that we can now pair each digit and form ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Need help proving this integration If $a>b>0$, prove that :
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
| First step: $\cos(\theta+\pi)=-\cos(\theta)$ and $\sin^2(\theta+\pi)=\sin^2(\theta)$ hence the integral $I$ to be computed is
$$
I=\int_0^\pi\frac{\sin^2\theta}{a+b\cos\theta}\mathrm d\theta+\int_0^\pi\frac{\sin^2\theta}{a-b\cos\theta}\mathrm d\theta=2aJ,
$$
with
$$
J=\int_0^\pi\frac{\sin^2\theta}{a^2-b^2\cos^2\theta}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Fibonacci Sequence in $\mathbb Z_n$. Consider a Fibonacci sequence, except in $\mathbb Z_n$ instead of $\mathbb Z$:
$$F(1) = F(2) = 1$$ $$F(n+2) = F(n+1) + F(n)$$
It is easy to see that each of these sequences must cycle through some sequence and repeat. Consider, for example, the sequence in $\mathbb Z_4$:
$$1,1,2,3,... | I found a way to do this with prime $n.$ Not sure how well it extends to prime powers.
We want to know what happens to the pair of consecutive entries $(x,y),$ shifting over one to get $(y,x+y).$ So we define the matrix
$$ A \; = \;
\left( \begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}
\right) ,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
How do I transform the left side into the right side of this equation? How does one transform the left side into the right side?
$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
$$
| Use the method of foil on the right hand side and then foil the left hand side. It's easier doing right hand side to the left hand side.
$$(ac-bd)^2=(a^2c^2-2abcd+b^2d^2)$$
$$(ad+bc)^2=(a^2d^2+2abcd+b^2c^2)$$
Then simplify.
For the right hand side:
$$(ac-bd)^2+(ad+bc)^2= (a^2c^2-2abcd-b^2c^2)+(a^2d^2+2abcd+b^2c^2)$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/330863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
How to solve this system of equations? How to solve this system of equations?
$$\begin{cases}
1+\sqrt{2 x+y+1}=4 (2
x+y)^2+\sqrt{6 x+3 y},\\
(x+1) \sqrt{2 x^2-x+4}+8 x^2+4
x y=4.
\end{cases}$$
| The trick here is to not use both equations at the same time.
The first equation screams to use the substitution $k=2x+y$. Substituting that, we get $1+\sqrt{k+1}=4k^2+\sqrt{3k}$. So we transform it to $(4k^2-1)+(\sqrt{3k}-\sqrt{k+1})=0$. So $(4k^2-1)(\sqrt{3k}+\sqrt{k+1})+(2k-1)=0$.
(this is because we multiplied bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral
$$\int \frac{1}{1+x^4} \mathrm dx.$$
The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.... | $$I =\int \frac{1}{{x^4+1}} dx$$
If we add and subtract $2x^2$ to $x^4 + 1$, we get:
$$\int \frac{1}{{x^4 + 2x^2 + 1 - 2x^2}}$$
We know $x^4 + 2x^2 + 1 = (x^2 + 1)^2$
$$\int \frac{1}{{(x^2 + 1)^2 - 2x^2}}$$
We know $a^2 - b^2 = (a - b)(a + b)$
Hence, $(x^2 + 1)^2 - 2x^2 = (x^2 + 1 - \sqrt{2}x)(x^2 + 1 + \sqrt{2}x)$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"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.