Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$ $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$
I believe I've done everything right; however, my answer does not resemble the answer in the book. I think it has something to do with my Algebra. Please tell me what I am doing wrong.
| $$\int\frac{1}{t^3\sqrt{t^2-1}}\space\space\text{d}t=$$
Substitute $t=\sec(u)$ and $\text{d}t=\tan(u)\sec(u)\space\space\text{d}u$. Then $\sqrt{t^2-1}=\sqrt{\sec^2(u)-1}=\tan(u)$ and $u=\sec^{-1}(t)$:
$$\int\cos^2(u)\space\space\text{d}u=$$
$$\int\left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)\space\space\text{d}u=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
| In general case whenever you have something like $1-a^2x^2$. Remember $\sin^2x+\cos^2x=1$ and so we should multiply it by $a$. Thus take $\sin y=\frac{1}{a}x$. Now we have
$\sqrt{1-a^2x^2}=\sqrt{1-\sin^2y}=\cos y$.
Hence one can see that
$\int\sqrt{1-a^2x^2}dx=\int\sqrt{1-\sin^2y}\times a\cos y \,dy=\int a\cos^2 y dy$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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A quotient of trigonometric expressions in complex analysis How is $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}\ \ \ ?$$
I think that I am having trouble understanding a fundamental concept in complex analysis, but cannot ... | $$\frac{\cos\left(\frac{20\pi}{3}\right)+\sin\left(\frac{20\pi}{3}\right)i}{\cos\left(\frac{15\pi}{4}\right)+\sin\left(\frac{15\pi}{4}\right)i}=\frac{e^{\frac{20\pi}{3}i}}{e^{\frac{15\pi}{4}i}}=e^{\left(\frac{20\pi}{3}-\frac{15\pi}{4}\right)i}=e^{\frac{35\pi}{12}i}=\cos\left(\frac{35\pi}{12}\right)+\sin\left(\frac{35\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving degree three inequality
$$\frac{2}{(x^2+1)}\geq x.$$
I thought to do this:$$2-x^3-x\geq 0$$
But we haven't learnt how to solve equations of third grade. Could you help me maybe by factorizing sth?
| \begin{align}
\frac{2}{x^2+1} &\ge x \iff \\
2 &\ge x(x^2+1) = x^3 + x \iff \\
0 &\ge x^3 + x - 2
\end{align}
Trying some easy roots one gets $x=1$:
\begin{align}
0 &\ge (x-1)(x^2+x+2) \quad (*)
\end{align}
Then we have
$$
x^2 +x+ 2 = (x+1/2)^2 + 7/4 \ge 7/4
$$
which means no real roots for this part, no change of sign... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the probl... | So you have
$$
\lvert x \rvert \leq 2\lvert x+2\rvert.
$$
If $x \geq 0$, then $x + 2 > 0$. And then the equation reads $x \leq 2x + 4$ implying that $x \geq -4 $. This in all gives you the solutions $x \geq 0$.
If $x\in [-2, 0)$, then the equation becomes $-x \leq 2x + 4$. So $-4\leq 3x$. So $\frac{-4}{3} \leq x$. In a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Solve $z^3=(sr)^3$ where $r,s,z$ are integers? Let $x,y,z$ be 3 non-zero integers defined as followed:
$$(x+y)(x^2-xy+y^2)=z^3$$
Let assume that $(x+y)$ and $(x^2-xy+y^2)$ are coprime
and set $x+y=r^3$ and $x^2-xy+y^3=s^3$
Can one write that $z=rs$ where $r,s$ are 2 integers?
I am not seeing why not but I want to be s... | Yes, there exist such integers $r$ and $s$. It is simplest to use the Fundamental Theorem of Arithmetic (Unique Factorization Theorem). The result is easy to prove for negative $z$ if we know the result holds for positive $z$. Also, the result is clear for $z=1$. So we may assume that $z\gt 1$.
Let $z=p_1^{a_1}p_2^{a_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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A Basic Probability Question - I am getting the wrong answer Problem:
In a deck of $52$ cards there are $4$ kings. A card is drawn at random
from the deck and its face value noted; then the card is returned. This
procedure is followed $4$ times. Compute the probability that there are
exactly $2$ kings in the $4$ select... | $p_2$ the probability that exactly two of four draws will be kings (one of thirteen cards in a suit) is:$$p_2= \binom{\color{red}{4}}{2}\left(\frac{1}{13}\right)^2\left(\frac{12}{13}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Limit of a sum of infinite series How do I find the following?
$$\lim_{n\to\infty} \frac{\left(\sum_{r=1}^n\sqrt{r}\right)\left(\sum_{r=1}^n\frac1{\sqrt{r}}\right)}{\sum_{r=1}^n r}$$
The lower sum is easy to find. However, I don't think there is an expression for the sums of the individual numerator terms... Nor can I... | HINTS:
*
*Summations:
$$\sum_{r=1}^{n}\sqrt{r}=\text{H}_n^{-\frac{1}{2}}$$
$$\sum_{r=1}^{n}\frac{1}{\sqrt{r}}=\text{H}_n^{\frac{1}{2}}$$
$$\sum_{r=1}^{n}r=\frac{1}{2}n(1+n)$$
With $\text{H}_n^r$ is the generalized harmonic number.
*
*Fraction:
$$\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \th... | An easy way is to use Euler formula:
$$\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Then use the binomial expansion:
$$\cos^6\theta = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - k)\theta}e^{-ik\theta} = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - 2k)\theta}.$$
If $k = 3$, $\int_0^\pi e^{i(6 - 2k)\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Estimating the number of integers relatively prime to $6$ between $1$ and some integer $x$? I am trying to understand the standard way to estimate the number of integers relatively prime to $6$ where we don't know which congruence class $x$ belongs to.
For a given $x$, if we know the congruence class, it is straight fo... | The number of integers relatively prime to $6$ for each $x$ is given by the expression
$$\left\lfloor \frac{x-1}{6}\right\rfloor +\left\lfloor \frac{x+1}{6}\right\rfloor +1$$
which is clearly $\sim x/3.$
Check:
rp6[x_] := Floor[(x - 1)/6] + Floor[(x + 1)/6] + 1
{Length@Select[Range@#, GCD[#, 6] == 1 &] & /@ Range@40, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | Assume the statement is true
\begin{align}
\sqrt{x + 2} - \sqrt{x + 1} &= \sqrt{x + 1} - \sqrt{x}\\
\sqrt{x + 2} + \sqrt{x} &= 2\sqrt{x + 1}\\
2x + 2 + (2\sqrt{x^2 + 2x}) &= 4x + 4\\
\sqrt{x^2 + 2x} &= x + 1\\
x^2 + 2x &= x^2 + 2x + 1\\
0 &= 1\\
\end{align}
Which is a contradiction. Now I am not sure if I'm allowed to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$? $$
\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx
$$
A bit confused with how to integrate this question. I though it was partial fractions but was unsure about the what to do after that.
| The integral can be written as
$$I=\frac{1}{4}\int\frac{4x^3}{x^4-1}+5\int\frac{x^2}{(x^2-1)(x^2+1)}+\int\frac{2x}{(x^2-1)(x^2+1)}-4\int\frac{dx}{(x^2-1)(x^2+1)}=I_1+I_2+I_3+I_4$$
$$I_1=\frac{1}{4}ln|x^4-1|+C$$
$$I_2=\frac{5}{2}\int\frac{1}{x^2-1}+\int\frac{1}{x^2+1}$$
$$I_3=\int\frac{dt}{(t-1)(t+1)}$$ where $t=x^2$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx}... | In the same spirit as Leg's answer, starting with the partial fraction decomposition $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$ and using harmonic numbers definition $$\sum_{n=p}^m \dfrac1{n+5}=H_{m+5}-H_{p+4}$$ $$\sum_{n=p}^m \dfrac1{n+4}=H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
question on right angle triangle Let ABC and DBC be two equilateral triangle on the same base BC,a point P is taken on the circle with centre D,radius BD. Show that PA,PB,PC are the sides of a right triangle.
|
First let's assume $AB=1$ and call $PB=a;PC=b;PA=c$, we have to prove that $a^2+b^2=c^2$. Applying the law of cosines to $PBC$ with respect to angle $\angle BPC=\pi/6$ we get
$$a^2+b^2-\sqrt3ab=1$$
Now, call $\angle PCB=y$. Applying the law of cosines to $PBC$ with respect to angle $y$ we get
$$ 1+b^2-2b\cos y=a^2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to use boundary condition for Euler-Bernoulli equation How to use enter boundary condition into a Euler-Bernoulli fourth order equation
| OK, first integrate step by step to obtain
$$\begin{array}{*{20}{l}}
{\frac{{{d^4}y}}{{d{x^4}}} = F} \\
{\frac{{{d^3}y}}{{d{x^3}}} = Fx + A} \\
{\frac{{{d^2}y}}{{d{x^2}}} = \frac{F}{2}{x^2} + Ax + B} \\
{\frac{{dy}}{{dx}} = \frac{F}{6}{x^3} + \frac{A}{2}{x^2} + Bx + C} \\
{y = \frac{F}{{24}}{x^4} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How many numbers have unit digit $1$?
Let $f(n)$ be the number of positive integers that have exactly $n$ digits and whose
digits have a sum of $5$. Determine, with proof, how many of the $2014$ integers
$f(1), f(2), . . . , f(2014)$ have a units digit of $1.$
HINTS ONLY
EDIT:
for $f(2)$ we have: $x + y = 5$ with... | If you want to solve $$\frac{n(n+1)(n+2)(n+3)}{24}\equiv 1\pmod{10},$$
You can multiple both sides by $3$, since it is relatively prime to $10$, but when you multiply by $8$, you have to apply that to the modulus, to. So, you are trying to solve:
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{80}$$
Solve this in pairs, using Chinese... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so fa... | Hint:
Use differential binomial. Note that $$\sqrt{1+\frac{1}{4x}}=0.5x^{-0.5}(1+4x)^{0.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book. Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.
My method:
$\frac{9x}{(2x+1)^2(1... | Spotted mistake. $A(2x+1)$ should be $A(2x+1)(1-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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calculate $\lim\limits_{x \to 1}(1 - x)\tan \frac{\pi x}{2}$ I need to calculate $$\lim_{x \to 1}\left((1 - x)\tan \frac{\pi x}{2}\right)$$.
I used MacLaurin for $\tan$ and got $\frac{\pi x} {2} + o(x)$. Then the full expression comes to $$\lim_{x \to 1}\left(\frac {\pi x} {2} - \frac {\pi x^2} {2} + o(x)\right) = 0$$B... | We use that $\tan\theta =\cot\left (\frac{\pi}{2}-\theta\right)$.
Letting $y=\frac{\pi(1-x)}{2}=\frac{\pi}{2}-\frac{\pi}{2}x$, we get that we are seeking:
$$\lim_{y\to 0} \frac{2}{\pi}y\cot y = \frac{2}{\pi}\lim_{y\to 0}\left(\frac{y}{\sin y}\cdot \cos y\right)$$
From there, it should be easy.
Alternatively, note tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that there is a common line of intersection of the three given planes. Let $x-y\sin\alpha-z\sin\beta=0,x\sin\alpha-y+z\sin\gamma=0$ and $x\sin\beta+y\sin\gamma-z=0$ be the equations of the planes such that $\alpha+\beta+\gamma=\frac{\pi}{2}$,(where $\alpha,\beta,\gamma\neq0$).Then show that there is a common line ... | Rewrite the required result as
$$ \cos^2\gamma - \sin^2 \alpha - \sin^2 \beta \overset{?}= 2\sin \alpha \sin \beta \sin \gamma $$
We'll try to show that the LHS equals the RHS. Note that $\cos \gamma = \sin (\alpha + \beta)$ and vice versa, as $\gamma = \pi/2 - (\alpha + \beta)$
\begin{equation}
\sin^2 (\alpha + \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A general approach for this type of questions I've come across multiple questions like these. $\left(\iota=\sqrt{-1}\right)$
If $f\left(x\right)=x^4-4x^3+4x^2+8x+44$, find the value of $f\left(3+2\iota\right)$.
If $f\left(z\right)=z^4+9z^3+35z^2-z+4$, find $f\left(-5+2\sqrt{-4}\right)$.
Find the value of $2x^4+5... | Ill give a guide line we have $x=-2-\sqrt{3}i$ now bring two on lhs and square both sides. So we get $x^2+4x+4=-3$ so $x^2+4x+7=0$ ..(1).now just divide the given expression by this qyadratic so we have . $\frac{2x^4+5x^3+7x^2-x+41}{x^2+4x+7}$ . I hope now you know this type of division . If not look up on net. Sowe ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Converting a sum of trig functions into a product Given,
$$\cos{\frac{x}{2}} +\sin{(3x)} + \sqrt{3}\left(\sin\frac{x}{2} + \cos{(3x)}\right)$$
How can we write this as a product?
Some things I have tried:
*
*Grouping like arguments with each other. Wolfram Alpha gives $$\cos{\frac{x}{2}} + \sqrt{3}\sin{\frac{x}{2}... | $$\cos{\frac{x}{2}} +\sin(3x) + \sqrt{3}\left(\sin\frac{x}{2} + \cos(3x)\right)$$
$$=\cos{\frac{x}{2}} + \sqrt{3}\sin\frac{x}{2} +\sin(3x) + \sqrt{3}\cos(3x)$$
$$=2\left(\frac{1}{2}\cos\frac{x}{2} + \frac{\sqrt{3}}{2}\sin\frac{x}{2} +\frac{1}{2}\sin(3x) + \frac{\sqrt{3}}{2}\cos(3x)\right)$$
Note that $\frac{1}{2}=\si... | {
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"url": "https://math.stackexchange.com/questions/1540940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how can i find the matrix $P$ that diagonalizes the matrix $A$? I want to find matrix $P$ that diagonalizes the matrix $A$:
$$
\begin{bmatrix}
4 & 0 & 1 \\
2 & 3 & 2 \\
1 & 0 & 4 \\
\end{bmatrix}
$$
| You need to calculate the eigenvectors which are (1,2,1), (-1,0-1) and (0,1,0). So we have the passage matrix
\begin{equation}P=\left(\begin{array}{ccc}
1 & -1& 0\\
2 & 0 & 1\\
1 & 1 & 0
\end{array}\right)
\end{equation}
\begin{equation}P^{-1}=\left(\begin{array}{ccc}
1/2 & 0& 1/2\\
-1/2 & 0 & 1/2\\
-1 & 1 & -1
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\rig... | If you want to compute this limit by making use of known basic limits then
use the following
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}=-\frac{1}{6},\ \ \ \ and\ \ \ \
\ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}
\end{equation*}
as follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| $$\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} =\dfrac{(x^2+1)(x^8+1)}{(x^4+1)(x^6+1)}$$
$$=\dfrac{\left(x+\dfrac1x\right)\left(x^4+\dfrac1{x^4}\right)}{\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right)}$$
$$=\dfrac{\left(x^2+\dfrac1{x^2}\right)^2-2}{\left(x^2+\dfrac1{x^2}\right)\left(x^2+\dfrac1{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$. First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$. Solving this gives $a_n^h =1$. The particular solution, as I understand, will be $a... | Let $f(x) = \sum_{n=0}^\infty a_nx^n$. Multiplying the recurrence by $x^n$ and summing over $n\geqslant 2$ we find that the LHS is
$$\sum_{n=2}^\infty a_nx^n = f(x)-1-x $$
and the RHS is
$$3\sum_{n=2}^\infty a_{n-1}x^n - 2\sum_{n=2}^\infty a_{n-2}x^n + 3\sum_{n=2}^\infty x^n = 3x(f(x)-1) - 2x^2f(x)+\frac{3x^2}{1-x}. $... | {
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"timestamp": "2023-03-29T00:00:00",
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The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question
| *
*check if the statement holds true for $n=1$:
$$1^2=1=\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{6}{6}=1$$
*Inductive step: show that if statement holds for $k$, then it also holds for $k+1$. This is done as follows:
$$1^2+2^2+3^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}$$
$$1^2+2^2+3^2+\ldots+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k... | {
"language": "en",
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Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$ Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$
Is there some intuitive method or formula for finding GCD of $n$ integers?
| Fisrtly, $A_0 = 35$, so $d:=\gcd(A_0\ldots A_{2015})\mid 35$, that means that $d \in \{1,5,7,35\}$.
Now let's try to factorize the $\gcd$. Starting with $5$, notice that $2^{4n} \equiv 3^{4n} \equiv 1\pmod{5}$ for every $n\geqslant 0$ (Fermat's little theorem). So:
\begin{align}A_n = 2^{3n}+3^{6n+2}+5^{6n+2} \equiv 2^{... | {
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How can I find $\cos(\theta)$ with $\sin(\theta)$?
If $\sin^2x$ + $\sin^22x$ + $\sin^23x$ = 1, what does $\cos^2x$ + $\cos^22x$ + $\cos^23x$ equal?
My attempted (and incorrect) solution:
*
*$\sin^2x$ + $\sin^22x$ + $\sin^23x$ = $\sin^26x$ = 1
*$\sin^2x = 1/6$
*$\sin x = 1/\sqrt{6}$
*$\sin x =$ opposite/hypoten... | You know that
\begin{equation}
\textrm{sin}^2x + \textrm{sin}^2 2x + \textrm{sin}^2 3x = 1 \qquad (\textrm{Eq.} 1)
\end{equation}
Notice that
\begin{equation}
\begin{aligned}
\textrm{sin}^2x + \textrm{cos}^2 x = 1 \Rightarrow \textrm{sin}^2x = 1 - \textrm{cos}^2 x
\end{aligned}
\end{equation}
Substituting the identi... | {
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How to eliminate $\theta$? While doing a sum I was stuck in a particular step:
$$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$
How to eliminate $\theta$ ?
| $$r_1\cdot r_2 = \frac{(4a)^2}{\sin \theta \cos \theta}$$
$$\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}=\frac{\sin \theta}{\cos \theta}$$
$$\frac{\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}}{r_1\cdot r_2 }=\frac{\sin^2 \theta}{(4a)^2}=r_1^{-\frac{4}{3}}r_1^{-\frac{2}{3}}$$
$$\frac{1}{\left(\frac{r_2}{r_1}\right)^{\frac{1... | {
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Prove that $ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $ For all $a, b, c>0$ and $a+b+c=3$ Prove that $$ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $$
I tried cauchy-schwarz inequality for the L. H. S like and I get
$ [\left( \sqrt{a... | First, we use Cauchy-Schwartz inequality:
$$
A^2=\left(\sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} }\right)^2 =\\
\left(\sqrt{\frac{a}{(a+3)(b+3)}(a+3) } +\sqrt{\frac{b}{(b+3)(c+3)} (b+3)} +\sqrt{\frac{c}{(c+3)(a+3)}(c+3) }\right)^2 \leq \\
\left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{... | {
"language": "en",
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Min and max of a two variables function
I consider the function
$$f:\mathbb{R}^2 \to \mathbb{R}, \
(x,y) \mapsto \max\left(1-\sqrt{x^2+y^2},2-\sqrt{(x-6)^2+y^2},0\right).$$
I have to find $\max$ and $\min$ (local and global).
I calculate where $1-\sqrt{x^2+y^2}=2-\sqrt{(x-6)^2+y^2}$ is an ellipse but I don't unde... | Note that $1 - \sqrt{x^2 + y^2} > 0$ inside a circle of radius $1$ about the origin, and is $< 0$ outside that circle. Meanwhile $2-\sqrt{(x-6)^2+y^2} > 0$ inside a circle of radius $2$ about the point $(6, 0)$ and is $< 0$ outside. Since the centers of these circles are a distance of $6$ apart while the sum of the rad... | {
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Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:
Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations
What I've Tried
*
*$ \frac{dx}{dt} = -5\sin(t) $
*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of... | The user kccu already found your mistake. Here is a shorter way to do the question.
The curve is $(5\cos t, 3\sin t\,\cos t)$. With $0\leq t<2\pi$, it goes through the origin two times: when $t=\pi/2$ and when $t=3\pi/2$. Then derivative is, as you say,
$$
(-5\sin t, 3\cos 2t).
$$
At the two values of $t=\pi/2$ and $... | {
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Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further... | Notice, $$\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt{x+2}}\ dx=\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt 2\sqrt{\frac{x}{2}+1}}\ dx$$
now, let $\frac{x}{2}=\tan^2\theta\implies dx=2\tan\theta\sec^2\theta \ d\theta$, $0\le \theta\le \pi/2$
$$=\frac{1}{\sqrt2}\int \frac{\tan^{-1}\left(\tan\theta\right)}{\sqrt{\tan... | {
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Prove that functional equation doesn't have range $\Bbb R.$ Prove that any solution $f: \mathbb{R} \to \mathbb{R}$ of the functional equation
$$ f(x + 1)f(x) + f(x + 1) + 1 = 0 $$
cannot have range $\mathbb{R}$.
I transformed it into
$$ f(x) = \frac {-1} {f(x + 1)} - 1 = \frac {-1 - f(x + 1)} {f(x + 1)} $$
I tried to... | You should always be careful when dividing, because you'll run into trouble if you try and divide by 0.
We have that for all $x$, $$f(x+1)(f(x) + 1) = -1$$
Notice that $f$ can never be zero, then, because that would mean $0$ on the left-hand side was $-1$ on the right-hand side. This justifies your division step, and p... | {
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Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$. Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
So far I have done the following proof, but I am unsure how to complete... | An essential property of primes is the following:
If $p \mid ab$ then $p \mid a$ or $p \mid b$.
In your case, $n \mid (x+1)(x-1)$, so $n \mid (x+1)$ or $n \mid (x-1)$.
| {
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Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more c... | $$\int \frac{x+1}{\left(x^2-x+8\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x+8\right)}{\left(x^2-x+8\right)^3}+\int\frac{3}{\left(x^2-x+8\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x+8\right)^2}+96\int \frac{d(2x-1)}{\left((2x-1)^2+31\right)^3}\right)$$
Let $2x-1=\sqrt{31}\tan t$. Then ... | {
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Find all values of $n$ for which the Diophantine equation $n=a^2-b^2$ has a solution Let $n$ be an integer. Find all values of $n$ for which the Diophantine equation $n=a^2-b^2$ has a solution for integers $a$ and $b$. For those values of $n$ found in the previous part find all solutions of $n=a^2-b^2$ for integers $a$... | The equation $$n=a^2-b^2=(a-b)(a+b)$$ is solveable over the integers if and only if there are numbers $u,v$ with $uv=n$, such that the system $$a-b=u$$ $$a+b=v$$
is solveable over the integers.
The system is solveable if and only if $u$ and $v$ have the same parity. We can find suitable $u$ and $v$, if and only if $n\n... | {
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Volume bounded by elliptic paraboloids Find the volume bounded by the elliptic paraboloids given by $z=x^2 + 9 y^2$ and $z= 18- x^2 - 9 y^2$.
First I found the intersection region, then I got $x^2+ 9 y^2 =1$. I think this will be area of integration now what will be the integrand. Please help me.
| You probably got something looking like this
where the required volume is the dark volume inside the two parabolas. Of course there are many ways to find that volume; I'll stick to the slice method: essentially you fix the third variable $z$ and calculate the area of desired region intersecting the plane $z = z_0$, the... | {
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | Well, then $n \cdot 0=0\; \forall \mathbb{R} \implies \frac{0}{0}=n=\frac{1}{n}$ so, $\frac{0}{0}$ is pretty much any real number. This is definitely not the case. This is why it is undefined.
| {
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How to show that $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex and has a minimum value I want to show that $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex on the interval $]-1,1[$. How do I have to proceed?
I did take the derivative of the function which is
$f'(x) = \frac{2x\sqrt{\cos x }}{(1-x^2)^2}-\frac{\sin x}{2\... | Let $f$ be given by
$$f(x)=\frac{\sqrt{\cos x}}{1-x^2}$$
for $|x|<1$. Then, the first derivative of $f$ is
$$f'(x)=\frac{\sqrt{\cos x}}{2(1-x^2)}\left(\frac{4x}{1-x^2}-\tan x\right) \tag 1$$
From $(1)$ we can write
$$\frac{f'(x)}{f(x)}=\frac12\left(\frac{4x}{1-x^2}-\tan x\right) \tag 2$$
whereupon differentiating bo... | {
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Help me evaluate this infinite sum I have the following problem:
For any positive integer n, let $\langle n \rangle$ denote the integer nearest to $\sqrt n$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n \rangle = k$.
(b) Show that $$\sum_{n=1}^\infty{\frac{2^{\langle n \rang... | Direct computation seems to work. I recall this problem being an old Putnam problem.
We have $\langle n \rangle = k$, iff $n \in [k^2 - k + 1, k^2 + k]$.
Now, let's compute this sum.
$$
\sum_{n = 1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n}
= \sum_{k=1}^\infty \sum_{i=k^2 - k + 1}^{k^2 + k} \fr... | {
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A question on summation of series...
If $$\frac{\left(1^4+\frac14\right)\left(3^4+\frac14\right)\ldots\left((2n-1)^4+\frac14\right)}{\left(2^4+\frac14\right)\left(4^4+\frac14\right)\ldots\left((2n)^4+\frac14\right)}=\frac1{k_1n^2+k_2n+k_3}$$ then $k_1-k_2+k_3$ is equal to . . . ?
What I basically tried was simply pu... | Use the Sophie-Germain identity: $$a^4+4b^4=(a^2+2b^2)^2-(2ab)^2=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$$Using this, we can factor an expression of the form $m^4+\frac{1}{4}$ into $$m^4+4\bigg(\frac{1}{2}\bigg)^4=(m^2-m+\frac{1}{2})(m^2+m+\frac{1}{2})$$Therefore, the LHS of your equation is equal to $$\frac{\displaystyle\prod_{i... | {
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Easy way of memorizing or quickly deriving summation formulas My math professor recently told us that she wants us to be familiar with summation notation. She says we have to have it mastered because we are starting integration next week. She gave us a bunch of formulas to memorize. I know I can simply memorize the lis... | Perhaps reorganizing them this way :
Class 1 : clear pattern (without end in fact!)
\begin{align}
\sum_{k=1}^n k&=\frac{n(n+1)}2\\
\sum_{k=1}^n k(k+1)&=\frac{n(n+1)(n+2)}3\\
\sum_{k=1}^n k(k+1)(k+2)&=\frac{n(n+1)(n+2)(n+3)}4\\
&\cdots\\
\end{align}
$$\boxed{\displaystyle\sum_{k=1}^n k^{(p)}=\frac{n^{(p+1)}}{p+1}}$$
(th... | {
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Family of Circles A system of Circles pass through $(2,3)$ and have their centers on the line $x+2y-7=0$.
Show that the chords in which the circles of the given system intersects the circle $S_1:x^2+y^2-8x+6y-9=0$ are concurrent and also find the point of concurrency.
ATTEMPT:
The circles of the given system must inter... | The center of the system of circle lies on $x+2y-7=0$, if the center is $(h,k)$ then $h+2k-7=0$
And as the circles passes through $(2,3)$, therefore the radius is $\sqrt{(h-2)^2+(k-3)^2}$
Therefore, the equation of the circle is
$(x-h)^2+(y-k)^2=(h-2)^2+(k-3)^2$ i.e., $x^2+y^2-2xh-2yh+4h+6k+13=0$ we call this $S$
An... | {
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Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\g... | $(a^2+b^2)+(2ab)\ge c^2\implies $ at least one of the two terms in the LHS must be $\ge c^2/2$. Evidently this must be the larger one, viz $(a^2+b^2)$.
Now the second and third inequalities follow from the first (replace $a,b,c$ with $a^2, b^2, c^2/2$ etc).
| {
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Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$ $$f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$$ then how to find limit of $\dfrac{3f(n)}{(n+1)(n+2)}$ as $n\to\infty$?
I do... | Recall the expansion: $$\begin{align}\tan nx &= \dfrac{\binom{n}{1}\tan x - \binom{n}{3}\tan^3 x + \cdots }{1-\binom{n}{2}\tan^2 x + \binom{n}{4}\tan^4 x + \cdots } \\&= \frac{\binom{n}{1}\cot^{n-1} x - \binom{n}{3}\cot^{n-3} x + \cdots }{\cot^{n} x - \binom{n}{2}\cot^{n-2} x + \binom{n}{4}\cot^{n-4} x + \cdots }\end{a... | {
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"answer_id": 0
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Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playi... | First, simplify the difference between the two functions:
$$
\begin{eqnarray}
f(x) - g(x) &=& \frac{2x}{x-3} - x(x - 4) \\
&=& \frac{-x^3 + 7x^2 - 10x}{x-3} \\
\end{eqnarray}
$$
Then, integrate by substituting $u = x-3$:
$$
\begin{eqnarray}
\int_0^2 \! \frac{-x^3 + 7x^2 - 10x}{x-3} \, \textrm{d}x &=& \int_{-3}^{-1} \! ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\7... | Must write more correctly. $Z$ representation as a sum of squares gives the standard approach.
$$X^2+Y^2=Z^{k}$$
Uses the standard formula of the Brahmaputra. View the sum of the squares piece.
$$X^2+Y^2=(ab+cd)^2+(cb-ad)^2=(a^2+c^2)(b^2+d^2)$$
So the sum of squares is converted until
$$(b_1^2+d_1^2)...(b_{k}^2+d_{k}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \inf... | Once you've reduced to
$$\sum_{n = 1}^{\infty} \big((-1)^n + n \left(-\frac 1 3\right)^n\big)$$
you can simply note that the $n$th term of this series does not tend to zero; the right-hand portion tends to $0$ (and hence, is eventually less than $1/2$ in absolute value), while the left hand portion is always $1$ in abs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Express the following complex numbers in standard form $$\left(\frac{\sqrt 3}{2}+\frac{i}{2}\right)^{25}$$
I know that you have to put it in the form $\cos\theta+i\sin \theta$ but I'm not sure how to go about it.
| Applying De-moivre's theorem, $(\cos\theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta$
hence, $$\left(\frac{\sqrt 3}{2}+\frac{i}{2}\right)^{25}$$
$$=\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^{25}$$
$$=\cos\frac{25\pi}{6}+i\sin\frac{25\pi}{6}$$
$$=\cos\left(4\pi+\frac{\pi}{6}\right)+i\sin\left(4\pi+\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help me with the result of this determinant.. $$
D =
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
2 & 1 & 1 & \dots & 1 & 0 \\
3 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n-1 & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=n*1*(-1)^\frac{n(n-1)}{2}
$$
I do... | First do an elementary transformation that does not change the determinant, namely add to the first column the sum of all the others. You are left with
$$
\begin{vmatrix}
n & 1 & 1 & \dots & 1 & 1 \\
n & 1 & 1 & \dots & 1 & 0 \\
n & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n & 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
prove inequality $2(x+z)^3<(3x+z)(x+3z)$ Let $0<x<1,0<z<1$. Then
$$
2(x+z)^{3}<(3x+z)(x+3z)
$$
This checks out numerically, but I don't know why.
|
Expanding, our claim is that
$$ 2x^3 + 6xz^2 + 6x^2z + 2z^3 < 3x^2 + 3z^2 + 10xz$$
$0 < x < 1, 0 < z < 1$.
Note that we can write the RHS as $$2x^2 + 2z^2 + x^2 + z^2 + 10xz $$
We see that
$$ 2x^3 < 2x^2 \\ 2z^3 < 2z^2 \\ 6xz^2 < 5xz + z^2 \\ 6x^2z < 5xz + x^2$$
By adding these together our claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determinant of Tridiagonal matrix I'm a bit confused with this determinant.
We have the determinant
$$\Delta_n=\left\vert\begin{matrix}
5&3&0&\cdots&\cdots&0\\
2&5&3&\ddots& &\vdots\\
0&2&5&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&\ddots&\ddots&0\\
\vdots& &\ddots&\ddots&\ddots&3\\
0&\cdots&\cdots&0&2&5\end{matrix}
... | Let prove the theorem. Suppose the determinant of tri-diagonal matrix as $\Delta_{n}$, and operate the following calculation.
$$
\begin{align}
\Delta_{n}=& \det
\begin{bmatrix}
a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\
c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\
0 & c_{2} & a_{3} & \ddots & a_{n-2} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Express the real root of $x^3-3x+7$ using radicals I want to express the real root of $x^3-3x+7=0$ using radicals.
My attempt is contained in the answer below.
| To express the real root of $x^3-3x+7$ using radicals, we can apply Cardano's method:
\begin{align*}
x^3-3x+7=0&,\text{ let } x=u+v,\quad 3uv=3\implies uv=1\\
\implies& (u+v)^3-3(u+v)+7=0\\
\implies& (u^2+2uv+v^2)(u+v)-3(u+v)+7=0\\
\implies& u^3+2u^2v+uv^2+u^2v+2uv^2+v^3-3(u+v)+7=0\\
\implies& u^3 + v^3+uv(3u+3v)-(3u+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Estimate of n factorial: $n^{\frac{n}{2}} \le n! \le \left(\frac{n+1}{2}\right)^{n}$ on our lesson at our university, our professsor told that factorial has these estimates
$n^{\frac{n}{2}} \le n! \le \left(\dfrac{n+1}{2}\right)^{n}$
and during proof he did this
$(n!)^{2}=\underbrace{n\cdot(n-1)\dotsm 2\cdot 1}_{n!} \c... | Not equal, but by a standard inequality: $\sqrt{ab} \le \frac{a+b}{2}$, so $ab \le \frac{(a+b)^2}{4}$.
So all products $i\cdot ((n+1)-i)$ are estimated above by $\frac{(i + ((n+1)-i))^2}{4} = \frac{(n+1)^2}{4}$. So no equality, but upper bounded by. Because we have $(n!)^2$ we get rid of the square roots again so $n! \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by
$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left|
\begin{array}{c}
\frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\
\frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\fra... | You can do this using this empirical formula to find perimeter of general super-ellipse
$$L=a+b\times\left(\frac{2.5}{n+0.5}\right)^\frac{1}{n}\times \left( b+a\times(n-1)\times\frac{\frac{0.566}{n^2}}{b+a\times\left(\frac{4.5}{0.5+n^2}\right)}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Using Demoivre's Theorem prove that $ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ . $ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ .
Demoivre's Theorem
$$ \{\cos \theta + i \sin \theta \}^n = \cos n\theta + i\sin n\theta $$
Where n is an integer .
I tried
$$ ... | Notice, $$\cos5\theta=(\cos\theta+i\sin\theta)^5-i\sin 5\theta$$
$$=(\cos\theta+i\sin\theta)^2(\cos\theta+i\sin\theta)^3-i\sin 5\theta$$
$$=(\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta)(\cos^3\theta-3\sin^2\theta\cos\theta-i\sin^3\theta+3i\sin\theta\cos^2\theta)-i\sin 5\theta$$
comparing the real parts on both the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from
$\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$.
For some $\theta$ between $a-1$ and $a$ and odd $n\in\mathbb{N}$ we have equality... | Direct solution
$\int_a^{a+1} \ln(x)\ dx = [ x \ln(x) - x ]_a^{a+1}$
$ = (a+1) \ln(a+1) - a \ln(a) - 1$
$ = \ln(a) + (a+1) \ln(1+\frac{1}{a}) - 1$
$ = \ln(a) + (a+1) ( \frac{1}{a} - \frac{1}{2a^2} + \frac{1}{3a^3} - \frac{1}{4a^4} + \cdots ) - 1$
$ = \ln(a) + \frac{1}{2a} - \frac{1}{6a^2} + \frac{1}{12a^3} - \cdots$ [w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of ordered pairs that have a product of $10!$ and a least common multiple of $9!$ Problem
How many ordered pairs $(a,b)$ of positive integers have a product of $10!$ and a least common multiple of $9!$?
I was told my answer of $16$ to this problem was wrong. I don't see how since we just have to distr... | You can rephrase it as: how many ordered pairs $(a,b)$ of positive integers satisfy $ab=10!$ and $\gcd(a,b)=10$, because $ab=\gcd(a,b)\text{lcm}(a,b)$.
I.e. let $a=10a_1, b=10b_1$, and you're asking how many ordered pairs $(a_1,b_1)$ of positive integers satisfy $a_1b_1=\frac{10!}{100}=2^6\cdot 3^4\cdot 7$ and $\gcd(a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show by induction that $(n^2) + 1 < 2^n$ for intergers $n > 4$ So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$.
for $n = k + 1$ I get into a bit of a kerfuffle.
I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently:
$(k + 1)^2 + 1 < 2^k * 2$.
A b... | First, show that this is true for $n=5$:
$5^2+1<2^5$
Second, assume that this is true for $n$:
$n^2+1<2^n$
Third, prove that this is true for $n+1$:
$(n+1)^2+1=$
$n^2+\color\green{2}\cdot{n}+1+1<$
$n^2+\color\green{n}\cdot{n}+1+1=$
$n^2+n^2+1+1=$
$2\cdot(\color\red{n^2+1})<$
$2\cdot(\color\red{2^n})=$
$2^{n+1}$
Please... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$ Let $x,y \in \mathbb{R} $ such that;
If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1 \leq y^2 +1 -1 < 2\d... | You came close. We have
$$\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|=\frac{|x^2-y^2|}{\sqrt{1+x^2}+\sqrt{1+y^2}} =|x-y|\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}. $$
Now
$$\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\le \frac{|x|+|y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\lt 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Division in finite fields Let's take $GF(2^3)$ as and the irreducible polynomial $p(x) = x^3+x+1$ as an example. This is the multiplication table of the finite field
I can easily do some multiplication such as $$(x^2+x)\cdot(x+1) = x^3 + x^2 + x +1 = x+1+x^2+x+2 = x^2$$
I am wondering how to divide some random fields s... | Since any element of this field can be written as $ax^2+bx+c$ we write:
$\frac{x}{x^2+x+1}=ax^2+bx+c$
$x=(ax^2+bx+c)*(x^2+x+1)$
$x=ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c$
Use the fact that $x^3=x+1$ to further simplify.
$x=a(x^2+x)+(a+b)(x+1)+(a+b+c)x^2+(b+c)x+c$
$x=(b+c)x^2+(c)x+(a+b+c)$
Now solve for $a$, $b$, and $c$ by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\... | Note that the series $\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^n$ is valid only for $-1<x\le 1$. We can write for $|x|>1$,
$$\begin{align}
\log(1+x)&=\log x+\log \left(1+\frac1x\right) \\\\
&=\log x +\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^{-n} \tag 1
\end{align}$$
Using $(1)$, we can write for $a>1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1... | All we need to facilitate analysis is the double angle formula
$$\sin(2a)=2\sin(a)\,\cos(a) \tag 1$$
the Prosthaphareis Identity
$$\cos (a-b)-\cos(a+b)=2\sin (a)\,\sin (b) \tag 2$$
along with the Reverse Prosthaphareis Identity
$$\sin (a)+\sin (b)=2\sin\left(\frac{a+b}{2}\right)\,\cos\left(\frac{a-b}{2}\right) \tag 3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$? How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( shou... | The slope of the plane $x+2z=3$ is smaller than the slope of the cone, so their intersection curve is an ellipse, drawn with brown in the figure. The plane $z=0$ passes through the apex. Therefore, the region we are interested in is a ''hat''.
The projection of the surface region onto to the $x-y$ co-ordinate plane is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for Rationals $p,q,r$ Satisfying $\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1$. Find all rational solutions $(p,q,r)$ to the Diophantine equation
$$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,.$$
At least, determine an infinite family of $(p,q,r)\in\mathbb{Q}^3$ satisfying thi... | If you combine the system into a single equation, you find that $r$ satisfies the quadratic equation
\begin{equation*}
r^2-\frac{4(p^2q+p(q^2-4q-1)-q)}{p^2(q^2-1)-4pq-q^2+1}r-1=0
\end{equation*}
For $r \in \mathbb{Q}$, the discriminant must be a rational square, so there must exist $d \in \mathbb{Q}$ such that
\begin{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | i) $k=2n$:
\begin{align}
\sum_{m=1}^{2n}(-1)^{m-1}m^2&=\sum_{m=1}^{2n}m^2-2\sum_{m=1}^{n}(2m)^2\\
&=\sum_{m=1}^{2n}m^2-8\sum_{m=1}^{n}m^2\\
&=\frac{2n(2n+1)(4n+1)}{6}-8\frac{n(n+1)(2n+1)}{6}\\
&=\frac{2n(2n+1)(4n+1-4n-4)}{6}\\
&=\frac{2n(2n+1)(-1)}{2}=-\frac{k(k+1)}{2}=(-1)^{k-1}\frac{k(k+1)}{2}
\end{align}
ii)$k=2n+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I se... | The usual way to do something like this is to square to get rid of one radical, rearrange, and square again to get rid of the remaining radical, as @vhspdfg says in a comment. I won’t detail the intermediate steps, but the octic polynomial I got was
$$
2000 - 1600x - 3480x^2 + 2960x^3 + 825x^4 - 1340x^5 + 510x^6 - 140... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of function $ f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ Where $x>0$
Find the range of the function $\displaystyle f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ where $x>0$
$\bf{My\; Try::}$ Let $\sqrt{x}=t\;,$ Then $\displaystyle f(t) = t^3+\frac{1}{t^3}-4\left(t^2... | HINT: $$f'(x)=3/2\,\sqrt {x}-3/2\,{x}^{-5/2}-4+4\,{x}^{-2}$$
solving for $x$ we obtain
$$1,1/4\, \left( \sqrt {5}-3 \right) ^{2},1/4\, \left( \sqrt {5}+3 \right) ^{2}$$
plugging $1$ in $f(x)$ we obtain $-6$ plugging the other two terms in $f$ we get $$-10$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right... | Your solution is almost correct. You should note that $x=0$ is neither a point of maximum or minimum, because $f(0)=0$, but $f(2)>0$ and $f(-2)<0$. So considering
$$
f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}
$$
for $x\n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:
Let $x$, $y$, $z$ be real
numbers. Prove that
$$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$
The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square ... | Its just a matter of factorization:
$$
(x^2+1)(y^2+1)(z^2+1)+8-2(x+1)(y+1)(z+1)=(x^2y^2z^2-2xyz+1)+\left(\sum_{cyc}x^2y^2-2xy+1\right)+\left(\sum_{cyc}x^2-2x+1\right)=(xyz-1)^2+\left(\sum_{cyc}(xy-1)^2\right)+\left(\sum_{cyc}(x-1)^2\right)≥0
$$
With equality only if $x=y=z=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving Fourier series of $ f(x)=\begin{cases} x+1 ;-1Please take a look at below Fourier series :
$ f(x)=\begin{cases} x+1 &-1<x<0\\ 1-x & 0<x<1 \end{cases} $
I tried to solve it as follows :
$
a_n=\displaystyle \dfrac{1}{1}\int_{-1}^1f(x)\cos (n\pi x) dx=2\int_{0}^{1} (1-x)\cos( n\pi x) dx
$
solving $a_n$ comes to :
... | As asked in a comment, I compute the Fourier coefficients :
\begin{align}
a_0(f) & = \int_{-1}^{1}f(x)\mathrm{d}x \\
& = \int_{-1}^{0}\left(x+1\right)\mathrm{d}x + \int_{0}^{1}\left(1-x\right)\mathrm{d}x \\
& = \left[\frac{x^2}{2}+x\right]_{-1}^{0} + \left[x-\frac{x^2}{2}\right]_{0}^{1} \\
& = \left[-\frac{1}{2}+1\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What will be the remainder when $2^{31}$ is divided by $5$? The question is given in the title:
Find the remainder when $2^{31}$ is divided by $5$.
My friend explained me this way:
$2^2$ gives $-1$ remainder.
So, any power of $2^2$ will give $-1$ remainder.
So, $2^{30}$ gives $-1$ remainder.
So, $2^{30}\times 2$ or ... | We have that $5$ is a prime, so, $2^4 \equiv 1 \pmod 5$ (see Fermat's little theorem).
Then, $(2^4)^{7} = 2^{28} \equiv 1 \pmod 5$. Multiplying this by $2^3$, we get $2^{31} \equiv 8 \equiv 3 \pmod 5$.
Your friend is saying that:
$$2^{30} \equiv -1 \pmod 5 \implies 2^{31} \equiv -1 \times 2 \equiv -2 \equiv -2 + 5 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$
Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality.
$$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1)... | Note that $\displaystyle \sum_{i = 1}^n \dfrac{1}{1+a_i} = 1 \implies \sum_{i = 1}^n \dfrac{a_i}{1+a_i} = n-1$. Then, $$\displaystyle \sum_{i = 1}^n \sqrt{a_i}-(n-1)\sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \sum_{i = 1}^n \dfrac{1}{1+a_i}\sum \sqrt{a_i} - \sum_{i = 1}^n \dfrac{a_i}{1+a_i} \sum_{i = 1}^n \dfrac{1}{\sqrt{a_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k ... | Use modular arithmetic (this is not the most simple way but it also works very well for similar, more complicated problems). Notice that for any odd $a$ we have $a^2 \equiv 1 \ \text{(mod 4)}$ and $a^2 \equiv 1 \ \text{(mod 8)}$. So $a^2 + 3 \equiv 0\ \text{(mod 4)}$, which means that $a^2 + 3$ is divisable by $4$, and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent expansion on an annulus problem I have $f(z)= \frac{1}{\sin(z)}$ and am required to show that on the disc {$0<|z|<\pi$} the Laurent expansion is equal to:
$$c_{-1}z^{-1}+\sum_{n=0}^\infty{c_nz^n}$$
My plan is use the expansion of $$\sin(z)= z-\frac{z^3}{3!}+\frac{z^5}{5!}...$$ and rewrite $\frac{1}{\sin(z)}$ as... | $$
z\frac{1}{\sin(z)}
$$
has no singularities in $0 < |z| < \pi$ and has a limit at $z=0$ of $1$, which means that this function has a removable singularity at $z=0$. So you have a power series expansion of the following in $|z| < \pi$:
$$
z\frac{1}{\sin(z)} = 1 + a_1z + a_2 z^2+ a_3z^3... | {
"language": "en",
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A formula which gives the maximum of a series of numbers This formula gives the maximum of 3 numbers:
$$\frac{a}{2} + \frac{b}{4} + \frac{c}{4} + \frac{|b-c|}{4} + \frac{1}{2}\left|a -\frac{b}{2} - \frac{c}{2} - \frac{|b-c|}{2}\right| = \max(a,b,c)$$
I've found this over the internet, I have no idea how can one develop... | Let's start with $2$ numbers. The best way to see this is to imagine the numbers on the number line.
We find the midpoint of the $2$ numbers: $\frac{a+b}{2}$.
Next, we can find half the distance of the $2$ numbers: $\frac{|a-b|}{2}$
Adding them up, we have $\max(a, b)=\frac{a+b}{2}+\frac{|a-b|}{2}$.
The formula you hav... | {
"language": "en",
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"source": "stackexchange",
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General solution for matrix inverse I don't know if anyone know about this,
but solving gpcm(generalized partial credit model) requires the inverse of the matrix of the form below.
in Mathetmatica langauge,
{{b1, -1, 0},{0, b2, -1},{1,1,1}} ^-1
{{b1, -1, 0, 0},{0, b2, -1, 0},{0,0,b3,-1},{1,1,1,1}} ^-1
cf) http://ww... | Denote your matrix by $B$ and the inverse matrix by $B^{-1} = A$. Let us write
$$ B \begin{pmatrix} x_1 \\ \vdots \\ x_n \\ x_{n+1} \end{pmatrix} = \begin{pmatrix} b_1 x_1 - x_2 \\ \vdots \\ b_n x_n - x_{n-1} \\ \sum_{i=1}^{n+1} x_i \end{pmatrix} = \begin{pmatrix} y_1 \\ \vdots \\ y_n \\ y_{n+1} \end{pmatrix}. $$
We al... | {
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | First I'll prove a Lemma: $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$.
Proof: it follows from the Rearrangement Inequality, because $(a,b,c)$ and $(a,b,c)$ are similarly sorted.
Or notice that it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or add the following inequa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Infinite primes proof There is a proof for infinite prime numbers that i don't understand.
right in the middle of the proof:
"since every such $m$ can be written in a unique way as a product of the form:
$\prod_{p\leqslant x}p^{k_p}$. we see that the last sum is equal to: $\prod_{\binom{p\leqslant x}{p\in \mathbb{P}}}... | Suppose the prime numbers not exceeding $x$ are $2$, $3$, and $5$. Then
\begin{align}
& \prod_{\begin{smallmatrix} p\in\mathbb P \\ p\le x \end{smallmatrix}} \sum_{k\ge 0} \frac 1 {p^k} \\[10pt] = {} & \left( 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots+ \frac 1 {2^k} + \cdots \right) \left( 1 + \frac 1 3 + \frac 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$ The equation $x^2+bx+c=0$ has distinct roots .If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$
Let $\alpha$ and $\beta$ are the root... | Hint...$\alpha^2-2\alpha-1=0\Rightarrow b=-2, c=-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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To find constant term of polynomial defined recursively
Let $\{f_n(x)\}$ be a sequence of polynomials defined inductively as
\begin{align*}
f_1(x) &= (x-2)^2, \\
f_{n+1}(x) &= (f_n(x)-2)^2, \quad n \geq 1.
\end{align*}
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f... | To expand on the comment by Vinod, for any polynomial, we get the constant term by plugging 0 in for $x$. In fact, that's pretty much the definition of the "constant term". Doing this we see that we will get
$$\begin{align*}
(\cdots (0-2)^2 - 2)^2 - \cdots - 2)^2 &= (\cdots ((-2)^2-2)^2 - 2)^2 - \cdots - )^2 \\
&= (\cd... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integral $\int \frac{1}{x^3\sqrt{x^2+x-1}}\mathrm dx$ What substitution to use in this integral? I tried to factorize $x^2+x-1$ and use a substitution $u=\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.
| $$I=\int \frac{1}{x^3\sqrt{x^2+x-1}}dx$$
Euler Substitution:
Let $x=\frac{u^2+1}{2u+1}$, $$I=2\int \frac{(2u+1)^2}{(u^2+1)^3}du$$
Partial fraction get $I=2\int \frac{4u-3}{(u^2+1)^3}du+8\int\frac{1}{(u^2+1)^2}du$
| {
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Evaluate the integral $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$ $$x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}=\frac{\sqrt{(-\sqrt[3]{x^2}+1)}}{\sqrt[3]{x^4}}$$
Is it necessary to simplify the function further? What substitution is useful?
$u=\sqrt[n]{\frac{ax+b}{cx+d}}$ doesn't work.
| Notice, $$\int x^{-4/3}\left(-x^{2/3}+1\right)^{1/2}\ dx$$$$=\int \frac{1}{x}\left(-x^{2/3}+1\right)^{1/2}(x^{-1/3}\ dx)$$
Let $-x^{2/3}+1=\sin^2\theta\implies -\frac{2}{3}x^{-1/3}\ dx=2\sin\theta\cos\theta\ d\theta$ or $x^{-1/3}\ dx=-3\sin\theta\cos\theta\ d\theta $
$$=\int\frac{1}{(1-\sin^2\theta)^{3/2}}(\sin\theta)(... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{i = 1}^n \frac{x_i}{i^2} \geq \sum_{i = 1}^n \frac{1}{i}.$
Let $x_1,x_2,\ldots,x_n$ be distinct positive integers. Prove that $$\displaystyle \sum_{i = 1}^n \dfrac{x_i}{i^2} \geq \sum_{i = 1}^n \dfrac{1}{i}.$$
Attempt
I tried using Cauchy-Schwarz and I got that $$(x_1^2+x_2^2+\cdots+x_n^2) \left (\d... | Let $0 < y_1 < y_2 < \ldots < y_n$ be an rearrangement of $x_1, x_2, \ldots x_n$ sorted in ascending order.
Since $y_k$ are distinct positive integers, one can show $y_k \ge k$ for $k = 1,\ldots, n$ by induction.
Since the finite sequence $\frac{1}{1^2}, \frac{1}{2^2}, \ldots, \frac{1}{n^2}$ is ascending, by rearrang... | {
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Why is $(1+\frac{3}{n})^{-1}=(1-\frac{3}{n}+\frac{9}{n^2}+o(\frac{1}{n^2}))$ and how to get around the Taylor expansion?
Let be $(u_n)$ a real sequence such that $u_0>0$ and that $\forall n \in \mathbb{R}$:
$$\frac{u_{n+1}}{u_n}=\frac{n+1}{n+3}$$
Let be $(v_n)$ a real sequence such that $\forall n \in \mathbb{R}$:
$$v... | As jordan's comment notes, the sum is telescoping: $$\begin{align*} \sum_{n=1}^N \log \frac{v_{n+1}}{v_n} &= \sum_{n=1}^N \log v_{n+1} - \log v_n \\ &= \log v_{N+1} + \sum_{n=2}^{N} \log v_n - \sum_{n=2}^N \log v_n - \log v_1 \\ &= \log v_{N+1} - \log v_1 \\ &= \log \frac{v_{N+1}}{v_1}. \end{align*}$$ Next, it is eas... | {
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A trigonometric identities with the ratio of four terms like $1+(\frac{\tan x}{\sin y})^2$ Prove:
$$\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}$$
I started by opening the brackets and squaring... | Notice, the given equality can be easily proved by simplifying $LHS$ $$LHS=\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}$$
$$=\frac{1+\left(\frac{\sin x}{\sin y\cos x}\right)^2}{1+\left(\frac{\sin x}{\sin z\cos x}\right)^2}$$
$$=\frac{\sin^2z(\sin^2 y\cos^2 x+\sin^2 x)}{\sin^2y(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Line integral of a vector field $$\int_{\gamma}ydx+zdy+xdz$$
given that $\gamma$ is the intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ and its projection in the $xz$ plane is taken clockwise.
In my solution, I solved the non-linear system of equations and I found that $x^2+y^2+z^2 = 4$. Given the projection in the $... | Let us just rewrite some equations first
$$\begin{align}
x^2+y^2+z^2&=2(x+y)\\
(x^2-2x+1)+(y^2-2y+1)+z^2-2&=0 \\
(x-1)^2+(y-1)^2+z^2&=2 \\
\end{align}$$
So this is a sphere of radius $R=\sqrt{2}$ centered at $(1,1,0)$. It's parametric equations are
$$\begin{align}
x&=1+\sqrt{2}\sin u \cos v \\
y&=1+\sqrt{2}\sin u \sin ... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
\ge 1$.
My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of ... | Based on the inequality
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
we have shown in This Question, for any positive $x$, $y$, $z$, $a$, $b$, and $c$, we have that
\begin{align*}
a^2 + 3b^2 + 5c^2 &\ge 3a^2 + 3 b^2+3c^2\\
&\ge \frac{(a+b+c)^2}{\frac{1}{3} + \fr... | {
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Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$
Do you know if there are other integer solutions to
$$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$
besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?
... | Here's a solution when $a$, $b$, $c$, and $d$ are positive; I'm not sure how to tackle the more general problem though (although something tells me it's probably similar).
Note that substituting $b=cd-a$ into the second equation yields $$a(cd-a)=c+d\implies a^2-cad+c+d=0.$$ By the Quadratic Formula on $a$, we have $$a=... | {
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Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+................ | Notice $$
\begin{align}
& \frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \cdots\\
= & \frac12 \left[ \frac{3-1}{6} + \frac{(3-1)(6-1)}{6^2 \cdot 2!} + \frac{(3-1)(6-1)(9-1)}{6^33!} + \cdots\right]\\
= & \frac12 \left[ \frac{\frac23}{2} + \frac{\frac23(\frac23+1)}{2^2\cdot 2!}
+ \frac{\frac23... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| The easiest way is by brute checking.
\begin{align*}
a^2&=\frac{(t^2-1)^2}{(t^2+1)^2}\\
b^2&=\frac{4t^2}{(t^2+1)^2}\\
a^2+b^2&=\frac{t^4-2t^2+1+4t^2}{(t^2+1)^2}\\
&=\frac{t^4+2t^2+1}{(t^2+1)^2}\\
&=\frac{(t^2+1)^2}{(t^2+1)^2}\\
&=1
\end{align*}
So we know that if we can find a $t$ everything works out fine. Now observe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$
For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$
Attempt
Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \g... | Use Cauchy–Schwarz's inequality as follows :
$$(a(b+2c)+b(c+2a)+c(a+2b)) \left (\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \right ) \geq (a+b+c)^2$$ and then :
$$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$
Now it's easy to see that $$\frac{(a+b+c)^2}{3(ab+bc+ca)} \geq 1$$ this b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$.Find $a,b,c$.
The... | Let $C(3,4),D(1,2)$. Also, let $E(X,Y)$ be the midpoint of $AB$.
$\qquad\qquad\qquad$
Since $\triangle{CAE}$ is a right triangle with
$$|AC|=6,\quad |CE|=\sqrt{(X-3)^2+(Y-4)^2}$$
we have
$$|AE|^2=|AC|^2-|CE|^2=36-(X-3)^2-(Y-4)^2\tag1$$
Also, since we can see that $D$ is on the circle whose diameter is the line segment ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I pref... | Hint do you know about Maclaurin series?
You can write
$$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right) = \lim _{x\to 0}\left(\frac{x\left(\sqrt{3 \cdot (1 + x + \frac{x^2}{2} + O(x^2)) +(1 + 3x^2 + O(3x^2))}-2\right)}{4-\left((1 - \frac{x^2}{2} + O(x^2))+1\right)^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$g(Z^3,Z^2) = 0$ implies $g(X,Y) = (X^2 - Y^3)h(X,Y)$ Let $F$ be a field and $g(X,Y) \in F[X,Y]$. Suppose that $g(Z^3,Z^2) = 0$ where $g(Z^3, Z^2) \in F[Z]$. I want to show that $g(X,Y) = (X^2 - Y^3) h(X,Y)$ for some $h(X,Y) \in F[X,Y]$.
My first idea is that $g(X,Y)$ can be expressed as $g(X,Y) = X^2 p(X,Y) +Y^3 q(X,... | It is better to start with (by division algorithm)
$$g(X,Y)=(X^2-Y^2)h_1(X,Y)+X\cdot h_2(Y)+h_3(Y)$$
Plugging $X=Z^3, Y=Z^2$ gives:
$$g(Z^3,Z^2)=(Z^6-Z^6)h_1(Z^3,Z^2)+Z^3\cdot h_2(Z^2)+h_3(Z^2)$$
which implies
$$Z^3\cdot h_2(Z^2)+h_3(Z^2)=0\quad\quad \quad \text{for all }Z $$
The first component contains odd order term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you find the value of $m$ and $n$ if $x+y+z=\frac{m}{\sqrt n}$ given certain conditions on x,y,z? Problem:
Let $x,y$ and $z$ be real numbers satisfying:
$$x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$y=\sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}$$
$$z=\sqrt{x^2 - \frac{1}{36}} + \sqrt{y^... | Working on The hint of joey we can make this triangle and take x,y,z as the sides of triangle as follows without the loss of generality....
Now just to make a detailed figure ....
Just to sum these up ...from the figure we have ...
$$DC=x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$BD=y=\sqrt{z^2 - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it ... | Using the identity $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$, we get
$$
\begin{align}
&\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}\\
&=\frac{\left(n^3+n^{1/2}\right)-\left(n^3-1\right)}{\left(n^3+n^{1/2}\right)^{2/3}+\left(n^3+n^{1/2}\right)^{1/3}\left(n^3-1\right)^{1/3}+\left(n^3-1\right)^{2/3}}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How can I evaluate $\sum_{i=0}^\infty \frac{1}{k^i} \binom{2i}{i}$ Evaluate $$\sum_{i=0}^\infty \left(\frac{\binom{2i}{i}}{k^i}\right),$$
where $k$ is a whole number.
I can't figure out how to approach this question, as no binomial series has such coefficients.
| Note, that
$$\binom{2i}{i}=(-4)^i\binom{-\frac{1}{2}}{i}$$
So, we can write OPs series as binomial series
\begin{align*}
\sum_{i=0}^{\infty}\binom{2i}{i}\frac{1}{k^i}
&=\sum_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}\left(-\frac{4}{k}\right)^i\\
&=\frac{1}{\sqrt{1-\frac{4}{k}}}\\
&=\sqrt{\frac{k}{k-4}}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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