Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1... | \begin{align}
\sum_{n=5}^\infty \frac{8}{n^2 -1}
&=\sum_{n=5}^\infty \frac{8}{(n -1)(n+1)}\\
&=\sum_{n=5}^\infty \left(\frac{4}{n-1}-\frac{4}{n+1} \right)\\
&=\left(\frac 44 - \color{red}{\frac 46} \right)
+ \left(\frac 45 - \color{blue}{\frac 47} \right)
+ \left(\color{red}{\frac 46} - \color{green}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Finding a sequence of elementary matrices So I have this matrix
A = $\begin{bmatrix}2 & 4\\1 & 1\end{bmatrix}$
I am tasked with finding all the elementary matrices such that Ek...E2E1A = I. Use this sequence to write both A and A-1 as products of elementary matrices/
I ended up getting four elementary matrices
E1 = $\... | A matrix is elementary if it differs from the identity matrix by a single elementary row or column operation.
See for example, Wolfram MathWorld Elementary Matrix
The Gauss-Jordan reduction process is expressed as a sequence of elementary operations:
$$
%
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}
\end{array}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex conjugate and their product I have two complex numbers that are non real, $k$ and $z$. $k$ and $z$ are going to be complex conjugates if and only if the product $(x-k)(x-z)$ is a polynomial with real coefficients.
Here is my answer :
$$k=a+bi\\
z=c+di\\
(x-k)(x-z) = x^2-(k+z)x+kz$$
After that, I'm not sure what... | Okay, look at it this way.
Let $ax^2 + bx + c$ be a 2nd degree polynomial with real coefficients and no real roots. But it does have complex roots. What are the roots?
Well by the quadratic equation the are
$z = \frac {-b}{2a} + \frac {\sqrt{4ac - b^2}}{2a} = \frac {-b}{2a} + \frac {\sqrt{b^2 - 4ac}}{2a}i$ (as there ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Mistake in my NLP using Lagrange Multipliers?
I have the following NLP
\begin{matrix}\text{Minimize:} &x^2+y^2+z^2 \\
\text{Subject To:} & x+2y+z-1=0 \\ &2x -y -3z-4=0\end{matrix}
I need to solve this using the Lagrange Multiplier Method.
My attempt:
$$L(x,y,z,\lambda_1, \lambda_2) = x^2+y^2 +z^2 +\lambda_1(x+y+z-1) ... | The right values (solving the system by row reduction) are
$$
x=\frac{16}{15},\ y=\frac13\, z=-\frac{11}{15},\ \lambda_1=-\frac{52}{75},\ \lambda_2=-\frac{18}{25}.
$$
Plugging the values you can check that you made a mistake in your second equation for the lambdas. It should have been
$$
3\lambda_1-14\lambda_2-8=0.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Algebra with complex numbers Let $m$ be the minimum value of $|z|$, where $z$ is a complex number such that
$$ |z-3i | + |z-4 | = 5. $$
Then $m$ can be written in the form $\dfrac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find the value of $a+b$.
I cannot seem to get rid of the absolute values, ... | Given $|z-4|+|z-3i|=5\;,$ Now Put $z=x+iy\;,$ We get $$\displaystyle \sqrt{(x-4)^2+y^2}+\sqrt{x^2+(y-3)^2}=5$$
Now Using Minkowski inequality, We get
$$\sqrt{(4-x)^2+y^2}+\sqrt{x^2+(3-y)^2}\geq \sqrt{[(4-x)-x]^2+[y-(3-y)]^2}=5$$
And equality Hold when $$\frac{4-x}{x}=\frac{y}{3-y}\Rightarrow 12-3x-4y+xy=xy$$
So we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving the equation $z^7=-1$
Solve the equation $z^7=-1$
My attempt:
$$z=x+yi$$
$$(x+yi)^7+1=0$$
$$(x^2+2yi-y^2)^3(x+yi)+1=0$$
but now it's start to look ugly.
I'm sure that there is a simple way
| We can say that $$z^7=-1$$
or,$$z^7=\cos\pi+i\sin\pi$$
or,$$z^7=\cos(4n+2)\pi+i\sin(4n+2)\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
or,$$z=\left[\cos(4n+2)\pi+i\sin(4n+2)\pi\right]^{\frac{1}{7}} \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
or,$$z=\cos(\frac{4n+2}{7})\pi+i\sin(\frac{4n+2}{7})\pi \,\ \text{where} \,\ n=0,1,2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
| We have
$$9^{2n-2} = (7+2)^{2n-2} = \sum_{k=0}^{2n-2} \dbinom{2n-2}k 7^k2^{2n-2-k} = 2^{2n-2} + 7M$$
Hence, we have
\begin{align}
4^n+10 \cdot 9^{2n-2} & = 4^n + 10 \cdot (2^{2n-2}+7M) = 4^n+10\cdot 4^{n-1} + 70M = 4^{n-1}(4+10)+70M\\
& = 14(5M+4^{n-1})
\end{align}
Hence, in fact we have that $14$ divides $4^n+10 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
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Evaluating definite integral - Let's say we want to evaluate $$ \int_0^{2\pi} \frac{1}{a^2\cos^2x+b^2\sin^2x}dx$$
With substitution, one obtains $$ \frac{1}{ab} \arctan\left(\frac ba \tan x\right) $$
as antiderivate. For more details on how to do this, see this question.
Now my question is, why do I receive $0$ if I in... | Notice, the property of the definite integral $\int_{0}^{2a}f(x)\ dx=2\int_{0}^{a}f(x)\ dx\ \ \ \ \ \ \forall \ \ \ f(2a-x)=f(x)$
$$\int_{0}^{2\pi} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$$$=2\int_{0}^{\pi} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$ $$=4\int_{0}^{\pi/2} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$
$$=\frac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to integrate a 4th power of sine and cosine? I'm having some trouble figuring out the right substitutions to make to integrate
$$\int \sin^4(\theta)d\theta$$
and
$$\int \cos^4(\theta)d\theta$$
Any hints or suggestions are welcome.
Thanks,
| Notice that you can write $\sin^4(x)$ as follows:
\begin{align}
\sin^4(x) = &\ (\sin^2(x))^2 = \Big(\frac{1 - \cos(2x)}{2}\Big)^2 \\
= &\ \frac{1 -2\cos(2x) + \cos^2(2x)}{4} \\
= &\ \frac{1}{4} - \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8}.
\end{align}
For $\cos^4(x)$ we procede as above:
\begin{align}
\cos^4(x) = &\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(... | Seeing it as a polynomial of one variable might help.
$$\begin{align}&-a^3b+a^3c+ab^3-ac^3-b^3c+bc^3\\&=(c-b)a^3+(b^3-c^3)a+bc^3-b^3c\\&=(c-b)a^3+(b-c)(b^2+bc+c^2)a+bc(c-b)(c+b)\\&=(c-b)(a^3-(b^2+bc+c^2)a+bc(c+b))\\&=(c-b)((-a+c)b^2+(-ac+c^2)b+a^3-ac^2)\\&=(c-b)((c-a)b^2+c(c-a)b+a(a-c)(a+c))\\&=(c-b)(c-a)(b^2+bc-a(a+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How many ways can $3$ regular polygons meet at a vertex? This is equivalent to the positive integer solutions to $$\frac{a-2}{a} + \frac{b-2}{b} + \frac{c-2}{c} = 2$$ with $3 \le a \le b \le c$.
Small solutions like $(6, 6, 6)$ and $(4, 8, 8)$ can be guessed, but other solutions such as $(4, 5, 20)$ exist.
| Well, basically ${1\over a}+{1\over b} +{1\over c}={1\over2}$ is what you want.
Let $a\geq b\geq c$ we know since ${({1\over2})\over3}={1\over6}$, ${1\over c}\geq{1\over6}$ and hence $c\leq 6$.
This leaves us four possibilities.
(1) $c=3$ then ${1\over a}+{1\over b}={1\over 6}\implies ab=6a+6b\implies (a-6)(b-6)=36$
(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the Integral: $\int\frac{5x+1}{(2x+1)(x-1)}dx$ $$\int\frac{5x+1}{(2x+1)(x-1)}dx$$
$$\frac{A}{(2x+1)}+\frac{B}{(x-1)}$$
$$\frac{A(x-1)+B(2x-1)}{(2x+1)(x-1)}$$
I put $x=1$ to attain B: $5\ (1)+1=A(1-1)+B\ (2(1)+1)$
$$6=0+3B$$
$$B=2$$
However, I am unable to utilize the same method to attain A.
What should I do? ... | You had a mistake in your computations. See the first line below. You have written $B(2x-1)$ instead of $B(2x+1)$.
Consider the following steps
$$\eqalign{
& {{5x + 1} \over {(2x + 1)(x - 1)}} = {A \over {2x + 1}} + {B \over {x - 1}} = {{A\left( {x - 1} \right) + B\left( {2x + 1} \right)} \over {(2x + 1)(x - 1)}} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proof: $f(x)=x^2+x-4$ is continuous at $x=2$ Proof: Let $\epsilon>0$ and $\delta =\min\{1,\epsilon/6\}$.Then if $|x-2|<\delta$, $|x-2|<1$, so $|x+3|<6$.
Thus, $$|f(x)-f(c)|= |x^2+x-4-(2^2+2-4)| = |x^2+x-6|=|x-2||x+3|<6\delta \leq \epsilon$$
Doing this problem as practice. Thoughts? comments?
| If $x \in \Bbb{R}$, then $|f(x) - f(2)| = |x^{2}+x-6| = |x-2||x+3|$; we have $|x-2| < 1$ only if $|x| - 2 \leq |x-2| < 1$, only if $|x| + 3 < 6$, only if $|x+3| \leq |x|+3 < 6$, and only if $|x-2||x+3| < 6|x-2|$; given any $\varepsilon > 0$, we have $6|x-2| < \varepsilon$ if $|x-2| < \varepsilon/6$. Hence we have prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{... | Hint :
$$3^{3n+4}+2^{n+2}=27\times 3^{3n+1}+2\times 2^{n+1}=2\times(3^{3n+1}+2^{n+1})+25\times 3^{3n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
$x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}$. Then what is the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\}$?
If the equation $x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}\;,$ Where $x_{1}<x_{2}<x_{3}$.
Then the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\} = \;,$ Where $\{x\}$ Represent fractional part of ... | Note that
$$\{x\}=x-\lfloor x\rfloor$$
and that
$$x_1+x_2+x_3=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the probability of getting exactly 3 two's OR three's when a die is rolled 8 times? What is the probability of getting exactly $3$ two's OR three's when a die is rolled $8$ times?
I know that $P(E) = |E| / |S|$.
I believe that $|S| = 36$, since there are $36$ different combinations when rolling a die. I am not... | Probability of exactly $3$ twos is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ threes is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ twos and exactly $3$ threes is $\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to solve the equation $t = \sqrt{x^2 - 1} - x$ for $x$?
Let the equation $t = \sqrt {x^2 - 1} - x$. Find $x$.
So I've tried the following:
$$t^2 = x^2 - 1 -2\sqrt {x^2-1}x + x^2 = 2x^2 - 2\sqrt{x^2-1}x - 1 $$
What should I do next?
| Note that $$\left(\sqrt{x^2-1}-x\right)\left(\sqrt{x^2-1}+x\right) = -1,$$ so $$\frac{1}{t}=-\sqrt{x^2-1}-x$$ Adding we get $$t+\frac{1}{t}=-2x$$ Or:
$$x=\frac{-1}{2}\left(t+\frac{1}{t}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I complete the proof of proving the $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $ $$\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $$
Proof: Let $\epsilon > 0$
Then, $$ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $$
$$\Longleftrightarrow \le... | Rewrite it as
$$
\lim_{x\to\infty}\frac{1+\dfrac{1}{x^2}}{1+\dfrac{2}{x}+\dfrac{1}{x^2}}
=1
$$
For $x>0$, the denominator is $>1$ and it's not restrictive to assume it.
The inequality to be solved becomes then
$$
\left|-\frac{2}{x}\right|<
\varepsilon\left(1+\dfrac{2}{x}+\dfrac{1}{x^2}\right)
$$
which is certainly sati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the fo... | Hint for an intuitive solution:
$$\left|\frac{\sqrt{x+1}}{\sqrt{x}+1}-1\right|<\left|\frac{\sqrt{x+1}}{\sqrt{x}}-1\right|=\left|\sqrt{\frac{x+1}{x}}-1\right|=\left|\sqrt{1+\frac{1}{x}}-1\right|.$$
Now by letting that $x\to\infty$ then the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find the expected area of a randomly chosen triangle. The set of numbers $(x,y)$ are positive natural numbers such that $x+y=n$. 2 points are chosen from this set. What is the expected area of the triangle formed by the origin and the two points?
| Note that the $x$ coordinate is supported on $[0:n]$. I'll assume that the points are selected uniformly and independently, and you should edit the question in case this is not true.
Suppose the first point is $(X_1, Y_1)$, and the second $(X_2, Y_2)$. It's easy to see that the area is
\begin{align}
A &= \frac{1}{2}|X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5? What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5?
I think that the only way to solve this would be to applying to the proposition that “the sum/product of congruence classes is equal to the congruence cla... | We have $p=5$ is prime, then by Fermat's little theorem: for every integer $a$ such that $p \nmid a$ we have $$ a^{p-1} \equiv 1 \mod{p}$$
Now for any $x$ integer in $[1,100]$, if $5$ divides $x$ then the remainder is zero, if $5\nmid x$ , then $x^{4} \equiv 1 \mod{5}$, and so as $$y=1^6+ 2^6+3^6+...+99^6+10... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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The derivative of $ \ln\left(\frac{x+2}{x^3-1}\right)$ I know it is a simple question, but what would be the next few steps in this equation to find the derivative?
$$f(x)= \ln\left(\frac{x+2}{x^3-1}\right)$$
| I would NOT use the chain rule before doing this:
\begin{align}
f(x) = \ln\frac{x+2}{x^3-1} & = \ln (x+2) - \ln(x^3-1) \\[10pt]
& = \ln(x+2) - \ln((x-1)(x^2+x+1)) \\[10pt]
& = \ln(x+2) - \ln(x+1) - \ln(x^2+x+1).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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Find the highest and the lowest value of the function $y=1+\sin x \cos x$
Please dont use the formula $2\sin x \cos x=\frac{1}{2}\sin 2x$. We have't learnt it yet.
The key is to return it into an inequation for example $-1 \le \sin x \le 1$
Then multiply with something to give an inequality
For example find the highes... | We want to maximize/minimize $(\sin x)(\cos x)$, or equivalently $2(\sin x)(\cos x)$. Note that
$$2\sin x\cos x=(\sin^2 x+\cos^2 x)-(\sin x-\cos x)^2=1-(\sin x-\cos x)^2.$$
This attains a maximum when $(\sin x-\cos x)^2$ is as small as possible, namely $0$.
For the minimum, use the same idea, and
$$2\sin x\cos x=(\s... | {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$
Show that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$.
My reasoning is as follows:
Since the only ways $X^2$ and $X^3$ can be factored in $K[X^2,X^3]$ are $X^2 = X^2 * 1$ and $X^3=X^3*1$ it follows that $X^2$ and $X^3$ are ir... | You can do as Thomas said, by appealing to unique factorization in the larger ring, but here it is simple enough to do it directly. By definition of product, the degrees of the factors must sum to the degree of the product. Now you can prove by induction that $K[X^2,X^3]$ does not have any element with a nonzero coeffi... | {
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Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $... | $\sqrt x=\cos2t~$ seems like the most natural substitution, since there are well-known trigonometric formulas for $1\pm\cos2t,~$ and it is fairly obvious that $x\in\Big[0,1\Big]$.
| {
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Minimize $x^2+y^2$, subject to... (optimal points, KKT conditions, dual theories) I am new to this. I am self learning to get ahead of my next years course and came across this question. I thought it would be a good question to look at due to it touching an many different aspects of optimization.
Minimize $x^2+y^2$
su... | I'm too lazy to use KKT, so I'm providing a non-KKT solution. Maybe, someone else will help with the KKT requirement.
With only the first constraint,
from $(x-1)^2+(y-1)^2\leq 1$, we have by AM-GM that $x^2+y^2+1\leq 2x+2y\leq 2\sqrt{2}\sqrt{x^2+y^2}$. Thus, if $r:=\sqrt{x^2+y^2}$, then $r^2-2\sqrt{2}r+1\leq 0$, whi... | {
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"source": "stackexchange",
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On an exercise with a series: $\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$ The series in question is the following:
$$\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$$
I want to study for which $\alpha >0$ does the series converge.
I tried the r... | You can indeed do Taylor expansions for $\frac{1}{n^\alpha}$, as long as $\frac{1}{n^\alpha} \xrightarrow[n\to\infty]{} 0$. (That is, for any $\alpha > 0$. Note that the case $\alpha \leq 0$ is straightforward). Doing so, you obtain:
$$
\sin \frac{1}{n^\alpha} = \frac{1}{n^\alpha} + o\left(\frac{1}{n^{2\alpha}}\right)
... | {
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Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\fr... | Use induction and a little cheat!
Consider $F_5=\dbinom50+\dbinom41+\dbinom32=1+4+3=8$
Now $\dbinom{n}{k-1}+\dbinom{n}{k}=\dbinom{n+1}{k}$, so for example $\dbinom41+\dbinom42=\dbinom52$.
So adding these like terms gives us the next Fibonacci number, except for the misplaced leading term.
And we have $\dbinom{n}{0}=1$ ... | {
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probability of getting at-least 2 heads Let assume, I through a fair coin three times, and I want to get at least two heads. I want to find the the probability.
To find that,
At first I evaluated the probability of getting 1 head which is:
$${3\choose1} \left(\frac 12\right)^{1} \times \left(\frac 12\right)^{3-1}= \f... | At least two heads means "two or three heads". Let $X$ denote the number of heads in three tosses. In particular, $X\sim Binom(3, 1/2)$, hence
\begin{align}
P(X\ge 2) = P(X=2) + P(X=3) = \binom{3}{2}\frac{1}{2^3} + \binom{3}{3}\frac{1}{2^3} = \frac{1}{2} .
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square.
For $n=2$ and $n=11$, $2^8+2^{12}+2^n$ is a perfect square.
How to find a closed form?
| If $n\in\{1,2,\ldots,7\}$, then $n=2$ is the only solution. Let $n\ge 8$. Then $2^8\left(1+2^4+2^{n-8}\right)$ is a square iff $17+2^{n-8}$ is a square. Let $n-8=m$. You're solving $17+2^m=k^2$ in non-negative integers.
All the solutions are $m=3,5,6,9$. Now, the solutions for $m$ are relatively large, and assuming $m... | {
"language": "en",
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Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
| Multiplying $x$ to both sides, the identity is equivalent to
$$ \sum_{k=1}^{\infty} \frac{kx^k}{1+x^k} = \sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}}. $$
Expanding and rearranging, each series can be written as
\begin{align*}
\sum_{k=1}^{\infty} \frac{kx^k}{1+x^k}
&= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (-1... | {
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Simplify $\sqrt[3]{162x^6y^7}$ The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
| $162x^6y^7=(3x^2y^2)^3\cdot6y$, such that
$$\sqrt[3]{162x^6y^7}=\sqrt[3]{(3x^2y^2)^3\cdot6y}=\sqrt[3]{(3x^2y^2)^3}\sqrt[3]{6y}=3x^2y^2\sqrt[3]{6y}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrate. $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How should i solve this integral?
i know that it is the same question like here
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$
but I've tried solve it for more then 3 hours and i still have no idea ho to solve it. Thank for help.
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
I tried use $$... | HINT:
$$\int\frac{x+1}{(x^2+7x-3)^3}\space\text{d}x=$$
$$\int\left(\frac{2x+7}{2(x^2+7x-3)^3}-\frac{5}{2(x^2+7x-3)^3}\right)\space\text{d}x=$$
$$\frac{1}{2}\int\frac{2x+7}{(x^2+7x-3)^3}\space\text{d}x-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$
Substitute $u=x^2+7x-3$ and $\text{d}u=(2x+7)\space\text{d}x$:... | {
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Olympiad problem algebra inequality I'm having trouble solving the following inequality problem:
If $n$ is positive integer greater than $1$, and $x>y>1$, then show that:
$\frac{x^{n+1}-1}{x(x^{n-1}-1)} > \frac{y^{n+1}-1}{y(y^{n-1}-1)}$
Any hints? Thanks.
| Since $x^{n-1} > x^{n-2} > \dots > 1$, and $y^{n-1} > y^{n-2} > \dots > 1$,
we have from the rearrangement inequality
$$
x^{n-1} y^{n-1} + x^{n-2} y^{n-2} + \dots + 1
>
x^{n-1} + x^{n-2} y + \dots + y^{n-1}.
$$
Multiplying both side by $(xy - 1)(x - y)$, we get
$$
(x^n y^n - 1)(x - y) > (x^n - y^n)(xy - 1).
$$
Or
$$
(x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
... | You have
$$4x^2+20x+23-4K\le 0.$$
So,
$$\begin{align}K&\ge x^2+5x+\frac{23}{4}\\&=\left(x+\frac 52\right)^2-\frac 12\\&\ge -\frac 12\end{align}$$
The equality is attained when $x=-\frac 52,y=4$.
| {
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Finding two non-congruent right-angle triangles The map $g: B \to A, \ (x,y) \mapsto \left(\dfrac {x^2 - 25} y, \dfrac {10x} y, \dfrac {x^2 + 25} y \right)$ is a bijection where $A = \{ (a,b,c) \in \Bbb Q ^3 : a^2 + b^2 = c^2, \ ab = 10 \}$ and $B = \{ (x,y) \in \Bbb Q ^2 : y^2 = x^3 - 25x, \ y \ne 0 \}$.
Given the Py... | Update: It turns out that for rational $a,b,c$, if,
$$a^2+b^2 = c^2$$
$$\tfrac{1}{2}ab = n$$
then $n$ is a congruent number. The first few are $n = 5, 6, 7, 13, 14, 15, 20, 21, 22,\dots$
After all the terminology, I guess all the OP wanted was to solve $(1)$. He asks, "... find two non-congruent right-angled triangles... | {
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"timestamp": "2023-03-29T00:00:00",
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Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\le... | Notice, $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\frac{1}{\sqrt2}\sin\frac{\pi}{8}+\frac{1}{\sqrt2}\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\sin\frac{\pi}{8}\cos\frac{\pi}{4}+\cos\frac{\pi}{8}\sin\frac{\pi}{4}\right)^2$$
Using trig identity $\sin A\cos B+\cos A\sin B=\sin(A+B)$ $$=2\sin^2\left(\f... | {
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Determinants: divisibility by 6 without remainder and $n\times n$ matrix
*
*Compute the determinant of \begin{pmatrix}
1 & 2 & 3 & ...& n\\
-1 & 0 & 3 & ...& n\\
-1 & -2 & 0 & ...& n\\
...& ...& ...& ...& \\
-1 & -2 & -3 & ...& n
\end{pmatrix}
after some elementary raw operation, one can reach:
\begin{pmatrix}... | *
*Add the first row to the other rows:
$$
\pmatrix{
1 & 2 & 3 & 4 & 5 \\
-1 & 0 & 3 & 4 & 5 \\
-1 & -2 & 0 & 4 & 5 \\
-1 & -2 & -3 & 0 & 5 \\
-1 & -2 & -3 & -4 & 5 \\
}
\to
\pmatrix{
1 & 2 & 3 & 4 & 5 \\
0 & 2 & * & * & * \\
0 & 0 & 3 & * & * \\
0 & 0 & 0 & 4 & * \\
0 & 0 & 0 & 0 & 2\times 5 \\
}
$$
So the det... | {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $... | Using logarithms and Taylor expansions can help a lot. $$A=\Big(\frac{3x^2+2}{5x^2+2}\Big)^{\frac{3x^2+8}{x^2}}$$ $$\log(A)=\frac{3x^2+8}{x^2}\,\log\Big(\frac{3x^2+2}{5x^2+2}\Big)=\frac{3x^2+8}{x^2}\,\log\Big(1-\frac{2x^2}{5x^2+2}\Big)\approx -\frac{3x^2+8}{x^2}\times\frac{2x^2}{5x^2+2}$$
I am sure that you can take fr... | {
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Show that $\lim\limits_{x\rightarrow 0}f(x)=1$
Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$
Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}... | $$
a + \frac 1 a = \left( a - 2 + \frac 1 a \right) +2 = \left( \sqrt a - \frac 1 {\sqrt a} \right)^2 + 2 = \text{square} + 2.
$$
So this is $\ge 2$ unless the square is $0$.
$a+ \dfrac 1 a$ cannot get close to $2$ unless $a$ gets close to $1$.
Even if you allow complex numbers (so that the inequality above doesn't hol... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is .... Problem :
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is
(a) $2^{105}-2^{121}$
(b) $2^{121}-2^{105}$
(c) $2^{120}-2^{104}$
(d) $2^{110}-2^{108}$
My approach :
Tried to... | Given $$(1-x)(1-2x)(1-2^2x)...........(1-2^{15}x) = 2^{0+1+2+3+....+15}x^{16}\cdot \left(\frac{1}{x}-1\right)\left(\frac{1}{2x}-1\right).......\left(\frac{1}{2^{15}x}-1\right)$$
So $$=2^{120}x^{16}\left[\left(1-\frac{1}{x}\right)\cdot \left(1-\frac{1}{2x}\right)\cdot \left(1-\frac{1}{2^2x}\right)..........\left(1-\frac... | {
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"timestamp": "2023-03-29T00:00:00",
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Roots of unity filter, identity involving $\sum_{k \ge 0} \binom{n}{3k}$ How do I see that$$\sum_{k \ge 0} \binom{n}{3k} = (1 + 1)^n + (\omega + 1)^n + (\omega^2 + 1)^n,$$where $\omega = \text{exp}\left({2\over3}\pi i\right)$? What is the underlying intuition behind this equality?
| The first thing we need is the binomial theorem which says that
$$(1+x)^n = \sum_{k\geq 0}^n{n\choose k} x^k = \sum_{k\geq 0}{n\choose k} x^k$$
where the last equality follows since ${n\choose k} \equiv 0$ for all $k > n$.
Next we have that for any sequence $a_n$ we can split the sum of it over all the integers into a... | {
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Evaluate the integral $\int \frac{dx}{2 \sin x - \cos x + 5}$ I'm trying to evaluate the following integral:
$$ \int \frac{dx}{2 \sin x - \cos x + 5}.$$
This is in a set of exercises following a chapter on partial fractions, so I imagine there is a substitution we can make to get this into a rational function where we ... | Notice, $$\int \frac{1}{2\sin x-\cos x+5}\ dx$$
$$=\int \frac{1}{2\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}-\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+5}\ dx$$
$$=\int \frac{1+\tan^2\frac{x}{2}}{6\left(\tan^2\frac{x}{2}+\frac{2}{3}\tan\frac{x}{2}+\frac{2}{3}\right)}\ dx$$
$$=\int \frac{\sec^2\frac{x}{2}}{6\left... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | Once you arrive at $$A(x+2) + B(x+2)x + C(x^2) = 5x^2 + 3x - 2,$$ this equation must hold true for all $x$. If you substitute $x = -2$, then $$C(-2)^2 = 5(-2)^2 + 3(-2) - 2 = 5(4) - 6 - 2 = 12,$$ or $4C = 12$, or $$C = 3.$$ You seem to be under the impression that you can disregard any expressions involving $x$ on t... | {
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Could someone check my solution for finding constant of a difference quotient? So the question was,
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be three times differentiable and $f'''$ is bounded, find constants $a,b,c$ such that $$f''(x) = \lim_{h\rightarrow 0} \frac{af(x-h)+bf(x)+cf(x+2h)}{h^2}$$
My Solution:
Expand... | We find the first three terms, in powers of $h$, of the Taylor expansion of the numerator.
The first term is
$$af(x)+bf(x)+cf(x).\tag{1}$$
The second term is
$$\left(-af'(x)+2cf'(x)\right)h.\tag{2}$$
The third term is
$$\frac{af''(x)+4cf''(x)}{2}h^2.\tag{3}$$
We will arrange for the first two terms to vanish, and fo... | {
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Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ ... | Use the Distributive Property.
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2$$
$$=\underbrace{(x+2)(x-1)^2}_{\text{common factor}}\left(2(x-1)\right)-\underbrace{(x+2)(x-1)^2}_{\text{common factor}}(3(x+2))$$
$$=(x+2)(x-1)^2\left(2(x-1)-3(x+2)\right)$$
$$=(x+2)(x-1)^2(-(x+8))$$
| {
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Is $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}}) $a Galois extension? I am wondering if the number field $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}})=\mathbb{Q}(\alpha)$ is a Galois extension. I think it is and I have the following argument, but it feels like a circulair argument:
I know that we have the subfield $\mathbb{Q}(\sqrt{5})$ so w... | Let's start with a choice of radicals $x = i \sqrt{5 - \sqrt{5}}$. Note that
$x^2 = - (5-\sqrt{5})$ so $x^2 + 5 = \sqrt{5}$, $(x^2 + 5)^2 = 5 $, and finally
$$x^4 + 10 x^2 + 20 =0$$
an irreducible equation with roots $\pm i \cdot \sqrt{5 \pm\sqrt{5}}$. We see that the different choices of radicals gives isommorphic ext... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$ I want to calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$.
I tried to separate it into a sum of real numbers ($n=0,2,4,\dots$) and complex numbers that are not real numbers ($n=1,3,5,\dots$) but it didn't work.
So I did it another way, using Ca... | Taking the derivative of the geometric series twice and dividing by $z$ gives
\begin{align}
\frac{1}{1-z} &= \sum_{n=0}^\infty z^n
\\
\frac{1}{(1-z)^2} &= \sum_{n=1}^\infty nz^{n-1}
\\
\frac{2}{(1-z)^3} &= \sum_{n=2}^\infty n(n-1)z^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) z^n
\\
\frac{2}{z(1-z)^3} &= \sum_{n=0}^\infty (n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Convergence of infinite series question The sum of the series
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$ equals?
The answer is $\frac{3e}{2}$.
But I dont Know How?
I have tried following:
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$=$\sum\limits_{n=1}^{\infty} \frac{n(n+1)}{2n!}
$
I know that sequence of partial sum $S_n$=$... | Hint:
$$\frac{n(n+1)}{n!}=\frac{n+1}{(n-1)!}=\frac{(n-1)+2}{(n-1)!}=\frac{(n-1)}{(n-1)!}+2\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+2\frac{1}{(n-1)!}$$
the both last ones with $n\to\infty$ gives us $e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,... | We have $f(x) = x(x+k)$. This means
$$f(f(x)) = f(x)(f(x)+k) = x(x+k)(x^2+kx+k)$$
Hence, we see that $0$ and $-k$ are solutions to both. For $x^2+kx+k$ not to have a real solution, we need $k^2-4k < 0 \implies 0 < k < 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following e... | You can also do it with almost no computation if you make the right combinatorial argument. Let’s count the ways to choose $6$ numbers from the set $S=\{0,1,\ldots,12\}$. Let $r$ be the third-smallest of the $6$; how many ways are there to choose the other $5$ numbers?
There are $r$ members of $S$ less than $r$, so the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Why series expansion in evaluating $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$ is not working here? Prove that $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$
I can evaluate this limit using L Hospital rule but i wonder why Series expansion method is failing here and L Hospital rule is working here.
$\lim_{x\t... | As $x\to0,1-x^2>0$
$\arccos(1-x^2)=\arcsin\sqrt{1-(1-x^2)^2}=\arcsin\sqrt{2x^2-x^4}$
$$\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}x=\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}{\sqrt{2x^2-x^4}}\cdot\lim_{x\to0^+}\dfrac{x\sqrt{2-x^2}}x$$
Set $\arcsin\sqrt{2x^2-x^4}=y$ in the first
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $A$ when $A^2$ is a $2\times 2$ zero-matrix and $A$ is symmetric.
$A^2 =
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$ and $A$ is symmetric. Find the matrix $A$.
Well, I know that $A = A^T$ and $A^2 =
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
$... | $\left(\matrix{a & b\\ b & d}\right)^2=\left(\matrix{a^2+b^2 & b(a+d)\\b(a+d) & b^2+d^2}\right)=\left(\matrix{0 & 0\\ 0 & 0}\right)\to b=0$ and $a^2=d^2=0$ or $a=-d$ and $b^2=-a^2$, ie, A is zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Reducibility of polynomials modulo p
How do we show that $X^4-10X^2+1$ is reducible modulo every prime $p$?
I've managed to show it for all primes less than 10, for primes greater than 10 we have $X^4+(p-10)X^2+1$. Where do I go from here?
| Write
$$\begin{align}
x^4-10x^2+1&=(x^2-1)^2-2(2x)^2\\
&=(x^2+1)^2-3(2x)^2\\
&=(x^2-5)^2-6\cdot 2^2
\end{align}$$
Now if $2$ is a quadratic residue $\pmod{p}$ then the first identity makes our polynomial the difference of two squares.
If $3$ is a quadratic residue $\pmod{p}$ then the second identity makes our polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$
Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$
I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric substitution of variables, but ... | Use C-S we have
$$(ax+by)^2\le (a^2+b^2)(x^2+y^2)\le 1
\Longrightarrow ax+by\le 1$$
we can rewrite the inequality
$$(1-ax-by)^2\ge (1-a^2-b^2)(1-x^2-y^2)$$
$$\Longleftrightarrow 1-(ax+by)\ge \sqrt{(1-a^2-b^2)(1-x^2-y^2)}$$
Let $A=a^2+b^2,B=x^2+y^2$,since
$$1-(ax+by)\ge 1-\sqrt{AB}$$it is enought to show that
$$1-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x)... | Hint: The following perspective with focus on operator methods might also be useful.
We can successively apply the $\left(x\frac{d}{dx}\right)$-operator to a generating function
\begin{align*}
A(x)=\sum_{n=0}^{\infty}a_nx^n
\end{align*}
to obtain
\begin{align*}
\left(x\frac{d}{dx}\right)A(x)&=\sum_{n=0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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How do I get $ \int_0^1 \frac{dz}{\sqrt{z(z - 1\,)(z+1\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})}$? While reading physics papers I found a very interesting integral so I decided to write it down. Let $p(z) = z^ 3 - 3\Lambda^ 2 z$ where $\Lambda$ could be any number. If you want $\Lamb... | Since:
$$\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx = B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\tag{1}$$
through a change of variable we get:
$$ \int_{0}^{1} z^{-1/2}(1-z^2)^{-1/2}\,dz = \frac{1}{2}\int_{0}^{1}z^{1/4}(1-z)^{-1/2}\,dz = \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$ If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denomin... | In general, the equation of a tangent is $y-f(x_0)=f'(x_0)(x-x_0)$.
Rearrange $4x-y-3=0$, we have $y-5=4(x-2)$,
i.e. $f(2)=5$ and $f'(2)=4$.
As $t\rightarrow 2$, $f(x^2-2)-f(f(x)-3) \rightarrow f(4-2)-f(5-3)=0$.
Now
$\lim_{x\rightarrow 2} \frac{f(x^2-2)-f(f(x)-3)}
{(x-2)^2}
=\lim_{x\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$
What is the least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$:
$$ (2^n)_{10} = d_m \, d_{m-1} \cdots
d_{q+1} \, (n)_{10} \\ (n)_{10} = d'_{q} \cdots d_1' \\ d_i, d'_j \in
\{0,... | Here is a less brute-force method:
In order for the digit string of $2^n$ to end with the digit string of $n$, it is necessary (but not sufficient) that $2^n$ and $n$ have the same last digit, i.e. $2^n \equiv n \pmod{10}$.
It is easy to check that $\begin{cases}2^n \equiv 2\pmod{10} & \text{if} \ n\equiv 1 \pmod{4} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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How do you find the value of $\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $? How do you find the value of:
$$\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $$
I tried to apply partial fractions/divide numerator and denominator by $6^r$ and use the fact that $a^x \rightarrow 0$ as $x \righ... | Hint:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
= \frac{2^r}{3^r-2^r} - \frac{2^{r+1}}{3^{r+1}-2^{r+1}}.$$
Edit:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
= \frac{A}{3^r-2^r} + \frac{B}{3^{r+1}-2^{r+1}}
= \frac{(3A+B)3^r - (2A+B)2^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}.$$
One obvious way to make the numerators e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but... | HINT: i think your inequality must be reversed, try $a=1,b=2,c=3$
We can prove it by BW
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) ... | Use Bill's observation that
Theorem: If $a \mid b^2 $ and $ a = \pm c^2 \pm kb^2$, then $ a = \pm (c, b) ^2$
Proof: $ a = \gcd(a, b^2) = \gcd(\pm c^2 \pm kb^2, b^2) = \gcd(c^2 , b^2 ) = \gcd(c, b)^2$.
Now,
*
*$ x -y > 0$ by assumption.
*OP has showed that $x-y \mid y^2$.
*$x-y = 5y^2 - 4x^2 = -(2x)^2 + 5y^2$ .
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Solve this equation $xy-\frac{(x+y)^2}{n}=n-4$ Let $n>4$ be a given positive integer. Find all pairs of positive integers $(x,y)$ such that
$$xy-\dfrac{(x+y)^2}{n}=n-4$$
What I tried is to use
$$nxy-(x+y)^2=n^2-4n\Longrightarrow (n-2)^2+(x+y)^2=nxy+4$$
|
Note: For symmetry, it is best to do the minor change of variables $n=m+2$.
When dealing with integer solutions to quadratic forms, there may be a Pell equation lurking nearby. For any integer $m>2$, the equation,
$$xy-\frac{(x+y)^2}{m+2}=m-2$$
has an infinite number of positive integer solutions. Define $D = m^2-4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5... | Without loss of generalization, let us consider $x_1,x_2,\ldots,x_k \geq C$.
Then let
$$y_i = \begin{cases}x_i - C &i\leq k\\ x_i & i > k \end{cases}$$
Solving
$$\left\{\begin{array}[lll] xx_1 + x_2 + \ldots + x_n &=& N\\ x_i \geq 0& & \\ x_1,x_2,\ldots,x_k& \geq & C\end{array}\right.$$
is equivalent to just solving... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | You can prove this through two separate steps:
*
*$5^n \mod 8$ is 5 if $n$ is odd and 1 if $n$ is even.
*$2 \cdot 3^{n-1} \mod 8$ is 2 if $n$ is odd and 6 if $n$ is even.
Once you have the above two statements you have then concluded that in either case of $n$ odd/even it's equivalent to 0 modulo 8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Solve $x^n+y^n=2015$ Determine the natural numbers $x,y,n$ matching equality
$$x^n+y^n=2015.$$
I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.
For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$ we come to contradiction.
What must be done to $n\geq3$?
| Piggybacking off user's answer that we need only concern ourselves with a few odd exponents, with $x\lt y$ also pretty small, let's use the fact that $x+y$ divides $x^n+y^n$ for odd $n$ and $x^p\equiv x$ mod $p$ for prime $p$.
Note that $2014$ is not an $n$th power for any $n\gt1$, so we can assume $1\lt x\lt y$. No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How to compute $\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\ri... | I thought it might be instructive to present a development that goes back to "the basics" and relies on elementary tools only. To that end, we recall from geometry that the sine function satisfies the inequalities
$$z \cos z \le \sin z\le z \tag 1$$
for $0\le z\le \pi/2$.
Rearranging $(1)$ we immediately see that $z\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Maclaurin Series Representation for $f(z)=\frac{z}{z^4+9}$ I need help finding the Maclaurin series representation for $$f(z)=\frac{z}{z^4+9}$$
I first tried to factorize $z^4+9$, but am I missing something? I could not figure out how to factorize this. Is there another approach to this? I am open to any approach. Jus... | The geometric series $1-x+x^2-x^3+\cdots$ converges to $\frac{1}{x+1}$ for all $|x|<1$.
Substituting $x=\frac{y^4}{9}$ tells us that
$$1-\frac{1}{9}\,y^4 + \frac{1}{9^2} \, y^8 - \frac{1}{9^3}\, y^{12} + \cdots = \frac{1}{\frac{y^4}{9}+1} \equiv \frac{9}{y^4+9}$$
for all $\left|\frac{y^4}{9}\right|<1$, i.e. for all $|y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$ Find all pairs $(x,y)$ of real numbers that satisfy the equation $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$
I supposed $a=\sin^2x$ and $b=\cos^2x$
So the equation b... | Expanding $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 = 12 + \sin y$
will get us $a^2 + 2 + \frac{1}{a^2} + b^2 + 2 + \frac{1}{b^2} = 12 + \sin y$.
And we can subtract 4 from both sides.
$a^2 + b^2 + \frac{1}{a^2} + \frac{1}{b^2} = 8 + \frac{1}{2}\sin y$
Now make y as a function of x:
$y = asin(2a^2 + 2b^2 + \frac{2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $ $$ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $$
I know that the answer is $\frac{2\sqrt{7}}{21} arctg \frac {2x+1}{\sqrt{7}}-\frac{1}{3}arctg(x+2)+C$, but I have no idea what to do.
| We have: $$ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $$
$$(x^2+x+2):$$
$$\bigtriangleup = 1-4*2=-7$$
$$(x^2+4x+5):$$
$$\bigtriangleup = 16-4*5=-4$$
So: $$\frac{x+1}{(x^2+x+2)(x^2+4x+5)} = \frac{ax+b}{ x^2+x+2} +\frac{cx+d}{x^2+4x+5}$$
$$ x+1= (ax+b)(x^2+4x+5)+(cx+d)(x^2+x+2)$$
$$X+1= ax^3+4Ax^2+5ax+bx^2+4bx+5b+cx^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc} \leq \frac{1}{abc}.$
Let $a,b,$ and $c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc} \leq \dfrac{1}{abc}.$$
This question seems hard since we aren't given any information on $a,... | By AM-GM or Muirhead, $\frac{abc}{a^3+b^3+abc}\leq \frac{abc}{a^2b+ab^2+abc}=\frac{c}{a+b+c}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | $$\int \frac { dx }{ 2x^{ 2 }+x+3 } =\int { \frac { dx }{ 2\left( { x }^{ 2 }+\frac { x }{ 2 } +\frac { 3 }{ 2 } \right) } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+\frac { x }{ 2 } +\frac { 1 }{ 16 } -\frac { 1 }{ 16 } +\frac { 3 }{ 2 } } } } =$$ $$\frac { 1 }{ 2 } \int { \frac { dx }{ { \left( x+\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
| Hint: Let $a=\sqrt[3]{35-x^3}$. Then you're solving $ax(a+x)=30$.
$(a+x)^3=\left(a^3+x^3\right)+3ax(a+x)=35+3\cdot 30=5^3$.
$\iff a+x=5$. Then $(5-x)x(5)=30$. Solve this quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | $$
\begin{align}
&\left(\frac{a^3}b+\frac{b^3}c+\frac{c^3}a\right)-(ab+bc+ca)\\
&=\frac ab\left(a^2-b^2\right)+\frac bc\left(b^2-c^2\right)+\frac ca\left(c^2-a^2\right)\\[3pt]
&=\left(\frac ab-1\right)\left(a^2-b^2\right)+\left(\frac bc-1\right)\left(b^2-c^2\right)+\left(\frac ca-1\right)\left(c^2-a^2\right)\\
&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem :
Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$
My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8... | Method $2$
Here are hints to another approach.
$1.$ You can observe from the De Moivre's formula that
$$\sin (8x) = 128\sin \left( x \right)\cos^7 {\left( x \right)} - 192\sin \left( x \right)\cos^5 {\left( x \right)} + 80\sin \left( x \right)\cos^3 {\left( x \right)} - 8\sin \left( x \right)\cos \left( x \right)$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Problems & Solutions on Fermat Theorem of Multiple of 3 I am working on an assignment in elementary number theory, in which I have to come up with original problems and then work out their solutions on Fermat theorem of multiple of 3, that is, the equation
$$x^3 + y^3 = z^3$$
does not have integer solution for $xyz \ne... | Solve in integers $\left(a-b\right)\left(a^2-b^2\right)\left(a^3-b^3\right)=3c^3$.
To solve it, notice $(3c)^3=(a^2-2b^2+ab)^3 + (2a^2-b^2-ab)^3$,
so either $c=0$ or $a^2-2b^2+ab=0$ or $2a^2-b^2-ab=0$.
Here's where I found the problem.
$$(a-b)\left(a^2-b^2\right)\left(a^3-b^3\right)=(a-b)^3(a+b)\left(a^2+ab+b^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for positive real numbers $a,b,c$ we have $\frac{a}{b+c}+ \frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.$
Prove that for positive real numbers $a,b,c$ we have $$\dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}.$$
Attempt
I tried using AM-GM and got $ \dfrac{a}{2\sqrt{bc}}+\dfrac{b}{2\sqrt{ac}... | We have by AM-GM inequality twice: $\displaystyle \sum_{\text{cyclic}} \dfrac{a}{b+c}=\dfrac{1}{2}\displaystyle \sum_{\text{cyclic}} (a+b)\displaystyle \sum_{\text{cyclic}} \dfrac{1}{a+b}-3\geq \dfrac{1}{2}\cdot 3\cdot 3-3=\dfrac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | $$f\left( x \right) =f\left( 0 \right) +{ f\left( 0 \right) }^{ \prime }x+\frac { { f\left( 0 \right) }^{ \prime \prime } }{ 2! } { x }^{ 2 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime } }{ 3! } { x }^{ 3 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime \prime } }{ 4! } { x }^{ 4 }+...\\ f\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
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Walk me through some basic modular arithmetic? The problem is as follows:
Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?
The solution is as follows:
First we factor $abc... | In this case you can think mod as an abreviate way for write the conditions. Concretely, you need $3|bc+b+1$. Now, as $3|3$, then $3|bc+b+1-3=bc+b-2$ (it means $b(c+1)\equiv 2(mod3)$). Now, $b$ only can be of the form $b=3n$ or $b=3n+1$ or $b=3n+2$. For the first one, non $c$ exists in order to get $3|bc+b-2$. If $b=3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\... | Hint:
$$
\sin (2x)=2 \sin x \cos x
$$
so your equation becomes :
$$
2\sin x \cos x + \cos x=0 \iff \cos x(2\sin x+1)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Nested Radicals Involving Primes How do you evaluate $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots } } } } } $ ?
This question appears to be rather difficult as there is no way to perfectly know what $p_{ n }$ is , if $p_{ n }$ denotes the $n$th prime.
It is simple to show that the value above is conve... | For the primes $(p_k):=(2,3,5,\dots)$ we have
$$\tag{1}0\le\dfrac{p_{n+1}}{p_1p_2\cdots p_n}\le2.$$
The inequality is trivial if $p_n\lt p_{n+1}\lt p_1p_2\cdots p_n$. Otherwise the number $1+p_1p_2\cdots p_n$ is a prime and the inequality holds.
For $a,x\gt0$ we have
$$0\lt\sqrt{a+x}={\sqrt{a}+\int_a^{a+x}\dfrac{d\xi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But thi... | you write the integral as $$I=\int \frac{1-\frac{1}{x}}{x^2\sqrt{1+(1-\frac{1}{x})^2}}$$ now put $1-\frac{1}{x}=y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Now am I doing induction correctly? Recursion: $L_n = L_{n-1} + n$ where $L_0 = 1$.
We guess that solution is $L_n = \frac{n(n+1)}{2} + 1$.
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1... | The induction schema is (base) $P(0)$ is true (might start elsewhere too), (induction) if $P(n)$ is true, we prove $P(n + 1)$ is true, (conclusion) $P(n)$ is true for all $n \in \mathbb{N}_0$.
You have to work from $P(n)$ to $P(n + 1)$ somehow.
In your case, the claim is that $L_n = n (n + 1) / 2 + 1$.
Base: For $n = 0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum number of terms resulting from the product of two polynomials with a given number of terms Given two integers ($n$, $m$), what is the smallest number of terms that could result from the product of two polynomials with $n$ and $m$ non-zero terms respectively?
That is, what is the smallest number of non-zero term... | The minimum is 2 for all cases. For even number of terms, there is an elegant factorization solution, but for odd number of terms, my solution made use of complex roots.
Let $A$ have $a$ terms, $B$ have $b$ terms. $C=A\cdot B$.
We prove that for any $a,b\geq2$, we can always find $A,B$ so that $C$ has $2$ terms, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the values of a and k from the curve
The diagram below shows a curve with equation of the form ${y = kx(x +
a)^2}$, which passes through the points (-2, 0), (0, 0) and (1, 3).
What are the values of a and k.
I know my roots are x = -2, x = 0 and x = 3.
But as the y intercept is 0, I don't see how I can get a... | Be careful, $x=3$ is not a root. The curve does not cross the $x$-axis when $x=3$. There are only two roots: $x=-2$ and $x=0$.
You know that $(-2,0)$, $(0,0)$ and $(1,3)$ are all points on the curve.
You should substitute $x=-2$ and $y=0$ to get one equation. Then substitute $x=0$ and $y=0$ to get a second equation. F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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To find area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ How to find the area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ ? I am completely stuck ; please help . Thanks in advance
| First, lets see where both surfaces intersect. In polar coordinates, we have
$$
z=3-r = 1+ r^2\quad \Rightarrow r=1.
$$
So the intersection is the circle $r=1$ at height $z=2$.
The area of your surface is the area $A_1$ of $z=3-\sqrt{x^2+y^2}$ above this circle (let $S_1$ be this surface), plus the area $A_2$ of $z=1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$.
Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$.
I think that using AM-GM might work. Thus we have $\dfrac{a+b+c}{3}+\sqrt{(1-a)(1-b)(1-c)} \geq \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}$ but I am not sure how to proceed. ... | Note that $c<1$ and $1-c<1$ so :
$$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt{ab}+\sqrt{(1-a)(1-b)}$$
Now use AM-GM :
$$\sqrt{ab}+\sqrt{(1-a)(1-b)} \leq \frac{a+b}{2}+\frac{(1-a)+(1-b)}{2} =1$$ and so the conclusion follows .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equatio... | The easiest method seems to be to use Lagrange multipliers.
$$
L = \sqrt{6} xy + 4 yz - \frac{1}{2} \lambda (x^2 + y^2 + z^2 - 1) \, ,
$$
for some $\lambda$. Looking at derivatives
$$
\frac{\mathrm{d} L}{\mathrm{d} x} = \sqrt{6} y - \lambda x = 0
$$
$$
\frac{\mathrm{d} L}{\mathrm{d} y} = \sqrt{6} x + 4z - \lambda y = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ? Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ? What I have tried is that I ha... | The answer is Yes.
I. (Update)
The solution to,
$$\begin{aligned}
kn+1 &= x^2\\
(k+1)n+1 &= y^2
\end{aligned}$$
is given by,
$$n = \frac{ -(\alpha^2 + \beta^2) + \alpha^{2(2m+1)}+\beta^{2(2m+1)} }{4k(k+1)}$$
where,
$$\alpha = \sqrt{k}+\sqrt{k+1}\\
\beta = \sqrt{k}-\sqrt{k+1}$$
For example, for $m=1,2,3,\dots$ we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Given complex $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ : when and what is $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$
Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
... | $$\dfrac S4=3\cos(A-B)+6\cos(B-C)+4\cos(C-A)$$
$$=\cos A(3\cos B+4\cos C)+\sin A(3\sin B+4\sin C)+6\cos(B-C)$$
$$=\sqrt{25+24\cos(B-C)}\cos\left(A-\arccos\dfrac{3\cos B+4\cos C}{3\sin B+4\sin C}\right)+6\cos(B-C)$$
$$\le\sqrt{25+24\cos(B-C)}+6\cos(B-C)$$
If $\sqrt{25+24\cos(B-C)}=y, 1\le y\le7$ and $\cos(B-C)=\dfrac{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | You need to analyze each case separately, and it is always wrong to round off in mathematics without explicitly stating that it is an approximation. Do not forget that to find the solutions you need to take into account the cases! Also, you missed the case of $x = 0$.
$\def\eq{\Leftrightarrow}$
Take any real $x \ne -4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(... | We can show that $f(x)=-f(-x)$, hence $f(x)$ must be an odd function. If $f(x)$ is a polynomial function then we can write it as:$$f(x)=\sum_{i=0}^\infty a_ix^{2i+1}\tag{1}$$
We can also show that:$$f(x^2)=x\,f(x)\tag{2}$$
Using (1) we get:$$f(x^2)=\sum_{i=0}^\infty a_ix^{4i+2}=a_0x^2+a_1x^6+a_2x^{10}+\cdots\tag{3}$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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To obtain the condition for vanishing of the given determinant If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes.
$$\left|
\begin{array}{cc}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{array}
\right|$$
My answer: After a little calculation I was ab... | This is NOT a one line solution you are expecting but it is an idea which is very useful in solving such determinants which may remind one of Vandermonde matrix.
I'm replacing $a$ by $x$ (for the sake of clarity). Using the linearity of determinants we get
$$\det=
\begin{vmatrix}
x & x^2 & 1+x^3\\
b & b^2 & 1+b^3\\
c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| One of your examples is that $2^{11} + 1$ is divisible by $3$. We investigate as follows:
Let us consider instead raising $2$ to an even power and subtracting $1$. And then let us factor.
Example: $2^{10} - 1 = (2^5 - 1)(2^5 + 1)$.
Among any three consecutive integers, exactly one of them must be divisible by $3$.
Clea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 6
} |
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n... | Here's a proof of infinitely many solutions $(x_1, x_2, x_3, \dots, x_n, z)$:
For $n = 1$ we have $(1, 1)$. For $n = 2$ we have $(3^3, 2 \cdot 3^2, 3^2)$.
For $n \ge 3$ we have $(2, 3, 1, 2, 2, 2, 2, \dots, 2)$.
(To verify: $2^2 + 3^3 + 1^4 + 2^5 + 2^6 + \dots + 2^n = 32 + 2^5 (1 + 2 + 4 + \dots + 2^{n-5}) = 32 + 2^5 (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdo... | You have obtained that $f'(x)=\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}.$ Note that this expression doesn't exist if $x=\pm 1.$ I don't think that critical point is the most adequate word but your teacher was saying that $f'(-1)$ and $f'(1)$ don't exist.
Now, $$f_+'(1)=\lim_{x\to 1^+}f'(x)=\lim_{x\to 1^+} \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Convert $f(x)=(\cos(x))^3$ to powers of x and find if converges. I started out by writing the Taylor series for $x_0=0$ (Maclaurin series) of $f(x)=(\cos(x))^3$.
If my calculations are correct $$f(x)=1-\frac{3x^2}{2!}+\frac{21x^4}{4!}-\frac{183x^2}{6!}+...$$
and after some simplifications $$f(x)=\sum\limits_{n=0}^{\inf... | By using the identity $\cos(3x)=4\cos^3 x-3\cos x$ it follows
\begin{align}
\cos^3x&=\frac{1}{4}\cos (3x)+\frac{3}{4}\cos x\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(3x)^{2k}}{(2k)!}+\frac{3}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(x)^{2k}}{(2k)!}\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{3(3^{2k-1}+1)}{(2k)!}x^{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{... | Hint: $\sin x = 1 - \sin^2 x = \cos^2x \Rightarrow \sin^2 x = \cos^4 x \Rightarrow \cos^4 x = 1 - \cos^2 x$. Thus you can express $\cos^{10}x , \cos^8 x, \cos^6 x, \cos^4x$ in term of $\cos^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Probability of almost correct order Let's say I'm a teacher handing tests back to seven students.
If I do it with my eyes closed, what's the probability I hand exactly 5 of the tests back to the correct students?
There are many possible ways this could happen, and if my understanding of probability is correct, the su... | Or as another way way of reaching @Zubin Mukerjee correct answer:
There are $\binom{7}{5}$ ways to pick the 5 students who get their proper papers, and only one way to give the remaining two the wrong papers...
Edit to respond to OP's comment...
Well, consider the first way you enumerate.. Your first two terms give the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Value of $\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$ To find the value of:
$\displaystyle\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$
The answer says: $1/2$
My attempt:
$=(\displaystyle\sum_{k=1}^\infty 1/{k^2})-1-(\sum_{k=1}^\infty 1/{k^3})+1+(\sum_{k=1}^\infty 1/{k^4})-1-\cdots$
$=(1+1/2^2... | So I made calculation error when using the equation on geometric series.
$S_2=2/3-1+1/2=1/2-1/3$ (with z=-1/2)
$S_3=3/4-1+1/3=1/3-1/4$ (with z=-1/3)
$S_4=4/5-1+1/4=1/4-1/5$ (with z=-1/4)
Sum them up gives the value of original equation:
$=1/2-1/3+1/3-1/4+1/4-1/5+...=1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.