Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Best linear unbiased estimates Problem is : Model given $$Y_i=ib+ e \quad; \, i=1,2,3$$
Mean is $0$ and variance is $\sigma, 2\sigma, 3\sigma$. What will be best linear unbiased estimator for $b$?
Answer given in book is $(y_1+y_2+y_3)/6$? this is unbiased I know, but is it correct?
| (W)OLS estimator is the BLUE by the Gauss-Markov theorem, which is given by the normal equation (weighted by the inverse of the noise covariance matrix)
$\hat{b}=(X^T\Sigma^{-1}X)^{-1}X^T\Sigma^{-1}y$
Here we have
$X=
\left[ {\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
\end{array} } \right]
$
,
$y=
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ for positive real $a,b,c$, $abc\ge 1$ I have been able to simplify the problem. Since $a^{4}\ge a^{3}$ iff $a\ge 1$ (and analogously for $b,c$), $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ holds iff $a+b+c\ge 3$. It is thus sufficient to prove $a+b+c\ge 3$.
I am not sure how t... | I think maybe the question should be $a^4+b^3+c^2\ge a^3+b^2+c$.
By AM-GM,
$a^4+a^4+a^4+1\ge 4a^3\Rightarrow a^4\ge \frac{4a^3-1}{3}$
Similarly, $b^3\ge \frac{3b^2-1}{2}$ , $c^2\ge 2c-1$
Hence, $a^4+b^3+c^2\ge \frac{1}{6}(8a^3+9b^2+12c-11)$
So we only need to prove that
$8a^3+9b^2+12c-11\ge 6a^3+6b^2+6c$ , which is equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the expectation and variance of X A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}
*
*Find the value of C.
*Find the expectation and variance of X.
*Find the expectation of random variable $Z=\frac{X}{2X+3}$... | Your $C=1/5$ and $E(x)=17/15$ are correct, then $$E(x^2)=\frac{1}{5}\int_{0}^{2} x^2(x+3/2)dx=\frac{8}{5}$$
$$Var=E(x^2)-(E(x))^2=\frac{8}{5}-(\frac{17}{15})^2=\frac{71}{225}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$ Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn... | Formulae for $\sin4x,\,\sin5x$ provide another solution. Let $S:=\sin^220^\circ\ne\sin^260^\circ=\tfrac34$ so$$\begin{align}0&=\frac{\sin^2100^\circ-\sin^280^\circ}{S}\\&=(5-20S+16S^2)^2-16(1-S)(1-2S)^2\\&=256S^4-576S^3+432S^2-120S+9\\&=(4S-3)(64S^3-96S^2+36S-3)\\\implies64S^3-96S^2+36S-3&=0\\\implies\frac{4}{S}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $ Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{ar... | The combined values of the maximum value for the $x$'s is 25. We can then take four part ordered partitions of $5$ (including zeros) and subtract each of those from the maximum.
The first set of partitions is $(5,0,0,0)$ with the $5$ being in four potential locations.
You cannot subtract $5$ from $x_1$ or $x_4$ so ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to so... | To show that $15$ divides $n(3n^4+5n^2+7)$, show that $3$ and $5$ do.
$3$ does because either $3$ divides $n$ or $n^2\equiv1\pmod3$ by Fermat's little theorem,
in which case $3n^4+5n^2+7\equiv5n^2+7\equiv5+7=12\equiv0\pmod3.$
$5$ does because either $5$ divides $n$ or $n^4\equiv1\pmod5$ by Fermat's little theorem,
in w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{... | I made a mistake in my solution with the remainders. Below is the correct approach and answer :-
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 36 +13+22+69 = 140 \Rightarrow Rem[\frac{140}{70}] = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$ Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$
My solution goes as follows:
Factorizing we have:
... | Step by step, we have:
$$9(x^3+x^2y-5x^2)=4(y^3+xy^2-5y^2)$$
$$\Rightarrow 9x^2(x + y - 5) = 4y^2 (y + x - 5)$$
and moving everything to one side gives $(9x^2-4y^2)(x + y - 5) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Question of rank of a matrix $4\times 6$ Let $a,b,c,d$ be complex numbers. Find the rank of the following matrix
\begin{align*}
\begin{pmatrix}
2& -3& 6& 0& -6& a\\
-1& 2& -4& 1& 8& b\\
1& 0& 0& 1& 6& c\\
1& -1& 2& 0& -1& d\\
\end{pmatrix}
\end{align*}
Idea: Proceed the elementary row operations... | Guide:
Your idea is fine.
The question is what is the rank of the matrix, the answer should be a number, of which you have not provided.
Consider two cases:
*
*$c-b-2d=0$.
*$c-b-2d \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the square root function behave the same when calculating complex roots? This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:
$
z^4 = \sqrt 9
$
Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed w... | $z^4=\sqrt9$ which implies: $z^4=3$ and $z^4=-3$
Solving: $z^4=3$
Let $u=z^2$, then $u^{2}=3 \rightarrow u=\pm \sqrt 3$
Now, $z^2=\sqrt3$ and $z^2=-\sqrt3$. Thus, there are four roots $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3}$$
Solving $z^4=-3$
Let $u=z^2$, thus $u^{2}=-3 \rightarrow u=\sqrt3i \,\,\,\&\,\,\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| Differentiate $f(n) = n^3+12n^2+48n+64$ to get $f'(n) = 3n^2 + 24n + 48 = 3(n^2 + 8n + 16) = 3(n+4)^2$.
Hence $\int f'(n) \ dn = 3 \int (n+4)^2 \ dn = 3 \cdot \left(\frac{(n+4)^3}{3} + C \right) = (n+4)^3 + C' = f(n)$, by the fundamental theorem of calculus. Since $f(-4) = C' \Rightarrow 0 = C'$, $f(n) = (n+4)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
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Sequence queue - converges or diverges? I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?
My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^... | I continue from where you stopped,
Easy to show that :
$$\frac{n^2-n-1}{n^4+n^2+1} < \frac{n^2-1}{n^4+1} < \frac{n^2}{n^4} < \frac{1}{n^2} $$
When we know that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges.
Therefore, by comparison test :
$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{conve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration using digamma functions I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:
$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$
where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Diga... | Just to add a bit to the evaluation provided by @giobrach and @Roman - another way to get Digamma function explicitly.
$$I=\int_0^x \frac{x^r}{x^r+y^r}\,dy\overset{z=\frac{y}{x}}=x\int_0^1 \frac{1}{1+z^r}\,dz=x\int_0^1 \frac{1-z^r}{(1+z^{r})(1-z^r)}\,dz=x\int_0^1 \frac{1-z^r}{1-z^{2r}}\,dz$$
Next we make regularisation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Harvard MIT math question Let P be a point selected uniformly at random in the cube $[0, 1]^3$. The plane parallel to x+y+z = 0
passing through P intersects the cube in a two-dimensional region R. Let t be the expected value of
the perimeter of R. If $t^2$ can be written as $\frac{a}{b}$, where a and b are relatively p... | If you take a unit cube and try to cut it along a plane $x + y + z = r$, then you will notice that the two-dimensional region can be two different shapes.
More precisely, if $r \leq 1$ or $r \geq 2$ (which are symmetric via $x, y, z \mapsto 1 - x, 1 - y, 1 - z$), then the region $R$ is a triangle with side length $\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed $x^2-|5x-3|-x<2$
For this problem I took two cases where in one case $5x-3 \geq 0$ and in second case $5x-3 \lt 0$.
$1st$ Case when $5x-3 \geq 0$ :-
For this case I got the solution set as $x \in [0.6,3+2\sqrt{2}]$.
$... | Let me forcus on when $5x-3<0$, we have $x < \frac35$ and
$$x^2-|5x-3|-x<2$$
becomes
$$x^2-(3-5x)-x<2$$
$$x^2+4x-5<0$$
$$(x+5)(x-1)<0$$
Hence we have $-5 < x < 1$ and $x < \frac35$, hence taking intersection, we have $-5<x < \frac35$.
To answer the question, why did we exclude $x=-5$, consider $(x+5)(x-1)<0$, if we sub... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
... | For every $n\geqslant 1$, $\cos \dfrac{1}{n}$ gets closer and closer to $1$.
That is,
$$\cos \dfrac{1}{n}\approx 1-\dfrac{1}{2n^2}+\dfrac{1}{24n^4}-\dfrac{1}{720n^6}$$ and for large enough $n$, the larger terms go to zero, making the approximation more accurate; as $n$ approaches $\infty$, our polynomial, and the cosin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$ Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\ f_{y}= 4y^{3} \\ f_{xx} = 12x... | $$f(x,y)=x^4+y^4-x^3\\\qquad\quad ~~ ~=x^3(x-1)+y^4$$
Apply $x\to\ +\infty$ and $y=0$, you get $f(x,y)\to +\infty.$
So, global maxima doesn't exist.
Now, we will prove that global minimum exists and is equal to
$$\text{min}[x^4-x^3+y^2]=-\frac{27}{256}.$$
In this case , of course, $y=0$ must be. Then observe that,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis? Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
Basis step: The statement is true for $n=1$ s... | Here's a way you can see it. Suppose $n$ is a counterexample, so that $y^n+\frac{1}{y^n} \notin \Bbb Z$ for some $n$. Then because every non-empty set of natural numbers has a least element, there must be some least $n$ that works as a counterexample. Therefore, for each $k \lt n, y^k+\frac{1}{y^k} \in \Bbb Z$. Mor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solving $|2x-5| + |7-2x| =2 $ $$|2x-5| + |7-2x| =2 $$
I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-
*
*1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.
*2nd case, where $2.5<x\le3.5... | You have the info about $7-2x$ completely backwards. I $x \le 3.5$ then $7-2x \ge 0$ and $|7-2x| = 7-2x$.
And if $x >3.5$ then $7-2x < 0$ and $|7-2x| = 2x - 7$.
So case 1: $x <2.5$ then $|2x - 5|+|7-2x| = (5-2x)+(7-2x) = 2$ and $12-4x = 2$ and $x =2.5$ but $x < 2.5$ so that isn't a solution.
Case 2: $2.5 \le x \le 3.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove the solution of an inequality is lower bounded? Let $x_{n,k}\in (0,1]$ be any solution of the following inequality
\begin{align}
n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4 \leq 0,
\end{align}
where $n, k$ are parameters such that $n\geq k\geq 1$.
I want to prove that there exists a strictly positive constant $c$ ... | Let $f(x) = n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4$.
We have, for all $x$ in $(0, 1]$,
\begin{align}
f(x) &\ge k(1-x)^2 + 0 - x^2 - k(1-x)^4 \\
&= k[(1-x)^2 - (1-x)^4] - x^2\\
&\ge (1-x)^2 - (1-x)^4 - x^2 \\
&\ge (1-x)^2 - (1-x)^3 - x^2 \\
&= x(x^2 - 3x + 1).
\end{align}
Thus, if $x\in (0, 1]$ satisfying $f(x) \le 0$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots... | Some nice answers have been posted, but I'm still adding this answer since it gives a straightforward non-recursive way to get a representation.
That the expression is in $[{1 \over 2},1]$ is easy to see.
For the other way, consider ${x \over y} \in [{1 \over 2},1]$ and let $m=y-x$. Then $x-m \geq 0$.
Express $x-m=(a_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 1
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Trigonometric integral $\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}$ The function
$f(x_1,x_2)=\frac{e^{-\sqrt{x^2_1 + x^2_2}}}{\sqrt{x^2_1+x^2_2}}$, $(x_1,x_2)\neq(0,0)$ is integrable in $\mathbb{R}^2\setminus\{(0,0\})$ as $f>0$ and
\begin{align}
\int_{\mathbb{R}^2}f(x_1,x_2)\,dx_1dx_2=2\pi\int^\infty_0e^{-r}\,dr
\end{... | \begin{aligned}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}&=\frac{1}{2}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}+\frac{1}{2}\int_{\pi}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\left(2\pi -\theta\right)}}}\\ &=\frac{1}{2}\int_{0}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082555",
"timestamp": "2023-03-29T00:00:00",
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Find $(a, b, c)$ that satisfies the system of equations $2a + 3b -4c = 7, a-b+2c=6$ $\begin{align}
2a + 3b -4c = 7, \\
a - b + 2c = 6
\end{align}$
I haven't seen a three-variable system of equations with only two equations before. I try to make the problem into a two-variable system.
Solving for $a$ in the first equat... | First, see my comment, following your query.
Since you are trying, not only to solve the equations, but also to express
$b$ and $c$ in terms of $a$, I would have approached it differently.
$$2a + 3b - 4c = 7. \tag1$$
$$a - b + 2c = 6. \tag2$$
Multiplying equation (2) above by $(3)$ gives
$$3a - 3b + 6c = 18.\tag3$$
Add... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$ I have a sequence $\{r_n\}$ of positive numbers defined recursively by
$$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$
for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by indu... | First, reproving the lower bound: we have, for all $n\geq 1$,
$$
r_{n-1}^2 +K = r_n r_{n-1} \leq \frac{r_n^2+ r_{n-1}^2}{2} \tag{1}
$$
using the AM-GM inequality; from which
$$
r_n^2 \geq r_{n-1}^2 + 2K \tag{2}
$$
leading to $r_{n}^2 \geq 2K n + r_0^2$ for all $n\geq 0$.
Then, the upper bound: We know, using the lower ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
ho... | A calculation by foot. We obtain from
\begin{align*}
\color{blue}{0}&\color{blue}{=\frac{1}{a+\omega}+\frac{1}{b+\omega}
+\frac{1}{c+\omega}+\frac{1}{d+\omega}-\frac{2}{\omega}}\tag{1}\\
\end{align*}
by multiplication with the common denominator $(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega$:
\begin{align*}
\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Solving $\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$
Find all $x$ that satisfy the equation $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$$
Attempt:
Taking tangent on both sides, and using the id... | This should be an exercise for high school students who just learned inverse trig function.
Denote $\alpha=\tan^{-1}((2x+1)/(x+1)), \beta=\tan^{-1}((2x-1)/(x-1))$ and $\gamma=\tan^{-1}(1+x)$. The original equation is
\begin{equation}
(1)\qquad \alpha+\beta=2\gamma.
\end{equation}
The equation you solved is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3... | Like Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
using $\cos(A+B)$ and the definition of principal values
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}) &\mbox{if }\cos^{-1}x+\cos^{-1}y\le\pi \\
2\pi-\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}) & \mbox{otherwise } \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is div... | If $a_2>2$, then
$$ \sum_{k=1}^4\frac1{a_k}\le1+\frac13+\frac14+\frac15=\frac{107}{60}<\frac{11}{6}.$$
Hence we must have $a_2\le 2$ and therefore $a_1=1$, $a_2=2$.
Solving for $a_4$ in terms of $a_3$, we now find
$$ a_4=\frac1{\frac{11}6-1-\frac12-\frac1{a_3}}=\frac{3a_3}{a_3-3}$$
which forbids $a_3=3$, is no integer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$\angle DAB = 10^\circ$, $\angle CAD = 20^\circ$, $\angle CDB = 140^\circ$, $CD = BD$. Find $\angle ABD$ $\angle DAB = 10^\circ$
$\angle CAD = 20^\circ$
$\angle CDB = 140^\circ$
The triangle $\triangle BCD$ is isosceles. $CD = BD$
Find the angle $x$
The answer is $30º$.
Playing with GeoGebra, I found that $AC = AD$. B... | Construct an equilateral triangle $\triangle DCE$ inside of $\triangle ADC$ as follows:
We can see that $\angle ADB = 180^\circ - \angle DAB - \angle ABD = 170^\circ - x$ because $\angle DAB = 10^\circ$. Now, because $\angle CDB = 140^\circ$, we now know that $\angle ADC = 360^\circ - \angle ADB - \angle CDB = 50^\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$ when $a$ is at most $6$. Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$.
We can see that $a+b+c+d+e \leq10$ and so there are $10+5 \choose 5$ solutions.
Now, what if $a$ is at m... | The shortcut is that $a + b + c + d + e \leq 10$, where $a \geq 7$ bijects to $a + b + c + d + e \leq 3$, where $a,b,c,d,e$ are non-negative integers.
Therefore, the number of solutions to deduct is
$$\binom{7}{4} + \binom{6}{4} + \binom{5}{4} + \binom{4}{4} = \binom{8}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Understanding proof of $(m+n+2)^2 \neq 3(mn + 1)$ The problem is: Prove that there are no integers $m$ and $n$ such that:
$$(m+n+2)^2 = 3(mn + 1)$$
I found the explanation for the problem solution, but I still don't understand it.
Here's the solution I found:
If there are numbers $m$ and $n$ such that: $(m+n+2)^2 = 3(... | Alternative proof
$$(m+n+2)^2-3mn-3=0$$
$$\iff m^2 - m (n -4) + n^2 + 4 n + 1=0 $$
$$\begin{align}\Delta &=-3 n^2 - 24 n + 12\\
&=60 - 3 (n + 4)^2\end{align}$$
$$60 - 3 (n + 4)^2≥0, ~ 60 - 3 (n + 4)^2=T^2, T\in\mathbb Z$$
This means, such $n\in\mathbb Z$ doesn't exist.
Small supplement:
$$\begin{align}0&≤60 - 3 (n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $ \sum_{k=0}{n\choose 4k+1}$ Calculate $$\sum_{k=0}{n\choose 4k+1}$$
This should be an easy and short result but I'm messing up somewhere. What I've done so far is take $f(x)=(1+x)^n$ and with the binomial theorem expand $f(1), f(-1), f(i), f(-i)$ but it leads to taking cases for n's division rest to 4 (whi... | In case there is interest we can also do this one with residues, by way
of enrichment, and demonstrate some complex arithmetic. Start with
$$\sum_{k\ge 0} {n\choose 4k+1}
= \sum_{k\ge 0} {n\choose n-4k-1}
= [z^{n-1}] (1+z)^n \sum_{k\ge 0} z^{4k}
\\ = [z^{n-1}] (1+z)^n \frac{1}{1-z^4}
= \mathrm{Res}_{z=0} \frac{1}{z^n}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
I wonder if calculus can be used to solve this problem? If not how to solve this? I am thinking about $(3\cos x + 2\cos y)^2=9\cos^2x+12\cos x\cos y + 4... | Hint:
$\begin{align}(3\cos x + 2\cos y)^2=&9\cos^2x+12\cos x\cos y + 4\cos^2y\\=&13+12\cos x\cos y +12\sin x\sin y -(9\sin^2x+4\sin^2y+12\sin x\sin y)\\=& 13+12\cos(x-y)-(3\sin x+2\sin y)^2\\=&-3+12\cos(x-y)\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove $f(x)$ is differentiable at a point.
Let $f(x)$ be a funciton defined over $(0,1)$ such that
$$f(x)=\begin{cases} \dfrac{1}{p^3},&x=\dfrac{q}{p},\text{where} ~p,q
\in \mathbb{N}, \text{and}~ {\rm GCD}(p,q)=1,\\ 0, &\text{otherwise}.
\end{cases}$$
Prove $f(x)$ is differentiable at $x=\dfrac{n\sqrt{2}}{m}$ for ... | The differential quotient
$$
\frac{f\left(x\right)-f\left(\frac{n\sqrt{2}}{m}\right)}{x - \frac{n\sqrt{2}}{m}}
$$
is zero for irrational $x$, therefore it remains to show that it tends to zero for rational numbers $x\to \frac{n\sqrt{2}}{m}$.
For $x=q/p$ is
$$
\left|\frac{f\left(x\right)-f\left(\frac{n\sqrt{2}}{m}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which... | Write $(b+1/y)(y+1/a)(a+1/b)$ as $\frac{(by+1)(ya+1)(ab+1)}{yab}$ by finding a common denominator for each bracket. Expanding this gives:
$$\frac{(aby)^2+aby(a+b+y)+ab+by+ya+1}{aby}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the volume of the region bounded by a sphere and a paraboloid using cylindrical and spherical coordinates. The sphere is $x^2 + y^2 + z^2 = 4$ and the paraboloid is $x^2 + y^2 = 3z$
I've already done for the cylindrical coordinates and got:
$$\int^{2\pi}_0\int_0^1\int_{\frac{r^2}{3}}^\sqrt{4-r^2}rdzdrd\theta = \fr... | At the intersection of paraboloid and sphere,
$x^2 + y^2 + z^2 = 4, x^2+y^2 = 3z \implies z^2 + 3z - 4 = 0$.
i.e, $z = 1, x^2+y^2 = 3$.
So in cylindrical coordinates, $0 \leq r \leq \sqrt3$.
Now in spherical coordinates,
$x = \rho \cos\theta \sin\phi,y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
At the intersection ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum of $\sum_{k=1}^n x_k^4$ Given $x_k\in\mathbb{R}$ satisfying
$$\sum_{k=1}^n x_k=0\quad\text{and}\quad\sum_{k=1}^n x_k^2=n^2-n$$
Find the maximum of
$$\sum_{k=1}^n x_k^4$$
I have notice that $\displaystyle\sum_{k=1}^n x_k=0$ implies that
$$x_m^2=\bigg(\sum_{\substack{k=1\\ k\not=m}}^n x_k\bigg)^2\leq (n-... | It looks smart at the first glance to use $(x_k^2-(n-1)^2)(x_k+1)^2\leq0$, but it would not get you to anywhere. In this case the equity holds at $x_1=1-n$, and $x_2=\cdots=x_n=1$, therefore we obtain $x_m\geq1-n$, hence we have
\begin{align*}\begin{split}&\qquad\qquad\sum_{k=1}^n (x_k+n-1)^2(x_k-1)^2\geq 0\\ \therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Help with $ \lim \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) $ I have proved that $$\lim_{n\to \infty} \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4},$$
by using the Riemann's sum of $\arctan$ on $[0,1]$. Now... | By Taylor's formula
$$
\frac{1}{{1 + (x/n)^2 }} = \frac{1}{{1 + (k/n)^2 }} - \frac{1}{n}\frac{{2k/n}}{{(1 + (k/n)^2 )^2 }}(x - k) + \frac{{(3(\xi /n)^2 - 1)}}{{(1 + (\xi /n)^2 )^3 }}\frac{1}{{n^2 }}(x - k)^2 .
$$
for $0<x<n$, $0\leq k\leq n$ and some $0<\xi<n$ depending on $x$, $k$ and $n$. Since
$$
\left|\frac{{(3(\x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Given $\sin{x} = \frac{1 + \sqrt{5}}{4}$, find $x$ If I have been given that
$$\sin{x} = \frac{1 + \sqrt{5}}{4}$$ is there any way using which I can find out what the value of $x$ is? Sure I can remember the values, but I was curious whether if I knew the values of $\sin{\frac{\pi}{2}}, \sin{\frac{\pi}{4}}, \sin{\frac... | Let $\cos y = \frac{\sqrt5+1}4$. Then
$$\cos 2y =2\cos^2 y-1= 2\left(\frac{\sqrt5+1}4\right)^2-1=\frac{\sqrt5-1}4$$
$$\cos 4y =2\cos^2 2y-1= -\frac{\sqrt5+1}4= -\cos y= \cos(\pi\pm y)$$
The $+$ sign leads to $y=\frac\pi3$ while the $-$ sign leads to $y=\frac\pi5$. But then the value above for $\cos 2y$ is positive so w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Image under a Mobius Transformation I am studying the Mobius transformation
$$f(z) = \frac{z(1+i)}{z - i}$$
I need to find where the transformation maps the real axis, the imaginary axis and the unit circle. I know that Mobius transformations map lines and circles to lines or circles.
So I let $z = x + iy$, so we have ... | Start with
$$
\frac{z}{z-i} = \frac{x+iy}{x+(y-1)i} = \frac{(x+iy)(x-(y-1)i)}{x^2+(y-1)^2} = \\
= \frac{x^2 + y(y-1) + (xy - x(y-1))i}{x^2+(y-1)^2} =
\frac{x^2 + y(y-1)}{x^2+(y-1)^2} + i\frac{x}{x^2+(y-1)^2}
$$
So, for $z = x + i 0$, $x\in\mathbb{R}$:
$$
f(z) = (1+i)\left(
\frac{x^2}{x^2+1} + i\frac{x}{x^2+1}
\right)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the following differential equation by undetermined coefficients $(1-x^2)y''-2xy'+6y=0$ I'm trying to obtain a recursion formula relating $a_{n+2}$ to $a_n$ using the method of undetermined coefficients. Then determining $a_n$ explicitly for each $n$ and finding the sum of the series.
$$(1-x^2)y''-2xy'+6y=0$$
Tak... | This is a special case of Legendre's differential equation at $\ell=2$.
The series expansion should be $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\color{blue}-\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-2\sum_{n=1}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n=0$$ from which $$(n+2)(n+1)a_{n+2}-(n+3)(n-2)a_n=0.$$ Note that $a_2=-3a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating $\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$ Please help me calculate the:
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$$
My attempt is: First method: for $t>0$ we have $e^t\geq \frac{t^3}{3}$ now i have:
$$(x^2+y^2)e^{-(x+y)}\leq\frac{3(x^2+y^2)}{(x+y)}\rightarrow 0, (x,y)\r... | Using $ e^t > \frac{t^4}{4!}$, we have for $x,y >0$
$$e^{x+y} > \frac{(x+y)^4}{4!} = \frac{(x^2+y^2+2xy)^2}{4!} > \frac{(x^2+y^2)^2}{4!} $$
$$(x^2+y^2)e^{-(x+y)} < 4! \frac{x^2+y^2}{(x^2+y^2 )^2}=\frac{4!}{(x^2+y^2)} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches)
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$
It is a problem from a timed exam so I'm looking for alte... | One approach could be to parameterize the curve $x^2+y^2=100$. Define the functions $x$ and $y$ by $x(t)=10\cos(t)$ and $y(t)=10\sin(t)$ on the interval $(0,\pi/2)$ (this constraint ensures that $x$ and $y$ are bijective and strictly positive, since $\cos(t)>0$ and $\sin(t)>0$ for all $t\in(0,\pi/2)$). The quantity $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$ I'm looking for proof of the following identity
$$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$$
I have worked on this problem for quite some time, however since I'm not much comfortable with beta fu... | $$I=\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx$$ $$=\frac{1}{4}\int_0^1\frac{\ln t\ln^2(1-t)}{\sqrt t\sqrt{1-t}}dt=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}\int_0^1t^{a-1/2}(1-t)^{b-1/2}dt|_{a=b=0}$$
$$=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}B(a+1/2;b+1/2)|_{a=b=0}$$
$$\frac{1}{4}B(a+1/2;b+1/2)=\frac{1}{4}\frac{\Ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Angle between tangents to the curve $x^2+3y^2=9$
Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.
My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation fo... | Hint: $\alpha^4+\alpha^2-12=(\alpha^2-3)(\alpha^2+4)$ making the angle $\pi/2$ for $\alpha=\pm\sqrt{3}.$
Now check for say $\alpha\approx 2.2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Double Integral of $\sqrt{x^2+y^2}$ I want to calcule the following integral,
$$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}}\sqrt{x^2+y^2}dxdy$$
what I have is ...
Let $x=y\tan\theta$ where $\theta\in(-\pi/2,\pi/2)$, then, $dx=y\sec^2\theta d\theta$, so,
$$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}... | That integral is equal to$$2\int_0^{1/\sqrt2}\int_0^x\sqrt{x^2+y^2}\,\mathrm dy\,\mathrm dx.$$In polar coordinates, that is equal to$$2\int_0^{\pi/4}\int_0^{1/(\sqrt2\cos(\theta))}\rho^2\,\mathrm d\rho\,\mathrm d\theta=\frac{\sqrt{2}+\log \left(1+\sqrt{2}\right)}{6\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-{\sqrt{4n^2+n+9}}$ Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.
My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain
$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac... | $$
\begin{align*}
&\lim_{n\rightarrow \infty} \left( \sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9} \right)
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right)
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128775",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| Apply the trinomial theorem: $$(1+x+x^2)^6=\sum_{a+b+c=6,\:0\le a,b,c\le 6} \binom{6}{a,b,c}(1)^a(x)^b(x^2)^c$$
To get the $x^6$ term, the possibilities are $(a,b,c)=(3,0,3),(2,2,2),(1,4,1),(0,6,0)$. Hence the $x^6$ term is $$\binom{6}{3,0,3}x^6+\binom{6}{2,2,2}x^6+\binom{6}{1,4,1}x^6+\binom{6}{0,6,0}x^6=141x^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 0
} |
Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfe... | $$
\begin{array}{l}
a + b + c = 5\quad \left| {\,1 \le a,b,c\left( { \le 3} \right)} \right.\quad \Rightarrow \\
\Rightarrow \quad \left( {6 - a} \right) + \left( {6 - b} \right) + \left( {6 - c} \right)
= x + y + z = 13\quad \left| {\,3 \le x,y,z, \le 5} \right.\quad \Rightarrow \\
\Rightarrow \quad \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Infinite fraction's derivative
$f(x)=x+\dfrac{1}{x+\dfrac{1}{x+\dfrac{1}{x+\ldots}}}$
$f'\left(\dfrac{3}{2}\right)=?$
I tried to make equation like $y^2=xy+1$ but I can't made it clearly.
| Using implicit differentiation,
$f^2-xf-1=0
$
gives
$0
=2ff'-f-xf'
=(2f-x)f'-f
$
so
$f'
=\dfrac{f}{2f-x}
$.
Since
$f(x)
=\dfrac{x+\sqrt{x^2+4}}{2}
$
(need the positive root),
$f(\frac32)
=\dfrac{\frac32+\sqrt{\frac94+4}}{2}
=\dfrac{\frac32+\sqrt{\frac{25}{4}}}{2}
=\dfrac{\frac32+\frac52}{2}
=2
$
so
$f'(\frac32)
=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the limit of the $a_{n+1}=\frac{1+a_n+{a_{n-1}}^3}{3}$? $a_{n+1}=\frac{1+a_n+a_{n-1}^3}{3}$ with $n \geq 2$ ( $a_1 =0$, $a_2 = 1/2$)
By mathematical induction, the $0 \leq a_n \leq 1$ and the $a_n$ is monotonically increasing. If the limit exists, then $1$ or $\frac{-1 + \sqrt{5}}{2}$ would be value.
What i... | If the limit exists, then let $L = \lim\limits_{n \to \infty} a_n$. We must have $L = \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n + 1} = \lim\limits_{n \to \infty} \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + (\lim\limits_{n \to \infty} a_n) + (\lim\limits_{n \to \infty} a_{n - 1})^3}{3} = \frac{1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding the bounds to solve for the volume using cylindrical coordinates My goal is to set up the triple integral that will solve for the volume inside the sphere $x^2 + y^2 + z^2 = 2z$ and above the paraboloid $x^2 + y^2 = z$ using cylindrical coordinates.
Upon solving, I know that the intersection of the sphere and t... | Yes you are correct. Paraboloid $z = x^2 + y^2 = r^2$ and Sphere $r^2 + z^2 = 2z$
To find $z$ at intersection, plug in $r^2 = z$ from equation of paraboloid into the equation of sphere. We get $z^2 - z = 0 \implies z = 0, 1$. So at intersection above origin, $z = 1$. We also find that at intersection $r = 1$ which is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve... | Go step by step:
$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\
\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\
\frac{1}{(x-2)(x-1)} - \frac{1}{x(x+1)} \le 0 \\
\frac{x(x+1) - (x-2)(x-1)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{x^2+x - (x^2-3x+2)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{4x-2}{(x-2)(x-1)x(x+1)} \le 0$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that $\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$
The goal is to prove for equality: \begin{equation*}
\sum_{n = 1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6} \end{equation*} To do this, follow the steps below:
*
*Study the poles of the function $\phi(z) = \frac{\pi\cos(\pi z)}{\sin(\pi z)}$.
... | You had the right idea using the series expansion to calculate the residue at 0. You can do the following:
In our case we have that $\cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{(\pi z)^3}{45} + o(z^5)$. Then:
\begin{eqnarray*}
\text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)
& = & \lim_{z\to0}\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Maximum value of $x+2 x y+4 x y z$ Given that $x,y,z \in R^+$ such that $x+y+z=1.5$. Without using calculus find the maximum value of $x+2 x y+4 x y z $
My try:
We can write $S=x+2 x y+4 x y z =x(1+2y+4yz)$
$\implies$
$$S=x((1+2y)(1+2z)-2z)$$
$\implies$
$$S=\frac{1}{2}(1+2 x-1)(1+2 y)(1+2 z)-2 x z$$
$\implies$
$$S=\fra... | The idea is to reduce the number of variables.
EG We could use the substitution $ z = 3/2 - y - x $ to make this a 2 variable inequality.
How can we make this a 1 variable inequality?
Work through the following. If you're stuck, explain what you've tried.
*
*(In hindsight, because) The maximum occurs at $ (1, 0.5, 0... | {
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"source": "stackexchange",
"question_score": "5",
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What values of $n$ is $A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ an integer?
Suppose $a_i\in \mathbb{N} $ such that $a_i\ne a_j$ and $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ for $i \ne j$ $($take $a_{n+1}=a_1)$ and $\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+... | $$ \frac{81}{2} + \frac{36}{81} + \frac{2}{36} = 41 $$
$$ \frac{98}{12} + \frac{63}{98} + \frac{12}{63} = 9 $$
$$ \frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6 $$
$$ \frac{6}{5} + \frac{5}{3} + \frac{3}{10} + \frac{10}{12} + \frac{12}{6} = 6 $$
$$ \frac{3}{12} + \frac{12}{9} + \frac{9}{10} + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$ Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$.
The area of $ABCD$ is given by $$S_{ABCD}=\dfrac{a+b}{2}.h=\dfrac{a+b... | The trapezoid is isoceles, hence $\,\overline{AC} = \overline{BD}\,$ in:
Given the above coordinate system, we have for the diagonals, with a bit of analytic geometry:
$$
\overline{AD} \; : \; \begin{cases} x = -a/2 + (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases}
\quad \Longrightarrow \quad x-(a/2+b/2)y... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $7^{2n}-1$ is divisible by 48 Trying to prove $7^{2n}-1$ is divisible by 48.
I began by taking the base case $n=1$ so $7^{2}-1=48$ which is true.
Now from here taking the inductive step to get
$7^{2(n+1)}-1=48k$
$7^{2n+2}-1=48k$
$7^{2}*7^{2n}-1=48k$
Not quite sure where to go from here. Thanks.
| Let S = $1 + 49 + 49^2 + \cdots + 49^{n-1} $
\begin{align}
S &= 1 + 49 + 49^2 + \cdots + 49^{n-1} + 49^n - 49^n\\
S &= 1 + 49(1 + 49 + 49^2 + \cdots + 49^{n-1}) - 49^n \\
S &= 1 + 49S - 49^n \\
49^n - 1 &= 48S\\
48 &\mid 49^n - 1 \\
48 &\mid 7^{2n} - 1 \\
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Rate of change of radius of spherical balloon
A spherical balloon is filled with gas at a rate of $4 \text{ cm}^3/\text{s}$. What rate is the radius $r$ changing with respect to the time when the vol $V=36π \text{ cm}^3$?
[ans:$\frac 19, \frac{\pi}{9} $or 9$\pi$]
I tried this one with implicit differentiation:
$\frac... | $\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^3]$
*
*$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$
Volume formula: 36 π = 4/3 π r^3
*
*$r^3= 27$
*$r= 3$
Substituting r into the implicit differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculating $ \tan w$ for $ w = \dfrac{1} {i} \log\dfrac{1-iz} {1+iz}$ Let $ w = \dfrac{1} {i} \log\left(\dfrac{1-iz} {1+iz}\right)$ and calculate $\tan w$.
I begin by using the fact that
$$ \tan w = \frac{\sin w}{\cos w} = \frac1i\frac{e^{iw} - e^{-iw}}{e^{iw}+e^{-iw}}$$
I now replace $ w = \dfrac{1} {i} \log(\dfra... | Use
$$e^{\log (a)} = a,$$
and
$$e^{-\log(a)} = \frac{1}{a}.$$
Then:
$$\tan w = \frac{1} {i} \left(\frac{{e^{\log\left(\frac{1-iz} {1+iz}\right)} - e^{-{\log\left(\frac{1-iz} {1+iz}\right)}}}} {e^{\log\left(\frac{1-iz} {1+iz}\right)} + e^{-\log\left(\frac{1-iz} {1+iz}\right)}}\right) = \\
= \frac{1}{i} \frac{\frac{1-iz}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If $b$ is a divisor of $a^2 - a + 1$, can $a$ be a divisor of $b^2 - b + 1$ I observed that for $a > 1$, if $b$ is a divisor of $a^2 - a + 1$ then $a$ is not a divisor of $b^2 - b + 1$. This implies that their divisors are mutually exclusive. Is this true in general?
$$
b \mid a^2 - a + 1 \implies a \nmid b^2 - b + 1
$... | We restrict our attention to positive integers.
This is a Vieta jumping solution, where we use the existing solution set to find another solution set, and hope to reach the base case or find a contradiction.
We call $(a, b)$ a valid set if $ a \geq b > 0$, $a \mid b^2 - b + 1, b \mid a^2 - a + 1$, and $ a > 1$.
There i... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to solve $\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz$? I need to solve the integral $$\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz\,,$$ where $T$ is the domain blocked by the cylinder $x^2+y^2=2x$, the cone $z=\sqrt{x^2+y^2}+1$ and plane $z=0$.
The solution:
In order to understand exactly the relationship betw... | In $T$, you have $0<z<\sqrt{x^2+y^2}+1$ and then
$$\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz=\iint_D\int_0^{\sqrt{x^2+y^2}+1}\frac{x}{\sqrt{x^2+y^2}}\,dz\,dx\,dy=$$
$$\iint_D\frac{x(\sqrt{x^2+y^2}+1)}{\sqrt{x^2+y^2}}\,dx\,dy=\iint_D x+\frac{x}{\sqrt{x^2+y^2}}\,dx\,dy$$
where $D$ is the projection of $T$ over $XY$ p... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$
How can I prove this?
$$\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$$
Here... | By using integration by parts,
Differentiating ln(1+x) and integrating the remaining function gives an inverse tangent function integral in second part ,now set 1+x =z×(a+b)/ab this substitution reduces the second integral to integration of arctan(z)/z under respective limits and observe that this integral is solvable... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points... | Since the goal is to describe the relationship between the two lines quickly, I will give a one-sentence way to do this:
The difference between the graphs of $3x-2y=4$ and $3x-2y=10$ is that the graph of the latter equation is a dilation, centered at the origin, of the graph of the former by a factor of $10/4$.
That me... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Why does the triangle inequality seem to be false when rewritten based on $|x| = \sqrt{x^{2}}$? The triangle inequality is $$|x + y| \leq |x| + |y|.$$
Also, we know that $|x| = \sqrt{x^{2}}$. Then,
\begin{align*}
\sqrt{x + y} &\leq \sqrt{x} + \sqrt{y} \\
x + y &\leq x + y + 2\sqrt{xy} \\
0 &\leq 2\sqrt{xy} ... | The problem is not with writing $|x|=\sqrt{x^2}$: that statement is true as long as $x$ is a real number. But the correct translation of $|x+y|\le|x|+|y|$ is
$$
\sqrt{(x+y)^2} \le \sqrt{x^2} + \sqrt{y^2} \tag{*}\label{*} \, ,
$$
not what you wrote. To prove that $\eqref{*}$ is true, we can use the fact that if $a$ and ... | {
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"source": "stackexchange",
"question_score": "3",
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Is a symmetric matrix a subspace of nxn matrices? Let $M$ be a vector space for all 2x2 matrices.
Show that the set of all the symmetric matrices $M2$={$B ∈ M2 : B=B^t$} is a subspace of M.
My solution:
*
*The null matrix is symmetric
$$B=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
*Let $B$, $C$ $∈$ $M2$, where $B = ... | Your answer is fine.
Or you could have noticed that there is a basis for your space:
$$m_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, m_2=\begin{pmatrix}0&1\\1&0\end{pmatrix},
m_3=\begin{pmatrix}0&0\\0&1\end{pmatrix}$$
Show a $2\times 2$ matrix is symmetric if and only if it can be written as $$am_1+bm_2+cm_3$$
for some rea... | {
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"source": "stackexchange",
"question_score": "4",
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Solve the system of equations $x + y^2 + z^3 = 3; y + z^2 + x^3 = 3; z + x^2 + y^3 = 3$ (where x,y,z are positive real numbers) It seems that $x=y=z=1$ is the only solution, but I can't find how to prove it. I've tried defining $a=x-1; b=y-1; c=z-1$, but with no success. Any ideas?
Geometrically, each equation defines ... | The system can be written as:
$$
\begin{align}
\begin{cases}
\;\; (x-1) &+\;\; (y-1)\,(y+1) &+\;\; (z-1)\,(z^2+z+1) &= 0
\\ \;\; (x-1)\,(x^2+x+1) &+\;\;(y-1) &+\;\; (z-1)\,(z+1) &= 0
\\ \;\; (x-1)\,(x+1) &+\;\;(y-1)\,(y^2+y+1) &+\;\; (z-1) &= 0
\end{cases}
\end{align}
$$
Regarding it as a linear homogeneous system in $... | {
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Solutions of $ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$ For equation $$ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$$
is there any nontrivial solution?
Easy to find $x=0,1$ are the trivial solutions, and also, easy to figure out that $ x>1 $ or $ x<0 $ would not satisfy the equation.
B... | We prove there is no solution in $(0,1)$. The equation is equivalent to
$$\iff 5^x\left(1-\left( \frac{4}{5} \right)^x \right) = 5^{x^2}\left(\left( \frac{6}{5} \right)^{x^2} -1\right)$$
$$\iff 5^{x-x^2}\left(1-\left( \frac{4}{5} \right)^x \right) = \left( \frac{6}{5} \right)^{x^2} -1 \tag{1}$$
We observe that, for $x ... | {
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How would one solve $z^6+z^4+z^3+z^2+1=0$? I recently came upon the following equation, which is supposed to be solvable using complex numbers:
$$z^6+z^4+z^3+z^2+1=0$$
The problem is that it is a degree $6$ polynomial with many other exponential terms, which I don't know how to deal with. I try substituting $w=z^2$ to ... | Note that neither $z = 1$ nor $z = -1$ are roots. Then our equation becomes equivalent to
$$0 = (z^2 - 1)(z^6 + z^4 + z^3 + z^2 + 1) \text{ and } z \neq \pm 1.$$
This equation simplifies to
$$z^8 + z^5 - z^3 - 1 = 0.$$
We can factor this by:
$$z^8 + z^5 - z^3 - 1 = z^5(z^3 + 1) - 1(z^3 + 1) = (z^5 - 1)(x^3 + 1) = 0.$$
... | {
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Fraction/Binomial Coefficient Estimate Let $k \leq k+x <n.$ Prove
$$(1)\ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{n-k-x}{n-x}\right)^x \leq \frac{{n-x \choose k}}{{n \choose k}} \leq \left(\frac{n-k}{n}\right)^x.$$
(This is part of formula 1.23 in Stasys Jukna's extremal combinatorics.)
What I have tried:
(A) I tried to appl... | It may be helpful to adjust the $\dfrac{{n-x \choose k}}{{n \choose k}}$ expression so each of the three items is the product of $x$ terms:
*
*$\dfrac{{n-x \choose k}}{{n \choose k}}= \frac{n-x}{n}\frac{n-x-1}{n-1}\frac{n-x-2}{n-2}\cdots \frac{n-x-k+1}{n-k+1} = \frac{n-k}{n}\frac{n-k-1}{n-1}\frac{n-k-2}{n-2}\cdots \f... | {
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How to evaluate $\int_0^{\infty}\frac{\arctan{x}}{x^{1/4}(1+x)}\mathrm{d}x$? I want to evaluate
$ \displaystyle \int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } \mathrm{d}x \tag*{} $
I know that it equals
$\displaystyle \frac{\pi^2}{2 \sqrt{2} } + \frac{\pi}{\sqrt{2}} \ln \left( \frac{2+\sqrt{2-\sqrt{2}} }{2-\sqrt{2... | Let $I(a)=\int_{0}^{\infty} \frac{\arctan{(ax)}}{x^{1/4}(1+x) } dx $ and evaluate
\begin{align}
I’(a)=& \int_{0}^{\infty} \frac{x^{3/4}}{(1+x)(1+a^2 x^2) } dx \\
=& \>\frac1{1+a^2}\int_0^\infty \left(\frac{a^2 x^{3/4}}{1+a^2 x^2} + \frac{x^{-1/4}}{1+a^2 x^2} - \frac{x^{-1/4}}{1+x}\right)dx\\
=& \>\frac1{1+a^2}\left(\f... | {
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Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $ab... | $bc > 4 \implies abc > 4a = a+a+a+a > a+b+c+2$ since wlog $a \geq b \geq c \geq 2$.
$bc = 4 \implies b,c = 2 \implies a=2 \implies abc=a+b+c+2 \implies bc=4$ using the contrapositive of the above statement and all is proved here.
| {
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Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + ... | Consider
$$
x=\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}
$$
Then you know that $x>0$. Then
$$
x^2(\sqrt{2}+\sqrt{10})^2=3+\sqrt{5}
$$
that becomes
$$
x^2(2+10+2\sqrt{20})=3+\sqrt{5}
$$
and therefore, owing to $2\sqrt{20}=4\sqrt{5}$,
$$
4x^2=1
$$
so $x=1/2$.
| {
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Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$.
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points.
For increasing function, $$f'(x)\gt0\\2\cos(2x) - 8(a... | By your work we need $$\cos^2x-2(a+1)\cos x+a^2+2a-4>0.$$
We have three cases:
*
*$a+1>0$. In this case we need $$1^2-2(a+1)\cdot1+a^2+2a-4>0.$$
*$a+1<-1$. In this case we need $$(-1)^2-2(a+1)\cdot(-1)+a^2+2a-4>0.$$
*$$(a+1)^2-(a^2+2a-4)<0.$$
Can you end it now?
| {
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Find $\sin^5 x + \cos^5 x$
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$
My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\si... | This is my try.. I'm pretty sure that there are shorter methods to solve this problem. And it's not complete as I have ignored some tedious cases.
Let
$$A=\sin x + \cos x=\sqrt{2}\cos(\pi/4-x)\tag{1}$$
$$B=1-\sin x\cos x=1-\frac{\sin 2x}{2}=1-\frac{\cos[2(\pi/4-x)]}{2}=\frac{3}{2}-\cos^2(\pi/4-x) \tag{2}$$
Multiplying... | {
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Find all possible solutions of $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$ The Question: Find all possible solutions $x$ to the equation $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$.
My answer:
Let $x^2 + x - 1 = y$.
Then, $\sqrt[3]{y} + \sqrt[3]{9 - y} = 3.$
So, $\sqrt[3]{9 - y} = 3 - \sqrt[3]... | Alternatively, let $x^2+x=t+\frac{11}2$ to get the symmetric equation in $t$
$$\sqrt[3]{\frac92-t} + \sqrt[3]{\frac92+t} = 3 \implies
\sqrt[3]{\left(\frac{9}2\right)^2-t^2}=2\implies t=\pm \frac72
$$
which leads to $x^2+x=9$ and $x^2+x=2$.
| {
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Patterns in indefinite integral (1) So far, I've found out this:
$\int \arcsin x \mathrm d x = x \arcsin x + \sqrt{1-x^2}+C$
$\int x\arcsin x \mathrm d x = \dfrac{x^2}{2}\arcsin x +\dfrac {x}{4}\sqrt{1-x^2} -\dfrac{1}{4}\arcsin x + C$
$\int x^2 \arcsin x \mathrm d x = \dfrac {x^3}{3}\arcsin x+\dfrac{x^2}{9}\sqrt{1-x^2}... | Here is what Wolfram alpha thinks about the solution of
$$\int x^n \arcsin (x) \;\text d x$$
Now, the hypergeometric function is of course a polynomial. So, your guess for the pattern was quite right.
Now, Wolfram doesn't give you the steps. But, once you have the solution, it's very easy to prove that using induction.... | {
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Finding $x_1+x_2$, where the $x_i$, and $y$ and $z$, are positive integers satisfying $x^2+y^2+z^2=2(xy+1)$ and $x+y+z=2022$
Let $x$, $y$, and $z$ be positive integers satisfying
$$(x)^2+(y)^2+(z)^2=2(xy+1) \quad\text{and}\quad x+y+z=2022$$
If $x_1$ and $x_2$ are 2 distinct solutions for $x$, what is the value of $x_1... | So we have , $$(x-y)^2+z^2=2$$
Since $(x,y,z)\in\mathbb Z^+$ we can say $|x-y|=1$ and $z=1$ [ It is because only $1^2 ,1^2$ are the positive integers that sum upto $2$ ]
We have also given that , $$x+y+z=2022$$ $$x+y=2021$$
Now : $$x+y=2021$$ $$x-y=1$$
From these two linear equations in two variable you'll get , $$... | {
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Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges
Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges.
$Proof.$ by simplifying the integral we would get: $$\int _{2}^{\infty }\left( x^{5} \cdot \ln\left( x^{5} -1\right) -5x^{5}\ln( x)... | As you started writing it $$x^5 \log(x^5-1)-5x^5\log( x)=x^5 \log(x^5-1)-x^5\log( x^5)=x^5\log \left(1-\frac{1}{x^5}\right)$$ When $x$ is large
$$x^5\log \left(1-\frac{1}{x^5}\right)=-1-\frac{1}{2 x^5}+O\left(\frac{1}{x^{10}}\right)$$
$$\int x^5\log \left(1-\frac{1}{x^5}\right)\,dx=-x+\frac{1}{8 x^4}+O\left(\frac{1}{x^... | {
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modulo in a 24 h clock Suppose we have a clock with $24$ hours and it's $12$ o'clock. I need to calculate what time it will be after $100^{100}$ hours.
Obviously I just have to calculate $100^{100}=4 \pmod {24}$
So I started by $100=4 \pmod {24}$ but I cannot go anywhere from here.
I also tried typing it as $100=-20 \... | Many ways to approach this sort of problem. Mine may be a bit long, but it works.
You want $4^{100} = 2^{200} \pmod{24}$.
First note that $2^7 = 128 \equiv 8 = 2^3 \pmod{24}$
This allows you to write $2^{7k} \equiv 2^{3k} \pmod{24}$.
Now $200 = 7(28)+4$ so you can write $2^{200} \equiv 2^{3(28) + 4} \\= 2^{7(12)+4} \eq... | {
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Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive s... | Let $$f_n(x) = \frac{1}{x+1} + \cdots + \frac{1}{n(x+n)}.$$
Observe that $f_n(x)$ is strictly decreasing in $x$ for $x\ge 0$. Further, for $n\ge 2$, $f_n(0) > 1$ and $\lim_{x\to\infty} f_n(x) = 0$. It follows that there is a unique positive solution $x_n$ to $f_n(x) = 1$.
Let $f$ be defined by
$$f(x) = \sum_{i=1}^\inf... | {
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Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
... | An alternative approach is using Lagrange multipliers. Construct the Lagrangian auxiliary function
$$\mathcal L=x^{2} + y^{2} + z^{2}+\lambda\left[1-\left(x^{2} - yz\right)\right]$$
and solve the first order conditions
$$\begin{aligned}
\frac{d\mathcal L}{d x}=2x-2x\lambda=0\\
\frac{d\mathcal L}{d y}=2y+z\lambda=0\\
\f... | {
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Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequa... | Using triangle inequality,
$a(b+c)+b(c+a)+c(a+b) \gt a^2 + b^2 + c^2$
$a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b) \gt 2 (a^2 + b^2 + c^2)$
$(a+b+c)^2 \gt 2(a^2 + b^2 + c^2)$
$\cfrac{a^2 + b^2 + c^2}{(a+b+c)^2} \lt \cfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
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show that series $\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n+x^2}$ is uniformly convergent for all real values of x I tried using weirestrass M test .
My method : let $$\ f(x)=\frac {1}{(n+x^2)}\ $$
$$f'(x)= \frac {-2x}{(n+x^2)^2}=0 $$
this gives $x=0 $
$$f''(x)= -2[\frac{(n+x^2)^2 -2(n+x^2)(2x.x)}{(n+x^2)^4}]\ = \frac {-2... | You can rewrite your sum as follows:
\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{2n-1 + x^2} - \frac{1}{2n + x^2} & = \sum_{n=1}^{\infty} f(2n-1) - f(2n) \\
\end{align*}
where $$f(t) = \frac{1}{t+x^2}$$
The Mean Value Theorem says there is some $c \in (t-1,t)$ for which we have $$f(t-1) - f(t) = (\frac{d}{dt}f)(c) = ... | {
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Determine $\sin(x + y)$ and $\cos(x - y)$, if you know $\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5},\quad y\in(\pi/2,\pi)$ I need advice on this example.
Determine $\sin (x + y)$ and $\cos (x - y)$, if you know that
$\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5}, \q... | You have a sign error.
$y$ is in the second quadrant, hence $\cos(y) = -\frac{\sqrt{24}}{5}$.
\begin{align}
\sin(x+y) = -\frac{\sqrt{24}}{7} \cdot \frac{\sqrt{24}}{5} + \frac57 \cdot \frac15 = -\frac{19}{35}
\end{align}
| {
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Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
| The following is a somewhat more laborious, but also more generic, way to solve the problem.
*
*The minimal (rational) polynomial of $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11}\,$ is
$$
P(x)=x^8 - 88 x^6 + 1696 x^4 - 10560 x^2 + 14400
$$
which can be determined similar to Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$.
... | {
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"timestamp": "2023-03-29T00:00:00",
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solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$. I want to solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$.
I started by saying $\cos (z) = \frac {e^{iz} + e^{-iz}}{2}$.
I am therefore led to solve:
$$e^{iz} + e^{-iz} = 2 + 2i $$ this implies that $$e^{2iz}+1 =(2 + 2i) e^{iz}$$
By setting $x = e^{iz}$
Consider ... | At this point
$x^2-(2 + 2i) x + 1 = 0.$
Applying the quadratic formula:
$x = 1+i \pm \sqrt {2i - 1}:$
$(1+i + \sqrt {2i - 1})(1+i - \sqrt {2i- 1}) = 2i - (2i-1) = 1$
This means we can let $e^{iz} = 1+i + \sqrt {2i - 1}$ and $e^{-iz} = 1+i - \sqrt {2i - 1}$
$iz = \ln (1+i + \sqrt {2i - 1})\\
z = -i\ln (1+i + \sqrt {2i -... | {
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Evaluating integral $\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$ I come across a problem, which is to evaluate: $$\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$$
My attempt so far is:
$$\begin{align}\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx
&= \int_0^{2\pi}\frac{1}{2-\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}\,dx\\
&= \int_0^{2\p... | Note that because of periodicity, $\displaystyle \int_0^{2\pi} = \int_{-\pi}^{+\pi}.$
If you let $u = \tan\tfrac x2$ so that $\sin x= \dfrac{2u}{1+u^2}$ and $du = \dfrac{2\,du}{1+u^2},$ then as $x$ goes from $-\pi$ to $+\pi,$ $u$ goes from $-\infty$ to $+\infty.$
\begin{align}
& \int_{-\pi}^{+\pi} \frac 1 {1+\tan^2(\fr... | {
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Irreducibilty of polynomial $x^9-6x^6+282x^3-8$ over $\mathbb {Q} $ While trying to deal with the final parts of this answer I found that one needs to establish $$a=\sqrt[3]{1+\sqrt{11}}\notin\mathbb{Q} (\sqrt{11},\sqrt[3]{10})=F$$ Since both $a, F$ are of degree $6$ over $\mathbb {Q} $ the tower theorem does not help... | The polynomial $P(x)$ factors $\!\!\!\mod \!\!13$ as $(x^3+6) ( x^6+ x^3 + 3) $ (link).
Assume that $P(x)$ is reducible. We conclude that $P(x)$ is a product of two polynomials $P_1(x)$, $P_2(x)\in \mathbb{Z}[x]$. with $P_1(x)\equiv x^3 + 6 \!\!\!\mod \!\!13$, $P_2(x)\equiv x^6 + x^3 + 3 \!\!\!\mod\!\! 13$.
Now, $P... | {
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Contradiction in arranging words in two different ways
In how many ways can the letters in "WONDERING" be arranged with exactly two consecutive vowels ?
The solution is $18 \times 7!$ by classical counting technique
(because firstly choose $2$ adjacent vowels out of $3$ vowels by $3$ ways.Lets say we chose $O,E$ Afte... | Your first calculation is correct.
Method 1: First, arrange the six consonants N, N, D, G, R, W. This can be done in
$$\binom{6}{2}4! = \frac{6!}{4!2!} \cdot 4! = \frac{6!}{2!}$$
ways since we must select two of the six positions for the Ns, then arrange the remaining four distinct letters in the remaining six positi... | {
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"url": "https://math.stackexchange.com/questions/4215554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that addition modulo 2 is associative I learned that |a⊕(a⊕b)⟩=|(a⊕a)⊕b⟩ as addition modulo 2 is associative. However, I am wondering why addition modulo 2 is associative?
| *
*Try them all; there are only eight cases
a) $(0\oplus0)\oplus0=0\oplus0=0;0\oplus(0\oplus0)=0\oplus0=0$
b) $(0\oplus0)\oplus1=0\oplus1=1;0\oplus(0\oplus1)=0\oplus1=1$
c) $(0\oplus1)\oplus0=1\oplus0=1;0\oplus(1\oplus0)=0\oplus1=1$
d) $(0\oplus1)\oplus1=1\oplus1=0;0\oplus(1\oplus1)=0\oplus0=1$
e) $(1\oplus0)\oplus0=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=... | Using that :
*
*$x^2=-x+5$
*$x^3=x\cdot x^2=-x^2+5x=x-5+5x=6x-5$
Then:
$$
\require{cancel}
y = \frac{1}{(x+1)^3}=\frac{1}{x^3+3x^2+3x+1}=\frac{1}{(6x-5)+3(-x+5)+3x+1} = \frac{1}{6x+11}
\\ \iff\;\;\;\; 6x = \frac{1}{y}-11
$$
Substituting the latter in $36x^2 + 36x - 180 = 0$ and multiplying by $y^2$ gives:
$$
(1-11y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find all primes $r$ and $s$ such that $ r^7+s=s^3+r $
Find all primes $r$ and $s$ such that $\;r^7+s=s^3+r\;.$
I have tried to factorise the expression since we are dealing with primes so:
$$ s(s^2-1)= r(r^6-1)=r(r^2-1)(r^4+r^2+1).$$
Since $ s^3>r^7$, then $\;s^2-1>r^4-1$.
That is all what I have found, thank you in ... | Trying to milk more out of the factorization
$$
s(s-1)(s+1)=s^3-s=r^7-r=r(r-1)(r+1)(r^2-r+1)(r^2+r+1).\qquad(*)
$$
The OP already observed that $s^3>r^7$. Therefore $s>r^2$. Let's remember that $s$ is a prime. Staring at $(*)$ for a few seconds tells us that the only possible prime factor of the RHS larger than $r^2$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Evaluate by polynomial expansion the following integral with an error less than 0.000001 My work till now:
$\int_0^{0.2} \frac{1}{1+x^5} dx=\int_0^{0.2}(1+x^5)^{-1} dx$
$(1+x)^n=1+nx+ \frac{n(n-1)x^2}{2}...$
$\int_0^{0.2} \bigl( 1+(-1)x^5+ \frac{(-1)(-2)x^{10}}{2}- \frac{1(-2)(-3)x^{15}}{3!}+...\bigr)dx$
Now, I could i... | I think that the approximation works well: since
$$1+x^{15}=(1+x^5)(1-x^5+x^{10})$$
it follows that for any $x\in [0,1]$
$$\frac{1}{1+x^5}=1-x^5+x^{10}-\frac{x^{15}}{1+x^5}.$$
Hence
$$\begin{align}
\left|\int_{0}^{0.2}\frac{1}{1+x^5}\,dx-\int_{0}^{0.2}(1-x^5+x^{10})\,dx\right|&\leq\int_{0}^{0.2} \frac{x^{15}}{1+x^5}\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solving $\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$ I’m having a hard time with this radical equation.
$$\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$$
I subtracted square root of x^2-3 then I got ^3 square root of -x+4 + square root of x^2-3 = 1. Next I cube both sides which I got -x^2-3= (1- square root of x^2-3) ^3. ... | HINT
We can let $u = x^{2} - 3$, where $x^{2} - 3 \geq 0$. Then we get that
\begin{align*}
\sqrt[3]{1 - u} + \sqrt{u} = 1 & \Longleftrightarrow \sqrt[3]{1-u} = 1 - \sqrt{u}\\\\
& \Longleftrightarrow 1 - u = 1 - 3\sqrt{u} + 3u - \sqrt{u^{3}}\\\\
& \Longleftrightarrow \sqrt{u^{3}} - 4u + 3\sqrt{u} =0\\\\
& \Longleftrig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate a coordinate of the center of mass of the solid $A$ with constant density $1$ Find the $y_G$ coordinate of the center of mass $G$ of the solid $A=\{(x,y,z) \in \mathbb{R}^3 \ | \ 3x^2 \leq y^2+z^2 \leq 3-x^2, yz \leq 0, z \geq |y|\}$ with constant density $\rho(x,y,z)=1$.
Can someone help me with the integrati... | We are talking about solid bound outside cone $3x^2 = y^2 + z^2$ and inside sphere $x^2+y^2+z^2 = 3$ along with constraints, $y z \leq 0$ and $z \geq |y|$.
First of all there is symmetry about $x = 0$, so $\overline x = 0$.
We are using $y = r \cos\theta, z = r \sin\theta$ and as we have $y z \leq 0$ and $z \geq |y|$,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by ... | As implicitely pointed out in the given answer, your mistake is due to the fact that for the sum of the squares of even natural numbers up to $n-1$ we need to consider the following:
$$2^2 + 4^2 .... (n-1)^2=2^2\left(1+2^2+\dots+\left(\frac{n-1}2\right)^2\right) $$
and not
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proving $\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}$ using the extraction/absorbtion identity. I want to use the extraction/absorbtion identity to prove the equality
$$
\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}
$$
Followed from this identity, we could directly obtain
$$
\binom{n+1}{2k+1}=\frac{n+1}{2k+1}... | Since $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, we have
$$\binom{n+1}{1} + \binom{n+1}{3} + \binom{n+1}{5} + \dots = \left[\binom{n}{0} + \binom{n}{1}\right] + \left[\binom{n}{2} + \binom{n}{3}\right] + \cdots$$
which is $2^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Deducing an inequality from a Taylor series expansion Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$
by... | We have a function
$$f(x) =\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128 }- \frac{7 x^5}{256}- \cdots = \\=1 - a_1 x - a_2 x^2 -a_3 x^3 \cdots $$
The series converges for every $x\in [0,1]$, so we have
$$a_1 + a_2 + \cdots = 1$$
Therefore we can write
$$1 - \frac{x}{2} - \frac{x^2}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals.
Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$
The problem is from an inequality handout. Here is my attempt to solve the problem:
I first rew... | By AM-GM, C-S, Schur and Muirhead we obtain:
$$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}=\sum_{cyc}\frac{4a^2}{2\sqrt{(b+c)^2(4b^2-4bc+4c^2)}}\geq\sum_{cyc}\frac{4a^2}{(b+c)^2+4b^2-4bc+4c^2}=$$
$$=\sum_{cyc}\frac{4a^2}{5b^2-2bc+5c^2}=\sum_{cyc}\frac{4a^4}{5a^2b^2-2a^2bc+5a^2c^2}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
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