Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is $ \frac{x^2 -4}{x-2} $ a polynomial? The expression $ \frac{x^2 -4}{x-2}$ doesn’t looks like a polynomial because of $ x $ in the denominator. But it can be factorized. After factorizing and canceling the common factor $$\frac{x^2 -4}{x-2} = \frac{(x+2)(x-2)}{x-2} = x+2, $$ and $ x+2 $ is a polynomial which satisfie... | It is not because it is not equivalent to $x+2$. You can evaluate $x+2$ for any real $x$ if you are working in the reals. You cannot evaluate $\frac {x^2-4}{x-2}$ at $x=2$ because of the division by $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Angle chasing: Finding the $\angle{NRQ}$ in a quarter circle
Let $MON$ be a quarter circle. $P$ is the midpoint of $OM$ and $PQ$ is the angle bisector of $\angle{OPN}$. If $QR\parallel{OM}$, find $\angle{QRN}$.
Here is an angle-chasing problem which I am trying to solve. I couldn't make a good progress solving the p... | Here is a solution which tries to not use the algebraic knowledge of some trigonometric function (sine, cosine, tangent) of the angle of $36^\circ$ or some related angle (half or double of it). However, since there are "only a few" angles in the picture which are in measure a rational multiple of $\pi$, mainly those in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$ $$\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$$
I can reduce it to the general term,
$$\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)2^r}$$
I don't know how to go a... | Here is a longer and more painstaking way that uses calculus:
For $|x|\lt 1$, $$\sum_1^{\infty}(2r^2 +7r+4) x^{r-1}\\ = 2\sum_1^{\infty} r^2 x^{r-1} +7\sum_1^{\infty} rx^{r-1} +4\sum_1^{\infty}x^{r-1} \\ = 2\frac{d}{dx} \left( x \frac{dS}{dx}\right ) +7\frac{dS}{dx} +4S \\ = \frac{-2(1+x)}{(1-x)^3} +\frac{7}{(1-x)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\l... | Alternatively
$$I(n)= \int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=\frac4{n^4}\int_{0}^{\pi /2}\frac{dx}{\left ( a-\cos^2 x \right )^2},\>\>\>\>\>a= 1+\frac1{n^2}$$
Note that
$$J(a)= \int_{0}^{\pi/2}\frac{dx}{ a- \cos^2 x }
= \int_{0}^{\pi/2}\frac{d(\tan x)}{ a\tan^2x +(a-1)}dx= \frac\pi{2\sqrt{a(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :-
$$\rightar... | $$\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$$
$$=\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}\times\frac{2^{3/2}}{2^{3/2}}$$
$$=\frac{[8+2\sqrt{15}]^{3/2} + [8-2\sqrt{15}]^{3/2}}{[12+2\sqrt{35}]^{3/2} - [12-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Matrix Multiplication - Express a row as a linear combination $$ Let \ A \ = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 3 & 6 \\ \end{bmatrix} and
\ let \ B = \begin{bmatrix} 0 & 1 & -3 \\ -3 & 1 & 4 \end{bmatrix} $$
Express the third row of AB as a linear combination of the rows of B
$$ AB \ = \ \begin{bmatrix} -6 & 3 & 5 \\... | From the definition of matrix multiplication it follows that:
$$\underbrace{\begin{bmatrix}-18&9&15\end{bmatrix}}_{\text{third row of }AB} = 3\cdot \underbrace{\begin{bmatrix}0&1&-3\end{bmatrix}}_{\text{first row of }B}+6\underbrace{\begin{bmatrix}-3&1&4\end{bmatrix}}_{\text{second row of }B}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving that for integers $a \ge 5, a > b, b \ge 1$, $2^a - 27 \nmid 2^b + 15$ I am trying to show that for integers $a \ge 5, a > b, b \ge 1$:
$$2^a - 27 \nmid (2^b + 15)$$
Is the following inductive argument valid?
Please let me know if I made any mistake or if there is a more standard way to prove the same point.
(1... | Your proof appears to be basically correct. Nonetheless, a minor point is your inductive case doesn't actually technically involve induction as there's no actual use of the inductive hypothesis, i.e., that $2^a - 27 \not\mid 2^b + 15$, so induction was not required.
Alternatively, using size comparisons similar to wha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
$A^ {1/A} =B^ {1/B} =C^ {1/C} ,A^ {BC} +B^ {AC} +C^ {AB} =729$ If $A^
{1/A}
=B^
{1/B}
=C^
{1/C}
,A^
{BC}
+B^
{AC}
+C^
{AB}
=729$
Which of the following equals $A^
{1/A}$?
I tried solving it and in the end got $A^{BC}= B^{AC}=C^{BA}= 3^5$ Thus $A^{1/A}=\ ^{ABC}\sqrt {3^5}$
Now I have two doubts. In some of the s... | Let $A^{1/A} = B^{1/B} = C^{1/C} = k$. Now, raise each term to the $ABC$ power and you get:
$A^{BC} = B^{AC} = C{AB} = k^{ABC}$ We know that $A^{BC} + B^{AC} + C^{AB} = 729$.
So
it becomes,
$k^{ABC}\times k^{ABC}\times k^{ABC}=729$
$(k^{ABC})^3 = 729$
$k^{ABC} = 9$
$k = 9^{\frac{1}{ABC}}$
So, $A^{1/A} = 9^{1/ABC}$
Hope... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the rank of $\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$ Found this exercise in Serge Lang's Introduction to Linear Algebra:
Find the rank of the matrix $$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$
So my process to solve it is as follows. First, I set a syst... | Thanks to the comments I found that I just have to reduce it to row echelon form, I guess I misunderstood ranks, the answer is
$$\begin{aligned}
\begin{pmatrix}
1 &1 &0 &1 \\
1 &2 &2 &1 \\
3 &4 &2 &3 \end{pmatrix} &\overset{(2) - (1)}{\implies} \\
\begin{pmatrix}
1 &1 &0 &1 \\
0 &1 &2 &0 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The derivative of the cumulative distribution of $\min(X_1, a_1) + \min(X_2, a_2)$ Assume $X_1, X_2$ are continuous random variables in $L^{\infty}$. Let $a_1, a_2$ be real numbers and
$$Y := \min(X_1, a_1)+ \min(X_2, a_2).$$
The dependency structure between $X_1$ and $ X_2$ is not known, but we assume that the joint a... | We can write, for the first term
$$
\eqalign{
& \Pr \left( {\min \left( {X_{\,1} ,a_{\,1} } \right) \le t_{\,1} } \right)
= F_{\,m} \left( {t_{\,1} ;\,a_{\,1} } \right) = \cr
& = \left[ {t_{\,1} < a_{\,1} } \right]\Pr \left( {X_{\,1} \le t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right]1 = \cr
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many ways are there to color a $3 × 3$ grid in the same way if five squares have to be red and four squares have to be blue? When configurations after rotations and flips are considered the same, how many ways are there to color a $3 × 3$ grid in the same way if five squares have to be red and four squares have to ... | Enumerating the symmetries of the grid for the cycle index so that we
may apply PET we obtain
The identity, $a_1^9.$
Two rotations by $90$ and $270$ degrees which fix
the central slot, $2 a_1 a_4^2$
One rotation by $180$ degrees which also fixes the central slot,
$a_1 a_2^4$
A vertical and a horizontal flip, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4118792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $3^{n-2} \geq n^5$ for $n\geq20$ I am new to induction proofs and wanted to know if my reasoning behind the following proof is correct. Here are my steps :
*
*$n=18 \Rightarrow 3^{18} \geq 20^5$
*Given that $3^{n-2} \geq n^5$ is true for n, show $3^{n-1} \geq (n+1)^5$
$3^{n-1} = 3\times(3)^{n... | It might be easier to bypass the second induction and instead show directly that, for $n\ge20$, we have
$$\begin{align}
(n+1)^5&=n^5+5n^4+10n^3+10n^2+5n+1\\
&\le n^5+40n^4\\
&=n^5+2(20n^4)\\
&\le n^5+2(n\cdot n^4)\\
&=3n^5
\end{align}$$
where the first inequality (introducing the $40n^4$) holds for $n\ge1$ while the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | Considering the more general case $$S_a=\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^2 - a^2}$$
$$\frac{\log (k)}{k^2 - a^2}=\sum_{n=0}^\infty \frac{\log (k)}{k^{2n+2}} a^{2n}$$
$$\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^{2n+2}}=-\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$
$$S_a=-\sum_{n=0}^\infty a^{2n}\,\text{HurwitzZeta}^{(1,0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 0
} |
Rewrite $\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$ in a common base then solve for $x$ I am to rewrite $\frac{125}{(\frac{1}{625})^{-x-3}}=5^3$ and then solve for x. My textbooks solutions section says the solution is -3. I gave it a shot and got 3.25. Here is my working:
$$\frac{125}{\left(\frac{1}{625}\right... | $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large -}x-3}}=5^3$
Here you have $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large\text{negative sign}}x-3}}=5^3$
Here you correctly surmised that we can "pull" the $\frac 1{5^4}$ "up" to get that $\frac 1{(\frac 1{5^4})^M} = (5^4)^M$. That is good. So you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
I need a sanity check to compute the limit of a function in two variables So I have $g(x,y) = \tfrac{f(x,y) - y}{\sqrt{x^2 + y^2}}$, where $f(x,y) = \tfrac{y^3 -x^8y}{x^6 + y^2}$ for $(x,y) \neq (0,0)$.
I want to find $\lim_{(x,y) \to (0,0)} g(x,y)$ if it exists.
My question is if after doing some algebraic manipulatio... | call the ratio $G,$ I got
$$ |G| = \frac{|y| x^6 (1+x^2)}{r (x^6 + y^2)} $$
Demand $c = | \cos \theta | $ and $s = | \sin \theta | $
$$ |G| = \frac{r^7 c^6 s (1+r^2 c^2 )}{r^3 (r^4c^6 +s^2)} $$
$$ |G| = \frac{r^4 c^6 s (1+r^2 c^2 )}{ r^4c^6 +s^2} $$
We are allowed to demand $r < 1,$ so that $1+r^2 c^2 < 2,$
$$ |G|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | $$\lim_{(x,y)\to(0,0)}\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}} \\
= \lim_{(x,y)\to(0,0)}\frac{(xy)^3 }{(x^4+y^4)\sqrt{x^2+y^2}}\frac{\sin(xy^2)}{xy^2}$$
Let $y=x\tan\theta$
$$\lim_{(x,y,x\tan\theta)\to(0,0,0)}\frac{(x^{{6} }\tan^3\theta)}{x^4(1+\tan^4\theta)x\sqrt{1+\tan^2\theta}}\frac{\sin(xy^2)}{xy^2}\\
=\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Simplifying the sequence defined by $A_0=6$ and $A_n=\frac{8}{9}A_{n-1}+\frac{6}{5}(\frac{20}{9})^n$, using sigma notation
I have $A_0 = 6$ and
$$A_n = \left(\frac{8}{9}\right)A_{n-1} + \left(\frac{6}{5}\right) \left(\frac{20}{9}\right)^n$$
and I want to simplify it with sigma notation.
I have it up to
$$\begin{align... | For $ n\ge 1$,
$$A_n=aA_{n-1}+b_n$$
$$aA_{n-1}=a^2A_{n-2}+ab_{n-1}$$
$$a^2A_{n-2}=a^3A_{n-3}+a^2b_{n-2}$$
$$a^kA_{n-k}=a^{k+1}A_{n-k-1}+a^kb_{n-k}$$
...
$$a^{n-1}A_1=a^nA_0+a^{n-1}b_1$$
the sum gives
$$A_n=6a^n+\sum_{k=0}^{n-1}a^kb_{n-k}$$
$$=6(\frac 89)^n+\frac 65\sum_{k=0}^{n-1}(\frac 89)^k(\frac{20}{9})^{n-k}$$
$$=6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+... | The natural log of this expression is $$ K= n\ln( 1+ (x-1) )$$ where $$x=\sum_{r=1}^n \frac{1}{\sqrt{n^2+n+r}} $$ Due to the bounds you have shown, $K$ must tend to a finite non-zero number, which can only happen if $x\to 1$. Now, $$K = n(x-1) +n O((x-1)^2)$$ The first term is $$n \sum_1^n \left( \frac{1}{\sqrt{n^2+n+r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
Perfect Square With Two Integer Variables I am trying to solve a number theory problem in general form. However, I got stuck in the following step:
$a,b,n \in \mathbb Z^{+}$ for which values of $n$, this equation is solvable $\frac{(n+1)(n+2a)}{2} = b^2$ ?
Can we make a general statement about $n$ ? By the way I have t... | $\frac{1}{2}(n+1)(n+2a) = b^2$
$(2n+1+2a)^2-8b^2 = (2a-1)^2$
Let $x = \frac{\large{2n+1+2a}}{\large{2a-1}}, y = \frac{\large{2b}}{\large{2a-1}}$ then we get
$x^2-2y^2 = 1\tag{1}$
This equation is Pell equation.
The fundamental solution of equation $(1)$ is $(x,y)=(3,2).$
Thus, all of the solutions are given by $x+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the Taylor series for $\frac{1}{1-x^2}$ Where I'm stuck
$f(x)=\frac{1}{1-x^2}=\frac{1}{1-x}\cdot\!\frac{1}{1+x}=(1-x)^{-1} \cdot(1+x)^{-1}$
$(1+x)^{-1}= 1-x+x^2-x^3+...+(-1)^{2n}x^{2n}+o(x^{2n})$
$(1-x)^{-1}= 1+x+x^2+x^3+...+x^{2n}+o(x^{2n})$
At this point I have no idea how to multiply all this together.Could ... | For every $x\in(-1,1), \frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$. So we replace $x$ with $x^2$, and since $x\in(-1,1)\rightarrow x^2\in(-1,1)$ we get $\frac{1}{1-x^2}=\sum_{i=0}^{\infty}x^{2i}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ beha... | Set $x=\sqrt{1+y}-1$. Then $y=(x+1)^2-1=x^2+2x$, and your expression simplifies to
$$ \bigl(x^2+2x+\tfrac23 \bigr)\log\bigl(1+\tfrac2x \bigr) - 2x - 2 $$
Expanding the logarithm in powers of $2/x$ gives us
$$ \bigl(x^2+2x+\tfrac23 \bigr)\Bigl(\frac2x-\frac2{x^2}+\frac{8}{3x^3}-\frac{4}{x^4}+\frac{32}{5x^5}+o(x^{-5}) \B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\lim\limits_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist I want to prove that $\displaystyle \lim_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist
I denoted $f(x, y):=\frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$. I tried looking at $f(\frac{1}{n}, \frac{1}... | As Maxim pointed out the limit actually exists and is equal to 0. I give an alternative proof for that now. It suffices to show, that for |x|,|y| sufficiently small there holds
$$\left| \frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\right| \le \sqrt{2}\max(\sqrt{|y|},|x|^{\frac{1}{3}}).$$
This we see at follows. First by Cauch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ Using the generating function of Legendre polynomials, show
\begin{equation}
\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln
\left(\frac{1+x}{1-x}\right)
\end{equation}
My attempt
I know the generating function of ... | Write \begin{equation}
f(x,y)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(y)
\end{equation}
Differentiate this to obtain $f_x(x,y) = \sum_{n=0}^{\infty} x^n P_n(y)$. This is known to be $\frac{1}{\sqrt{1 - 2 x y + x^2}}$. Integrating we arrive at $f(x,y) = \log(\sqrt{-2yx+x^2+1} -y+x)$. Subtitute $x=y$ to obtain the an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | You can use Young's inequality to prove it
$$(\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}}(\frac{2}{1})^{\frac{1}{2N+2}}\leq \frac{1}{\frac{2N+2}{2N+1}}((\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}})^{\frac{2N+2}{2N+1}}+\frac{1}{2N+2}((\frac{2}{1})^{\frac{1}{2N+2}})^{2N+2}=1$$
the two sides are equal iff $((\frac{2N}{2N+1})^{\frac{2N+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
A closed expression for this sequence of integers Say we have $a_k=n-p$ for $n,p \in \mathbb{Z}_{+}$ where $n-p >0$ and $a_1=n-p-1$, and in general:
$a_k=n-p$
$a_1=n-p-1$
$a_{k-1}=n-p-2$
$a_2=n-p-3$
$a_{k-2}=n-p-4$
$a_{3}=n-p-5$
$\vdots$
and so on. The idea is that if the sequence $(a_1,a_2,...,a_k)$ is ordered from gr... | It is convenient to split the problem into two cases with even and odd $k$. Here we consider the case $k=2K$ even:
\begin{align*}
\left(a_j\right)_{1\leq j\leq 2K}=\left(a_1,a_2,\ldots,\color{blue}{a_K,a_{K+1}},\ldots,a_{2K}\right)
\end{align*}
We list the elements $a_j, 1\leq j\leq k=2K$ in a detailed way to better se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction if $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ then $H_{2^n}\ge 1+\frac{n}{2}$? Suppose $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ and we want to prove that $H_{2^n}\ge 1+\frac{n}{2}$?
*
*Since $\sum\limits_{i=1}^{2^1}\frac{1}{i}\ge1+\frac{1}{2}$ note that $H_{2^n}\ge 1+\frac{n}{2}$ for $n= 1$
*Assume $H_... | You see
$$\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{2^{n+1} } = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $~~a=b=c~~$ if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are integers.
Prove that $~~a=b=c~~$ if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are integers.
My first attempt was to add $~\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $~\fr... | Note that this is false when $a,b,c$ aren’t integers, so we need to use that fact.
Let $p$ be prime. Let $x,y,z$ be the largest so that $p^x|a,p^y|b,p^z|c$.
If $a,b,c$ are all different then there’s some $p$ such that the corresponding $x,y,z$ aren’t the same. Either one is larger than the other two or one is smaller t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expressing $\cos(\frac{\pi}{2}+i)$ and $\sin(\frac{\pi}{2} +i)$ in the exponential function
Is $\cos(\frac{\pi}{2}+i)$ equal to $i\frac{-1+e^2}{2e}$ and $\sin(\frac{\pi}{2} +i)$ to $\frac{1+e^2}{2e}$?
Trying to answer a test online. Apparently the answer in the question is incorrect or I am not expressing it correctl... | You can use the formulas
$$
\cos x = \frac 1 2 (e^{ix}+e^{-ix}) \qquad \sin x = \frac{1}{2i}(e^{ix}-e^{-ix})
$$
After inserting the values $\frac \pi 2 + i$ into both equations you get the results
$$
\cos \left(\frac \pi 2 + i\right) = \frac 1 2(e^{i\pi/2 + i^2}+e^{-i\pi/2-i^2} ) \\
=\frac 1 2\left(\frac i e + \frac{e}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| $$1+x+x^2+x^3+\dots$$
$$=(1+x)(1+x^2+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4)\dots$$
$$=\frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^8}{1-x^4}\dots$$
$$=\frac{1}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$
This is a problem ... | Since the vertices have integer coordinates we can use Pick's theorem.
In the border of the triangle there are the three vertices, and also $(4,4)$ on $AB$, the points $(6,2)$, $(5,1)$, $(4,0)$ and $(3,-1)$ on $BC$ and no other points in $AC$. These make $8$ points in the border.
On the other hand the interior of the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form? I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$
$$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\fr... | It seems that a series solution is possible
$$I=\int_0^{2\pi} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx=4\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx$$ Expanding the second logarithms
$$I= -4\sum_{n=1}^\infty \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac ... | You have\begin{align}\frac{1-\sin(x)}{1+\sin(x)}=4&\iff\frac{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}=4\\&\iff\left(\frac{\sin\left(\frac x2\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Probability that the sum of two integers in $\{1,\dots,n\}$ equals a perfect square I found the following question in MIT OCW's Mathematical Problem Solving and I'd like to know if my solution is ok:
"Let $p_n$ be the probability that $c+d$ is a perfect square when the integers $c$ and $d$ are selected independently at... | When you choose two integers uniformly at random in $\{1, 2, ..., n\}$, the probability that the sum is $i$ is
$$
P(i) = \frac{n - |n+1 - i|}{n^2} = \frac{1}{n^2}\begin{cases}i -1 & 2 \le i \le n+1 \\ 2n+1-i & n+1 \le i \le 2n \\ 0 & \mathrm{else}\end{cases}
$$
So you get
$$P(sq) = \sum_{k = 2}^{\lfloor\sqrt{n}\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Find the value of $ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2}\right )^{\frac{k}{n^2} +1} $ Compute limit of sum :
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2} \right)^{\dfrac{k}{n^2} +1} $$
My Attempt :
$$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} ... | Trying to solve my problem moving along with Winther's idea (thanks a lot) :
$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} +1 \Big) \log\Big(\frac{k}{n^2}\Big)} = \Big( e^{\Big( \frac{k}{n^2} +1 \Big) }\Big)^{\log\Big(\frac{k}{n^2}\Big)} = \dfrac{1}{n} \cdot \dfrac{k}{n} \Big( e^{\frac{k}{n^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Complex integration $\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx$ I have integral
$$I_1=\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx.$$
I tried something like this:
$$f\left(z\right)=\frac{\sqrt[3]{4z^{2}\left(1-z\right)}}{\left(1+z\right)^{3}}=\frac{\left(... | You already showed that the integral over the three circles vanishes in the limit, and we are left with (since the argument of $1 - z$ increases by $-2\pi$ when moving from the upper to the lower edge)
$$I_{1}\left(1 - e^{-2\pi i / 3}\right) = 2\pi i\operatorname{Res}(f,-1).$$
Since we are dealing with the third root, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
| Observe,
$$a^2+b^2=2\implies (a+b)^2=2(ab+1)$$
It remains to show that
$$3(a+b)+(ab+1)+4\ge0\;\Longleftrightarrow\;3(a+b)+\frac{(a+b)^2}{2}+4\ge0$$
Let $a+b=x$, we have,
$$3x+\frac{x^2}{2}+4\ge0\;\Longleftrightarrow\;x^2+6x+8\ge0\;\Longleftrightarrow\;(x+2)(x+4)\ge0$$
$$\therefore\;x\le-4 \;\;\text{or}\;\;x\ge -2$$
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Diagonalizing the matrix $A$ then finding $A^{10}$ Diagonalize the matrix
$$A= \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}$$
Then find $A^{10}.$
We have the characteristic polynomial of $A:$
$$\left | A- \lambda I \right |= \left | \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -... | $A^{10}=(SJS^{-1})^{10}=SJ^{10}S^{-1}$ and it is easy to find $J^{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log... | Yes go ahead
$$S=\int_{0}^{1} x^{-x} dx= \int_{0}^{1} e^{-x\ln x} dx =\int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-x \ln x)^k}{k!}dx.$$
Let $x=e^{-t}$,and use $\int_{0}^{\infty} t^n e^{-at}dt=\frac{n!}{a^{n+1}}.$
$$\implies S=\sum_{k=0}^{\infty} \int_{0}^{\infty}\frac{t^ke^{-(k+1)t}}{k!} dt=\sum_{k=0}^{\infty}\frac{1}{(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For $a>0$, $\displaystyle\int_{\arctan a}^{\frac{\pi}{2}}\int_{0}^{a\csc\theta}\frac{r}{(1+r^2)^2}\,dr\,d\theta=?$ I found an answer, $\displaystyle\int_{\arctan a}^{\frac{\pi}{2}}\int_{0}^{a\csc\theta}\frac{r}{(1+r^2)^2}\,dr\,d\theta=\frac{a}{2\sqrt{a^2+1}}\arctan\left(\frac{1}{\sqrt{a^2+1}}\right)$ for $a>0$.
But I a... | So, your second integral is
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2 \csc^2 \theta}{1 + a^2\csc^2\theta}d\theta$$
Multiplying the numerator and denominator by $\sin^2\theta$, you get
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2}{\sin^2\theta + a^2}d\theta$$
The next step is very generic for so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the area of the part of the sphere $x^2+y^2+z^2=2$ that lies inside the cylinder $y^2+z^2=1$ This is what I have tried so far. I am just not sure whether I got it right or not.
We have $y^2+z^2=1$, then $r=1,\quad x=\sqrt{2-y^2-z^2}\quad\text{and}\quad x=\sqrt{2-r^2}$
$$\frac{\partial{x}}{\partial{y}}=-\frac{y}{\s... | My calculations are easy, but I use the
arc length of parametric curves. The question is to
find the area of the sphere $x^2+y^2+z^2=2$ inside the cylinder $x^2+y^2=1.$ (The changed variables give the same area.)
The total area of two caps equals $8A,$ where $A$ is one-quarter of the area
of the top cap. I use cylindr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equatio... | I encourage you to keep working on your solution using derivatives. However, solving this equation does not need derivatives. Here I write a hint and if you need more help I'll give you another one!
Hint1:
Rearrange the equation: $(x+2)^8 - 1 = 1 - x^8$
Edit: Ah! Now that you posted your work I see you don't need he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
Here is a procedure how to do it analiticaly.
*
*If $T(x_0,y_0)$ is ... | We have the coordinates of any point on the ellipse centered at the origin $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in parametric form as $P(a\cos\theta,b\sin\theta)$, where $\theta$ is a parameter. Now, by using basic calculus, we get the equation of tangent to the ellipse at this point as $\frac{x}{a^2}a\cos\theta+\frac{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that $3 \mid n^4-n^2 \forall n \in \mathbb{Z^+}, n \ge 2$. Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$
My attempt
Lemma: $3 \mid (m-1)m(m+1)$
Proof.
Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.
We want... | Note that $n^4 - n^2 = (n^2 - n) (n^2 + n) = (n - 1) n^2 (n + 1)$, one of the three factors is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$5^x - y^2 = 4$ Diophantine Equation I wrote a Diophantine equation and tried solving it. Then I got stuck at a stage of the solution.
Problem: Find all $(x,y)$ positive integer pairs that satisfy the equation $5^x - y^2=4$.
My Partial Solution: If $x$ is an even number then $x=2n\quad$ ($n \geq 1$ an integer). Therefo... | $$5^{x} - y^2 = 4\tag{1}$$
We take the three cases $x=3a, x=3a+1$, and $x=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ x=3a$
Let $X=5^{a}$, then we get
$y^2 =X^3 - 4.$
According to LMFDB, this elliptic curve has integral solutions $(X,y)=(2,\pm 2), (5,\pm 11).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Minimum percentage of a random variable within two bounds
Random variable $Z$ has a mean of $15$ and a standard deviation of $2$. What is the minimum percentage of $Z$ values that lie between $8$ and $17$?
I have tried the following:
Here on the right side the value is $17$ that is $1$ sd, on the left side it is $(8-... |
$\def\F{\mathscr{F}}\def\R{\mathbb{R}}\def\d{\mathrm{d}}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$Proposition: For any $σ > 0$, define$$
\F_σ = \brace{ F: \R → [0, 1] \,\middle|\, F\ \text{is c.d.f.}, \int_\R x \,\d F(x) = 0, \int_\R x^2 \,\d F(x) = σ^2 }.
$$
If $a, b > 0$, then$$
\min_{F \in \F_σ} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Triple integral between two spheres Evaluate
$$
\iiint_{E}(2.1 z) d V
$$
where $E$ is bounded between two spheres:
$$
x^{2}+y^{2}+z^{2}=8.5^{2} \text { and } x^{2}+y^{2}+(z-8.5)^{2}=8.5^{2} .
$$
Region between two spheres
I am supposed to convert the integral to spherical coordinate $(\rho, \phi, \theta)$ where $\phi$ ... | Your working is correct for cylindrical coordinates. If you were to do it in spherical coordinates,
$x = \rho \cos\phi \sin \theta, y = \rho \sin\phi \sin\theta, z = \rho \cos\theta$
$x^2 + y^2 + (z-8.5)^2 = 8.5^2 \implies x^2 + y^2 + z^2 = 17 z$
So $ \ \rho = 17 \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$
The sphere... | {
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Confused about order of operations to simplify $\frac15\div\frac15\div\frac15\div\frac15\div\frac15\div5\div5\div5$ (and others) Here is the question that confused me:
$$\text{What is the value of}\;\frac15\div\frac15\div\frac15\div\frac15\div\frac15\div5\div5\div5\;? \tag1$$
If the signs stay the same, is the operatio... | As far as I know, when dealing with consecutive multiplication and division, one can simply go from left to right. This means that your example problem is solved as follows:
$$
\begin{array}
\div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5}\div 5 \div 5 \div 5 &= 1 \div \frac{1}{5} \... | {
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How to evaluate $\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$ As seen in the title I'm interested in a way to evaluate
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$
But I'm not sure what to do, I did attempt some... | $$\frac 1{n^2\left(k^2-2n^2\right)}=\frac 1{n^2\left(k-n \sqrt 2\right)\left(k+n \sqrt 2\right)}$$
Using partial fracion decomposition
$$\frac 1{n^2\left(k^2-2n^2\right)}=\frac{1}{2 \sqrt{2} n^3}\left(\frac{1}{k-\sqrt{2} n}-\frac{1}{k+\sqrt{2} n}\right)$$
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k-\sqrt{2} n}=\frac{1}{2}
... | {
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What does the tilde ("$\sim$") mean in $\tan^{-1}\frac{M_1\sim M_2}{M_1+M_2}$?
$$ \theta_{} \leq \pi/2 $$
$~ H_{x} ~$ can be positive or negative but $~ H_{y} ~$ can be assumed which takes only upward vertical value.
$$ \tan \left( \theta_{} \right) = \frac{ H_{x} }{ H_{y} } = \frac{ \left( M_{1}-M_{2} \right... | $$ \tan \left( -\theta_{} \right) =-\tan \left( \theta_{} \right) $$
$$ \tan \left( \theta_{} \right) =\frac{ M_{1}-M_{2} }{ M_{1}+ M_{ 2 } } $$
$$ \therefore ~~ \theta_{} = \tan^{-1} \left( \frac{ M_{ 1 } -M_{ 2 } }{ M_{ 1 } +M_{ 2 } } \right) $$
$$
\theta_{} =
\begin{cases}
\tan^{-1} \... | {
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Fibonacci generating function - quick way to see formula I know that
$$
\frac{1}{1-z-z^2} = \sum_{k = 1}^\infty a_nz^n
$$
in $B_{\varepsilon}(0) \subseteq \mathbb{C}$ where $(a_n)_{n \in \mathbb{N}}$ denotes the Fibonacci series and $\varepsilon > 0$ is small. This can be obtained by simply comparing coefficients.
Then... | Let $\sum_{n=0}^\infty a_n z^n$ be the Macluarin series for $\frac{1}{1 - z - z^2}$. We can easily prove that $a_n$ is the Fibonacci sequence. Note that, for sufficiently small $z$,
\begin{align*}
1 &= (1 - z - z^2)\sum_{n=0}^\infty a_n z^n \\
&= \sum_{n=0}^\infty a_n z^n - \sum_{n=0}^\infty a_n z^{n+1} - \sum_{n=0}^\i... | {
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Find the $\frac{k}{8}$ when the rest of the lengths are given 6 points $A,B,C,D,E,F$ are on a circle in this order and line segments $AD,BE,CF$ intersect at one point. If $AB=1,BC=2,CD=3,DE=4,EF=5,FA=\frac{k}{8}$ find the value of $k$.
I tried to sum the total lengths of the arc:
$1+2+3+4+5+\frac{k}{8}=2\pi r$
$15+\fra... | I think they are chord lengths and not arc lengths.
By Intersecting Chords theorem or by Inscribed Angle theorem, $\triangle GAB \sim \triangle GED$
So, $GD = 4y, GE = 4x$
Similarly, $\triangle GBC \sim \triangle GFE$
So, $GE = \frac{5z}{2}, GF = \frac{5y}{2}$
$\triangle GCD \sim \triangle GAF$
So, $GF = \frac{5y}{2} ... | {
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Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$? I was trying to solving this question:
If roots of the equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, find the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$
I was not able to solve it so I... | Notice that $y$ cannot be equal to $-1$ in the first place, if $y = \dfrac{1-\alpha}{1+\alpha}$
Range 0f $f(\alpha) = \dfrac{1-\alpha}{1+\alpha}$ is $\mathbb{R} \setminus \{-1\} ~\forall ~\alpha \in \mathbb{R}$
This is because if $\dfrac{1-\alpha}{1+\alpha} = -1$ then $$ 1-\alpha = -1-\alpha$$ $$ 1=-1$$
Which is clearl... | {
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"source": "stackexchange",
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Find x in the figure (Answer: 40 degrees)
I found the following mathematical relationships but some are still missing.
$\triangle ABC: 80^o+4\theta+2\alpha = 180^o \rightarrow \boxed{2\theta+\alpha=50^o}(i)\\
\triangle CAH:40^o+4\theta+m = 180^o\rightarrow \boxed{m=140^o - 4\theta}(II)\\
\triangle CBG: n+4\theta+\alpha... | $\angle BFE = 90^0 - 2 n = \angle BAF + \angle ABF = 40^0 + \alpha $
So, $\alpha + 2n = 50^0$ ...$(i)$
Now $\angle AFC = 140^0 - 2\theta$, $\angle BFC = 180^0 - \alpha - 2\theta \ $ and
$\angle AFB = 140^0 - \alpha$
But they should add to $360^0$,
So we get $100^0 = 4 \theta + 2\alpha \ $ or $\alpha + 2 \theta = 50^0$... | {
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"question_score": "1",
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Curvature and Torsion for a Space Curve Calculate the curvature and torsion of
x= $\theta - \sin \theta, y = 1 - \cos \theta, z = 4 \sin (\theta/2)$
$\vec r = (θ - \sin θ)i+(1-\cos θ)j + 4 \sin (θ/2) \ k$
$\vec dr/dt = (1- \cos θ)i + \sin θ \ j+2 \cos (θ/2) \ k$
$d^2\vec r/dt^2 = \sin θ \ i + \cos θ \ j - sin (θ/2) \ k... | $\vec r (\theta) = (\theta - \sin \theta, 1 - \cos \theta, 4 \sin \frac{\theta}{2})$
There is another formula for curvature that is easier here.
$r'(\theta) = (1 - \cos \theta, \sin\theta, 2 \cos \frac{\theta}{2})$
$||r'(\theta)|| = \sqrt{(1-\cos\theta)^2 + \sin^2\theta + 4 \cos^2 \frac{\theta}{2}}$
$ = \sqrt{(1 + \cos... | {
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How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$? How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$ ?
Here is my work:
By expanding the expression we get, $$(x^2-x-2)(x^2+15x+54)+108=x^4+14x^3+37x^2-84x$$
So I got $x(x^3+14x^2+37x-84)=... | The midpoints of $-2,9$ and $1,6$ are the same: $3.5$. Thus let $x = t - 3.5$:
$$(t-5.5)(t+5.5)(t - 2.5)(t + 2.5) + 108 = 0$$
$$(t^2 - 30.25)(t^2 - 6.25) + 108 = 0$$
$$t^4 - 36.5t^2 + \text{constant term} \ (297.0625) = 0$$
Since $t \to -t$ results in the same polynomial, it is symmetric across the line $x = -3.5$. Al... | {
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With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y... | If $x^2+y^2=1$, then $|x|$ and $|y|$ are each no larger than $1$. Consequently, $|x|^5\le x^2$ and $|y|^5\le y^2$. It follows that
$$|x^5+y^5|\le|x|^5+|y|^5\le x^2+y^2=1$$
| {
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If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $\vert a-b\vert$ (where $a$ & $b$ are relatively prime) The following question is taken from the practice set of JEE exam.
If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $|a-b|$ (where $a$ & $b$ are ... | a very common trick among JEE summation questions is: pair up the first and last terms
$$\sum_{n=1}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+\sum_{n=1, n\not=200}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+ \sum_{n=1}^{199} \frac{5^{200}}{5^{200}+5^n}+\frac{5^{200}}{5^{200}+5^{400-n}}$$ and now an amazing thing... | {
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"source": "stackexchange",
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how to find greatest and least values of $|z-2-3i|$ if $|z-5-7i|=9$ Context: conceptual question in jee
I got the answer by using properties of modulus but i don't know if the procedure is correct
If $|z-5-7i|=9$ then find the greatest and least values of $|z-2-3i|$
Answer: $14$, $4$
here's what I did:
given,
$$||... | You could also treat this as an extremization problem in which you want the minimum and maximum values of $ \ (x - 2)^2 \ + \ (y - 3)^2 \ $ subject to the constraint $ \ (x - 5)^2 \ + \ (y - 7)^2 \ = \ 9^2 \ \ . $ The method of Lagrange multipliers (for example) gives us the equations
$$ 2·( x \ - \ 2) \ - \ \lambda... | {
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I don't know how to exactly compute this determinant I've tried to compute this determinant by row transformations and column transformations, but it gives me a formula that doesn't work. The determinant is:
\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & ... | Notice that the sum of every row equals to $(a+b+c+d+x).$ You can take this approach:
$$\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix} \xrightarrow{\text{$C_1=C_1+C_2+C_3+C_4+C_5$}} $$
$$
\begin{vmatrix}
(a+b+c+d+x) & a & b & c & d\\
(a+... | {
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of i... | Try using Cauchy-Schwarz on Attempt 2. You wanted only AM-GM, so you can use this post: Prove Cauchy-Schwarz with AM-GM for three variables
| {
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Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}}dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^4}{\... | There is a typo, though it was not critical:
$$sin2\theta=2sin\theta cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}x^2-\color{red}a^2}}{a} \text{ must be } \\
\sin2\theta=2\sin\theta \cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}a^2-\color{red}x^2}}{a}$$
You made some minor errors while substituting:
$$\frac{a^4}{8}... | {
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Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^... | Let $x=a\cos\theta$ then ${dx/d\theta} = -a\sin\theta$ and $\sqrt{a^2-x^2}=a(1-(x/a)^2)^{1/2}=a\sin\theta$, so the integral
$$
\int {x^4\over\sqrt{a^2-x^2}} dx = -a^4\int\cos^4\theta \,d\theta.
$$
Repeated substitution of the formula $\cos\phi = 1/2 (\cos 2\phi+1)$ gives
$$
\int\cos^4\theta\,d\theta = {1\over 8}\int(\c... | {
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"source": "stackexchange",
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Is multiplying and dividing inequalities valid? I am facing certain problems in solving inequalities. I do not know whether multiplying and dividing inequalities works or not but it has helped me solve many problems. Here are some problems which cannot be solved using multiplying and dividing. Why does the rule not app... | By AM-GM $$(a+1)^7(b+1)^7(c+1)^7=\prod_{cyc}\left(4\cdot\frac{a}{4}+3\cdot\frac{1}{3}\right)^7\geq$$
$$\geq\left(7^7\right)^3\prod_{cyc}\left(\left(\frac{a}{4}\right)^4\left(\frac{1}{3}\right)^3\right)=\frac{7^{21}}{4^{12}3^9}a^4b^4c^4>7^7a^4b^4c^4.$$
The last inequality we can prove by the following way.
$$\frac{7^{21... | {
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if $x, y, z$ are $3 \leq x \leq y \leq z$ and they are odd numbers, show that $1 < (xyz - 1) / (x-1)(y-1)(z-1) < 3$ There are three question. Maybe there are related.
(1) if x, y, z are integers greater than 2, showing that
$$ \frac{xyz - 1}{(x-1)yz} < \frac{xyz}{(x-1)(y-1)(z-1)}$$
(2) if $x, y, z$ are $3 \le x \le y \... | Presumably part 3 wants to use part 2 to conclude that $(C+2)/(A+B+C+1)=2$.
If so, I believe that part 2's condition should be $ 3 \leq x < y < z$, reflecting the "three different integers in their roots". As such, I will make this change to the question, and proceed accordingly.
Hints/guide. If you're stuck, show wha... | {
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Is this proof regarding the nonexistence of odd perfect numbers correct? Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow.
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)... | In your mathoverflow post, you write
$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \gcd(n^2, \sigma(n^2))$$
because $\gcd(\sigma(q^k), q^k) = 1.$ Why is this true? I do not see any reason it should hold.
| {
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Help with $\int_{0}^{\infty} \frac{\cos^k (x)}{a^2 + x^2} \, dx$ I'm having trouble evaluating the following integral:
$$\int_{0}^{\infty} \frac{\cos^k(x)}{a^2+x^2} \, dx \,\text{ where } k \in \mathbb{N} \text{ and } a >0$$
I've tried to follow this question which approaches tan instead.
$$I = \frac{1}{2} \int_{-\inft... | At least, for odd values of the exponent, the expansion given by @aaaaaaaaabbbbbbbbbcccccc in comments is very useful since
$$\int_0^\infty \frac {\cos \big[ (2 n+1-2k)x\big] }{x^2+a^2}\,dx=\frac \pi{2a}\,\exp\big[-(2n+1-2k)a\big]$$ Summing from $k=0$ to $k=n$ then gives for
$$I_n=\int_{0}^{\infty} \frac{\cos^{2n+1} (x... | {
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Find minimum value of $\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$
Let $x,y,z>$ and $x+y+z=xy+yz+zx$
. Find the minimum value of $$P=\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$$
My solution: I know the minimum value is $\frac{3}{2}$ whe... | We have
\begin{align*}
P &= \frac{x^3}{\sqrt{2(y^4+1)}} + \frac{y^3}{\sqrt{2(z^4+1)}} + \frac{z^3}{\sqrt{2(x^4+1)}}\\
&\ge \frac{x^3}{2(y^2 - y + 1)} + \frac{y^3}{2(z^2 - z + 1)} + \frac{z^3}{2(x^2 - x + 1)} \tag{1}\\
&\ge \frac{(x^2 + y^2 + z^2)^2}{2x(y^2 - y + 1) + 2y(z^2 - z + 1) + 2z(x^2 - x + 1)} \tag{2}\\
&= ... | {
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"source": "stackexchange",
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How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so tha... | Let $f(x) = a_kx^k + a_{k-1}x^{k-1} + ..... + a_0$.
Then $f(f(x) = a_k(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^k + a_{k-1}(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^{k-1} + ..... + a_0$
If we expand that out we will get a very large polynomial of degree $k^2$.
But $f(f(x)) = 4x + 1$ which is of degree $1$.
So $k^2 = 1$. A... | {
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A difficult Vietnamese university entrance exam problem minimize/maximize $P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}}+\frac{z}{{z + x}}$? A difficult problem from the Vietnamese Entrance exam of 2011 for Natural Science and Technology group.
Note that the original problem only involve minimization.
How to minimize an... | The maximization problem:
Since $P$ is homogeneous, WLOG, assume $x = 4$.
If $y = 4$, then $P = \frac{6}{5}$ for any $z$.
Indeed, $\frac65$ is the maximum of $P$, since
\begin{align*}
P &= \frac{4}{8 + 3y} + \frac{y}{y + z} + \frac{z}{z + 4}\\
&= \frac{6}{5} - \frac{(4 - y)[3(z - 1)(4y - z) + 25z]}{5(8 + 3y)(y + z)(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
calculate cubic equation discriminant Let $p(x)=x^3+px+q$ a real polynomial.
Let $a,b,c$ the complex root of $p(x)$.
What is the easiest way to calculate $\Delta=(a-b)^2(b-c)^2(a-c)^2$ in function of $p,q$?
The result is $\Delta=-4p^3-27q^2$
Ps. $f=(x-a)(x-b)(x-c)$ and so $a+b+c=0$; $ab+bc+ac=p$ and $abc=-q$; how to ... | Note that
$$f(x)=(x-a)(x-b)(x-c)$$
$$f'(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$$
$$\boxed{f'(a)=(a-b)(a-c)}$$
and its cyclic variants.
We have that
$$\prod_\text{cyc}f'(a)=\prod_\text{cyc} (a-b)(a-c)$$
$$\prod_\text{cyc} f'(a)=-(a-b)^2(b-c)^2(a-c)^2$$
Hence, our expression is equivalent to $-f'(a)f'(b)f'(x)$
Since $f'(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$, then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\s... | $\alpha+\beta+\gamma=20$
$\implies (3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5) = 75$
By AM-QM inequality,
$\cfrac {\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \gamma + 5}}{3} \leq \sqrt{\cfrac{(3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5)}{3}}$
leads to $\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate $ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx $ (Using line integral) (complex analysis) I want to calculate the integral $$ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx $$ using a line integral $\varGamma $ which is the boundary of an arc of a circle of radius $ R $ and $ 0\leq Arg(z) \leq 2\pi/3 $
Such a ... | I had a few calculations mistakes: the correct calculation would be the following:
The residue at the poles I mentioned given by
$$ \text{Res}\left(\frac{1}{z^{6}+z^{3}+1};e^{i\frac{2\pi}{9}}\right)=\frac{1}{\left(z^{6}+z^{3}+1\right)'}|_{z=e^{i\frac{2\pi}{9}}}=\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{4\pi}{9}}} $$
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solution Verification for Problem Finding constant $a$ and $b$ such that $\lim\limits_{x\to\frac{\pi}{2}}\dfrac{a(2x-\pi)\cos x+b}{\sin x-1}=1.$
Find $a$ and $b$ such that
$$\lim\limits_{x\to\frac{\pi}{2}}\dfrac{a(2x-\pi)\cos x+b}{\sin
x-1}=1.$$
I try answer as follows.
The denumerator is zero when $x=\dfrac{\pi}{2}... | Without l'Hospital, to verify we can also proceed by standard limits, by $y=\frac \pi 2-x \to 0$
$$\dfrac{a(2x-\pi)\cos x+b}{\sin x-1}=\dfrac{-2ay\sin y+b}{\cos y-1}=\frac{y^2}{1-\cos y}\dfrac{2ay\sin y-b}{y^2}$$
from which we obtain the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a technique in how to write certain expression in certain times of integration as in case of $\int \frac {(3x + 2)} {(5x + 1)^2}dx$
I can solve this integration $$\frac{3} {5}\int\frac{1}{5x + 1}dx + \frac{7} {5}\int\frac{1}{(5x + 1)^2}dx$$ but thing is I dont under how that online calculator wrote $$3x + 2$$... | One way to get at such things is to use a substitution. Let $u= 5x +1$ in your example, then $x=(u-1)/5$ and your integrand goes like this:
$$\frac{3x+2}{(5x+1)^2} = \frac{ 3\left(\frac{u-1}{5}\right)+2}{u^2} = \frac{ \frac{3}{5}u - \frac{3}{5}+2}{u^2} = \frac{ \frac{3}{5}u+\frac{7}{5}}{u^2}. $$
Now put $5x+1$ back in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is integral of $(x+y)^2 \,dx$ different than integral of $(x^2+ 2xy + y^2) \,dx$? I calculated the integral of $(x + y)^2 \,dx$ using the substitution method and got $\frac{1}{3}(x+y)^3$ as result.
Then I applied the distributive property to $(x + y)^2$, calculated the integral of $(x^2+2xy+y^2) \,dx$ and I got $\f... | The meaning of the word "constant" gets too little attention in our pedagogy. "Constant" means not changing as something else changes, but what is the thing in the role of "something else"?
\begin{align}
& \int (x^2 + 2xy+y^2)\, dx \\[10pt]
= {} & \frac{x^3}3 + x^2 y + xy^2 + \text{constant} \\[10pt]
= {} & \frac 1 3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$ I was trying to simplify
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$$
I change all $\tan, \cot, \sec, \operatorname{cosec}$ into $\cos$ and $\sin$ using conversion formulas (I find it easier). Here, I get this:
$$\frac{\sin A + 1 - \cos A}{\sin A - 1 ... | Another way.
$$LS=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}+2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}-2\sin^2\frac{A}{2}}=\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}-\sin\frac{A}{2}}=$$
$$=\frac{\left(\cos\frac{A}{2}+\sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2}-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If $a$ is a root of $x^2(x-5)+2$, then what is $[a^4]$? If $a$ is the largest root of $f(x)=x^2(x-5)+2$, then what is $[a^4]$ where $[\ ]$ is a Gauss bracket ? (i.e. $[x]$ is a largest integer not strictly greater than $x$)
Here $f(5)=2,\ f(4)=-14$. By considering $f'(x)$ we can know that $a$ is in an open interval $(4... | Combining two ideas from the comments.
Let $a,b,c$ be the zeros with $a$ the largest. By expanding we see that
$$f(x)f(-x)=4-20x^2+25x^4-x^6=-g(x^2),$$
where
$$g(x)=x^3-25x^2+20x-4$$
has $a^2,b^2$ and $c^2$ as its zeros. By the Vieta relations it follows that
$$
s_1:=a^2+b^2+c^2=25
$$
and the second symmetric polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that uncorrelatedness implies independence for bivariate joint normal random variables Question
Let $X$ and $Y$ be independent standard normal random variables and consider the following linear transformations, $$U = aX + bY$$ and $$V = cX + dY,$$ where $a, b, c, d \in \mathbb{R}$. Find the joint density of $U$... | I did not check your calculations but, assuming that
*
*$U,V$ are jointly Gaussian,
*EDIT: $a,b,c,d \ne 0$
Your covariance calculation is right, thus $\rho^2=\frac{(ac+bd)^2}{(a^2+b^2)(c^2+d^2)}$ and the joint density is the following
$$f_{UV}(u,v)=\frac{1}{2\pi\sigma_u\sigma_v\sqrt{1-\rho^2}}\text{exp}\left\{-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric elimination between two variables Eliminate $\theta$ and $\phi$ between the following equations: $$\begin{cases}\sin \theta + \sin \phi = x \\ \cos \theta + \cos \phi = y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} = z\end{cases}$$
What I've done so far
I've established that $$\tan \left(\frac {\theta+... | By sum to product and product to sum formulas we have
$$\begin{cases}
\sin \theta + \sin \phi = 2\sin\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=x\\
\cos \theta + \cos \phi = 2\cos\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=y \\
\tan \frac {\theta}{2} \tan \frac {\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ &... | WLOG, it can be assumed that $z_1$ is on Imaginary axis(or y-axis).
Since $\left|\frac{z_1+z_2+z_3}{3}\right|=\frac{2}{3}$
It means that distance of the Centroid of triangle formed by joining $z_1,z_2,z_3$ from the centre of circle is $\frac{2}{3}$ and consequently distance of centroid from $z_1$ is $\frac{4}{3}$.
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Completeness of continuous functions space under $L^{1}$ norm Prove that the space of continuous functions $C([0,1])$ is not complete under the norm
$\|\cdot\|_{L^{1}([0,1])}$.
I have tried the following sequence of functions
$f_{n}(x)=\left\{\begin{array}{cl}1 & \text { if } & 0 \leq x \leqslant \frac{1}{2} \\ 1-n\lef... | Assuming $m<n$, we first show $(f_n)$ is cauchy's sequence.
$\|f_n-f_m\|= \int_0^1 |f_n-f_m|dx$
RHS is equal to
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}|m(x-\frac{1}{2})-n(x-\frac{1}{2})|dx+$
$\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}|1-m(x-\frac{1}{2})|$
Try to simplify, if needed you can use $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Nicolaus Mercator's extension of logarithmic power series from $(-1,1) \to \mathbb{R}$ I was reading this Quora answer in which the expansion of $\log(1+x)$ was extended in such a way that one could approximate any real number's natural logarithm quickly. The following was written in the answer as an identity:
$$ \ln(Z... | Hint:
For $|x|\lt 1$,
$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+O(x^5)$
and $\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+O(x^5)$
Subtract these to get: $\ln(1+x)-\ln(1-x)=2(x+\frac{x^3}3+O(x^5))$
Put $x=Q$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the closed form of the following series $$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
The answer given is $$\frac{1}{3^r}$$ I tried expanding the expression so it becomes $$\sum_{m=r}^{\infty}\frac{(m-1)!}{(r-1)!(m-r)!}\frac{1}{4^m}$$but I do not know how to follow.
Any help will be appreciated, thanks... | Consider the series
$$S := \displaystyle \sum_{m=r}^{\infty}\binom{m-1}{r-1}x^m $$
We have the following factorial relation: $$\displaystyle \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$$
So that $\displaystyle \binom{m-1}{r-1} = \frac{r}{m}\binom{m}{r}. $ Hence we have
$$S := \sum_{m \ge r}\binom{m-1}{r-1}x^m = \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\sum_{k=1}^{n-1} k\cdot3^k$$
I have a calculus-based solution: As a more general result, I replace $3$ with $x$ and denote the sum by $f(x)$. Then
$$f(x) = \sum_{k=1... | Let $$S=\sum_{k=1}^{n-1} k\cdot3^k$$
Then $$3S=\sum_{k=1}^{n-1} k\cdot3^{k+1}$$
Subtracting , we get $$2S=\sum_{k=1}^{n-1} k\cdot3^{k+1}-\sum_{k=1}^{n-1} k\cdot3^k$$
$$2S=(1.3^2+2.3^3+3.3^4+\cdots+(n-1)3^n)-(3+2\cdot3^2 + 3\cdot3^3+\cdots+(n-1)3^{n-1})$$
$$-2S=(3+2\cdot3^2+3\cdot3^3+\cdots+(n-1)3^{n-1})-(1\cdot3^2 + 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int\frac{3\cot3x-\cot x}{\tan x-3\tan3x}\,dx$ To evaluate:
$$I= \int_{ }^{ }\frac{3\cot3x-\cot x}{\tan x-3\tan3x}dx$$
My approach:
Convert $\cot x$ and $\cot 3x$ terms into $\tan x$ and $\tan3x$ respectively and use $$\displaystyle \tan3x=\frac{\left(3\tan x-\tan^{3}x\right)}{1-3\tan^{2}x}$$
Further simpl... | Lets take $\int\tan(x)\cot(3x)\, dx$.
We write $\tan(x) \cot(3 x)$ as $\frac{\sin(4 x)-\sin(2 x)}{\sin(2 x)+\sin(4 x)}$ which leads to $\int\frac{\sin(4 x)-\sin(2 x)}{\sin(2 x)+\sin(4 x)}\, dx$
Expanding the integrand leads to:
$$
\int\frac{\sin(4 x)}{\sin(2 x)+\sin(4 x)}-\frac{\sin(2 x)}{\sin(2x)+\sin(4 x)}\, dx
=
\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^\infty \frac{e^{-x}}{x}\ln\big(\frac{1}{x}\big) \sin(x)dx$ I'm having trouble with this integral
Evaluate $$\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx$$
$$I=\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx=\Im\left[\int_0^\infty \frac{e^{-x+ix}}{x}\ln\left(... | We'll use the following two crucial facts to calculate the integral. The first is that:
$$\boxed{\tan^{-1}x\ln(1+x^2)=-2\sum_{ n \ge 1}\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}}$$
which is shown in here. The second is that
$$\displaystyle \boxed{\Gamma'(2k-1) = (2k-2)! \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n} \bigg)} $$
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Compressibility factor (Z) of the Redlich-Kwong Equation I've been having a problem trying to calculate the compressibility factor of the Redlich-Kwong equation:
\begin{equation}
P = \frac{RT}{v-b}-\frac{a}{\sqrt{T} \cdot (v^2+vb)}
\end{equation}
The compressibility factor is calculated as:
\begin{equation}
z = \frac{P... | \begin{align*}
P &= \frac{RT}{v-b} - \frac{a}{\sqrt{T}(v^2+vb)}\\
&=\frac{RT\sqrt{T}(v^2+vb)-a(v-b)}{\sqrt{T}(v^2+vb)(v-b)}\\
\implies P\sqrt{T}(v^3 - b^2 v)
&-(RT\sqrt{T}(v^2+bv)-av+ab)=0\\
\implies P \sqrt{T} v^3 - b^2 P \sqrt{T} v
&-a b + a v - b R \sqrt{T^3} v - R \sqrt{T^3} v^2=0 \\
\implies P \sqrt{T} v^3- R \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
... | $x, y$ and $z$ are roots of the polynomial $x^3-x+1=0$
In the usual notation ,we have $$\Sigma x^2=(\Sigma x)^2-2\Sigma xy=0-2(-1)=2$$
Then since the relation $x^3=x-1$ holds for all three roots, summing these gives:
$$\Sigma x^3=\Sigma x - \Sigma 1=0-3=-3$$
Likewise,
$$\Sigma x^4=\Sigma x^2-\Sigma x=2$$
$$\Sigma x^5=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 3
} |
Integrate $\int^{\pi}_{0}\left\{\frac{\tan^2\left(\frac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$ $\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$
$\sf{\color{blue}{My\,\,approach\,}:}$
$=\displaystyle\int^{\pi}_{0... | As @person commented, using $x=2\tan^{-1}(t)$,ws have to compute
$$I=2^{7/9}\int_0^\infty\frac{dt}{\left(1-t^2\right)^{2/9} \left(1+t^2\right)^{5/9}}$$
The antiderivative is not so bad
$$\int_0^ u\frac{dt}{\left(1-t^2\right)^{2/9} \left(1+t^2\right)^{5/9}}=u\, F_1\left(\frac{1}{2};\frac{2}{9},\frac{5}{9};\frac{3}{2};u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
| Maybe this identity for ${}_{2}F_{1}$ helps:
$\displaystyle I=\int_{0}^\frac{\pi}{2} \frac{\sqrt{\cos\theta}}{1+\cos^2 \theta}d\theta = \sum_{n=0}^{\infty} (-1)^n\int_{0}^{\frac{\pi}{2}} \cos^{2n+\frac{1}{2}}\theta d\theta = \sum_{n=0}^{\infty} (-1)^n\frac{\sqrt{\pi}\Gamma\left(n+\frac{3}{4}\right)}{2\Gamma\left(n+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding a closed formula for a simple integer sequence sum I'm trying to compute the average path lengths of a path graph.
I made the following observations:
There are $n$ nodes and $n-1$ edges and:
$n-1$ paths of length $1$
$n-2$ paths of length $3$
$n-3$ paths of length $4$
...
$1$ path of length $n-1$
I'd like to co... | The answer that's already been given is no doubt the "right" one, but maybe you still find the generating function approach interesting. It's a quite simple idea that you can apply to these kinds of problems very often and that may yield results even if you can't directly find a solution via known formulas as you're ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \rig... | 1)Integral evaluation applying Differentiation Under the Integral Sign:
First, let's consider the following function:
$$f( \alpha ) =\int _{0}^{\infty }\frac{1-x^{2}}{\left( 1+x^{2}\right)^{2}}\frac{\tanh( \alpha x)}{x} dx$$
Then similarly to OP's series expansion for $\operatorname{sech}^2\left(\frac{\pi x}{2}\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Explicit estimate of sum of primes Let $p$ be prime. In this post, Charles gave the following answer
$$\sum_{p\leq x}p=\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{3x^2}{8\log^4 x}+O\left(\frac{x^2}{\log^5 x}\right). $$
I have two problems which I don't understand well. The first is how can we ... | I quote from Christian Axler, New bounds for the sum of the first $n$ prime numbers, https://arxiv.org/abs/1606.06874
Let $\pi(x)$ denote the number of primes not exceeding $x$. de la
Vallée-Poussin estimated the error term in the Prime Number Theorem by showing that $$\pi(x)={\rm li}(x)+O(xe^{-a\sqrt{\log x}})\tag1$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$ I'm trying to evaluate the following expression$$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$
I'm not really used to these types of problems, so I first tried using logarithms but I'm no... | Hint: Multiply by $(3^1-2^1)=1$
Explanation-
$$(3^1-2^1)(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\ =(3^2-2^2)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\=(3^4-2^4)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})$$
Complete the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3... | WLOG assume $a\leq 1.$ Then, let $f(a,b,c) = a^8+b^8+c^8 +2(a-1)(b-1)(c-1)$ and $\dfrac{b+c}{2} = t$ consider:
$$f(a,b,c) - f\left(a,t,t\right) = \left(b^8+c^8 - 2\left(\dfrac{b+c}{2}\right)^8\right)+\dfrac{1}{2}(1-a)(b-c)^2\geq 0.$$
This reduces your inequality into single variable case:
$$g(t) = (3-2t)^8+2t^8-4(t-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is so... | Let $k\in\mathbb{R}$. It is enough to find the $(x,y)$ in the domain of $f$ (all $\mathbb{R}^2$) for which the equality is true
$$\frac{x-y}{1+x^2+y^2}=k$$
It is clear that the equality is true when $k=0$ (in this case $\frac{x-y}{1+x^2+y^2}=0\Longleftrightarrow x-y=0\Longleftrightarrow x=y$). Suppose that $ k\neq 0$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"solved" Confusion calculating $\mathop {\lim }\limits_{x\to 0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}$ I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason,... | It should be noted that
$\color{blue}{\left(1+x\right)^{\frac{1}{2}}=1+\dfrac{x}{2}+\cdots}$
is true for $\color{orange}{|x|\le1}$
In your assumption, $\mathtt{\big(1+cos(2x)-1\big)^{\frac{1}{2}}=1+\dfrac{cos(2x)-1}{2}},$
$\mathtt{\color{violet}{|cos(2x)-1|\le2}}$
So, I think it should not be used.
Instead, use binomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many sequences are there? I saw this problems in an combinatorics exam:
\begin{array}{l}For\;all\;n\geq0,\;find\;the\;number\;of\;integer\:sequences\;a_0,...,a_{2n}\;,\;a_0=a_{2n}=1\;\\and\;for\;all\;1\leq i\leq2n\;we\;have\;\frac{a_i}{a_{i-1}}\in\{\frac16,\frac23,\frac32,6\}\end{array}
It reminds me similar proble... | The hint of @TeresaLisbon is nice. We generalise the set
\begin{align*}
\left\{\frac{1}{2\cdot3},\frac{2}{3},\frac{3}{2},2\cdot 3\right\}
\end{align*}
and denoting with $[z^n]$ the coefficient of $z^n$ of a series we consider
\begin{align*}
\color{blue}{[x^0y^0]}&\color{blue}{\left(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\lim_{n→∞} \left( \frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1} \right)$ where $k∈\mathbb{N}$ I had to find the limit of the sequence
$$ a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}$$, where $ k $ is a natural number.
After applying the Stolz theorem, I was able to get here.
$$\lim ... | We have
$$a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=a_n =\frac{(k+1)(1^k+2^k+\cdot \cdot \cdot +n^k)-n^{k+1}}{(k+1)n^k}$$
and then by Stolz-Cesaro
$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k}$$
and since by binomial expansion
*
*$(k+1)(n+1)^{k}=(k+1)n^k+k(k+1)n^{k-1}+O(n^{k-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Inequality without BW and Uvw, just AM GM Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$
I guess equality holds: two of them equal $1$ and one equal to $0$.
My approach: The equality equivalent to
$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}... | $uvw$ helps!
Let $\sqrt{ab}=z$, $\sqrt{ac}=y$ and $\sqrt{bc}=x$.
Thus, $x^2+y^2+z^2=1$ and we need to prove that:
$$\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\right)(3-x-y-z)+2(xy+xz+yz)\geq4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $w^3$, but
$$\frac{xy}{z}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.