Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Unsure if I have solved/proven this trigonometric inequality. This is my first post here. I apologize if it goes against any guidelines for posting. I study math as a hobby and am currently dealing with trigonometry on a high school level. I have so far learned the formulas for trigonometric addition and subtraction an... | Use $AM-GM$ and $\sin^2 A + \cos^2 A = 1$.
.....
I start by simply doing.
$(1 + \frac {1}{\sin A})(1 + \frac 1{\cos A}) = $
$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}$.
Now $0 < \sin A < 1$ so $ \frac 1{\sin A} > 1$ and $0 < \cos A < 1$ so $ \frac 1{\cos A} > 1$.
So $(1 + \frac 1{\sin A} + \frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$... | Let $f(x)=x^k,$ where $k>1$ or $k<0$.
Thus, $f$ is a convex function and since $(11,7,5)\succ(9,8,6),$ by Karamata we obtain:
$$f(11)+f(7)+f(5)>f(9)+f(8)+f(6).$$
Also, for $0<k<1$ we see that $f$ is a concave function.
Thus, by Karamata again
$$f(11)+f(7)+f(5)<f(9)+f(8)+f(6).$$
Thus, it remains to check, what happens f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, tha... | Solution
$$\frac{4x^2+8x+13}{6(x+1)}=\frac{4(x+1)^2+9}{6(x+1)}=\frac{2}{3}(x+1)+\frac{3}{2(x+1)}\geq 2$$
with the equality holding if and only if $\dfrac{2}{3}(x+1)=\dfrac{3}{2(x+1)}$,namely, $x=\dfrac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How can these two equations be solved by elimination? In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.
$$
\frac{3x(3x^... | If you want to implement the resultant yourself, you can use the Sylvester matrix (see wikipedia).
In M2:
R=ZZ/23[x,y]
f=3*x*(3*x^2+9)*(4*y^2)-(3*x^2+9)^3-(y+6)*8*y^3;g=(3*x^2+9)^2-(2*x+12)*(2*y)^2;
sylvesterMatrix(f,g,y)
yields
$${\tiny \begin{pmatrix}{-8}&
{-2}&
-10 x^{3}-7 x&
0&
-4 x^{6}+10 x^{4}+7 x^{2}+7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$ Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\... | Using continued fraction:
\begin{align}
f+\frac{1}{n+f} &= 1 \tag{$n>1$, $0<f<1$} \\[5pt]
n+f &= n+1-\frac{1}{n+f} \\[5pt]
x &= n+\frac{n+f-1}{n+f} \\[5pt]
&= n+\frac{1}{\dfrac{n+f}{n+f-1}} \\[5pt]
&= n+\frac{1}{1+\dfrac{1}{n+f-1}} \\[5pt]
x &= n+\frac{1}{1+\dfrac{1}{x-1}} \tag{$\star$} \\[5pt]
&= \left[ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Plot $|\frac{z+i}{z-1}|<1$ Let $D=\{|\frac{z+i}{z-1}|<1\}$ plot $D$
$$\frac{z+i}{z-1}|<1\iff \frac{|z+i|}{|z-1|}<1\iff |z+i|<|z-1|\iff |x+(y+1)i|<|(x-1)+yi|\iff\\ \iff \sqrt{x^2+(y+1)^2}<\sqrt{(x-1)^2+y^2}\iff {x^2+(y+1)^2}<{(x-1)^2+y^2}\iff \\ \iff x^2+y^2+2y+1<x^2-2x+1+y^2\iff x<-y$$
Which is the are under the line $... | $$\frac{|z+i|}{|z-1|}<1$$
$$|z-(-i)| < |z-1|$$
We are describing points that are closer to $(-i)$ than $1$.
The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$? I recently came upon the following expression in an attempt at getting a closed-form solution for a recursive relation:
$$\prod_{i=1}^n \cos(a^i\theta)$$
where $a<1$. Is there a way to make this product into a sum or otherwise make it simpler, or a... | from
$$
\cos z = {{e^{\,i\,z} + e^{\, - \,i\,z} } \over 2} = {1 \over 2}e^{\,i\,z} \left( {1 + e^{\, - \,i\,2z} } \right)
$$
we get
$$
\eqalign{
& \ln \cos z = \ln {1 \over 2} + i\,z + \ln \left( {1 + e^{\, - \,i\,2z} } \right) = \cr
& = \ln {1 \over 2} + i\,z + \ln \left( {1 + 1 - i2z - 2z^2 + O\left( {z^3 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x... | I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y \leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably... | I think the constant $16$ in the question should be $16/3$. Also, the AM-HM inequality implies
$$\frac{\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}}{4} \ge \frac{4}{3(a+b+c+d)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Relationship between roots
Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, \ldots, x^2 + (2 + k) x + 1, k \in \mathbb{N}$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length... | Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.
When $f_k(x)=0$,
$$
x=\frac{-(2+k)\pm\sqrt{(2+k)^2-4}}{2}
$$
Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-\frac{(2+k)+\sqrt{(2+k)^2-4}}{2}$$
Define $$r_k:=-\frac{(2+k)+\sqrt{(2+k)^2-4}}{2}$$
Define $$l_k:=r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problems with Recurrence Relations as a form of Counting I have been having trouble trying to understand how to do the following problem
Solve by unfolding: $a_0=3$, and for $n\geq1$, $a_n=5a_{n-1}+3$. Hint: This will involve the geometric sum formula.
This is my work so far:
$$a_n=5a_{n-1}+3$$
$$a_n=5(5a_{n-2}+3)+3... | \begin{align}
a_n
&= 5a_{n-1}+3\\
&= 5(5a_{n-2}+3)+3\\
&= 5^2a_{n-2}+3(5^1+5^0)\\
&= 5^2(5a_{n-3}+3)+3(5^1+5^0)\\
&= 5^3a_{n-3}+3(5^2+5^1+5^0)\\
&= \dots\\
&= 5^na_0+3(5^{n-1}+\dots+5^0)\\
&= 5^n(3)+3\cdot\frac{5^n-1}{5-1}\\
&= 3\cdot5^n+\frac34(5^n-1)\\
&= \frac{15}4\cdot5^n-\frac34\\
&= \frac34(5^{n+1}-1)
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} $ I was trying to evaluate the following series, which I know converges:
$$\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} \tag{1}\label{1} $$
In a telescoping fashion, I began writing out the terms in hopes to find a pattern:
$$= (2\ln{2... | Here is the answer of the series. With expansion of $\ln$ we have
\begin{align}
\sum_{k=1}^{\infty} 2\ln(2k)-\ln(2k-1)-\ln(2k+1)
&= \sum_{k=1}^{\infty} -\ln\left(1-\dfrac{1}{2k}\right)-\ln\left(1-\dfrac{1}{2k}\right) \\
&= \sum_{k=1}^{\infty} \sum_{n\geqslant1} \dfrac{1}{n}\left(\dfrac{1}{2k}\right)^{2n} \\
&= \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(... | $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
Substitute $k=s+2$
$$F(k) = \frac{1}{(k-3)(k+1)k}$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 {(k+1)k} \right)$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 k-\frac 1 {k+1} \right)$$
$$F(k) = \frac{1}{(k-3)}\frac 1 k-\frac{1}{(k-3)}\frac 1 {(k+1)} $$
$$F(k) = \frac 13 \left (\frac{1}{(k-3)}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!$ Let $a_{n}=n+\dfrac{1}{n}$ for $n \in \mathbb{N}$. Find the sum of series $$\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{a_{n+1}}{n!}.$$
This becomes:
$$\begin{align}
\sum_{n=1}^{\infty}(-1)^{n+1}\Big[\dfrac{(n+1)+\dfrac{1}{n+1}}{n!}\Big] &\implies
\sum_{n=1}^{\infty}(-1)^{n... | $\begin{array}\\
\sum_{n=1}^{\infty} (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n + \frac{1}{n}}{(n-1)!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2 + 1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2}{n!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summati... | The general term is $\frac{(2k-1)^2}{2^k}$. Using that, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{(2k-1)^2}{2^k}
&=\sum_{k=1}^\infty\frac{4k(k-1)+1}{2^k}\tag1\\
&=\sum_{k=1}^\infty8\binom{k}{k-2}\left(\frac12\right)^k+\sum_{k=1}^\infty\binom{k-1}{k-1}\left(\frac12\right)^k\tag2\\
&=\sum_{k=1}^\infty8\binom{-3}{k-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Trigonometry : Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$ Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
| I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $\tan$ and $\cot$ need to reinforce.
First, some preparation.
We all know that
$$
\tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}.
$$
Then
$$
\cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic series problem Given $\left\{a_n\right\}$ arithmetic progression, $a_1=2$, $a_{n+1}=a_n+2n$ $\left(n\:\ge \:1\right)$. $a_{50}=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_{50}=2+d\left(n-1\right)$$
$$a_{50}=2+2\left(n^2-n\right)$$
$$a_{50}=2+2\cdot 2450$$
$$a_{50}=4902$$
But this is wrong. Answers:
$$A=2452... | \begin{align}a_1&=2\\a_{n+1}&=a_n+2n\\&=a_{n-1}+2[n+(n-1)]\\&=a_{n-2}+2[n+(n-1)+(n-2)]\\
&= \ldots\\
&=a_1+2[n+(n-1)+\ldots +1]\end{align}
Hence \begin{align}a_{50}&=a_1+2(49+\ldots + 1)\\&=2+2\cdot \frac{49(50)}{2}\\&=2+49(50)\\&=2+(50-1)(50)\\&=2+2500-50\\&=2452 \end{align}
Remark:
This is not an AP, if it is an AP, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\lim_{x \to 0} \frac{\sin(x+a) -\sin(a)}{\sin2x}$. I am having trouble proving out the below:
$$\lim_{x \to 0} \frac{\sin(x+a) - \sin(a)}{\sin2x}$$
I have got as far as below using $\sin(a) + \sin(b)$ in numerator and $\sin(2a)$ in the denominator but am not sure how to expand further, or if these are ... | Use the sum to product formula
$$\sin(x)-\sin(y) = 2\cdot \cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$\begin{align}
\sin(x+a)-\sin(a)~&=~2\cdot \cos\left(\frac{(x+a)+a}2\right)\sin\left(\frac{(x+a)-a}2\right)\\
&=~2\cdot\cos\left(\frac{x}2+a\right)\sin\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Showing that $\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4}$
I am trying to show that if $|z|=r<1$, then
$$\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
I have shown the inequality $$\left|\frac{1}{z^3+1}\right|\leq\frac{1}{1-r^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ holds under t... | Recall that
$$\left|\frac{1}{z^4+1}\right|=\frac1{\left|z^4+1\right|}$$
and since
$$\left|z^4+1\right|\ge \left|r^4-1\right|=1-r^4\ge 0$$
the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Polynomial division : finding the unknown parameters in polynomial via divisibility (Division statement) Here is a question that involves unknowns to be found out in a polynomial and in the divisor:
Find a relation between the constants $m$, $p$ and $q$, such that:
$x^4 + px^2 +q\space$ is divisible by $x^2+mx+1$.
Th... | If the quartic polynomial $x^4 + px^2 + q$ is divisible by a quadratic of the form $x^2 + mx + 1$, so that
$x^4 + px^2 + q = f(x)(x^2 + mx + 1) \tag 1$
for some polynomial $f(x)$, severe restrictions are placed upon $x^4 + px^2 + q$, $x^2 + mx + 1$, and $f(x)$, as is seen below:
First of all, the relation (1) implies t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
[Integral][Please identify problem] $\displaystyle\int \cfrac{1}{1+x^4}\>\mathrm{d} x$ Here is my attempt. The result is not right. Please help identify the issue(s).
$\displaystyle f(x)=\int\cfrac{1}{x^4+1}\>\mathrm{d}x$, let $x=\tan t$, we have $ \mathrm{d}x = \sec^2 t\>\mathrm{d}t,\> t=\tan^{-1} x\in\left(-\cfrac{\p... | Comparing to the method I used in the following, maybe the issue occurs when computing
$$
\int \frac {\mathrm dt} {2-\sin^2(2t)}.
$$
Then from now on $t$ cannot take the value $\pm \pi /4$ if we want to devide the numerator and the denominator by $\cos^2(2t)$. Now to compute the improper integral, we should take the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{x\to 0} \frac{1 + 1/x}{1 + 1/x^2}$ using definition of a limit I need to either compute $$\lim_{x\to 0} \dfrac{1 + 1/x}{1 + 1/x^2}$$ using definition of a limit or prove it doesn't exist.
My attempt:
Given $\epsilon > 0$, we wish to find $N$ such that $\forall x \geq N$,
$$\left|\dfrac{1 + 1/x}{1 + 1/x^... | Let $f:\mathbb{R}{\setminus}\{0\}\to \mathbb{R}$ be defined by
$$f(x)=\frac{1 + 1/x}{1 + 1/x^2}$$
Claim:$\;\lim_{x\to 0}f(x) = 0$.
Let $\epsilon > 0$, and let $\delta=\min\bigl(1,{\large{\frac{\epsilon}{2}}}\bigr)$.
Suppose $|x| < \delta$, $x\ne 0$. We want to show that $|f(x)| < \epsilon$.
Since $|x| < 1$, it foll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| Correct me if wrong:
Consider $(a,b,c)$ , then
1) $a^2+b^2+c^2=$
$ (a,b,c)\cdot(a,b,c)=||(a,b,c)||^2$.
2) $ab+bc +ca =$
$(a,b,c) \cdot (b,c,a) =$
$||(a,b,c)|| ||(a,b,c)|| \cos \phi =$
$||(a,b,c)||^2 \cos \phi$.
Combining:
$a^2+b^2+c^2 +2\rho (ab+bc+cd) \ge$
$0$, or
$||(a,b,c)||^2 + 2\rho ||(a,b,c)||^2 \cos \phi \ge 0.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdo... | Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $\dbinom{10}{7}$
Second we get $8$ heads and $2$ tails in $\dbinom{10}{8}$
Third we get $9$ heads and $1$ tail in $\dbinom{10}{9}$
Fourth we $10$ heads and $0$ tails in $\dbinom{10}{10}$
Now total number of permutations $=\dbinom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Find the value $\int_{0}^{1}f(x) \,\mathrm dx$ without words
Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,\quad f(0.5)=5, \quad f(1)=15$$
Find: $$\int_{0}^{1}f(x) \,\mathrm dx$$
It is said that it can be solved without words.
| $f(0) = 1\implies d = 1, f(0.5) = 5 \implies \dfrac{a}{8} +\dfrac{b}{4} + \dfrac{c}{2} = 4, f(1) = 15 \implies 2a+2b+2c = 28\implies \displaystyle \int_{0}^{1} f(x)dx = \dfrac{a}{4}+\dfrac{b}{3}+\dfrac{c}{2}+d= \dfrac{3a+4b+6c}{12}+1=\dfrac{32+28}{12}+1 = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to proof the reason? I have this statement:
If $\frac{a}{b} = \frac{c}{d},$ prove that
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised... | $$\frac{a}{b} = \frac{c}{d}$$
write $$\frac{a}{b}+1 = \frac{c}{d}+1$$
$$\frac{a+b}{b} = \frac{c+d}{d}$$
also if $$\frac{a}{b} = \frac{c}{d}$$
then $$\frac{b}{a} = \frac{d}{c}$$
write $$1-\frac{b}{a} =1- \frac{d}{c}$$
$$\frac{a-b}{a} = \frac{c-d}{c}$$
we first obtained that $$\frac{a+b}{b} = \frac{c+d}{d}$$
$$\frac{(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Natural parameterization of the following curve The given curve is: $ 27 x^{2} = 4 y^{3} $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = \pm \sqrt{\frac{4}{27} y^3} $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x ... | If we write $x=2t^3$ then we have $$\dfrac{y^3}{27}=\dfrac{x^2}{4}=t^6\implies y=3t^2.$$
Thus we get $$\alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$? Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$ ?
For $a$ and $b$ are both positive or both negative,I proved this.
But I am not able to prove for ... | Suppose $a \ge 0$
and $b < 0$.
Writing $-b$ for $b$,
this becomes
$\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}-\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}
$
with $a \ge 0, b > 0$.
The right-hand inequality
is obviously true.
The left-hand one is
$ab/3 \le (a^2+b^2)/6$
or
$2ab \le a^2+b^2$
or
$0 \le (a-b)^2$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos... | Note that
$$
1+2\sin\theta\cos\theta=\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=
(\cos\theta+\sin\theta)^2
$$
Therefore
\begin{align}
\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}
&=\frac{(\cos\theta+\sin\theta)^2}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}
\\[6px]
&=\frac{\cos\theta+\sin\theta}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$
I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get... | $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{\sin (2(\cdot40^\circ))}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{2 \cdot \sin 40^\circ \cdot \cos 40^\circ }$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{1}{2\cdot \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluate the sum $\sum\limits_{k=1}^n {\binom{n}{k} f(k)}$.
Evaluate the following sum $:$
$$\sum\limits_{k=1}^n \binom{n}{k} f(k),$$ where $f(k)=0$ if $k$ is even and $f(k)=(-1)^{\frac {k-1} {2}}$ if $k$ is odd.
How do I proceed? Please help me in this regard.
Thank you very much.
| As already shown, complex numbers will come very handy. Here's one trick:
$$(1+i)^{n} = \binom{n}{0} + i\binom{n}{1} -\binom{n}{2} - i \binom{n}{3}... + i^{n} \binom{n}{n}$$
$$(1-i)^{n} = \binom{n}{0} - i\binom{n}{1} -\binom{n}{2} + i \binom{n}{3}... + (-i)^{n} \binom{n}{n}$$
It is easy to check now, (by multiplying bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $n$ for which $8n^3+16n^2+72n+64$ is a perfect cube
What is the maximum value of $n$ for which
$$8n^3+16n^2+72n+64$$
is a perfect cube?
I know how this could be done for a quadratic but how to extend it for a cubic.
| I shall find all integers (not necessarily positive) $n$ such that $$8n^3+16n^2+72n+64=2^3\,\left(n^3+2n^2+9n+8\right)$$
is the cube of an integer. Clearly, $p(n):=n^3+2n^2+9n+8$ must be a perfect cube. Note that
$$p(n)-(n-1)^3=5n^2+6n+9=\frac{(5n+3)^2+36}{5}>0$$
and
$$(n+3)^3-p(n)=7n^2+18n+19=\frac{(7n+9)^2+52}{7}>0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
My (attempted) proof:
Suppose that $a^2 \mid b$ and $b^3 \mid c$, where $a$, $b$, and $c$ are integers.
Therefo... | I keep your notations and write $a^2 k_1 = b$ and $b^3 k_2 = c$. Then we have
$$c^3=b^9k_2^3=b^2b^7k_2^3=(a^4k_1^2)b^7k_2^3$$
by squaring the first equation and replacing $b^2$. This can be written as
$$c^3=(a^4b^5)k_3$$
where $k_3=b^2k_1^2k_2^3$ is an integer. Hence, $a^4b^5 \mid c^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I ... | Another approach is
$$\sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\cdots$$
then
$$(x\sinh x)'\Big|_{x=1}=2\left(1+\dfrac{2}{3!}+\dfrac{3}{5!}+\dfrac{4}{7!}+\cdots\right)$$
which gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
find angle between two lines between $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6$ how to find to angle between these two lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$
i tried so far like this
$y-\sqrt{3}x-5=0$
$y=\sqrt{3}x+5$
in the form of $y=mx+b$
got the value for $m_1=\sqrt{3}$
and for
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6}... | Your calculation is very good so far.
You have found the tangent of the angle between two line to be $$ \tan \theta = \frac {-1}{\sqrt 3}$$
Thus $$ \theta = \tan ^{-1} \frac {-1}{\sqrt 3}=-\pi /6$$
You want a positive value for your angle so you add $\pi$ to get $$ \pi -\pi/6 = 5\pi/6$$ which is $150$ degrees.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\, dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\, dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 ... | $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+C$$
Can be written as
$$\frac 1 2 \log (\frac 4 3) +\frac 1 2\log( x^2-x+ 1)+C $$
Where first term can be added to the constant
$$\frac 1 2 \log (\frac 4 3) +C $$ write it as new constant $C_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Form a quadratic equation with the following details. If $\alpha, \beta$ are the roots of the equation
$
x^2 - px + q = 0
$
and $\alpha_1, \beta_1$ are the roots of the equation $x^2 - qx + p = 0$,
Form the quadratic equation whose roots are
$$
\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} and \frac{1}{\alpha_1... | By Vieta's formulas:
$$\alpha+\beta=p; \alpha\beta=q;\\
\alpha_1+\beta_1=q; \alpha_1\beta_1=p.$$
The equation to be found: $X^2+BX+C=0$, whose roots must be:
$$\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} \ \ \text{and} \ \ \ \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}$$
The sum:
$$-B=\frac{1}{\alpha_1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Multiplicative inverse of cosets I wish to find the multiplicative inverse of the coset $(1 + x) + (x^2 + x + 1)Q[x]$ in the field $Q[x]/(x^2 + x + 1)Q[x]$ but am not sure how. Do I start with Euclidean Algorithm?
| Note that every coset of $(x^2+x+1) Q[x]$ is of the form $(ax + b) + (x^2+x+1)Q[x]$ by the division algorithm.
The product of two cosets $p + (x^2+x+1)Q[x]$ and $q + (x^2+x+1)Q[x]$ ($p,q \in Q[x]$) is just $pq + (x^2+x+1)Q[x]$.
Therefore, your task is to find $a$ and $b$ such that $(ax + b) \times (x+1)$ leaves a remai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why is $y^2 = 1+x^4$ an elliptic curve? I saw in a document that $y^2 = 1+x^4$ is (the affine equation of) an elliptic curve. Why is it the case? Typically, SAGE tells me it is isomorphic to $y^2 = x^3 - 4x$, which is an elliptic curve with Weierstrass equation, but I don't know how to prove this.
Thank you!
| A very concrete answer to your question can be found in Exercise 1.15 on page 31 of Silverman and Tate's Rational Points on Elliptic Curves (2nd Edition).
You ask in a comment to your question:
Do all equations of the form $y^2 = \text{quartic}$ give elliptic curves?
The answer is clearly no, since for example the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$... |
Minimize $f(a,b,c,d)=(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3)$ subject to $a+b+c+d=4$.
The Lagrange function: $L(a,b,c,d,t)=f(a,b,c,d)+t(4-a-b-c-d)$.
FOC:
$$\begin{cases}L_a=2a(b^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_b=2b(a^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_c=2c(a^2 + 3)(b^2 + 3)(d^2 + 3)-t=0\\
L_d=2d(a^2 + 3)(b^2 + 3)(c^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
zero Jacobian matrix determinant and local inverse Consider the mapping $f: \mathbb R^2 \backslash \{(0,0)\} \to \mathbb R^2$ given by
$$\begin{aligned}
f(x,y) = \begin{pmatrix}
(x^2-y^2)/(x^2+y^2) \\
xy/(x^2+y^2)
\end{pmatrix}
\end{aligned}$$
Does $f$ have a local inverse at every p... | This map is not locally injective for all points in your domain. For $x = y=t$ we have
\begin{align}
f(t,t)=&
\begin{pmatrix}
0/(t^2+t^2) \\
t^2/(t^2+t^2)
\end{pmatrix}
=
\begin{pmatrix}
0 \\
\frac{1}{2}
\end{pmatrix}
\end{align}
Therefore, this function is not lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find angle UFO in the picture attached I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.
*
*I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success.
*I als... | Without loss of generality, let $OE=1$. By the Law of Sines on the triangle $ONE$, $ON=\dfrac{\sin(40^\circ)}{\sin(60^\circ)}$. Thus, using the Law of Sines on the triangle $ONF$, we get
$$NF=ON\,\left(\frac{\sin(20^\circ)}{\sin(100^\circ)}\right)=ON\,\left(\frac{\sin(20^\circ)}{\sin(80^\circ)}\right)=\frac{\sin(20^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
} |
For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?
For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?
Since the denominator can be factored into $(xy-1)(xy+1)$ and then the $(xy-1)$'s in both the numerator and denominator can be cancelled, we are lef... | Consider $X=x-1$ and $Y=y-1$, then we have
$$
\begin{array}{c}
\lim_{(x,y)\to (1,1)}\frac{xy-1}{x^2 y^2-1}=
\lim_{(X,Y)\to (0,0)}\frac{XY-Y-X}{ \left( XY-Y-X+2 \right) \left( XY-Y-X \right)}
=\lim_{(X,Y)\to (0,0)}\frac{1}{ \left( XY-Y-X+2 \right)}=\frac{1}{2}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$ Find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$.
I approached this question first by expanding the the left-hand side to get:
$$2p^2 +2\sqrt{2}pq+q^2 = 34-24\sqrt{2}$$
The problem becomes intractable for me at this point
| We want to find integers $p$ and $q$ such that
$$(p\sqrt{2} + q)^2 = 34 - 24\sqrt{2}$$
Expanding the expression on the left-hand side yields
$$2p^2 + 2pq\sqrt{2} + q^2 = 34 - 24\sqrt{2}$$
Matching rational and irrational parts yields
\begin{align*}
2p^2 + q^2 & = 34 \tag{1}\\
2pq\sqrt{2} & = -24\sqrt{2} \tag{2}
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I simplify $\frac{\log_7 32}{\log_7 8\cdot\sqrt2}$? So far I have got $\log_7 2^5 - \log_7 2^3 + \log_7 2^{1/2}$ and am unable to proceed. Am I, on the right track so far? and how do I proceed? thank you!
| You can use the formula
$$\frac{\log_{a}b}{\log_{a}c}=\log_{c}b$$
$$\frac{\log_{7}32}{\log_{7}8}\cdot \sqrt{2}=\left(\log_{8}32\right)\cdot \sqrt{2}=\frac{5\sqrt{2}}{3}$$
Or If you are looking for this
$$\frac{\log_{7}32}{\log_{7}8\sqrt{2}} =\left(\log_{8\sqrt{2}}32\right)=\frac{5}{\frac{7}{2}}=\frac{10}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial ... | You can use the Cauchy condensation test to show convergence:
$$\sum_{n=1}^\infty \frac{2^n}{2^n(2^n+1)} = \sum_{n=1}^\infty \frac{1}{2^n+1} \le \sum_{n=1}^\infty \frac{1}{2^n} = 1$$
because the latter is a geometric series.
Hence $\displaystyle\sum_{n=1}^\infty \frac1{n(n+1)}$ also converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the length of the curve y = $\sqrt{-x(x+1)} - \arctan \sqrt{\frac{-x}{x+1}}$ I used the formula for this example:
$\displaystyle L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$
And I start by computing the derivative:
\begin{align*}
y' &= \left(\sqrt{-x^2-x}\right)' - \left(\arctan \sqrt{\frac{-x}{x+1... | I think that you had problems with the simplifications of derivatives. Let $y=A-B$
$$A=\sqrt{-x (x+1)}\implies A'=\frac{-2 x-1}{2 \sqrt{-x (x+1)}}$$
$$B=\tan ^{-1}\left(\sqrt{-\frac{x}{x+1}}\right)\implies B'=\frac{\frac{x}{(x+1)^2}-\frac{1}{x+1}}{2 \sqrt{-\frac{x}{x+1}}
\left(1-\frac{x}{x+1}\right)}=\frac{\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | As an alternative, following the idea by Count Iblis, we have that by Taylor expansion
$$\sin x = x-\frac16 x^3+o(x^3) \implies \frac1{\sin x}=\frac 1x\left(1-\frac16x^2+o(x^2)\right)^{-1}=\frac1x+\frac16x+o(x)$$
therefore
$$\left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)
=\left( {\frac{1}{x}} + {\frac{1} {\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 2
} |
How to prove this inequality using AM-GM? Suppose $a,b,c$ are positive real numbers. Then prove that $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(\frac{a+b+c}{3}\Big)\Big(abc\Big)^\frac{2}{3}\tag{*}$$
My approach: From AM-GM $$\Big(\frac{a+b}{2}\Big)\ge\Big(ab\Big)^\frac{1}{2}\tag{1}$$... | The inequality
$$ \sum_{sym} a^2 b^1 c^0 \geq \sum_{sym} a^1 b^1 c^1 $$
holds by bunching / Muirhead's inequality and it is equivalent to
$$ \frac{a+b}{2}\cdot\frac{a+c}{2}\cdot\frac{b+c}{2}\geq \frac{a+b+c}{3}\cdot\frac{ab+ac+bc}{3}.$$
The given inequality is weaker than the latter, since $\frac{ab+ac+bc}{3}\geq\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt = \frac{1}{2n} + O\left( \frac{1}{n^2}\right)$
The integral $\int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt$ pops up when estimating the remainder of the series $\sum_{k} \frac{(-1)^k}k$.
Indeed, by Taylor's theorem with integral remainder, $$\log(1+x) = \sum_{k=1}^n (-1)^{k+1}\fr... | Apply the substitution $u = \frac{1-t}{1+t}$ to notice that
$$ \int_{0}^{1} \frac{(1-t)^n}{(1+t)^{n+1}} \, dt
= \int_{0}^{1} \frac{u^n}{1+u} \, du. $$
Now integration by parts proves the desired estimates:
\begin{align*}
\int_{0}^{1} \frac{u^n}{1+u} \, du
&= \left[ \frac{u^{n+1}}{n+1} \cdot \frac{1}{1+u} \right]_{0}^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving an equation with Lambert's W function? Or by any other means? I am trying to solve the following equation for x in terms of $y$ and $c$ (with $x,y \in [0,1]$)
\begin{equation}
\log\left(\frac{x}{1-x-y}\right) + \frac{x}{1-x-y} + \frac{y}{1-x-y} = c
\end{equation}
I can solve this easier equation
\begin{equati... | \begin{align}
\ln\left(\frac{x}{1-x-y}\right)
+\frac{x}{1-x-y} + \frac{y}{1-x-y}
&= c
\tag{1}\label{1}
\end{align}
\begin{align}
\ln\left(\frac{x}{1-x-y}\right)
&= c+\frac{1-x-y-1}{1-x-y}
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c+1-\frac{1}{1-x-y}
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c-\left(\frac{1}{1-x-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
| Since
$$\dfrac x{(1-3x)(1+x)} = \dfrac14\cdot\dfrac1{1-3x}-\dfrac14\cdot\dfrac1{1+x},$$
then
$$I(x)=\int\dfrac{x\,\mathrm dx}{(1-3x)^{^{^3/_2}}(1+x)^{^{^3/_2}}}
=I_1(x)+I_2(x),\tag1$$
where
$$I_1(x)=\dfrac14\int\dfrac{\mathrm dx}{(1-3x)^{^{^3/_2}}\sqrt{1+x}},\quad
I_2(x) = -\dfrac14\int\dfrac{\mathrm dx}{(1+x)^{^{^3/_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Is the sequence $a_{n} = 1 + \frac14 + \frac{2^2}{4^2} + \cdots +\frac{n^2}{4^n}$ Cauchy? I think that it is Cauchy (but I am not sure of this) and this is my proof:
$$|a_{m} - a_{n}| = \left|\frac{n+1}{4^{n+1}} + \frac{n+2}{4^{n+2}} + ..... + \frac{m^2}{m}\right| =\sum_{k=n+1}^{m} \frac{k^2}{4^k}$$
And then knowing t... | Prove by induction that $k^2 \le 2^k$ for $k \ge 4$ so
$$|a_{m} - a_{n}| \le \sum_{k=n+1}^m \frac{k^2}{4^k} \le \sum_{k=n+1}^\infty \frac1{2^k} = \frac1{2^{n}}\xrightarrow{m,n\to \infty} 0$$
Hence $(a_n)_n$ is Cauchy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$ for any $x, y, z, w \geq 0$ Prove that
$\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$
For any $x, y, z, \geq 0$
And prove the AM-GM inequality for three numbers, $\sqrt[3]{xyz} \leq \frac{x+y+z}{3}$ where $x, y, z \geq 0$, by using the previous proof with $w= (xyz)^{1/3}$
| You are on the good track, indeed let consider by $AM-GM$
$$u=\frac{x+y}{2}\ge \sqrt{xy} \quad v=\frac{z+w}{2}\ge \sqrt{zw}$$
then
$$\frac{u+v}{2}= \frac{x+y+z+w}{4}\ge \sqrt{uv}\ge \sqrt{\sqrt{xy}\sqrt{zw}}=\sqrt[4]{xyzw}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solution of a quadratic equation to satisfy a constraint I have the roots of a quadratic equation as
$$x = \frac{1 \pm \sqrt{1-4\theta^2}}{2\theta}$$ I know that $|\theta| < 1/2$. Among these two roots, I want to find the one which has a value $|x| < 1$.
My attempt is:
It can't be $\frac{1 + \sqrt{1-4\theta^2}}{2\theta... | You can use Vieta's formula. We can easly see that $x$ is a solution to $$\theta x^2-x+\theta=0$$ Since you figer out $|x_1|>1$ you can use now:$$x_1\cdot x_2 =1\implies |x_2| = {1\over |x_1|} <1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that
$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$
My try:
First I wrote the inequality as
$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac... | $$a^3+b = {a^3\over 4}+{a^3\over 4}+{a^3\over 4}+{a^3\over 4}+b\geq 5\sqrt[5]{a^{12}b\over 2^8}$$
$$b^3+c = {b^3\over 4}+{b^3\over 4}+{b^3\over 4}+{b^3\over 4}+c\geq 5\sqrt[5]{b^{12}c\over 2^8}$$
$$c^3+a = {c^3\over 4}+{c^3\over 4}+{c^3\over 4}+{c^3\over 4}+a\geq 5\sqrt[5]{c^{12}a\over 2^8}$$
So $$\left(a^3+b\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Asymptotics of $\sum _{n \leq x}\sigma_{-2}(n)$ It is known that the sum of the squared inverses of the divisors satisfies $$\sum _{n \leq x} \sigma_{-2}(n) = \zeta(3)x + \mathcal O(1).$$ On the other hand, an alternate calculation gives me another answer:
$$\begin{align*}
\sum _{n \leq x} \sigma_{-2}(n) &= \sum _{... | You are tacitly assuming that a constant is $1$, while that is not the case.
$$ \sum_{b\leq N}\frac{1}{b^2} = \zeta(2)-\frac{1}{N}+\frac{1}{2N^2}-\frac{1}{6N^3}+\frac{1}{30N^5}+\ldots $$
hence by replacing $N$ with $\frac{x}{a}$ we get
$$ \sum_{b\leq\frac{x}{a}}\frac{1}{b^2} = \zeta(2)-\frac{a}{x}+\frac{a^2}{2x^2}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $a$ so that a line is tangent, secant or external from a sphere I am given the following problem:
Given the line $$r \{ R = (1,0,a) + \lambda [a \quad a \quad 0]$$ and the sphere $$S \{ 8x^2 + 8y^2 +8z^2 - 16x +24y -8z + 19 = 0$$ find, relating to values of $a$, when the line is external, tangent and secant to th... | The distance squared,
writing $c$ for $\lambda$, from
$R
= (1,0,a) + c [a \quad a \quad 0]
= (1+ca,ca,a)
$
to
$C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right)
$
is
$\begin{array}\\
d^2
&=(1+ca-1)^2+(ca+\frac32)^2+(a-\frac12)^2\\
&=c^2a^2+c^2a^2+3ca+\frac94+a^2-a+\frac14\\
&=2c^2a^2+3ca+\frac52+a^2-a\\
\end{array}
$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$? $$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried ... | If you want to go beyond the limit it self, use Taylor series and binomial expansion
$$\cos(x^2)=1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)$$
$$1-\cos(x^2)=\frac{x^4}{2}-\frac{x^8}{24}+O\left(x^{12}\right)$$
$$\sqrt{1-\cos \left(x^2\right)}=\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right)$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Solve this inequality with nested radicals (possibly by induction) I tried to solve this problem by induction but didn't succeed. Given the series
$$ a_n = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}$$
Prove that $a_n < 2 (\forall n \in \mathbb{N^*}) $
Now I thought that maybe I could find a reccurence formula. I... | Fix $x \ge 0$, and define $(a_n)$ by
$$
a_n
=
\sqrt{x^2 +
\sqrt{x^4 +
\sqrt{x^8 +
\sqrt{\cdots +
\sqrt{x^{2^n}}
}
}
}
}
$$
Then for all $n > 1$, we have the relation
$$a_n^2=x^2+xa_{n-1}$$
Now let $x = \sqrt{5}-1$.
Claim:$\;a_n < 2$, for all $n$.
Proceed by induction on $n$.
For $n=1$, we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Application of Euler's totient function to find last digits Q:what are the last five digits of the number $2018^{2017^{.^{.^{.^{2^{1}}}}}}$.
My Approach:I know how to find the last two digits of $N=2018^{2017^{k}} $ by$N=2018^{2017^{k}\pmod{\phi(25)}}\pmod {25}\equiv 2018^{2017^{2016^b\pmod{8}}\pmod{20}}\pmod {25}\equi... | Let $\psi(0)=1$ and $\psi (n)=n^{\psi (n-1)} $.
We have to find $\psi (2018)\pmod {10^5} $.
Clearly $\psi (2015)\equiv 0\pmod {5^2} $ and $\psi (2015)\equiv 1\pmod 4$, from which $\psi (2015)\equiv 25\pmod {4\cdot 5^2} $, by chinese remainder theorem.
Consequently, $\psi (2016)\equiv 2016^{25}\equiv 16^{25}\equiv 2^{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$ \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$ and ... To find limits:
$(a) \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$
$(b) \lim_{n \to \infty} \frac{n^3}{n!}$
For the first one the sequence is oscillating so it does not converge.
For the 2nd one I used ratio test: Let $a_n = \frac{n^3}{n!} $, then $\frac{a_{n... | Another way for the second is
$$\dfrac{n^3}{n!}<\dfrac{n}{n}\dfrac{n}{n-1}\dfrac{n}{n-2}\dfrac{1}{n-3}<1\dfrac{n}{\frac{n}{2}}\dfrac{n}{\frac{n}{2}}\dfrac{1}{\frac{n}{2}}=\dfrac{8}{n}<\varepsilon$$
for $n>6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Point of negative inflection
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < \frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
t
| $f(x) =\frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$\frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = \frac{-(a+2) +- \sqrt{4a + 5}}{a}$
Ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $ \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Finding $\displaystyle \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Try: Let $$\displaystyle I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$ (Function is even )
$$I = 2\int^{1}_{0}\bi... | Since $\int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac{1}{3}(2+x^2)\sqrt{1-x^2}$, integration by parts yields
\begin{align*}
\int_{-1}^{1} \frac{x^3}{\sqrt{1-x^2}} \log\bigg(\frac{1+x}{1-x}\bigg) \, dx
&= \int_{-1}^{1} \frac{2(2+x^2)}{3\sqrt{1-x^2}} \, dx.
\end{align*}
Now the substitution $x = \sin\theta$ quickly yields ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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first order Ordinary differential equation help $T''+T'+\lambda T=0$ I stuck with these ODE question.
solve the ODE
$T$ is function of $t$
$T''+T'+\lambda T=0$
$T(0)=0,T'(0)=0,\lambda=n^2 $
when $n\in \Bbb N$
My sulution So far:
Characteristic polynomial
$r^2+r+\lambda=0$
sol:
$r_{1,2}= \frac{-1\pm \sqrt {1-4\lambda} }... | I would say, if
$T(t)=C_1e^{t\frac{-1+ \sqrt {1-4n^2} }{2}}+C_2e^{t\frac{-1-\sqrt {1-4n^2}}{2}}$ then is for $n=1,2,3\cdots$
$T(t)=e^{-t/2}\left(C_1e^{\frac{it\sqrt {4n^2-1
} }{2}}+C_2e^{\frac{-it\sqrt {4n^2-1}}{2}}\right), $ i is the imaginary unit
$=e^{-t/2}\left(C_1\cos \frac{t\sqrt {4n^2-1
} }{2}+iC_1\sin \frac{t\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A choice of Lyapunov function for this 2D system? I am thinking of a choice of a suitable Lyapunov function$V(x_{1},x_{2})$ which can make the system stable around the fixed point $x_{1}=1,x_{2}=1$
$\dot{x_{1}} = x_{1}x_{2} - x_{1}^2 $
$\dot{x_{2}} = x_{2} - x_{1}x_{2} + 2 -2x_{2}^2$
I thought of using $V(x_{1},x_{2}... | You only need that $\dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
\begin{align}
\dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
\end{align}
Thus for $a=b=1$ you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to ... | Let$$
z^2=(x+yi)^2=−1+4\sqrt3i,
$$
i.e.$$
(x^2-y^2)+2xyi=−1+4\sqrt3i.
$$
Compare real parts and imaginary parts, $$
\begin{cases}
x^2 - y^2 = -1&\qquad\qquad(1)\\
2xy = 4\sqrt3&\qquad\qquad(2)
\end{cases}
$$
Now, consider the modulus: $|z|^2 =|z^2|$, then
$$x^2 + y^2 = \sqrt{\smash[b]{(-1)^2+(4\sqrt3)^2}} = 7\tag3$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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Find $xyz$ if $x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0$ The question is
Find the sum of all possible values of $xyz$ given that $x, y, z\in \Bbb Z$ satisfying
$$x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0.$$
Some thought so far:
Obviously $x$ is even. I assume $x=2k$, but I don't know how to proceed. I tried to complete squ... | Completing the squares:
$$(x+y-1)^2+(y+z-2)^2+(z-3)^2=0 \Rightarrow \\
\begin{cases}x+y-1=0\\ y+z-2=0\\ z-3=0\end{cases} \Rightarrow \\
(x,y,z)=(2,-1,3) \Rightarrow xyz=-6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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First year college question on divisibility of integers I'm having a hard time with a practice question.
Given $n$ is an integer, prove $2$ divides $(n^4 -3)$ iff $4$ divides $(n^2 +3)$.
So I know since it's an iff statement, I have to show the implication going both ways. Let's start with the left side first.
There ... | Proof:
Since the statement is biconditional we must prove the following two statements:
$(a)$ If $2|(n^4 − 3)$ then $4|(n^2 + 3)$
$(b)$ If $4|(n^2 + 3)$ then $2|(n^4 − 3)$
We will begin with statement $(b) ($since it should be the easiest to prove$)$
We will prove this statement directly. Assume that $4|(n^2 + 3)$. Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sq... | WolframAlpha understands the expression $\sqrt[3]{x}$ for negative x in a different way than you expect.
Try this: lim\frac{\sqrt{1-x}-3}{2+surd(x,3)} as x to -8
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 0
} |
$\frac{a^{3}+1}{b+1}+\frac{b^{3}+1}{a+1}$ an integer $\Rightarrow \frac{a^{3}+1}{b+1}$ and $\frac{b^{3}+1}{a+1}$ are integers. I want to show that if the natural numbers $a,b \in \mathbb{N}$ are such that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, then, necessarily, $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\f... | Hint $\ r\! +\! s,\, rs \in \Bbb Z\,\Rightarrow\, r,s \in \Bbb Z\ $ by applying the Rational Root Test to $\,(x\!-\!r)(x\!-\!s)\in\Bbb Z[x]$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove central binomial coefficient upper bound I am trying to prove that $\binom{2n}{n} < \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) < 4\sqrt{n(n+1)}$ which is false)
| $ n^n e^{-n}\sqrt{2\pi n} < n! < n^n e^{-n} \sqrt{2\pi n} (1+\frac{1}{8n}) $ this is Stirling approximation
So $\binom{2n}{n} \leq \frac{(2n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{(n^n e^{-n} \sqrt{2\pi n})^2} = \frac{4^n (n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{n^{2n} e^{-2n}* 2\pi n} = \frac{4^n (1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that the sequence space is a metric space in $\mathbb{R}$ I need to show that $(X, d)$ is a metric space:
$X$ is the sequence space where its elements are sequences of real numbers, and $d : X \times X \rightarrow \mathbb{R}$ where d is defined as:
$$d(x,y) = \sum_{j=1}^\infty\frac{1}{2^j} \frac{|x_j - y_j|}{1 ... | Let $f(x) = \frac{x}{1 + x}$. We wish to show that, for $x, y \ge 0$, we have $f(x + y) \le f(x) + f(y)$ (i.e. the function is subadditive over $[0, \infty)$). I hope you can see how this would prove triangle inequality for the proposed metric.
Suppose $x, y \ge 0$. Consider
\begin{align*}
f(x + y) \le f(x) + f(y) &\if... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\df... | You could try like this (with help of triangle inequality):
$$|z-{1\over z}|= |{z^2-1\over z}| = {|z^2-1|\over 2} \geq {|z^2|-1\over 2} ={3\over 2}$$
clearly this can be achieved at $z = 2$ and $$ {|z^2-1|\over 2} \leq {|z^2|+1\over 2} = {5\over 2}$$ which can be achieved at $z=2i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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How to find inverse of general curvilinear coordinates Lets say I have a curvilinear coordinate system
$A=A(x,y,z) = \frac{x^2+y^2+z^2}{2z} $, $B=B(x,y,z)= \frac{x^2+y^2+z^2}{2\sqrt{x^2+y^2}}$, $C=C(x,y,z)=\tan^{-1}(y/x)$
How do I find the inverse of those i.e
$x=x(A,B,C)$, $y=y(A,B,C)$, $z=z(A,B,C)$
I know that in cy... | I will assume that $x,y,z>0$. You will need to check if it make sense to have them negative.
The first equation we are going to use is $$\tan C=\frac{y}{x}$$
Since everything else we have squares, I will write this as $$y^2=x^2\tan^2C\tag{1}$$
The second equation can be written from the ratio $A/B$ or in fact we can us... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I integrate $\int \frac{x^2}{(x^2+9)^2}\, dx$? How do I integrate $\displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx$
?
I tried doing using algebra and solving the question a bit. But it didn't become something that looks solvable.
How should I do this?
| I think the best thing to do in such a question, should be suitable substitution.
So let
$x=3\tan{\theta}$
$dx=3\sec^2(\theta) d\theta$
$\begin{align} \displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx & = \int \frac{9 \tan^2(\theta) }{(9 \tan^2(\theta) + 9)^2} 3\sec^2(\theta)\, d\theta = \int \frac{27 \tan^2(\theta)\sec^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$
If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_j|=2\space\space\space\forall j\in\{1,2,3,4,5,6\}$
I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=\sqrt{a^2+b^2... | $$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z \neq 1$
Then $\frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Sum of $\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - ...$ I have to find the sum of :
$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$
So far I have :
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$
which is very close to $\ln(1+x)$... but I just can't figure o... | Note that $$\frac {\mathrm d^2}{\mathrm dx^2}\frac{(-1)^nx^n}{n(n-1)}=\frac {\mathrm d}{\mathrm dx}\frac{(-1)^nx^{n-1}}{n-1}=(-1)^nx^{n-2}=(-x)^{n-2}$$
so that we expect $f(x)_=\sum_{n=2}^\infty\frac{(-1)^nx^n}{n(n-1)}$ to be a function with $f''(x)=\sum_{n=0}^\infty (-x)^{n} =\frac1{1+x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Problems with an exact differential eqution Consider the following differential equation
$$
\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx=\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not... | We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M \equiv \frac{1}{x} - \left(\frac{y}{x-y}\right)^2, N \equiv \left(\frac{x}{x-y}\right)^2 - \frac{1}{y}.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M \Rightarro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find polynomial $p(n)$ such that $\displaystyle \sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$ converges
Find a polynomial $p(n)$ such that series:
$$\sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$$
converges.
Attempt. If $p(n)=q^3(n)$ for some polynomial $q(n)$ then
$$\sqrt[4]{n^4+... | $$\sqrt[4]{n^4+2n^2} = n\sqrt[4]{1+\frac{2}{n^2}}= n\left(1+\frac{1}{2n^2}-\frac{3+o(1)}{8n^4}\right) = n+\frac{1}{2n}-\frac{3+o(1)}{8n^3} $$
and $\left(n+\frac{1}{2n}\right)^3 = n^3+\frac{3n}{2}+o(1)$, hence by taking $\color{red}{p(n)=n^3+\frac{3n}{2}}$ we are fine, since
$$\sqrt[3]{n^3+\frac{3n}{2}}=n\sqrt[3]{1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$ I wanted to know if this equation could be solved any further please.
$$2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$$
I have gone this far:
$$4 \sin x \sin(60^\circ-x)= \sqrt{2}- 1$$
Thank you
| Hint:
$$2\sqrt{3} \sin x \cos x- 2\sin^2x = \sqrt3\sin 2x+\cos2x-1=\sqrt{2} - 1.$$
Then
$$3\sin^22x=3(1-\cos^22x)=(\sqrt2-\cos 2x)^2.$$
This is a quadratic equation in $\cos 2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^2}$ Prove or disprove each of the follow function has limits $x \to a$ by the definition
$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^2}$
Given $\epsilon > 0$. Choose $\delta = \epsilo... | You can also use this.
$$\left|{xy^2\over x^2 + y^2 }\right| = |x| {y^2\over x^2 + y^2 } \le |x|. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine whether $\sum_{k=1}^{\infty}\frac{k+2}{\sqrt{k^5+4}}$ converges or diverges Question
Determine if
$$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}}$$
converges or diverges.
I'm looking for a proof that this sum converges in a simpler way than I've shown.
My (ugly) work
I have the following chain of implication... | $$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}} \le \sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5}} \le 3+\sum_{k=2}^\infty\frac{2k}{k^{2.5}}=3+2\sum_{k=2}^\infty\frac1{k^{1.5}}$$
Hence it converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960292",
"timestamp": "2023-03-29T00:00:00",
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find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
| Let $f(x) = x^4+x^3-1$ and $g(x) = x^6+x^4 + x^3 - x^2 - 1$.
Let $a, b, c, d$ be the roots of $f(x)$. It is easy to see these four roots are
distinct and differ from zero. If we set $\lambda = a + b$ and $\mu = ab$, we will have $\mu \ne 0$.
Since $a \ne b$ are roots of $f(x)$, $b$ is a root of
$$\begin{align}A(x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3? 3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?
| For any integers $B>A>1,$ to gauge "how much bigger is $B$ compared to $A,$" consider the following questions :
*
*How many $A$s do we need to add together to reach $B$?
Ans: The least $x$ such that $\underbrace{A+A+...+A}_{x\ A\text{s}}\ge B$, i.e., $A\times x \ge B: \quad x=\left\lceil{B\over A}\right\rceil.$
*Ho... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n... | The sum in the question can be re-written as
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^n\frac1{2k}\right)\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Finding dependencies such that $0 > \frac{2b^2r^2}{z}-\left(2r ^2-2br\sqrt{1-\frac{b^2}{z^2}}\right)z$ I'm trying to solve an inequality with 3 variables.
$$0 > \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Basically, I want to know under which dependencies the formula is less than ze... | The final solution comes out to be very nice, so I suspect this is a homework problem. As such, I will just give you an outline.
$$0 < \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Simplifying terms, dividing both sides by $z$, letting $\frac{b}{z}=\alpha$ and pulling the square root ... | {
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"url": "https://math.stackexchange.com/questions/2967619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | Less Than $\boldsymbol{3}$
The inequality
$$
1+\frac1{n^3}\lt\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\tag1
$$
can be verified by cross-multiplying and then multiplying both sides by $2n^5(n-1)^2$; that is,
$$
2n^7-4n^6+3n^5\underbrace{-3n^3+3n^2-2n+1}_\text{$-(3n^2+2)(n-1)-1\lt0$ for $n\ge1$}\lt2n^7-4n^6+3n^5\tag2
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Proof for a formula of a special real determinant For some $x_1,\cdots,x_n$ in $\mathbb{R}$ we define:
$$A_n:=
\left(\begin{matrix}
1 &x_1 &x_1^2 &\cdots &x^{n-1}_1\\
1 &x_2 &x_2^2 &\cdots &x_2^{n-1}\\
\vdots &\vdots &\vdots & &\vdots\\
1 &x_n &x_n^2 &\cdots &x^{n-1}_n
\end{matrix}\right)
\in \mathbb{R}^{n\times n}
$$
... | Take the case $n = 4$ as an example. Let $(x_1,x_2,x_3,x_4) = (x,y,z,t)$, we have
$$\begin{align}
\left|\begin{matrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3 \\
1 & t & t^2 & t^3 \\
\end{matrix}\right|
&\stackrel{\color{blue}{[1]}}{=}
\left|\begin{matrix}
0 & x-t & x^2-t^2 & x^3-t^3 \\
0 & y-t & y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to prove Fibonacci recurrence holds mod p? Let $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)},$$ with $c$ and $c^{-1}$ integers such that $c^2 \equiv 5 \ \text{(mod p)}$ and $cc^{-1} \equiv 1 \ \text{(mod p)}$. And $c$ is an odd integer.
It is easy t... | Write $a=\frac12(1+c)$ and $b=\frac12(1-c)$. Then $a+b=1$ and
$$ab=\frac14(1-c^2)\equiv\frac14(1-5)\equiv-1\pmod p.$$
Therefore
$$a^2-a-1\equiv a^2-(a+b)a+ab\equiv0\pmod p$$
and
$$b^2-b-1\equiv b^2-(a+b)b+ab\equiv0\pmod p.$$
Then
$$c(J_n-J_{n-1}-J_{n-2})\equiv a^n-b^n- a^{n-1}+b^{n-1}-a^{n-2}+b^{n-2}
\equiv(a^2-a-1)a^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Gaussian curvature of a given metric The exercise is from do Carmo, Differential Forms and Application, p.97. Consider $\mathbb{R}^2$ with the following inner product: if $p=(x,y)\in\mathbb{R}^2$ and $u,v\in T_p\mathbb{R}^2$, then $$\langle u,v\rangle_p=\frac{u\cdot v}{(g(p))^2},$$where $u\cdot v$ is the canonical inne... | I'm not sure if there's a simpler way to do it, but here's a way that only involves the basic theory of surfaces.
Your first fundamental form is $E = G = \frac{1}{g^2}, F=0.$ Hence you can compute the Christoffel symbols by solving
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two different series for $\frac{1}{x + 2}$
Find the appropriate power series for $f(x) = \frac{1}{x + 2} $
First method : $\frac{1}{x + 2} = \frac{1}{2(1 - (-\frac{x}{2}))} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ , $|x|\lt2$
Second method : $\frac{1}{x + 2} = \frac{1}{1 - (-x-1)} = \sum_{n=0}^\infty (-x-1)^n... | The first one is the series centered at $x=0$ while the second one is the series centered at $x=-1$ therefore the radius of convergence are different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that $\int_0^{\pi} \dfrac{\cos2\theta \ d \theta}{1-2 a \cos {\theta} +a^2}=\dfrac{a^2 \pi}{1-a^2}; \ \ (-1For $a=0$ the equality holds. For $a \ne 0$, by change of variable and letting $C_0$ be the unit circle about the origin, we have:
\begin{align*}
\int_{-\pi}^{\pi} \dfrac{\cos2\theta \ d \theta}{1-2 a \cos {... | From the second line, it should be
$$\int_{C_0} \dfrac{(1+z^4)dz}{2iz^2(z-a)(1-az)}
=\pi\left(\dfrac{ 1+a^4}{a^2(1-a^2)}-\frac{1+a^2}{a^2}\right)=\dfrac{2a^2 \pi}{1-a^2}$$
where we evaluated the residues at $a$ AND at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$
Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$ at the nodes $x_0=0, x_1 = 1, x_2=2, x_3 = 3$.
Here are my workings below
The basic Lagrange polynomials are:
$$L_0(x) = \frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)... | Your calculations are correct.
The final result is
$P_3(x) = 0.0 + 0.5 (x-3) (x-2) (x+0)-4 (x-3) (x-1) (x+0)+4.5 (x-2) (x-1) (x+0) = x^3$
The formula for the error bound is given by:
$$E_n(x) = {f^{(n+1)}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
So, we have
$$E_3(x) = {f^{(4)}(\xi(x)) \over 4!} \times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 ... | First, Note that
$$342342 = 34234*10+2= 34234*2*5 +2$$
So you have
$$342342 \equiv 2 \pmod 5$$
Then, remember that
$$\forall a,b,c,n \in \mathbb N, a \equiv b \pmod n \implies a^c \equiv b^c \pmod n$$
Therefore,
$$342342^{1001} \equiv 2^{1001} \pmod 5$$
Finaly, note that $1001 = 1000+1 = 4*250 +1$ and try to conclude... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Elementary Level Algebra Question I'm trying to solve the following homework problem:
If $a\neq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
\begin{align}
\text{(1) }& a = 3x \\
\text{(2) }& a = 3x \text{ or } a = -2x \\
\text{(3) }& a = -3x \text{ or } a = 2x \\
\text{(4) }& a = 3x... | Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$\implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $\color{purple}{a-b = x}$.
$$\implies \color{purple}{x}(a^2+ab+b^2) = 19x^3 \implies a^2+ab+b^2 = 19x^2$$
Set $\color{blue}{b = a-x}$.
$$\implies a^2+a\color{blue}{(a-x)}+\color{blue}{(a-x)}^2 = 19x^2$$
Move $19x^2$ to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$? Wolfram Alpha shows that $$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$
I tried to use the Fourier series
$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$
I am not sure how to continue from this point. I need some he... | Thanks mrtaurho, I am able to finish it.
$$\begin{align}
\int_{0}^{2\pi}\pi^2\ln(1-\cos x)dx
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\left[-x^2\frac{\sin(2nx)}{2n}+\frac{2x\cos(2nx)}{4n^2}-\frac{2\sin(2nx)}{8n^3}\right]_{0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$ if $x_n$ satisfy $x^{-n}=\sum_{k=1}^\infty (x+k)^{-n}$? Suppose $x_n$ is the only positive solution to the equation $x^{-n}=\sum\limits_{k=1}^\infty (x+k)^{-n}$,how to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$?
It is easy ... | For any $n \ge 2$, consider the function $\displaystyle\;\Phi_n(x) = \sum_{k=1}^\infty \left(\frac{x}{x+k}\right)^n$.
It is easy to see $\Phi_n(x)$ is an increasing function over $(0,\infty]$.
For small $x$, it is bounded from above by $x^n \zeta(n)$ and hence decreases to $0$ as $x \to 0$.
For large $x$, we can appro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
For any positive integer $n$, prove the following inequality. Prove that for any positive integer n, the following inequality is true.
$$\left(1+ \frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}$$
Attempt
Not a good attempt but this is my thinking
It will not be a problem when checking whether this is true for... | I know this question is already answered, but I want to give an answer that only uses the binomium of Newton. This is certainly not the easiest way to show the inequality, but it is an "artisanal" way.
By the binomium we get
$$
\begin{align*}
\left(1+ \frac{1}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Algebraic manipulation with indices The question is:
For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$
So I first squared the given equation and got:
$$a+\frac{1}{a}+2=9$$
$$a+\frac{1}{a}=7$$
Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:
$$(a+\frac{1}{a})(\sqrt{a}... | Let $x=\sqrt{a}$ then $x+\frac{1}{x}=3$ and
$$
\\a\sqrt{a}+\frac{1}{a\sqrt{a}}=x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3x\cdot\frac{1}{x}(x+\frac{1}{x})=27-3\cdot3=18
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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.