Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof involving polynomial roots From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit l... |
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $\,\ldots$
Following up on OP's approach, let $\,ab=p\,$ and, for symmetry, $\,a+b=s\,$. To begin with:
$$
\begin{cases}
\begin{align}
a^4 + a^3 - 1 &= 0 \\
b^4 + b^3 - 1 &= 0
\end{align}
\end{cases} \tag{1}
$$
Subtracting the two equations in $\,(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
... | It seems easier to use $\log_{12}$ rather than $\log$.
$\log_{12}(36)=\log_{12}(3)+1=k$, and
$$2\log_{12}(2)+\log_{12}(3)=\log_{12}(12)=1,$$ so $2\log_{12}(2)=2-k$,
or $$\log_{12}(2)=1-\frac{k}{2}$$
Thus
$$\log_{24}(48)= \frac{\log_{12}(48)}{\log_{12}(24)}=\frac{2\log_{12}(2)+1}{\log_{12}(2)+1}$$
$$=\frac{2-k+1}{2-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find eigenvalues & eigenvectors for an integral. Can anyone please explain me how to solve it?
Find the nonzero eigenvalues and the corresponding eigenvectors:
$T:[-1,1]\rightarrow[-1,1]$
$$T((f(x))=\int_{-1}^1(x^2 y + y^2 x) f(y) \, dy$$
| Since $$Tf\left(x\right)=x^2\int_{-1}^{1} yf\left(y\right)dy+x\int_{-1}^{1} y^2f\left(y\right)dy$$
we have that $Tf\left(x\right)=ax^2+bx$ for some $a,b$. Hence, if $f\left(x\right)$ is an eigenvector it must take this form. Plugging this in we get $$Tf\left(x\right)=x^2\int_{-1}^{1} y\left(ay^2+by\right)dy+x\int_{-1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| Checking other answers:
Note that $\frac{x^2-1}{x}\ge 0;\frac{x-1}{x}\ge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 \ \ (1)$, then:
$$ x^2-1=x \Rightarrow \frac{x-1}{x}=\frac{1}{x+1},\\
{\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x \ \ \ \ \ \ \ \ \ \ (2) \ \ \ \ \ \Rightarrow \\
{\sqrt {(x+1)- 1\over x}} + {\sqrt{1\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1... | $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= $$
$$2^{2x}-2^x + 2^{2x-2}-2^{x-1} + .... + 2^{2x-198}-2^{x-99}= $$
$$(2^{2x}+ 2^{2x-2} .... + 2^{2x-198})-(2^x +2^{x-1} + .... + 2^{x-99})= $$
$$2^{2x-198}(2^{198}+ 2^{196} .... + 2^{0})-2^{x-99}(2^{99} +2^{98} + ...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$ Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$
There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$
By trigonometry, we know t... | Because $$\frac{2t}{1+t^2}=\frac{2\tan\frac{\theta}{2}}{\frac{1}{\cos^2\frac{\theta}{2}}}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\sin\theta.$$
You can use your formula, but we need to write before:
$$\frac{1-t^2}{1+t^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\cos^2\frac{\theta}{2}-\sin^2\frac{\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Confusing problem determining the exact formula for area of 'complex' shape Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at th... | In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $\triangle PGB$ and $\triangle FQC$, less the area of $\triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\in... | Note that, if $f(x)=\arctan(xi)$, then
$$
f'(x)=i\frac{1}{1+(xi)^2}=\frac{i}{1-x^2}
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-\frac{1}{x}-i\arctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
\frac{1}{x^2(1-x)^2}=
\frac{a}{x}+\frac{b}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding integer solutions to $ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $ I was browsing through facebook and came across this image:
I was wondering if we can find more examples where this happens?
I guess this reduces to finding integer solutions for the equation
$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for... | Let $S$ be the set of allowed values of $a$, $b$, and $c$ ($S=\mathbb{Z}$ in this OP's setting, but $S$ can be something else like $\mathbb{Q}$, $\mathbb{Q}_{>0}$, $\mathbb{R}$, or even $\mathbb{F}_p$, where $p$ is a prime natural number). If $b=-a$, then $c$ can be any number not equal to $-a$. That is, $(a,b,c)=(a,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$ Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$
Things I have done so far:
$$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$
$$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\fra... | $$\sum_{cyc}\dfrac{1}{(b+c)^2+a^2} \le \dfrac{2}{25}(a+b+c)+\dfrac{9}{25}=\dfrac{3}{5}$$
$$\dfrac{1}{(b+c)^2+a^2}=\dfrac{1}{(3-a)^2+a^2}\le \dfrac{2}{25}a+\dfrac{3}{25}\ , (a\ge 0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does $L$ have to be lower triangular in the LU factorization? I was studying LU factorization, when I didn't understand a particular phrase, or rather, how it works or why it works.
During LU factorization L is said to be a lower triangular matrix and U is said to be an upper triangular matrix, for A=LU.
All... | There is a slightly different LU decomposition called the Cholesky decomposition that creates two upper triangular matrices. The reason primarily is how it constructs the matrices. Gaussian elimination is simple to illustrate and I will show you. I recommend Trefethan and Bau personally.
Gaussian Elimination without P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. If it's converge... | It all boils down to showing that $$0<a_n<7\implies a_n<\sqrt{7a_n}=a_{n+1}<7,$$ which is fairly obvious (geometric average).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Simplifying expression: $1 - \frac{1}{ (1 + a) / (1 - a)}$ Correct Answer is $a$
My attempt:
$$1 - \frac{1}{ (1 + a) / (1 - a)}= 1 - \frac{1 - a}{1 + a} $$
multiply $(1 + a)$ on num and dem for the first term.
$$=\frac{1 + a}{1 + a} - \frac{1 - a}{1 + a}$$
combine and subtract.
$$=\frac{2a}{1 + a}$$
How does this s... | $$1 - \dfrac{1}{\dfrac{1 + a}{1 - a}} = 1 - \dfrac{1 - a}{1 + a} = \dfrac{1 + a}{1 + a} - \dfrac{1 - a}{1 + a} = \dfrac{(1 + a) - (1 - a)}{1 + a} = \dfrac{2a}{1 + a} \ne a;$$
I agree with user582949; our OP is right at the answer he/she was given is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy=0$
Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy$
This is not one of the standard forms and neither is it an exact form. How do I go about doing this question?
| $$
(3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0
$$
$$
\implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$}
$$
We know that:
$$
d\left(\frac{x^2}{y}\right) = \frac{2xydx - x^2dy}{y^2}
$$
Substituting this back into $(1)$:
$$
\implies 3x^2y^4dx + 2x^3y^3dy + d\left(\frac{x^2}{y}\right)y^2 = 0
$$
Assuming $y \not= 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating... | Using the Mean Value Theorem, the following statement holds:
For $f(z) = 2z\tan^{-1}{z}$
$$f'(x+\xi)=\frac{2(x+2)\tan^{-1}{(x+2)}-2x\tan^{-1}{x}}{x+2-x}$$
For some $\xi \in (0,2)$
The left hand side simplifies to:
$$f'(x+\xi)=2\tan^{-1}{(x+\xi)}+\frac{2(x+\xi)}{1+(x+\xi)^2}$$
The right hand side simplifies to the limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Need some help finding the equation (nth term) of a sequence The sequence goes like this
Sequence = 1 , 4 , 14 , 58 , 292
Each number is being multiplied by (n+1) and then 2 is added to them.
Should i proceed to find the equation from
$ 2(n+1)^n $
this? . Thank you very much in advance!
| An exponential isn't quite right here, since you are multiplying by a different amount each step. At the first step, $n+1=2$, while at the second step, $n+1=3$, etc. It seems like factorials will be a much better fit for this sequence, since they have the same sort of "multiply by an increasing sequence of numbers beha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple limit with asymptotic approach. Where's the error? Simply calculus question about a limit.
I don't understand why I'm wrong, I have to calculate
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}
$$
Using asymptotics, limits and De l'Hospital rule I would write these passages...
$$... | You are just replacing $\sin x$ by $x$ and $\cos \sqrt{x^3}$ with $1-(x^3/2)$. Both these replacements are wrong for the very simple reason that $\sin x\neq x$ and $\cos \sqrt{x^3}\neq 1-(x^3/2)$ unless $x=0$.
The right approach is to use the standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =1,\,\lim_{x\to 0}\frac{1-\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find two $2\times2$ real matrix $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$ Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$
Tried to write the matrices as
$$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g... | Using the formula from
Inverse of the sum of matrices
we can take, for arbitrary $a,b$ with $a+b\neq 0$,
$$
A=\begin{pmatrix} -1 & 1 \cr 0 & 2 \end{pmatrix},\quad
B=\begin{pmatrix} a & b \cr \frac{2(a^2-a+1)}{a+b} & \frac{-2(b-ab+1)}{a+b} \end{pmatrix}.
$$
All of $A$, $B$ and $A+B$ have determinant $-2$. It is easy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How does $\frac{x + \frac{1}{2}}{\frac{1}{x} + 2} = \frac{x}2$? How does simplifying $\dfrac{x+\frac12}{\frac1x+2}=\dfrac{x}2$
Plugging and chugging seems to prove this, but I don’t understand the algebra behind it.
How would you simplify $\dfrac{x+\frac12}{\frac1x+2}$ to get $\dfrac{x}2$?
I tried multiplying the expre... | If $x \neq \frac{-1}{2}$, then $\frac{(x+1/2)}{(1/x+2)} = \frac{x^2+x/2}{1+2x} = \frac{2x^2+x}{2+4x}$. This is multiplying by 2 on top and on the bottom. Note that we can write $x = \frac{2x+4x^2}{2+4x}$. Therefore $\frac{x}{2} = \frac{x+2x^2}{2+4x}$, and the expressions are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
$112x\equiv392\pmod{91}$ has $7$ solutions? True or false? $112x\equiv392\pmod{91}$ has exactly $7$ solutions.
A necessary and sufficient condition for a linear congruence equation to have solution says that $\gcd(112,91)\mid392$, and the solutions are $\gcd(112,91)=7$. Since $7\mid392$ then the statement is true.
Is ... | Since $91=13\cdot 7$ we have that
$$112x\equiv392\pmod{91} \iff 21x\equiv 28\pmod{91}$$
is equivalent to
$$21x\equiv 28\pmod{7} \iff 0\equiv 0 \pmod{7}$$
$$21x\equiv 28\pmod{13} \iff 8x\equiv 2\pmod{13}\iff 4x\equiv 1\pmod{13}$$
and for the latter since $10\cdot4-3\cdot13=1$ we have
$$10\cdot 4x\equiv 10\cdot 1\pmod{13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$ By the Binomial expansion for integer powers,
$$ (A+B)^n = \sum_{k=0}^n {n\choose{k}} A^{n-k} B^{k}$$
(I'm assuming $A,B\ge 0$) and so we get the easy estimate $A^n + B^n \le (A+B)^n$ for any positive integer $n$.
Now what happens when we take ... | Alt. hint: $\;\displaystyle \left(\frac{A}{A+B}\right)^n + \left(\frac{B}{A+B}\right)^n \le \frac{A}{A+B}+\frac{B}{A+B}=1\,$ when $\,n \ge 1\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \c... | another solu tion
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt \frac{x^3+x}{x+1}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ x^2 - x +2 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ (x - 1/2)^2+7/4 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x- x +\frac12 )) = \frac {1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim _{x\to \infty \:}|\frac{x+1}{x-2}|^{\sqrt{x^2-4}}$ I need help solving the following tough limit , please.
$$\lim _{x\to \infty} \left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}$$
| Since eventually $x-2>0$ we have
$$\lim _{x\to \infty \:}\left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}=\lim _{x\to \infty \:}\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}$$
then we have
$$\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}=\left(\frac{x-2+3}{x-2}\right)^{\sqrt{x^2-4}}=\left[\left(1+\frac{3}{x-2}\right)^{\frac{x-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
p is prime. Show that there is at most one $(m,n) \in \mathbb Z^2$ with $0I've been trying to solve this question without great success.
I've found that the problem is correct for $p \equiv 1 \mod 4$.
Thank you,
| Let $p=2$, then $p=1^2+1^2$. So assume $p$ is an odd prime s.t. $p=a^2+b^2$, hence neither $a$ nor $b=0$. Now $a$ has a multiplicative inverse $c$ modulo $p$: $ac\equiv1\pmod{p}$.
$$(ac)^2+(bc)^2\equiv1+(bc)^2\equiv0\pmod{p}\implies (bc)^2\equiv-1\pmod{p}$$
and so $-1$ is a quadratic residue modulo $p$. Let $bc\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.
| Set
$$ I = \int \frac{\sin x}{\sin x - \cos x} dx = \int 1 + \frac{\cos x}{\sin x - \cos x} dx$$
Therefore:
$$ 2I = \int 1 + \frac{\sin x +\cos x}{\sin x - \cos x} dx $$
$$ 2I = x + \log(\sin x - \cos x) + C$$
$$ I = \frac{x}{2} + \frac{1}{2} \log(\sin x - \cos x) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
$\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
Clearly, we can see that $$T_r=\frac{n-r+1}{r(r+1)(r+2)}$$
Now, somehow, we have to make this telescoping.
But we do not have $n$ as a factor o... | We have that
$$T_r=\frac{n-r+1}{r(r+1)(r+2)}=\frac{n+1}{r(r+1)(r+2)}-\frac{r}{r(r+1)(r+2)}=$$$$=(n+1)\left(\frac{1}{2r}+\frac{1}{2(r+2)}-\frac{1}{r+1}\right)+\frac1{r+2}-\frac1{r+1}$$
and by telescoping we can see that the first term gives
$$(n+1)\left(\frac12+\frac14+\frac12\cdot 2\sum_{r=3}^n\left(\frac1r\right)+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction step in proof that $\binom{s}{s} + \binom{s + 1}{s} + \cdots + \binom{n}{s} = \binom{n + 1}{s + 1}$ Prove by induction that (binomial theorem)
$$
\binom{s}{s} + \binom{s+1}{s} + \dotsb + \binom{n}{s} = \binom{n+1}{s+1}
$$
for all $s$ and all $n>s$.
I used base case $s=0$, and I got my base case to work. Howev... | Alternatively, note that:
$${n\choose m}={n\choose n-m}; \ \text{so}:\\
\begin{align}&\binom{s}{s} + \binom{s+1}{s} + \binom{s+2}{s} +\binom{s+3}{s} +\dotsb + \ \ \ \binom{n}{s} \ \ \ \ = \binom{n+1}{s+1} \iff \\
&\color{red}{\binom{s}{0} + \binom{s+1}{1}} + \color{blue}{\binom{s+2}{2}} +\color{green}{\binom{s+3}{3}} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
| By definition of the limit
$$f(x)-3=\frac{x^3-1}{x-1}-3=x^2+x-2=(x-1)^2+3(x-1)$$
\begin{align}
|f(x)-3|
&= |(x-1)^2+3(x-1)| \\
&= |x-1|^2+3|x-1| \\
&<\delta^2+3\delta \\
&= \varepsilon
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Solve polynomial equation with real parameter Solve the equation $x^4-(2m+1)x^3+(m-1)x^2+(2m^2+1)x+m=0,$ where $m$ is a real parameter.
My work: So far I've been able to factor the polynomial to $(-x^2+x+m)(-x^2+2mx+1)=0$. Then after using the quadratic formula with each of the factors I'm here: $x=\frac{-1 \pm \sqrt{1... | Take the negatives out. It will look tidier.
So:
$$(x^2-x-m)(x^2-2mx-1)=0$$
Then:
$$x=\frac{1\pm\sqrt{1-4m}}{2}=\frac 12\pm\frac{\sqrt{1-4m}}{2}$$
And:
$$x=\frac{2m\pm\sqrt{4m^2+4}}{2}=m\pm\sqrt{m^2+1}$$
You can't do anything more here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } ... | You multiply by $1$ in a fancy fashion:
$$\sqrt{2-x^2}=\sqrt{2-x^2}\cdot 1=\sqrt{2-x^2}\cdot \frac{\sqrt{2-x^2}}{\sqrt{2-x^2}}=\frac{(\sqrt{2-x^2})^2}{\sqrt{2-x^2}}=\frac{2-x^2}{\sqrt{2-x^2}}$$
This of course only holds if $2-x^2\neq 0$, as otherwise also $\sqrt{2-x^2}=0$ and the last expression(and the intermediate on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How can I determine general formula of this sequence? I am trying to find general formula of the sequence $(x_n)$ defined by
$$x_1=1, \quad x_{n+1}=\dfrac{7x_n + 5}{x_n + 3}, \quad \forall n>1.$$
I tried
put $y_n = x_n + 3$, then $y_1=4$ and
$$\quad y_{n+1}=\dfrac{7(y_n-3) + 5}{y_n }=7 - \dfrac{16}{y_n}, \quad \forall... | The chracteristic equation of the given sequence is
$$y=\dfrac{7y+5}{y+3} \Leftrightarrow y_1 = 5 \lor y_2 = -1.$$
Let us consider the sequence
$$b_n = \dfrac{x_n-y_1}{x_n - y_2}=\dfrac{x_n-5}{x_n+1}.$$
We note that
$$b_{n+1}=\dfrac{x_{n+1}-5}{x_{n+1}+1}=\dfrac{\dfrac{7x_n+5}{x_n+3}-5}{\dfrac{7x_n+5}{x_n+3} + 1}=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
what would the value of determinant of a matrix be if a specific entry changed?
What will the value of determinant of matrix $A=\pmatrix{1&3&4\\5&2&a\\6&-2&3}$ be change if we change $a$ to $a+2$.
This is an easy problem because $|A|=-127+20a$ and if we did that changing, we would get $-87+20a$. My question here is:
... | The property of the matrix (in general, for any size $n\times n$):
$$\begin{vmatrix}
a+b&c+d \\ e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ e&f
\end{vmatrix}+\begin{vmatrix}
b&d \\ e&f
\end{vmatrix};\\
\begin{vmatrix}
a+b&c \\ d+e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ d&f
\end{vmatrix}+\begin{vmatrix}
b&c \\ e&f
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\int \frac{x^4+1}{x^{12}-1} dx$ So I found this problem:
Calculate $$\int \frac{x^4+1}{x^{12}-1} dx$$ where $x\in(1, +\infty)$
and I don't have any ideea how to solve it. I tried to write $$x^{12}-1=(x^6+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)$$ but with $x^4+1$ I had no idea what to do, an obvious thing would be t... | You can use that $$x^4+1=(x^2+1)^2-2x^2=…$$
You will get $$\frac{x^4+1}{x^{12}-1}=\frac{-2 x-1}{12 \left(x^2+x+1\right)}-\frac{1}{3 \left(x^2+1\right)}+\frac{2 x-1}{12
\left(x^2-x+1\right)}+\frac{-x^2-1}{6 \left(x^4-x^2+1\right)}+\frac{1}{6
(x-1)}-\frac{1}{6 (x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous... | The Euclidean algorithm steps are
\begin{align*}
x^3+1&=x (x^2+1) + (-x+1)\\
x^2+1&=-x (-x+1)+ (x+1)\\
-x+1&=-(x+1)+2
\end{align*}
So the GCD is $1$.
We can go back up
\begin{align*}
2 &= (-x+1) +(x+1)\\
&= (-x+1) + (x^2+1)+x (-x+1)\\
&= (-x+1) (x+1)+(x^2+1)\\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\\
&= (x+1) (x^3+1) + (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum
$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$$
I have tried to make this series in the form $1 + n x + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (... | Let $a_n=\frac{1}{2}\prod_{k=1}^{n}(3k-1)$ and $b_n=\prod_{k=1}^{n}(6k)$. Then
$$ \frac{a_n}{b_n} = \frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\Gamma(n+1)\Gamma\left(\frac{2}{3}\right)}=\frac{\pi}{2^n \sqrt{3}} B\left(n+\frac{2}{3},\frac{1}{3}\right)=\frac{\pi}{\sqrt{3}}\int_{0}^{1}\frac{x^{n-1/3}(1-x)^{-2/3}}{2^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)... | Alternatively:
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{(1-(1-x))(1 + (1 - x) + (1 - x)^2)}{1-(1-x)} + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{1^3-(1-x)^3}{x} + \frac{(1 - x)^3}{x},$$
where it was used:
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prob. 7, Sec. 6.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: A necessary and sufficient condition for the existence of $\big(|f|\big)^\prime$ Here is Prob. 7, Sec. 6.1, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Suppose that $f \colon \mathbb{R} \to \mathbb{R... | Of course everything you did is correct. But such a simple statement deserves a shorter proof, which then also shows what's going on here.
Assume that $c=0$, $f(0)=0$, and $f'(0)=a$. Then
$$m(x):={f(x)\over x}\to a \quad (x\to0)\ ,$$
and therefore
$${\bigl|f(x)\bigr|\over x}={\rm sgn}(x)\,\bigl|m(x)\bigr|\to\left\{\eqa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... |
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
No, since $\frac10$ is not defined.
*
*Does $\infty\cdot\frac{1}{2}$ equal to $\infty$?
There are the "extended real numbers" $\mathbb{R}\cup\{\pm\infty\}$
Where you can define this.
Note, that $\infty$ is in general not a number. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
} |
Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is ... | In that case a possible trick to use induction is to prove the stronger condition
$$\frac{1}{2} + \frac{2}{2^2} + \ldots +\frac{n-1}{2^{n-1}} + \frac{n}{2^{n}} <2-\frac{n+1}{2^{n-1}}<2$$
and the induction step becomes
$$\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} + \frac{n+1}{2^{n+1}}\stackrel{Ind. Hyp.}<2-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Closed form solution for logarithmic inequality We have the following inequality that was very hard to be solved in a closed form. Yet, someone solved it this way, and I can't get to fully understand what is done.
eq: $\sqrt{x-2\sqrt{x-1}} + \sqrt{x+2\sqrt{x-1}} +log_2(x-1)=0$
Solution:
$x\in Df \iff x \geq 1$
We have... | Suppose you have, instead,
$$
\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}+\log_2x=0
$$
Then the substitution with the absolute values would be sensible, but there's a better method.
Let $y=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}$; then $y>0$ and
$$
y^2=2x+2\sqrt{x^2-4(x-1)}=2x+2\sqrt{(x-2)^2}
$$
However, $\log_2x$ shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The number of positive integral solutions of $abc =$ $30$ is
The number of positive integral solutions of $abc =$ $30$ is
My attempt
Factors of $30$ are $3 \cdot 5 \cdot 2$.
Therefore $a$ will have three choices, similarly $b$ & $c$ will have $2$ choices & $1$ choice.
So,it's answer must be $3 \cdot 2 \cdot 1$
But an... | The texbook is right.
Start with $1\cdot1\cdot1$ and assign $2$, say to $b$ and get $1\cdot2\cdot1$.
Now assign $3$, say to $a$ and get $3\cdot2\cdot1$.
Then assign $5$ to $a$ again, giving $15\cdot2\cdot1$.
On every stage, you have three choices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$
Calculate $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$
First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t... | Long method
One possible method is the following. Using
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^n \, x^n}{4^n} &= \frac{1}{\sqrt{1 + x}} \\
\gamma(s,x) &= \int_{0}^{x} e^{-t} \, t^{s-1} \, dt
\end{align}
then:
\begin{align}
I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\
&= \int_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1... | here is another approach:
Consider a set with $n$ objects. There are $\binom nk$ ways to choose $k$ of them. That way gives $\sum_{i=0}^n\binom ni$ ways to choose objects from the set. On the other hand, we can choose to choose each item or not.That gives $2^n$. And of course they are equal, since they both represent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Convergence of $\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$? $a_n=\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$
I think that $\sqrt[n]{1^n+2^n+...+(2n)^n}\rightarrow 2n+1$
So $a_n=\frac{2n+1}{2n}\rightarrow 1$
Any ideas if that's correct? And if so how do I prove it?
| We have
\begin{align}
a_n=&
\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}
\\
=&\frac{1}{2n}\sqrt[n]{(2n)^n\left[\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1\right]}
\\
=&\sqrt[n]{\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1}
\\
=&\sqrt[n]{\left(\frac{1}{2n}\right)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
He... | I would rather approach on a very different way...I want to solve for y.
So,here we are given,
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
Now,assume that $g$ is our variable and $x$ and $y$ are constant.then equalling the coefficients we can say that,
$$\dfrac{y}{x}=\dfrac{y}{y^2} ~~and~~\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sort those 3 logarithmic values without using calculator I found this problem interesting, namely we are given three values: $$\log_{1/3}{27}, \log_{1/5}{4}, \log_{1/2}{5}$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
Fi... | For clarity, we take the inverses of the bases, which will reverse the order ($\log_{1/x}y=-\log_x y$).
We have
$$\log_3{27}=3,$$
$$\log_54<1,$$
because $4<5^1$ and
$$1<\log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$\log_54<1<\log_25<3=\log_327$$
and
$$\log_{1/5}4>\log_{1/2}5>\log_{1/3}27.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Modular Multiplicative Inverses I'm in a cryptography and struggling with understanding how the Euclidean Algorithm necessarily works for finding multiplicative inverses. We haven't actually covered the algorithm yet, just a way to brute force finding the inverses but I like working with an algorithm instead of just gu... | You are right indeed by Euclidean algorithm we have
*
*$13=5\cdot 2+\color{red}3$
*$5=\color{red}3\cdot 1+\color{red}2$
*$3=\color{red}2\cdot 1+1$
and therefore starting from the last one
$$1=3-2=3-(5-3)=2\cdot 13-2\cdot 5\cdot 2-5 \implies 2\cdot 13-5\cdot 5=1 $$$$\implies -5\cdot 5=1-2\cdot 13$$
and then
$$5x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Induction proof of a series Suppose we have the series
$$f(n) = \sum_{i=1}^n \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}.$$
As a hint it was said that this series "telescopes". I observed the pattern to be $f(n)=\frac{n^2}{n^2 +1}$. I wish to prove this via induction. The base case holds, since $f(1)=\frac{1}{2}$ which... | For completeness sake I want to finish my own proof, without telescoping and simply by induction as I intended to:
Now notice that:
$$(k^2+1) (k^2 +2k +2) =k^4 +2k^3 +3k^2+2k +2$$
We can thus write:
$$f(k+1)=\frac{k^2}{k^2 +1}+ \frac{2k+1}{(k^2+1)(k^2+2k+2)} = \\ \frac{k^2 (k^2+2k+2)+ (2k+1)}{(k^2+1)(k^2+2k+2)}=\frac{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the ... | Hint: $$\frac{1}{\sqrt{a^3+4}} \le \frac{1}{\sqrt{x^3+4}} \le \frac{1}{\sqrt{0^3+4}}$$
for $0 \le x \le a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by insp... | Let $\,a=\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}=\dfrac{1+i}{\sqrt{2}}\,$ and $\,b=\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}=\dfrac{-1+i\sqrt{3}}{2}\,$, so:
$$
ab = \dfrac{(1+i)(-1+i\sqrt{3})}{2 \sqrt{2}}=\dfrac{\left(-1-\sqrt{3}\right)+i\left(-1+\sqrt{3}\right)}{2 \sqrt{2}}
$$
Then $\,\cos\left(\dfrac{\pi}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Implication of equations Just an interesting question I saw online. :)
Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be ... | If $a,b,c$ are all positive then we have by Am-Gm:
$$\dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} \geq 3\sqrt[3]{\dfrac{ab}{c^2} \cdot \dfrac{bc}{a^2} \cdot \dfrac{ca}{b^2}} = 3$$
With eqaulity iff $ \dfrac{ab}{c^2} = \dfrac{bc}{a^2} = \dfrac{ca}{b^2}$ which is iff $a=b=c$ so the value of expression is $6$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On the eigenvalue of a particular kind of matrix Let $x+y=1$. Consider the matrix $A=\begin{pmatrix} x^5+y^5&5x^5&10x^5&10x^5&5x^5\\5y^5&x^5+y^5 &5x^5&10x^5&10x^5 \\10y^5&5y^5&x^5+y^5&5x^5&10x^5\\10y^5&10y^5&5y^5&x^5+y^5&5x^5\\5y^5&10y^5&10y^5&5y^5&x^5+y^5 \end{pmatrix} $
Is $1$ an eigenvalue of $A$ ?
It is not obviou... | If $x = \frac{g}{g+h}$ and $y = \frac{h}{g+h},$ then
$$
\left(
\begin{array}{c}
g^4 \\
g^3 h \\
g^2 h^2 \\
g h^3 \\
h^4
\end{array}
\right)
$$
is an eigenvector with eigenvalue $1.$ So is the multiple
$$
\left(
\begin{array}{c}
x^4 \\
x^3 y \\
x^2 y^2 \\
x y^3 \\
y^4
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's Find limit of
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$
I started by defining $\ x = y + \frac{\pi}{4} $
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$
T... | As an alternative without derivatives, let $x=\frac{\pi}{4}-y$ then
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4}=\lim_{y \to 0} \frac{\tan \left(\frac{\pi}{4}-y\right)-1}{-y}=\lim_{y \to 0} \frac{1-\frac{\cos \left(2y\right)}{1+\sin \left(2y\right)}}{y}=\lim_{y \to 0} \frac{1-\cos \left(2y\right)+\sin \left(2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are... If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are...
I've been trying to find the limit as usual:
$\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}... | $\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}$ so $b$ must be 1 and $c$ must be $1/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\i... | $$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin^2(u)}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$
$$\int \frac{-(1 - \cos(u))(1 + \cos(u)) }{1-\cos(u)}du=-\int (1 + \cos(u)) du=-u-\sin u+C$$
Can you replace the value of $u$ by using $\cos u =x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Calculate $\int_0^1{x·\lceil1/x\rceil dx}$ I am trying to calculate following integral:
$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$
I tried usual change t=1/x but not able to further advance.
Thanks!
| $$
\begin{aligned}
\int_0^1x\cdot\left\lceil \frac{1}{x}\right\rceil\; dx
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
\underbrace{\left\lceil \frac{1}{x}\right\rceil}_{\in (k,k+1)\text{ a.e}} \;dx
\\
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
(k+1)\;dx
\\
&=
\sum_{k\ge 1}
(k+1)
\left[\ \frac 12 x^2\ \right]_{1/(k+1)}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Simplify $\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$
Simplify
$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$$
Found in a book with tag "Moscow 1982", the stated answer is $1+\sqrt[4]{5}$. Used all tricks that I know but without success. The answer appears to be correct, checked in Wolfram ... | $$\begin{align}\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125})(\sqrt{1+\sqrt[4]{5}})}}\cdots \cdots(1)\\&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{5^3})+(4\cdot \sqrt[4]{5}-3\sqrt{5}+2\sqrt[4]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime
Let $p>3$ be a prime number. How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$?
When $p=3$, you also have $p+1$ as a divisor of $\left(p^2... |
Proposition. For a positive integer $p$, $\left(p^2+1\right)^2$ is divisible by some positive integer $d\equiv 1\pmod{p}$ such that $$d\notin\left\{1,p^2+1,\left(p^2+1\right)^2\right\}$$ if and only if
$$p=t_s(b)\text{ for some integers }b\geq 1\text{ and }s\geq 2\,,$$
where
$$t_j(b):=\frac{\left(\frac{\left(b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Understanding how $x^2+2x+y^2-6y-15=0$ becomes $(x+1)^2+(y-3)^2=25$
I am looking at how this equation was simplified, $15$ was pushed to the other side and $10$ I believe was added by adding $1$ to the first bracket equation and $9$ to the other part.
The next part where to get $x² + 2x + 1$ to $(x + 1)^2$ there $2x$ ... | Yes, you’re right. The equation was simplified by completing the square. Remember that $(m+n)^2 = m^2+2mn+n^2$ and vice versa.
$$x^2+\frac{b}{a}x = -c \implies x^2+\biggr(\frac{b}{2a}\biggr)^2+\frac{b}{a}x = -c+\biggr(\frac{b}{2a}\biggr)^2$$
$$\implies \biggr(x+\frac{b}{2a}\biggr)^2 = -c+\biggr(\frac{b}{2a}\biggr)^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Showing that the inequality $\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}$ stands.
If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that
$$\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}.$$
I have already proven the second inequa... | $\sqrt{t+1}$ is approximately constant on $(0,1)$, hence the following application of Cauchy-Schwarz
$$ I_n=\int_{0}^{1}t^n\sqrt{1+t}\,dt\leq \sqrt{\int_{0}^{1}t^{n}\,dt\int_{0}^{1}t^n(1+t)\,dt}=\color{red}{\frac{1}{n+1}\sqrt{\frac{2n+3}{n+2}}}<\frac{\sqrt{2}}{n+1}$$
is expected to provide a tight upper bound. Indeed,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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A proof on multinomial roots If $x_1,x_2,...,x_{n-1},x_n$ be the roots of the equation $$1 + x + x^2 + ... + x^n = 0$$ and $y_1,y_2,...,y_{n},y_{n+1}$ be those of equation $$1 + x + x^2 + ... + x^{n+1} = 0$$ show that $$(1-x_1)(1-x_2)...(1-x_n)=2\left[\frac{1}{1-y_1}+\frac{1}{1-y_2}+...+\frac{1}{1-y_{n+1}}\right]$$
I c... | The hint:
$$x^{n+1}-1=(x-1)(x-x_1)(x-x_2)...(x-x_n),$$ which gives
$$x^n+x^{n-1}+...+x+1=(x-x_1)(x-x_2)...(x-x_n),$$ which for $x=1$ gives
$$n+1=(1-x_1)(1-x_2)...(1-x_n).$$
By the similar way you can calculate the second expression.
Indeed, $\frac{1}{y_1-1}$, $\frac{1}{y_2-1}$,...,$\frac{1}{y_{n+1}-1}$ they are roots o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$ So I got this problem:
Determine the following limit value:
$$\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$$
What I tried is:
$\large{\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}\cdot{\sqrt{x^2+16}+4\over\sq... | Hint:
$$f(a)=\lim_{x\to0}\dfrac{\sqrt{x^2+a^2}-a}{x^2}=\lim...\dfrac{x^2+a^2-a^2}{x^2(\sqrt{x^2+a^2}+a)}=\dfrac1{\sqrt{a^2}+a}$$
For $a>0,$ $$f(a)=\dfrac1{2a}$$
$$\dfrac{f(1)}{f(4)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ Accordin... | $$z=\frac{3+4i}5=\frac{(1+2i)^2}5=\frac{1+2i}{1-2i}.$$
The ring $\Bbb Z[i]$ is a UFD and $1+2i$ and $1-2i$ are non-associate
primes therein. So in $\Bbb Z[i]$, $(1+2i)^n$ is never divisible by $1-2i$
so that
$$z^n=\frac{(1+2i)^n}{(1-2i)^n}$$
is always in lowest terms and cannot cancel to equal $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
What does $\lim \inf _ { n \rightarrow \infty }$ mean? I'm new into Mathematical Analysis, and my textbook says:
Consider the series: $\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac {
1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 2 ^ { 3
} } + \frac { 1 } { 3 ^ { 3 } } + \frac { 1 } { 2 ^ { 4 } } + \f... | You can refer to your notes or to the wiki page Limit superior and limit inferior for the definition.
It is just a generalization for the limit of a sequence, in particular we consider
$$I_n=\inf\{a_n:k\ge n\}$$
and since $I_n$ is an increasing sequence the limit always exist (finite or infinite) and then we can consid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4 + y^2}$ Prove or disprove each of the follow function has limits $x \to a$ by the definition
$$\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4 + y^2}$$
Let $y = mx^2$
$$\frac{x^2y}{x^4 + y^2}=\frac{x^4m}{... | Yes that correct, indeed we can consider for example
*
*for $x=0,\, y\neq 0 \implies \frac{x^2y}{x^4 + y^2}=0$
*for $x=t,\,y=t^2, \,t\to 0 \implies \frac{x^2y}{x^4 + y^2}=\frac{t^4}{t^4 + t^4}\to \frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $4\alpha^2–5\beta^2+6\alpha+1=0$.Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. If $4\alpha^2–5\beta^2+6\alpha+1=0$. Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. I tried to solve this question by taking a Gene... | We have $\beta(\alpha)=\pm\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}$ so you want to find the envelope of the family $f_\alpha(x,y)=x\alpha+y\beta(\alpha)+1=0$.
In other words, $F(x,y,\alpha):=x\alpha\pm y\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}+1=0$, $\dfrac{\partial}{\partial\alpha}F(x,y,\alpha)=0$,
i.e.,
$$
\left\{
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac... | Let $$f(x)=\sum^\infty_{k=1} \frac{(-1)^k}{k+1}x^k$$
Then, $G(x)=e^f$.
To expand $G$ in Taylor series up to the cube term, we have to compute up to the third derivative of $G$.
This isn’t too difficult.
$$G(0)=1$$
$$G’(0)=f’(0)=-\frac12$$
$$G’’(0)=f’’(0)+f’^2(0)=2\cdot \frac13+\frac14$$
$$G’’’(0)=f’’’(0)+2f’(0)f’’(0)+f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{... | $$x_n=\frac{(n^3+1)^2-(n^2-1)^3}{\sqrt[3]{(n^3+1)^6}+\sqrt[3]{(n^3+1)^5}\sqrt{n^2-1}+...}\rightarrow0,$$
which says that $\{x_n\}$ is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How to find the average value of $f(x,y)$ on $R=\{(x,y): -1\le x\le1,0\le y\le3\}$ Given $R=\{(x,y): -1\le x\le1,0\le y\le3\}$
For the function $f(x,y)$ evaluate $\int\int_R f(x,y)dA$ and find the average value of $f(x,y$ on $R$
$f(x,y)=x^3+xy^3-3y$
My Try:
$=\int_{-1}^{1}\int_0^3x^3+xy^3-3ydydx$
$=\int_{-1}^{1}(x^3y+\... | So, not quite!
First, we need to know how exactly the average value of a two variable function $f(x,y)$ is computed. Typically, the following equation is utilized to compute the average value of a function $f(x,y)$ over a rectangle $R$:
$$f_{ave} = \frac{1}{\text{Area}(R)} \iint_R f(x,y) dxdy$$
Since we're given a rect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R... | Supposing $$(x-1)(x-2)...(x-(p-1))\equiv x^{p-1}\pm1\pmod{p}$$
The left-hand side is $$(x^2-1^2)(x^2-2^2)...(x^2-({p-1\over2})^2)\pmod p$$
Replace $x^2$ by $y$; and then replace $y$ by $-1$. The left-hand side is now a square-root of your product $\pmod p$, give or take a sign. The right-hand side is $1\pm1$.
Can you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proving the given inequalities Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used... | First of all, you should have $\geq $ or $\leq$ instead of $>$ and $<$ in your inequalities. In each of them, the equality occurs if and only if $a=b=c$.
(i) Use the Weighted AM-GM Inequality to show that
$$\frac{bc+ca+ab}{a+b+c}=\frac{b}{a+b+c}c+\frac{c}{a+b+c}a+\frac{a}{a+b+c}b\geq c^{\frac{b}{a+b+c}}a^{\frac{c}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove following inequality Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$
What I tried was to use AM-GM for the left side of this inequality, what I got was
$3(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 3$ and $(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 1$ or ju... | Hint:
WLOG you may set $a+b+c=3$ and show instead that for $x\in[0,3]$,
$$\left(\frac{2x}{3-x}\right)^{2/3}\geqslant x$$
But this is equivalent to the obvious $x^2(4-x)(x-1)^2\geqslant0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Question about the dimension of elements in quotient ring Here is the question:
Let $\def\Z{\mathbb Z}\Z[X]$ be the polynomial ring over $\Z$ with variable $X$. Let $f(X) = X^3+X^2+2X+2$. Let $I$ be the ideal of $\Z[X]$ generated by $f(x)$ and $5$. Put $A =\Z[X]/I$.
*
*Find the number of elements $a$ in $A$ such tha... | Note that $X^3+X^2+2X+2 = (X+1)(X^2+2)$. So we have, by the Chinese reminder theorem:
$$
\Bbb Z_5[X]/(X^3+X^2+2X+2) \cong \Bbb Z_5[X]/(X+1)\times \Bbb Z_5[X]/(X^2+2)
$$
We can find the number of solutions to $a^2 = 1$ and $b^{18} = 1$ in each of those separately. This is helpful because each of those two factors in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Assuming that $|z|<1$, calculate: $\sum_{n=2}^{\infty}(n^2-3n+2){z^{n-1}}$ The answer is: $\frac{-2z^2}{(1-z)^{3}}$
When I do it, I get the answer without the minus sign.
So far, I got:
$$
\begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^n
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^n \\
&= z^3 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \... | \begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^{n-1}
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\
&= z^2 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-3} \\
&= z^2 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}\\
&= z^2 \frac{d^2}{dz^2} \sum_{n=0}^{\infty}z^{n}\\
&= z^2 \frac{d^2}{dz^2} \frac{1}{1-z}\\
&= z^2 \frac{2}{(1-z)^3}
\end{split... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Irrational equation
Solve over the real numbers: $$(x^2+x+1)^{1/3}+(2x+1)^{1/2}=2$$
I know for the second radical to be defined $x≥-0,5$ and I've attempted various methods I've solved other such equations with but to no avail; if I could write $x^2+x-7$ in terms of $2x+1$ to use a convenient notation in $x^2+x-7-6(... | If we put $$a =(x^2+x+1)^{1/3}\;\;\;{\rm and}\;\;\;b= (2x+1)^{1/2}$$
So $$a^3= x^2+x+1 \;\;\;{\rm and}\;\;\;b^2= 2x+1\;\;\;{\rm and}\;\;\; a+b=2$$
and thus $$4a^3 = 4x^2+4x+4 = (2x+1)^2+3 = b^4+3$$
and finally $$ 4(2-b)^3 = b^4+3$$ ...
$$ b^4+4b^3-24b^2+48b-29=0$$
$$ (b-1)(b^3+ \underbrace{ 5b^2-19b+29}_{f(b)})=0$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Congruence with Lucas sequence I want to show that for each $n \geq 2$ the following congruence holds,
$$L_{2^n} \equiv 7 \pmod{10}.$$
According to my notes,
$$L_n=\left( \frac{1+\sqrt{5}}{2}\right)^n+\left( \frac{1-\sqrt{5}}{2}\right)^n$$
and
$$L_n=F_{n-1}+F_{n+1},$$
where $F_n$ is the $n$-th Fibonacci number.
Do we u... | The following helps :
$$L_{2n}=L_n^2-2(-1)^n\tag1$$
(the proof is written at the end of the answer.)
This is a proof by induction.
For $n=2$, $L_{2^2}=7\equiv 7\pmod{10}.$
Suppose that $L_{2^n}=10k+7$ for some $k\in\mathbb N$.
Then, using $(1)$, we get
$$\begin{align}L_{2^{n+1}}&=L_{2^n}^2-2(-1)^{2^n}
\\\\&=(10k+7)^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the limit (Squeeze Theorem?) I have to calculate the limit of this formula as $n\to \infty$.
$$a_n = \frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$\frac{1}{\sqrt{2}}\leftarrow\frac{n}{\sqrt{2n^2}}\le\frac{1}{\sqrt{... | for a decreasing function such as $1/\sqrt x$ with $x$ positive, a simple picture shows
$$ \int_a^{b+1} \; f(x) \; dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b} \; f(x) \; dx $$
$$ \int_{n+1}^{2n+1} \; \frac{1}{\sqrt x} \; dx < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < \int_{n}^{2n} \; \frac{1}{\sqrt x} \; dx $$
getting ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+k^2 \sin^2 \theta}} \,d \theta$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+k^2 \sin^2\theta}}
\,d \theta$
I wang to let $k=-ai \,\,\,\,\,$ ,then :$$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-a^2 \sin^2\theta}}
\,d \theta$$
But I am not sure whether the elliptic... | The following manipulations assume $k^2<1$. Since for any $x\in(-1,1)$ we have
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^n \tag{1}$$
and the central binomial coefficients also appear in
$$ \int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n}\tag{2} $$
by Maclaurin se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can pro... | Hint:
Numerator:
$a^3-b^3= (a-b)(a^2+ab+b^2)$.
$a=(x+5)^{1/3}$; $b=2.$
Denominator:
$c^4-d^4=(c^2-d^2)(c^2+d^2)=$
$(c-d)(c+d)(c^2+d^2);$
$c=(x-2)^{1/4};$ $d=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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How to calculate $\sum_{n=1}^{\infty}A_n(1/2)^{2n+2}(n+2)/(2n+2)$ This is a follow-up to a previous problem, which was answered as follows:
$$S=\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\frac{\pi}{4},$$
where $C_n$ is the $n$th Catalan number.
A related problem has arisen ... | Assuming that $A_n$ is defined as $\frac{3(2n)!}{(n+\color{red}{2})!(n-1)!}$ (otherwise the associated series is divergent, since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$) we have that $T$ equals
$$ \frac{1}{2}+\frac{3}{8}\sum_{n\geq 1}\frac{1}{4^n}\binom{2n}{n}\frac{n}{(n+1)^2} $$
which due to $\sum_{n\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\bi... | $$\sin(x)(1+\sin^2(x))=1-\sin^2(x)$$
Square both sides & replace $\sin^2(x)$ with $1-\cos^2(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Calculate the limit $\lim \limits_{x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although $x\neq 2$ I should calculate the Limit $\lim \limits_ {x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although I noticed, that $x\neq 2$ must apply. Is the limit undefined? Otherwise, with which steps should I go on to calculat... | Given $$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{\color{red}{(x-2)}(x+4)}{x\color{red}{(x-2)}}=\lim_{x\rightarrow2}\dfrac{x+4}{x}=\lim_{x\rightarrow2}1+\dfrac4x = 3$$
OR
You could also use L'Hopital's rule
$$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{2x+2}{2x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2} \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.
| Let $x = \tan\theta$ and make use of
\begin{align}
1 + \tan^{2}\theta &= \frac{1}{\cos^{2}\theta} \\
dx &= \frac{d\theta}{\cos^2\theta} \\
\sin(\tan^{-1}\theta) &= \frac{x}{\sqrt{1+x^2}} \\
\cos(\tan^{-1}\theta) &= \frac{1}{\sqrt{1 + x^2}}
\end{align}
to obtain:
\begin{align}
I &= \int \frac{dx}{(1+x^2)^2} \\
&= \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$
Prove convergence of the following sequence: $$\frac{n+3}{n^2 -3} \rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $\frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value s... | Yup, the proof is correct.
Also, "we write $n-3>\color{blue}{\frac1{\epsilon}}$"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Integral $\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$ I was playing around with some integrals and noticed that some integrals of the form: $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form.
I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is probl... | The simplest approaches are sometimes the right ones. Therefore choose any arbitrarily numbers $a,b$ and $c$ and apply your algorithm. So by firstly setting $t=x-a$ we get
$$\begin{align}
I(a,b,c)=\int_a^{\infty}\frac{\arctan(x+b)}{x^2+c}dx=\int_0^{\infty}\frac{\arctan(t+a+b)}{t^2+2at+(c+a^2)}dt
\end{align}$$
Now set ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Evaluate the integral $\int \frac{x^2 + 1}{x^4 + x^2 +1} dx$ Evaluate the integral $\int \frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
| It turns out that we must compute the integral in the following way:
$$I=\int\frac{x^2+1}{x^4+x^2+1}\mathrm{d}x$$
$$I=\int\frac{x^2+1}{(x^2-x+1)(x^2+x+1)}\mathrm{d}x$$
After a fraction decomposition,
$$I=\frac12\int\frac{\mathrm{d}x}{x^2+x+1}+\frac12\int\frac{\mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=\int\frac{\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding a formula of a power of a matrix Part of a solution I came across of calculating the following matrix:
$$\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to ... | Idea
$$
\\\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{pmatrix}=\begin{pmatrix}1 & -1\\
1 & 1
\end{pmatrix}\cdot\frac{\sqrt{2}}{2}
\\\begin{pmatrix}1 & -1\\
1 & 1
\end{pmatrix}^n\cdot\Big(\frac{1}{\sqrt{2}}\Big)^n=\begin{pmatrix}-i & i\\
1 & 1
\end{pmatrix}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to evaluate $\int_{0}^{2\pi}\frac{1}{1+\sin^{2}(\theta)}d\theta$ I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = \int_{0}^{2\pi}\frac{1}{1+\sin^{2}(\theta)}d\theta$$
We take $C = \{z \in \mathbb{C} : |z| = 1\}$, which is positively oriented... | How about using Weierstrauss substitution.
$$I=\int_0^{2\pi}\frac{1}{1+\sin^2(x)}dx$$
$t=\tan\frac{x}{2}$ so it becomes:
$$I=4\int_0^\infty\frac{1}{1+\left(\frac{2t}{1+t^2}\right)^2}\frac{dt}{1+t^2}=4\int_0^\infty\frac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4\int_0^\infty\frac{t^2+1}{t^4+6t^2+1}dt=4\int_0^\infty\frac{t^2+1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to proceed with this integral?
Let $ G:= \left\{ (x,y) \in \mathbb{R}^2 : 0 < y,\: x^2 + \frac{y^2}{9} <1\: ,\: x^2+y^2 > 1 \right\} $.
I want to calculate this integral:
$ \displaystyle\int_G x^2\,dxdy $.
I want to try with polar coordinates:
so I set $ (x,y) = (r\cos\phi,r\sin\phi)$, but
I am not sure how to ge... | Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.
Since $y>0$, we are integrating over the first two quadrants, so that $0 \leq \phi \leq \pi$.
Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
$$ r^2\cos^2\theta + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a point lies on a ellipse
Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $\frac{x... | Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y... | With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\\
y=0.4377195786259503\\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\\
2033z^4-19696z^3-84456z^2+176704z-46464=0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \co... | $$
\cos(a+b)=\cos a\cos b-\sin a\sin b
$$
and hence
$$
\cos\left[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}\right]=\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\cos\left(\frac{\pi}{3}\right)-\sin\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\sin\left(\frac{\pi}{3}\right)\\=\frac{3}{5}\cdot\frac{1}{2}-\sqrt{1-\frac{3^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $\int\frac{2x-3}{(x^2+x+1)^2}dx$ $\int\frac{2x-3}{(x^2+x+1)^2}dx$
$\int\frac{2x-3}{(x^2+x+1)^2}dx=\int\frac{2x+1}{(x^2+x+1)^2}dx-\int\frac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
| $$\dfrac{d\left(\dfrac{ax^2+bx+c}{x^2+x+1}\right)}{dx}=\dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0\iff a=b$
$b-c=-3\iff c=b+3$
$2(a+c)=2\iff1=a+c=b+b+3\iff b=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \fr... | In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $\dfrac2x -\dfrac5{\sqrt{x}}=1$ as
$$2 \left(\dfrac{1}{\sqrt x}\right)^2 - 5\left(\dfrac{1}{\sqrt x}\right)-1 = 0$$
So $$\dfrac{1}{\sqrt x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find the derivative of $y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$
What is the derivative of
$$y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$$
What I did here was:
$$\begin{align}y'&=({1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x)'\\&={1\over2}\cdot4\tan^3x(\tan x)'-{1\over2}\cdot2\tan{x}(\tan x)'-{1\over\cos x}... | You made a mistake, the first denominator should be $\cos^2 x \times \cos^3 x = \cos^5 x$ and you should be able to do more factoring then
UPDATE
Note after fixing sine power mistake in the last step you have
$$
\cos^4 x - \cos^2 x + 2\sin ^2 x
= \cos^4 x - 3\cos^2 x + 2
= \left(\cos^2 x - 1\right)\left(\cos^2 x - 2\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and
$$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$
Find the smallest possible value of
$$a^2 + b^2 + c^2+ d^2$$
I thought of adding $7$ to the fi... | Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$
$$2ac+2ad+2bc+2bd+2ab+2cd \leq 3a^2+3b^2+3c^2+3d^2$$
So $$a^2+b^2+c^2+d^2\geq 154$$
But, unfortenuly this is not the minumum, since variables can not be all the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}... | Observe that if $y < 0$ then $$\sqrt{y^2} = |y| = -y.$$
Take $y=-3$ for example. In your case, $y = x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove this algebra question? If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1$$
| $$x^2+y^2+z^2=x(x+1)\implies\dfrac x{x+1}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A nice relationship between $\zeta$, $\pi$ and $e$ I just happened to see this equation today, any suggestions on how to prove it?
$$\sum_{n=1}^\infty{\frac{\zeta(2n)}{n(2n+1)4^n}}=\log{\frac{\pi}{e}}$$
| Since
\begin{align*}
\zeta(2n) &= \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!},
\end{align*}
for integers $n > 0$, the given sum $S$ is
\begin{align*}
S &= \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n}
= \sum_{n=1}^\infty (-1)^{n+1} \frac{B_{2n}}{(2n)(2n+1)}\frac{\pi^{2n}}{(2n)!}
\end{align*}
But
\begin{align*}
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Trying to prove an equation I would like to receive some help about the next problem.
The problem:
I'm trying to prove the next equation:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = 0 \quad, n = 1, 2, ...$$
My work until now:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = \sum_{k = 0}^{n} \frac{1}{(-1)^{k}k!(n... | HINT
Actually you were thinking a bit to complicated. Therefore consider the series expansion of $(1-x)^n$ which is due the Binomial Theorem given by
$$\sum_{k=0}^n\binom nk (-x)^k(1)^{n-k}=\sum_{k=0}^n\binom nk (-x)^k$$
Further notice that the binomial coefficient can be written as
$$\binom nk = \frac{n!}{k!(n-k)!}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.