Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Determine appropriate $c$ and $x_0$ for Big-O proofs. "Prove that $f(x)$ is $O(x^2)$:"
$$f(x) = \frac{x^4+2x-7}{2x^2-x-1}$$
Let $c=10$ (addition of coefficients of the numerator less the addition of coefficients of the denominator), and $x_0 = 1$ (the lowest coefficient among all elements in the function).
$$\frac{x^4+... | I will try to give you a more intuitive approach.
Theoretical Approach
The basic idea is the following: when $x$ get sufficiently large, the highest order term dominates over all the rest.
$x^4 + 2x - 7$ will eventually look like $x^4$.
$2x^2 -x - 1$ will eventually look like $2x^2$.
So
$\dfrac{x^4 + 2x - 7}{2x^2-x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Tips for solving linear equations Is there an easier way to solve linear equations than trial and error?
$$2(3b -1) - (4b -6) = 3$$
| To solve $2(3b - 1) -(4b -6) =3$ for the unknown $ b $, it is best to simplify the equation's left side. To do that:
*
*Distribute: $2(3b - 1) -(4b -6) = (2 \cdot 3b - 2 \cdot 1) - 1 \cdot 4b - 1 \cdot {-6} = 6b - 2 - 4b + 6$
*Combine like terms: $ 6b - 2 - 4b + 6 = 6b - 4b - 2 + 6 = 2b + 4 $
Thus, $$ 2(3b - 1) -(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Coefficient of $x^4$ in multinomial expansion What is the coefficient of $x^4$ in $(1 + x - 2x^2)^7$? What is a quick way to solve this problem using the binomial theorem (I have not learned multinomial theorem)?
| By factorizing the inner expresion,
$$(1+x-2x^{2})^{7}=((1+2x)(1-x))^{7}=(1+2x)^{7}(1-x)^{7}$$
Using the Binomial Expansion on both, we get:
$$\begin{align}(1+x-2x^{2})^{7} &= \left(1 + \binom{7}{1}(2x) + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \binom{7}{4}(2x)^4\right) \\
&\times\left(1 + \binom{7}{1}(-x) + \binom{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to solve for matrix $A$ in $AB = I$ Given $B$ = $\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 1
\end{bmatrix}$
I know that $B$ is equal to inverse of $A$, how can I go backwards to solve for $A$ in $AB = I$?
| Start by writing $[B|I]$ and apply row operations until you end up with $[I|A]$. This works because row operations are multiplications with suitable matrices from the left, and if this suitable matrix turns $B$ into $I$ it must be $B^{-1}=A$, which of course turns $I$ into $A$:
$$\begin{align}\begin{bmatrix}
1 & 0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometry problem about angles and triangles I've been working on this problem for a while. It doesn't seem to hard, but I cannot reach a satisfying solution.
The triangle $ABC$ is isosceles with base $\overline{AC}$. A point $O$ is also given.
Knowing: $\overline{OA}=R$, $\overline{AB}\equiv \overline{BC}=r$ and the a... | I'm going to take as starting data the following, using your first formulation of things.
\begin{align}
\newcommand{\uvec}{{\bf u}}
\newcommand{\wvec}{{\bf w}}
P &= (x, y), \text{the coordinates of your point $O$, which I'll call $P$}\\
B &= (s, t), \text{the coordinates of $B$}\\
A &= (a, b), \text{the coordinates of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/654488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find possible minimal polynomial for $L^3 + 2L^2 + L + 3 \cdot I_v$ Given:
$L: V \rightarrow V ; L^2 + I_v = 0$
Find:
possibilities for the minimal polynom of $L^3+2L^2+L+3\cdot I_v$
| $x - 1$. The one and only minimal polynomial of $L^3 + 2L^2 + L + 3I_V \;$ when $L^2 + I_V = 0$ is $x - 1$.
Observe that
$x^3 + 2x^2 + x + 3 = (x + 2)(x^2 + 1) + 1, \tag{1}$
which in turn implies, since $L^2 + I_V = 0$,
$L^3 + 2L^2 + L + 3I_V = (L + 2I_V)(L^2 + I_V) + I_V = I_V; \tag{2}$
the one and only minimal poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/658036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^\infty\frac{\cos(xt)}{1+t^2}dt$ Let $x \in \mathbb R$
Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ .
Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without resorting to Fourier serie... | We can also find the differential equation by just differentiating under the integral sign, after rewriting a little:
$$F(x) = \int_0^\infty \frac{\cos (xt)}{1+t^2}\,dt$$
is continuous by the dominated convergence theorem, with $F(0) = \frac{\pi}{2}$ known and $\lvert F(x)\rvert \leqslant \frac{\pi}{2}$. For $x \neq 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/658397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdo... | If $a$ is divisible by $4$, then $b$ is divisible by $4$ if and only if $b-a$ is divisible by $4$.
So, you can assume $6\cdot 7^n-2\cdot 3^n$ is divisible by $4$ and try for
\begin{align}
(6\cdot 7^{n+1}-2\cdot 3^{n+1})-(6\cdot 7^n-2\cdot 3^n)
&=
(6\cdot 7^{n+1}-6\cdot 7^{n})-(2\cdot 3^{n+1}-2\cdot 3^{n})
\\
&=
(6\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Why does this outcome change depending on infinity Why does the outcome of the limit as x approaches infinity of $$\sqrt{x^2+2x}- \sqrt{x^2-2x}$$
which simplifies to
$$\dfrac {4x}{x \left(\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}\right) }= \dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$
change depending on wh... | The equation $$\sqrt{x^2+2x}=x\sqrt{1+\frac{2}{x}}$$
is true only for $x>0$. The correct equation is $$\sqrt{x^2+2x}=\sqrt{x^2}\sqrt{1+\frac{2}{x}}=|x|\sqrt{1+\frac{2}{x}}$$
The intuition is that for $x>0$ you are subtracting a smaller number from a larger one (i.e. the two square roots); however for $x<0$ you are sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Formula help with this equation I don't know what the answer to this formula is, can someone please help me.
I've tried lots of things but getting no where.
If $x=\dfrac56+\dfrac{15}{18}-\dfrac{10}{12}$, then $(x-1)3=$ ?
| The expression $x=5/6+15/18-10/12$ may be simplified considerably:
\begin{align*}
x &= \frac{5}{6}+\frac{15}{18}-\frac{10}{12}\\
&= \frac{5}{6}+\left(\frac{30}{36}-\frac{30}{36}\right)\\
&= \frac{5}{6}.
\end{align*}
Thus, $x-1=-\frac{1}{6}$ and $3(x-1)=-\frac{1}{2}$.
As @ParthKohli has pointed out in a comment, you ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$. How Find this sum
$$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My idea: since
$$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$
so
$$I=\sum_{n=1}^{\infty}H^3_{n}\int_{0}^{-1}x^ndx$... | We will use the combinatorial identity, which can be proved through induction
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n$ and summing both sides ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 2
} |
$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles.
$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles.
And we can't use calculus, logarithms,... | Brute-force solution:
$$\frac{a+b}{a+c}=2\cos B-1=\frac{a^2+c^2-b^2-ac}{ac}=\frac{b^2+c^2-bc+c^2-b^2-ac}{ac}=\frac{2c-a-b}{a}$$
$$
0=a(a+b)+(a+c)(a+b-2c)=2a^2+2ab-ac+bc-2c^2=2(b^2+c^2-bc)+2ab-ac+bc-2c^2=(a+b)(2b-c)
$$
Therefore, $c=2b$, and consequently $a=\sqrt{3}b$. The angles follow easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Two very difficult induction proofs; having trouble with the inductive step $$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+1}\frac{n-2k-1}{k+1} = n-2 + \frac{1}{n+1}\binom{2n}{n}$$
$$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+2}\frac{n-2k-1}{k+1} = -n + \frac{n}{(n+2)(n+1)}\binom{2n}{n}$$
The two are clearly related in some ... | Let me present an algebraic proof of the first equality and a proof by
induction will perhaps appear.
Suppose we seek to verify that
$$\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1}
\frac{n-2k-1}{k+1} = n-2+\frac{1}{n+1}{2n\choose n}.$$
The left has two pieces which are, first, piece $A$,
$$\sum_{k=0}^{n-2} {n\choose k} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/666615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Did I solve these limits right? I have some limits that I tried solving:
$\lim_{x\rightarrow2}(4x^5+2x^3-1)=4(2)^5+2(2)^3-1=143$
$\lim_{x\rightarrow1}(\frac{5x}{x-1})=\frac{5}{0}=does~not~exist$
$\lim_{x\rightarrow(-3)}(\frac{x^2-5x-24}{4x+12})=\frac{(x+3)(x-8)}{4(x+3)}=\frac{x-8}{4}=-\frac{11}{4}$
$\lim_{x\rightarrow4... | Your approach to these is spot on, in all cases. You've done very well! Nice work!
Note, that when canceling common factors in the numerator and denominator, for example, say, the factor $(x-a)$ in a limit for which $x \to a$, you should specify $x \neq a$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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High degree polynom divisibility The problem is:
Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$
I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.
| HINT:
Both are in Geometric Series
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$ and $$1+x^{11}+x^{22}+x^{33}+x^{44}=\frac{1-x^{55}}{1-x^{11}}$$
So, $$\frac{1+x^{11}+x^{22}+x^{33}+x^{44}}{1+x+x^2+x^3+x^4}=\frac{(1-x^{55})(1-x)}{(1-x^{11})(1-x^5)}$$
Now using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,
$1-x^{55}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/670555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Given, $2\le aGiven, $2\le a<b<c<d\le 16$ and $(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2$,$(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2$. Find all $a,b,c,d$
My work:
If $(a,b,c,d)\in \mathbb{N}$, for that I could show this
$$(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2\implies d^2-2d=(a^2+b^2+c^2)-2(a+b+c)+2$$
Now, From second equation we get,
$$d^2+2d... | Start as you did (i.e. by expanding, simplifying, and “equating” the two given equalities), then solve for $a$ to obtain
$$a = \sqrt{2(b+c-d)-1}.$$
Note that $b+c \le 29$, so $b+c-d \le 24$, so $2(b+c-d)-1 \le 47$. Since $2(b+c-d)-1$ must be an odd square, and $a \ge 2$, we conclude $2(b+c-d)-1 \in \{9,25\}$. Hence $a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor 9 terms with 3 variables into 4 expression I just got the determinant from a 4x4 matrix and the simplified version is below.
$$
det(M) = \begin{vmatrix}
2k-mw^2 & -k & 0 & 0 \\
-k & 2k-mw^2 & -k & 0 \\
0 & -k & 2k-mw^2 & -k \\
0 & 0 & -k & 2k-mw^2\\
\end{vmatrix}
$$
the polynomial I got after 1 hr is:
$$
5k... | First, let's make things a bit nicer to look at. Let's substitute $x = 2k - mw^2$:
$$
det(M) = \begin{vmatrix}
x & -k & 0 & 0 \\
-k & x & -k & 0 \\
0 & -k & x & -k \\
0 & 0 & -k & x\\
\end{vmatrix}
$$
I calculate the determinant to be $x^4 - 3k^2x^2 + k^4:$
$$
det(M) = x\begin{vmatrix}
x & -k & 0 \\
-k & x & -k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the general formula for the partial sum? $\sum 2^{(n-1)} \cdot n$ from $n=1$ to $k$ I'm having a bit of trouble finding the general formula for the following partial sum
$$ \sum_{n=1}^k 2^{n-1} \cdot n $$
| First note that $2^{n-1}+2^{n-1}=2^n$. By using this formula for different values of $k$ we can derive a formula for the sum without the $n$ factor:
\begin{eqnarray*}
2^{k} &=& 2^{k-1} + 2^{k-1}\\
&=& 2^{k-1} + 2^{k-2} + 2^{k-2}\\
&=& 2^{k-1} + 2^{k-2}+2^{k-3}+2^{k-3} \\
&=& \cdots\\
&=& (2^{k-1}+2^{k-2}+\cdots+1) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Transform $\int \frac{x^2}{x - 2}$ to $ \int x +\frac{4}{x-2} + 2 $ I was trying to solve the following integral and I failed so I went to wolframalpha to see the step by step solution, but the following transformation is confusing me.
$\int \frac{x^2}{x - 2} = \int x +\frac{4}{x-2} + 2 $
I am unable to do the transfo... | First you want to reduce the rational function by polynomial division to get
$$
\frac{x^2}{x - 2} = \frac{x^2 - 2x + 2x}{x - 2} = x + \frac{2x}{x - 2}
$$
Then you want to perform partial fractions to reduce the resulting rational function to get
$$
\frac{2x}{x - 2} = \frac{ 2x - 4 + 4 }{ x - 2 } = 2 + \frac{4}{x - 2}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Given two determinants, find the det of this matrix? If det $ =\begin{bmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{bmatrix} = -2$
and If det $ =\begin{bmatrix} a & 1 & d\\b & 2 & e\\c & 3 & f\\\end{bmatrix} = -1$
what is the determinant of $\begin{bmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{bmatrix}$
I'm pretty... | (This is essentially a restatement of the earlier answers, but expanding on how the multilinearity can be viewed.)
If we compute these determinants using an expansion along the second column, we find
$$\begin{vmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{vmatrix} = -1 \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+1\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Nice Arctan Identity Prove that $ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $ if $a, d, c$ are positive reals satisfying $$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$
(credit: bobthesmartypants)
| Try to arrive at the equation that needs to be proved from the given equation.
$$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$
$$a^2(a^2+d^2+2ad)=c^2d^2-a^2c^2$$
$$(a^2)(a+d)^2+c^2(a^2-d^2)=0$$
$$(a+d)(a^2(a+d)+c^2(a-d))=0$$
$(a+d)$ cannot be equal to zero as $a,d$ are positive real numbers.So:
$$a^2(a+d)+c^2(a-d)=0$$
$$\frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Agai... | We consider four cases for $(3- 2a),(3-2b),(3-2c)$ : The case when all of them are non-negative, the case when exactly one of them is negative, the case when exactly two of them are negative, and the case when all of them are negative.
For the first case when all of them are non-negative, notice that by the AM-GM inequ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Don't know how to continue modulo reduction Ok , so the question is to find the last three digits of $2013^{2012}$. After some reduction using Euler's Theorem I got $13^{12}$(mod 1000). I tried dividing it in 8 and 125 and later use the CRT , but this didn't help (or maybe I did it wrong) .
What is the shortest way to ... | HINT:
$$13^{12}=(10+3)^{12}=3^{12}+12\cdot3^{11}10+\binom{12}2\cdot3^{10}10^2\pmod{1000}$$
Now, $\displaystyle3^{10}=9^5\equiv(-1)^5\pmod{10}\equiv-1$
$\displaystyle\implies\binom{12}2\cdot3^{10}10^2\equiv\binom{12}2\cdot(-1)10^2\pmod{100}$
and $\displaystyle3^{11}=3\cdot9^5=3(10-1)^5\equiv3(-1+\binom51\cdot10^1)\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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determining order of $x+1$ given the $x$ has order three I was trying to expand $(x+1)^n$, then plug $x^3$ in to the expansion of the $(x+1)^n$, keep trying it until I get the order, are there any other ways?
So if $x^3\equiv 1\pmod y$, how would I determine order of $x+1$?
Can I brute force $x$ and $y$ for the condit... | Clearly $x\neq1$ and hence
$$
x^3 \equiv 1 \Rightarrow x^2+x+1 \equiv 0 \tag1$$
Now consider
$(x+1)^6$ and show that
$$
(x+1)^6 - 1 \equiv x\,\left(x+2\right)\,\left(x^2+x+1\right)\,\left(x^2+3\,x+3\right)\tag2$$
and conclude that the order is $6$ by showing no smaller power works.
Added in response to comments
Using (... | {
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Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| Using Cauchy-Schwarz Inequality twice:
$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Checking irreducibility I have the polynomial $f(X)=X^{2n}-2X^{n}+1-p$ where $p$ is a prime number and $n\in\mathbb{N}$. I want to check whether it is irreducible or not over $\mathbb{Q}[X]$.
If $2^{2}\nmid1-p$ then $f(X)$ is irreducible by Eisenstein's Criterion. However, I can't make any progress when I consider the ... | You are correct, it is always irreducible.
Your polynomial factors as
$AB$ where $A=X^n-(1+\sqrt{p})$ and $B=X^n-(1-\sqrt{p})$. It will suffice
to show that $A$ (and therefore $B$ also) is irreducible over $K={\mathbb Q}[\sqrt{p}]$.
Thanks to Karpilovsky’s theorem (many thanks to Bill Dubeque for
quoting it here), it w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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What is the maximum height of a snowman (consisting from 3 snowballs) when I have $\frac{4}{3}\pi$ cubic meters of snow Let balls $1,2,3$ have radia $r_1, r_2, r_3$. Then
$V_1=\frac{4}{3}\pi r_1^{3}$
,$V_2=\frac{4}{3}\pi r_2^{3}$
and $V_3=\frac{4}{3}\pi r_3^{3}$.
I know that $V_1 + V_2 + V_3 =\frac{4}{3}\pi$.
From thi... | Really, the problem is to maximize $h=2(r_1+r_2+r_3)$ given that $r_1^3+r_2^3+r_3^3=1$. You can find critical points by expressing
$$h(r_1,r_2) = 2 [ r_1+r_2 + (1-r_1^3-r_2^3)^{1/3})]$$
Consider
$$f(x,y) = x+y+(1-x^3-y^3)^{1/3}$$
$$f_x = 1 -\frac{x^2}{(1-x^3-y^3)^{2/3}}$$
$$f_y = 1 -\frac{y^2}{(1-x^3-y^3)^{2/3}}$$
Set... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Combinatorics on letters How many "words" of length n is it possible to create from {a,b,c,d} such that a and b are never next to each other?
| Let the total number of such words on $n$ letters be $s_n$. Let the number of words which end with $a$ be $a_n$ and the number of words which end with $b$ be $b_n$.
Now consider a word on $n-1$ letters. We may add $c$ or $d$ to the end of the word to obtain a new word on $n$ letters. Conversely, all words on $n$ letter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$
In a triangle $ABC$, it's given that the following two equations are satisfied:
$$ 3\sin A + 4\cos B = 6 $$
$$ 3\cos A + 4\sin B = 1 $$
Source: ISI B-math UGA 2017
We have to find the angle $ C$. Now, it's easy to see that $ \... | $ 3\sin A + 4\cos B = 6 $ ......(i)
$ 3\cos A + 4\sin B = 1 $.... (ii)
Squaring (i) we get :
$9 \sin^2 A +16\cos^2 B +24 \sin A \cos B = 36 $ ....(iii)
Now Squaring (ii) we get :
$9 \cos^2 A +16\sin^2 B +24 \sin B \cos A =1 $...(iv)
Now adding (iii) and (iv) we get :
$9(\sin^2 A +\cos^2 A) +16(\cos^2 B + \sin^2 B... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Interesting and unexpected applications of $\pi$ $\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$
Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
and the generalization of $\zeta (2k)$, my perception of $\pi$ has ... | $\pi$ as a matrix eigenvalue
For a given positive integer $m$, let us form an $(m-1)\times(m-1)$ dimensional matrix $M$. The terms on the diagonal of $M$ will be $-2m^2$, the terms on the off diagonals of $M$ will be $m^2$ and all other terms will be zero. Thus,
$$M=m^2\left(
\begin{array}{cccccc}
-2 & 1 & 0 & \cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "76",
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Howto prove that $\sum\limits_{cyc}\cos\frac{A}{2}\cos\frac{B}{2}\le\frac{1+2\sqrt{2}}{2}+\frac{7-4\sqrt{2}}{R}r$ let $ABC$ is a triangle with inradius $r$ and circumradius $R$. Show that
$$\cos\frac{A}{2}\cos\frac{B}{2}+\cos\frac{C}{2}\cos\frac{B}{2}+\cos\frac{A}{2}\cos\frac{C}{2}\le\frac{1+2\sqrt{2}}{2}+\frac{7-4\sqr... | your inequality isn't true, try $a=4,b=3,c=5$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Difficulties evaluating the endpoints of the radius of convergence for a particular power series. I am having difficulties evaluating the endpoints of the radius of convergence for the following power series.
$$\sum_{k=0}^{\infty}\frac{(k!)^2 x^k}{(2k)!}$$
Using the ration test we get |x|<4. However, evaluating $x=4$ a... | $\begin{align*}
a_k = \frac{(k!)^2 4^k}{(2k)!} &= \frac{(k!)^2 4^k}{(2k)(2k-1)(2k-2) \cdots 3\cdot 2 \cdot 1} \\
&=\frac{(k!)^2 4^k}{((2k)(2k-2) \cdots 4 \cdot 2)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{(k!)^2 4^k}{(2^k k!)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{k! \cdot 2^k}{(2k-1)(2k-3) \cdots 3 \cdot 1} \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider t... | Still not true. If $m = 1, n = 2,$ then $1$ isn't divisible by $2.$ You need $n$ to be odd as well.
| {
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"source": "stackexchange",
"question_score": "3",
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Finding the Absolute Maximum and Minimum of a 3D Function Find the absolute maximum and minimum values of the function:
$$f(x,y)=2x^3+2xy^2-x-y^2$$
on the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.
| We first take the partial derivatives with respect to each variable and set them to zero:
$$\frac{\partial f}{\partial x}=6x^2 + 2y^2 - 1=0$$
$$\frac{\partial f}{\partial y}=4xy - 2y=0$$
From the second equation, we have either: $y=0$ or $x=\frac{1}{2}$. Now we substitute each value into the first equation to get the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to show that this sequence is monotonic We are given that $(s_{n})$ is a bounded sequence. My goal is to show that $(s_{n})$ is monotonic, and thus it converges. It has the following property $$s_{n+1} \geq s_{n} - \frac{1}{2^n}$$
I have tried induction, because intuitively the sequence monotonically increasing, b... | $s_n$ is not necessarily monotonic given the information provided.
Example.
Let $s_n = \frac{2}{3} \left(\frac{-1}{2}\right)^n $.
Then $s_n$ is bounded, and
\begin{align*}
s_{n+1} - s_n
&= \frac{2}{3}\left(\frac{-1}{2}\right)^{n+1} - \frac{2}{3}\left(\frac{-1}{2}\right)^n \\
&= \frac{2}{3}\left(\frac{-1}{2}\right)^n \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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PMF and CDF Calculating I've got the following homework problem that I'm stuck on.
Two fair six-sided dice are tossed independently. Let M= the maximum of the two tosses.
a. What is the PMF of M? [Hint: first determine P(1), then p(2), and so on.]
b. Determine the CDF of M.
I am just now learning how about PMF/CDF s... | First Question: Let us find the probability that the maximum is $3$, first a slightly tedious way, and then a less tedious way.
Imagine that the dice are Christmas dice, green and red. We record the result of the tossing as an ordered pair $(a,b)$, where $a$ is the number on the green, and $b$ is the number on the red.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral $\int_0^1\frac{\ln x}{x-1}\ln\left(1+\frac1{\ln^2x}\right)dx$ Is it possible to evaluate this integral in a closed form?
$$
I \equiv \int_{0}^{1}{\ln\left(x\right) \over x - 1}\,
\ln\left(1 + {1 \over \ln^{2}\left(x\right)}\right)\,{\rm d}x
$$
Numerically,
$$I\approx2.18083278090426462584033339029703713513\dot... | Substitute $x=e^{-y}$:
$$
I=\int^\infty_0\frac{y}{e^y-1}\log(1+y^{-2})dy.
$$
We Start from Binet's Second Formula:
\begin{align}
\log\Gamma(z) &= (z-\frac12)\log z-z+\frac12\log(2\pi)+2\int^{\infty}_0\frac{\tan^{-1}(t/z)}{e^{2\pi t}-1}dt\\
&=(z-\frac12)\log z-z+\frac12\log(2\pi)+\frac1\pi\int^{\infty}_0\frac{\tan^{-1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/710175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
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Definition of matrix exponential Is there an alternative definition of a matrix exponential so I can use it to prove that $$e^{A}=\sum_{m=0}^{\infty} \frac{1}{m!}(A)^m \;?$$
Thanks a lot in advance!
| $$\begin{align}&\lim_{n \to \infty} \left(I + \frac{1}{n} A \right)^n = \\ &\lim_{n \to \infty} \left(I^n+(n)I^{n-1}\frac{A}{n}+\frac{(n)(n-1)}{2}I^{n-2}\frac{A^2}{n^2} + \frac{(n)(n-1)(n-2)}{6}I^{n-3}\frac{A^3}{n^3} +\;...\right) = \\ &\lim_{n \to \infty}\left(I + A+\frac{(n)(n-1)}{2n^2}A^2 + \frac{(n)(n-1)(n-2)}{6n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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How to evaluate this definite integral $ \int_{-\pi/3}^{\pi/3} \frac{(\pi +4x^3)\,dx}{2-\cos(|x|+ \frac{\pi}{3})} $ $$ \int_{-\pi/3}^{\pi/3} \frac{(\pi +4x^3)\,dx}{2-\cos(|x|+ \frac{\pi}{3})} $$
I have separated the integral into two parts, then expanded using $\cos(a+b)$ formula, after that I am lost. Can someone pro... | You don't need to separate the integrand. First of all, the $x^3$ piece vanishes; this follows immediately from the fact that this is an odd integrand over an even interval. That leaves the first piece, which I can reduce to
$$2 \pi \int_{\pi/3}^{2 \pi/3} \frac{dx}{2-\cos{x}}$$
This may be evaluated using a substitut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating polynomial fraction of same degree I'm to integrate:
$$ \int {\frac{x^2}{x^2+x-2} }dx$$
If there was a constant in the numerator then I could solve this but I'm not sure with $x^2$. Can I use long division? I tried and it didn't seem to work out (but my long division is very rusty).
| $$\int {x^2\over x^2+x-2}dx=\int {x^2+x-2-x+2\over x^2+x-2}dx=\int1+{-x+2\over x^2+x-2}dx$$
$$=x-{1\over2}\int{2x-4\over x^2+x-2}dx=x-{1\over2}(\int{2x+1\over x^2+x-2}-{5\over x^2+x-2}dx)$$
$$=x-{1\over2}ln( x^2+x-2)+{5\over2}\int{1\over x^2+x-2}dx $$
After that you need to deal with $$\int{1\over x^2+x-2}dx=\int{1\ove... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? I have never seen this problem in my homework.. I need help right before final!!
Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the ... | Observe that via integration, it follows that
$$
y' = - \frac{x^4}{4} + C.
$$
Hence, at $(1,1)$, $y'(1) = C - \frac{1}{4}$, and the tangent line is of the form $y - 1 = ( C - \frac{1}{4} )(x - 1)$, that is ( using the knowledge of the tangent curve at (1,1) ) that
$$
3 - 2x = y = \left(C- \frac{1}{4} \right)x + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}... | Since the numerator and denominator is zero at $1$, let's factor out $(x-1)$ from both of them to get an idea how the function behaves around $1$.
The fraction equals $\dfrac{(3x^3-5x^2-5x-5)(x-1)}{(x^2-1)(x-1)}=\dfrac{3x^3-5x^2-5x-5}{x^2-1}$.
At $x=1$, the numerator equals $-12$. So for values around and very close to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$? I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of t... | Step 1.
For n = 1 it's true that 1 = 2*1-1
Step 2.
We suppose that 1+3+5+...+(2n-1) = n2
and want to prove that: 1+3+5+...+(2(n+1)-1) = (n+1)2
We add (2(n+1) -1) to this:
1+3+5+...+(2n-1) = n2
and get:
1+3+5+...+(2n-1) + (2(n+1) -1) = n2 + (2(n+1) -1)
so:
1+3+5+...+(2(n+1) -1) = n2 + 2n+2 -1
but n2+2n+1 = (n+1)2
so we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 11,
"answer_id": 7
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One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your ... | You can consider another equation:
$2(a^2+y^2+c^2+d^2+u^2)=(a+y+c+d+u)^2$
And write the formula to solve this equation.
$a=-(k^2+2(q+t+b)k+b^2+q^2+t^2+bq+bt+qt)$
$y=k^2+2(q+t+b)k+2b^2+q^2+t^2+2bq+2bt+qt$
$c=k^2+2(q+t+b)k+b^2+2q^2+t^2+2bq+bt+2qt$
$d=k^2+2(q+t+b)k+b^2+q^2+2t^2+bq+2bt+2qt$
$u=2k^2+2(q+t+b)k+b^2+q^2+t^2$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding Power Series Representations For f(x), find a power series representation centered at the given value of a and determine the radius of convergence.
$$ f(x) = \frac {4x} {x^2-2x-3} ; a=0.$$
How would i begin this?
And what does it mean at the given value of a?
Thanks!
| Hint: Start with
$$
\frac {4x} {x^2-2x-3} = \frac {4x}{(x-1)^2-4}
=\frac {x-3+3(x+1)}{(x-3)(x+1)}.
$$
details:
then you get to
$$
\frac {4x} {x^2-2x-3} = \frac 1{x+1} + \frac 3{x-3}
= \sum_{n=0}^\infty (-1)^nx^n + \sum_{n=0}^\infty -3^{-n}x^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/729554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral $I=\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx$ Hi I am stuck on showing that
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx=\pi G-\frac{3\zeta(3)}{8}
$$
where G is the Catalan constant and $\zeta(3)$ is the Riemann zeta function. Explictly they are given by
$$
G=\beta(2)=\sum_{n=0}^\infty \frac{(-1... | I have a simple solution for this problem. Let
$$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x^{-2})}{x}dx. $$
Then $I(0,0)=I(\alpha,0)=I(0,\beta)=0$ and $I(1,1)=I$. It is easy to check
\begin{eqnarray*}
\frac{\partial^2 I}{\partial \alpha\partial\beta}&=&\int_0^\infty\frac{1}{(1+\alpha x)(x^2+\beta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
Finding the coefficient of a generating function Given $f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4$. This is the generating function $f(x)$ of $a_n$, which is the number of ways to get $n$ as the sum of the upper faces of four thrown dice.
How do I calculate a coefficient from said generatin... | You want:
\begin{align}
[z^{17}] z^4 (1 - z^6)^4 (1 - z)^{-4}
&= [z^{13}] (1 - 4 z^6 + 6 z^{12} - 4 z^{18} + z^{24})
\cdot \sum_{k \ge 0} \binom{-4}{k} (-1)^k z^k \\
&= [z^{13}] (1 - 4 z^6 + 6 z^{12}) \cdot \sum_{k \ge 0} \binom{k + 3}{3} z^k \\
&= \binom{16}{3} - 4 \cdot \binom{10}{3} + 6 \cdot ... | {
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"url": "https://math.stackexchange.com/questions/733731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} +... | You have to find the (complex) pairs of roots of the polynomial $x^6+1$, and then use the uncertain coefficients method as you did.
Notice that using the substitution $t=x^2$ leaves you with $x=\sqrt{t}$ and $dx=dt/\sqrt{t}$ which doesn't look quite nicely integrable.
When you have the polynomials $t^2+pt+q$ with compl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof, wheather a subset of a Group is a Subgroup I have to check, weather the following subset of a group is also a subgroup:
$$U = \left\{ \begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \in GL(2, \mathbb{C}) \bigg\vert |a|^2 + |b|^2 = 1, \ a,b \in \mathbb{C} \right\} \subset (GL(2,\mathbb{C}), \c... | For part (1):
$$\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \cdot \begin{pmatrix} a' & -b' \\ \overline{b'} & \overline{a'} \end{pmatrix} = \begin{pmatrix} aa'-b\ \overline{b'} & -ab'-b\ \overline{a'} \\ \overline{b}a'+\ \overline{ab'} & -\overline{b}b' +\overline{a'a} \end{pmatrix}$$
You want ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hard Olympiad Inequality Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$.
I tried to square expand homogenize then majorize. But I couldn't make it work. Any help would be much appreciated.
| problem: since $a,b,c>0$,and such $ab+bc+ac=1$, show that
$$\sqrt{a^3+a}+\sqrt{b^3+b}+\sqrt{c^3+c}\ge 2\sqrt{a+b+c}$$
Poof:
Using Holder inequality,we have
$$\left(\sum\sqrt{a^3+a}\right)^2\left(\sum\dfrac{a^2}{a^2+1}\right)\ge\left(\sum a\right)^3$$
it remains to prove that
$$\left(\sum a\right)^2\ge 4\sum\dfrac{a^2}{... | {
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Why does $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverge? $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverges, and I have no idea why? It would seem using L'hop, $\frac{n^3 + n^2 + n}{n^3 + 1}$ goes to 1. So it should end up just being $1\cdot1\cdot1\cdots$, which makes me feel it converges. Is the p... | Intuitively, as $\frac{n^3 + n^2 + n}{n^3 + 1} \sim 1$, if we erase $n$ from the top and $1$ from the bottom, the fraction should get smaller. Lets prove it:
$$\frac{n^3 + n^2 + n}{n^3 + 1}-\frac{n+1}{n}=\frac{n^2-n-1}{n(n^3+1)}$$
Thus, for $n \geq 2$ we have
$$\frac{n^3 + n^2 + n}{n^3 + 1} > \frac{n+1}{n}$$
Now, since... | {
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"url": "https://math.stackexchange.com/questions/745936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of eight dice showing sum of 9, 10 or 11 Suppose we roll eight fair dice. What is the probability that:
1) The sum of the faces is $9$
2) The sum of the faces is $10$
3) The sum of the faces is $11$
I'm thinking that we start with $8$ dice, each showing $1$. Then we think of the problem as assigning one $1$... | Sum of 9 : 1, 1, 1, 1, 1, 1, 1, 2 = $$\frac{8!}{7!} = 8$$
Sum of 10: 1, 1, 1, 1, 1, 1, 2, 2 =$$\frac{8!}{6!2!} = 28$$
1, 1, 1, 1, 1, 1, 1, 3 =$$\frac{8!}{7!} = 8$$
$$ 8+28 = 36$$
Sum of 11: 1, 1, 1, 1, 1, 1, 1, 4 :$$\frac{8!}{7!} = 8$$
1, 1, 1, 1, 1, 1, 2, 3 : $$\frac{8!}{6!} = 56$$
1, 1, 1, 1, 1, 2, 2, 2 : $$\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \fra... | The middle term, $\dfrac{(n+1)^{p+1} - n^{p+1}}{p+1}$, equals $\int_n^{n+1} x^p dx$. On the interval $(n,n+1)$, which has length $1$, $n^p<x^p<(n+1)^p$, from which the inequality follows immediately.
| {
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Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$... | Firstly we can make some simple remarks about the problem:
*
*let $ax^2-bx+c = f(x)$
*$a,b,c \in \Bbb{N} \Rightarrow a,b,c \ge 0$
*there are $2$ distinct roots, so the parabola is not a degenerate one $\Rightarrow a \ge 1$
*$x_{1,2} \in ]1,2[ \Rightarrow x_1 \cdot x_2 > 1\cdot 1 = 1 \Rightarrow \frac{c}{a} > 1 \R... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$.
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$
It kind of seems useful to begin with a division of both sides by $cd$:
$$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$
It seems like a simple Cauchy-Schwarz would... | Holder's inequality for sums states that for $p, q \in (1, \infty)$ with $\frac{1}{p}+\frac{1}{q}=1$, $$\sum_{k=1}^{n}{|a_kb_k|} \leq \left(\sum_{k=1}^{n}{|a_k|^p} \right)^{\frac{1}{p}}\left(\sum_{k=1}^{n}{|b_k|^q} \right)^{\frac{1}{q}}$$
(For $p=q=2$ this reduces to Cauchy Schwarz)
Applying this for $p=\frac{4}{3}, q=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding all Points on a Edwards curve I need to find all affine points on the Edwards curve:
$x^2 + y^2 = 1 - 5x^2y^2$ over $F_{13}$
I tackle this by transforming the equation to:
$y^2 = \frac{1-x^2}{1+5x^2}$
I then go from x = 0 to $\frac{p-1}{2}$ in this case x from 0 to 6.
If you can take the square root of $y^2$ yo... | The congruence class of $10$ is a square in $\mathbb{F}_{13}$. Indeed,
$$6^2\equiv 36\equiv 10 \bmod 13$$
and similarly $(-6)^2\equiv 7^2\equiv 49\equiv 10\bmod 13$. Thus, for $x\equiv 3$ and $-3\equiv 10\bmod 13$, there are two possibilities for $y$, namely $6$ and $7\bmod 13$.
Notice that your algorithm could have ru... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$ let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$
My Try $$A=\sqrt{a^2+4a+4}+\sqrt{b^2+2b+1}=$$then I can't
Please just a hint.
| Simply write $b$ in terms of $a$ (and mind the conditions on $a$ and $b$: $a \leq −2$ so $a+2\leq0$ and $b\leq-1$ so $b+1\leq0$) and you get
$$A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}=\sqrt{(a+2)^2}+\sqrt{(a-6+1)^2}=|a+2| + |a-5|=-a-2-a+5=-2a+3$$
since $a+2\leq0$ and $b+1\leq0$ ($\Rightarrow |a+2| = -a-2 \text{ and } |a-5| = -... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Problem with LU decomposition I have this matrix:
$$
A =\begin{bmatrix}1 & -2 & 3\\ 2 & -4 & 5 \\ 1 & 1 & 2\end{bmatrix}
$$
After I decomposit it, I get:
$$
L = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\\
U = \begin{bmatrix} 1 & -2 & 3\\ 0 & 3 & -1\\ 0 & 0 & -1\end{bmatrix}\\
P = \begin{bmatrix}1 & ... | All of your work is correct for $A = PLU$.
Now, we want to solve $Ax = b$ for $b_1$ and $b_2$, so we use forward and back substitution as:
$$Ly = Pb, Ux = y$$
Forward substitution yields:
$$Ly = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=Pb = \begin{bmatrix}... | {
"language": "en",
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"source": "stackexchange",
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${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100 \cdot6\over 4}$ Find the value of
$${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100\cdot6\over 4}$$
My approach:
$${2\over 4^{100}}+{2\over 4^{99}}+\cdots+{2\ove... | The second piece is simply
$$6 \sum_{k=1}^{100} \frac{k}{4^{101-k}}= \frac{6}{4^{101}} \sum_{k=1}^{100} k \, 4^k$$
To evaluate, let $S = 4 + 2 \cdot 4^2 +3 \cdot 4^3 + 4 \cdot 4^4+\cdots+100\cdot 4^{100}$ Then
$$4 S = 4^2+ 2 \cdot 4^3 +3 \cdot 4^4 + 4 \cdot 4^5+\cdots+100 \cdot 4^{101}$$
$$\implies S-4S=-3 S = 4+4^2+4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
| Use the fact that:
$$\frac{ (x^3-4x^2+5x-6)}{(x^2-x-6)}=(x-3)+\frac{8}{(x+2)} $$
Therefore your solving:
$$ (x-3)+\frac{8}{x+2}=4 \\ x^2-5x-6+8=4x+8\\ x^2-5x-6=0$$
Can you solve this?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \fr... | Proper sign to be considered for scalars after coming so far with normal curvature. For a negatively curved (Gauss curvature $K<0$ ) surface
$ k_n = -k_{+} \cos^2{\theta} + k_{-} \sin^2{\theta} = 0 $ and for real $\theta$,
$ \tan{\theta} = \sqrt {(k_{+} / k_{-})} $
For example, asymptotic lines on rotationally symm... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent? Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent ?
I use it to compare with $1/n^2$, and then I used LHôpitals rule multiple times. Finally , I can solve it. However,I think we have other methods! somebody can help me?
| By using the integral test:
The function $\frac{1}{3^{\sqrt{x}}}$ is continuous, positive and decreasing on $[1,\infty)$. We then evaluate the integral:
\begin{align}
\int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx = \frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.15918311754471
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $ Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$
Determine $n$.
The memorandum says the answer is 2011 but how is that so? Where did I go wrong?
| Numerator :
\begin{eqnarray*}
T_n &=& 1+3+5+...+(2n-1)\\
&=& 2+4+6+...+2n - n\\
&=& 2(1+2+3+...+n) - n\\
&=& 2\frac{n(n+1)}{2} - n\\
&=& n^2
\end{eqnarray*}
Similarly for denominator (check) :
\begin{eqnarray*}
B_n &=& 2+4+6+...+2n\\
&=& n(n+1)
\end{eqnarray*}
But it back in the initial equation and you get :
$$\frac{n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2 \cos^2 x − 2 \cos x− 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$
Find the solutions of $$2 \cos^2 x − 2 \cos x− 1 = 0$$ for all $0^\circ ≤ x < 360^\circ$.
For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.
Is this correct? Thanks!
| Let $y = \cos(x)$. Then we have the following:
$$2y^2 - 2y - 1 = 0$$
Using the quadratic formula, we get the roots to be:
$$y = \cos(x) = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$$
And so:
$$x = \arccos(\frac{1}{2} \pm \frac{\sqrt{3}}{2}) $$
Since $\frac{1}{2} + \frac{\sqrt{3}}{2}>1$, we can ignore this solution. Since bot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of $\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}$? After simplification using sum to product transformation equations I keep ending up with
$$4\cos36^\circ\cdot\cos7^\circ\cdot\cos18^\circ$$
How do I simplify this to a single term?
| Using Prosthaphaeresis Formulas
$$\sin47^\circ+\sin61^\circ=2\sin54^\circ\cos7^\circ$$
$$\sin25^\circ+\sin11^\circ=2\sin18^\circ\cos7^\circ$$
$$\implies\sin47^\circ+\sin61^\circ-(\sin25^\circ+\sin11^\circ)=2\cos7^\circ(\sin54^\circ-\sin18^\circ)$$
$$S=\sin54^\circ-\sin18^\circ$$
Method $\#1:$
$$S=\sin54^\circ-\sin18^\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Testing for convergence of a series Test the following series for convergence
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k}$$.
I am just wondering if the following method is ok:
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k} = \sum_{k=1}^\infty \sin (\frac{1}{k} - (\frac{1}{k} -k)) \sin \frac{1}{k} \\ \leq \sum_{k=1}^\infty... | By Taylor's theorem:
$$\begin{align}
\sin k\sin \frac 1k&=\frac 12\left[\cos\left(k-\frac1k\right)-\cos\left(k+\frac1k\right)\right]=\\
&=\frac 12\left[\cos k-\frac1k\sin\xi_k-\cos k+\frac1k\sin\xi'_k\right]=\\
&=\frac1{k}\sin\left(\frac{\xi_k-\xi'_k}2\right)\cos\left(\frac{\xi_k+\xi'_k}2\right)
\end{align}
$$
where $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding a holomorphic function with a prescribed real part. I am to find a holomorphic function on $\mathbb{C}\setminus\{0\}$ with $u(x,y)=\dfrac{x+y}{x^2+y^2}$ and $f(1)=1$. Preceding this, I had to show we could write $f'(z) = \frac{\partial u}{\partial x} -i\frac{\partial u}{\partial y}$, which is clearly supposed t... | Since $u(x,y) = \dfrac{x+y}{x^2+y^2}$,
$$\frac{\partial u}{\partial x} = \frac{1}{x^2+y^2}-\frac{2x(x+y)}{(x^2+y^2)^2} = \frac{x^2+y^2-2x^2-2xy}{(x^2+y^2)^2} = \frac{-x^2-2xy+y^2}{(x^2+y^2)^2}.$$
Likewise, by symmetry,
$$\frac{\partial u}{\partial y} = \frac{-y^2-2xy+x^2}{(x^2+y^2)^2}.$$
Putting these together shows th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/770668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A logic problem. No need for calculation Three people were told to go to a cave and each pick up a hat in pitch dark. The three then came out of the cave with the hat on their heads. There were five hats in the cave. Three of them are black, the rest is white. When the first one is asked to tell the color of his hat, h... | Initially there are $7$ possibilities:
$$
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 \\ \hline
B & B & B \\ \hline
B & B & W \\ \hline
B & W & B \\ \hline
B & W & W \\ \hline
W & B & B \\ \hline
W & B & W \\ \hline
W & W & B \\ \hline
\hline
\end{array}
$$
First says that he doesn't know, so one of $2-3$ has a black,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/770732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 6
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Integral of $\int\frac{dx}{x^2+1}$ I know the formula$$\int\frac{dx}{x^2+1} = \tan^{-1}(x) + C$$
But, when integrating by parts:
$$u = x^2+1$$
$$u' = 2x$$
$$v' = 1$$
$$v = x$$
$$\int\frac{dx}{x^2+1} = uv - \int vu' dx = x(x^2+1) - \int 2x^2dx$$
$$ \int 2x^2dx = \frac{2x^3}{3} + C$$
So, finally: $$\int\frac{dx}{x^2+1} =... | Look this is an easier approach: Put $x = \tan \alpha $ . then $dx = \sec^2 \alpha d \alpha $. Hence
$$ \int \frac{dx}{1 + x^2} = \int \frac{\sec^2 \alpha}{1 + \tan^2 \alpha} d \alpha = \int d \alpha = \alpha = \arctan x + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\i... | Another solution could use the tangent half-angle substitution. If you define $t=\tan \frac{x}{2}$, you have $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$. So, the equation becomes $$\cos x+8\sin x-7=\frac{16 t+2}{t^2+1}-8=0$$ that is to say $$-8 t^2+16 t-6=0$$ the roots of which being $t_1=\frac{1}{2}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Simplifying $\frac{a}{(a-b)(a-c)(x-a)}+\frac{b}{(b-c)(b-a)(x-b)}+\frac{c}{(c-a)(c-b)(x-c)}$ We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$
The biggest problem is that the above expression has four variables.I transformed the expression into $$\dfrac{a}{-(a-b)(c... | $$
\frac{a}{(a-b)(a-c)(x-a)}=\frac{a((a-c)-(a-b))}{(b-c)(a-b)(a-c)(x-a)}=\frac{a}{(b-c)(a-b)(x-a)} - \frac{a}{(b-c)(a-c)(x-a)}
$$
Thus,
$$
\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-a)(b-c)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}=\frac{1}{(b-c)(b-a)}(\frac{b}{x-b}-\frac{a}{x-a})+ \frac{1}{(c-a)(c-b)}(\frac{c}{x-c}-\frac{a}{x-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int\frac{1}{\sin(x-a)\sin(x-b)}\,dx$ I'm stuck in solving the integral of $\dfrac{1}{\sin(x-a)\sin(x-b)}$. I "developed" the sin at denominator and then I divided it by $\cos^2x$ obtaining $$\int\frac{1}{\cos(a)\cos(b)\operatorname{tan}^2x-\cos(a)\sin(b)\operatorname{tan}x-\sin(a)\cos(b)\operatorname{tan}x+\... | Looks like I am late in the race. Let me present you a slight different way to integrate it.
let $\displaystyle x = y + \frac{a+b}{2}$, and also let $\displaystyle \frac{a-b}{2} = c$
$$\int \frac{1}{ \sin\left(y - \frac{a-b}{2}\right)\sin\left(y + \frac{a-b}{2}\right)} \, dx = \int \frac{1}{\sin(y-c) \sin(y+c)} \, dy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Are there nice ways to solve $(2+x)^{0.25}-(2-x)^{0.25}=1$ Are there nice and elegant ways to solve this equation?
$(2+x)^{0.25}-(2-x)^{0.25}=1$
Thanks.
| Let $t = 2 + x$, and $u = 2 - x$, then $u + t = 4$, and $t^{1/4} - u^{1/4} = 1$. Again let $p = t^{1/4}$, and $q = u^{1/4}$, then $p - q = 1$, and $p^4 + q^4 = 4$. So $(q+1)^4 + q^4 = 4$ and this gives: $2q^4 + 4q^3 + 6q^2 + 4q - 3 = 0$. Solving this we have:
$q = \dfrac{-1 + \sqrt{2\sqrt{10} - 3}}{2}$, and $u = q^4 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/780193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do you show the limit of a function? Using the epsilon-delta definition of a limit, how would you show that $\lim_{x \to 0} \frac{3+7x}{5+5x} = \frac{3}{5}$? The lecturer's definition and solution are nothing to shout about...
| By definition, for a given $\epsilon > 0$, you want to $\delta > 0$ such that if $|x| < \delta$, then
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| < \epsilon $$
We want to find a bound for $\left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right|$. Lets see
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| = \left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Diophantine equations problem/exercise 3 Find all the pythagorean tripples (x,y,z) with x=40. Well I started with the known formulas for the pythagorean tripples but got me nowhere. Or I was not able continue the thought process required. I do not have a lot of experience on Diophantine equations so I'm asking for a li... | The formula
$$
2mn, m^2 - n^2, m^2 + n^2
$$
where $m, n$ are relatively prime and of diferent parity,
and $m > n > 0$,
generates all primitive triples.
Thus, to get a triple including $x = 40$, we must solve for when
$2mn \; \mid 40$ or $m^2 - n^2 \; \mid \; 40$.
Case 1: $\boldsymbol{2mn \; \mid 40}$, i.e. $mn \; \mid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$.
It seems that the infinity product is not 0. This is related to the post Se... | Consider the corresponding finite product containing $n+1$ factors. Multiplying the last factor (with $n$ square roots) by
$$2-\underbrace{\sqrt{2+\sqrt{2+\ldots}}}_{x_n},$$
the product telescopes to $4^{-n}$.
On the other hand, since
$x_n^2-2=x_{n-1}$,
writing $x_n=2\cos\varphi_n$ we get $2\varphi_n=\varphi_{n-1}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum possible area of the rectangle ABZP Let ABZP be a rectangle. D and C lie on AP and BZ respectively such that ABCD is a square. X and Y lie on CD and ZP respectively such that CXYZ is a square.
Given: ar(ABCD) + ar(CXYZ) = 1
Find the maximum possible area of the rectangle ABZP.
| Let $x=AD$ (that is, its length). Let $y=CX$.
Clearly $y<x$ and $x^2+y^2=1$. We are to maximize $f(x)=x(x+y)$. But
$$f(x)=x(x+y)=x^2+xy=x^2+x\sqrt{1-x^2}=x^2+\sqrt{x^2-x^4}$$
so
$$f'(x)=2x-\frac{x-2x^3}{\sqrt{x^2-x^4}}=2x-\frac{1-2x^2}{\sqrt{1-x^2}}$$
The value $x_0$ of $x$ that maximizes $f(x)$ satisfies $f'(x_0)=0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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number of arrangements of $n$ zeroes and $n$ ones so that each zero has a neighbor zero and each one has a neighbor one this question is similar to a question in whitworth's choice and chance.
he gives the answer as $1 + (C(n-2,0) + C(n-3,1))^2 + (C(n-3,1) + C(n-4, 2))^2 + \ldots$
i found a combinatorial "proof" by di... | To get the generating function for the number of ways to arrange $m$ $0$s and $n$ $1$s, I generated all blocks of at least $2$ $x$s and at least $2$ $y$s, starting with an optional block of at least $2$ $y$s and ending with an optional block of at least $2$ $x$s
$$
\begin{align}
&\left(1+\frac{y^2}{1-y}\right)\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$ Find all solutions for $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$ if $0\leq x< \pi$
Can you verify my work? Thanks!
$$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$$
$$\cos(2x+x)=\frac{1}{\sqrt{2}}$$
$$\cos(3x)=\fr... | Right idea, but after reducing, you'll get cos(3x) = 0 (reference circle), so you'll be left with x=π/6, x=π/2 and x=5π/6 after simplifying.
Realize that 5π/4 and 23π/12 are >π, all larger than π will not be included with your solution.
EDIT: SEE COMMENT BELOW.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/784603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to integrate $\sqrt{1-\sin 2x}$? I want to solve the following integral without substitution:
$$\int{\sqrt{1-\sin2x}} \space dx$$
I have:
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$
but this can be written in two ways: $... | hint: $1- \sin (2x) = (\sin x - \cos x)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Solving $315 x \equiv 5 \pmod {11}$ I have to solve this: $$315 x \equiv 5 \pmod {11}$$
Isn't it like that?
$$315 \equiv (22+8) \cdot 10+15 \equiv 8 \cdot 3+4 \equiv 5+8+4 \equiv 6$$
Or have I done something wrong?
| $315\equiv 7\pmod{11}\\7x\equiv5\pmod{11}\\-4x\equiv-6\pmod{11}\\2x\equiv3\pmod{11}\\2x\equiv-8\pmod{11}\\x\equiv-4\pmod{11}\\x\equiv7\pmod{11}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solving a non-exact differential I started off solving the differential equation $$(xy^2 + 3e^{x-3})dx - x^2ydy = 0$$
It's a non-exact first order equation whose integrating factor is $1/x^4$.
Finally I got to the equation where I needed to integrate $$\frac{e^{x-3}}{x^4} dx$$ But I can't seem to proceed.
Is there any... | First, we rearrange the given equation (which is a Bernoulli equation):
$$
x^2y'y - xy^2 = 3e^{x-3} \Rightarrow 2y'y - \frac{2y^2}{x} = \frac{6e^{x-3}}{x^2}.
$$
Notice that we can simplify this result by letting $z(x) := y^2(x)$:
$$
z' - \frac{2z}{x} = \frac{6e^{x-3}}{x^2}.\tag{1}\label{eq:z}
$$
Let $\mu(x) := e^{\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Roots of polynomial $x^3-3\sqrt 5x^2+13x-3\sqrt 5$ given the factor $x-\sqrt 5$
Given that $x-\sqrt 5 $ is a factor of the cubic polynomial
$x^3-3\sqrt 5x^2+13x-3\sqrt 5$, find all the values of
the polynomial
After the long division method I get $x^2-2\sqrt 5x+3$.
Now how to split the middle term to find the all... | As an alternate method, you can use Vieta's formulas: if $a, b, c$ are the three roots, you have
$$\begin{eqnarray}
(i) & a+b+c&=&3\sqrt{5}\\
(ii) & ab+bc+ca&=&13\\
(iii) & abc&=&3\sqrt{5}\end{eqnarray}$$
One of the roots, say $c$, is $\sqrt{5}$, so
*
*from $(i)$, $a+b=2\sqrt{5}$,
*from $(iii)$, $ab=3$,
So
$$(a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another Inequality Let $a$, $b$ and $c$ be positive reals such that $$\dfrac{a^3+2a}{a^2+1}+\dfrac{b^3+2b}{b^2+1}+\dfrac{c^3+2c}{c^2+1}=\dfrac 92,$$
then is it true that $\dfrac 1a+\dfrac1b+\dfrac1c\ge3$ ?
| $$\sum_{cyc}\frac{1}{a}-3=\sum_{cyc}\left(\frac{1}{a}-1\right)=\sum_{cyc}\left(\frac{1}{a}-1+\frac{a^3+2a}{a^2+1}-\frac{3}{2}\right)=$$
$$=\sum_{cyc}\frac{(a-1)^2(2a^2-a+2)}{2a(a^2+1)}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/794533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Maximum and Minimum value of an inverse function
Find the maximum and minimum value of
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$.
given that $-1\le x\le 1$
I have solved the problem but i am just curious to know if there are any other ways to solve this particular problem other than the method i used belo... | Over the interval $[-1,1]$ the functions $\arcsin(x)$ and $\arccos(x)$ have a constant sum, $\frac{\pi}{2}$.
It follows that the problem is equivalent to finding the minimum and maximum value of
$$ g(t) = t^3+\left(\frac{\pi}{2}-t\right)^3 $$
over the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. By setting $t=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/795642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the locus of $2/z$ given that $|z-(1+i)| = 2$
If complex numbers $z$ satisfy the equation $|z-(1+i)| = 2$ and $\displaystyle \omega = \frac{2}{z}$, then locus traced by $\omega$ in complex plane, is ...
My try
I want to solve it geometrically. Here $|z-(1+i)| = 2$ Represent a Circle whose center is at $(1,1)$ an... | Some ideas:
$$|z-(1+i)|=2\iff|z|\left|1-\frac{1+i}z\right|=2\iff|w|=\left|\frac2z\right|=\left|1-\frac{1+i}z\right|$$
If $\;z=x+iy\;$ , then
$$\frac{1+i}z=\frac{(1+i)\overline z}{|z|^2}=\frac{x+y}2+\frac{x-y}2i\implies 1-\frac{1+i}z=\frac{2-(x+y)}2-\frac{x-y}2i$$
But
$$2=|z-(1+i)|=|(x-1)+(y-1)i|\implies (x-1)^2+(y-1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/797849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Why does $\frac{4}{2} = \frac{2}{1}$? I take for granted that $\frac{4}{2} = \frac{2}{1}$.
Today, I thought about why it must be the case. My best answers amounted to $\frac{4}{2}=2$ and $\frac{2}{1}=2$; therefore $\frac{4}{2}=\frac{2}{1}$. However, that explanation seems circular:
*
*one can express $2$ as $\frac... | If $c \neq 0$, do you agree that $\frac{c}{c} = 1$? Do you also agree that 1 is a unit and a multiplicative identity element?
Given any rational number $\frac{a}{b}$, if we multiply both the numerator and the denominator by the same number, we're computing $\frac{a}{b} \times \frac{c}{c} = \frac{a}{b} \times 1$. So, if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/798687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Finding a Jacobian determinant of transformation I want to find the Jacobian determinant of the transformation:
$$\begin{align}
\left(\begin{array}\\x\\y\\z\end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \left(\b... | $\def\d{\partial}$
To find the determinant of the transformation matrix, we need to complete the Jacobian matrix:
$$\begin{align}
\left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find all $m,n$ such that $mn|m^2+n^2+1$? How to find all positive integers $m,n$ such that $m|n^2+1$ and $n|m^2+1$ ?
My work:- Let $n^2+1=mk , m^2+1=ng$ , then $mkg=gn^2+g=m^2n+n+g$ , hence
$m|n+g$ i.e. $m|n+\dfrac{m^2+1}n$ i.e. $\dfrac{n^2+m^2+1}{mn}$ is an integer i.e. $mn|m^2+n^2+1$ . Now conversely
assume $... | Write a formula and although it is possible to draw some more.
Solutions of the equation: $\frac{X^2+aX+Y^2+bY+c}{XY}=j$
If a root. $t=\sqrt{(b+a)^2+4c(j-2)}$
Then using the equation Pell: $p^2-(j^2-4)s^2=1$
Solutions can be written:
$X=\frac{(b+a\pm{t})}{2(j-2)}p^2+(t\mp{(b-a)})ps-\frac{(b(3j-2)+a(6-j)\mp{(j+2)t})}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
We c... | Let us make the change of variables
$$v=\frac{x^3}{1+x^3}\iff x=\left(\frac{v}{1-v}\right)^{1/3}$$
This transforms the integral $I$ to the following form
$$
I=-\frac{1}{3}\int_0^1\log(v)\,v^{-1/3}(1-v)^{-2/3}dv
$$
Now, If
$$f(\alpha):=B(\alpha,\frac{1}{3})=\int_0^1v^{\alpha-1}(1-v)^{\frac{1}{3}-1}dv=\frac{\Gamma(\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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How prove $((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$ let $x,y,z>0$, and such $$x^2+y^2+z^2=x^2y^2+y^2z^2+x^2z^2$$
show that
$$((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$$
My try: let
$$x-y=a,y-z=b,z-x=c\Longrightarrow a+b+c=0$$
then we only prove
$$2[a(a+b)^2+b(b+c)^2+c(c+a)^2]\g... | Indeed, we can prove the following sharper inequality holds
$$2\big[(x-y)(y-z)(z-x)\big]^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}$$
Without loss of generality, assume that $x \geq y \geq z > 0,$ then
\begin{aligned} 2\big[(x-y)(y-z)(z-x)\big]^{2} \ & \leq 2(x-y)^2x^2y^2 \leq 2(x-y)^2(x^2y^2+y^2z^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integral involving the spherical Bessel function of the first kind: $\int_{0}^{\infty} x^{-n}j_{n+1}(x) \, dx$ Let $j_{n}(x)$ be the spherical Bessel function of the first kind of nonnegative integer order $n$.
How can I prove the equation below using the spherical Bessel function recurrence relation?
$$ \int_0^\inf... | EDIT:
I think the OP was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x). \tag{1} $$
From $(1)$ it follows immediately that $$\begin{align}\int_{0}^{\infty}x^{-n} j_{n+1}(x) \, dx &= -x^{-n} j_{n}(x) \Bigg|^{\infty}_{0} \\ &=-x^{n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are ... | Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
when the transformation $x = \tanh(t)$ is made. The resulting integral is given by
\begin{align}
I &= \int_{0}^{\infty} \ln\left( \frac{1+\tanh(t)}{1-\tanh(t)}\right) \frac{\cosh^{2}(t)}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
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How find this sum $ \frac{1}{1999}\binom{1999}{0}-\frac{1}{1998}\binom{1998}{1}+\cdots-\frac{1}{1000}\binom{1000}{999}$
prove or disprove :
$$S=\dfrac{1}{1999}\binom{1999}{0}-\dfrac{1}{1998}\binom{1998}{1}+\dfrac{1}{1997}\binom{1997}{2}-\dfrac{1}{1996}\binom{1996}{3}+\cdots-\dfrac{1}{1000}\binom{1000}{999}=\dfrac{1}{1... | This sum can also be done using Wilf / generatingfunctionology. Let
$$a_n = \sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k}$$
and introduce the generating function
$$f(z) = \sum_{n\ge 1} a_n z^n
= \sum_{n\ge 1}
\sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n
\\ =
\sum_{n\ge 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I thi... | For $t \in (0,\frac{\pi}{n})$, we have $\sin kt > 0$, for $k \in 1,\cdots,n-1$.
Thus from AM-GM inequality, $\dfrac{\sin (k+1)t}{\sin kt}a_k^2+\dfrac{\sin kt}{\sin (k+1)t}a_{k+1}^2 \ge 2a_ka_{k+1}$,
with equality iff $a_{k+1}\sin kt = a_k\sin (k+1)t$.
Summing over we have, $\displaystyle \sum\limits_{k=1}^{n-1} \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
} |
Measure of an Interior Angle Triangle $ABC$ has $AC = BC$ , $\angle ACB = 96^\circ$ . $D$ is a point in $ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$ . What is the measure (in degrees) of $\angle ACD$ ?
| If we put $AB=1$ and apply the low of sines, we get
$$AC=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\sin 42^\circ}{\cos 6^\circ},
\\ AD = \frac{\sin 30^\circ}{\sin 132^\circ}=\frac{1}{2\cos 42^\circ}.$$
But then it follows that $AC=AD$ from $2 \sin 42^\circ \cos 42^\circ=\sin 84^\circ = \cos 6^\circ$.
So $\triangle ADC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to calculate this double integral I have this double integral:
$$ \int_0^\frac{\pi}{4}\int_0^\frac{\pi}{2}(\cos x + \sin y) \, dy \, dx $$
Is this correct?
$$\begin{align*} \int_0^\frac{\pi}{2}(\cos x + \sin y) \, dy &= \int_0^\frac{\pi}{2}\cos x \, dy + \int_0^\frac{\pi}{2}\sin y \, dy \\ &= \cos x\int_0^\frac{\... | You are doing fine just be careful of the antiderivative of $\sin x$ which is $-\cos x$. So the only mistake that I am seeing is in the first line: $$\int_{0}^{\pi/2}\sin ydy=[-\cos y]_{0}^{\pi/2}=1-0$$ not $-1$.
You have written $[-\cos \dfrac{\pi}{2}-1]$ instead of $[-\cos \dfrac{\pi}{2}+1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it do... | Note that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$, and $\sin^2\theta+\cos^2\theta=1$. Hence
$$
\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\dfrac{1+\dfrac{\sin^2\theta}{\cos^2\theta}}{1+\dfrac{\cos^2\theta}{\sin^2\theta}}=\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
} |
proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without... | Part I: Consider $f(x) = tan^{-1}\left(\dfrac{x}{1-x^2}\right) - tan^{-1}x - tan^{-1}(x^3)$, taking derivative of $f$:
$f'(x) = \dfrac{1+x^2}{1-x^2+x^4} - \dfrac{1}{1+x^2} - \dfrac{3x^2}{1+x^6} = \dfrac{(1+x^2)^2 - (1-x^2+x^4) - 3x^2}{1+x^6} = \dfrac{1+ 2x^2 + x^4 - 1 + x^2 - x^4 - 3x^2}{1+x^6} = 0$. Thus $f$ is const... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do I find the equation of a circle, given radius and centre coordinates? Say I am asked to find, in expanded form without brackets, the equation of a circle with radius 6 and centre 2,3 - how would I go on about doing this?
I know the equation of a circle is $x^2 + y^2 = r^2$, but what do i do with this information... | The general solution for the circle with centre $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2.
$$
Now, we have the centre $(2,3)$ and the radius $6$, therefore the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=6^2\\
x^2-4x+4+y^2-6x+9&=36\\
x^2+y^2-4x-6y+4+9-36&=0\\
\large\color{blue}{x^2+y^2-4x-6y-23}&\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
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.