Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proof that $y^2=x^3+x$ has a unique integer solution Prove that the equation $y^2=x^3+x$ has only one integer solution, namely $x=y=0$.
| We see that $x$ divides $y^2$. But $x^2+1$ is prime with $x$. Therefore any prime dividing $x$ must appear with an even power. Therefore $x=a^2$ and $a$ divides $y$.
Putting this into the equation we get $y=ay_1$ and
$$a^2y_1^2=a^2(a^4+1).$$
If $a\neq 0$, then $$y_1^2=a^4+1$$
or $$(y+a^2)(y-a^2)=1.$$
Therefore $a=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What converges the series? $$\sum _{n=4}^{+\infty \:}\frac{4^n\sqrt{n+1}-4^n\sqrt{n}+2^{n-1}\sqrt{n}\sqrt{n+1}}{4^n\sqrt{n}\sqrt{n+1}}$$
I tried different ways to figure this out and at first i did integrals
$$\int _4^{\infty \:}\frac{4^x\sqrt{x+1}-4^x\sqrt{x}+2^{x-1}\sqrt{x}\sqrt{x+1}}{4^x\sqrt{x}\sqrt{x+1}}dx$$
THEN... | $$\sum _{n=4}^{+\infty \:}\frac{4^n\sqrt{n+1}-4^n\sqrt{n}+2^{n-1}\sqrt{n}\sqrt{n+1}}{4^n\sqrt{n}\sqrt{n+1}} =\sum_{n=4}^{\infty} \left(\frac{1}{\sqrt{n}} -\frac{1}{\sqrt{n+1}}\right) +\sum_{n=4}^{\infty}\frac{1}{2^{n+1}} =\frac{1}{\sqrt{4}} +\frac{1}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/737845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we solve this using stars and bars? The number of ways of distributing 12 identical oranges among 4 children so that every child gets at least one and no child more than 4 is 31.
My try: First of all give each child 1 orange and we are left with 8 oranges.
Then
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2
and permuting each, w... | A simple solution by generating functions, ellipses cover terms that don't influence the result:
\begin{align}
[z^{12}] (z + z^2 + z^3 + z^4)^4
&= [z^{12}] \left( \frac{z (1 - z^4)}{1 - z} \right)^4 \\
&= [z^8] (1 - z^4)^4 (1 - z)^{-4} \\
&= [z^8] (1 - 4 z^4 + 6 z^8 - \cdots)
\sum_{n \ge 0} \binom{-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
| $(x+1)^4+(x-1)^4=16\\
\implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\
\implies (2x^4+12x^2+2)-16\\
\implies 2x^4+12x^2-14=0\\
\implies 2(x^4+6x^2-7)=0\\
\implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$
Now let $x^2=a$:
$2((a)^2+6(a)^1-7(a)^0)=0\\
\implies 2(a^2+6a-7)=0\\
\implies a^2+6a-7=0\\
\implies (a+7)(a-1)=0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$
| $$\begin{align}
\int_0^1 \frac{1+x^2}{1+x^4} \operatorname d \!x &=
\int_0^1 \frac{1+x^4-x^4+x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1 + x^4 - x^4-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \frac{1}{3} -\int_0^1 x^4\frac{1+x^2}{1+x^4} \oper... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that $ a^2 + b^2 + c^2 +2abc < 2$ let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that
$ a^2 + b^2 + c^2 +2abc < 2$
I used the fact that $ a+b-c>0 $ and multiplied the 3 cyclic equations but coudn't reach the final equation.
| Just continue your approach, focusing on reaching the goal:
$$(a+b-c)(b+c-a)(c+a-b) > 0 \iff (1-c)(1-a)(1-b) > 0 $$
$$\iff 1-(a+b+c)+(ab+bc+ca)-abc > 0$$
$$\iff 1-(2)+\tfrac12 \left((a+b+c)^2-a^2-b^2-c^2 \right)-abc > 0$$
$$\iff -2+ \left(4-a^2-b^2-c^2 \right)-2abc > 0$$
$$\iff a^2+b^2+c^2 + 2abc < 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Inductions and proofs Let $(h_n)$ be a sequence defined by $h_0 =1 , h_1 = 2, h_2 =3$ and $h_n = h_{n-1} + h_{n-2} + h_{n-3}$, for all $n\ge 3$. Prove that $h_n\le 2^n$ , for all $n\ge 0$
Not sure how to go with this problem?
| Let's prove it by strong induction on $n$.
Base case: $n = 0$. $h_0 = 1 = 2^0$. So assume it is true for $n < k$. We prove it is true for $n = k+1$.
$h_{k+1} = h_k + h_{k-1} + h_{k-2} \leq 2^k + 2^{k-1} + 2^{k-2} = \dfrac{2^{k+1}}{2} + \dfrac{2^{k+1}}{4} + \dfrac{2^{k+2}}{8} = \dfrac{7}{8} \cdot 2^{k+1} < 2^{k+1}$.
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Coefficients of this generating function For the first part of a problem, I solved the generating function to be $F(x) = \frac{x^3}{(1-x)^2}$
Now it's the easy part that has me a little confused. What would the coefficients be in this case?
I have computed so far:
$$\frac{x^3}{(1-x)^2} = x^3 (1+2x+3x^2 + \dots) =x^3 +... | There is the binomial theorem, which says that
$$(1+x)^n = \sum_{k \ge 0} \binom{n}{k} x^k$$
This is true even when $n$ is not necessarily a positive integer; we just have to define the binomial coefficient properly: for nonnegative integer $k$ and any $n$, we define
$$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}$$
W... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac{-2}{\sqrt{16-x^2}}$ So I am asked to find the anti-derivative of
$$\frac{-2}{\sqrt{16-x^2}}$$
First step I took was to make it easier for me to visualise
$$\int{-2(16-x^2)^{-\frac{1}{2}}}dx$$
I let $u = 16-x^2$
So $\frac{du}{dx} = -2x du$
So substituting in $u$ we have
$$=\int{-2(u)^{-\frac{1}{2}}}dx... | trigonometric substitution isn't that difficult. It just requires practice.
try expressing
$$\frac{-2}{\sqrt{16-x^2}}$$
as
$$\frac{-2}{\sqrt{16(1-(\frac{x}{4})^2)}} = \frac{-2}{4\sqrt{(1-(\frac{x}{4})^2)}}=\frac{-1}{2\sqrt{(1-(\frac{x}{4})^2)}}=-\frac{1}{2\sqrt{(1-\sin^2(\sin^{-1}(\frac{x}{4})))}}$$
so for the integral... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How do I evaluate this limit with an integral function? Could anybody give me some pointers on how do I evaluate this limit:
$$\lim_{t \to +\infty} \frac{\int_0^t x^9e^{-x^2}\,dx}{\int_0^t x^7e^{-x^2}\,dx}$$
I'm not asking for the complete solution, just a hint to point me in the right direction. I appreciate your help... | $\frac{\int_0^tx^9e^{-x^2}dx}{\int_0^tx^7e^{-x^2}}=\frac{-\frac{1}{2}(e^{-t^2}(t^8+4t^6+12t^4+24t^2+24)+24)}{-\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6)+6)}=\frac{\frac{1}{2}e^{-t^2}(t^8+4t^6+12t^4+24t^2+24+24e^{t^2})}{\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6+6e^{t^2}))}=\frac{t^8+4t^6+12t^4+24t^2+24+24e^{t^2}}{t^6+3t^4+6t^2+6+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/746706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Area of triangle inscribed in a parabola How can u prove that the area of the triangle inscribed in a parabola is twice the area of the triangle formed by the tangents at the vertices?
| Claim: Given any 2 points on the parabola, the area that is bounded by the line between these 2 points and the parabola, is twice the area that is bounded by the 2 tangents and the parabola.
Proof. Standard calculus techniques work.
WLOG, we normalize the quadratic to be of the form $ y = x^2$. Let the points be $ (a, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/750501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I find integers $n \gt 1$ such that the average of $1^2,2^2,3^2...n^2$ is itself a perfect square.
$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$
so we would like to solve
$6k^2=(n+1)(2n+1)$
here we see that $6|(n+1)(2n+1)\implies 2|n+1$
hence we can set $n=2j-1$
$2j(4j-1)=6k^2\implies j(4j-1)=3k^2$
how can i ... | We are looking for $n,k\in\mathbb{N}$ such that $k^2=\frac{1}{n}\sum^n_{i=1}i^2=\frac{(n+1)(2n+1)}{6}$. Set $n_0=n+1$.
The equation becomes $n_0(2n_0-1)=6k^2$. So $n_0$ is even.
Set $n_0=2n_1$ to get $n_1(4n_1-1)=3k^2$. Now suppose $3|n_1$ and set $n_1=3n_2$. Then $n_2(12n_2-1)=k^2$. Since $n_2$ and $12n_2-1$ are cop... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find sequence function and general rule the function $$a_{n+2}=3a_{n+1}-2a_n+2$$
is given, and $$a_0=a_1=1, (a_n)_{n\ge0}$$
multiplying everything by $$/\sum_{n=0}^\infty x^{n+2}$$
also adding $$\sum_{n=0}^\infty (a_{n+2}x^{n+2}+a_1x+a_0)-a_1x-a_0=\sum_{n=0}^\infty (3a_{n+1}x^{n+2}+a_0)-a_0-\sum_{n=0}^\infty 2a_nx^{n+2... | $a_{n+2}=3a_{n+1}-2a_n+2$, $a_0=a_1=1$
Doing it the more pedestrian way, you would first find the basis solutions of the homogeneous equation that are geometric progressions, $a_n=q^n$. This is the case if
$$
q^2=3q-2\iff 0=(q-1)(q-2),
$$
so that the general homogeneous solution is
$$
a_n=C+D\,2^n
$$
Since the inhomo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/753074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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find sum of first 2002 terms if $\left \{ a_n \right \}$ is sequence of Real Numbers for $n \ge 1$ such that
\begin{equation}
a_{n+2}=a_{n+1}-a_n \tag{1}
\end{equation}
\begin{equation}
\sum_{n=1}^{999} a_n=1003 \tag{2}
\end{equation}
\begin{equation}
\sum_{n=1}^{1003}a_n=-999 \tag{3}
\end{equation}
Then Find the val... | Anaother way to solve the problem is to use the characteristic equation. The characteristic equation is
$$ r^2=r-1 $$
whose solution is $r_1=e^{\frac{\pi}{3}i}, r_2=e^{-\frac{\pi}{3}i}$. So $a_n$ can be expressed as
$$ a_n=C_1e^{\frac{n\pi}{3}i}+C_2e^{-\frac{n\pi}{3}i} $$
where $C_1, C_2$ are constants which will be d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help... | There's another way to get to answer if we take $a=c+d$ and $b=c-d$, then using binomial expansions we have
$$
a^2+b^2=2(c^2+d^2)=6\\
a^3+b^3=2c^3+6cd^2=14
$$
We have $d^2=3-c^2$ from first equation and using this in second equation
$$
c(9-2c^2)=7
$$
We can easily see that one of the solutions is c=1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from sol... | Let $\delta x = y-x$ and $x\ge y$. Then
$$4^x(1+4^{\delta x}) = 2^{x+1}(1+2^{\delta x})$$
then from factorization we can claim that:
$$2x = x+1 ~\text{and}~~ 4^{\delta x} = 2^{\delta x}$$
So, we get $x = y = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 6
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Probability of number formed from dice rolls being multiple of 8
A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multip... | Let's use Zhuli's notation.
$c$ has a probability of $\frac{1}{2}$ to be even.
Now if $\frac{c}{2}$ is even (i.e., $4|c$), then $4|2b+c \iff b$ is even. If $\frac{c}{2}$ is odd, then $4|2b+c \iff b$ is odd. Both conditional probabilities are $\frac{1}{2}$ therefore the marginal probability $4|2b+c$ is $\frac{1}{4}$.
Ne... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Normalizer of a subgroup of $GL_2(\mathbb{R})$ I have following subgroup of $GL_2(\mathbb{R}):$ $$A=\Bigg\{\left( \begin{array}{cc}
1 & 0 \\
0 & 1 \end{array} \right),\left( \begin{array}{cc}
0 & -1 \\
1 & 0 \end{array} \right),\left( \begin{array}{cc}
0 & 1 \\
-1 & 0 \end{array} \right),\left( \begin{array}{cc... | $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in N_{GL_2(\Bbb R)}(A)\implies $$
$$\begin{align*}
\bullet&\begin{pmatrix}b&\!\!-a\\d&\!\!-c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\!\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating absolute value and argument of a complex number I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$.
In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:
$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{... | I would start by finding the polar form of $z = \sqrt{3} - i$. This corresponds to a nice triangle, the so called 30-60-90 triangle. Here one leg is of length $\sqrt{3}$ and the other is of length 1 (in the negative imaginary direction). The hypotenuse of this triangle is the magnitude of $z$, so $|z|=2$.
The argument ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polar Equation to Rectangular? The equation is:
$$r = \frac{4}{1+2sin(\theta)}$$
I'm confused about how to convert it into rectangular form.
This is what I have so far, although I'm not sure it's correct:
$$r = \frac{ 4(1-2sin(\theta)) }{ 1-2sin^2(\theta) } $$
$$r = \frac{ 4-8sin(\theta) } {( cos^2(\theta) - sin^2(\th... | $$r = \dfrac{4}{1+2 \sin(\theta)} \rightarrow r(1 + 2 \sin(\theta)) = 4$$
A polar plot shows:
We have:
*
*$r = \sqrt{x^2 + y^2}$
*$x = r \cos(\theta)$
*$y = r \sin(\theta)$
So,
$$r(1 + 2 \sin(\theta)) = r + 2 r \sin(\theta) = \sqrt{x^2 + y^2} + 2 y = 4$$
We can re-write this as:
$$3y^2 - 16y - x^2 + 16 = 0$$
| {
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"url": "https://math.stackexchange.com/questions/765262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int_0^\pi\frac{dx}{\sqrt{3-\cos(x)}}$?
How to integrate $\displaystyle\int_o^\pi\frac{dx}{\sqrt{3-\cos(x)}}$ ?
If I take $y=\sin\left(\frac{x}{2}\right)$ then,
$\displaystyle dx=\frac{2dy}{\cos\left(\frac{x}{2}\right)}=\frac{2dy}{\sqrt{1-y^2}}$
and with $\displaystyle\cos(x)=\cos^2\left(\frac{x}{2}... | The integral can be written as an elliptic integral of the first kind with complex elliptic modulus:
$$I=\int_{0}^{\pi}\frac{d\phi}{\sqrt{3-\cos{\phi}}}\\
=\int_{0}^{\pi}\frac{d\phi}{\sqrt{2+2\sin^2{\left(\frac{\phi}{2}\right)}}}\\
=\frac{1}{\sqrt2}\int_{0}^{\pi}\frac{d\phi}{\sqrt{1+\sin^2{\left(\frac{\phi}{2}\right)}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving the area of an equilateral triangle How do you prove that How do you prove that for any equilateral triangle with side length s, area is $\frac{s^2 √3}{4}$ ? I tried using an equilateral triangle in a square, but I keep coming up with a $2x^2√3$ , as shown below. What am I doing wrong?
I started with the follow... | Another approach would be using calculus. For instance, consider the line $f(x) = \sqrt{3}x $. Notice the line from $(0,0)$ to $(\frac{s}{2}, f( \frac{s}{2}) ) $ is the hypothenuse of half the equilateral triangle. The other sides are $\frac{s}{3}$ and $\sqrt{3} \frac{s}{2} $. So, the area of half of the equilateral tr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Last two digit of number raised to exponents.
Find the last two digits of $3^{3^{2014}}$.
Attempt:
First I try to work with $3^{2014}$. So we can work on $\text{mod 10}$.
Then, $$\begin{align}&3^1 \equiv 3\pmod{10}\\ &3^2 \equiv 9 \pmod{10}\\ &3^3 \equiv 27 \equiv 7\pmod{10}\\ &3^4 \equiv 81 \equiv 1\pmod{10}\end{a... | Hint: by Euler's totient theorem, $3^{40} \equiv 1 \pmod{100}$. Alternatively, note (by playing with a calculator) that $3^{20} \equiv 1 \pmod{100}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Formula for $\sum_{k=0}^n k^d {n \choose 2k}$ If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$
We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = \frac{n(n+1)2^n}{32}$.
Note that ${n \choose 2k} = 0$ when $2k > n$.
| This is an half-answer. We have
$$(1+x)^{n}=x^{0}\binom{n}{0}+x^{1}\binom{n}{1}....+x^{n}\binom{n}{n}$$
If we take derivative of this with respect to x.
$$n(1+x)^{n-1}=1x^{0}\binom{n}{1}+2x^{1}\binom{n}{1}....+nx^{n-1}\binom{n}{n}$$
For $x=1$ we have $$n2^{n-1}=1\binom{n}{1}+2\binom{n}{2}....+n\binom{n}{n}$$
Now if we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Problem of Number Theory Can you give a tip for this question?
For all $ n $ odd positive integer, let be
$ F(n) = \# \{ a($mod $ n): a^{n-1} \equiv 1($mod $n) \} $
that is the number of elements of the set.
*
*Show that
$$ F(n) = \prod_{p|n} gcd(p-1,n-1) $$
*Let be $ F_0(n) $ the numbers $ a($mod $n) $ such that $... | I have skimmed over some steps; you can try filling in the details.
1) Write $n=\prod_{i=1}^{m}{p_i^{a_i}}$, where $p_i$ are odd primes.
2) There exists a primitive root $\pmod{p_i^{a_i}}$.
3) Using 2), conclude that the number of solutions $\pmod{p_i^{a_i}}$ of $a^k \equiv 1 \pmod{p_i^{a_i}}$ is $\gcd(\varphi(p_i^{a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How do I evaluate this integral by hand? TL;DR how do I evaluate $\int_0^{2 \pi } \frac{1}{\cos ^2(\theta )+1} \, d\theta$ by hand?
I'm trying to solve this problem:
Find the volume of the region defined by $x^2+xy+y^2+yz+z^2\le1$.
(The answer is $\frac{4 \sqrt{2} \pi }{3}$)
I have tried several tactics. First of all... | Here is an approach. First evaluate the indefinte integral using the substitution $\theta=\arctan(u)$ which gives
$$ I=\int \frac{d\theta}{1+\cos^2(\theta)} = \int \frac{du}{u^2+2} =\frac{1}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right).$$
Then, split the interval of integration as $[0,\pi/2]\cup[\pi/2,\pi]\cup [\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $(2x^3+3x^2y+y^2-y^3)dx+(2y^3+3xy^2+x^2-x^3)dy=0$
Solve the differential equation
$$(2x^3+3x^2y+y^2-y^3)dx+(2y^3+3xy^2+x^2-x^3)dy=0$$
I don't know how to solve it, and I have tried many methods,but they didn't work!
| Hint: Let:
$$y = v x \rightarrow \dfrac{dy}{dx} = v + x \dfrac{dv}{dx}$$
Substitute this into
$$(2x^3+3x^2y+y^2-y^3)+(2y^3+3xy^2+x^2-x^3)~\dfrac{dy}{dx} = 0$$
It looks nasty!
Anyway, you will get a solvable equation after you use an integrating factor to make it an exact equation and end up with:
$v(x) = \dfrac{\sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
= $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
= $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
= $(k+1)^2(k^2+2k+2) k (k+2)$
=... | If $f(n)=n^2(n^4-1)=n^6-n^2$ is divisible by $60$ for $n=k$
$$f(k+1)-f(k)=(k+1)^6-(k+1)^2-\{k^6-k^2\}$$
$$=6k^5+15k^4+20k^3+15k^2+6k-2k$$
which is $\displaystyle6(k^5-k)+15k^4+20k^3+15k^2+10k$ divisible by $5$ as $k^5-k$
Again, $\displaystyle6k^5+15k^4+20k^3+15k^2+4k\equiv2(k^3-k)\pmod3\equiv0$
finally, $\displaystyle6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think... | $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle\cos4x=-\cos3x=\cos(\pi+3x)\implies4x=2n\pi\pm(\pi+3x)$ where $n$ is any integer
'+' $\displaystyle\implies x=(2n+1)\pi\equiv\pi\pmod{2\pi}\implies\cos x=-1$
'-' $\displaystyle\implies x=\frac{(2n-1)\pi}7$
Now $\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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If $\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$ then $x^2 - 44x - 36 = 0$ holds for $x=\sin 2A$ If
$$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$
then prove that
$$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$
I have no idea how to solve it. Plz help.
| I think you have the problem wrong, you can show $\sin 2A$ is a root of $x^2-44x+36$ using Awesome's hint.
EDIT:
If $\sin 2A$ is a root of $x^2-44x+36$, then
$$\sin 2A = \frac{44 \pm \sqrt{1936 - 144}}{2}=22\pm 8\sqrt{7}$$
Since $|\sin 2A|\leq 1$, we have that $\sin 2A = 22-8\sqrt{7}$.
Using the hint by Awesome,
\beg... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt$
Calculate the following integral for $n \in \mathbb{Z}$ with the
residue theorem $$\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt$$
So far I have tried two approaches. Firsty, for $n\geq 0$:
$$\begin{align*}\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt ... | Notice that
$$z^{2(2n+1)} + 1 = (z^2)^{2n+1} + 1^{2n+1} = (z^2 + 1) \sum _{k=0} ^{2n} (z^2)^k (-1)^{2n-k} \ ,$$
and the terms with $k < n$ are killed by the derivative of order $2n$, while the ones with $k > n$ survive it but are killed by the evaluation at $0$, leaving $z^{2n}$ (corresponding to $k=n$) as the sole sur... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $ If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2... | Noting $ab+bc+ca \le a^2+b^2+c^2$ and $a^4+b^4+c^4 \ge \frac13(a^2+b^2+c^2)^2$, it is sufficient to show:
$$\sum_{cyc} \frac{a^3}{a^2+b^2+c^2-bc} \le \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc}$$
Now $a^2+(b^2+c^2-bc) \ge a^2+bc$. So it is enough to show
$$ \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc} \ge \sum_{cyc} \frac{a^3}{a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it I'm writing some software that takes a group of users and compares each user with every other user in the group. I need to display the amount of comparisons needed for a countdown type feature.
For example, this group [1,2,3,4,5] would be analyse... | $$N=2:\ 1 + 2 = (1 + 2) = 1\times3$$
$$N=4:\ 1 + 2 + 3 + 4 = (1 + 4) + (2 + 3) = 2\times5$$
$$N=6:\ 1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) + (2 + 5) + (3 + 4) = 3\times7$$
$$N=8:\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= (1 + 8) + (2 + 7) + (3 + 6)+ (4 + 5) = 4\times9$$
More generally, $N/2\times(N + 1)$.
For odd $N$, sum the $N-1$ fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Integral $\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx$ $$I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx=\frac{(1+\alpha)\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.$$
This one looks very nice. It has ... | Assuming the $3\alpha x$ term is in fact $3\alpha x^2$ (otherwise the numerical results do not match).
$$\begin{align*}
I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx\\
&=\int_0^\infty \ln x\, d\left(-\alpha x^{-3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that
$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $
( don't apply Schur's in... | Let $x=a^2$, $y=b^2$, $z=c^2$ and $f(x,y,z)=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}$ and g(x,y,z)=x+y+z. Consider the restriction $\{(x,y,z) \in \mathbb{R}^3;g(x,y,z)=1\}$. Applying the method of Lagrange multipliers, we have
$$\frac{y}{1+z}+\frac{z}{1+y}-\frac{yz}{(1+x)^2} = \lambda$$
$$\frac{z}{1+x}+\frac{x}{1+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How is this automorphism an inversion? And how is this group abelian? Let $G$ be a finite group, and let $T$ be an automorphism of $G$ which sends more than three-quarters of the elements of $G$ onto their inverses. Then how to demonstrate that $T(x) = x^{-1}$ for all $x$ in $G$? And, how to show that $G$ is abelian?
... | I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
These problems are Problem 12 and Problem 13 on p.71 in Herstein's book.
I solved Problem 13 as follows:
I used GAP and I found the following example.
Let $H:=\{(), (1\text{ }2), (3\text{ }4), (1\text{ }2)(3\text{ }4), (1\text{ }3)(2\text{ }4), (1\text{ }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove that the following function is convex? I want to prove convexity of the following function:
$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$
for any fixed $a, b \in (0, 1)$ and:
*
*$x\in(0,1)$
*$x\in(1, \infty)$
I'm trying to solve it for a very long time, tried to investigate the sign ... | We have $$f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)
= \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x}
=\ln \left(\left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)^{1/\ln x}\right)$$
Then we may show Log-Convexity:
For $f(x)=\log g(x)$ (on convex domains $x\in(0,\infty)/1$):
$$ f(x) \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it sp... | Here is another solution. Remark that
$$
\sqrt{n^2+n}-n=\sqrt{n^2+n}-\sqrt{n^2}=\frac{(\sqrt{n^2+n}-\sqrt{n^2})(\sqrt{n^2+n}+\sqrt{n^2})}{\sqrt{n^2+n}+\sqrt{n^2}}
$$
Then
$$
\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}.
$$
Since $\lim\limits_{n\to \infty} \sqrt{1+\frac{1}{n}}=1$, for $\delta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 0
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Rational polynomials I'm not very familiar with algebra and was wondering if there are any results regarding the effective order of rational polynomials (i.e. rational functions).
Specifically: given $P(z)$ and $Q(z)$ as polynomials in $z$ with real coefficients of order $p$ and $q$ respectively, is there a way to kno... | Euclid's GCD algorithm can be used to find common factors between two polynomials. For instance:
$$
\begin{align}
\gcd(x^3 + 1, x^4 + x^2 + 1) & = \gcd(x^3 + 1, x^4 + x^2 + 1 - x(x^3 + 1))
\\ & = \gcd(x^3 + 1, x^2 - x + 1)
\\ & = \gcd(x^3 + 1 - x(x^2 - x + 1), x^2 - x + 1)
\\ & = \gcd(x^2 - x + 1, x^2 - x + 1)
\\ &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$ ab-1|a^2+ab+b^2 $ I hava a number theory problem. I think on it yestarday night and today, afternoon.
The problem :
$ a,b $ are two natural numbers such that : $ ab>1 $
how many pairs $ (a,b) $ is there such that : $ ab-1|a^2+ab+b^2 $
we have $ ab-1|(a^2+ab+b^2)+(ab-1) = (a+b)^2 - 1^2 = (a+b+1)(a+b-1) $
so $ ab-1|(a+... | $a + b + 1$ and $a + b - 1$ are either relatively prime or have $2$ as a common divisor. In the first case, we must have $ab - 1$ divides $a + b + 1$ or $ab - 1$ divides $a + b - 1$. In the second case, there are a few more options to consider.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove Fibonacci sequence with matrices? How do you prove that:
$$
\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}^n
=
\begin{pmatrix}
F_{n+1} & F_n\\
F_{n} & F_{n-1}
\end{pmatrix}$$
| Let
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
And the Fibonacci numbers, defined by
$$\begin{eqnarray}
F_0&=&0\\
F_1&=&1\\
F_{n+1}&=&F_n+F_{n-1}
\end{eqnarray}$$
Then, by induction,
$$A^1=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix}$$
And if for $n$ the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
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To solve $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ How do we find all positive integers $(m,n)$ such that $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ ?
| Since you need $\frac 1m+\frac 1n\gt \frac 34$ you need either $\frac 1m\ge \frac 38\gt \frac 13$ or alternatively $\frac 1n \gt \frac 13$
The number of cases is small and they can be checked quickly.
To be more concrete about the cases involved, we must have either $m\leq 2$ or $n\leq 2$.
$m=1$ gives $\frac 1n-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Need a generalized way to solve $127=1+2+2^2+...$ How can I solve this equation generally? I can solve it by checking in my calculator. But I don't know any generalized way.
$$127=1+2+2^2+2^3+....+ 2^{x-1}$$
| The most intuitive way:
\begin{align*}
&\;\underbrace{1 + 1} +2+2^2+2^3+ \cdots + 2^{x-1} \\
&= \underbrace{2 + 2} + 2^2 + 2^3 + \cdots + 2^{x-1} \\
&\;\;\;= \underbrace{4 + 2^2} + 2^3 + \cdots + 2^{x-1} \\
& \; \\ & \; \\ %vertical space
&\quad \quad\quad \quad\quad \; \; \; \; \; \cdots \\
& \; \\ & \; \\ %vertical s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Indefinite Integral $\int \frac{dx}{\sqrt {ax^4-bx^2}}$ I am trying to Integrate
$$
I=\int \frac{dx}{\sqrt {ax^4-bx^2}}, \qquad a,b\in \mathbb{R}.
$$
Thanks.
I tried to do $x=\sin \phi$
$$
\int \frac{\cos \phi\, d\phi}{\sqrt{a\sin^4 \phi-b\sin^2 \phi}}=\int \frac{\cot \phi \, b\phi}{\sqrt{a\sin^2\phi-b}}
$$
but get stu... | $\displaystyle\int\dfrac{dx}{\sqrt{ax^4-bx^2}}=\displaystyle\int\dfrac{dx}{x\sqrt{ax^2-b}}=\dfrac{1}{\sqrt{a}}\displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}$
$\therefore x^2-\dfrac{b}{a}=t^2 \implies x\ dx=t\ dt$
$\therefore \displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}=\displaystyle\int\dfrac{dt}{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/787377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$\int_0^\infty x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$ This is my third question following the previous post.
Prove that \begin{equation} \int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)... | Examining the original integral, use the substitution $u=x^{-1}$ so that
$$\frac{du}{dx}=-x^{-2}\Rightarrow dx=-(u^{-2})du$$ and
$u\rightarrow 0$ when $x\rightarrow\infty$, $u\rightarrow \infty$ when $x\rightarrow0$
Therefore
$$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=-\int_{\infty}^0u^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/787703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Evaluate $a+b+c+d$ If $a$, $b$, $c$, and $d$ are distinct integers such that
$$(x-a)(x-b)(x-c)(x-d)=4$$
has an integral root $r$, what is the value of $a+b+c+d$ in terms of $r$?
I tried to analyze graphically by shifting the graph of
$f(x)=(x-a)(x-b)(x-c)(x-d)$ four units downward but couldn't infer anything due t... | Plugging in $x=4$ shows that $4$ can be written as product of four distinct integers. The only divisors of $4$ are $\pm1, \pm2,\pm4$.
The only way to obtain $4$ is as product $(-2)\cdot(-1)\cdot 1\cdot 2$. Then $(r-a)+(r-b)+(r-c)+(r-d)=-2+(-1)+1+2=0$, i.e. $a+b+c+d=4r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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To evaluate $\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx $ and $\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx$ How to evaluate the following integrals?
$$\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx $$
$$\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx$$
| Using what Lucian answered $$\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx=\left(\text{Ci}\left(2+\frac{\pi }{2}\right)-\text{Ci}(2)\right) \cos
(2)+\left(\text{Si}\left(2+\frac{\pi }{2}\right)-\text{Si}(2)\right) \sin (2)$$
$$\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx=\Big(\text{Ci}(10)-\text{Ci}(2)\Big) \cos (2)-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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To solve $1+2^mp^2=q^5$ How do we find all posible solutions of $1+2^mp^2=q^5$ for positive integer $m$ and primes $p,q$ ? $m=1,q=3,p=11$ is a solution , is there any other solution ?
| Following Ivan Loh, $2^mp^2=(q-1)(q^4+q^3+q^2+q+1)$ implies $(q^4+q^3+q^2+q+1)$ equals either $p$ or $p^2$ (because it's odd). In either case $p\gt q^2$, which implies $p$ cannot divide $q-1$, hence we must have $q^4+q^3+q^2+q+1=p^2$ and thus $q-1=2^m$. Writing now $q=2^m+1$, we wind up with
$$2^{4m}+5\cdot2^{3m}+5\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof for inequality with $a,b,c,d$ with $d =\max(a,b,c,d)$
Let $a,b,c,d$ positive real numbers with $d= \max(a,b,c,d)$. Proof
that
$$a(d-c)+b(d-a)+c(d-b)\leq d^2$$
*
*I believe that the GM-AM inequality with $n=4$ variables might be helpful.
$$\sqrt[n]{x_1 x_2 \dots x_n} \le \frac{x_1+ \dots + x_n}{n}$$
We... | Divide both sides by $d^2$ to get an equivalent inequality:
$\dfrac{a}{d}\cdot \left(1 - \dfrac{c}{d}\right) + \dfrac{b}{d}\cdot \left(1 - \dfrac{a}{d}\right) + \dfrac{c}{d}\cdot \left(1 - \dfrac{b}{d}\right) \leq 1$.
Now let $x = \dfrac{a}{d}$, $y = \dfrac{b}{d}$, and $z = \dfrac{c}{d}$, then : $0 \leq x, y, z \leq1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/798721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Volume of the intersection two elliptical cylinders How can I find the volume of the intersection of the following elliptical cylinders?
$$\dfrac{x^2}{9} + \dfrac{y^2}{4} \leq 1$$
$$\dfrac{y^2}{4}+\dfrac{z^2}{9} \leq 1$$
In the first octant.
This is what I have done:
Let $A$ be the projection of the intersection in th... | That is right, but most of the work is actually to split $A$ into the proper bounds for each variable, and then do the actual integration. So write $A$ as \begin{equation} A = \{ (x, y) \in \mathbb{R}^2 : 0\leq x \leq 3\sqrt{1-y^2/4}\;,\; 0\leq y \leq 2\}, \end{equation} and then we have \begin{eqnarray} V&=&\int_0^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple... | Something is amiss. We have
$$\int_0^{\pi/2} \frac{dx}{2+\sin (nx)} = \frac{1}{n}\int_0^{n\pi/2} \frac{dy}{2+\sin y}.$$
Using
$$\int_0^{2\pi} \frac{dy}{2+\sin y} = \frac{2\pi}{\sqrt{3}},$$
writing $n = 4k + r$ with $0 \leqslant r \leqslant 3$, we find
$$\int_0^{\pi/2}\frac{dx}{2+\sin (nx)} = \frac{k}{4k+r}\cdot \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
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Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to d... | Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 9,
"answer_id": 8
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For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$? This question comes from the 1980 Putnam exam. My work is shown below.
For all integers $n \ge 1$, \begin{align}
(2n)!&=n!\cdot\underbrace{(n+1)(n+2)(n+3)\cdots(2n-2)(2n-1)(2n)}_{n \text{ terms}} \\
&\ge n! \cdot \underbrace{2... | In fact we have the following:
*
*The method presented in the proposal of the problem proves that any $c\geq \frac{1}{2}$ works.
*The solution of Leucippus proves that no $c< \frac{1}{2}$ works. So, these are not alternative solutions, but complementary ones. and the conclusion is that
$$
\left\{c:\forall\, x,~\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Can you find a ellipse so that its image is a circle? This is a "fun" question and I have already a solution. I asked this question so that I may see a different approach or an elegant solution.
Let $P$ be a plane with equation $x+y+z=1$. Find an ellipse on $x,y$ plane so that its "shadow" on $P$ is a circle.
Edit: By ... | The circle with radius 1 and center $(0,0,1)$ in the plane $x+y+z=1$ can be parametrized by
\begin{align}
\vec{p}(t) &= \langle 0,0,1 \rangle + \cos(t)\langle 1,-1,0 \rangle/\sqrt{2} + \sin(t) \langle 1,1,-2 \rangle/\sqrt{6} \\
&= \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos
(t)}{\sqrt{2}},\frac{\sin (t)}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How do you evaluate this limit? $\ln({x^3+2x^2+x})+ \frac{2}{x}$ How do you evaluate the following?
$$\lim_{x \to 0^+} \left [ \ln({x^3+2x^2+x})+ \frac{2}{x} \right ]$$
If I plug in $x$, I get $\infty-\infty$, which is undetermined, and I haven't been able to get the limit at a more manageable form. Can you please help... | $\ln(x^3 + 2x^2 + x) + \dfrac{2}{x} = -ln\left(\dfrac{1}{x}\right) + 2\ln(x+1) + \dfrac{2}{x} = \dfrac{1}{x} + 2\ln(x+1) + y\left(1 - \dfrac{lny}{y}\right) \to +\infty$ as $y = \dfrac{1}{x} \to +\infty$ when $x \to 0^+$, and $\dfrac{lny}{y} \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Probability Bertsekas Question Question:
We have two jars each containing an equal number of balls. We perform four successive ball exchanges. In each exchange, we pick simultaneously and at random a ball from each jar and move it to the other jar. What is the probability that at the end of the four exchanges all the b... | I can tell you that probability, $\frac{1 + 8(n-1)^2}{n^4}$, is wrong for sure as I wrote a quick simulation to check it and it does not agree (by orders of magnitude). Are you sure you copied it down correctly?
There are essentially two ways in which the jars can have the same configuration after $4$ steps:
*
*The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that existsts $\delta>0$ such that $2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$ I need to prove that exists:
$\delta>0$ such that $1-\delta < x < 1+ \delta\implies 2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$
What I did:
$1-\delta < x < 1+ \delta\implies |x-1|<\delta$ and then I suppose that:
$$|x+2|<\delta$$
So:
$$|x-1||... | We continue on the path you started. We want $|x-1||x+2|$ to be "small." The idea is to make $|x-1|$ small, that is, to choose $x$ close to $1$.
Unfortunately, the $x+2$ term could spoil things. But for example if we make $|x-1|\lt \frac{1}{2}$, then $x+2$ will be between $2+\frac{1}{2}$ and $2+\frac{3}{2}$, and in pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\f... | It is very much easier to note that $$n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)=4n(n+1)(n+2)$$ from which you get a telescoping series.
This is related to the binomial identity $\binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ with $r=3$, and the pattern clearly generalises.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$?
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$ ?
The numerator is a irreducible polynomial so I can't use partial fractions.
I tried the substitutions $t = x^2, t=x^4$ and for the formula $\int u\,dv = uv - \int v\,du$ I tried using: $u=\frac{x^4 + 1 }{x^6 + 1} , \,dv=\,d... | $$\int \frac{x^4 + 1 }{x^6 + 1}=\int \frac{x^4 -x^2+ 1 }{x^6 + 1} + \int \frac{x^2 }{x^6 + 1}=\int \frac{1}{x^2 + 1} + \int \frac{x^2 }{x^6 + 1}$$
The first integral is known while the second is easy by $u=x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Integrals of a particular form Recently, I was given a question sheet with a lot of integrals, and I could solve all of them except for a particular type of them:
$$I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=\int \frac{dx}{x^2\sqrt{x^2+1}}=\frac{-\sqrt{x^2+1}}{x}+C$$
$$I_2=\int \frac{dx}{(x-1)^2\sqrt{x^2-x+1}}$$
$$I_3=... | for the type of $\displaystyle \int \frac{1}{(ax+b)^2 \sqrt{cx^2 + dx + e}}dx$, $\displaystyle \frac{1}{ax+b} = t$ transforms into $\displaystyle \frac{at}{\sqrt{bt^2 + ct + d}}$ which can be reduced into standard integrals.
The same seems to work in linear case too.
$$\frac{1}{(ax+b) \sqrt{cx^2+ dx + e}} = \frac{at}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solving exponential equations using logarithms This is the equation that I am having troubles with:
$$\large x^{\large\log_{10}5}+5^{\large\log_{10}x}=50$$
So the first thing I do, I logarithm the whole expression with $\log_{10}$.
So I get:
$ {\log_{10} 5} \times {\log_{10} x} + {\log_{10} 5} \times {\log_{10} x} ... | Let $y=x^{\large\log_{10}5}$, then
$$
\log_{10}y=(\log_{10}5)(\log_{10}x)=\log_{10}5^{\large\log_{10}x}\color{red}{\quad\Rightarrow\quad} y=5^{\large\log_{10}x}.
$$
Hence
\begin{align}
x^{\large\log_{10}5}+5^{\large\log_{10}x}&=50\\
5^{\large\log_{10}x}+5^{\large\log_{10}x}&=50\\
2\times5^{\large\log_{10}x}&=50\\
5^{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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A little Problem in Trigonometry (Multiple Angle) If $\tan^2 \theta = 1 + 2\tan^2 \phi$, show that $\cos 2\phi = 1 + 2\cos2\theta$.
What I have done..
$$\implies \tan^2 \theta = 1 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2(1 + \tan^2 \phi)\\
\implies \sec^2 \theta = ... | Apply Componendo & Dividendo on $$\frac{\tan^2\theta}1=\frac{1+2\tan^2\phi}1$$
$$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-(1+2\tan^2\phi)}{1+(1+2\tan^2\phi)}$$
$$\implies\cos2\theta=-\frac{\tan^2\phi}{1+\tan^2\phi}=-\sin^2\phi$$
$$\implies\cos2\theta=-\frac{1-\cos2\phi}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve $x^{3}-3x=\sqrt{x+2}$ Solve for real $x$
$$x^{3}-3x=\sqrt{x+2}$$
By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe... | Two preliminary remarks.
If $x$ is solution, then $x(x+\sqrt{3})(x-\sqrt{3}) = x^3-3x \ge 0$, so $-\sqrt{3} \le x \le 0$ or $x \ge \sqrt{3}$.
Observe also that $2$ is solution and no real number larger than $2$ can be solution since for $\ge 2$,
$$\frac{d}{dx}(x^3-3x) = 3x^2-3 \ge 9 > \frac{1}{4} \ge \frac{1}{2\sqrt{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
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Skanavi 2.003, Difference in answers, which is right? In Skanavi book i have a exercise to simplify an equation
$$\begin{align}
((\sqrt[4]{p} - \sqrt[4]{q})^{-2} + (\sqrt[4]{p} + \sqrt[4]{q})^{-2}) : \frac{\sqrt p + \sqrt q}{p-q}
\end{align}$$
Solving:
Replacing:
$$ a =\sqrt[4]{p} $$
$$ b=\sqrt[4]{q} $$
Equation is ch... | What Mark said is exactly right.
$$
\frac{2(\sqrt p +\sqrt q)}{(\sqrt p-\sqrt q)} \\ =
\frac{2(\sqrt p +\sqrt q)(\sqrt p + \sqrt q)}{(\sqrt p-\sqrt q)(\sqrt p + \sqrt q)} \\ =
\frac{2(\sqrt p +\sqrt q)^2}{(\sqrt p)^2 - (\sqrt q)^2} \\ =
\frac{2(\sqrt p +\sqrt q)^2}{p-q}
$$
Remember $(a+b)(a-b)= a^2 - b^2$. It's very... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find volume of revolution Find the volume:
$$x=y^2; x=1-y^2; \text{ rotated around }x=3$$
Use the disk/washer method.
My inner radius is $3+y^2$ and outer radius is $3-y^2$ and the limits of integration are $\frac1{\sqrt2}$ and $\frac{-1}{\sqrt2}$
My final answer is $14\pi\frac{\sqrt2}3$
The answer is supposed to be $1... | The inner radius is $3-(1-y^2)$, that is, $2+y^2$. By symmetry the volume is
$$2\int_{y=0}^{1/\sqrt{2}}\pi\left((3-y^2)^2-(2+y^2)^2\right)\,dy.$$
Expand and integrate.
Remark: The inner radius of $3+y^2$ used in the OP does not yield volume $14\pi\frac{\sqrt{2}}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cd... | $$y + (z + 1) = \frac{1}{2} (z + 1) (z + 2)$$
$$2y+2(z+1)=(z+1)(z+2)$$
$$2y=(z+1)(z+2)-2(z+1)$$
$$2y=(z+1)(z+2-2)$$
$$2y=z(z+1)$$
$$y=\frac{1}{2}z(z+1)$$
Where is your error? From
$$y + (z + 1) = \frac{1}{2} (z^2 + 3z + 2)$$
you can say
$$y+(z+1)=\frac{1}{2}z^2+\frac{3}{2}z+1$$
where instead you only multiplied $z^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Primes as a difference of powers Find the smallest prime that cannot be written as
$$|3^a - 2^b|$$
EDIT: I forgot to mention that $a$ and $b$ are whole numbers.
I tried to expand $3^a$ as $(2+1)^a$ using binomial theorem but I couldn't infer much. Please help. Thanks in advance!
| $2$. The second subtrahend is always $0~ (\text{mod}~ 2)$ and the first never is, so their difference can never be $\pm2$.
Conveniently, there are no smaller primes to check.
If on the other hand, $a$ and $b$ can be zero, then you break your problem into eight Pell equations and solve the set sequentially over primes ... | {
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"url": "https://math.stackexchange.com/questions/821346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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$\sum^{\infty}_{n=1} \log(\frac{n+a+b}{n+a} \times \frac{n+b}{n+b+1})$ Prove $$\sum^{\infty}_{n=1} \log(\frac{n+a+1}{n+a} \times \frac{n+b}{n+b+1})=\log\frac{1+b}{1+a}$$
Hints/Answers are appreciated
| I think you mean $n+a+1$ not $n+a+b$.
If we put aside the issue of convergence, then using the property $\log(A)+\log(B) = \log(AB)$, we have
$$\sum_{n = 1}^{\infty} \log \left( \frac{n+a+1}{n+a} \cdot \frac{n+b}{n+b+1} \right) = \log \left( \prod_{n = 1}^{\infty} \frac{(a+1+n)(b+n)}{(n+a)(n+b+1)} \right).$$
Note that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the explicit notation of $f(n)$, based on it's recursive description. I came across this problem on a HackerRank challenge.
The function $f(n)$ is
*
*$1$ if $n = 0$
*$2f(n - 1)$, if $n$ is odd
*$f(n -1) + 1$, if $n$ is even
I solved the problem using a recursive function and it worked just well. However,... | First I would start by writing out the sequence: $$f(0)=1, f(1) = 2, f(2) = 3, f(3) = 6, f(4) = 7, f(5) = 14, f(6) = 15, ...$$
Now notice that $f(0) = 2 - 1 = 2^{0+1} - 1$, $f(2) = 4-1 = 2^{1+1} - 1$, $f(4) = 8-1 = 2^{2+1} - 1$, and $f(6) = 16 - 1 = 2^{3+1}-1$. So we conjecture that $$f(2n) = 2^{n+1} - 1$$
Now it is al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to calculate the Nth member of this Generating function? The series 012012012... matches the following generating function:
$$T=\frac{x+2x^2}{1-x^3}$$
How could i find a closed expression of the nth member of this series?
| For a bloated answer:
\begin{align*}
T &= \frac{x+2x^2}{1-x^3} \\
&= \frac{1}{1-x}+\frac{\frac{1}{2}-\frac{\sqrt{3}}{6}\, i}{-\frac{1}{2}+\frac{\sqrt3}{2}\, i - x}-\frac{\frac{1}{2}+\frac{\sqrt{3}}{6}\, i}{\frac{1}{2}+\frac{\sqrt3}{2}\, i + x}
\end{align*}
Extracting $[x^n]$ now gives:
\begin{align*}
a_n &= 1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Matrix Power Formula
Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$:
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$
How would we prove this? Right now we are doing work with proofs by induction... but how ... | By induction:
for $n = 1 $ it's true.
For $n +1 $ $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix} = \begin{pmatrix}a^{n+1} & a^n + na^n\\0 & a^{n+1} \end{pmatrix} = \begin{pmatrix}a^{n+1} & (n+1)a^n\\0 & a^{n+1} \end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Solving combinatorical problem using characteristic polynomial
How many $6$ length strings above $\left\{1,2,3,4\right\}$ are there such that $24$ and $42$ aren't allowed.
The suitable recurrence relation for this problem is: $a_{n+2} = 2a_{n-1} + 4a{n-2}$. Hence, the characteristic polynomial is: $x^2 -2x -4 = 0$.... | Let $a_n$ be the number of strings of length n,
$b_n$ be the number of strings of length n starting with 1 or 3,
$c_n$ be the number of strings of length n starting with 2, and
$d_n$ be the number of strings of length n starting with 4.
Then $a_n=b_n+c_n+d_n$, $b_n=2a_{n-1}$, $c_n=b_{n-1}+c_{n-1}$, and $d_n=b_{n-1}+d_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
| $$
\begin{align}
\ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
\ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A problem of sum floors let $n$ be a positive integer, prove that $$\sum_{i=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\left\lfloor\frac{n-3i}{2}\right\rfloor=\left\lfloor\frac{n^2+2n+4}{12}\right\rfloor.$$
It looks like we have to consider lots of cases to solve the problem. Are there any simple solutions?
| I think this is an easy way to handle it. We induct on $n$. See we use a step argument. Call
$$S_n=\sum_{i=0}^{\lfloor{n/3}\rfloor}\left\lfloor{\frac{n-3i}{2}}\right\rfloor$$
We have to prove $S_n=\left\lfloor{\dfrac{(n+1)^2+3}{12}}\right\rfloor \tag{1}$
Now $S_{n+3}-S_{n}=\left\lfloor{\dfrac{n+3}{2}}\right\rfloor$ To... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Please someone help with this nearly impossible integral $$ \int \frac{4x^5 -1} {(x^5 + x +1)^2} dx $$
So this is the integral and I have been stuck on it for ages without getting anywhere at all. Nothing I tried has gotten me anywhere so I'm basically stuck on nowhere with this.
I would really love for someone to go t... | $$\int \frac{4x^5-1}{(x^5+x+1)^2}\,dx=\int \frac{4x^5+5x^4-5x^4-1}{(x^5+x+1)^2}\,dx$$
$$=\int \frac{4x^5+5x^4}{(x^5+x+1)^2}\,dx-\int \frac{5x^4+1}{(x^5+x+1)^2}\,dx=J_1-J_2$$
$J_2$ can be handled by using the substitution $x^5+x+1=t$ and comes out to be $\dfrac{-1}{x^5+x+1}$. For $J_1$, factor out $x^{5}$ from the expre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What's the integral of $\int_0^\infty \frac{dx}{(x^4+1)^5}$ $$\int_0^\infty \frac{dx}{(x^4+1)^5}$$
My answer would be : $\dfrac{\Gamma(\tfrac{1}{4})\Gamma(\tfrac{19}{4})}{4\Gamma(5)}$
Solution: You can use this technique. – Mhenni Benghorbal
| \begin{align}
x^4+1 & = (x^4 +2x^2 + 1)-2x^2 \\[8pt]
& = (x^2+1)^2 - (\sqrt{2}\, x)^2 \\[8pt]
& = (x^2 - \sqrt{2}\, x +1)(x^2 + \sqrt{2}\, x + 1)
\end{align}
Then go to partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/847905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the ... | Note that
$$\sum 1=n=C_n^1$$
$$\sum\sum1=\sum i=\frac{n(n+1)}{2}=C_{n+1}^2$$
$$\sum\sum\sum1=\sum\frac{i(i+1)}2=\frac{n(n+1)(n+2)}{2.3}=C_{n+2}^3$$$$\sum\sum\sum\sum1=\sum\frac{i(i+1)(i+2)}{2.3}=\frac{n(n+1)(n+2)(n+3)}{2.3.4}=C_{n+3}^4.$$
These are diagonals in Pascal's triangle.
From there, you derive
$$\sum i^2=2\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Values of $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ Let $f$ and $a$ such that $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$.
I need to find the values of $f$ and $a$ that satisfies this condition.
For this i tried:
$F(x) = 6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ then $F´(x) = \frac{f(x)}{x^2} = \frac{1}{... | Put $x=a$ in $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$. This gives you that
$2\sqrt a=6$ and so $a=9$. Differentiating $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ we get $\frac{f(x)}{x^2}=\frac{1}{\sqrt x}$ which implies that $f(x)=x^{3/2}$. Note that $\int_a^a \frac{f(t)}{t^2} dt=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Series $\sum \frac{1}{n^2\sin^3n}$ Question : Show that series $\sum \cfrac{1}{n^{2}\sin^{3}n}$ is divergent.
Hint:
Show that $$\sum \frac{1}{n|\sin(n)|}$$ is divergent.
I am interested in other possible proofs for this question.
| Since $\pi$ is irrational, by Hurwitz's theorem, there are infinitely many pairs of relative prime integers $(n,m)$ such that
$$\left|\pi - \frac{n}{m}\right| < \frac{1}{\sqrt{5}{m^2}}
\quad\implies\quad|n - m\pi| < \frac{1}{\sqrt{5}m}
$$
We can assume both $n, m > 0$ and for any such pair, we have
$$
|\sin n | = |\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \fra... | $ f(x) = \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ x+3 &, -3 \leq x \leq -2 \\ 0 &, -2 \leq x \leq \phantom{-{}}2 \\ 3-x &, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0 &, \phantom{-{}}3 \leq x\end{cases}$
\begin{align}
F(x) = \int_{-\infty}^x f(t) \,\mathrm{d}t &= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold si... | Yet another way, let $p = x+y+z, q = xy+yz + zx$. Then $x^2+y^2+z^2 = p^2-2q=1$ and $x^3+y^3+z^3-3xyz = p^3-3pq = \frac12p(3-p^2)$ . Also $x^2+y^2+z^2 = 1 \implies p^2\le 3$.
So we want the maximum of $\frac12p(3-p^2)$, subject to $p^2 \le 3$, which easily$^\dagger$ gives $p=1$.
$^\dagger$ Equivalently this is wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of $ \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$ Evaluation of $\displaystyle \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$
$\bf{My \; Try::}$ Let $\displaystyle I = \int \frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\c... | Let
\begin{align}
I = \int \frac{\sin(x+a)}{\cos^3(x)}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx
\end{align}
then
\begin{align}
I &= \int (\sin(x) \cos(a) + \cos(x) \sin(a))(\cos(x) + \sin(x)) \ \frac{dx}{\cos^{3}(x) \sqrt{\cos(2x)}} \\
&= \cos(a) \ \int \frac{\sin^{2}(x) \ dx}{\cos^{3}(x) \sqrt{\cos(2x)}} +
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving a sharp inequality After experimenting, I've come to the conclusion that if $x\geq y\geq z\geq 0:$
$$\sum_{x,y,z}\frac{x}{\sqrt{x+y}}\geq \sum_{x,y,z}\frac{y}{\sqrt{x+y}}$$
(the sums are cyclic)
Does anyone know how to prove this? I've tried variants of Cauchy Schwarz, and the rearrangement inequality, but thes... | The condition $x^2+y^2+z^2=1$ is redundant.
Let $a=\sqrt{y+z},b=\sqrt{z+x},c=\sqrt{x+y}$ we have $c\ge b \ge a \ge 0$ and $$x=\frac{b^2 + c^2 - a^2}{2},y=\frac{c^2 + a^2 - b^2}{2},z=\frac{a^2 + b^2 - c^2}{2}.$$
Replacing into the original inequality, it becomes
$$\sum_{a,b,c} \frac{b^2 + c^2 - a^2}{2c} \ge \sum_{a,b,c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that function is continuous at $x_0$ I want to show that $f: \mathbb R \rightarrow \mathbb R$,
$f(x) = \frac{x-1}{x^2+1}$ is continuous at $x_0=-1$. So if $|x+1|<\delta$, then $|f(x)+1| < \epsilon$.
I rearranged $|f(x)+1|=|\frac{(x-1) +(x^2+1)}{x^2+1}|=|\frac{x+x^2}{x^2+1}|=|\frac{1+x}{x+ \frac{1}{x}}|$, but now I... | You can use that $\displaystyle\left|\frac{x+x^2}{x^2+1}\right|=\frac{|x||x+1|}{x^2+1}<|x+1|$ for all x, since $|x|<x^2+1$
because $x^2<(x^2+1)^2=x^4+2x^2+1$ for all x.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/854680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ et... | \begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
&& \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}\\
&=& \frac{2(x^3+y^3+z^3)-(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)}{2 (x + y) (x + z) (y + z)}\\
&=& \frac{(x+y)(x-y)^2+(y+z)(y-z)^2+(z+x)(z-x)^2}{2 (x + y) (x + z) (y + z)}\geq0
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
} |
The limit of a sequence $\lim_{n\rightarrow \infty}\prod_{k=0}^{n-1}( 2+\cos \frac{k\pi }{n})^{\pi/n}$. $$
\lim_{n \to \infty}\prod_{k = 0}^{n - 1}
\left[\,2 + \cos\left(k\pi \over n\right)\right]^{\pi/n}\ =\ ?
$$
Please give some hints.
| For any $a \in (1,\infty)$, consider the equation
$$\frac{x+x^{-1}}{2} = a\quad\iff\quad x^2 - 2ax + 1 = 0$$
It has two roots and one of it, $x = a + \sqrt{a^2-1}$, is outside the unit circle.
Notice
$$z^{2n}-1
= (z^2 - 1)\prod_{k=1}^{n-1}\left(z-e^{i\frac{k\pi}{n}}\right)\left(z-e^{-i\frac{k\pi}{n}}\right)
= (z^2 - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Modular Arithmatic - Solving congruences I'm sure this is pretty basic but I'm struggling to understand how to go about solving this problem for my homework. The question states "Solve the following congruences for x". The first problem is $2x+1\equiv 4\pmod 5$.
| There are many ways to solve the problem. The conceptually simplest, but most tedious, is to test one by one the possibilities $x\equiv 0\pmod{5}$, $x\equiv 1\pmod{5}$, and so on up to $x\equiv 4\pmod{5}$. Quickly we find that $x\equiv 4\pmod{5}$. (This approach would become quite unpleasant if $5$ were replaced by $97... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $aI am trying to prove $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ if $a<b$. I am having trouble with the left inequality:
$a^2<\frac{1}{3}(a^2+ab+b^2) \Rightarrow 2a^2<ab+b^2$.
If $a>0$, then all is well and adding $a^2<b^2$ to $a^2<ab$ will yield the result I want, but I do not know how to deal with the case $a<0$, or if th... | Is true only if $0< a<b$, then for the left side:
$$a^2<b^2
$$
$$a^2+a^2+ab<b^2+ab+a^2
$$
$$a^2+a^2+ab<b^2+ab+a^2
$$
If $a<b$ then how $a>0$ then $a^2<ab$
$$a^2+a^2+a^2<a^2+a^2+ab<b^2+ab+a^2
$$
$$3a^2 <b^2+ab+a^2
$$
For the rigth side
$$a^2<b^2
$$
$$a^2+b^2+ab<b^2+b^2+ab
$$
If $a<b$ then how $b>0$ then $ab<b^2$
$$a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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inequality method of solution Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$
I've tried first determining when the absolute value will be positive or negative etc, and than giving it the signing in accordance to range it is... | When $t \equiv \frac{x-3}{x+1} > 0$, the solution to $ t^2 - 7t + 10 <0$ is $ 2<t<5$. (Factor the quadratic.)
When $t < 0$, the solution to $ t^2 + 7t + 10 <0$ is $ -2<t<-5$.
On the $t < 0$ branch, we need to solve $-2 < \frac{x-3}{x+1} < 5$. We break this up into two possibilities, $x < -1$ and $x > -1$, because when... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to prove that the number $1+4a_{n}a_{n+1}$ is a perfect square.
A sequence of integer $\{a_{n}\}$ is given by the conditions $a_{1}=1,
a_{2}=12,a_{3}=20$,and $$a_{n+3}=2a_{n+2}+2a_{n+1}-a_{n}$$
show that
for every postive integer $n$, the number $1+4a_{n}a_{n+1}$ is a perfect square.
since the
$$r^3=2r^2+2r-1$$
t... | Given the sequence $a_{n+3} = 2 a_{n+2} + 2 a_{n+1} - a_{n}$ it is evident that a solution of the form $a_{n} \approx r^{n}$ leads to $r^{3} - 2 r^{2} - 2r + 1 = 0$, or $(r+1)(r^{2} - 3r + 1) = 0$ has roots $r_{1} = -1$, $r_{2,3} = (3 \pm \sqrt{5})/2$. Now it is seen that $r_{2,3} = (1 \pm \sqrt{5})^{2}/4$ and
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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$\lim_{n \rightarrow \infty} f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$
Let
$$f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$$
Let M be the set of x s.t. $\lim_{n \rightarrow \infty} f_n(x)$
exists. For each $x \in M$ let $f(x) = \lim_{n \rightarrow \infty} f_n(x)$. Then
*
*M is bounded ... | it is easy to see that
$\cos \frac{x^{3}-1}{n} = 1 - \frac{(\frac{x^{3}-1}{n})^{2}}{2} + ... = 1 - \frac{x^{6} - 2x^{3} + 1}{2n^{2}} + ...$ so that $f(x) = n^{2} (1-\cos \frac{x^{3}-1}{n}) \rightarrow \frac{x^{6}}{2} - x^{3} + \frac{1}{2}$
so 1 - false, 2 - false, 3 - false, 4 - false
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find the minimum of the $|w^3+z^3|$,if $|z+w|=1,|z^2+w^2|=14$ let complex $z,w$ such
$$|z+w|=1,|z^2+w^2|=14$$
find the minimum of the value
$$|w^3+z^3|$$
My idea: let
$$z=a+bi,w=c+di\Longrightarrow z+w=(a+c)+(b+d)i,z^2+w^2=(a^2+b^2+c^2+d^2)+2(ab+cd)i$$
then we have
$$(a+c)^2+(b+d)^2=1,(a^2+b^2+c^2+d^2)+4(ab+cd)^2=1... | Let $z+w = a$ and $z^2+w^2 = b$. Then, $zw = \dfrac{(z+w)^2-(z^2+w^2)}{2} = \dfrac{a^2-b}{2}$.
Thus, $z^3+w^3 = (z+w)^3-3zw(z+w) = a^3 - 3 \cdot \dfrac{a^2-b}{2} \cdot a = \dfrac{3ab-a^3}{2}$.
If $|a| = |z+w| = 1$ and $|b| = |z^2+w^2| = 14$, then using the reverse triangle inequality, we have
$|z^3+w^3| = \left|\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/868669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An improper integral : $\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx$ How to evaluate the following improper integral:$$\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ where $a,b>0$.
I tried to suppose $$f(a)=\int_0^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ based on the convergence theorem, and then I tried $${df(a... | Letting $y=\frac{x}{b} $ yields
$$
\begin{aligned}
I &=b \int_0^{\infty} \frac{\ln \left(a^2+b^2 y^2\right)}{b^2+b^2 y^2} d y \\
&=\frac{1}{b} \int_0^{\infty} \frac{\ln \left(b^2 y^2+a^2\right)}{1+y^2} d y
\end{aligned}
$$
By my post,
$$
\begin{aligned}
\boxed{I =\frac{1}{b} \pi \ln \left(\sqrt{a^2}+\sqrt{b^2}\right)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| Product of 5 consecutive integer numbers is of course divisible by $5$, so $5|(n-2)(n-1)n(n+1)(n+2)$. If $5\not|n,$ then from primality of $5$ we have
\begin{align*}
5|(n-2)(n-1)(n+1)(n+2) & = (n^2-1)(n^2-4)\\
& = n^4 - 5n^2 +4\\
& = n^4 - 1 - 5(n^2 - 1)
\end{align*}
so $5|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 12
} |
If $a+b+c=1$, then $\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$ Someone can to help me with a hint in the following problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that:
$$\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq... | Homogenising, we have to show:
$$\frac{2\sqrt{abc(a+b+c)}}{a(a+b+c)+bc} + \frac{2\sqrt{abc(a+b+c)}}{b(a+b+c)+ca}+\frac{ab-c(a+b+c)}{ab+c(a+b+c)} \le \frac32$$
Now it motivates the triangle substitution $x=a+b, y=b+c, z = c+a$, so we have a triangle with area say $\Delta$, to get the equivalent:
$$\frac{\Delta}{2xz}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c... | $$a^2+b^2=c^2\iff \left(\frac ac\right)^2=1-\left(\frac bc\right)^2\le1$$
$$\implies\left(\frac ac\right)^n<\left(\frac ac\right)^2$$ for $n>2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac... | You have a very bad habit of repeating variables when you should not. This is a grave mistake and causes errors like you have made. I'm going to get a series expansion for the Bessel function because the integrand is fairly difficult to work with, in my opinion. Series are quite nice though-especially with respect to d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \i... | Let $x,y$ be any two real numbers. Then $\lfloor x\rfloor+\lceil y\rceil>x+y-1$, since $\lfloor x\rfloor>x-1$ and $\lceil y\rceil\geq y$. Similarly $\lfloor x\rfloor+\lceil y\rceil<x+y+1$. So $\lfloor x\rfloor+\lceil y\rceil$ differs from $x+y$ by strictly less than $1$. Since $\lfloor x\rfloor+\lceil y\rceil$ is alway... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Polar form of a complex number Question:
Write the polar form of $$\frac{(1+i)^{13}}{(1-i)^7}$$
Well its obviously impractical to expand it and try and solve it.
Multiplying the denominator by $(1+i)^7$ will simplify the denominator, and a single term in the numerator.
Answer I got:
$$(\frac{1}{\sqrt2}(cos(\frac{\pi}{... | You can also convert numerator and denominator into polar form immediately to write
$$ \frac{ [ \ \sqrt{2} \ cis(\frac{\pi}{4}) \ ]^{13} \ }{[ \ \sqrt{2} \ cis(-\frac{\pi}{4}) \ ]^7} \ \ . $$
DeMoivre's Theorem for powers gives us
$$ = \ \frac{ (\sqrt{2})^{13} \ cis(\frac{13\pi}{4}) }{(\sqrt{2})^7 \ cis(-\frac{7\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral of trignometric function What is the trick to integrate the following
$$\int \frac{1-\cos x}{(1+\cos x)\cos x}\ dx$$
| Hint :
\begin{align}
\int\frac{1-\cos x}{\cos(1+\cos x)}\ dx&=\int\frac{\color{red}{1+\cos x}-2\cos x}{\cos(\color{red}{1+\cos x})}\ dx\\
&=\int\frac1{\cos x}\ dx-\int\frac{2}{1+\cos x}\ dx\\
&=\int\frac{\cos x}{\cos^2 x}\ dx-\int\frac{2}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}\ dx\\
&=\int\frac{d(\sin x)}{1-\sin^2 x}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.