Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply... | Let's start here and simplify the problem.
$$\frac{(x+3)}{(x-1)(x+1)} - \frac{(x-2)}{(x+1)(x+1)}$$
I'm going to call $(x+1)=a$ and $(x-1)=b$
$$\frac{(x+3)}{(b)(a)} - \frac{(x-2)}{(a)(a)}$$
So, on the right side we have (a)(a) in the denominator, and only one in the denominator on the right. So we have to multiply the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a ... | The cube of a binomial is
$$
(u+v)^3=u^3+v^3+3uv(u+v)
$$
Let us observe that in this case
\begin{align}
&u^3+v^3&&=2a\\
&uv&&=[a^2-(a^2+b^3)]^{1/3}=-b\\
&u+v&&=x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
What is the expected value of the product of randomly selected balls
An urn contains four balls numbered 1, 2, 5, and 7. If a person
selects a set of two balls at random, what is the expected value of
the product of the numbers on the balls?
My thoughts:
$E(X)=\sum_{k=1}^{n}a_{k}b_{k}$
$E(X) = E_{1}(X)*E_{2}(X)$... | There's only 6 different ways to draw two balls, and each of those are equally likely, so I would just calculate it as:
$$
\frac{1\cdot 2+1\cdot 5+1\cdot 7+2\cdot 5+2\cdot 7+5\cdot 7}{6}
=\frac{2+5+7+10+14+35}{6}
=\frac{73}{6}
=12\frac{1}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the sum of the $81$ products in the $9 \times 9$ multiplication grid? What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
| Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:
First, show that this is true for $n=1$:
$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
Solve $z^5 +32 =0$
Solve $z^5 +32 =0$
My attempt : $$z^5 = -32$$
Multiply the powers on both sides by $\frac{1}{5}$
we get $$z = 2 * (-1)^\frac{1}{5}$$
Now I'm stuck at this step I don't know how to proceed. Kindly help.
| We have $$\left(-\dfrac z2\right)^5=1$$
Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
If $x^5-1=0,$
either $x-1=0\iff x=1$
or $x^4+x^3+x^2+x+1=0$
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
as $x\ne0,$ divide both sides by $x^2$ to find $$0=x^2+\dfrac1{x^2}+x+\dfrac1x+1=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \... | The solution to this system of equations there. http://www.artofproblemsolving.com/community/c3046h1046718__4
$$\left\{\begin{aligned}&x^2+qy=z^2\\&y^2+qx=v^2\end{aligned}\right.$$
Use this decision.
$$x=2psb^2-a^2p^2$$
$$y=2abp^2-b^2s^2$$
$$q=as(4bp-as)$$
$$z=a^2p^2+2psb^2-abs^2$$
$$v=2abp^2-psa^2+b^2s^2$$
In our ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many words of length $n$ can we make from $0, 1, 2$ if $2$'s cannot be consecutive? How many words we can make from $0,1,2$? The restriction is we can't put the digit $2$ after the digit $2$.
My solution:
I tried to solve it with Inclusion-Exclusion Principle. Count the number of the words without restriction and ... | This answer is based upon the Goulden-Jackson Cluster Method which is a convenient method to derive a generating function for problems of this kind.
We consider words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1,2\}$$ and the set $\mathcal{B}=\{22\}$ of bad words, which are not allowed to be part of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A variation of Ahmed's integral $\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $ Given that the closed form exist, evaluate the following Integral:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
| Let
$J(a)=\int_0^\infty\frac1{1+t^2} \ln \frac{t^2+2t\sin 2a+1}{t^2-2t\sin 2a+1}dt
$
and establish $J’(a)=8a\csc(2a)$. Note\begin{align}
&I=\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx
=\frac14\int_{0}^{1}\sin^{-1}x \ d\left(\ln \frac{4+2x-x^2}{4-2x-x^2} \right)
\end{align}
Integrate by parts and then sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Algebraic manipulation of a limit. What are the algebraic manipulations and steps that makes the limit
\begin{equation}
\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
\end{equation}
equal to
\begin{equation}
\lim_{x\to2}(x^2+2x+4)
\end{equation}
It's probably trivial, I just don't seem to be able to see it.
| Based on @Kushal Bhuyan hint
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ formula
In our case
$$\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
=\lim_{x\to2}\left(\frac{x^3-2^3}{x-2}\right)
$$
Then $x^3-2^3=(x-2)(x^2+2x+2^2)$, therefore:
$$\require{cancel}\lim_{x\to2}\left(\frac{\cancel{(x-2)}(x^2+2x+2^2)}{\cancel{x-2}}\right)
=\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}$ I have a problem with finding the following limit. I suspect that it should be easy, but really I don't have a clue.
$$
\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}
$$
| Note that if $(n-1)x \leqslant u \leqslant nx \leqslant v \leqslant (n+1)x,$ then
$$\frac{1}{4 + v^2} \leqslant \frac{1}{4 +n^2x^2} \leqslant \frac{1}{4 + u^2}, $$
and
$$\int_{nx}^{(n+1)x}\frac{dv}{4 + v^2} \leqslant \frac{x}{4 +n^2x^2} \leqslant \int_{(n-1)x}^{nx}\frac{du}{4 + u^2}, $$
Summing we get
$$\int_x^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after ge... | This is simply false.
As $\lim_{x \to 0} x + n = n$ for all $n \in \mathbb{N}$ and the square root is continuous, we have $\lim_{x \to 0} \sqrt{x+n} = \sqrt{n}$.
Then, you must show
\begin{align*}
a + \sqrt{2} b + \sqrt{3}c = 0 \iff a+b+c =0.
\end{align*}
Take $a = 2$, $b = -\sqrt{2}$ and $c=0$, then the LHS is satisf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, whi... | Your method is correct, but note that, as noted in the comment, your derivative is wrong.
In general: if a line passes thorough the points $P=(x,f(x))$ and $Q=(x_Q,y_Q)$ than its slope is :
$$
m=\frac{f(x)-y_Q}{x-x_Q}
$$
and if the line is tangent to the function $y=f(x)$ at $P$ than this slope is the value of the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1... | Short answer: yes, it does: the denominator doesn't cancel and the asymptotic behavior is $\sim x^{-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Question regarding roots of a cubic polynomial If $\alpha$, $\beta$ and $\gamma$ are the roots of a cubic equation with
$$\alpha + \beta + \gamma = 1$$
$$\alpha^2 + \beta^2 + \gamma^2 = 2$$
$$\alpha^3 + \beta^3 + \gamma^3 = 3$$
Then find the value of
$$\alpha^4 + \beta^4 + \gamma^4$$
| I will use the notation of this question of mine. The $\text{EXP}$ map gives:
$$ \exp\left(-\sum_{m\geq 1}\frac{p_m}{m}x^m\right) = \sum_{r=0}^{3}(-1)^r e_r x^r \tag{1}$$
hence a polynomial having $\alpha,\beta,\gamma$ as roots is:
$$ p(x)=x^3-x^2-\frac{x}{2}-\frac{1}{6} \tag{2} $$
and the $\text{LOG}$ map gives:
$$ p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac {\sin x}{1-\sin x}-\frac {\sin x}{1+\sin x}\equiv 2\tan^2 x$ I need assistance with proving the following identity:
$$\frac {\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x} \equiv 2\tan^2 x$$
What I have done so far is expanded them:
$$\frac {\sin x\;(1+\sin x)}{(1-\sin x)(1+\sin x)}-\frac {\sin x\;(1-\sin ... | As $\cos^2x=(1-\sin x)(1+\sin x)$
$$\dfrac{\sin x}{1-\sin x}=\dfrac{\sin x-1+1}{1-\sin x}=-1+\dfrac{1+\sin x}{\cos^2x}=\tan^2x+\sec x\tan x$$
Similarly, $$\dfrac{\sin x}{1+\sin x}=\sec x\tan x-\tan^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
evaluate if integral converge & determine antiderivative The problem is i need to
study the convergence of A and B
and find the antiderivative of C
$$A=\int_0^\infty \frac{\sin(x) +x}{\sqrt x + x^3}dx$$ $$B=\int_0^\infty \frac{1}{\sqrt {e^x-1}(x^2+x^{1/3} )}dx$$
$$C=\int\frac{1}{x}\sqrt\frac{2-x}{2+x}dx$$
in C i t... | hi sami i look if the problem C u need to use substitution 5 time
$$u=\dfrac{1}{x+2}$$
& then after that
$$X=\sqrt{4u-1}$$
&
$$w=X+1$$
&
....
but the final answer is $$\ln\left(\left|\sqrt{-\dfrac{x-2}{x+2}}-1\right|\right)-\ln\left(\sqrt{-\dfrac{x-2}{x+2}}+1\right)+2\arctan\left(\sqrt{-\dfrac{x-2}{x+2}}\right)+C$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$
Could someone please explain to my why this is th... | Another approach based on Taylor expansions.
Consider $$y=x(\sqrt{x^2+1} - x)=x^2\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ and remember that, for small $y$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$. Replace $y$ by $\frac{1}{x^2}$ to get $$\sqrt{1+\frac{1}{x^2}}-1=\frac{1}{2 x^2}-\frac{1}{8 x^4}+O\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
\end{bmatrix} $
is at the same time $3D$... | This answer is not an independent one, it's just a supplement to Yves'
answer to have made his formula for generating rational values of required matrix more explicit (what is important for me and maybe useful for others)
So we have Yves' equation $ p^2+3g^2=1$ what can be written as $\left({\dfrac{a}{d}}\right)^2+3{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Antiderrivative of ${d^2 y \over dx^2} = 1-x^2$
At any point $(x,y)$ on a curve, ${d^2 y \over dx^2} = 1-x^2$, and an equation of the tangent line to the curve at the point $(1,1)$ is $y=2-x$. Find an equation of the curve.
This is what I've done
$${d^2 y \over dx^2} = 1-x^2
\\ \int dy' = \int (1-x^2)dx
\\y' = x- {x^... | Hint:
Since $y=2-x$ is the tangent to the curve at $(1,1)$ we have $y'(1)=-1$, so
\begin{align*}
1-\frac13+c_1&=-1\\
\frac12-\frac1{12}+c_1+c_2&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
I'm looking for a matrix $M$ with $\det(M)=a^2+b^2+c^2+d^2$ In order to show that $(a^2+b^2+c^2+d^2)(A^2+B^2+C^2+D^2)= \alpha^2+\beta^2+\gamma^2+\delta^2$ with $a,b,c,d,A,B,C,D,\alpha,\beta,\gamma,\delta \in \mathbb Z$.
I would like to find a matrix with determinant $a^2+b^2+c^2+d^2$ in order to multiply the matrix wi... | The matrix, where $i = \sqrt{-1}$,
\begin{pmatrix}
a+ib & -c+id\\
c+id & a-ib
\end{pmatrix}
works. When we multiply two matrices of the above form, we get a matrix of the same form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove the sum of squares of 3 rationals cannot be 7
Prove there isn't $r_1, r_2,r_3 \in \mathbb{Q}$ so that ${r_1}^2 +
{r_2}^2 + {r_3}^2=7 \tag1$
From (1) we get $a^2 + b^2 + c^2=7n^2 \tag2$ where $a,b,c,n \in \mathbb{N}$. I have tried playing with parity of these numbers, without success.
UPDATE
Suppose $n$ is eve... | This is about 2-adic restrictions. First, odd squares of integers are $1 \pmod 8.$ Integer squares can only be $0,1,4 \pmod 8$ in any case. Therefore the sum of three integer squares cannot be $7 \pmod 8.$
Next, if the sum of three squares is divisible by $4,$ so $x^2 + y^2 + z^2 = k$ with $k \equiv 0 \pmod 4,$ then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$5$ numbers add up to 3231.What is the $6$th number? This is Q27 from Australian Maths 2013.
$3$ different non-zero digits are used to form $6$ different $3$-digit numbers.The sum of $5$ of them is $3231$.What is the $6$ th number?
What I tried:
Let $a,b,c$ be the different digits.
$(100a+10b+c)+(100a+10c+b)+(100b+10a+... | With three digits $a,b,c$, You should be able to get at most six different $3$-digits numbers, and they are: $abc,acb,bac,bca,cab,cba$So when you add them up, the equation should be$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)=3231+n$$where $n$ is the unknown $6$th number.
By quick estimation you can find that,
When $a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$
Prove that
$$
\begin{vmatrix}
1 & a^2 + bc & a^3 \\
1 & b^2 + ac & b^3 \\
1 & c^2 + ab & c^3 \\
\end{vmatrix}
=(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$
My work
I am getting - sign extra , please tell me why I ... | Alternatively, it is easily seen that the determinant of the given matrix is a homogeneous polynomial $P(a,b,c)$ in $a$, $b$, and $c$ of degree $5$. Furthermore, $P(a,b,c)=0$ if any two of the three inputs are equal. Therefore, $P(a,b,c)$ is divisible by $(b-c)(c-a)(a-b)$. Hence, $$P(a,b,c)=(b-c)(c-a)(a-b)\,Q(a,b,c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inverse of matrix with very structured submatrix Does this matrix admit an easy analytic expression for its inverse?
$$\begin{bmatrix}
a_1 & 0 & 0 & 0 & 0 &0&\dots&0 \\
a_2 & 1 & -b & 0 & 0&0&\dots&0 \\
a_3 & 0 & 1 & -b & 0 & 0&\dots&0\\
a_4 & 0 & 0 & 1 & -b & 0 &\dots&0\\
\vdots \\
a_{N-1} & 0 & 0 & 0 & 0 & 0 & \dots ... | If we call your matrix $A$, then
$$
A^{-1} =
\begin{pmatrix}
\frac{1}{a_1} & 0 & 0 & 0 & 0 & \dots \\
-\frac{1}{a_1b^2} \sum_{k=2}^Na_kb^k & 1 & b & b^2 & b^3 & \dots \\
-\frac{1}{a_1b^3} \sum_{k=3}^Na_kb^k & 0 & 1 & b & b^2 & \dots \\
-\frac{1}{a_1b^4} \sum_{k=4}^Na_kb^k & 0 & 0 & 1 & b & \dots \\
\vdots & \vdots & &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Limit involving the Sine integral function $$
\mbox{Prove that}\qquad
\lim_{x \to \infty}\left[\vphantom{\large A}%
x\,\mathrm{si}\left(x\right)+ \cos\left(x\right)\right]
= 0
$$
where we define
$$\mathrm{si}\left(x\right) =
- \int^{\infty}_{x}\frac{\sin\left(t\right)}{t}\,\mathrm{d}t
$$
I have no clue how to start. I... | Define $$Si(x) = \int^x_0 \frac{sin t}{t} dt$$
By the relation $Si(x) = \frac{\pi}{2} + si(x) $
And the asymptotic series expansion of $si(x)$:
$$si(x) = -\frac{\cos x}{x} (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\frac{\sin x}{x} (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7)) $$
Therefore
$$xsi(x)+\cos x ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| We notice this looks a bit like $(x^2-y^2)^2$, so we write
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= (x^4 - 2x^2y^2+y^4) - 9x^2y^2 \\
&= (x^2-y^2)^2 - (3xy)^2 \\
&= (x^2 + 3xy - y^2)(x^2 - 3xy - y^2).
\end{align*}
It factors further, but no longer over the integers (there will be square roots involved).
We also could have... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Points on the elliptic curve for Ramanujan-type cubic identities Given the rational Diophantine equation,
$$t^3 - t^2 - \tfrac{1}{3}(n^2 + n)t - \tfrac{1}{27}n^3=w^3\tag1$$
Two points are,
$$t_0 = 0\tag2$$
$$t_2 = \frac{-(1 + 2 n) (1 + 11 n + 42 n^2 + 14 n^3 + 13 n^4)}{9 (7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4)}\tag3$$
Q... | The cubic
\begin{equation*}
u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0
\end{equation*}
can be shown to be equivalent to the elliptic curve
\begin{equation}
y^2=x^3+1296n^2(n^2+n+1)^2
\end{equation}
by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient ... | {
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"source": "stackexchange",
"question_score": "2",
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is... | We can obtain a better bound. We need first some preliminary result:
Proposition 1:$$\sum_{k=1}^{N}\sqrt{k}\in\left(\frac{2}{3}N\sqrt{N},\frac{4N+3}{6}N\sqrt{N}\right).
$$ Proof: Let $$S=\sum_{k=1}^{N}\sqrt{k}
$$ using Abel's summation and the bound $t-1\leq\left\lfloor t\right\rfloor \leq t
$ we have $$S=N\sqrt{N}-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
... | There are no solutions. For $x\to196^-$ the function value is $12342$, and for $x=196$ it is $12348$, with no values in between.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $... | By Fermat's little theorem, $3^6\equiv 1 \bmod 7$, so that $3^{35}\equiv 3^5\equiv 5\bmod 7$. Hence $7\mid (3^{35}-5)$, and explicitly, $3^{35}=7\cdot 7147363585571386.$
Edit: "shifting interest to the factorisation": $$3^{35}-5=2\cdot 7\cdot 1729363\cdot 2066472911,$$ using integer factorisation algorithms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| $2xyz$ even $\rightarrow$ $x$ even and $y$,$z$$\in$ odd OR $x$,$y$,$z$$\in$ even
*
*$x=2k$ $y$,$z$$\in$odd:
$4k^2 + y^2 + z^2 = 4kyz$
$4k(k-yz) +y^2 + z^2 = 0$
We have this problem: $4|y^2 + z^2$ so $y$ or $z$ must be even but are odd.
*$x=2k, y=2l, z=2m$:
$4k^2 + 4l^2 + 4m^2 = 16klm$
$k^2 + l^2 + m^2 = 4klm$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Regarding irrationality of $\sqrt{5}$ In the proof of irrationality of $\sqrt{5}$, Hardy and Wright define $x=\frac{\sqrt{5}-1}{2}$. From that I know $1-x=x^2$. But then the authors say that when $1$ is divided by $x$ the remainder is $1-x=x^2$. But I am unable to follow how ?
| When $1$ is divided by $x$, you get $$\frac1x = \frac{x+1-x}{x} = \
\frac xx + \frac{1-x}{x} = 1 + \frac{1-x}{x}$$
The quotient of the division is $1$ and the remainder is the numerator $1-x$ of the "fractional part". Note that $0\leq \frac{1-x}{x} < 1$ here, so this makes sense.
This is analogous to $$5\div 3 = \frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$3\times 3$ matrix with eigenvalues are given
If $\displaystyle P=\begin{bmatrix}
0 & -2 & 3\\
-1& 1& -1\\
a & 2 & b
\end{bmatrix}$ for some $a,b\in \mathbb{R},$ suppose $1$ and $2$ are eigen values of $P$
and $\displaystyle P\begin{bmatrix}
1\\
0\\
-3
\end{bmatrix} =\begin{bmatrix}
3\\
0\\
-3
\end{bmatri... | $$tr(P)=\lambda_1+\lambda_2+\lambda_3$$
$$b+1=1+2+\lambda_3\implies\lambda_3=b-2$$
$$det(P)=\lambda_1\lambda_2\lambda_3$$
$$-a-2b-6=2(b-2)\implies a+4b=-2$$
on the other hand
$$a-3b=-3$$
if
$$\displaystyle \begin{bmatrix}
0 & -2 & 3\\
-1& 1& -1\\
a & 2 & b
\end{bmatrix}\begin{bmatrix}
1\\
0\\
-3
\end{bmatrix}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)... | It is true that $x=0$ is not a root of $ax^2+bx+c$, since if so then $c$ would be zero, and the second polynomial $cy^2+by+a$ is linear with at most one root. Similarly, $y=0$ is not a root of $cy^2+by+a$.
Consider
$$cy^2+by+a = y^2 \left(c+\frac by + \frac a{y^2}\right)$$
If $x=0$ is not a root of $ax^2+bx+c$, i.e. $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question on the coefficients of $(1+x+x^2+x^3+x^4)^{496}$
Consider the expansion $$(1+x+x^2+x^3+x^4)^{496} = a_0+a_1x+\cdots+a_{1984}x^{1984}.$$
$\quad$ (a) Determine the greatest common divisor of the coefficients $a_3,a_8,a_{13},\ldots,a_{1983}$.
$\quad$ (b) Prove that $10^{340} < a_{992} < 10^{347}$.
Is there an... | About point $(b)$, we may notice that $992=\frac{1984}{2}$, hence $a_{992}$ is the largest coefficient (we are dealing with palyndromic polynomials) and
$$ a_{992} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(e^{-2iz}+e^{-iz}+1+e^{iz}+e^{2iz}\right)^{496}\,dz $$
is a real integral not so difficult to approximate:
$$ a_{992} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A couple of series questions that I just can't figure out (Calc 2)
Show that
$$
\begin{align}
\left(\frac{\pi}{2}\right)^2\left[\int_0^{\pi/2}\cos^{2n}t\ dt-\int_0^{\pi/2}\cos^{2n+2}t\ dt\right]&=\frac{\pi^3}{8}\left[\frac{(2n-1)!!}{(2n)!!}-\frac{(2n+1)!!}{(2n+2)!!}\right]\\[10pt]
&=\frac{\pi^3}{8}\frac{(2n-1)!!}{(2... | I'm assuming that you already show that the closed form of this Wallis' integral is $$\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\frac{\pi}{2}\tag{1}
$$ hence $$\frac{\pi^{2}}{4}\left(\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx-\int_{0}^{\pi/2}\cos^{2n+2}\left(x\right)dx\right... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| $123+4+5+6+7+8+9=162>144$. Since any other way of having a three-digit number used in the summation will be strictly larger, we know none using a three-digit number exist. Similarly, we know no arrangement with a number greater than three digits will yield a sum of $144$.
Using only one two-digit number, the largest ... | {
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Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.
Find the smallest $\alpha\in\mathbb{R}$ such that, for all $x,y,z\in\mathbb{R}$, the following inequality holds $$\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+... | If $x=y=z=1$ we get $\alpha\geq3$.
But for $\alpha=3$ it's enough to prove our inequality for non-negatives $x$, $y$ and $z$.
Since $3(x^2-x+1)^3-(x^6+x^3+1)=(x-1)^4(2x^2-x+2)\geq0$,
our inequality follows from Holder:
$$3\prod_{cyc}(x^2-x+1)\geq\sqrt[3]{\prod\limits_{cyc}(x^6+x^3+1)}\geq x^2y^2z^2+xyz+1$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers!
I also tried to view it as a quadratic equation in ter... | Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1\implies b ={a^2+1\over a+2}$$
Write $c=a+2$ and then $$b = {(c-2)^2+1\over c} = {c^2-4c+5\over c}= c-4 +{5\over c}$$
so $c\mid 5\implies a+2\in \{-5,-1,1,5\} \implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and... | It is correct to me. In short,
$$
\begin{align}
(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\frac1{\sqrt{n+1}+\sqrt{n}}
\\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+\sqrt{1+1/n}}
\\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+1+O(1/n)}
\\\\&=\frac{(-1)^n}{2\sqrt{n}}\left(1-O(1/n) \right)
\\\\&=\frac{(-1)^n}{2\sqrt{n}}+O\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of differential equation $\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$ Solve the given differential equation.
$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$
I have been trying to solve given differential equation using elementary approaches but no manipulation is a leading to a solvable form. Could someone help m... | $$y'(x)=\frac{1}{xy(x)\left(x^2\sin(y(x)^2)+1\right)}\Longleftrightarrow$$
Write the differential equation in terms of $x$.
Notice: $\frac{\text{d}y(x)}{\text{d}x}\cdot\frac{\text{d}x(y)}{\text{d}y}=1$
$$\frac{1}{x'(y)}=\frac{1}{yx(y)\left(x(y)^2\sin(y^2)+1\right)}\Longleftrightarrow$$
$$x'(y)=y\sin(y^2)x(y)^3+yx(y)\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$? The primary question was:
What is the minimum value for $(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
$\color{red}{\text{But sorry guys! I messed it up! my qu... | Given $$ \frac{(1-a)(1-b)(1-c)}{abc}$$
Now Using $a+b+c=1\;,$ We get
$$\frac{(b+c)(c+a)(a+b)}{abc} = \left[\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \left(\frac{c}{b}+\frac{a}{b}\right)\cdot \left(\frac{b}{c}+\frac{a}{c}\right) \right]$$
Now Using $\bf{A.M\geq G.M},$ We get $$\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to decompose $x^3-1$ I need to decompose $x^3-1$, I know the Binomial theorem, and finding roots of a polynomial, how should I approach this?
| You have $(x-1)(x^2+x+1)$ now we need a root for the second polynomial.
We have $x^2+x+1=0\iff x^2+x+\frac{1}{4}=\frac{-3}{4}\iff(x+\frac{1}{2})^2=\frac{-3}{4}\iff x+\frac{1}{2}=\frac{\pm\sqrt{-3}}{2}$
$\iff x=\frac{\pm\sqrt{-3}-1}{2}$
So the polynomial is equal to:
$(x-1)(x+\frac{1+\sqrt{3}i}{2})(x+\frac{1-\sqrt{3}i}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equ... | Alternatively, observe that the given equation can be parametrized as follows: $$(x(t), y(t)) = (a \cos t, b \sin t).$$ Then the curvature is given by $$\kappa(t) = \frac{|x'(t) y''(t) - x''(t) y'(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}.$$ We compute $$(x'(t), y'(t)) = (-a \sin t, b \cos t), \\ (x''(t), y''(t)) = (-a \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using CS inequality to find maximum of a function I am trying to us Cauchy-Schwarz inequality to find the maximum of:
$$|(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)|$$
Where $a$, $b$, and $c$ are real numbers, and $a+b+c=0$ and $a^2+b^2+c^2=2$.
What I've tried:
$$a_1=a,\ a_2=b,\ a_3=c$$
$$b_1=b\sqrt{a-b}, \ b_2=c\... | Let $a=1$, $b=-1$ and $c=0$. Hence, we get a value $2$.
We'll prove that it's a maximal value.
Indeed, since the condition gives $ab+ac+bc=-1$, we obtain:
$|\sum\limits_{cyc}(a^3b^2-a^3c^2)|=|(ab+ac+bc)(a-b)(a-c)(b-c)|=|(a-b)(2a+b)(a+2b)|$.
Since the condition gives $a^2+ab+b^2=1$, it remains to prove that
$$4(a^2+ab+b... | {
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Prove the Inequality $\frac{1}{1-x}-\frac{x(3-x)(2-x)(13x^4-50x^3+89x^2-84x+36)}{4(1-x)(2x(1-x))^2}<1$ Can anyone suggest any hints to prove the following inequality:
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
for all $x \in (0,1)$?
| $$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{1}{1-x}-1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{x}{1-x}.$$
For $x\in(0,1)$ we can prove that
$... | {
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"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated
| $$\tan ^{ -1 } \left( \frac { \sqrt { 4+x } +\sqrt { 4-x } }{ \sqrt { 4+x } -\sqrt { 4-x } } \right) =\tan ^{ -1 } \left( \frac { { \left( \sqrt { 4+x } +\sqrt { 4-x } \right) }^{ 2 } }{ \left( \sqrt { 4+x } -\sqrt { 4-x } \right) \left( \sqrt { 4+x } +\sqrt { 4-x } \right) } \right) =\\ =\tan ^{ -1 }{ \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Combinatorics problem involving binomial coefficient I found this interesting problem in a Romanian mathematical magazine while preparing for the USAMO. Let $k$ be a non-zero natural number. Determine $x,y,z \in \Bbb N$ such that
$$\binom {z+k}{x+y} - \binom {z}{x} \le k \space and \space 2x+y \le z.$$
| I suspect you are looking for less of a brute force argument than those above.
Firstly, note that $(x,y,z)=(0,0,n)$ is a solution for $n \geq 0$ (for any appropriate convention for $0$ choose $0$), since the LHS of the first inequality is just $0$. Therefore, assume $x+y \geq 1$.
A useful identity is obtained as follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
power series representation of a function by differentiation Find power series representation for $\frac{1}{(7+x)^2}$
What I tried...
$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
$$\frac{1}{7(1-(-{x \over7}))}=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{7^{n+1}}$$
$${d\over dx}(\frac{1}{7+x})={d\over dx}(\sum_{n=0}^{\infty}\frac{(... | Hint: Instead of differentiation you could also apply the formula for the binomial series expansion with $\alpha=-2$
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1\tag{1}
\end{align*}
We obtain
\begin{align*}
\frac{1}{(7+x)^2}&=\frac{1}{7^2}\cdot\frac{1}{(1+\frac{x}{7})^2}\\
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| using the formula $$a^2-b^2=(a-b)(a+b)$$ we obtain
$$(x-1)^2-(x-5)^2=(x-1-x+5)(x-1+x-5)=4(2x-6)=8(x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
APICS Mathematics Contest 1999: Prove $\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$ is a constant function of $x$ This is question 3 from the APICS Mathematics Competition paper of 1999:
Prove that $$\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ i... | Let
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$
Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$.
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Finding true roots of $x^2 + 10^9x +1$ using the Quadratic Formula I've been given a hint to rearrange the quadratic formula to:
$$b\sqrt{1-\frac{4ac}{b^2}}$$
but am still dealing with a numerical error at $4/10^{18}$
Can I solve this numerically with the Taylor Series?
| Almost as H. H. Rugh answered,
consider the case of $$x^2+a x+1=0$$ where $a$ is a huge number.
The roots are given by $$x_{1,2}=-\frac{1}{2} \left(a\pm\sqrt{a^2-4}\right)=-\frac{a}{2} \left(1\pm\sqrt{1-\frac 4{a^2}}\right)$$ Now, using Taylor series for small $y$ (or the generalized binomial expansion)$$\sqrt{1-y}=1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all functions that satisfy $f(\frac{x+4}{1-x}) + f(x) = x$ I found the following task in a book and I would be interested if someone has an idea to solve it:
Find all the functions $f$ that satisfy $f(\frac{x+4}{1-x}) + f(x) = x$.
My ideas:
Assuming that $f$ is a power series or making a substitution.
I tried seve... | Hint:
Consider $T(x+1)=\dfrac{T(x)+4}{1-T(x)}$ ,
Let $T(x)=U(x)+1$ ,
Then $U(x+1)+1=\dfrac{U(x)+5}{-U(x)}$
$U(x+1)+1=-1-\dfrac{5}{U(x)}$
$U(x+1)=-2-\dfrac{5}{U(x)}$
Let $U(x)=\dfrac{V(x+1)}{V(x)}$ ,
Then $\dfrac{V(x+2)}{V(x+1)}=-2-\dfrac{5V(x)}{V(x+1)}$
$\dfrac{V(x+2)}{V(x+1)}=-\dfrac{2V(x+1)+5V(x)}{V(x+1)}$
$V(x+2)+2V... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $x^6+x^4+x^3+x^2+1=0$? There is a hint in the question, use the factorization of $x^5+x+1$.
| How about first multiplying by $(x-1)$ ?
$$(x-1)(x^6+x^4+x^3+x^2+1)= x^7-x^6+x^5 - x^2+x-1= (x^2-x+1)(x^5-1)$$
and carry on from there?
Added: Not sure if worth mentioning, but doing the same on $x^5+x+1$ requires an extra little trick:
$$ (x-1)(x^5+x+1)= x^6-x^5 (+x^3 - x^3) +x^2-1=(x^3-x^2+1)(x^3-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Proving that $\lim_{(x,y) \to (0,0)} (x^2 +y^2 -x^3 y^3)/(x^2 +y^2) =1$ How can I go about proving that $$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1 ?$$
I checked some lines along $x, y$ and $x=y$ and it all gave $1$
| The first step,
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$
Now study this and find $a$
$1-a=1-\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2}$
$a=\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2}$
$\frac {1}{a}=\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that... | If we MUST use the binomial coefficients, this becomes rather tedious:
$$\frac{1+x^2+x}{(1-x)^2}=\frac{(1-x)^2+3x}{(1-x)^2}=1+\frac {3x}{(1-x)^2}$$$$=1-3x(1-x)^{-2}=1-3(1-x)^{-1}+3(1-x)^{-2}$$
$$=1-3\sum\binom{-1}{r}(-1)^rx^r+3\sum\binom{-2}{r}(-1)^rx^r$$
$$=1+3\sum \left(-\binom{-1}{r}+\binom{-2}{r}\right)(-1)^rx^r$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.
My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int ... | Using the substitution $x=\sec(\theta)$, we get
$$
\begin{align}
\int\frac{x\log(x)}{\left(x^2-1\right)^{3/2}}\,\mathrm{d}x
&=-\int\log(x)\,\mathrm{d}\left(x^2-1\right)^{-1/2}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\frac{\mathrm{d}x}{x\sqrt{x^2-1}}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\mathrm{d}\theta\\
&=-\frac{\log(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Real function with complex antiderivative $\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}$? Consider this indefinite integral (I'm interested in the interval $x>0$):
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx$$
By substitution:
$$u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}, \qquad x=\fra... | Consider
$$
u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}>
\sqrt{\frac{1}{2}+\frac{1}{2}}=1
$$
Assuming your substitutions are correct, the integral becomes
$$
\sqrt{2}\int\frac{1}{1-u^2}\,du=
\frac{\sqrt{2}}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\,du=
\frac{\sqrt{2}}{2}\log\left|\frac{1+u}{1-u}\right|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| For $n=3$ reminder is $5$.
If $n$ is even, then $2^n+n^2$ is also even, hence not prime.
Therefore, we should only consider odd numbers $n$.
In fact, we should consider only cases $n=6k\pm 1$, $n=6k\pm 3.$
Since $$2^n+n^2=2^n+1+(6k\pm1)^2-1=3(2^{n-1}-2^{n-2}+\dots-1)+6T=0\mod 3,$$ this number is not prime for $n=6k\pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The indefinite integral $ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$ $$ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$
$$ \int \frac{\s... | $$I=\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha}\int \frac{\sin x}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx+\sin{\alpha}\int \frac{1}{\cos^2 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha} I_1+\sin{\alpha}I_2,$$ whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct? Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.
Proof. By strong inductio... | Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Does the limit exist for $\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$? $$\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\le \lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3}}=\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\s... | Use equivalents:
*
*$\arctan x\sim_{+\infty}\dfrac\pi 2$,
*$\sqrt{2x^3+4x^2+}x\sim_{+\infty}\sqrt{2x^3}$,
hence $\; \dfrac{x^{\tfrac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\sim_{+\infty}\dfrac{\pi x^{\tfrac{3}{2}}}{2\sqrt{\mathstrut 2} x^{\tfrac{3}{2}}}=\dfrac{\pi\sqrt{\mathstrut2}}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^1\frac{\ln^2(1-x)}{x}\ dx$ In solving $\displaystyle\int_0^\frac{\pi}{4}\dfrac{\ln(\sin x)\ln(\cos x)}{\sin x\cos x}\ dx,$ I have found that this is equal to $\dfrac{1}{16}\displaystyle\int_0^1\dfrac{\ln^2(1-x)}{x}\ dx.$ WolframAlpha says that the desired value is $\dfrac{\zeta(3)}{8},$ so I suspect a conv... | Here is an approach that makes use of an Euler sum.
We will first find a Maclaurin series expansion for $\ln^2 (1 - x)$. As
$$\ln (1 - x) = - \sum_{n = 1}^\infty \frac{x^n}{n},$$
we have
\begin{align*}
\ln^2 (1 - x) &= \left (- \sum_{n = 1}^\infty \frac{x^n}{n} \right ) \cdot \left (- \sum_{n = 1}^\infty \frac{x^n}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$The... | Here is a 3rd solution (inspired by yours, @Christian Blatter, but using different variables).
Let $$Z_1=\frac{z^2_{1}}{z_{2}z_{3}}, \ Z_2=\frac{z^2_{2}}{z_{3}z_{1}}, Z_3=\frac{z^2_{3}}{z_{1}z_{2}} \ \ \ (*)$$
Let us remark that all $Z_k \in \mathbb{U}$ (unit circle).
Thus the initial condition is replaced by $Z_1+Z_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The significance of the number in front of a series? I believe I have solved the series to $-4/9$ however I'm not sure of the significance of the 9 in front of sigma. Is this simply multiplying the final result or is it something different?
The series in question is
$$9\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n$$
| Your solution of the series is correct. According to the formula for geometric power series
$$\sum_{n=1}^\infty q^n=\frac{q}{1-q}\qquad\qquad |q|<1$$
we obtain
\begin{align*}
\sum_{n=1}^\infty\left(-\frac{4}{5}\right)^n=\frac{-\frac{4}{5}}{1+\frac{4}{5}}=-\frac{4}{9}
\end{align*}
We do not explicitely need to wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to derive $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ Looking at an proof of $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ that argues that $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ but I can't see how this is accomplished since th... | So there is a post in @Robert Z's link in the comments that proves
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$
So I won't restate what has already been done, but will just show that
$$ \int_0^{1} \int_0^1 \frac{1}{1 - xy} dx dy = \sum_{n=1}^{\infty} \frac{1}{n^2} $$
Observe that
$$ 1 + x + x^2 + x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that the expression has infinitely many solutions
Does $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$ have infinitely many solutions in positive integers $(m,r)$?
Does $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$ have infinitely many solutions in positive integers $(m,r)$ where $m \equiv 3 \pmod{4}$?
I w... | The first has no solutions with positive integer $m$. However, the second case,
$$\begin{aligned}
m&=\dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)\\
&=\dfrac{1}{18}\left(x-1\right)\\
&\equiv {3\pmod{4}}
\end{aligned}$$
has infinitely many given by,
$$x= \frac{ (7+\sqrt{48})^{6k+3}+(7-\sqrt{48})^{6k+3}}{2}=1351,\; 98633821... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why in the binomial expansion are the powers of x positive? In Binomial expansion with $n$ as a rational, positive number then,
$$(x+y)^n =\binom{n}{0}y^n + \binom{n}{1} y^{n-1} x +\binom{n}{1} y^{n-2}x^2 +\cdots+\binom{n}{n} x^{n}$$
While if $\alpha$ is a rational, non-positive number and $\lvert x \rvert< 1$ then... | Here is one way to see this - from the theory of GP, it may be familiar to write for a common ratio $|x| < 1$,
$$\frac1{1-x} = 1+x+x^2+x^3 + \cdots$$
Using $x \to -x$, we get
$$\frac1{1+x} = 1-x+x^2-x^3 + \cdots$$
If you differentiate that, you get
$$-\frac1{(1+x)^2} = -1+2x-3x^2+4x^3 -\cdots$$
$$\implies \frac1{(1+x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Expand$\frac{ x^{k+1}}{k+1} -\frac{ (x-1)^{k+1}}{k+1} $using the binomial theorem The required is to expand
$$\frac{x^{k+1}}{(k+1)} - \frac{(x-1)^{k+1}}{(k+1)}$$
using the binomial theorem.
Here is my solution, is it correct?
$$x^k -\frac{kx^{k-1}}{2} + \frac{\binom{k}{2}x^{k-2}}{3} - \frac{\binom{k}{3}x^{k-3}}{4} + \... | The binomial theorem states:
$$
\left(a + b\right)^n = \sum_{m=0}^n \binom{n}{m} \cdot a^m \cdot b^{n-m}
$$
Applying it for $(x-1)^{k+1}$, with $a = x$, $b=-1$ and $n=k+1$ we have:
$$\begin{eqnarray}
\left(x-1\right)^{k+1} &=& \sum_{m=0}^{k+1} \binom{k+1}{m} \cdot x^m \cdot \left(-1\right)^{k+1-m} \cr &=& x^{k+1} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| Since the OP admitted that they can't continue from Dr.MV's hint, I'll do it myself:
$$\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}=\frac{1}{1-b} \left(\sum_{n=0}^\infty a^n-b \sum_{n=0}^\infty (ab)^n \right)=$$
Here we use geometric series, so we need to have $|a|<1$ and $|ab|<1$: *
$$=\frac{1}{1-b} \left(\frac{1}{1-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
How can I integrate $\int \arctan(\sec x + \tan x) dx$ I got this problem in my homework exercise:
$$\int \arctan(\sec x + \tan x) dx$$
I simplified it to
$$\int \arctan\left(\dfrac{1+\sin x}{\cos x}\right) dx$$
$$=\int \arctan\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right) dx$$
now, I tried putting $\sqrt{\dfrac{1+\si... | By integration by parts we obtain
$$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C.$$
Since we have the identity (see Qwerty's answer):
$$\sec x + \tan x={1+\sin(x)\over \cos(x)}=\tan(\pi/4+x/2),$$
the primitive can be simplified (!?) to
$$x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C=x\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| There is another inductive way, I don't know if better or worse:
Our goal is to prove that $2^{n+1}>(n+1)^4$, assuming that $2^n>n^4$ and, as noted, $n\ge 17$. Let's estimate $2^{n+1}-(n+1)^4$:
$$2^{n+1}-(n+1)^4=\big(2^n-[(n+1)^4-n^4]\big)+(2^n-n^4)>(2^n-n^4)+(2^n-n^4)>0$$
To show the first inequality we have to check:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to show that this identity holds? By comparing the first terms of the Taylor expansion at $0$, it seems that
for $|x|<4/27$ the following identity holds:
$$\ln\left(\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^{n-1}}{2n+1}\right)=
\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^n}{n}.$$
I tried by differentiating both sides, ... | Let's try to use your idea of looking at derivatives: For $x=0$ both sides result in $0$, so indeed, if the derivatives are the same then we have the equality.
Let $g(x)$ be the left-hand side and $f(x)$ the right-hand side. The derivative of $g(x)$ is
$$g'(x)=\frac{1}{\sum_{n=1}^\infty\binom{3n}{n}\frac{x^{n-1}}{2n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Solve $\binom{x+2}{5} = 126$ for $x$ How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one solution exists?
| Use brute force approach, We can easily find x=7 is the solution.
And x=7 is the only one solution. Because:
There's no solution for x<3, because $\binom{x+2}{5}=0$ for x<3
for x>=3:
$\binom{x0+2}{5} < \binom{x+2}{5} < \binom{x1+2}{5}$ for every x1 > x > x0 >=3; and x,x1,x0 is real number
so
$\binom{x0+2}{5} < \binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
I determined that since $2a^2-1 \equiv 1,7 \pmod{8}$ we must have $b^2+2 \equiv 3 \pmod{8}$ in order for the fraction to be an integer. I didn't see how to find a contradiction from he... | Hint $\ $ If a prime $\,p\,$ divides numerator and denominator then $\,{\rm mod}\ p\!:\ b^2\equiv -2,\,$ and $\, (2a)^2\equiv 2.\,$ Since both $\,2\,$ and $\,-2\,$ are squares mod $\,p\,$ it follows by reciprocity that $\,p\equiv 1\pmod 8.\,$ But $\,b^2+2\,$ must be divisible by a prime $\,q\not\equiv 1\pmod 8,\,$ els... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it ... | $$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$
Because the $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx $ is convergent so the integration can be linearly seperated.
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx - \int_0^\infty(\sqrt{x^4+b^4}-x^2) dx$$
Ou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Finding a set of 3 positive prime numbers that satisfy a polynomial equation What is the set of positive primes p,q,r that satisfy the equation: $p^{4}+2p+q^{4}+q^{2}=r^{2}+4q^{3}+1$. It can be easily shown (using parity concepts) that either p or r must be 2, but it is not clear to me how to advance any further since ... | We have
$$ r^2=p^4+2p+q^4-4q^3+q^2-1$$
*
*If $q\ge 5$, then $q^4-4q^3+q^2-1\ge q^3+q^2-1>0$ and hence $r^2>p^4\ge 16$, so $r\ge 5$.
*If $q=3$, then $r^2=p^4+2p-19$, so $r=2$ leads to $(p^3+2)p=23$, contradiction
*If $q=2$, then $r^2=p^4+2p-13$, so $r=2$ leads to $(p^3+2)p=17$, contradiction.
We conclude that $r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Taylor Expansion of a Trig Function Did a problem in a lecture today with Taylor Expansions.
The question is:
Find the Taylor Expansion of
$
z=\sin(x^2+y^2)
$
To the second order at the point (1, -1)
I am happy with the solution until she gets to
$
\frac{d^2(f)}{d(y)^2}(1,-1)=-4\sin(2)=-3,637
$
According to my own ... | Your computations are fine. We have
$$
f(x,y):=\sin(x^2+y^2)
$$ giving by the chain rule
$$
\frac {df}{dy}(x,y)=2y\cos (x^2+y^2), \qquad \frac {d^2f}{d^2y}(x,y)=2\cos(x^2+y^2)-4y^2 \sin(x^2+y^2)
$$ and
$$
\left.\frac {d^2f}{d^2y}(x,y)\right|_{(1,-1)}=2\cos(2)-4 \sin(2)=-4.469483380\cdots.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| Let $a_n=x^n+y^n$.
Then
$$3a_n=(x+y)a_n=x^{n+1}+xy^n+yx^n+y^{n+1}=a_{n+1}+xya_{n-1}=a_{n+1}+3a_{n-1}$$
and so
$a_{n+1}=3a_n-3a_{n-1}$.
Also, $a_0=x^0+y^0=2$, $a_1=x+y=3$.
So $a_2=3a_1-3a_0=3$, $a_3=3a_2-3a_1=0$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Find $x+y+z$ From Equations Including Radicals Let $(x,y,z)$ be the real solution to the system of equations
\begin{align*}
x+y &= \sqrt{4z-1} ,\\
y+z &= \sqrt{4x- 1} , \\
z+x &= \sqrt{4y-1} .
\end{align*}
Find $x+y+z.$
I could add all the equations up but that doesn't do any good. Thanks in advance!
| To initiate Arthur's hints in the comments:
$$x^2 + 2xy + y^2 = (x + y)^2 = 4z - 1$$
$$y^2 + 2yz + z^2 = (y + z)^2 = 4x - 1$$
$$z^2 + 2zx + x^2 = (z + x)^2 = 4y - 1$$
Rather than adding all three equations, we subtract them pairwise. Then assuming $x \neq y \neq z$, we have:
$$2y(x - z) + (x + z)(x - z) = 4(z - x) \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Limit of the difference between two exponential functions
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$
I did nothing by now.
| We have that
$$\lim_{x \to +\infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right)=
\lim_{x \to +\infty}\left((x+3)\exp\left[\frac{\ln(x+3)}{x}\right]-x\exp\left[\frac{\ln(x)}{x+3}\right]\right).
$$
Moreover
$$(x+3)\exp\left[\frac{\ln(x+3)}{x}\right]=(x+3)\cdot\left(1+\frac{\ln(x+3)}{x}+O(\ln^2(x)/x^2)\right)\\
=x+3+(x+3)\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the number of 3-element subsets from the set {1,2,3,...,11,12,13} such that the sum of the 3 elements is divisible by 3 I found this problem from an old math questionnaire.
How many 3-element subsets of {1,2,3,...,11,12,13} are there for which the sum of the 3 elements is divisible by 3?
At first, I tried listi... | A more general approach is given by generating functions.
We may consider the bivariate polynomial
$$ q(x,y)=(1+yx)(1+yx^2)\cdot\ldots\cdot(1+yx^{12})(1+yx^{13}) $$
and the coefficient of $y^3$ in $q(x,y)$, that is
$$ [y^3]\,q(x,y) = r(x) = x^6+x^7+2 x^8+3 x^9+4 x^{10}+5 x^{11}+7 x^{12}+8 x^{13}+10 x^{14}+12 x^{15}+14 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A beautiful triangle inequality Prove that in any triangle with side lengths $a, b, c$ the inequality:
$$(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\geq \sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
For demonstration, I tried to use substitutions Ravi or algebraic calculation but without su... | Since $\sqrt{b}+\sqrt{c}-\sqrt{a}=\sqrt{b+c+2\sqrt{bc}}-\sqrt{a}>\sqrt{a}-\sqrt{a}=0$, we need to prove that
$$(a+b-c)^2(a+c-b)^2(b+c-a)^2\geq(a^2+b^2-c^2)(a^2+c^2-b^2)(b^2+c^2-a^2)$$
We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$ and from here we can assume that
$a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| Let
\begin{align}
y&=\sin ^{-1}\left(\frac{2^{x+1}3^x}{1+36^x}\right)\\
&=\sin ^{-1}\left(\frac{2^x\cdot2\cdot3^x}{1+36^x}\right)\\
&=\sin ^{-1}\left(\frac{2\cdot6^x}{1+6^{2x}}\right)\\
&=2\tan ^{-1}6^x
\end{align}
Thus
\begin{align}
\frac{dy}{dx}&=\frac{2}{1+6^{2x}}\cdot6^x\log 6\\
\implies \frac{dy}{dx}&=\frac{2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How to compute this limit $\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}$ $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$
such that $2\alpha n\le n$
I tried to use Stirling formula and we get
$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\li... | Assuming that $$\binom{m}{n} = \frac{\Gamma(m + 1)}{\Gamma(n + 1)\Gamma(m - n + 1)}\tag{1}$$ whenever the RHS is defined we can see that the sequence $$a_{n} = \binom{n}{2\alpha n}$$ satisfies $$\frac{a_{n + 1}}{a_{n}} = \frac{\Gamma(n + 2)}{\Gamma(2\alpha n + 2\alpha + 1)\Gamma(n + 2 - 2\alpha n - 2\alpha)}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Are there other good ways to look at this differential equation or is this it? \begin{align}
\tan \frac \alpha 2 & = u \\[4pt]
\alpha & = 2\arctan u \\[4pt]
d\alpha & = \frac{2\,du}{1+u^2} \\[4pt]
\sin\alpha & = \sin(2\arctan u) = 2(\sin\arctan u) (\cos \arctan u) \\[4pt]
& = 2\,\frac u {\sqrt{1+u^2}} \cdot \frac 1 {\s... | From the first time I heard about Weierstrss substitution, it also became my favored way for solving things such as $$\int \frac{dx}{a+b \sin(x)+c\cos(x)}=-\frac{2 }{\sqrt{b^2+c^2-a^2}}\tanh ^{-1}\left(\frac{(a-c) \tan
\left(\frac{x}{2}\right)+b}{\sqrt{b^2+c^2-a^2}}\right)$$
For things such as $$I=\int \frac{dx}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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A question related to compute infinite summation How to solve this: $$ 1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\cdots $$ So, the $n^{th}$ term of the sum can be written as $$a_n=\frac{2\cdot5\cdot8\cdots (2+3(n-1))}{6\cdot12\cdot18\cdots (6n)} = \frac{2\cdot5\cdot8 \cdots (2+3(n-1... | Consider
$$
1+\sum^\infty_{n=1}\prod^n_{i=1}\frac{2+3(i-1)}{6i}=1+\sum^\infty_{n=1}\prod^n_{i=1}\left(\frac{1}{2}-\frac{1}{6i}\right)=1+\sum^\infty_{n=1}2^{-n}\prod^n_{i=1}\left(1-\frac{1}{3i}\right)
$$
Product above can be expressed in terms of Gamma-functions:
$$
\prod^n_{i=1}\left(1-\frac{1}{3i}\right)=\frac{\Gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorial proof of a binomial identity: $\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$ I am to find a combinatorial argument for the following identity:
$$\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$$
For the right hand side, I was think that would just be nu... | Suppose we seek to verify that
$$\sum_{k=1}^r {2r\choose 2k-1} {k-1\choose s-1}
= 2^{2r-2s+1} {2r-s\choose s-1}$$
where presumably $s\ge 1$. The lower limit is set to $k=1$ as the
first binomial coefficient is zero when $k=0.$
Introduce
$${2r\choose 2k-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2r-2k+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| Suppose $a,b,c,d\geq 0$ and set $x=a^4,\,y=b^4,\,z=c^4,\,w=d^4$. Then
$$f(x,y,z,w)=x^{5/4}+y^{5/4}+z^{5/4}+w^{5/4}$$
is strictly convex and thus restricted to the domain
$$\{x,y,z,w\geq 0: x+y+z+w=16\}$$
it attains the maximum only in the extremal points, which are the four vertices $(16,0,0,0)$ and cyclicals.
If one n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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Evaluate $\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2}$ I want to evaluate the sum
$ \displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2} $
My approach so far has been the possible use of shifted Legendre polynomial. We know
$Q_n(x)=\displaystyle \sum_{j=0}^n(-1)^{n+j... | Suppose we seek a closed form of the sum
$$\sum_{q=0}^n (-1)^{n+q} {n\choose q} {n+q\choose q} \frac{1}{(q+1)^2}.$$
This is
$$\sum_{q=0}^n (-1)^{n+q} \frac{q+1}{n+1}
{n+1\choose q+1} {n+q\choose q} \frac{1}{(q+1)^2}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^{n+q}
{n+1\choose q+1} {n+q\choose q} \frac{1}{q+1}.$$
Observe t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to solve $\int\tfrac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx$? I am trying to find a solution for the following integration for a physics problem. Could anyone give hint on how to do this?
$$
\int\frac{\sqrt{x^2+1}}{\left(x^2+a\right)^\frac{3}{2}}dx
$$
Thanks for your time.
| Allow us to use generalized binomial expansion theorem on the $\sqrt{x^2+1}$ and the $u$ substitution $x^2+a=u$.
$$\sqrt{x^2+1}=\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}=x+\frac12x^{-1}-\frac18x^{-2}+\dots$$
$$\begin{align}
\int\frac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx & = \int\frac{\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$ Let $a, b, c, d \in \mathbb{R}_{>0}$, then prove that
$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$
Can this be done without using AM-GM inequality, or without using any identity/theorem of inequality? I don... | Let $\frac{a}{b}=x^4$, $\frac{b}{c}=y^4$, $\frac{c}{d}=z^4$ and $\frac{d}{a}=t^4$ for positives $x$, $y$, $z$ and $t$.
Hence, we need to prove that $x^4+y^4+z^4+t^4\geq4xyzt$, which is true because
$$x^4+y^4+z^4+t^4-4xyzt=(x^2-y^2)^2+(z^2-t^2)^2+2(xy-zt)^2\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number?
| There is no need to find an explicit formula for $\sum_{n=1}^{600}F_n^2$ (but this is doable and I'll show it in the second part of this answer). The Fibonacci sequence obeys the relation $F_{n+2}=F_{n+1}+F_{n}$ and $8$ is a Fibonacci number, hence the sequence $\{F_n\pmod{8}\}_{n\geq 1}$ has a short period and the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{... | Your first way is wrong and, unfortunately, a common mistake. With the same argument you would conclude that
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=0
$$
which is a big error (see below).
You can change $\sin x$ into $x$ when it's a factor, not a summand. For instance, if you have
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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square root in algebra Have a question about roots.
$$ \sqrt{9} = 3^2 = 3 \times 3 $$
$$ \sqrt{9a^6} = 3a^3 $$
But why does $ \sqrt{a^6} = a^3 $?
Is it correct vision?
$$ \sqrt{a^6} = \sqrt{a\times a \times a \times a \times a \times a} = \sqrt{2(a \times a \times a)} $$
Can anybody add more details about my questi... | -----$\sqrt{9} = 3^2 = 3 * 3$
Incorrect.
$\sqrt{9} = \sqrt{3^2} = 3$
====
----but why $\sqrt{a^6}=a^3$
Because $\sqrt{a^6} = \sqrt{(a^3)^2} = a^3$.
[Assuming $a \ge 0$. If $a < 0$ then $a^6 = |a|^6 =|a^6| > 0$ and $\sqrt{a^6} = |a|^3$. For the rest of the post I'm assuming $a \ge 0$.]
=====
----Is it correct vision... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a symmetric $7\times 7$-matrix Is there any simple method to calculate the determinant of the following symmetric $7\times 7$-matrix
$$M:=\begin{pmatrix}
1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\
0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\
0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\
0 & 0 & 0 & 1 & -a_{3} & a_... | The determinant of $M$ is given by
$$
\det(M)=(a_0 - a_1^2 - a_2^2 + a_3^2 - a_4^2)(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^2.
$$
Note that the matrix is not symmetric.
For the symmetric variant we have
$$
\det(M)=(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^3,
$$
where we could use rules of the determinant for block matrices (t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now... | When we do the Tayor series expansion of the numerator, we don't care about the $x^3$ and higher powered terms. They are all going to 0. I am going to drop them as I go.
$\lim_\limits{x\to 0}\frac {1 - (1 - \frac {x^2}{2})(1-2x^2)^\frac12(1-\frac {9x^2}{2})^\frac 13}{x^2}$
Now do the binomial expansion and continue t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square ... | The following is a shortcut that works for this particular equation.
Multiplying both sides by the sum of the radicals (which is always $\ne 0$) gives:
$$(3.5^2+x^2)-(3.0^2+x^2)=0.25(\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2})$$
$$\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2} = 13$$
Adding the latter to the original equation gives:
$$2 \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$ Let $x,y,z \in [0;1]$. Prove the inequality
$$(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$$
My work so far:
Let $f(x)=(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 -6$, where $x\in[0;1]$
$f(0)=(y^4+z^4)+(y^5+z^5)+(y)^6+(y... | @Alex Silva starts off with the reduction
$$\begin{align*} & (x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \leq \\& \leq 2(x^2+y^2+z^2)+(x-y)^2+(y-z)^2+(z-x)^2\end{align*}$$
Now note that
$$(x - y)^2 \leq (1 - x)^2 + (1 - y)^2$$
which is true because it is equivalent to
$$-2xy \leq -2x + 1 - 2y + 1 \implies x + y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\le... | We need to establish $$\sin2\theta\ge-\dfrac{1+3\cos^2\theta}2=-\dfrac{2+3(1+\cos2\theta)}4$$
$$\iff4\sin2\theta+3\cos2\theta\ge-5$$
Now
$4\sin2\theta+3\cos2\theta=5\sin\left(2\theta+\arcsin\dfrac45\right)$
Finally for real $\theta,$ $$-1\le\sin\left(2\theta+\arcsin\dfrac45\right)\le1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
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Fractions in Questions and Answers
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