Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Points on the hypotenuse of a right-angled triangle Points $K$ and $L$ are chosen on the hypotenuse $AB$ of triangle $ABC$ $(\measuredangle ACB=90^\circ)$ such that $AK=KL=LB$. Find the angles of $\triangle ABC$ if $CK=\sqrt2CL$.
As you can see on the drawing, $CL=x$ and $CK=\sqrt2x$.
I don't know how to approach the ... |
Draw horizontal lines from points K and L of side $\overline{AB}$ to points $K_A$ and $L_A$ of side $\overline{BC}$. Because points $K$ and $L$ trisect side $\overline{AB}$, then points $K_A$ and $L_A$ trisect side $\overline{BC}$. So $\triangle{CK_AK}$ and $\triangle{CL_AL}$ are right triangles.
Draw vertical lines f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$ If $f\colon\left[ { - 1,1} \right] \to \mathbb R$ be continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$, then $\mathop {\lim }\limits_{x \to 0} \frac{{f\le... | Notice that $f(1)=0$. Since the function is continuous, $\lim_{x\to1}f(x)=0$
Also, $4xf'(2x^2-1)=(x^3+x)f'(x)+f(x)(3x^2+1)\implies f'(1)=\lim_{\theta\to0}f'(\cos(\theta/2))=0$
Now
$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of $5.5+55.55+555.555..$ up till n terms?
Find the sum of $5.5+55.55+555.555..$ up till n terms?
My attempt: $ 5.5+55.55+555.555 ... $
$ 5(1.1+11.11+111.111...) $
$ \dfrac{5}{9} \times 9(1.1+11.11+111.111..) $
$ \dfrac{5}{9} (9.9+99.99+999.999...) $
$ \dfrac{5}{9} (9+0.9+99+0.99+999+0.999...) $
$ \dfrac{... | You have to make a general formula so you may use an already wxisting formula for sum of $n$ terms of a $G.P.$
You arrived at $ \dfrac{5}{9} [(10+100+1000 ... n)+(n-(0.1+0.01+0.001...)] $
For a $G.P$ of type $a,ar,ar^2,\cdots ar^{n-1}$ The sum is given as
$S=a\displaystyle\frac{r^n-1}{r-1}$
In the first bracket $a=10$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the value of $\sum_{n=1}^\infty \frac{1}{F_n F_{n+4}}$ where $F_n$ is nth Fibonacci number? I found this problem from my old note, with the memo 'The answer is very interesting."
But I forgot how to solve it!
I rummaged through all my bookshelf, but I couldn't find any clue.
The only thing I remember is
*
*... | It can be reduced to $\sum \frac1{F_n F_{n+2}}$ as follows.
Multiply the numerator and denominator by $F_{n+2}$, and write this in terms of $F_n$ and $F_{n+4}$ by using that the sequences $(F_{2n})$ and $(F_{2n+1})$ are linear recurrences.
You will obtain something like
$$\frac1{F_{n}F_{n+4}} = \frac{F_{n+2}}{F_{n}F_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving $(x^2 - y^2) = (x + y)(x-y)$ I am trying to prove from the field axioms that $(x^2 - y^2) = (x+y)(x-y)$. I am going to take for granted the definition of subtraction as the addition of a negation and that $x(-y) = (-x)y = -(xy)$. Here is my attempt.
We have:
\begin{align*}
(x+y)(x-y) & = x(x-y) + y(x-y) & & \te... | Looks good overall. I do have one nitpick:
At least for me, it is typically not given that the additive inverse is unique in fields, or that $x(-y) = -(xy)$ for sure. (Or the more general $(-1)(xy) = (-x)y = x(-y)$.)
I believe that technically needs to be shown if you want to go from the axioms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$
Compute the limit:
$$\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$$
I tried applying the sandwich rule, constructing an upper and lower bound for the sequence, but I ... | Let $S_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}$
The $k^{th}$ term = $t_k = \frac{kn}{n^3+k} \leq \frac{kn}{n^3} = \frac{k}{n^2}$, $\forall{k} \geq 1$
Then, $S_n=\sum\limits_{k=1}^{n}t_k \leq \sum\limits_{k=1}^{n}\frac{k}{n^2} = \frac{1}{n^2}\sum\limits_{k=1}^{n}k=\frac{1}{n^2}.\frac{n(n+1)}{2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $... | I'll provide a proof that do not use integrals (and will probabily make you apreciate FTC). We can add an extra $\sqrt{n}$ without changing the limit. Will show now that the following limit exists.
\begin{gather*}
L=\lim_{n \to +\infty}\frac{\sum_{i=1}^n \sqrt{i}}{n\sqrt{n}}
\end{gather*}
To show the limit exist, notic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find min $ P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$ under $x,y,z >0$ and $xyz=1$ Let $x,y,z >0$ and $xyz=1$
Find min: $P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$
My trying but it false :<<
We have: $VT=\frac{1}{\frac{x+2}{x}}+\frac{1}{\frac{y+2}{y}}+\frac{1}{\frac{z+2}{z}}$
$=\frac{1}{1+\frac{2}{x}}+\frac{1}{1+\f... | Let $ a = \ln x, b = \ln y, c = \ln z$
Want to minimize $ \sum \frac{ e^a}{e^a+2}$ subject to $ a + b + c = 0$.
Since the derivative is $ \frac{2e^a}{(e^a+2)^2} > 0$, hence we can apply Jensen's inequality to conclude that the minimum occurs at $ a = b = c = 0$, or when $ x = y = z = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant of an interesting Toeplitz matrix Let $ab=1$. Find
$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b & c& \dots &a^{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$
I tried to decompose by line however it doe... | Let $M$ be the matrix at hand. Its entries have the form
$$M_{ij} = (c-1)\delta_{ij} + b^{i-j} = (c-1)\delta_{ij} + b^i a^j$$
where $\delta_{ij} = \begin{cases}1, & i = j\\ 0,& i \ne j\end{cases}$ is the Kronecker delta. We can express $M$ as a rank-1 update of the diagonal matrix $(c-1)I_n$:
$$M = (c-1)I_n + uv^T
\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Alternative and more direct proof that an integral is independent of a parameter I'm looking for alternative ways to calculate the integral
$$
\int\limits_0^\infty\frac{\tanh(\alpha x)}{\tanh(\pi x)}\sin(2\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0.\tag {*}
$$
It was derived in a lengthy calculation using roundabout me... | Since no one has posted an answer, I'm going to use contour integration to show that the equation holds for the next "simplest" case, namely $\alpha = 2 \pi$.
$$\begin{align} \int_0^\infty \frac{\tanh(2 \pi x)}{\tanh(\pi x)} \sin (4 \pi x^2) \, \mathrm dx &= \int_{0}^{\infty} \frac{2 \cosh^2(\pi x)}{\cosh (2 \pi x)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 1,
"answer_id": 0
} |
Use Epsilon Definition of Limit to Prove Limit Exists We have the function
$$f(x,y)=
\begin{cases}
0 & \textrm{if } x=y=0 \\[2ex]
(x+y)\ln(x^2+y^2) & \textrm{otherwise}
\end{cases}$$
and need to prove $\lim\limits_{(x,y) \to (0,0)} f(x,y) = 0$.
We need to show that for any $\epsilon>0$ we can find a $\delta > 0$ such... | You can actually skip the second equality and note that $ |x+y| \leq \sqrt {1^{2}+1^{2}} \sqrt {x^{2}+y^{2}}$ by Cauchy-Schwarz inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r =... | Proceeding along the OP's idea (pqr method):
First of all, we have
\begin{align*}
p^2 &\ge 3q, \tag{1}\\
p^3 - 4pq + 9r &\ge 0. \tag{2}
\end{align*}
Note: (2) is just Schur's inequality $a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\ge 0$.
If $q \le \frac{p^2}{4}$, it suffices to prove that
$$p^2 - 2 \cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrating inverse of a polynomial using valid method gives wrong result The integral is:
$$\int \frac{1}{x^2+(x-1)^2}dx$$
One way to integrate inverses of polynomials is to find their roots and find the fractions which when added together, give the original fraction (with the roots being the denominators). For exampl... | consider this instead:
$$\begin{align}
\phantom{=}&x^2+(x-1)^2\\
=&x^2+(x^2-2x+1)\\
=&2x^2-2x+1\\
=&2\left(x-\frac12\right)^2+\frac12\\
=&\frac12\left[4\left(x-\frac12\right)^2+1\right]
\end{align}$$
you can take this half out of the integral (so it becomes two) now first think of the substitution $u=x-\frac12$ so you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further ... | The following approach is somewhat like taking a sledgehammer to crack a nut, but it can also conveniently used in more challenging situations. We use the coefficient of operator $[x^r]$ to denote the coefficient of $x^r$ of a series. This way we can write
\begin{align*}
\binom{2r}{r}&=[x^{r}](1+x)^{2r}\tag{1}\\
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I thi... | There is an assumption in the derivation of the quadratic formula that $A \ne 0$. You can see it does not directly apply when this assumption is violated due to the $A$ in the denominator.
However, the solution of the linear case can nevertheless be obtained from the quadratic formula on the understanding one of the tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Prove that $\sum{a^{2}bc}=\binom{n+3}{6}+\binom{n+2}{6}$ Question
We define $S$ as the set of positive integer triplets $(a,b,c)$ such that $a+b+c=n$. I want to prove that the following expression is correct:
$$
\sum_{S}{a^{2}bc}=\binom{n+3}{6}+\binom{n+2}{6}
$$
My Solution
Left Hand Side
Say that there is a line of $n... | Alternative solution: let us consider the following generating functions (power series with $\rho=1$):
$$ A(x)=\sum_{a\geq 1}a^2 x^a,\qquad B(x)=\sum_{b\geq 1} b x^b,\qquad C(x) = \sum_{c\geq 1} c x^c. $$
The sum $\sum a^2 bc$ over the triples of positive integers such that $a+b+c=n$ is exactly the coefficient of $x^n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \fra... | Since $$BM+GN=BN+MG,$$ in the standard notation we obtain:
$$\frac{c}{2}+\frac{1}{6}\sqrt{2b^2+2c^2-a^2}=\frac{a}{2}+\frac{1}{6}\sqrt{2a^2+2b^2-c^2}$$ or
$$3(a-c)+\sqrt{2a^2+2b^2-c^2}-\sqrt{2b^2+2c^2-a^2}=0$$ or
$$(a-c)\left(1+\frac{a+c}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2b^2+2c^2-a^2}}\right)=0,$$ which gives $a=c$ and our t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Validity of binomial expansion for any power I know that
$$(1+x)^n=1+nx+\frac {n(n-1)}{2!}x^2+\frac {n(n-1)(n-2)}{3!}x^3+\frac {n(n-1)(n-2)(n-3)}{4!}x^2+\\
\dots+\frac {n(n-1)\dots(n-r+1)}{r!}x^r\text.$$
I know that this formula is valid for any real number n, provided that $|x|<1$. The formula extends infinitely for a... | The series you have written is valid for any real value of $x$ if $n$ is zero or a natural number. In this case your expansion is an identity and you get a polynomial of degree $n+1$ in $x$.
But if $n$ is negative integer or non integral, your expansion is valid only for |x|<1 and the series is infinite and convergent.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $\int \frac{{\rm d}x}{1+e\cos x}$
Find $$\int \frac{{\rm d}x}{1+e\cos x}$$
*
*When $e$ lies between $0$ and $1$
*When $e$ is greater than $1$
I got $\int \frac{dx}{1+e\cos x}$ when $e$ lies between $0$ and $1$, by substituting $\tan\frac{x}{2} =t$.
$$\int \frac{dx}{1+e\cos x} = \frac{2}{\sqrt{1-e^2}} \tan^{-... | Substituting $t=\tan(\frac{x}{2})$ we obtain for $e\neq1$ (and the case $e=1$ gives $\int \frac{dx}{1+\cos x}=\tan(\frac{x}{2})+c_1$)
$$\int \frac{dx}{1+e\cos x}=\int \frac{2}{1+t^2+e(1-t^2)}dt$$
$$=\int\frac{2}{t^2(1-e)+(1+e)}dt=\int\frac{2}{(1-e)\left(t^2+\frac{1+e}{1-e}\right)}dt=\frac{2}{1-e}\int\frac{1}{t^2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Foci of Ellipse lies on Hyperbola and vice-versa Let the foci of the hyperbola $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$ , (A,B > 0) be vertices of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , (a, b > 0) and foci of ellipse be vertices of hyperbola. Let eccentricities of the ellipse... | In the given notation the focus of ellipse is $S_1(ae_1,0)$ and vertex is $V_1(a,0)$.
For the hyperbola $S_2=(Ae_2,0)$ and $V_2=(A,0)$. Given that $S_1=V_2$ and S_2=V_1$.
we get $e_1=A/a$ and $e_2=a/A$ \implies $e_1 e_2=1$, so by AM-GM, we get
$$\frac{e_1+e_2}{2}\ge \sqrt{e_1 e_2} \implies [e_1+e_2]\ge 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Given a prime number $p$ and integer numbers $x,y$. Prove that if a number $x(p-x)y(p-y)$ is a square of an integer then $x=y$.
Given a prime number $p>2$ and numbers $x,y \in \lbrace 1,2,...,\frac{p-1}{2}\rbrace$. Prove that if a number $x(p-x)y(p-y)$ is a square of an integer then $x=y$.
Firstly I noticed that $x$ ... | WLOG, Let $x,y \leq \frac{p-1}{2}$. Clearly, $p-x,p-y<p$. So, $k^2,l^2<\frac{p-1}{2}\times p<\frac{p^2}{2}$. This implies that $k,l<\frac{p}{\sqrt{2}}$ or equivalently, $k+l<\sqrt{2}p$ and $k-l<\frac{p}{\sqrt{2}}$.
Substracting the two equations, we get that $(x-y)p=k^2-l^2$. From the inequality, we get that $p=k+l$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find $\int x^2\arcsin(2x)dx$ Find $\int x^2\arcsin(2x)dx$
My work.
$\frac{1}{3}\int \arcsin(2x)dx^3=\frac{1}{3}(x^3\arcsin(2x)dx-\int x^3d(\arcsin(2x))$
This yields to finding $\int \frac{2x^3}{\sqrt{(1-4x^2)}}dx$ with which I have problem finding.
$Edit$
$x=\frac{1}{2}sin\theta$
$\frac{2}{8}\int\frac{sin^3\theta cos\t... | I'd like to show you my way to calculate $\int x^2 \arcsin 2x$.
First, let $t=arcsin 2x$. We get $$\int x^2 \arcsin 2x=\int \frac18 \sin^2 t \cdot t \cdot \cos t dt$$.
Use partical integration twice, we get the following equation
$$\int \frac18 \sin^2 t \cdot t \cdot t\cos t dt=\frac{1}{24}t\sin^3t-\int\frac{1}{24}sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving $\left\lfloor \sqrt{x^2+3x+3}\right\rfloor = x+1$
Let $$f(x)=\sqrt{x^2+3x+3}$$
Claim: $$\lfloor f(x)\rfloor = x+1$$
I came across this expression in a question and made this claim based on observations
$$⌊f(1)⌋=\lfloor\sqrt{7}\rfloor=2$$
$$⌊f(2)⌋=\lfloor\sqrt{13}\rfloor=3$$
$$⌊f(3)⌋=\lfloor\sqrt{21}\rfloor=4$... | The statement $[\sqrt{x^2 +3x + 3}] = x+1$ is true if and only if
*
*$x+1$ is an integer and
*$x+1 \le\sqrt{ x^2 + 3x + 3}^2 < x+2$.
The first is true if and only if $x$ is an integer. SO this not true if $x \not \in \mathbb Z$. But if $x \in\mathbb Z$ this might be true.
As for 2)
$(x+1)^2 = x^2 +2x + 1$ and $(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | Alternative
Make sure you only multiply with positive number because otherwise the inequality sign will reverse
$$
\begin{align}
\left|\frac{7}{x+4}\right|\cdot\left|\frac{7}{x+4}\right|&<1\cdot\left|\frac{7}{x+4}\right|\\
\\
&<1\cdot 1\\
\\
\\
7^{2}&<(x+4)^{2}
\end{align}
$$
From here we get the solutions:
$$
x<-11\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Question about calculating a series involving zeta functions On this page it had shown that the sum of $\frac{1}{n^3(n+1)^3}=10-\pi^2$. I'm curious about, what is the value of $$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
For some positive integer $k$.
According to partial fraction expansion, we can show that $$\frac1{n^3(n+... | Using partial summation and asymptotics, we have $$a_k=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}+\frac {\pi^2}{k^4}$$
$$a_k=\frac{1}{2k ^5}\Big[k^2 (\psi ^{(2)}(k+1)+2 \zeta (3))+\pi ^2 k-6 k \psi ^{(1)}(k+1)+12 \psi^{(0)}(k+1)+12 \gamma \Big]$$ which generate the sequence
$$\left\{10,\frac{21}{32},\frac{809}{5832},\frac{261... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Find a bijection between $[1,2)$ and $(1,2)$ I want to find a bijection between $[1,2)$ and $(1,2)$ and prove it.
My attempt:
$[1,2) = \{x \in \mathbb R | 1 \leq x <2\}$
$(1,2) = \{x \in \mathbb R | 1 < x < 2\}$
$f(x) = x$ if $x \ne 1\frac{1}{n}$ for $n = 1,2,3,...$
and
$f(x) = 1\frac{1}{x+1}$, if $x =1\frac{1}{n}$ fo... | In your question, you noted that card((1,2)) <= card([1,2))
Let $f: [1,2)\to(1,2)$
Fix $x \in (1,2)$. Let $\{A_i\}_{1}^{\infty}$ represent the non-terminating decimal expansion right of the decimal place (this is unique to each number).
Let $f(x) := 1 + \sum_{i=1}^{\infty}A_i*10^{-i-1}$. All values of f are of the form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Sequence $(u_n)$: $\left\{\begin{array} .u_1=1\\ u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots \end{array}\right.$ The sequence $(u_n)$ is defined by the formula:
$$\left\{\begin{array}
.u_1=1\\
u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots
\end{array}\right.$$
Find $\lim\limits_{n\to \infty}\sqrt[n]{u_n}$.
T... | If you don't insist on using a monotonicity argument, you can also derive an explicit form for $u_n$ and then calculate the limit:
Setting $v_n := \frac{1}{u_n}$ for all $n$, we get
\begin{alignedat}{2}
&&v_{n+1} &= 2^n + 3v_n\\
&\implies
&\frac{v_{n+1}}{3^{n+1}}-\frac{v_{n}}{3^{n}} &= \frac{2^n}{3^{n+1}}\\
&\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show the following inequality $_2F_1\left(5.5, 1, 5;-x^2\right)>0$? Consider the function $_2F_1\left(5.5, 1, 5;-|x|^2\right)$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the funct... | $$f(x)=\, _2F_1\left(\frac{11}{2},1;5;-x^2\right)=\, _2F_1\left(1,\frac{11}{2};5;-x^2\right)$$
As @Aaron Hendrickson already commented,
$$g(x)=\frac{315}{8} x^8 \left(x^2+1\right)^{3/2}\,f(x)=x^2 \sqrt{x^2+1} \left(35 x^6+5 x^4-6 x^2+8\right)-16 \left(\sqrt{x^2+1}-1\right)$$
$$g'(x)=315 x^7 \sqrt{x^2+1}$$ So, $g'(x)=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove or disprove that $\left(f'(x)+g'(x)\right)\geq 2$ It's a result found with geogebra .
Let us define :
$$f(x)=\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$$
And :
$$g(x)=\operatorname{W}\left(1-\frac{1}{x-1}\right)$$
Then it seems we have $2\leq x\leq \frac{215}{100}$ :
$$\left(f'(x)+g'(x)\right)\geq 2$$
My attemp... | Sketch of a proof:
Let
\begin{align*}
F(x) &= f(x) + \frac{27x^3 - 182x^2 + 394x - 316}{10} - 2x,\\
G(x) &= g(x) - \frac{27x^3 - 182x^2 + 394x - 316}{10}.
\end{align*}
We have
$f'(x) + g'(x) - 2 = F'(x) + G'(x)$.
It is not difficult to prove that $G'(x) \ge 0$
for all $x \in [2, 215/100]$.
It suffices to prove that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Suppose that $ab = cd \neq 0$. Then we get $\frac{a}{c} = \frac{d}{b} = k$.
From the above, we get $a = kc$ and $d = bk$. Substituting in $a+b = c+d$, we get $kc + b = c + bk$. Thus $(k-1)c - (k-1)b = (k-1)(c-b) = 0$. Thus, we must have that $b = c$ or $k = 1$. In the first case, we would then get $a=d, b=c$. In the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Minimum of $(xy+a)^2+(bx-cy)^2$ Let $a,b,c>0$. Consider the function
$$f:\mathbb R^2 \to \mathbb R,\quad f(x,y)=(xy+a)^2+(bx-cy)^2.$$
I'm interested in the globel minimum of this function. I could show that for $a \le bc$ the minimum is attained at $(x,y)=(0,0)$ and hence
$$f(0,0)=a^2.$$
For $a>2bc$ however it seems th... | From completing the square we get
$$
f(x, y) = (y^2+b^2)x^2 -2y(bc-a) x + (a^2+c^2y^2) \\
= (y^2+b^2) \left[ \left( x - \frac{y(bc-a)}{y^2+b^2}\right)^2 + \frac{a^2+c^2y^2}{y^2+b^2} - \frac{y^2(bc-a)^2}{(y^2+b^2)^2}\right]
$$
and therefore
$$
f(x, y) \ge a^2+c^2y^2 - \frac{y^2(bc-a)^2}{y^2+b^2}
= \frac{(c y^2 +ab)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the equation to tangent line to a function given certain info - Multiple answers?
The problem states: find the equation of the tangent line to the graph of continuous function $y = f(x)$ at $x=3$ given the following information.
$\lim_{x \to 3} {\dfrac{f(x)+7}{2x-6}}=8$ and $f(3) = -7$
How you are su... | As they have given the : $f(x=3) = -7$
And they have also given the value of $$\lim_{x\to3}\frac{f(x)+7}{2x-6} = 8$$
Now, this must be the undermined form($\frac{0}{0}$) $\implies$ you can use L'Hôpital's rule
Now,
$$\begin{align*} \frac{f'(x =3)}{2} =
8 \implies f'(x=3) = 8×2 = 16
\end{align*}$$
Which is the slope ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What's the measure of the segment $MN$ in the question below? For reference:(exact copy of the question) In the triangle $ABC$, $H$ is the orthocenter, $M$ and $N$ are midpoints of $AC$ and $BH$ respectively.
Calculate $MN$, if $AH=14$ and $BC=48$ (answer: $25$)
My progress..my drawing according to the statement and th... | $\triangle AFH \sim \triangle BFC$
$FC = \frac{48}{14} HF, BF = \frac{48}{14} AF$
$2 FN = BF+HF = \frac{24}{7} AF + HF \tag1$
$\triangle ABF \sim \triangle HCF$
$\frac{CF}{HF} = \frac{BF}{AF} \implies CF = \frac{48}{14} HF$
$2 FM = CF - AF = \frac{24}{7} HF - AF \tag2$
Squaring $1$ and $2$ and adding,
$4 (FN^2 + FM^2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the sum of the series $3x+8x^2+15x^3 + ....$ I'm trying to express the following series $3x+8x^2+15x^3 + ....$ as a sum and hope to find its sum for $|x| < 1$
Here is what I have so far:
To me the series looks to be the derivative of the following form:
$1 + \frac{3}{2}x^2+\frac{8}{3}x^3+\frac{15}{4}x^4 ...$
Given... | Hint $1$: First notice that the coefficients are $n^2-1$.
Hint $2$:
\begin{eqnarray*}
\sum_{n=1}^{\infty} n^2 x^n =\frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\int_0^\infty \frac{dx}{(x^2 + 1)^2} $ with the residue theorem I am solving this integral and am stuck on proving an inequality, I believe I have the rest worked out.
My work:
First observe that the integrand is even, hence $$\int_0^\infty \frac{dx}{(x^2+1)^2} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{(x... | The idea of your proof is correct, but there are some inaccuracies. First, it should be
$$
\frac{\pi}{2} = \int_R^R \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}
$$
for $R > 1$. You cannot replace the $R$ by $\infty$ in the first integral because the second integral depends on $R$. I would denote the half-circle w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If... | It is clear that $z=0$ is a solution. Hence we assume $z \ne 0$.
From $|z+1|^2=1$ we derive $ |z|^2= -(z+ \bar z) \quad (1)$.
From $|z^2+1|^2=1$ we derive $|z|^4= -(z^2+ \bar z^2) \quad (2)$.
Then $(1)$ gives: $|z|^4= z^2+ \bar z^2+2z \bar z=z ^2+ \bar z^2+2|z|^2$.
With $(2)$ we derive $|z|^4=-|z|^4+2|z|^2.$ Hence $|z|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Struggling with polynomial Long Division in $GF(256)$ I'm currently trying to build a QR Code generator, which has lead me to studying a bit of abstract algebra - more specifically, finite fields.
I've studied quite a bit on my own and I'm at the point where I'm trying to implement an algorithm for polynomial long divi... | I made a crash course for Pari and used an online version and got
? c=ffgen(c^8+c^4+c^3+c^2+Mod(1,2))
% = c
? divrem((c^2+1)*x^4+(c^5)*x^3+(c^4+c^3+c+1)*x^2+(c^2+c+1)*x+(c+1),(x^2+(c^2+1)*x+1))
% = [(c^2 + 1)*x^2 + (c^5 + c^4 + 1)*x + (c^7 + c^6 + c^5 + c^3 + c + 1),
(c^6 + c^4 + c^2 + c)*x + (c^7 + c^6 + c^5 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y =... | We want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x $
So, we want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2x}}}}=x \quad \quad (E)$
*
*We see that if $\sqrt{x+2x} =x$, then, $x$ is a solution of $(E)$.
Therefore, $0$ and $3$ are two solutions of $(E)$.
*Suppose that $x$ is a solution of $(E)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be writte... | $$\frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(1+\frac{x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(\frac{\sqrt{1+x²}+x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\frac{\left(x+\sqrt{1+x²}\right)^\frac{3}{2}}{\sqrt{1+x²}} \\ = \frac{3}{2}\frac{y}{\sqrt{1+x²}} \\ = \frac{3y}{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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The smallest value of the expression $4x^2y^2+x^2+y^2-2xy+x+y+1$ What is the smallest value that $4x^2y^2+x^2+y^2-2xy+x+y+1$ can take with real numbers $x$ and $y$?
I suspect the following transformation can be done: $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$.
| $(2xy−\frac12)^2+(x+\frac12)^2+(y+\frac12)^2+\frac14 \geq \frac14$.
Because $(2xy−\frac12)^2\geq 0,(x+\frac12)^2 \geq 0,(y+\frac12)^2 \geq 0$.
So this minimum is attained in the original expression when
$(2xy−\frac12)^2=0,(x+\frac12)^2=0,(y+\frac12)^2=0 \iff x=y=-\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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An uncommon continued fraction of $\frac{\pi}{2}$ I'm currently stuck with the following infinite continued fraction:
$$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$
There is an obscure clue on this: as one can derive the familiar Lord Brouncker’s fraction below
$$
\fra... | The result comes from a rather slight modification of Euler's continued fraction ,
$$a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+....=\cfrac{a_0}{1-\cfrac{a_1}{1+a_1-\cfrac{a_2}{1+a_2-\cfrac{a_3}{\cdots}}}}\label{1}\tag{1}$$
Note that for a product $a_0a_1a_2\cdots$; we can represent them as,
$$\begin{align}a_{1}a_{2}a_{3}...&=a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Decomposition of matrix occuring in problem of finding $n+1$ vectors in $\mathbb{R}^n$ with pairwise equal inner product I was toying with my intuition that there are always $n+1$ unit vectors in $\mathbb{R}^n$ such that every pair $v_i$ and $v_j$ ($i\neq j$) has the same angle between them. As those vectors are normal... | Adding to the existing answer, you can also extract your vectors by orthogonally diagonalizing $M$. The matrix $M$ has rank $n$ with two eigenvalues: The eigenvalue $\lambda := \frac{n+1}{n}$ of multiplicity $n$ and the eigenvalue $0$ of multiplicity one. Let $U$ be an orthogonal matrix such that
$$ U \begin{pmatrix} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Small clarification Spivak Chapter 1, 18(b)
Suppose that ${b}^{2} - 4c < 0$. Show that there are no numbers $x$ satisfying ${x}^{2} + bx + c = 0$; in fact, ${x}^{2} + bx + c > 0$ for all $x$. Hint; Complete the square.
There is one answer here.
Solution (per Spivak)
We have
${x}^{2} + bx + c = \left (x + \frac{b}{2}... | Since the assumption (given condition) was,
$$b^2-4c<0\implies c-\frac{b^2}{4}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y... | Let $r = [x, y]^T, F_1= [1, 2]^T , F_2 = [3, 4]^T $
Define a reference coordinate frame $O'uv$ with origin at $O' = \dfrac{1}{2}(F_1 + F_2) = [2,3]^T $, and let the $u$ axis extend along the vector $F_1 F_2$ while the $v$ vector is perpendicular to it and $+90^\circ$ away from it. If $r' =[u, v]^T $, then
$ r = O' + A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | $$\begin{align}
4x^2-2xy-4x+3y-3
&=(4x^2-4x-3)-(2x-3)y\\
&=(2x-3)(2x+1)-(2x-3)y\\
&=(2x-3)(2x+1-y)
\end{align}$$
If the quadratic $4x^2-4x-3$ hadn't had $2x-3$ as a factor, the polynomial in $x$ and $y$ would not have factored.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
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When is the sum of the products of any three distinct numbers less than $n$, divisible by $n$? Define $A$ as the set of all integer triples $(a,b,c)$, such that $0<a<b<c<n$.
$$A=\left\{(a,b,c)\in \mathbb{Z^3}:1\le a<b<c\le n-1 \right\}$$
Define $S$ as the sum of the product $abc$ for each triplet in $A$.
$$S=\displayst... | Suppose $n$ is even, then $(a,b,c)\in A\implies (n-c,n-b,n-a)\in A~~\forall (a,b,c)\ne (r,\frac n2,n-r)$,
here $(a,b,c)$ and $(n-c,n-b,n-a)$vare non identical.
So to calculate $S$ you only need to consider the cases where $(a,b,c)$ is of the form $(r,\frac n2,n-r)$
$$\begin{align*}S&\equiv \sum_{(a,b,c)=(r,\frac n2,n-r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6... | First, evaluate
\begin{align}\int_{0}^{\infty} \frac{1}{x^{6}+1}dx\stackrel{x\to\frac{1}{x}}{=}&\ \frac12 \int_{0}^{\infty} \frac{x^{4}+1}{x^{6}+1} d x\\
=&\ \frac12 \bigg(\int_{0}^{\infty} \frac{1}{x^{2}+1} d x
+ \int_{0}^{\infty} \frac{x^2}{x^{6}+1}\overset{x^3\to x}{ d x}\bigg)\\
=&\ \frac12\cdot \frac43\int_{0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
How to show $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$? I am trying to show that the function $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$, where $0.5 \leq L < H < 1$. Visual inspection in Mathematica suggests the conjecture is true. I tried showing this by differentiation, but am having trou... | You can write
$$
f(x) = \frac{(1-H)^x - (1-L)^x}{H^x - L^x} = \left(\frac{1-L}{H}\right)^x \cdot
\frac{1 - \left(\frac{1-H}{1-L}\right)^x}{1 - \left(\frac{L}{H}\right)^x}=
u(x)\cdot v(x).
$$
Then use the inequalities to obtain, $$(1-L)/H < 1 \quad\text{and}\quad
0 < \frac{1-H}{1-L}<\frac{L}{H}<1.$$
These inequalities... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$ I was trying to solve the following question:
\begin{align*}
\sinh x &= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\
\cosh x &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
\end{align*}
Usi... | We have
$$
\begin{align}
\cosh^2(x)&=\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)
\\&= \sum_{k=0}^\infty \sum_{l=0}^k \frac{x^{2l}x^{2(k-l)}}{(2l)!(2(k-l))!}
\\&= \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} \color{red}{\sum_{l=0}^k \binom{2k}{2l}}
\\&= 1 + \sum_{k=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
What is the series $ \dfrac{1}{\left( x-{y}^{2} \right) \left(1- yx \right) \left( {x}^{2}-y \right)} $ expansion? Consider the series expansion
$$
\frac{1}{\left( x-{y}^{2} \right) \left(1- yx \right) \left( {x}^{2}-y \right)}=P_0(y)+P_1(y)x+P_2(y)x^2+\cdots+P_n(y) x^n+\cdots.
$$
For small $n$ we have
\begin{gather*... | Your expression is a product of three geometric series (first two multiplied by $-1$ which does not affect the final product):
$$
\frac{1}{y-x^2}=\frac{1}{y}\frac{1}{1-\frac{x^2}{y}}=\frac{1}{y}+\frac{1}{y^2}x^2+\dots+\frac{1}{y^{n+1}}x^{2n}+\dots\\
\frac{1}{y^2-x}=\frac{1}{y^2}\frac{1}{1-\frac{x}{y^2}}=\frac{1}{y^2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$ $\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$
Normally the general factor is $a(n)=\cos\left(\fr... | $$S = \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right]$$
Consider vectors in complex plane:
*
*Analytical
$$\begin{align*}S
&= \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right] = \sum_{k =1}^ne^{i\frac {2k\pi}{n}}\\
& =\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of identity related to sum of binomial coefficients and powers of $2$ Given two positive integers $a,b$, prove that $\left(\sum_{n=a}^{a+b-1}\binom{n-1}{a-1}2^{-n}\right)+\left(\sum_{n=b}^{a+b-1}\binom{n-1}{b-1}2^{-n}\right)=1$
The context comes from this MSE problem, where I indirectly showed that this is true w... | We prove that $$\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n} = 2^{-a-b+1} \sum_{n=a}^{a+b-1} \binom{a+b-1}{n}$$ for all positive integers $a$ and $b$ by inducting on $b$.
When $b = 1$, the LHS and RHS evaluate to $2^{-a}$, so the statement holds in this case.
We proceed with the induction step. Using the induction hypothe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What's the measure of the segment $BC$ in the rhombus below? For reference: Given a rhombus $ABCD$, on $BC$
mark the point $P$ such that : $BP= 3PC$ and $AP^2+ 3DP^2 = 38$.
Calculate $BC$.(answer: $2\sqrt2$)
My progress:
$BP = 3CP\\
AP^2+3DP^2 = 38\\
AB=BC=CD=AD$
Th. Stewart:
$\triangle ABC:\\
AC^2.BP+AB^2.CP=AP^2BC+B... | $ \small AB^2 \cdot CP + AC^2 \cdot 3 CP = 4CP \cdot(AP^2 + 3 CP^2)$
As $ \displaystyle \small AB = BC \text { and } BC = 4 CP$,
$ \displaystyle \small BC^2 + 3 AC^2 = 4 (AP^2 + \frac {3 BC^2}{16})$
$ \displaystyle \small BC^2 + 12 AC^2 = 16AP^2 \tag1$
Similarly,
$ \displaystyle \small BD^2 \cdot CP + CD^2 \cdot 3CP = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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A remarkable Continued Fraction for $\pi$ How to analyze the Continued Fraction $3+\cfrac{\color{red}1^2}{5+\cfrac{\color{blue}4^2}{7+\cfrac{\color{red}3^2}{9+\cfrac{\color{blue}6^2}{11+\cfrac{\color{red}5^2}{13+\cfrac{\color{blue}8^2}{15+\cfrac{\color{red}7^2}{17+\ddots}}}}}}}$
Numerically, it seems to evaluate to $\p... | The difference between the even and odd squared coefficients is similar to that of the continued fraction of the ratio of two hypergeometric functions (DLMF):
\begin{equation}
\frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a,b+1;c+1;z\right)}=t_{0
}-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we solve inequality without expanding both sides if each side has perfect square? The problem
$$\begin{align} (x+5)^2 &< (x-3)^2\tag{1} \end{align}$$
Normal people (most people) will solve the problem like this:
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
x^2 + 10x + 25 &< x^2-6x +9\\
10x+6x + 25-9 &< 0\\
16x &< -16\\
x... | I see no big mistake and you were told wrong. The process is totally valid and who made that comment probably didn't bother to look at it (you just have a minor computational mistake).
The inequality $(x+5)^2<(x+3)^2$ is equivalent to
$$
(x+5)^2-(x-3)^2<0
$$
and the left-hand side can be factored to give
$$
(x+5-x+3)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is it possible to find $\tan\theta$ by squaring both sides of $\sec\theta+\tan\theta=5$ right away? From $\sec\theta+\tan\theta=5$, in order to find $\tan\theta $ we can rewrite it as $\sec\theta=5-\tan\theta$ then square,
$$1+\tan^2\theta=\tan^2\theta-10\tan\theta+25\quad\Rightarrow\quad \tan\theta=\frac{24}{10}$$
But... | The answer is no, not really.
In effect if $z=\tan \theta$ and $-\frac \pi 2 \lt \theta \lt \frac \pi 2$ then $\sec\theta+\tan\theta=5$ is equivalent to $\sqrt{1+z^2} +z=5$
If you just square both sides you then get $z^2 +z\sqrt{1+z^2}=12$ which does not take you forward; you still have the radical
while if you square ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$ Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$.
Prove $$xyz\geq27,\\xy+yz+zx\geq27,\\x+y+z\geq9.$$
Here's what I tried:
$$\begin{align} \frac{x^3+y^3+z^3}{3}&\ge\sqrt[3]{x^3y^3z... | By the AM-GM inequality, $$x^2 + y^2 + z^2 \ge 3 (x^2y^2z^2)^{1/3}$$
but $x^2 + y^2 + z^2 = xyz$, so $$xyz\ge 3(xyz)^{2/3} \implies (xyz)^{1/3} \ge 3 \implies xyz\ge 27$$
By two more applications of the AM-GM inequality and $xyz\ge 27$, you can show that $$x+y+z
\ge 3(xyz)^{1/3} \ge 9$$
and $$xy + yz + zx \ge 3(x^2y^2z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x... | I'd recommend this method. You already got $(x+1)(x+2)-2x\sqrt{3x+2}=0$. And by this, $x\ge0$ is induced right away, because the left hand is $>0$.
Then...
$$
(x+1)(x+2)=2x\sqrt{3x+2}\\
(x+1)^2(x+2)^2=4x^2(3x+2)\\
x^4-6x^3+5x^2+12x+4=0
$$
This is sure a quatric, so seems hard to solve. But...
Let $x^2-3x=y$, then thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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What's the rate of the stream's current? At his usual rate a man rows 15 miles downstream in five hours less than it takes him to return. If he doubles his usual rate, the time dowstream is only hour less than the time uqstream. What's the rate of the stream's current?
Let the man's rate in still water be $r$ and the r... | Or eliminate $r^2$ by combining $5(4)-4(3)$:
$$ 5(4r^2-c^2)-4(5r^2-5c^2)=5\cdot 30c-4\cdot 30c$$
which simplifies to
$$ 15c^2=30c,$$
$$ 15c(c-2)=0,$$
a quadratic in $c$ with solutions
$$ c=2\quad \text{or}\quad c=0.$$
As $c=0$ makes no sense with the original equations, we must have $c=2$.
(You divided by $c$ somewhere... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)... | $f(x,y)=x^2-2xy+6y^2-12x+2y+41$
$\frac{\partial f}{dx}=2x-2y-12$
$\frac{\partial f}{dy}=-2x+12y+2$
Stationary points are found when $\frac{\partial f}{dx}=0$ and $\frac{\partial f}{dy}=0$
$2x-2y-12=0$
$-2x+12y+2=0$
Adding the equations gives $10y-10=0\Rightarrow y=1$
Substitute to get $2x-2-12=0\Rightarrow 2x=14\Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Right... | (Not a completely rigorous proof, but too long for a comment.)
Let $\,x=\tan(a)\,$ and similar, then $\,\frac{x^2+1}{x}=\frac{2}{\sin(2a)}\,$, so the equations can be written as:
$$
\frac{3}{\sin(2a)}=\frac{4}{\sin(2b)}=\frac{5}{\sin(2c)} \tag{*}
$$
Also, $\,xy+yz+zx=1 \implies a+b+c=(2k+1)\frac{\pi}{2}\,$, by proving ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proving the derivative of a function of another function at a point Is the following equality true?
$$ \left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{x - x_0} $$
If it is true, how can I prove this?
It seems to work but I am unable to prove it. Take for example $ f(x) ... | To do a formal proof, we can just use the chain rule.
\begin{align}
\left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
&= \left[f'(x) \frac d{dx}(ax+b) \right]_{x=ax_0+b} \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{a(x-x_0)} a \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{(x-x_0)}
\end{align}
The first l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$ Let $\theta=2\pi/7$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$
I have found that $\cos(4\theta) = \cos(8\pi/7) = \cos(6\pi/7) = \cos(\theta)$ and,
similarly, $\... | If you can use the property that $\cos(x) = Re(e^{i x})$, where $Re(z)$ is the real part of $z$, then:
\begin{align*}
\sum_{k=0}^6 \cos(k\theta)
&= Re\left(\sum_{k=0}^6 e^{ik\theta}\right) \\
&= Re\left(\frac{1-e^{i7\theta}}{1-e^{i\theta}}\right) \\
&= Re\left(\frac{1-e^{i2\pi}}{1-e^{i\frac{2\pi}{7}}}\right) \\
&= 0
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\a... | Perhaps similar to epi163sqrt's answer in that this uses Vandermonde's Identity. However, this avoids the use of a derivative.
Using the power series
$$
\log\left(\frac1{1-z}\right)=\sum_{k=1}^\infty\frac{z^k}{k}\tag1
$$
and the Generalized Binomial Theorem with negative binomial coefficents
$$
\begin{align}
(1-z)^{-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Taking a limit inside an integral I would like to see a rigorous proof that:
$$\lim_{R\rightarrow \infty}\int_1^R \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x^2/R^2}}=\int_1^\infty \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$
Note that the second integral does converge (to $\log 2$). This ... | I finally was able to prove it, so I am posting the answer to my question. It was a case when doing it from scratch (the definition of limit) was easier (for me) than trying to find the right limit theorem that would apply. It is still possible it would follow right away from some limit theorem I don't know...
In my pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving an inflection point of a certain function
Show that the inflection points of the curve $y = \sin(x)/x$ lie on the curve $y^2(x^4+4)=4$
So, I know that to find the inflection points I have to set the second derivative of $f$ = $0$
I got $f''(x) = -((x^2-2)\sin(x)+2x\cos(x))/x^3 = 0$
Thus, $f''(x) = 2x\cos(x) ... | You know $2x\cos x = (x^2-2)\sin x$ and $y=\sin(x)/x$. Taking the square of the first implies $4x^2\cos(x)^2 = (x^2-2)^2\sin(x)^2$ and thus
$$ 4x^2(1-\sin(x)^2) = (x^2-2)^2\sin(x)^2$$
and thus
$$ \sin(x)^2 = \frac{4x^2}{(x^2-2)^2+4x^2} = \frac{4x^2}{x^4+4}$$
So $y^2(x^4+4) = \frac{\sin(x)^2}{x^2} (x^4+4) = \frac{4x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Check if $\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$ converges This is the series:
$$\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$$
We can drop the absolute value, because;
$$\frac{\ln(n+1)}{\ln(n^2+1)} > \frac{1}{2}$$
I've tried using inequalities below to form ... | One has
\begin{align*}
\frac{\ln(n+1)}{\ln(n^2+1)} -\frac{1}{2} &= \frac{\ln(n)+\ln\left(1+\frac{1}{n}\right)}{\ln(n^2)+\ln\left(1+\frac{1}{n^2}\right)} -\frac{1}{2} \\
&= \frac{\ln(n)+ \frac{1}{n}+ o\left( \frac{1}{n}\right)}{2\ln(n)+ o\left( \frac{1}{n}\right)} -\frac{1}{2} \\
&= \frac{ \frac{1}{n} + o\left( \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How many ways can $720$ be decomposed into a product of two positive integers?
How many ways can $720$ be decomposed into a product of two positive integers?
My solution: There are $5 \cdot 3 \cdot 2 = 30$ ways to choose the exponents a, b, c, such that $2^a \cdot 3^b \cdot 5^c= 720$. Soon there are $30$ dividers. As... | Since
\begin{align*}
720 & = 2 \cdot 360\\
& = 2 \cdot 2 \cdot 180\\
& = 2 \cdot 2 \cdot 2 \cdot 90\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 45\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 15\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5\\
& = 2^4 \cdot 3^2 \cdot 5
\end{align*}
each fact... | {
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The monotony of an implicit sequence Context of the exercice:
Let us define $(f_n)_n$ for any $n\geq 3$, $\left\{\begin{matrix}
f_n:]1,2]&\rightarrow & \mathbb{R} \\
x &\mapsto & f_n(x):=x^n-x-n
\end{matrix}\right.$
It's easy to verify that $f_n$ is a bijection between $]1,2]$ and $]-n,f_n(2)]$ (since $f_n$ has a s... | First of all, note that $u_n>1+\frac{1}{n}$ since
$$
{f_n(1+\frac{1}{n})=(1+\frac{1}{n})^n-1-n-\frac{1}{n}<e-1-n<e-4<0
\\\text{and $f_n(x)$ is increasing}
}
$$
To prove why $u_n^2+(n-1)u_n-(n+1)$ is always positive for $n\ge 3$ and $u_n>1+\frac{1}{n}$, note that the roots of $x^2+(n-1)x-(n+1)$ are
$$
{r_1=\frac{1-n-\sq... | {
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Find the roots of the polynomial $x^4+2x^3-x-1=0$. Problem
Find all $4$ roots of the polynomial: $$f(x)=x^4+2x^3-x-1.$$
My Attempt
Observe,
$$f(-2)=1,\; f(-1)=-1,\; f(0)=-1,\; f(1)=1.$$
Therefore, $f(x)$ has a root between $-2$ and $-1$, another root between $0$ and $1$.
Since $f(x)$ does not contain a second degree t... | The "basic" tools for dealing with a polynomial with integer coefficients tell us that the zeroes will not be simple to locate. The Rational Zeroes Theorem only suggests the candidates $ \ \pm 1 \ \ , $ which you have already shown are not correct. The Rule of Signs indicates one positive real and three or one negati... | {
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When is the projection of an ellipsoid a circle? Consider an ellipsoid in the three dimensional Euclidean space, say
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
where $a$, $b$, $c$ are positive reals.
I'm counting the number of planes through the origin so that the image is a perfect circle. There may be ... | Considering an enveloping cylinder, namely
$$
\left(
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1
\right)
\left(
\frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2}
\right)=
\left(
\frac{l x}{a^2}+\frac{m y}{b^2}+\frac{nz}{c^2}
\right)^2$$
The eigenvalues $\lambda$ are given by
$$0=\det
\begin{pmatrix}
\frac{... | {
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Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
| i don't think you made a mistake. continuing from where you stopped,
if $(y+1)(x+1)>0$ , then $y+1=\frac{c_1}{x+1}$.
otherwise if $(y+1)(x+1)<0$ , then $y+1=-\frac{c_1}{x+1}$.
| {
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Matrix written in terms of Kronecker delta What would be the matrix form of $P$ defined in equation (1.11) of this published work, where (I have chosen $n=2$, for simplicity)
$$P_{kl}=\begin{cases} \delta_{k,2l-1} ~~~ k \le 2\\\\ \delta_{2+k,2l} ~~~ l\le 2 \end{cases}$$
| From the context of the paper (namely the text below Equation (1.2) that defines $\mathbf R$ and the text below Equation (1.4) defining $\mathbf S$), it seems that the permutation matrix $\mathbf P$ being discussed for which $\mathbf P \mathbf R = \mathbf S$ is the commutation matrix $\mathbf P = \mathbf K^{(2,n)}$.
So... | {
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Computing a closed formula for a recurrent sequence using eigen -values and -vectors How would you use eigenvalues and eigenvectors to compute a closed formula for the following sequence: $$\{x_0=1, x_1=2, x_n=5x_{n-1} + 14x_{n-2}, n \ge 0 \}$$
I have come up with the following formula: $$ \begin{bmatrix}
0 & 1 \\
14 &... | Do you really need to use eigenvalues/eigenvectors? This is a linear difference equation for which you can easily identify a basis for the solution space. Looking at the roots of the characteristic polynomial $p(\lambda) = \lambda^2 -5 \lambda -14$, any solution is of the form
$$
x_n = c_1 \cdot (-2)^n + c_2 \cdot 7^n
... | {
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Determinant formula for coordinates of circumcenter and orthocenter of a triangle I've come across the following formulas for coordinates of circumcenter $O=(x_O,y_O)$ and orthocenter $H=(x_H,y_H)$ of a triangle, in a formula book, stated without derivation. For a triangle with vertices $(x_i,y_i),$ $i \in \{1,2,3\}$,
... | Consider instead triangle $A_1B_1C_1$ whose midtriangle is precisely triangle $ABC$.
A classical result says that the orthocenter $H$ of $ABC$ is the circumcenter of $A_1B_1C_1$.
It is easy to express the coordinates of points $A_1,B_1,C_1$ as functions of the coordinates of $A,B,C$ using relationships
$$\vec{AB}+\vec{... | {
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Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$
Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ such that
$$\left\{\begin{array}{l} a_{1}+a_{2}+\ldots+a_{n}>0 \\a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}<0 \\a_{1}^{5}+a_{2}^{5}+\ldots+a_{n}^{5}>0 \end{array}\right.$$
... | n = 3 seems to be fine. Let $c$ be a parameter to be determined later.
Put $a_1 =-1-6c^3$, $a_2=-1+6c^3$, and $a_3=6c^2$.
Clearly $a_1 + a_2 + a_3 = -1 +6c^2$ which is positive for $c$ large enough.
Now, $a_1^3 + a_2^3 +a_3^3 = (-1-6c^3)^3 + (-1 + 6c^3)^3 + (6c^2)^3 = -2 < 0 $
Also, $a_1^5 + a_2^5 + a_3^3 = (-1 - 6c^3)... | {
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Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a mu... | Let $x=a-11,y=b-11$ in $xy+11x+11y=0$ to get
$$ab=121$$
Since the change of variables is completely reversible in the integers, every divisor of $121$ corresponds one-to-one to a solution: $a=\pm1,\pm11,\pm121$, correspinding to
$$(x,y)=(0,0),(-22,-22),(-10,110),(110,-10),(-12,-132),(-132,-12)$$
| {
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$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second functi... | A complex analysis approach is to integrate the function $$f(z) = \frac{\left(\log(z+1) - \log(z-1)\right)^{\color{red}{2}}}{1-az}, \quad 0< a< 1, $$ where $0 \le \arg(z+1), \arg(z-1) < 2 \pi$, around a dog bone contour that goes around branch cut on $[-1, 1]$.
(The nicest picture I can find of the contour in this ans... | {
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Find all primes $m$ and $l$ such that $2m-1, 2l-1, 2ml-1$ are perfect squares Find all primes $m$ and $l$ such that the integers $2m-1$, $2l-1$ and $2ml-1$ are perfect squares.
I have been trying to solve this problem from the 2022 Moroccan Maths Olympiad training program, but I wasn’t able to find anything.
By some fe... | I don't know of any way to solve this using congruences. However, it can be done using the uniqueness, plus the format & number of representations, of sums of squares. First, the stated conditions mean there is a positive integer $a$ such that
$$\begin{equation}\begin{aligned}
2m - 1 & = (2a + 1)^2 \\
2m - 1 & = 4a^2 +... | {
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Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$ I need to prove that the sum $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals $ \ln(\frac{4}{3})$. I tried ex... | The issue with your calculation is that the sums are not individually convergent. Therefore, you cannot write
$$\sum_{n=1}^\infty \log \frac{3n+1}{3n+4} = \sum_{n=1}^\infty \log (3n+1) - \sum_{n=1}^\infty \log(3n+4),$$ because each of those sums on the RHS are infinite.
Instead, you need to look at the limiting behavi... | {
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Laurent series of $\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$ I want to expand the function following into Laurent series:
$$\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$$
for $2 < |z| < 3$
My solution
First thing to apply is partial function decomposition:
$$\frac{1}{(z - 2)(z +3)} = \frac{\frac 1 5}{z - 2} + \frac{-\frac 1 5}{z + ... | An idea for you: using Taylor's polynomial, or whatever, write the numerator as power of $\;z-2\;$ (first...), so you get:
$$f(z)=z^2+iz-1=f(2)+f'(2)(z-2)+\frac{f''(2)(z-2)^2}{2!}$$
and now multiply termwise the above by the development around $\;z=2\;$ to get an actual (Laurent) power series:
$$\left(f(2)+f'(2)(z-2)+\... | {
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Implicit differentiation: why I cannot multiple both sides by an expression? I was solving a problem on implicit differentiation and I was wondering why my answer does not match the given answer. Apparently I multiplied both parts of an equation by an expression to get rid of some fractions before differentiating. Here... | Consider a simpler example.
$$
f(x,y) = y-x=0,\qquad\text{implies}\qquad g(x,y)=y^2-xy = 0,
$$
so from $f$ we find $y'(x)=1$; from $g$ we find $y'(x)=\frac{y}{2y-x}$, which gives the same answer if $y=x$ but is a different function of $x,y$ in general.
The equation $\frac{d}{dx}f(x,y(x))=0$ allows you to find $y'(x)$ n... | {
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Find general form of eigenvalues of a circulant matrix I have an $n \times n$ matrix in a general form:
$$ A = \begin{pmatrix}
\alpha^{n-1} & 1 & \alpha & \alpha^2 & \dots & \alpha^{n-2} \\
\alpha^{n-2} & \alpha^{n-1} & 1 & \alpha & \dots & \alpha^{n-3} \\
\alpha^{n-3} & \alpha^{n-2} & \alpha^{n-1} & 1 & \dots & ... | Notably, $A$ is a circulant matrix. As such, its eigenvalues will be of the form
$$
\lambda_j = \alpha^{n-2}\omega^{(n-1)j} + \cdots + \alpha \omega^{2j} + \omega^j +\alpha^{n-1},
$$
where $\omega = e^{2 \pi i/n}$ and $j = 0,1,\dots,n-1$. We can simplify $\lambda_j$ to
$$
\lambda_j = \alpha^{-1}((\alpha \omega^j)^{n-1}... | {
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How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful tr... | Differentiation sometimes is easier than integration!
Inverse substitution rewrites the integral
$\displaystyle \int_{0}^{\infty} \frac{x^{r}}{x^{m}+a} d x \stackrel{x\mapsto\frac{1}{\sqrt[m]{a}}}{=}\frac{1}{a^{1-\frac{r+1}{m}}} \int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x \tag*{} $
where $0<a<2.$
Using the formula prov... | {
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Rhombus, find ratio of angles $KLMN$ is a rhombus with $\measuredangle MNK=120^\circ$. On the side $MN$ point $C$ is taken such that $CN=2CM$. Find the ratio $\dfrac{\cos\measuredangle CKN}{\cos\measuredangle CLM}$.
The Cosine Rule in triangle $CNK$ gives $$CK^2=KN^2+CN^2-2KN\cdot CN\cos120^\circ=\\=a^2+\dfrac49a^2+\d... | you have multiplied rather than divided the two ratios at the last step. If you divide them you get the right answer
| {
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Why $(\sec^2(x)+1)^{\frac{3}{2}}=\sec^3(x), x>0$? I was trying to find the intermediaries steps to this equality, considering $x$ is a positive number, but I couldn't. Using the identities $\sec^2(x)=1+\tan^2(x)$ and $\sin^2(x)+\cos^2(x)=1$, here some of my work:
First try:
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\lef... | It is not true in general. Indeed, one has:
\begin{align*}
\left(\sec^{2}\left(\dfrac{\pi}{3}\right)+ 1\right)^{3/2}= \dfrac{5\sqrt{5}}{8} \neq \frac{1}{8} = \sec^{3}\left(\dfrac{\pi}{3}\right)
\end{align*}
EDIT
As @JetfiRex has mentioned, you may have switched $\tan^{2}(x)$ by $\sec^{2}(x)$.
If this is the case, the c... | {
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Given $a,b,c>0$, $a+b+c=ab+bc+ca$, prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$ Given $a,b,c>0$, $$a+b+c=ab+bc+ca$$, prove $$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$$
I tried with derivatives but haven't solved it yet.
Is there a more natural and elementary proof?
My progress with derivatives:
First $a+b... | Remark: @Calvin Lin's comment reminds me that, in 2019, I proved some similar inequalities under the condition $a + b + c = ab + bc + ca$.
WLOG, assume that $c = \min(a, b, c)$.
Using AM-GM, we have
\begin{align*}
&\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - \sqrt{abc} - 2\\
\ge\,& \sqrt{ab}
+ 2\sqrt{\sqrt{bc}\sqrt{ca}} - ... | {
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$x^2\frac{dy}{dx}+xy+1=0$ Solve the following differential equation by the form of homogeneous equation. Letting $y=vx$
The equation:
$x^2\frac{dy}{dx}+xy+1=0$
I can’t separate variables
My solution steps are:
$x^2(v+x \frac{dv}{dx})+x^2v+1=0$
$(v+x \frac{dv}{dx})+v=\frac{-1}{x^2}$
$2v+x \frac{dv}{dx}=\frac{-1}{x^2}$
H... | Divide the proposed DE by $x$ in order to obtain
\begin{align*}
x^{2}y' + xy + 1 = 0 & \Longleftrightarrow xy' + y = -\frac{1}{x}\\\\
& \Longleftrightarrow (xy)' = -\frac{1}{x}\\\\
& \Longleftrightarrow xy = -\ln|x| + k\\\\
& \Longleftrightarrow y(x) = -\frac{\ln|x|}{x} + \frac{k}{x}
\end{align*}
and we are done.
Hopef... | {
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Given $a,b,c$ are real numbers such that $a-b=8$ and $ab + c^2 +16 =0$. The numerical value of $a^{2021} + b^{2021} +c^{2021}$ is? I saw this question and was intrigued by it. I wanted to see if what I have so far made sense.
So from the second equation $ab + c^2 +16 =0$ we get $c^2 = -(16+ab)$. Which implies $16+ab\le... | Yes, you are correct. The numerical value is $0$, with $a=4,b=-4$ and $c=0$.
How I did it [before opening this post]: First note that for there to be a real solution to the equation $ab+16+c^2=0$, the product $ab$ must satisfy $ab \le -16$. The only $(a,b)$ that satisfies $ab \le -16$ with $a-b=8$ is $a=4,b=-4$, and so... | {
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To prove: $\bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2 $ How could one show that
$$ \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2
$$
I have tried the below approach:
\begin{align*}
LHS & = \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr)\\
& = 1-\sin(A-B)+\sin(A+B)-\... | $$(1+\sin A\cos B+\cos A\sin B)(1-\sin A\cos B+\cos A\sin B)$$
$$=(1+\cos A\sin B)^2-(\sin A\cos B)^2$$
$$=1+2\cos A\sin B+\cos^2A\sin^2B-(1-\cos^2A)(1-\sin^2B)$$
$$=\cdots$$
$$=(1+\cos A\sin B)^2$$
| {
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Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $... | First write $f(n)=(n+1)(n+1)^2(n+2)^3(n+4)^4$ The gcd of the $f(n)$s; $n \in \mathbb{N}$, is of the form $2^b3^a$, for some positive integers $a$ and $b$. Indeed, no prime $p \ge 7$ divides $f(1)$, and $5$ does not divide $f(5)$. Thus indeed, the common prime divisors are restricted to $2$ and $3$, and thus the gcd of... | {
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"answer_id": 2
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Show this recursively defined integer sequence $a_{n+1} = (2b_1 - a_1)a_n + (a_1b_1 - b_1^2 - 2a_1^2)a_{n-1}$ is never $0$. Let $a_1, b_1 \in \mathbb{Z}$ be arbitary integers with $a_1 \neq 0$, and define sequences $\{a_n \}_{n \geq 1}$ and $\{b_n \}_{n \geq 1}$ to be solutions to the following linear system of recursi... | Set $a_0=0$ and $b_0=1$. Then the original recurrence works for $n\ge0$. Write it in matrix form:
$$
\begin{pmatrix}
a_{n+1} \\ b_{n+1}
\end{pmatrix}
=
\begin{pmatrix}
b_1 - a_1 & a_1 \\ -2a_1 & b_1
\end{pmatrix}
\begin{pmatrix}
a_{n} \\ b_{n}
\end{pmatrix}
=
A
\begin{pmatrix}
a_{n} \\ b_{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
show $k<2^m$ if $2^{2^m}+1=\binom{n}{k}$ let $m$ be give positive integer,and $n,k$ be positive integer ,such $n\ge 2k\ge 2$,and
$$2^{2^m}+1=\binom{n}{k}$$
show that: $k<2^m$
Some of my ideas and attempts:
$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}=\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}=\left(1+\dfrac{n-k}{k}\right)\left(1+\dfr... | If $k = 1$, clearly, $k < 2^m$.
In the following, assume that $k \ge 2$.
We have, for all $q > k \ge 0$,
$$\binom{q + 1}{k} =
\frac{(q + 1)!}{k! (q + 1 -k)!} = \frac{q!}{k! (q - k)!} \frac{q + 1}{q + 1 - k}
\ge \binom{q}{k}.$$
Thus, we have
$$\binom{n}{k} \ge \binom{2k}{k}.$$
Using Mathematical Induction, it is easy t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$
Let $x,y,z\ge0$ satisfy $\max\left \{ x,y,z \right \}\ge 1$. Prove that $$x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$$
My attempts:
From the condition we can deduce $x+y+z\ge 1$
The inequality can be written as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+(x+y+z)^2-2(x+y+z)\ge0$$ or $$x^2+y^2+z... | Continuing from OP's work. WLOG let $x \geq 1$, we have
$$q = xy+yz+zx = x(y+z) + yz \leq x(p-x) + (\frac{p-x}{2} ) ^2,$$
with equality iff $ y = z = \frac{ p-x}{2}$.
We WTS in the domain $ p \geq x \geq 1$,
$$p^2 + p - 2 - \frac{1}{4} ( p-x)(p+3x) \geq 0.$$
Conditioning on each value of $p$, we have a much simpler qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m... | Following on from $m^2+2m-2\sqrt{m+1}+1=0$, you already have $b=\sqrt{m+1}$, so substitute this in to get $b^4-2b=0 \implies b=\sqrt[3]2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solutions of Triangles inequalities - Duplicate A,B,C are angles of a triangle we are supposed to prove that $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$ $\leq$ $\frac{1}{8}$. I used trigonometric ratios of half angles which would give $\frac{(s-a)(s-b)(s-c)}{abc}$. How can I proceed after this step?
Any help wou... | Here is an alternative trigonometric proof of the inequality for fun. Denote $$
x = \frac{A}{2}, \qquad y = \frac{B}{2}, \qquad \{x, y, x + y\} \subset [0, 90^\circ].
$$
Then, using the product-to-sum formula
$$
\sin x \sin y = \frac{\cos (x - y) - \cos (x + y)}{2}
$$
as well as the AM-GM inequality,
we have
$$ \begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\int_{0}^{1} \frac{ \ln^n x}{1+x^{m}} d x$ =? Inspired by the post, I try to generalize the result to
$$
I(m, n)=\int_{0}^{1} \frac{ \ln ^nx}{1+x^{m}} d x
$$
Instead, I first investigate its partner integral $$
\begin{aligned}
I(a): &=\int_{0}^{1} \frac{x^{a}}{1+x^{m}} d x . \\
&=\sum_{k=0}^{\infty}(-1)^{k} \int_{0}^{... | Take the $n$th derivative of both sides of
$$\int_0^1\frac{y^{s}}{1+y}\mathrm{d}y=\frac12\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\quad \mathfrak{R}(s)>-1$$
w.r.t $s$, we get
$$\int_0^1\frac{y^{s}\ln^n(y)}{1+y}\mathrm{d}y=2^{-n-1}\left[\psi^{(n)}\left(\frac{s+2}{2}\right)-\psi^{(n)}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$. Let $p$ a prime number such that there exists $x,y$ positive integers such that $p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$.
Find $p$.
I found that $p=11$ . I noticed that $p|(x+y)$. I tried with quadratic residues but I am stuck.
| Subtracting the second problem equation minus the first one gives
$$p^2 - p = 2y^2 - 2x^2 \; \; \to \; \; p(p - 1) = 2(y - x)(y + x) \tag{1}\label{eq1A}$$
Note $p$ being an odd prime means $p \mid y - x$ or $p \mid y + x$. The first case gives $y - x \ge p$ and $y + x \gt p$, so $p^2 - p = 2(y - x)(y + x) \gt 2p^2$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\frac{\cos2x}{ \sin x + \sin 3x}$? I thought of using trig identities to get rid of $\cos2x$ and $\sin3x$, then use Weierstrass substitution, but I got myself into big trouble as the expression got too complicated.
Is there a simpler way to tackle this problem?
| The method using the Weierstass substitution isn't so bad.
$$t=\tan\left(\frac x2\right) \implies \begin{cases}\sin(x) = \frac{2t}{1+t^2} \\ \cos(2x) = 2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1 \\ \sin(3x) = \frac{6t(1-t^2)^2-8t^3}{(1+t^2)^3}\end{cases}$$
where we use the identities
$$\begin{cases}\sin(x) = 2\sin\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. A... | I'll do the change of variable $n-2 \to n$ to simplify the derivation. You can recover your original integral by reversing this substitution.
Using the multiple-angle formulas shown here we know that for a positive integer $n$
\begin{align}
\sin(nx) &= \sum_{k=0}^{\lfloor{\frac{n-1}{2}\rfloor}}(-1)^k \binom{n}{2k+1}\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
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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.