Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute $\int \frac {dx} {(a + b \cos x)^2} $ for $a > b.$ Let $a>b$ be real numbers. As the title suggests, I would like to compute $$\int \frac 1 {(a + b \cos x)^2} \, dx.$$
My attempt consisted of converting the cosine to half-angle tangents, substituting the half-angle tangent for $t$, and simplifying to $$\int \fr... | Here it is another way to approach it for the sake of curiosity.
Let us make the substitution:
$$\sinh(x) = t\sqrt{\frac{a-b}{a+b}} \Rightarrow \mathrm{d}t = \sqrt{\frac{a+b}{a-b}}\cosh(x)\mathrm{d}x$$
Then the denominator of the given expression can be written as
\begin{align*}
((a+b) + (a-b)t^{2})^{2} & = (a+b)^{2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The order of computing composition of permutation when using cycle notation Given two permutations $p,q$, say $p=(234)$ and $q=(123)$, we can compute the product permutation $pq$ by writing out $pq=(234)(123)$, then turn it into disjoint cycles.
In my understanding, we shall do the latter step in the backward direction... | One can understand this by writing out the tables explicitly. Note that permutation $pq$ is applying $q$ and then applying $p$.
\begin{array}{|c|c|c|}
\hline
\text{permutation} & \text{acts on} & \text{produces}\\
\hline
pq=(234)(123)&1&3 \\
\hline
pq=(234)(123)&2&4 \\
\hline
pq=(234)(123)&3&1 \\
\hline
pq=(234)(123)&4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1) \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ t... | Note that $\displaystyle\left|\sum_{k=1}^n \frac{\sin k}k - \frac12(\pi-1)\right| = \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|$.
We prove the following inequality:
$$
\left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|
\leq
\frac{1}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Question related to arithmetico-geometric series The question is $$\text{Find the sum of }0.2+0.004+0.00006+0.0000008+\cdots$$
My solution:
Given series can be written in the form $$S=\frac{2}{10}+\frac{4}{10^3}+\frac{6}{10^5}+\frac{8}{10^7}+\cdots(1)$$ Multiplying both sides by $10^{-2}$, we get
$$\frac{S}{10^2}=\frac... | Alternate method, per request:
Let $\displaystyle S = \frac{2}{10}
\left[1 + \frac{1}{10^2} + \frac{1}{10^4} + \cdots\right] =
\frac{2}{10}\times \frac{100}{99}.$
Then the desired sum is
$$ T = S\left[1 + \frac{1}{10^2} + \frac{1}{10^4} + \cdots\right]
= S \times \left(\frac{100}{99}\right) = \frac{2}{10} \times
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of this expression , was my inequality true or of any help? The problem which I encountered was a,b,c>0 and
$$f(a,b,c):= \left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor$$
where $a$, $b$ and $c$ are positive. Find minimum of $f(a,b,c)$... | Since $\lfloor x\rfloor\gt x-1$, we have, by AM-GM inequality,
$$\begin{align}f(a,b,c)&=\left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor
\\\\&\gt \frac{a+b}{c}-1+ \frac{b+c}{a}-1+ \frac{c+a}{b}-1
\\\\&=\bigg(\frac ba+\frac ab\bigg)+\bigg(\frac cb+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| The condition implies $xy = 1 - (x^2 + y^2)$, thus $x^2-3xy - 2y^2 = x^2 - 3 + 3(x^2 + y^2) - 2y^2$ $ = 4x^2 + y^2 - 3$.
Then as previously mentioned, the ellipse $x^2+xy+y^2 = 1$ can be parametrised as $(x,y) = \left(\frac{1}{\sqrt{3}}\cos t+\sin t,\frac{1}{\sqrt{3}}\cos t-\sin t\right)$. Substituting this into $4x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
calculate expected value of total number of offspring in second generation Suppose a plant has $X$ offspring per year with $P(X=k) = \frac{1}{4}$ for $k = 1,2,3,4.$
Independently, each offspring has from one to four offspring in the next year with the same discrete uniform distribution. Let $Y$ denote the total number ... | Hint: Start with the definition of expected value:
$$ E[X] = \sum_{i = 1}^n {x*P[X = x]} .$$
If we add in the condition, this becomes:
$$ E[Y | X = x] = \sum_{i = 1}^n {y*P[Y = y | X = x]} .$$
Now, start by asking what happens if the plant has $X = 1$ offspring? There are four possibilities: $Y = 1, 2, 3, 4$. Each has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\... | Note that
$$
\log (1 - x) < - x
$$
for $0<x<1$. Therefore,
\begin{align*}
0 & < \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)} = \exp \left( {\sum\limits_{k = 1}^n {\log \left( {1 - \frac{1}{{4k}}} \right)} } \right) \\ & < \exp \left( { - \sum\limits_{k = 1}^n {\frac{1}{{4k}}} } \right) < \exp \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$ \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\f... | Just for your curiosity since you already received @hamam_Abdallah's answer.
You can compute this integral musch faster with a different change of variable.
Because of the denominator, let $x=\sinh^2(t)$
$$\int \frac{x^3}{\sqrt{x^2+x}}\, dx=2 \int \sinh ^6(t)\,dt$$
Now, using the multiple angle formulae
$$\sinh ^6(t)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proof of $ d(n)\leq \sqrt {3n}$. Show that
$$ d(n)\leq \sqrt {3n}$$
and the equality is true if only if $n=12$,
where $d(n)$ is the number of positive divisors of $n$..
Here is a proof Proof of $ d(n)\leq \sqrt {3n}$.
Let $$ n=\prod_{k=1}^m p_k^{\alpha_k},$$
the proof given above considers the case when $\alpha_k\geq ... | We will use the following $\textbf{three facts}$:
(a) $$\max_{n\in\mathbb{N}^+}\frac{(n+1)^2}{2^n}=\frac94<3,$$
with equality if and only if $n=2$;
(b) Let $p\geq 3$ and $\alpha\geq 2$, then, by Bernoulli's inequality,
$$p^{\frac{\alpha}{2}}=(1+(p-1))^{\frac{\alpha}{2}}
\ge 1+\frac{p-1}{2}\alpha\ge1+\alpha,$$
with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Assistance: "A Gomory cut problem" Hope everything is operating smoothly for everyone here. I am working on a simple problem involving Gomory cutting planes and am having some trouble.
Problem
Consider the following LPP:
\begin{alignat*}{4}
\text{Maximize} \quad \quad \quad x_1 + x_2 & {}{} & & {}{} & & {}{} &\\
\text{... | So you have:
$$x_1 - \frac{1}{11}x_3 + \frac{2}{11} x_4 = \frac{14}{11}. \qquad (1)$$
Since all variables are nonnegative, we can round down the coefficients and obtain a valid inequality:
$$x_1 - x_3 \leq \frac{14}{11}.$$
Since $x_1$ and $x_3$ are integer, this inequality is the same as:
$$x_1 - x_3 \leq 1.$$
If you s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ .
What I Tried: I wrote :-
$$\rightarrow \frac{7}{2^{1/2} + 2^{1/4} + 1} = \frac{2^2 + 2 + 1}{2^{1/2} + 2^{1/4} + 1}$$
Then I ration... | Hint : $(2^{\frac{1}{4}}-1) \cdot (2^\frac{1}{2} + 2^\frac{1}{4} +1) = 2^\frac{3}{4} -1$
Hint : $7 = 2^3-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot\sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4 m^2-1}\Rightarrow \sum_{m=1}^{\infty}\frac{1}{16m^2-1} =$? By the equation $|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4m^2-1}$, how can I get the value of $\sum_{m=1}^{\infty}\frac... | Hint. One may observe that
$$
\frac{\cos \left(2 m \times 0\right)+\cos \left(2 m\times\frac \pi 2\right)}{2} =
\begin{cases}
1, & \text{if $m$ is even}, \\
0, & \text{if $m$ is odd}.
\end{cases}
$$ Then putting successively $\theta:=0$ and $\theta:=\frac \pi 2$ in the given identity and averaging may help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Length of cardiod $r=1+\sin\left(\theta\right)$ I'm pretty sure this is correct since I'm able to justify each of the steps but I'm just wondering if there's a more efficient way that isn't just skipping steps? Because I had to do 2 substitutions, one of which converted sine into cosine.
\begin{align*}
\text{Length }= ... | Everything is correct including the setup of the integral (using symmetry), but note that $1 + \sin x = (\cos x/2 + \sin x/2)^2$ as mentioned here.
When taking the square root you have $|\cos x/2 + \sin x/2|$, but since this equals $|\sqrt{2} \cos(x/2 - \pi/4)|$ which is non-negative for $-\frac{\pi}{2} ≤ x ≤ \frac{3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$ $$\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$$
My try:
I tried substituting $x= \tan\theta$ and got,
$\displaystyle\int e^\theta \tan^2\theta \sec\theta\ \mathrm{d}\theta$ and cannot move forward. so I thought of finding t... | Observe that:
$$(e^{\arctan(x)} \sqrt{1+x^2})' = \frac{e^{\arctan(x)}}{\sqrt{1+x^2}} + \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}}$$
$$x(e^{\arctan(x)}\sqrt{1+x^2})' = \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}} + \frac{x^2 e^{\arctan(x)}}{\sqrt{1+x^2}}$$
Let us integrate the left-hand side. So, we have:
$$J = \int_{-1}^{1} x(e^{\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise i... | Less elegant than @Quanto's answer, let us write
$$a^2+(x-t)^2=(t-r)(t-r)$$ with $r=(x+ia)$ and $r=(x-ia)$ and use partial fraction decomposition
$$\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }=\frac{r s+r t+s t-a^2}{\left(a^2+r^2\right) \left(a^2+s^2\right) \left(t^2+a^2\right)}+$$
$$\frac{1}{\left(a^2+r^2\right) (r-s) (t-r)}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $... | Consider $\lfloor n\log_{10}2\rfloor+\lfloor n\log_{10}5\rfloor+1$. Since $\log_{10}2$ and $\log_{10}5$ are irrational we can replace each floor by a ceiling and a $-1$, i.e. it is equal to
$$\lceil n\log_{10}2\rceil+\lceil n\log_{10}5\rceil-1$$
Now we have the following inequalities for any $x,y$:
$$\lfloor x+y\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and Jack are in the same group.
b-)Dennis , John and Jack are in different groups.
c... | While there is a simpler method as given in one of the answers, even going by your method the working can be simplified quite a bit.
a) Once you fix Dennis's group, there are $4$ more slots in the group to be picked from $24$ people. But if John and Jack are in the same group, you are left with choosing two spots from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solution Verification: $\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}$ I don't have the 'official' solution to this series so I wanted to verify mine.
\begin{equation*}
(s_n)=\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}
\end{equation*}
$\underline{\text{Claim:}}$ $(s_n)$ diverges.
$\u... | Yes. This works.
More directly, we note that, for $n\geq 1,$ $$\left(1+\frac 1 n\right)^n\geq 2$$
This follows by noting the first two terms of the binomial expansion of the left side are $1+n\cdot \frac1n=2.$
Then for even $n$:
$$2^{-n}\left(1+\frac{(-1)^n}n\right)^{n^2}\geq 2^{-n}\cdot 2^n=1.$$
So the terms do not co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find Two Rank 1 Matrices I'm looking for rank one matrices B and C such that A = B + C and BC = 0. Where
$$A = \begin{bmatrix}
0 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 0
\end{bmatrix}$$
with eigenvalues 0, -2, 6 and eigenvectors \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \begin{bmatrix... | You can diagonalize $A$ as $D_A=P^{-1}AP$ with:
$$P=
\begin{pmatrix}
1 & -1 & 1 \\
-1 & 0 & 2 \\
1 & 1 & 1 \\
\end{pmatrix} \text{, }
D_A=
\begin{pmatrix}
0 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 6 \\
\end{pmatrix}
=\text{diag}(0,-2,6)
$$
We try to find $B$ and $C$ with the same eigenv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\... | You have$$e^x\sin(x)=x+x^2+\frac{x^3}3+o(x^3),$$and therefore$$\lim_{x\to0}\frac{e^x\sin(x)-x-x^2}{x^3}=\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Condition that satisfies inequality involving fractions and the floor function I want to solve the following inequality for $a$ :
$$\left \lfloor \frac{a}{b} \right \rfloor > \frac{a}{b+1}$$
Where both $a$ and $b$ are strictly positive integers.
My end goal with this is, for a given $b$, to be able to find the smallest... | Let $a=bk+r$, where $0\leq r\leq {b-1}.$
Then we have, $\lfloor {\frac ab} \rfloor=k.$
So, $k>\frac {a}{b+1}.$
So that, $bk+k>bk+r.$
Which means, $k>r$. So, the range will be $a>b(r+1).$ So the range of $a$ for which your inequality is satisfied depends upon what $a$ mod $b$ is. But $a>b^2$ will invariably give safe va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \i... | Hint:
I found the expression $$=\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}-\dfrac{\sin(x-2)}{\cos(x-2)\cos(x-3)}$$
Now,
$$\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{\sin(x-1)}{\sin2}\cdot\dfrac{\sin(x-1-(x-3))}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{1-\cos^2(x-1)}{\sin2\cos(x-1)}-\dfrac{\sin(x-1)\sin(x-3)}{\sin2\cos(x-3)}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.... | Expanding both sides and moving positive terms to their respective sides, we get:
$a^4 +b^4+\dfrac{c^4}{2}\geqslant a^2 c^2 + b^2 c^2$
By AMGM we have:
$a^4 + \dfrac{c^4}{4}\geqslant a^2 c^2\;$ and $\;b^4 + \dfrac{c^4}{4}\geqslant b^2 c^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Recurrence Relation: $a_{n+2}+a_n=5\cos (n\pi/3)-7\sin(n\pi/4)$ Recurrence Relation: $a_{n+2}+a_n=5\cos \left(\frac{n\pi}{3}\right)-7\sin\left(\frac{n\pi}{4}\right)$
Attempt:
The solution of the associated homogeneous relation is
\begin{align}
a_n^{(h)}&=c_1\left(\cos \left(\frac{n\pi}{2}\right)+i\sin \left(\frac{n\pi}... | Here is our road map:
First we find a particular solution $p_n$, then a complementary solution $f_n$. Set
$a_n = f_n + p_n $, we are done.
It is easy to find a particular solution $p_n$.
For example, set $a_0$ = 0, $a_1$= 1 we get, for n = 2, 3, 4
$p_n$" />
To find the complete solution, we find the complementary func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generating functions in discrete mathematics - How to find coefficient from polynomial with multiplication So I encountered a certain problem when studying generating functions that I seem to be unable to fully solve.
The problem is as follows:
Carol is collecting money from her cousins to have a party for her aunt. If... | We have $1+2x^5+x^{10}$ and also two other complicated factors:
$$(1-x)^{-8}\quad=\quad1+{-8 \choose 1}(-x)+{-8 \choose 2}(-x)^2+\dots+{-8 \choose 4}(-x)^4+{-8 \choose 5}(-x)^5+{-8 \choose 6}(-x)^6+\dots+{-8 \choose 9}(-x)^{9}+{-8 \choose 10}(-x)^{10}+\dots{-8 \choose 14}(-x)^{14}+\dots,$$
$$(1-x^4)^8\quad=\quad1-{8 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the Diophantine equation $y(x^3-y)=z^2+2$ solvable?
Do there exists integers $a,b$ such that their sum is a perfect cube while their product minus two is a perfect square?
Equivalently, do there exist integers $x,y,z$ such that
$$
y(x^3-y)=z^2+2 \quad ?
$$
The source of this problem is the MathOveflow question "W... | Let $x$, $y$ and $z$ be integers such that $y(x^3-y)=z^2+2$. If $z$ is odd then
$$y(x^3-y)\equiv z^2+2\equiv3\pmod{8},$$
and so $y$ and $x^3-y$ are odd, meaning that $x$ is even. Then $x^3\equiv0\pmod{8}$ and so
$$y(x^3-y)\equiv-y^2\equiv7\pmod{8},$$
a contradiction. On the other hand, if $z$ is even then
$$y(x^3-y)\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom... | When I was writing the question I was hoping to get the steps to synthetic division. I thought I'd show the steps that I came up with from the discussion.
Step 1: is to get the divisor in (x-a) format. There cannot be a coefficient. It is done by factoring:
$$(2x-3)=2(x- \frac32)$$
The problem now looks like this $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I need help finding the values of the constants of a function $f\left(x,y\right)=ax^2y+bxy+2xy^2+c$
I need to find the values of the constants a, b, c so that the function has a point $P\left(\frac{2}{3},\frac{1}{3}\right)$ and and minimum value $-\frac{1}{9}$
My solution
$f´(x)=2axy+by+2y^2$
$f´(y)=2x^2+bx+4xy$
Evalua... | Plug in your values of $a$ and $b$ into $f$ to get your function of $x$ and $y$. Then you want the value of the function to be $-1/9$ at the point $P$, so solve the equation $f(P) = -1/9$ for $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A curious summation. Months ago I was interested in calculating new infinite summations and was able to make a nice summation and also find its closed form as follows :
$$\sum \limits_{n=0}^{\infty } \frac{(-1)^{0+1+2+..+..+n}}{(2n+1)^2} = \frac{\sqrt{2}\pi ^2}{16}$$
And also that
$$\sum \limits_{n=0}^{\infty } \frac{(... | 1. Let us first analyze the behavior of OP's sums. We find that
\begin{align*}
F(s)
&:= \sum_{n=0}^{\infty}\frac{(-1)^{1+2+\cdots+n}}{(2n+1)^s} \\
&= \sum_{n=0}^{\infty}\frac{1}{(8n+1)^s}
-\sum_{n=0}^{\infty}\frac{1}{(8n+3)^s}
-\sum_{n=0}^{\infty}\frac{1}{(8n+5)^s}
+\sum_{n=0}^{\infty}\frac{1}{(8n+7)^s}
\end{align*}
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions Problem: Find closed form expression for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions.
Hint: Calculate what function you need to multiply in order to get partial sums with alternating sig... | Introduce $A(x):=\sum_{n\geq 0}a_n x^{n+1}$ for
$$
a_n=1-2^4+3^4-...+(-1)^{n-1}n^4.
$$
Then observe that the relation
$$
a_{n+1}=a_{n}+(-1)^n(n+1)^4
$$
and multiply it by $x^n$
$$
a_{n+1}x^n=a_{n}x^n+(-1)^n(n+1)^4x^n
$$
and sum for $n\geq 0$ to get
$$
\frac{A(x)}x=A(x)+\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}
$$
yielding
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find a general term of a trinomial? I want to find the coefficient of $x^0$ in the expansion of $(x + 1 + 1/x)^4$.
Without an expansion, I keep getting "nested binomial" terms if I group two terms together:
$$\binom{4}{k}\binom{k}{m} (x)^{k-m}(1/x)^{m}$$
For which $k-2m = 0$. I cannot solve further without gues... | We can also apply the binomial theorem twice in order to derive the cofficient of $x^0$. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. This way we can write for instance
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{[x^0]}&\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Proof verification: Is my induction proof correct? Can anyone proof read my induction proof for this particular task?
For $$ n \in \mathbb{N_0} $$
$a_n$ is defined as $$ a_n= (-1)^{(n+1)}+ 1 + (-3n) + 2n^2 $$
and $b_n$ as:
$$ b_n =\begin{cases}
a_n \quad &\text{if} \, 0 ≤ n ≤ 2, \\
b_{n-1}+b_{n-2}-b... | Let me propose a bit cleaner alternative proof by strong induction.
You have already showed that these two sequences agree on first four terms $n=0,1, 2, 3.$ Then directly compute $a_{n-1}+a_{n-2}-a_{n-3}+8$ using the given formula, and make sure it is equal to $a_n$ for all $n\ge 3.$ Thus, both $(a_n), (b_n)$ satisfie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of ... | $$I=\int_{0}^{\infty} \frac{\ln x}{x^2+2x+4} dx$$
Take $x=az$, then,
$$I=\int_{0}^{\infty} \frac{a( \ln a + \ln z)dz}{a^2z^2+2az+4}.$$
Choose $a=2$, then,
$$I=\frac{a}{4}\int_{0}^{\infty}\frac{\ln 2~ dz}{z^2+z+1}+\frac{a}{4}\int_{0}^{\infty} \frac{\ln z~ dz}{z^2+z+1}=I_1+I_2$$
Put $z=1/t$ and $a=2$ in $I_1$, to get $I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate the following Limit $\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$ Calculate the Limit
$$\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$$
I've been working on trying to find the limit using Dominated Convergence.
If we let $f_n(x) = \frac{\sq... | Hint. By using
$$
1-\cos x=2\sin^2\frac x2
$$ and the change of variable
$$
x=n^{-1/6}u,\quad u^6=nx^6,\quad dx =n^{-1/6}du,
$$one gets
$$
\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx=\lim_{n\rightarrow \infty} 2n^{1/3}\int_{0}^{\infty} \frac{\sin^2\left(\frac{u}{2n^{1/6}}\right)}{1+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I evaluate the following limit $\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$? Evaluate this limit :
$$\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$$
I tried to simplify the function :
$$x^2e^{\frac{1}{x}}-x^2e^{\frac{1}{x+1}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x... | We have : \begin{aligned}\lim_{x\to +\infty}{x^{2}\left(\operatorname{e}^{\frac{1}{x}}-\operatorname{e}^{\frac{1}{x+1}}\right)}&=\lim_{x\to +\infty}{x^{2}\operatorname{e}^{\frac{1}{x+1}}\left(\operatorname{e}^{\frac{1}{x}-\frac{1}{x+1}}-1\right)}\\ &=\lim_{x\to +\infty}{\frac{x\operatorname{e}^{\frac{1}{x+1}}}{x+1}\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x... | A proof by someone posted here
\begin{align*} \ln x\ln(1-x)&=\sum_{n=1}^\infty\frac{1-x^n-(1-x)^n}{n^2}\\ &\le \sum_{n=1}^\infty\frac{1-2\left[\frac{x+(1-x)}{2}\right]^n}{n^2}\color{blue}{~~(\text{Jensen's inequality)}}\\ &=\sum_{n=1}^{\infty}\frac{1-\frac{1}{2^{n-1}}}{n^2}\\ &=\zeta(2)-2\rm{Li_2}\left(\frac{1}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac... | We can only "flip the sign" if what is inside the absolute value is negative. And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.
We could write it this way...
$-3<\frac {x+1}{x-2}<0$ or $0\le \frac {x+1}{x-2}<3$
Or say, the left-hand equation only applies when $-1<x<2$
Then we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inductive proof of series (sum of 1 / [i*(i+1)] ) I like to show that
$$
\sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1} \quad \forall n \in \mathbb{N}.
$$
This seems to be a possible solution:
Base case:
$$ n_0 = 1 $$
$$ \sum_{i=1}^1 \frac{1}{1*2} = 1 - \frac{1}{1+1}$$
$$ \frac{1}{2} = \frac{1}{2} $$
Induction hyp... | $$\frac{\color{red}{-1}}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{\color{red}{-1}(n+2)+1}{(n+1)(n+2)}=\color{orange}{-}\frac{(n+2)\color{orange}{-1}}{(n+1)(n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$ $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$
I have tried pinning it between $ \frac{(1-\... | We take the limit $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}}$. We can evaluate this by $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}} = \lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{1 - \sin^2{x}} = \lim\limits_{u \to 1} \frac{1 - u^n}{1 - u^2}$, which we evaluate to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality... | I give another idea.
WLOG, we may assume that $a \ge b \ge c$. It's easy to see that the original problem follows immediately by adding two following inequalities:
\begin{align} & \frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2} \ge \frac{4}{ab+bc+2ca+2b^2}
\ \ \ \ \ \ \ \ (1) \\ &\frac{4}{ab+bc+2ca+2b^2} + \frac{1}{c^2+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Linearly express a polynomial of a root Consider the equation $x^2-4x+7=0$ and $x_1,x_2$ its roots. Calculate:
$$\frac{x_1^4-2x_1^3+3}{x_1^3+x_1^2}+\frac{x_2^4-2x_2^3+3}{x_2^3+x_2^2}$$
Rules: cannot manually calculate the roots, must use Vieta's formulas.
Amplifying the fractions in this form seems like a bad idea so I... | from $x^2 = 4x-7$ I got $x^3 = 9x - 28$ and $x^4 = 8x-63$
Next, $x = 4 - \frac{7}{x},$ so $\frac{1}{x} = \frac{-x+4}{7}$ and so $\frac{1}{x^2} = \frac{-4x+9}{49}$
Then I wrote $(x+1)^2 = 6(x+1) - 12; \; \;$ then dividing gives $x+1 = 6 - \frac{12}{x+1}$ and
$ \frac{1}{x+1} = \frac{-x + 5}{12} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ \det(A^4 + A^2 B^2 + 2A^2 + I) \geq 0 $ Problem:
Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$
My Approach:
If $ A $ invertible, then
$$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \impl... | Since $AB=-BA$, so we have $A^2B^2=-(AB)^2$. Now consider the matrix
\begin{align*}
\det(A^4+A^2B^2+2A^2+I)&=\det((A^4+2A^2+I)\color{magenta}{-(AB)^2})\\
&=\det((A^2+I)^2-(AB)^2)\\
&=\det((A^2-AB+I)(A^2+AB+I))\\
&=\det((A^2+\color{blue}{BA}+I)(A^2+AB+I))\\
&=\det(A^2+BA+I)\det(A^2+AB+I)\\
&=\det((A+B)A+I)\det(A(A+B)+I)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- N... | It's true for any positives $a$, $b$ and $c$.
Indeed, we need to prove that:
$$a^2+b^2+c^2\geq\frac{36}{35}\left(\frac{(a+b+c)^2}{4}+\frac{2abc}{a+b+c}\right)$$ or
$$35(a^2+b^2+c^2)(a+b+c)\geq9(a+b+c)^3+72abc$$ or
$$\sum_{cyc}(13a^3+4a^2b+4a^2c-21abc)\geq0,$$ which is true by Muirhead or by AM-GM.
The best result in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | Hint: $ $ let $\ \begin{align}n &= 2k\!+\!1\\ a&=(-2)^k\end{align}. $ $\bmod 9\,$ it's $\:\!f(a)\equiv\, 2a^2\!+\!5a\!+\!2\,\equiv 2\,(a\!-\!1)^2\equiv 0\,$ by $\, 3\mid a\!-\!1\!\!$
Or, conceptually $\,a\equiv 1\,$ is double root of $\,f(a)\,$ by $\,f(1)\equiv 0\equiv f'(1)\,$ by $\,f'(a) \equiv 4a\!+\!5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
Linear algebra-Coordinate vector Given Base $A=(a_1,a_2,a_3) $
$$ B=B=\left(b_{1}=\left(\begin{array}{c}
1\\
-1\\
0
\end{array}\right),b_{2}=\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right),b_{3}=\left(\begin{array}{c}
0\\
-1\\
1
\end{array}\right)\right) $$
Given transformation matrix : $ \left[I\right]_{A}^{B}=\le... | I think $\begin{bmatrix}0\\ 0\\ 1\\\end{bmatrix}_{B} $ means the coordinate vector of $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} $ in the basis $B$. And since $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} = 0b_1 + 1b_2 + 1b_3$, it does seem right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Line through two given skew lines and origin Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connectin... | Your process and equations are correct, but I can't figure out how you arrived at your answers from them. Here's what I did:
$$a-2b+c=0 {\tag 1}$$
$$3a-10b+9c=0 {\tag 2}$$
Let $\frac ac=k$, $\frac bc=l$, then, from $(1)$:
$$k-2l+1=0$$
and from $(2)$:
$$3k-10l+9=0$$
Solving these gives $k=2$ and $l=\frac 32$.
Hence, $a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\iiint_{\mathbb{R}^3} \exp \left(-x^2 -y^2-z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z$ How to evaluate
$$
\iiint_{\mathbb{R}^3} \exp \left( -x^2 -y^2 -z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z,
$$
My Attempt:
If we use the transformation to spherical polar coordinates $R, \theta, \phi$, th... | The last integral is Gaussian integral
$$\int_0^{+\infty}e^{-x^2}dx = \dfrac{\sqrt{\pi}}{2}$$
There are many ways to evaluate this integral, and here is one way https://en.wikipedia.org/wiki/Gaussian_integral
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Minimum of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$
Let $a$ and $b$ be real numbers. Find the minimum of $$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$$
The problem is from an inequality book. While I was doing the problem myself, I got the follow... | It seems to me that you misapply Minkowski's inequality.
More accurately:
\begin{align*}
&\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\
&\geq\sqrt{2a^2+2(a-1)^2+2b^2+2(b-1)^2}\\
&=2\sqrt{\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}}.\end{align*}
Then it's just li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqr... | An uglier brute-force method by some trigonometric identities:
Let the green marked $\angle BCA$ be $\theta$. Then the corresponding angle at centre $\angle BEA = 2\theta$, and $\angle PEB = 90^\circ - 2\theta$.
By the laws of cosine,
$$\begin{align*}
BP^2 &= EB^2 + EP^2 - 2 EB\cdot EP \cos \angle PEB\\
&= 2R^2 - 2 R^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
How to simplify this equation? I'm trying to simplify equation
$$n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil$$
into
$$n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor$$
where $x$ is an integer. Indeed, both equations seem to output the same result when testing for some values. Thus far, I've done:
\begin{align*... | The first gives $$n-1<\sqrt{2x+\frac{1}{4}}-\frac{1}{2}\leq n$$ or
$$\left(n-\frac{1}{2}\right)^2<2x+\frac{1}{4}\leq\left(n+\frac{1}{2}\right)^2$$ or
$$\frac{n^2-n}{2}<x\leq\frac{n^2+n}{2}$$ and by the same way the second gives:
$$n\leq\sqrt{2x}+\frac{1}{2}<n+1$$ or $$\frac{n^2-n}{2}+\frac{1}{8}\leq x<\frac{n^2+n}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integration gives two different answers The question is :
\begin{array}{l}
\text { If } \int \frac{x+1}{\sqrt{2 x-1}} d x=\mathrm{f}(\mathrm{x}) \sqrt{2 x-1}+\mathrm{C}, \text { where } \mathrm{C} \text { is }\\
\text { constant of integration, then } \mathrm{f}(\mathrm{x}) \text { is equal to : }
\end{array... | $ \frac{1}{2}\left(\frac{1}{3} t^{3 / 2}+3 t^{1 / 2}\right) \\
=t^{1 / 2}\left(\frac{1}{6} t^{3}+\frac{3}{2}\right) $
This bit is wrong.
$t^{1/2} \times t^{3} \neq t^{3/2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin... | Let $a=\frac{\frac{1}{y}+\frac{1}{z}}{2},$ $b=\frac{\frac{1}{x}+\frac{1}{z}}{2}$ amd $c=\frac{\frac{1}{x}+\frac{1}{y}}{2},$ where $x$, $y$ and $z$ are positives,
Thus, we need to prove that:
$$\sum_{cyc}z\sqrt{x+y}\geq2\sum_{cyc}\frac{xy}{\sqrt{x+y}}$$ or
$$\sum_{cyc}\frac{y(z-x)-x(y-z)}{\sqrt{x+y}}\geq0$$ or
$$\sum_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y... | A SOS (Sum of Squares) solution with computer:
Denote $K = \frac{9(63+5\sqrt{105})}{32}$.
We have
$$
K\left(\frac{x + y + z}{3}\right)^4 - (x^3y + y^3z + z^3x)
= \alpha_1 f_1^2
+ \alpha_2 f_2^2
+ \alpha_3 f_3^2
$$
where
\begin{align*}
\alpha_1 &= \frac{567 - 16K}{13778100} > 0, \\[5pt]
\alpha_2 &= \frac{472K - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}... | The expression in your limit is $\frac{n^{1/n}-1}{\frac{\log n}{n}}$
Notice that $\lim_{n\rightarrow\infty}\frac{\log n}{n}=0$,
and $n^{1/n}=\exp(\tfrac{1}{n}\log(n))$.
Setting $h_n=\frac{\log n}{n}$ yields
$$\frac{n^{1/n}-1}{(\log n)/n}=\frac{\exp(\tfrac{\log n}{n})-1}{\frac{\log n}{n}}=\frac{e^{h_n}-1}{h_n}\xrightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Characterizing integer solutions to $a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr)$ I’m looking to characterize all integers $a$ and $b$ satisfying $$a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr), \qquad a > b \ge 1.$$
Brute force searches have so far turned up the two solutions $(a,b) = (13,7)$ and $(a,b)=(9601,5441)$. I’m gues... | This problem in a way is related to equation $x^2+(x+1)^2=y^2$ which has infinite solution.There is a proof in this site.I will try to show this relation by an examples:
$119^2+120^2=13^4$
$119=7\times 17$
$(7\times17)^2+(7\times17+1)=13^4$$\space\space\space\space\space\space(1)$
and we see that:
$7^4+(7^2+1)^2=29\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\c... | Change the variable:
$t=\tan{\frac{\theta}{2}}; t\in (0,1)$
$\tan{\frac{\theta}{2}}$ is one to one in this interval.
$\sin{\theta}=\frac{2t}{1+t^2}$
$\cos{\theta}=\frac{1-t^2}{1+t^2}$=>
$$f(\theta)=g(t)=2\frac{\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}}{\left(1+\frac{2t}{1+t^2}\right)\left(1+\frac{1-t^2}{1+t^2}\right)}=2t\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Compute $\frac 17$ in $\Bbb{Z}_3$
Compute $\frac 17$ in $\Bbb{Z}_3.$
We will have to solve $7x\equiv 1\pmod p,~~p=3.$
*
*We get $x\equiv 1\pmod 3.$
*Then $x\equiv 1+3a_1\pmod 9,$ so $7(1+3a_1)\equiv 1 \pmod 9$ basically lifting the exponent of $p=3,$ we get $1+3a_1\equiv 4\pmod 9\implies a_1\equiv 1\pmod 3.$
*So ... | lets use it 3-adic expansion here.
*
*The theorem says: A rational number with p-adic absolute value 1 has a purely periodic p-adic expansion if and only if it lies in the real interval [-1, 0). and 1/7 is periodic.
*we will use first ${-\frac{1}{7}}$ and then negate it.
*we know that ${\overline {n_0n_1...n_{k-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Transform ODE $(u-x)u_x + u + x = 0$ to polar coordinates According to Peter Olver in his book “Applications of Lie Groups to Differential Equations”, p. 104, Example 2.32, the ODE:
$$(u-x)u_x + u + x = 0 \tag{1}$$
is transformed in polar coordinates with $x = r\cos \theta$ and $u = r\sin\theta$ to:
$$\frac{dr}{d\theta... | You get by the chain rule, where it is applicable
$$
\frac{du}{dx}=\frac{\frac{du}{dθ}}{\frac{dx}{dθ}}
=\frac{r'(θ)\sinθ+r(θ)\cosθ}{r'(θ)\cosθ-r(θ)\sinθ}
$$
which on the other hand is equal due to the given equation to
$$
=\frac{\sinθ+\cosθ}{\cosθ-\sinθ}
$$
Now solve this by cross-multiplying
$$
(r'(θ)\sinθ+r(θ)\cosθ)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Exponent of Matrix A when A is not diagonalizable I want to compute the Matrix $A^{19}$ where
$$A = \begin{pmatrix}
6&1&0\\0&6&1\\0&0&6\end{pmatrix}$$
Since the only eigenvalue of the matrix is 6, it's not diagonalizable. How do you proceed to calculate the exponent matrix of A then?
| Let$$B=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$$Then $A=6\operatorname{Id}_3+B$. Besides, $6\operatorname{Id}_3$ and $B$ commute, and therefore\begin{align}A^{19}&=(6\operatorname{Id}_3+B)^{19}\\&=6^{19}\operatorname{Id}_3+19\times6^{18}B+\binom{19}26^{17}B^2\\&=\begin{bmatrix}6^{19}&19\times 6^{18}&\binom{19}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the measurement of the $CT$ segment in the square below? For reference The side of square $ABCD$ measures $4$. Calculate TC. (answer: 4)
My progress:
Por propriedade BTE é $90 ^ \circ$
G is the barycenter of the $\triangle ADC$
We have a 2:4 ratio in the $\triangle DEA$ so $\measuredangle EAD = \frac{53^\circ}... | Let's say $\angle ABF=\theta$. It's not hard to see that
$$BT=4\cos\theta$$
$$\angle TBC=90-\theta$$
By Law of Cosines on $\angle TBC$ in $\Delta TBC$, we have
$$TC^2=BT^2+BC^2-2\cdot BT\cdot BC\cos\angle TBC$$
$$TC^2=16\cos^2\theta+16-32\cos\theta\sin\theta$$
$$TC^2=16(1+\cos^2\theta-2\cos\theta\sin\theta)$$
Since $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers. Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers.
My Approach: Simplifying $$m^2n=20^{20}=(2^2 \times 5)^{20}=2^{40} \times 5^{20}$$
Hence one way is $n=5^... | Let $m=2^a5^b$ and $n=2^c5^d$.
So,
\begin{align*}
m^2n&=2^{2a}5^{2b}2^c5^d\\
&=2^{2a+c}5^{2b+d}=2^{40}5^{20}
\end{align*}
which implies
$$2a+c=40\\2b+d=20$$
Since $2a+c=40$, $c$ has to even. So, we can put $c=2,4,6\dots$ and find the corresponding values of $a$. We can do the same for $b$ and $d$.
Note that $40\leq c\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But i... | Surprised no one explicitly used L'Hopital's rule.
$\displaystyle \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2}.$
Derivative of the numerator is therefore
$\displaystyle \tan^{-1}(x) + \frac{x}{1 + x^2}.$
Derivative of the denominator is $\sin(x)$
Since both derivatives evaluate as $(0)$ at $(x = 0)$, L'Hopital's rule ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$ I was attempting this question: https://www.youtube.com/watch?v=600X-ZGNBbk
Which is the following:
$$\sum_{cyc (a,b,c)}{}\frac{a}{b^2+1} \geq \frac{3}{2}$$
Where $a+b+c = 3$ and $a,b,c > 0 $
The video itself used a solution where they added $3$ and then s... | To prove your inequality (not the original inequality), I would introduce the condition $a + b + c = 1$. Note that
\begin{align*}
\frac{1}{9b^2+1} &\geq 1 - \frac{3}{2}b\\
\iff 0 &\leq \frac{3}{2}b\left(1 - 3b\right)^2
\end{align*}
(this is the Tangent Line Trick). Hence, we have
\begin{align*}
\sum_{\text{cyc}} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Orbits of $z_{n+1} = z_n ^2 - 1$ Consider the sequence $z_{n+1} = z_n ^2 - 1$ defined for an arbitrary complex number $z_0$. I am trying to determine all $z_0$ such that the sequence eventually becomes periodic.
Here is my progress so far:
If $|z_0|> \frac{1+\sqrt{5}}{2}$ the sequence is absolutely increasing since
$$|... | Let us find the alternative solutions.
Firstly, let us consider the case of the exact periodicity.
Assuming the periods $1,2,3,\dots,$ easily to get
*
*$z=z^2-1,\quad z_{11}=\phi=\dfrac{1+\sqrt5}2,\quad z_{12}=-\dfrac1\phi=\dfrac{1-\sqrt5}2;$
*$z=(z^2-1)^2-1,\quad z(z+1)(z^2-z-1)=0,$
$\qquad z_{21}=z_{11},\quad z_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, ... | A proof by induction would express that if $f$ is increasing at $n$, then $f$ is increasing at $n+1$, or
$$\frac n{2n+1}<\frac{n+1}{2n+3}\implies\frac{n+1}{2n+3}<\frac{n+2}{2n+5},$$
which can be rewritten as
$$-\frac1{(2n+1)(2n+3)}<0\implies-\frac1{(2n+3)(2n+5)}<0.$$
This proposition is true, but a little nonsense as b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How many negative eigenvalues can a $3 \times 3$ symmetric matrix have? Suppose that I have a symmetric matrix of the form
\begin{equation}
M= \begin{pmatrix}
1 & B & C \\
B & 1 & E \\
C & E & 1\\
\end{pmatrix},
\end{equation}
where $B, C, E$, all belong to $[-1, 1]$. What is the most number of negative eigenvalues ... | Here's a strange application of the formula for the roots of a cubic equation. For a degree-3 polynomial in depressed form
$$
p(x) = x^3 + px +q,
$$
there is a formula for the roots, given by
$$
r_k = 2\sqrt{-\frac{p}{3}}\cos\left(
\frac{1}{3}\arccos\left(
\frac{3q}{2p}\sqrt{-\frac{3}{p}}
\right)
-\frac{2\pi k}{3}
\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the equation of a straight line joining the origin and the point of intersection of two straight lines Question:
Find the equation of the straight line joining the origin and the point of intersection of the straight lines $\frac{x}{a}+\frac{y}{b}=1...(i)$ & $\frac{x}{b}+\frac{y}{a}=1...(ii)$ (Answer: $x-y=0$)
... | Yes you work is correct but it will be good to state in the beginning that the lines intersect only if $a \ne b$. For $a = b$, both equations in fact represent the same line.
You can also simplify your work a bit. Say, the intersection point is $(x_1, y_1)$. Then you have,
$\frac{x_1}{a}+\frac{y_1}{b}=1...(i)$
$\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the mistake in my reasoning in calculating limit where $x \to -\infty$ I had this question in an exam today:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x}
$$
My question is: I have tried to solve it and I also found two other methods to solve it, one in my textbook and one from somebody el... | As noticed the second method is the correct one, to check the result we can make the limit simpler at positive $\infty$ by $u=-x\to \infty$, then
$$\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} =\lim \limits_{u \, \to \,\infty} \; \frac{\sqrt{u^2-4u}+2u}{-2u}=\lim \limits_{u \, \to \,\infty} \; \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Binomial Expansion for n < 1 Write Down the First 3 terms of $\sqrt[3]{1 + \frac{x}{8}}$
My First step : After Evaluation, I got $$\Bigl(\frac{1}{8}x + 1\Bigl)^\frac{1}{3}$$
but, $${{\frac{1}{3}} \choose {r} } = \text{undefined}$$
How am I supposed to do this or am I missing something crucial ?
| $$\sqrt[3]{1 + \frac{x}{8}}=1+\bigg(\frac{\frac{1}{3}}{1!}\bigg)\bigg(\frac{x}{8}\bigg)^1+\bigg(\frac{\frac{1}{3}\times(\frac{1}{3}-1)}{2!}\bigg)\bigg(\frac{x}{8}\bigg)^2+\cdots=1+\frac{x}{24}-\frac{x^2}{576}+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to find formula for the summation series I need a formula for the summation of the series $$
n+2(n−1)+2^2(n−2)+2^3(n−3)\cdots $$
Is there any way to simplify it further? I am guessing it will be of form
$$
\sum_{j=0}^{n} (n-j)a^j
$$
For the above summation series i have:
$
n + 2(n-1) + 4(n-2) + 8(n-3)
$
$
2^k... | Your sum is:
$\begin{align*}
S_n
&= \sum_{0 \le k \le n} 2^k (n - k) \\
&= n \sum_{0 \le k \le n} 2^k - \sum_{0 \le k \le n} k \cdot 2^k
\end{align*}$
First one is just a geometric sum:
$\begin{align*}
\sum_{0 \le k \le n} 2^k
&= \frac{2^{n + 1} - 1}{2 - 1} \\
&= 2^{n + 1} - 1
\end{align*}$
Second o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the area of a region in an isosceles triangle that contains squares of area $256$ and $49$ There was a quiz posted on F*cebook by someone. Here's the problem.
And here's my attempt:
First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green squar... | Let $\Delta ABC$ be our triangle, $AB=BC$, $PQRL$ be a square,
where $P\in AB$, $Q\in BC$, $R\in QC$, $L\in AR$, $PQ=16.$
Also, let $MRNK$ be a square, $M\in LR$, $N\in RC$ and $K\in AC$.
Let $AL=x$.
Thus, since $\Delta ALP\sim\Delta PQB,$ we obtain $$\frac{BQ}{16}=\frac{16}{x}$$ or $$BQ=\frac{256}{x}.$$
Also, since $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9... | I'd start by proving some algebraic relations between $a,b,c$ in $\mathbb{Z}/9\mathbb{Z}.$ For instance, by plugging in $x=0$ we have
$$abc\equiv 1 \pmod 9$$
so each of $a,b,c$ is a unit: congruent to either $1,2,4,5,7$ or $8$. It follows that the square of each of $a,b,c$ is either $1,4$ or $7$ (mod $9$).
Expanding $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the value of $ \angle x $ shown below? For reference: AB=CF
My progress:
This is a typical Peruvian exercise where the solutions are provided by auxiliary lines. I tried one way but I didn't find the solution
| Construct equilateral triangle $ABE$. Because $\angle AEB=2 \angle BCA$, $E$ is the circumcenter of $ABC$. Thus, $AB=FC=CE=BE=AE$. $\angle EAF=60^\circ-24^\circ=36^\circ, \angle AEC=180^\circ-72^\circ=108^\circ, \angle BEC=108^\circ-60^\circ=48^\circ, $ $\angle ECB=90^\circ-24^\circ=66^\circ, \angle ACE=36^\circ, \angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $... | Using the bisector’s property:
$$
\begin{align}
\frac{AB}{BC}&=\frac{AD}{DC}\\
\\
\frac{BC}{DC}&=\frac{BI}{ID}
\end{align}
$$
And $AD+DC=AC$ & $\frac{BI}{ID}=\frac{3}{4}$ can you get $\frac{3}{4}\left(AB+BC\right)=AC$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do I prove that $\sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge6(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$ given $x,y,z > 0$ and $x+y+z=3$? I tried to apply AM-GM inequality:
$$ \dfrac{1}{x^2-xy+y^2} + (x^2-xy+y^2) \ge 2 \implies \dfrac{1}{x^2-xy+y^2} \ge 2 - (x^2-xy+y^2) $$
Then,
$$ \sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}... | We need to prove that:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}+15\geq6(xy+xz+yz),$$ where $x$, $y$ and $z$ are positives such that $x^2+y^2+z^2=3$.
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $3u^2-2v^2=1$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}=\sum_{cyc}\frac{z^2}{x^4z^2-x^2y^2z^2+y^4z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that a certain Group is Abelian Let $G$ be a group with following two properties:
*
*for all $a, b \in G$ we have $(ab)^2=(ba)^2$
*every element $a \in G$ of order $2$, ie $a^2=e_G$, is already the neutral element $a=e_G$
question: is this group already Abelian?
what I tried: we can derive a nice identity of c... | As $\left(ab\right)^2 = \left(ba\right)^2$, then $e_G = b^{-1}a^{-1}b^{-1}a^{-1}baba$, for all $a,b \in G$.
Now
\begin{align}
\left[a,b\right]^2 &= \left(aba^{-1}b^{-1}\right)^2 \\
&= aba^{-1}b^{-1}aba^{-1}b^{-1} \\
&= ab\left(b^{-1}a^{-1}b^{-1}a^{-1}baba\right)a^{-1}b^{-1}ab\left(b^{-1}a^{-1}b^{-1}a^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Find all monic irreducible quadratics in $\mathbb{Z}_3[x]$ The general form for any monic quadratic in $\mathbb{Z}_3[x]$ is: $x^2+Ax+C$
We get the following list of possibilities:
$x^2$
$x^2+1$
$x^2+2$
$x^2+x$
$x^2+x+1$
$x^2+x+2$
$x^2+2x$
$x^2+2x+1$
$x^2+2x+2$
A polynomial is said to be irreducible if it's degree is gr... | $x^2+2$ is irreducible in for instance $\mathbb Z[x]$, but $\mathbb Z[x]/3\mathbb Z$ is a different domain. As you find in your answers $x^2+2\equiv(x+1)(x+2)\bmod 3$, and that settles reducibility. This factorization means while $x^2+2$ does not equal $0$ for any integer $x$, it can equal some other multiple of $3$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$.
My work-
$\frac{10^n}{x}-1<\lfloor \frac{10^n}{x}\rfloor≤\frac{10^n}{x}\... | Rewrite the equation as
$$\frac{10^n}{x} = 1989+\epsilon$$
where $0 < \epsilon < 1$ and $x$ is a positive integer.
Then $\dfrac{10^n}{1989 + \epsilon} = x$ and
$\dfrac{10^n}{1989}$ must be slightly larger than an integer.
The first few digits of $\dfrac{1}{1989}$ are $.00050276520864...$
$$\lfloor\dfrac{10^4}{5}\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}... | Another way:
$$\sum_{cyc}\left(\frac{a^2}{b}+a-\sqrt{2(a^2+b^2)}\right)=\sum_{cyc}\left(\frac{a^2}{b}-2a+b-\left(\sqrt{2(a^2+b^2)}-a-b\right)\right)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{1}{b}-\frac{1}{\sqrt{2(a^2+b^2)}+a+b}\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding all positive integers $n$ such that $\left \lfloor{\frac{2^n}{n}}\right \rfloor$ is a power of $2$. I am trying to figure out a question I have, I don't really know where to start. The question would be
Find all positive integers $n$ such that $\displaystyle \left \lfloor{\frac{2^n}{n}}\right \rfloor$ is a pow... | Given an integer $n$, by long division there are uniquely determined integers $q_n$ and $r_n$ with $0\leq r_n<n$ such that
$$2^n=q_n\cdot n+r_n.$$
Then $\lfloor\tfrac{2^n}{n}\rfloor=q_n$.
First, if $r_n=0$ then $q_n$ divides $2^n$ and so $q_n$ is a power of $2$. In this case $n$ also divides $2^n$, so $n$ is also a pow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$ I want to evaluate the series
$$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$$
I can rewrite this as
$$1+\sum_{n\geq 1} (-3)^{-3n}\left(\frac{3}{6n-1}+\frac{1}{6n+1}... | Consider the series,
$$x-\frac{x^5}{5\cdot 3^2}-\frac{x^7}{7\cdot 3^3}+\frac{x^{11}}{11\cdot 3^5}+\frac{x^{13}}{13\cdot 3^6}--++\cdots.$$
Differentiate this and get
$$1-\frac{x^4}{ 3^2}-\frac{x^6}{ 3^3}+\frac{x^{10}}{ 3^5}+\frac{x^{12}}{3^6}--++\cdots.$$
This is a geometric series, and sums to
$$3\frac{9-x^4}{27+x^6}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
help with $\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right)$ Is my solution right?
\begin{align}
\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right) &=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(1+2+3+\cdots+n-1\right)\right)\\
&=\lim_{n\to\infty}... | Correct solution. Another way is to use Riemann sums:
\begin{align*}
\sum_{k=1}^{n-1}\frac k{n^2}
&=\frac 1n\sum_{k=1}^n\frac kn-\frac1n
\end{align*}
and\begin{align*}
\lim_n\left(\frac 1n\sum_{k=1}^n\frac kn\right)=\int_0^1x\;dx=\frac12.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to show $\displaystyle a_{n}=\frac{4^{2n+1}}{n^{2n}}$ is a descending sequence? I have to show that the sequence
$$a_{n}=\frac{4^{2n+1}}{n^{2n}},\qquad n\geq1$$
is descending.
I thought about proving
$$\frac{a_{n+1}}{a_{n}}<1$$
However, since
$$\frac{a_{n+1}}{a_{n}}=\frac{16n^{2n}}{(n+1)^{2n+2}}$$
I have no clue at... | We have
\begin{eqnarray}
a_1&=&4^3=64\\
a_2&=&\frac{4^5}{2^4}=64
\end{eqnarray}
and for $n\ge 3$,
\begin{eqnarray}
\frac{a_{n+1}}{a_n}&=&\frac{4^{2n+3}\cdot n^{2n}}{(n+1)^{2n+2}\cdot 4^{2n+1}}\\
&=&\frac{16}{(n+1)^2}\cdot\left(\frac{n}{n+1}\right)^{2n}\\
&<&\frac{16}{(3+1)^2}\cdot 1^{2b}\\
&=&1
\end{eqnarray}
i.e. $a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Spherical integration of vector function $\frac{\vec{r}\times(\vec{\mathbb{e}_z}\times\vec{r})}{r^3} \exp^{-2 r}\mathrm{d}{\vec{r}}$ In the context of a problem about the magnetic response of the electron cloud in the hydrogen atom I need the solution of the following volume integral of a vector function:
\begin{align}... | By symmetry the $x$ and the $y$ component is $= 0$. For the $z$ component we obtain
\begin{align}
I_z & = \frac{1}{4\pi} \int_0^\infty\int_0^\pi \int_0^{2\pi} (r^2-z^2) \frac{\exp^{-2r}}{r} \sin{(\vartheta)} dr d\vartheta d\varphi \\
& = \frac{1}{4\pi} \int_0^\infty r \exp^{-2r} dr\int_0^\pi \sin^3(\vartheta)d\v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac1{12}(\sin^2A+\sin^2B+\sin^2C)\leq \sin^2\frac12B\sin^2\frac12C+\sin^2\frac12C\sin^2\frac12A+\sin^2\frac12A\sin^2\frac12B$ for $A+B+C=180^\circ$
If $A+B+C=180^{\circ}$, prove that
$$\begin{align}
\tfrac{1}{12}(\sin^2 A+\sin^2 B+\sin ^2C)
&\leq
\sin^2\tfrac{1}{2}B\sin^2\tfrac{1}{2}C \\[0.5em]
&+\sin^2\tfrac{1}{2... | This is only a special case
I managed to prove the inequality when $A,B,C$ are the angles of a triangle. I expect some kind of transformation to ensure that the general case $A+B+C = 180^\circ$ can be accommodated, but I acknowledge that I haven't investigated this.
If we do assume that $A,B,C$ are the angles of a tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Showing $\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$ by using multiplication of complex numbers Imagine three right triangles on top of each other. The legs of the first two triangles have the ratio $1/3$ and the third triangle $1/7$. The total angle of all three angles is $\pi/4$.
I can get that with the arctan, o... | Since
$$
\tan^{-1}x+\tan^{-1}y =\tan^{-1}[\tan(\tan^{-1}x+\tan^{-1}y )]
$$
i.e.
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)
$$
we have
$$
2\tan \frac{1}{3}=\tan^{-1}\left(\frac{\frac{2}{3}}{1- \frac{1}{9}}\right)=\tan^{-1}\frac{6}{8}=\tan^{-1}\frac{3}{4}
$$
and
$$
2\tan \frac{1}{3}+\tan^{-1}\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find inverse of a function $f(x)=x^3+3x^2+4x-1$ If $f(x)=x^3+3x^2+4x-1$, then $f(x)$ is one to one as it is increasing function for $x \in R$. I want to find $x$ such that
$\frac{-1}{4}x-\frac{7}{16}=2f^{-1}(\frac{-x}{2}-1)$
my attempt:
One option is that calculate $f^{-1}$ , inverse of $f$, and then changes $x$ to ... | Knowing that an inverse exists tells us that the equation $\frac{-x}{4} - \frac{7}{16} = 2 f^{-1}\bigl(\frac{-x}{2} - 1\bigr)$ is equivalent to
$$
f\Bigl(\frac{-x}{8} - \frac{7}{32}\Bigr) = \frac{-x}{2} - 1.
$$
Or $f(y) = 4 y - \frac{1}{8}$, where $y = \frac{-x}{8} - \frac{7}{32}$, which can be rewritten as
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The probability of sum 5 after n time’s Assuming the die is unbiased, what is the probability of getting sum of 5 after $n$ time (use only one cube at a time).
It's clear to me that more than 5 time is wrong.(Because the lowest number is $1$ so $1+1+1+1+1$ is $\max$)
I tried this way: $\frac{1}{6}+\frac{4}{6}*\frac{1}... | The number of cases corresponding to the sum $5$ for $n$ throws of a fair dice will be the coefficient of $x^5$ in the following polynomial:
$p(x)=(x^1+x^2+\ldots+x^6)^n$
$=x^n(1+x+\ldots+x^5)^n=x^n\frac{(1-x^6)^n}{(1-x)^n}=x^n(1-x^6)^n(1-x)^{-n}$
$=x^n(1-nx^6+\ldots+(-1)^nx^{6n})\left(1+{n \choose 1}x+{n+1 \choose 2}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that there is such integer $a, b, c, d$ for prime number $p$ which satisfies $p|n^4-1-(n^2+an+b)(n^2+cn+d)$ for all $n \in \mathbb{N}$.
Prove that there is such integer $a, b, c, d$ for prime number $p$ which satisfies $p|\Big(n^4-1-(n^2+an+b)(n^2+cn+d)\Big)$ for all $n \in \mathbb{N}$.
My attempt:
\begin{align... | Since $n^4 - 1 = (n^2 - 1)(n^2 + 1)$, this suggests using $b = -1$ and $d = 1$. Along with $a = ep$ and $c = gp$ for any integers $e$ and $g$, so those terms are congruent to $0$ modulo $p$, this gives
$$\begin{equation}\begin{aligned}
f(n) & = n^4 - 1 - (n^2 + an + b)(n^2 + cn + d) \\
& = n^4 - 1 - ((n^2 - 1) + epn)((... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By con... | If $\cos(5\theta) = 0$, then $z = e^{i\theta}$, must satisfy $z^5 = \pm i$. Note that the two possibilities correspond to conjugates of each other, which won't change the real part, so let's just assume $z^5 = i$. The roots of $z^5 = i$ are:
$$e^{i\pi/10}, e^{5i\pi/10} \color{red}{( = i)}, e^{9\pi i/10}, e^{13\pi i/10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from bo... | Trick question: there will not be any real-number solutions. If the original equation, $ \ x + 1 + \sqrt{4x + 1} \ = \ 0 \ $ is considered as a curve intersection equation, we would write $ \ \sqrt{4x + 1} \ = \ -x - 1 \ \ . $ The left side of the equation represents a square-root curve which resembles the "upp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using su... | Your recurrence formula has a little error on the first line in the 2nd term on the right should be $n-2$ not $n-3$ and the result should read:
$$
I_n=\frac{n-2}3 I_{n-3}
$$
Now we get:
$$
I_{11}=\frac93\cdot\frac63\cdot\frac33 I_2=2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What function does $\sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}\frac{x^n}{y^{2n}}$ correspond to? I have the following Taylor series, and although it looks familiar, I cannot figure out which function it corresponds to! Does anyone recognize this Taylor series?
$$1+\frac{x}{2y^2} - \frac{x^2}{8y^4} + \frac{x^3... | A good series to know offhand is
$$\frac{1}{\sqrt{1-4t}} = \sum_{n=0}^\infty {2n \choose n}t^n $$
which almost gets us the series we desire with $t = -\frac{x}{4y^2}$
$$\frac{1}{\sqrt{1+\frac{x}{4y^2}}} = \sum_{n=0}^\infty {2n \choose n}\left(-\frac{x}{4y^2}\right)^n $$
except for the offending term $1-2n$ in the denom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed expression for a continued fraction Does anyone know a closed expression for the following continued fraction?
$$G(x) = \cfrac{1}{x+1+\cfrac{1}{x+3+\cfrac{4}{x+5+\cfrac{9}{x+7+\cfrac{16}{x+9+\cdots}}}}}$$
All I know is that $G(0) = \frac{\pi}{4}$ and $G(x)$ converges for $x \geq 0$, while also $G(x) \sim \frac{1... | The form of the coefficients of the proposed continued fraction is rather similar to that of the ratio of two hypergeometric functions tabulated here:
\begin{equation}
\frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a+1,b+1;c+1;z\right)}={x
_{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_{2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Calculating a limit or proving it does not exist I have the following function:
$$\frac{\sqrt[3]{x+3}(\sqrt{x^2+1}-\sqrt{x^2+3x+4})}{\sqrt[6]{x^2+5}}$$
The limit definition we are using is
to $-\infty$ such that $L = \lim_{x\to -\infty}f(x)$ iff
for every $\epsilon > 0$ there exists a $M<0$ such that for every $x<M$ w... | You have\begin{multline}\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(\sqrt{x^2+1}-\sqrt{x^2+3x+4}\right)}{\sqrt[6]{x^2+5}}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(-3x-3\right)}{\sqrt[6]{x^2+5}\left(\sqrt{x^2+1}+\sqrt{x^2+3x+4}\right)}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}\lim_{x\to-\infty}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $\int_0^\infty \frac{\sin^2 x-x\sin x}{x^3} \, dx= \frac{1}{2} - \ln 2$
Prove that
$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$
Integration by parts gives
\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl(... | Alternatively, we could use the fact that $$\int_{0}^{\infty} t^{2} e^{-xt} \, \, \mathrm dt = \frac{2}{x^{3}}, \quad x >0,$$ and then apply Fubini's theorem.$$\begin{align} \int_{0}^{\infty} \frac{\sin^{2}x - x \sin x}{x^{3}} \, \mathrm dx &= \frac{1}{2}\int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) \int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
What's the measure of the radius of the circle below? For reference: In the figure, calculate $R$. If : $\overset{\LARGE{\frown}}{AB}= 120°$, $CM = 1$ and $MB = 2$
(Answer: $\frac{3}{4}(\sqrt{3}+1$))
My progress:
Draw OP
Th.Chords:
$CM.MP = BM.AM \implies 1.[(R-1)+(R)] = 2.AM\\
\boxed{2AM = 2R-1}: \boxed{R=AM+ \frac{1... | $\angle AOB = 120^o$
Law of cosines:
$AB^2 = 2R^2 + 2R^2 \cdot \frac 12 = 3R^2 \implies AB = R \sqrt3$
Power point M:
$1 \cdot (2R-1) = 2 \cdot (R \sqrt3 - 2) \iff 2R -1 = 2R\sqrt3 - 4 \iff R = \frac3{2(\sqrt3-1)} = \frac{3(\sqrt3+1)}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Playing tennis until one of the players wins 3 times (Binomial distribution) Question
Nadal and Federer playing tennis against each other.
Nadal's probability to win one match is $\frac{2}{3}$, independently from the previous results.
The two are playing until one of them wins 3 matches.
What is the probability for Nad... | $ \displaystyle {5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2} ~ $ considers that there are five matches and Nadal wins exactly three of the five matches.
The book's solution considers that Nadal may win three or more of the five matches which is the correct solution. On your point about matches ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x})$ for $x > 0$
Question: Let $x> 0$, show that
$$\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x}).$$
I tried the integration by parts, but somehow couldn't get the term involving natural logarithm of $x$. I am looking f... | $$I=\int_{0}^{\infty}\frac{e^{-xt}}{1+t}\,dt=e^x \,\Gamma (0,x)$$
If $x$ is large, we have
$$I=\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3}-\frac{6}{x^4}+O\left(\frac{1}{x^5}\right)$$ while
$$J=\frac{1}{2} \log \left(1+\frac{2}{x}\right)=\frac{1}{x}-\frac{1}{x^2}+\frac{4}{3 x^3}-\frac{2}{x^4}+O\left(\frac{1}{x^5}\right)$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.