Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped? The problem is as follows:
The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.
Prove this determinant
$\left|\begin{matrix}
1 & 1 & ... | $$\begin{align}
V &= bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2\\
&=bc^2 \color{fuchsia}{- bac} - cb^2 + ab^2 - ac^2 + ca^2 \color{fuchsia}{+ abc} - ba^2\\
&=b(c^2-ac-bc+ab)-a(c^2-ac+bc+ab)\\
&=(b-a)(c^2-ac-bc+ab)\\
&=(b-a)(c-b)(c-a)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, i... | I wrote my answer in spoiler blocks so you can decide when to see it.
It is defined that for $t = 3$, $l = a_{3}$.
For $t = 0$, $l = 0$. The remaining length should be $a_{3}$.
For $t = 1$, $l = \frac{a_{3}}{2}$. The remaining length should be $\frac{a_{3}}{2}$.
For $t = 2$, $l = \frac{a_{3}}{2} + \frac{a_{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Multiplication of matrices of the given form is a group. \begin{equation*}
A_{2,2} =
\begin{pmatrix}
a & b \\
0 & 1 \\
\end{pmatrix}
\end{equation*} for $a,b \in $ $\mathbf{Q}$. Find $Z(A)$ and all elements and subgroups of finite order in $A$, provided that $A$ is a group under multiplication of matrices.
My attempt ... | Note that the identity matrix is the identity element in this group. An element is of finite order if multiplying it finitely many times gives us the identity. Now note that
\begin{equation}
\begin{pmatrix}a&b\\0&1\end{pmatrix}\cdot\begin{pmatrix}a&b\\0&1\end{pmatrix} = \begin{pmatrix}a^2&b(a+1)\\0&1\end{pmatrix}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\frac{2\sqrt n + \frac{1}{\sqrt n}-3}{2\sqrt n -1}$ Consider the term $$\dfrac{2\sqrt n + \dfrac{1}{\sqrt n}-3}{2\sqrt n -1} \tag1$$
I simplified it to
$$\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1} \tag2$$
or also
$$\dfrac{2 n+1-3 \sqrt n}{2 n-\sqrt{n}} \tag3$$
but apparently one can simplify this ... | You were almost there:
I simplified it to $\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1}$
From there, the key is to spot the $4n-1$ in the numerator and pull it out:
$$\begin{align}
\ldots &= \frac{4n-1}{4n-1} + \frac{-4\sqrt n + \frac1{\sqrt n}}{4n-1}\\
&= 1 + \frac1{\sqrt n}\frac{-4n+1}{4n-1}\\
&= 1 - \frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the number of solutions in nonnegative integers of the equation $x_1+x_2+x_3+3x_4=7$. Find the number of solutions in nonnegative integers of the equation $x_1+x_2+x_3+3x_4=7$.
The term $3x_4$ will only be equal to $0,3$, and $6$. So $x_4\in\{0,1,2\}$. By letting $3x_4=0,3$, and $6$ we will have three cases and th... | Lets use generating functions for it such that :
$x_1 = \frac {1}{1-x}$ ,$x_2 = \frac {1}{1-x}$, $x_3 = \frac {1}{1-x}$ , $x_4 =\sum_{0}^{\infty} {x}^{3m}$
So , the result of multiplication is ${\frac {1}{1-x}}^3 \times x^{3m} = C(3+k-1,k)\times x^k \times x^{3m}$ ,because the coefficient of $x^{3m}$ is always $1$
We w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets... | We can count the total possible unique values of $4x-z$ by summing the size of the following 5 mutually disjoint sets of values:
*
*Non-positive values: corresponds to $x=0$, there are $21$ such values of $4x-z$ from $-20$ to $0$, corresponding to $z=0,1,2,\ldots , 20$.
*Positive values of form $\equiv 0 \pmod 4$: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Finding the solutions of an equation in positive intgers We are asked to find all positive integer solutions of the equation
$$x^7+7=y^2$$
or to show that it does not have any solutions. It is clear that $x$ can not be even. So $x$ is odd and $y$ is even (In fact $x$ is of the form $4k+1$). Adding $121$ to both sides ... | I’m turning my comment into an answer.
Let $x$ be an integer such that $11|x^7+128$. Then clearly $11$ and $x$ are coprime and $11|(x^7)^3+128^3=x^{21}+2^{21}$. By (Pierre) Fermat’s little theorem, $11|x^{21}-x$, $11|2^{21}-2$ and thus $11|x+2$. Then write $x=11y-2$, then $\frac{x^7+128}{x+2}=\frac{(11y-2)^7+2^7}{11y}=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $a^{b} +1\geq b(a+1)$ $a, b \in \mathbb{N} $ and $a>1$,$b>2$
Prove that $a^{b} +1\geq b(a+1)$
My attempt :
suppose $a=p+1$, ($ p \in \mathbb{N}$,$ p\geq 2$)
Remarks :(if $ p=1$ hence $ a=2$,in this case isn't difficult)
So let's show $(p+1)^b +1\geq bp+2b$
We know that $(p+1)^b=\sum_{k=0}^{b}\left( {\begin{a... | We can prove this by induction on $b$ when $b\gt2 ,a\gt 1$.
Step 1:
$b=3$, $a^3+1\geq 3a+3$ is easy to prove, note that $a(a^2-3)\geq2$ since $a \geq2$.
Step2:
Now suppose we have $a^b+1\geq ab+b$.
$a^{b+1}+1=a(a^b+1)-a+1$
$ \geq a(ab+b)-a+1=
ab(a+1)-a+1\geq2b(a+1)-a+1$,
and since $b(a+1)-a+1 \gt 2(a+1)-a+1 \gt a+1$,we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Get $f(x)=u_x\frac{x}{u}$ from ODE for $u$ Consider the Cauchy-Euler ODE
\begin{align*}
\frac{1}{2}x^2u_{xx}+xu_x-u=0.
\end{align*}
Guessing $u(x)=Cx^n$ gives
\begin{align*}
\frac{1}{2}n(n-1)Cx^n+nCx^n-Cx^n=0,
\end{align*}
which we can solve to get
\begin{align*}
n_1&=-2,\\
n_2&=1.
\end{align*}
Given initial conditions... | Your claim that $f$ is a constant for any $u$ which satisfies the given equation is false. Indeed,
$$u(x) = \frac{x^3 + 1}{x^2}$$
certainly satisfies the equation, but using that $u$ we have
$$ f(x) = \frac{x^3-2}{x^3+1} $$
Now, it's certainly true that some $u$ which satisfy the original equation make $f$ constant. In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4103216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$f_{n}$ uniformly converges to $f$ and $|f'_{n} (x)| ≤ C $, is $f$ necessarily differentiable? Assume $f(x)$ is the uniform limit of real differentiable functions $f_{n}(x)$ on $[−1,1]$. Assume that $|f'_{n} (x)| ≤ C $ for some $C$ independent of $n$ and $x ∈ [−1, 1]$. Is the function $f(x)$ necessarily differentiable?... | No. Choose $f_n(x) := \sqrt{x^2 + \frac{1}{n}}$. Then, for all $x \in [-1, 1]$:
$$
\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert = \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \leq \lvert x \rvert + \frac{1}{\sqrt{n}} - \lvert x \rvert = \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longright... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the limit of following sequence $\lim\limits_{n \to inf}\frac{n×1^r+(n-1)×2^r+...+1×n^r}{n^{r+2}}$
Ok so the only idea I got was to use Stholz theorem and then use $1^r+2^r+...+n^r$ ~ $
\frac{1}{(r+1)} (n^{r+1})$ somehow but that didn't lead anywhere (or at least I didn't see it)
| Igor Rivin's idea is great, but the limit is not $0$. We need to assume that $r\neq -1$ and $r\neq -2$. The general term of the given sequence is
\begin{align*}
S_n(r) & = \frac{1}{n^{r+2}} \sum_{k=1}^{n} (n-k+1)k^r \\
& = \frac{n+1}{n^{r+2}} \sum_{k=1}^{n} k^r - \frac{1}{n^{r+2}} \sum_{k=1}^{n} k^{r+1} \\
& =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all pair of primes $(p,q)$ such that both $p^2+q^3$ and $p^3+q^2$ are perfect squares. Let
$p^2+q^3=a^2$ and $p^3+q^2=b^2$. Let's suppose $ p \neq q$. When one of $p,q$ equals $2$, it yields system of equations with no solution, so $p,q \geq 3$.
Since any two primes numbers are coprime, then all $a,b,p,q$ are cop... | Your proof is way more complex than it needs to be. No need to first treat $2$ as a special case. No need to assume $p\ne q$ and arrive at contradictions.
$p^3+q^2=a^2 \Rightarrow (a-q)(a+q)=p^3$. Plainly $(a-q)\text{ and }(a+q)$ are factors of $p^3$ and multiply to $p^3$. Since $p$ is prime, the only factors of $p^3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
At $n=4$, $$5^4<4^5$$ which is indeed true.
By mathematical induction, we need to prove that $$(n+2)^{n+1} < (n+1)^{n+2}$$
$$\implies (n+2)^n\times (n+2) < (n+1)^n\times(n+1)^2 $$
I am not getting how to proceed further than th... | Using binomial expansion, we can write
\begin{align*}
\text{LHS} &= (n+1)^n\\
&= (n+1)^{n-1}\times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdots + {n-1 \choose n-2}n + {n-1 \choose n-1}\right] \times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Prove that $\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor=\lfloor nx\rfloor$ Let $x,y\in\Bbb R$ and $n\in\Bbb N$. Prove that, $$\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor=\lfloor nx\rfloor.$$
Try:
We have to $x<\lfloor x\rfloor+1$. By adding $k/n$ to $x$, could be that $x+(k/n)$ keep being... | Let's write $x = \lfloor x \rfloor + \{x \}$, there exists $u \in \Bbb N, u \in \{ 0,...,(n-1)\}$ such that $$ \frac{u}{n}\le \{x \} < \frac{u+1}{n}$$
We have
\begin{align}
\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor &= \sum_{k=0}^{n-1}\left \lfloor \lfloor x \rfloor +\{x \}+\dfrac{k}{n}\right \rfloor \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respective... | You can start from
$$
1+(a+b)^2=b^2+16 \\
\iff a^2+2ab+b^2+1=b^2+16 \\
\iff a^2+2ab-15=0 \\
\iff b= \frac{15-a^2}{2a}
$$
Now substitute in another equality
$$
a^2+9=b^2+16 \\
\iff a^2+9=\left(\frac{15-a^2}{2a}\right)^2+16 \\
\iff a^2+9=\frac{15^2-30a^2+a^4}{4a^2}+16 \\
\iff 3a^4+2a^2-225 = 0 \\
$$
Solve the quadratic f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115634",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometr... | $$\sin^2 A = 1 - \sin^4 A$$ $$ \sin^2A= (1 - \sin^2 A)(1 + \sin^2 A) $$ $$\sin^2 A= (\cos^2 A) (1 + \sin^2 A)$$ $$\therefore \tan^2 A = 1 + \sin^2 A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding minima from simultaneous equations We are given that a point $(x,y,z)$ in $\mathbb{R}^3$ satisfies the following equations
$x\cos\alpha-y\sin\alpha+z =1+\cos\beta$
$x\sin\alpha+y\cos\alpha+z =1-\sin\beta$
$x\cos(\alpha+\beta)-y\sin(\alpha+\beta)+z=2$
Where $\alpha,\beta\in\mathbb(0,2\pi)$
We need to find the Mi... | When $\alpha = \beta = \pi$ and $x = 1, y = 0, z = 1$, the system of equations is met. So there exists $(x, y, z)$, which is a solution of the system of equations for some $\alpha, \beta \in (0, 2\pi)$, such that $x^2 + y^2 + z^2 = 2$.
On the other hand, by Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximizing $x^2y$ given $x^2+y^2=100$, without using the AM-GM inequality and calculus tools Problem says:
Let $x^2+y^2=100$, where $x,y>0$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
I know these possible tools:
*
*AM-GM inequality
*Calculus tools
Here, I want to escape from all of the ... | Your approach is good and the easiest way without calculus.
An Approach that Exposes the Core Ideas
The following approach is a slight modification of yours that simplifies the algebra using Vieta's formulas.
If
$$
x^2+y^2=100\tag1
$$
then
$$
\begin{align}
x^2y
&=100\cos^2(\theta)\cdot10\sin(\theta)\tag{2a}\\
&=1000\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}$. My objective is to prove that:
$$\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}\text{ , where $$z is a co... | If you are unable to continue, you can look at the following steps:
$\dfrac{1}{2}-\dfrac{\cos\big((n+1)\theta\big)}{2}+\dfrac{\sin\theta\sin\big((n+1)\theta\big)}{2-2\cos\theta}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)(1-\cos\theta)+\sin\theta\sin\big((n+1)\theta\big)}{2(1-\cos\theta)}=$
$=\!\dfrac{1}{2}\!+\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | There is a connection to $\zeta'$ at positive even values
Using partial fractions, we have
$$\frac{1}{k^{2n}(k^2-a)}=\frac{1}{a^n(k^2-a)}-\sum_{j=1}^{n}\frac{1}{a^{n-j+1}k^{2j}},$$
so that
$$S_r(n,a)=\sum_{k\ge r}\frac{\ln k}{k^{2n}(k^2-a)}=\frac{1}{a^n}\left(\sum_{k\ge r}\frac{\ln k}{k^2-a}-\sum_{j=1}^{n}a^{j-1}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 4
} |
The ODE $y''(x)=\sinh(x)-3y'(x)-2y(x)$ I am trying to solve the differential equation that is in the title as a System of first order ode.
My Approach:
$\frac{d}{dx} \left(\begin{array}{c} y \\ y' \end{array}\right)=\left(\begin{array}{c} y' \\ \sinh(x)-3y'-2y \end{array}\right)=$
$\left( \begin{array}{rrr}
0 & 1 \\
-... | Rewrite equation as
$$y''+3y'+2y=\frac{e^x}{2}-\frac{e^{-x}}{2}$$
Characteristic equation is
$$\lambda^2+3\lambda + 2=0$$
Then $\lambda_1=-2,\lambda_2=-1$. Solution of $y''+3y'+2y=0$ is
$$y_h=c_1e^{-2x}+c_2e^{-x}$$
For particular solution use method of undetermined coefficients.
$$y_p=Ae^x+Bxe^{-x}$$
We get
$$6Ae^x+Be^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Minimizing $a_1x_1^2 + a_2x_2^2$ for positive $a_i$, where $a_1x_1+a_2x_2=B$
Find $$\min\{a_1x_1^2 + a_2x_2^2\}$$
Where $ a_1x_1 + a_2x_2 = B$, and $a_1>0$ and $a_2>0 $. Find $x_1$ and $x_2$.
Can we do it usig AG mean inequality?
Let's say we have the problem to find the minimum value of $ x_1^2 + x_2^2 $.
From: $ (x... | Going along your lines, we can write
\begin{align}
&a_1a_2(x_1-x_2)^2\ge0\\
\iff&a_1a_2(x_1^2+x_2^2)\ge 2a_1a_2x_1x_2\\
\iff&a_1^2x_1^2+a_2^2x_2^2+a_1a_2(x_1^2+x_2^2)\ge a_1^2x_1^2+a_2^2x_2^2+2a_1x_1a_2x_2\\
\iff&(a_1x_1^2+a_2x_2^2)(a_1+a_2)\ge(a_1x_1+a_2x_2)^2\\
\iff&a_1x_1^2+a_2x_2^2\ge\frac{B^2}{a_1+a_2}
\end{align}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Approximate $\log_{10}$ values without a calculator I've got this problem:
$1,000,000^{{1,000,000}^{1,000,000}} < n^{n^{n^n}}$
What is the first positive integer value of n for which this inequality holds?
I managed to reduce it to this:
$6+\log_{10}(6) < n\log_{10}(n)$
by using $\log_{10}$ three times (and discard... | Here is a solution with limited hand multiplications/additions. Starting from your condition that $6+\log_{10} 6< n\log_{10} n$, notice that
$$
10<6^2<100
\rightarrow 1< 2\log_{10} 6 < 2
\rightarrow 1/2 < \log_{10} 6 < 1
\longrightarrow \fbox{$6.5 < 6+\log_{10} 6 < 7$}.
$$
Also, $6^6=46,656$ (by hand!) therefore
$$
10^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Differentiate $x^{a^x}$ without logarithmic differentiation? Problem. Compute $\frac{d}{dx} x^{a^x}$.
Method 1: Logarithmic Differentiation (Correct):
\begin{align*}
y&= x^{a^x} \\
\ln(y)&=a^x \cdot \ln(x) \\
\frac{1}{y} \frac{dy}{dx} &=a^x \frac{1}{x} + \ln(x) \ln(a) a^x \\
\frac{dy}{dx} &= y\left(\frac{1}{x} a^x + \l... | As suggested, one can use the answer in Why the chain rule does not work for this question? to solve this problem.
\begin{align*}
\frac{d}{dx} x^{a^x}
&= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \frac{d}{dx} a^x \\
&= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \ln(a) \cdot a^x
\end{align*}
This is a spec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$ As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3.
My attempt: since $H_{n-1}=\sum_{k=1}^{n-1}\frac{1}... | Your bound is much tighter for sufficiently large $n$. Define $$f(n) = \frac{n^2 \log n}{8}, \\ g(n) = \frac{n^2 \log n}{2} - \frac{3n(n-1)}{4}.$$ Then the ratio $$\frac{g(n)}{f(n)} = 4 - \frac{6(n-1)}{n \log n}.$$ For $n \ge 5$, $\log n > 1.6$, hence $\frac{6(n-1)}{n \log n} < 3$, thus $g > f$.
In the case where $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to manipulate the Algebra in the parenthesis of this question? Consider the series $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ....$
Since the pattern of signs repeats every 3 terms, it is natural to consider only the partial sums that include... | $$s_{3n} = \sum_{k=1}^{n} \left(\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k}\right)\geqslant \sum_{k=1}^{n}\frac{1}{3k - 2}$$
as last diverged, then it have limit $+\infty$, so have $s_{3n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we use Vieta's formula in solving Trigonometric equations? **The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$
is ...
My Approach
I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\... | Let $\alpha = \dfrac{(2n+1)\pi}{11}$, which is the form of angles we are interested. Then $6\alpha = (2n+1)\pi - 5\alpha$ which implies $\cos 6\alpha +\cos 5\alpha = 0$. Now, expressing this in terms of $\cos \alpha$, we observe
$$32\cos^6\alpha + 16\cos^5\alpha- 48\cos^4\alpha - 20\cos^3\alpha + 18\cos^2\alpha + 5\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The area of $\triangle AMN$ where $MN$ is midsegment $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If the area of $\triangle ABC$ is $24$ $cm^2,$ what is the area of $\triangle AMN$?
We can say for sure that $MN\parallel BC$ and $MN=\dfrac12 BC$. I don't know what to do next. Any help would be appreciat... | We know that $$S_\triangle = \frac{1}{2}ab\sin\alpha$$
where $a$ and $b$ are sides and $\alpha$ the angle between them. Using this, we have
$$\frac{S_{ABC}}{S_{AMN}} = \frac{\frac{1}{2}\cdot \color{red}{AB}\cdot \color{blue}{AC}\cdot \sin\alpha}{\frac{1}{2}\cdot \color{red}{AM}\cdot \color{blue}{AN} \cdot \sin \alpha} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.... | Bring all terms to LHS$$\frac{a^3}b-\frac{a^3}a+\frac{b^3}a-\frac{b^3}b=(a^3-b^3)(1/b-1/a)$$
Now if $a\ge b$, both factors are $\ge0$. When $a<b$, both are negative. Can you see when the equality holds?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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solve gcd $n^3+3n^2-5$ and $n+2$ with bezout Hello I have to solve $\gcd (n^3+3n^2-5 , n+2)$ with bezout
Here’s how I did it:
bezout says $a$ and $b$ are co prime if and only if $au+bv = 1$.
Then I did :
$(n+2)(n^2+n-2) -(n^3+3n^2-5)$
I found one so their $\gcd$ is equal to $1$.
Did I get the right method ?
Sorry for ... | $$A(n^3+3n^2-5) + B(n+2) = 1$$
This isn't a blueprint for a general method of finding a solution. I am going to take advantage of the fact that the second polynomial is $n+2$, a first degree polynomial.
Let $m = n+2$. Then
$$n^3+3n^2-5 = (m-2)^3+3(m-2)^2-5 = m^3 - 3m^2 -1$$
So we need to solve
$$A(m^3 - 3m^2 -1) + B(m)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right... | For positive integers, we have the well known binomial formula
$$(1 + x)^n = \sum_{k\geq 0} \binom{n}{k} x^k = \sum_{k=0}^n \binom{n}{k} x^k$$
where
$$\binom{n}{k} = \frac{(n)_k}{k!} = \frac{n(n - 1)\cdots(n - k + 1)}{k!}$$
$(n)_k$ is the Pochhammer symbol. It is not difficult to show that if we replace $n$ by an arbit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find steps to solve : $\int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du $ $$ \int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du\quad\text{with}\quad x \in ]-1,1[ $$
$$ \int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du = \arctan \left( \frac{x \sin t}{1- x \cos t} \right) $$
I have this integ... | I'm not sure this helps but ...
\begin{align}\mathcal{I}&=\displaystyle\int \frac{x\cos u-x^2}{(-2x)\cos u+(x^2+1)}\mathrm du\\&=\underbrace{\displaystyle\int \frac{x\cos u}{-2x\cos u+(x^2+1)}\mathrm du}_{\mathcal{I_1}}-x^2\overbrace{\displaystyle\int \frac{\mathrm du}{-2x\cos u+(x^2+1)}}^{\mathcal{I_2}}\end{align}
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$f(a)=b,f(b)=c,f(c)=a$, find $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Let $f(x)=x^2-x-2$. It is given that $a, b,c \in \mathbb{R}$ such that
$$f(a)=b,f(b)=c,f(c)=a$$ Then find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
My attempt:
We have:
$$\begin{array}{l}
a^{2}-a-2=b \\
b^{2}-b-2=c \\
c^{2}-c-2=a
\end{arra... | The following uses the symmetry of the problem and "lucky" cancellations of terms to get the result with somewhat less brute work.
The case $a=b=c$ is obvious and will be left out.
Otherwise, with $u=a+b+c, v=ab+bc+ca, w=abc$, the OP showed that:
$$
uv - w + 1 = 0 \tag{1}
$$
A second relation can be derived by summing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | We present a proof that utilises a special case of the well known Bernoulli's Inequality.
Lemma:
$\forall \ x \geq 0, (1+x)^n \geq 1+nx, n\in \mathbb{N}.$
(This can easily be proven by applying the Binomial Theorem)
Now,
\begin{align}
&\left(\dfrac{2N}{2N+1}\right)^\dfrac{2N+1}{2N+2} \cdot \left(\dfrac{2}{1}\right)^\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Find all ordered pairs of integers$(x,y)$ which satisfy the equation $2(x^2+y^2)+x+y=5xy$
Find all ordered pairs of integers$(x,y)$ which satisfy the equation $2(x^2+y^2)+x+y=5xy$
We can transform this equation as
$$\begin{equation}
2(x^2+y^2-2xy)+x+y-xy=0\\
\implies2(x-y)^2+x+y-xy=0\\
\implies(xy-x-y)\ge0\end{equati... | First, let's expand $$2(x^2+y^2)+x+y=5xy$$ We get : $$2x^2+2y^2+x+y=5xy$$ Then, subtract $5xy$ from both sides and we get $$2x^2+2y^2+x+y-5xy=0$$ On factoring it we get : $$2x^2+(1-5y)x+2y^2+y=0$$ Now this is a quadratic equation with $a=2, b=1-5y, c=2y^2+y$ So we get :$$x=\frac{-(1-5y)\pm \sqrt{(1-5y)^2-4 \times 2(2y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do you find the shortest distance from the point $\left(-1,7\right)$ to a given $\mbox{curve}\ ?$ I find this difficult to solve. It involves maxima and minima of differential calculus.
Since it's requiring the shortest distance, I have to find $y'$ and set it to zero.
Here's how I did it, but I still can't seem to... | The ellipse equation is
$ Q_1(x,y) = 9 x^2 + 25 y^2 - 18 x + 100 y -116 = 0 $
The objective function (to be minimized) is the square of the distance between $(x,y)$ and $(-1, 7)$
$ f(x,y) = Q_2(x,y) = (x + 1)^2 + (y - 7)^2 $
So this is a quadratic optimization subject to a quadratic constraint problem.
Using Lagrange m... | {
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"source": "stackexchange",
"question_score": "2",
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Find a linear transformation $T\begin{pmatrix}-1\\-2\end{pmatrix}$ $T:\mathbb{R}^2 \Rightarrow\mathbb{R}^3, T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\4\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}4\\-1\\1\end{pmatrix}$. Find $T\begin{pmatrix}-1\\-2\end{pmatrix}$ and $T\begin{pmatri... | Note also that $$T \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} \implies T= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} I^{-1}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix}.$$
Next $$T\begin{pmatrix} -1 \\ -2 \end{pmatrix}=\begin{pmatrix} -7 \\ -2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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integration by integration by parts: $\cos^2(\frac{\pi}{L}\,x)$ (inner argument) Really simple integrale that can by solved much easier, but I wanna try straight integration by parts:
$\begin{array}{ccc}
&D&I \\
+&\cos(\frac{\pi}{L}\,x)&\cos\frac{\pi}{L}\,x \\ \\
-&-\dfrac{L}{\pi}\,\sin(\frac{\pi}{L}\,x)& \dfrac{L}{\pi... | We have
\begin{align*}
\int \cos^{2}\left(\frac{\pi x}{L}\right) \ dx &= \frac{L}{\pi}\cos\left(\frac{\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right) + \int \sin^{2}\left(\frac{\pi x}{L}\right) \ dx \\
&= \frac{L}{\pi}\cos\left(\frac{\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right) + \int\left(1-\cos^{2}\left(\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.
I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
| Square both sides to get
$$
x^2 = 2(a^2 + 1 - \sqrt{a^4 + a^2 + 1})
$$
We can then use $a^4 + a^2 + 1 > (a^2 + 1/2)^2$ to see that $x^2 < 1$. Now rearrange and square again to get
$$
(x^2 - 2a^2 - 2)^2 = 4(a^4 + a^2 + 1).
$$
Assuming $|x| < 1$, we can solve for $a$ to get
$$
a =x\sqrt{\frac{1- (x/2)^2}{1-x^2}}.
$$
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II The topic of odd perfect numbers likely needs no introduction.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect n... | On OP's request, I am converting my comment into an answer.
*
*$p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m
\\\\&\iff m^2-m-2^rt\lt 0
\\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$
*
*$(7)$ is better than $2m\lt 2^r+t+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction Squares Proof Prove that for every positive integer $n$ there exist positive integers $$a_{11}, a_{21}, a_{22}, a_{31}, a_{32}, a_{33}, \dots ,a_{n1}, a_{n2},\dots,a_{nn}$$
such that
$$
a_{11}^2 = a_{21}^2 + a_{22}^2 = a_{31}^2 + a_{32}^2 + a_{33}^2 = a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2.
$$
We're doing a ... | We proof by induction. You have already found the initial case.
$$5^2 = 3^2 + 4^2$$
We now use the formula
$$\left(\frac{x^2+1}2\right)^2=x^2+\left(\frac{x^2-1}2\right)^2$$ to find a sequence
$$z^2_n=x^2_n+y^2_n$$
such that $x_n=z_{n-1}$.
$$\begin{array}{r,r,r}
5^2&=&4^2&+&3^2\\
13^2&=&12^2&+&5^2\\
85^2&=&84^2&+&13^2\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
| let $a=x-y$ and $b=x+y$
$a^2+b^2=2 \implies x^2+y^2=1 $
substituting $y$ given that $x^2+y^2=1 \implies $
the problem now is to prove that $6x+x^2-y^2 \ge -5 \iff 2x^2+6x-1 \ge -5 $ (1)
$y^2=1-x^2 \ge 0 \implies -1 \ge x \ge 1$
it's enough to see that (1) $\iff (x-1)(2x-4) $
which is easy to check that it's positive wh... | {
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"url": "https://math.stackexchange.com/questions/4171501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Finding $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$ Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:
Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$
From the Taylor theorem we get that there exi... | Your solution is correct, but unnecessarily complicated. Also, there seems to be an error in your expression for $\left\lvert{R_N\left(\tfrac12\right)}\right\rvert,$ which doesn't invalidate your numerical answer, but adds to the complication of finding it.
You're using Taylor's theorem with the Lagrange form of the re... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Compute $\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$ How would you evaluate
$$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$$
Wolfram|Alpha says it equals $\;\pi/2-1/2.\;$ If the solution is complicated, ... I can handle complicated.
| $\displaystyle\sum_{n=1}^∞ \dfrac{\sin(nx)}{n} =\frac{1}{2i} \sum_{n=1}^∞ \dfrac{(e^{ix})^n}{n}-\dfrac{(e^{-ix})^n}{n}=-\frac{1}{2i}({\log(1-e^{ix})-\log(1-e^{-ix})})$
$= \displaystyle \dfrac{1}{2i} \log\left(\frac{(1-e^{-ix})}{(1-e^{ix})}\right)= \dfrac{1}{2i} \log(-e^{-ix})= \dfrac{\log(e^{iπ})-\log(e^{ix})}{2i} =\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge? Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge?
Given that $f^{n+1}(x) = f(f^n(x))$, where $f(x) = \sin x$, each "next" term grows by $f^{n+1}... | Given that, when $0<x<1$ you have $$x>\sin x >x\left(1-\frac{x^2}6\right)$$ you get, when $x=1/n$ for $n\geq 2,$ $$\begin{align}\sin\frac1n&>\frac{n^2-1/6}{n^3}\\&>\frac{n^2-1}{n^3+1}\\&=\frac{n-1}{n^2-n+1}\\&=\dfrac{1}{n+\frac{1}{n-1}}\\&\geq \frac1{n+1}
\end{align}$$
So $\sin(1/n)\geq \frac{1}{n+1}.$
So let $x_n=f^n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{ab... | You could define:
$$F(a,b,c) = (a+b+c)(ab+bc+ca) - abc$$
and check that $F(a, -a,c) = F(a,b,-b) = F(-c,b,c) = 0$ easily. This means that $F$ as a polynomial must contain the linear factors $a+b, b+c$ and $c+a.$ Then, you argue that $F$ must be divisible by all of them at the same time, which generates a cubic polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My a... | This is where things go wrong:
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$
The remainder inside the brackets can have a linear term as well, so you should have written it as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solution set to non-linear equation of $3$ variables I have the following trigonometric equation of $3$ variables:
$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi
)-(\cos (\theta )+3) \sin (\lambda ) \sin
(\phi )$$
$$-\sin (\theta ) \cos (\lambda )+\sin
(\theta ) \cos (\phi )+3 \cos (\thet... | Let $\;x=\tan\dfrac\theta2,\;y=\dfrac{\lambda+\varphi}2,\;z=\dfrac{\lambda-\varphi}2,\;$ then
$$g(x,y,z) = (1-x^2)(2\cos2y+\cos 2z+3)+4x\sin y\sin z\\
+(1+x^2)(-\cos2y+2\cos2z-7)\\
= (-3\cos2y+\cos2z-10)x^2+4x\sin y\sin z+\cos2y+3\cos2z-4\\
= (-6\cos^2y-2\sin^2z-6)x^2+4x\sin y\sin z - 2\sin^2y-6\sin^2z=0,$$
$$-6(1+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\t... | $$x\sqrt{100-x^2}=A$$
$$\begin{align}&\implies \left(\frac Ax\right)^2=100-x^2,~x≠0 \\
&\implies \frac{A^2}{x^2}+x^2=100 \\
&\implies\left(\frac Ax-x\right)^2+2A=100 \\
&\implies \left(\frac Ax-x\right)^2=100-2A\\
&\implies 100-2A≥0\\
&\implies A≤50.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $ab... | Another way,
The condition gives:
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1.$$
Thus, $$1=\sum_{cyc}\frac{1}{1+a}\leq\sum_{cyc}\frac{1}{1+2}=1,$$ which gives $a=b=c=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Does $ a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right) $ imply that $a_n$ is bounded? Let $(a_n)_{n\ge 1}$ be an increasing sequence of positive real numbers and suppose there is a constant $b > 0$ and $\varepsilon > 0$ such that
$$
a_{\lfloor n + \sqrt{n/2}\rfloor} \le a... | Write $f(n) = \bigl\lfloor n + \sqrt{n/2} \bigr\rfloor$ and consider the sequence $(n_k)_{k\geq 0}$ defined by
$$ n_0 \geq 2 \qquad\text{and}\qquad n_{k+1} = f(n_k). $$
Since $f(n) \geq n+1$ for $n \geq 2$, we find that $(n_k)_{k\geq 0}$ is strictly increasing. Moreover, for any sufficiently large $n$, we have
$$
\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving matrix equation with element-wise products I am wondering if there is a way to solve this equation for a:
$$(as^T ⊙ b)n = t.$$
where:
⊙ is element-wise multiplication
a is an unknown v x 1 vector
s is an i x 1 $\vec{1}$ vector
b is a known v x i matrix
n is a known i x 1 vector
t is a v x 1 $\vec{1}$ vec... | Yes. Multiply it out as elements. You will have a system of $v$ uncoupled linear equations in the components of $a$. Then solve each, by itself.
Example with $v=3$, $i=2$:
\begin{align*}
\left( \begin{pmatrix}a_1 \\ a_2 \\ a_3\end{pmatrix} (s_1 \, s_2) \odot \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for variable inside and outside of a log I came across this problem:
$$
0 = \ln(c_1 + c_2 \cdot x) + (c_3 + c_4 \cdot x)^{1/2} + c_5
$$
I simplified it down to:
$$
0 = \ln(c_1 + c_2 \cdot x) + c_3 \cdot x + c_4
$$
Is there a way to solve this analytically?
| You can isolate $x$ only if you make use of special functions (specifically Lambert's product log function $W$):
That may or may not be acceptable to you, depending of what you're trying to do.
For example:
$$
\begin{align}
\ln(5+6x)&=2+3x\\
\ln(5+6x)&=\frac{5+6x}{2}-\frac{1}{2}\\
5+6x&=e^{\frac{5+6x}{2}}\cdot\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For any $ a \in \mathbb{N}$ ; $f(x)=x^4+(4a+2)x^2+1$ is irreducible in $\mathbb{Z}[X]$ & reducible mod $p$ for all $p$. A.1 First we show $f(x)$ is reducible mod $p$ for all $p$.
Taking $y=x^2$ ; $f(x)$ becomes $y^2+(4a+2)y+1$. Solving the equation $y^2+(4a+2)y+1=0$ we get
$$
y=-(2a+1)\pm2\sqrt{a(a+1)}
$$
Now replacing... | The following gives a mild generalization, but the idea is essentially what has been discussed.
Proposition 1. Let $f(x)=x^4+ax^2+b,~a,b\in {\mathbb N},$ where $a^2-4b$ is a non-square, and either (1) $b$ is a non-square, or (2) $b$ is a square and $\pm2\sqrt{b}-a$ is a non-square. Then $f(x)$ is irreducible over ${\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Infinite series of matrix product
If $t$ is real and positive and matrix $A$ and $B$ are such that $A = \begin{pmatrix} \dfrac{t^2+1}{t} & 0 & 0 \\ 0 & t & 0 \\ 0 & 0 & 25 \end{pmatrix}$ and $B=\begin{pmatrix} \dfrac{2t}{t^2+1} & 0 & 0 \\ 0 & \dfrac{3}{t} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}$ and $X=(AB)^{-1} + (... | $$AB = diag(2,3,5)$$
$$(AB)^{-1}=diag\left(\frac12, \frac13, \frac15\right)$$
$$(AB)^{-n}=diag\left(\frac1{2^n}, \frac1{3^n}, \frac1{5^n}\right)$$
$$X=diag\left(\frac{1/2}{1-1/2}, \frac{1/3}{1-1/3},\frac{1/5}{1-1/5} \right)=diag(1, 1/2, 1/4)$$
Hence now, you can compute quantity about $X$ and $Y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do you write $4(x +1)^2 + 1$ in the form $(ax +b)^2 + c\;$? We can write $4x^2+8x+5$ in the form $a(x+b)^2+c$ as $4(x+1)^2+1$. However, the question I am doing asks me to write it in the form $(ax+b)^2+c$. How do I change it to that form?
| You can also solve the equation $(ax+b)^2+c\equiv 4x^2+8x+15$ (the $\equiv$ sign means "true for all values of $x$"—the coefficients on both sides of the equation have to be the same). If we expand the left-hand side, we get
$$
(ax+b)^2+c\equiv a^2x^2+2abx+b^2+c\equiv 4x^2+8x+5 \, .
$$
This means that $a^2=4$ and $2ab=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How we solve $0=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}-y\sin(t)-x\cos(t)$ for real $t$ ($\frac{4\pi}{3}I have a system of equation:
*
*$x^2+y^2=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}$
*$x^2+y^2=y\sin(t)+x\cos(t)$
I want to get an algebraic curve depending only from $x$ and $y$. If it is too complex, it would be already su... | You already isolated a linear equation in $x$ and $y$ for each value $t$:
$$ y \sin \frac{3t}{2} + x \cos \frac{3t}{2} - y \sin t - x \cos t = 0 $$
$$ y = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2} - \sin t} x $$
$$ y = m(t) \, x \quad\mathrm{\ where\ } m(t) = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$ let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$
My Approach:
so I thought, since this is a homogeneous relation, I expanded the expression and wrote the characteristic equation of the same, since that was... | We have $$\frac{a_n}{n} = a_1+a_2+\cdots +a_{n-1}= (a_1+a_2+\cdots +a_{n-2}) + a_{n-1} = \frac{a_{n-1}}{n-1}+a_{n-1} = n\cdot \frac{a_{n-1}}{n-1}$$
which is valid for $n\ge 3$ (notice that it's not valid for $n=2$ because we used $a_{n-2}$).
Then $$\frac{a_n}{n}= n\cdot \frac{a_{n-1}}{n-1}=n(n-1)\cdot \frac{a_{n-2}}{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solutions to $2^a3^b+1=2^c+3^d$
Find all $a,b,c,d$ positive integer such that:
$2^a3^b+1=2^c+3^d$
My progress:
One solution satisfying is $$\boxed{a=1,b=1,c=2,d=1} $$
We first take $\mod 3$ which gives $$ L.H.S\equiv 1\mod 3,~~R.H.S\equiv 2^c\mod 3$$
Hence we get $c$ even. So let $c=2k.$
We get $$2^a3^b-3^d=2^{2k}-1=... | This is not an answer, it is a comment for discussion:
We can construct similar equation as follows:
$$\begin{cases}3^m-2^n=1\\3^r-2^s=1\end{cases}$$
Multiplying both sides we get:
$$3^{m+r}-3^m\cdot 2^s-3^r\cdot 2^n +2^{n+s}=1$$
$$\big(\frac{3^m}{3^r}-\frac{2^n}{2^s}\big)2^s\cdot3^r+1=2^{n+s}+3^{m+r}$$
$$\Rightarrow \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
What is the order of magnitude of $\sum\limits_{n=1}^kn\binom{k}{n}\frac{(2k-2n-1)!!}{(2k-1)!!}$ as $k\to\infty$? What is the order of magnitude of the function $f(k)$ below in the limit as $k \rightarrow \infty$? Does it diverge, converge to a positive limit, or converge to zero?
$$
f(k) = \sum\limits_{n=1}^{k} n \bin... | We can write your $f(k)$ in the form
$$
\sum\limits_{n = 1}^k {\frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}} .
$$
Suppose that $k$ is large and fixed and consider
$$\tag{$*$}
\frac{1}{{(n - 1)!}}\frac{{\Gamma ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove that problem $\frac{\partial ^m f_n(x)}{\partial x^m}$? Let $n,N\in\mathbb{N}$ and $x\in\mathbb{R}$.
Let $f_{n}(x)=(1-\frac{x}{n})^n-(1-\frac{x}{N})^{N}$.
Prove the following:
If ${N}≤n$ then
$$\frac{\partial ^{M}f_{n}}{\partial x^M}(x)=(-1)^{M}\displaystyle\Pi_{k=0}^{M-1}\left(1-\frac{k}{n}\right)\left(1... | We put $f_n(x)=\sigma_1(x)-\sigma_2(x)$ where $\sigma_1(x)=(1-\frac{x}{n})^n$ and $ \sigma_2(x)=(1-\frac{x}{N})^N$
then we calculate $\frac{\partial ^M\sigma_1(x)}{\partial x^M}, \frac{\partial^M\sigma_2(x)}{\partial x^M }$
$\frac{\partial \sigma_1(x)}{\partial x}=(-1)^1(1-\frac{x}{n})^{n-1}$
$\frac{\partial^2\sigma_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$
Show that $$\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$$
My book wrote that
$$=\frac{\Gamma... | You made a small calculation error. In the second expression you wrote $1$ instead of $\frac12$:
$$\beta\left(\frac{\color{red}{\frac{1}{2}+1}}{2},\color{green}{\frac{1}{2}}\right)\beta\left(\frac{-\frac{1}{2}+1}{2},\frac{1}{2}\right)$$
$$\frac{\Gamma\left(\color{red}{\frac{3}{4}}\right)\Gamma\left(\color{green}{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is AM-GM giving 2 contradictory results for the same problem? The problem is the following:
given x and y such that $x+y=1$ and $x,y \geq 0$, find the maximum value of $x^2y$.
The point is, I tried to solve it in 2 different ways, which are shown below, but the second method I used gives a wrong result, and I can't... | By your second way the equality occurs for $x=y=\frac{1}{2}$, which gives $x^2y\leq\frac{1}{8},$ which is indeed the maximal value for $x=y$.
But why does $x=y$?
Maybe for $x\neq y$ we can get a greater value?
And indeed, your first solution does not depend on the assuming $x=y$ and it gives a right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $? Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $ ?
I have tried for the first few integers $\ n,m\ $ up until $\... | You can also use this (relies on Baker/Rhin): $2^n<2^l-3^n<3^n-2^n$, where the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{2\}$ and $l={\lceil n \log_23\rceil}$ is the smallest exponent of $2$ making $2^l-3^n$ positive.
As stated by blamethelag, we can write y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 2
} |
Solve $xy''+y'-xy=0$ using Frobenius method
Using Frobenius method solve $$xy''+y'-xy=0$$
Comparing the given equation with $y''+P(x)y'+Q(x)y=0$ we have,
$$P(x)=\frac{1}{x}\qquad Q(x)=-1$$
Here $x=0$ is a singular point of the given differential equation. Now
$$
\begin{aligned}
\lim_{x\rightarrow 0}(x-0)P(x)&=1\\
\... | The indicial equation has double root $r=0$. So the second solution will involve $\log x$.
$$
y_1(x)=\, \left(1+{\frac{1}{4}}{x}^{2}+{\frac
{1}{64}}{x}^{4}+\dots\right),
\\
y_2(x) =
\log(x) \left(1+{\frac{1}{4}}{x}^{2}+{\frac{1
}{64}}{x}^{4}+\dots \right) +\left(-{\frac{1}{4}}{x}^
{2}-{\frac{3}{128}}{x}^{4}+\dots\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exercise on limit of indeterminate form: $\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt2$ I'm not able to show that
$\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt{2}$
I proceeded as follows:
Substituting 1 to $x$ gives the indeterminate form ... | If you rationalize the denominator of your expression, the limit becomes$$\lim_{x\to1}\frac{\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)}{x-1}.\tag1$$Now define $f(x)=\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)$, and then $(1)=f'(1)=\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyl... | Here is yet another estimation of the $\lim_n\frac{(2n^{\frac 1n}-1)^n}{n^2}$.
Since
$$a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}=\Big(\frac{2n^{1/n}-1}{n^{2/n}} \Big)^n=\Big(\frac{-(1-2n^{1/n}+n^{2/n})+n^{2/n}}{n^{2/n}}\Big)^n=\Big(1-\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\Big)^n$$
we obtain that
$$1\leq a_n\leq \exp\left(n\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Find the volume of the region which lies inside both $x^2+y^2=r^2$ and $y^2+z^2=r^2$ Find the region inside two cylinders $x^2+z^2 \le r^2$ and $y^2+z^2 \le r^2$.
I attempted this question by combining the two inequalities $x^2+2z^2+y^2 \le 2r^2$ and I got the bounds for $z$ which is $-\sqrt{{(2r^2-x^2-y^2)}/{2}} \le ... | If you visualize it carefully, you might be able to see that for fixed $y$ the cross-section is a square. Then:
$$|x| \le \sqrt{r^2 - y^2},$$
$$|z| \le \sqrt{r^2 - y^2}.$$
Then, the cross-section is a square of area $4 (r^2 - y^2)$ and the volume should be:
$$\int_{-r}^r dy 4 (r^2 - y^2) = 16 r^3/3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$ Evaluate
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$$
I transformed it into:
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\ \rig... | Assuming $b^2-a^2>0$,
$$
\begin{align}
&\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\
& =b\sqrt{b^2-a^2}\left[\int_{0}^{\pi} \sin(\theta) \sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\theta+\int_{\pi}^{2\pi} \sin(\theta)\sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a polynomial $W(x)$. Find all pairs of integers $a,b$ that satisfy $W(a)=W(b)$ Given a polynomial $W(x) = x^4-3x^3+5x^2-9x$. Find all pairs of distinct integers $a,b$ that satisfy $W(a)=W(b)$. My approach was to factor the polynomial.
$$\begin{align*}
x^4-3x^3+5x^2-9x &= (x^4-3x^3+2x^2)+(3x^2-9x+6)-6\\
&= x^2(x^... | I would proceed from there as follows :
One can say that if $a\geqslant 3,b\leqslant 0$ and $|b|\color{red}{\geqslant}|a|$, then $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$
So one gets that $a\geqslant 3,b\leqslant 0$ and $|a|>|b|$.
Note that if $a\geqslant 3$ and $b\leqslant 0$, then $|a|\gt |b|$ is equivalent to $a+b\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sum_{i,j=1}^N \frac{1}{(i^2 + j^2)}$ (i..e. $1/r^2$ over the square lattice $i,j\in\{1,2,\ldots,N\}$) Does anyone have any insight as to how I might find a closed-form expression for
$\sum_{i=1}^N \sum_{j=1}^N\frac{1}{(i^2 + j^2)}$ ?
It feels like it should be straightforward because it's easy to do as an integral in... | An asymptotic formula for large $N$ may be derived as follows. First,
$$
\sum\limits_{m = 1}^N {\frac{1}{{m^2 + n^2 }}} = - \frac{1}{{2n^2 }} + \frac{\pi }{{2n}}\coth (\pi n) - \frac{{\operatorname{Im} \psi (1 + N + \mathrm{i}n)}}{n}
$$
where $\psi$ is the digamma function. From the known asymptotics of the digamma ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$ If $f\colon\left[ { - 1,1} \right] \to \mathbb R$ be continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$, then $\mathop {\lim }\limits_{x \to 0} \frac{{f\le... | First we note that $f$ is odd. The left hand side $f(2x^2-1)$ is even, and the factor $x^3+x$ in the right hand side is odd, so the other factor in the right hand side, $f(x),$ must also be odd. This implies that $f(0)=0.$
Taking $x=\sin\theta$ gives
$$
f(2\sin^2\theta - 1) = \sin\theta(\sin^2\theta+1) f(\sin\theta)
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Show that the following non-linear group of equations only have zero solution
If
$$\left\{\begin{array}{c}
(\lambda_1 + 1)(\lambda_2 + 1) \cdots (\lambda_n + 1) = 1 \\
(\lambda_1^2 + 1)(\lambda_2^2 + 1) \cdots (\lambda_n^2 + 1) = 1 \\
\vdots \\
(\lambda_1^n + 1)(\lambda_2^n + 1) \cdots (\lambda_n^n + 1) = 1
\end{array... | In the case of $n=2$ you have
$$
a+b+ab=0\\
a^2+b^2+a^2b^2=0
$$
The second equation transforms to
$$
0=(a+b)^2-2ab+a^2b^2=2(a^2b^2-ab).
$$
The solution $ab=0$ leads to $a=b=0$. The other solution $ab=1$ gives $a+b=1$, so that $a,b$ are the solutions of the quadratic $x^2-x+1=0$ which has non-zero solutions $\frac{1\pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | Another way to approach it for the sake of curiosity:
\begin{align*}
x^{2} - y^{2} & = (x^{2} - xy) + (xy - y^{2})\\\\
& = x(x-y) + y(x - y)\\\\
& = (x+y)(x-y)
\end{align*}
Hopefully this helps!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 $... | Using right Riemann sums, \begin{align*}
\frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {k - 1} } & = \frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt k } = \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt {\frac{k}{n}} } \\ & = \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What should be the winning strategy for Bob? Alice and Bob are playing a calculator game in which the calculator can only display positive integers and is used like this: starting with the integer $x$ one player types an integer $1<=y<=99$ in and if $y\%$ of $x$ (meaning $\frac{xy}{100}$) is again an integer the calcul... | Let $x = 2^a \cdot 5^b$ and $y = 2^c \cdot 5^d$, each round the calculator does $\frac{xy}{100} = 2^{a+c-2} \cdot 5^{b+d-2}$. The goal is to either reduce both exponents to zero so the opponent can't reach another integer or to reduce them just enough so the opponent can't win in the next move. Essentially there are $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $x=a/b$ and $y=b/c,$ where $a,$ $b$ and $c$ are positives.
Thus, $z=c/a$ and by C-S we obtain: $$\sum_{cyc}\frac{x}{x+2}=\sum_{cyc}\frac{a}{a+2b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}=1$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$0~$ division error occurs as I use $~ t= \tan^{}\left( \frac{\theta_{}}{ 2 } \right) ~$ of the integration. I have to integral the below formula of integration .
$$ a, d \in\mathbb{R}_{>0} ~~\wedge~~ d > a $$
$$ \alpha := \int_{0 }^{\frac{\pi}{2} } \frac{ d }{ d- a \cdot \cos^{}\left(\theta_{} \right) } \,d... | Use the below general equation to proceed the given integration .
$$ a \in \mathbb R_{> 0} $$
$$ \frac{ a }{ a ^{2} + x ^{2} } = \frac{ d }{ dx } \left( \tan^{-1} \left( \frac{ x }{a } \right) \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Wrong answer in Thomas Calculus 14th Edition textbook There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.
My Solution
$$
\theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi
$$
which gets reduc... | Differentiating your formula
$$\theta+\arctan\frac{6}{4-x}+\arctan\frac{3}{x}=\pi$$
with respect to time gives
$$\dot\theta+\frac1{1+\left(\dfrac6{4-x}\right)^2}\cdot\frac6{({4-x})^2}\dot x+\frac1{1+\left(\dfrac3x\right)^2}\cdot\frac{-3}{x^2}\dot x=0.$$
After substituting $\dot x=2$, this may be written
$$\dot\theta=6\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 =... | Another way.
Since $$\prod_{cyc}(a-1)^2=\prod_{cyc}((b-1)(c-1))\geq0,$$ we can assume $(b-1)(c-1)\geq0$, which gives $$a(b-1)(c-1)\geq0$$ or$$abc\geq ab+ac-a$$ and since $$b^2+c^2\geq\frac{1}{2}(b+c)^2,$$ it's enough to prove that
$$a^2+\frac{1}{2}(b+c)^2+\sqrt2(ab+ac-a)+2\sqrt2+3\geq(2+\sqrt2)(a+b+c)$$ or $$(\sqrt2a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral Involving Bessel function and trigonometric function Consider the following integral
\begin{equation}
\int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, I_0(B \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi
\label{1}
\end{equation}
where $A$ and $B$ are two constants and $I_0(.)$ is the modified... | In this paper by VNP Anghel, a result obtained by Glasser is generalized (eq. 32):
\begin{equation}
I_{\frac{m+1}{2}}(b)=\left( \frac{b}{2\pi \sin^m\alpha}\right)^{1/2} \int_0^\pi \exp(b\cos\alpha\cos\theta)I_{\frac m2}(b\sin\alpha\sin\theta)(\sin\theta)^{\frac{m+2}{2}}\,d\theta \tag{1}\label{eq1}
\end{equation}
Then,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Nonlinear ODE and $_2\text{F}_1$ Consider the differential equation $y'(x) = B \, y(x)^{-b} - A \, y(x)^a$ to which WA returns the following:
$$x - c_1 = \frac{y(x)^{b+1}}{B\,(b+1)} \; _2\text{F}_1 \left( 1, \frac{b+1}{a+b}; 1 + \frac{b+1}{a+b}; \frac{A}{B} \, y(x)^{a+b} \right)$$
Is there a way to turn this into an ex... | Firstly, the question should have your ideas on how to solve the problem.
The answer is yes, but it needs a limit since your hypergeometric function is a Lerch Transcendent. Here is the “closed form” solution using the Wolfram Language’s Incomplete Beta function $\text B_z(a,b)$ and Inverse Beta Regularized $\text I^{-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I thi... | When $A=0$, you do not have $A$ in the denominator.
You reached the quadratic formula assuming that $A\neq 0$. Didn't you?
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \fra... | Since $BNGM$ has an inscribed circle, it must follow pitot theorem i.e. the opposite sides must sum to the same value. Let $x,y,d$ be the lengths of $BM, BN, BG$ respectively.
By law of cosines, we have
$$MG=\sqrt{x^2+d^2-xd\sqrt{3}}$$
$$NG=\sqrt{y^2+d^2-yd\sqrt{3}}$$
By pitot theorem,
$$x+\sqrt{y^2+d^2-yd\sqrt{3}}=y+\... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
... | You got already that $$x^2+y^2+z^2=2.$$
Also, we have:
$$\sum_{cyc}x^2y^2=\left(\sum_{cyc}xy\right)^2-2xyz(x+y+z)=1$$ and $$x^2y^2z^2=1.$$
Thus, $$\sum_{cyc}x^8=\left(\sum_{cyc}x^4\right)^2-2\sum_{cyc}x^4y^4=$$
$$=\left(\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2\right)^2-2\left(\left(\sum_{cyc}x^2y^2\right)^2-2x^2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 7
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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... | Further you can write,
$e_1 = \sqrt{\dfrac{a^2-b^2}{a^2}} , e_2 = \sqrt{\dfrac{a^2}{a^2-b^2}}=\dfrac1{e_1} $
Now we have, $e_1+\dfrac1{e_1}$ where $0<e_1<1 $. For the Floor function to achieve minimum value, $e_1$ must be as close as possible to $1$.
For $0<e_1<1$, $~~2<e_1+\dfrac{1}{e_1}<\infty$
So, $\min\left(\Big{\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integer solutions to $2^{^{11}} a + 2^{^{11}} b + ab = 1$ $2^{^{11}} a + 2^{^{11}} b + ab = 1$
By guessing that $a+b = 0$, I was able to find the solutions (a, b) = (-63, 65), (65, -63). Is there any practical way of finding other solutions to the equation?
| Your equation is equivalent to the following
\begin{align*}
2^{22} + 2^{11}a + 2^{11}b +ab = 2^{22}+1 \\
(2^{11} + a)(2^{11} + b) = 2^{22} +1
\end{align*}
Now lets factor $2^{22} +1$ using Sophie Germain identitity
\begin{align*}
2^{22} + 1 =4(2^5)^4 +1^4 = (1 + 2^{11} - 2^6)(1 + 2^{11} + 2^6) \\
2^{22} + 1 = 1985 \cdo... | {
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"url": "https://math.stackexchange.com/questions/4247621",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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... | Using integration by parts:
$\int u\mathrm dv= uv-\int v\mathrm du$
$u=\arcsin {2x} \implies \mathrm du=\frac{2}{\sqrt{1-4x^2}}$
$\,dv=x^2 \mathrm dx \implies v=\frac{x^3}{3}$
$\int x^2 \arcsin{2x}\mathrm dx=\frac{x^3}{3}\arcsin {2x}-\int { \frac{2x^3}{3\sqrt{1-4x^2}}\mathrm dx}=\frac{x^3}{3}\arcsin {2x}+\frac{1}{72}(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with directional derivative
Suppose that $f$ is differentiable on $\mathbb{R}^2$, $\boldsymbol{l}_1$, $\boldsymbol{l}_2$ are given directions, and the intersection angle between them is $\varphi$($0<\varphi<\pi$). Prove that:
$$
\left( \frac{\partial f}{\partial x} \right) ^2+\left( \frac{\partial f}{\parti... | We have
$$\mathrm{RHS} - \mathrm{LHS}
= \frac{2}{\sin^2 \varphi}
\left[A\left( \frac{\partial f}{\partial x} \right) ^2 + B \left( \frac{\partial f}{\partial y} \right) ^2
+ C \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \right]
$$
where
\begin{align*}
A &= \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \... | {
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"url": "https://math.stackexchange.com/questions/4248350",
"timestamp": "2023-03-29T00:00:00",
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Proving $\left\lfloor \sqrt{x^2+3x+3}\right\rfloor = x+1$
Let $$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$... | I am assuming that $x\in\Bbb Z_+$.
Note that $x^2+3x+3=x^2+2x+1+x+2=(x+1)^2+x+2$. So, $x^2+3x+3>(x+1)^2$. Could it be that $x^2+3x+3\geqslant(x+2)^2$? No, because\begin{align}x^2+3x+3\geqslant(x+2)^2&\iff x^2+3x+3\geqslant x^2+4x+4\\&\iff3x+3\geqslant4x+4.\end{align}So$$(x+1)^2<x^2+3x+3<(x+2)^2$$and therefore$$x+1<\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}... | For $c\rightarrow0^+$ and $b\rightarrow+\infty$ we see that our expression is closed to $\frac{1}{3}$.
But $\sum\limits_{cyc}\frac{a}{b+3a}\geq\frac{1}{3}$ it's just $$\sum_{cyc}\left(9a^2b+\frac{53}{3}abc\right)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$... | Here's an elementary way to handle it. First put on the same denimator
$$\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{xy}}$$
then factor the numerator using the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$
$$\sqrt{x^3}+\sqrt{y^3}=(\sqrt{x}+\sqrt{y})(\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2})$$
but
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the follo... | your statement $$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$$
is not corrrect .
Note that $$1.2.3.4.5\equiv 0 \pmod 6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the solution to the differential equation $(1+x^2) \frac{dy}{dx} -2xy =2x$? I am using integrating factor method to solve $(1+x^2) \frac{dy}{dx} -2xy =2x$ but having some issues.
When I divide through by $(1+x^2)$ I get
$ \frac{dy}{dx}-\frac{2x}{(1+x^2)}y=\frac{2x}{1+x^2}$
Then I integrate $\frac{2x}{(1+x^2)}$ ... | As an alternative method, you can rearrange the given ODE to form a separable equation. We are given
$$(1+x^2) y' -2xy =2x$$
Rewrite as
$$y' = \frac{2x}{1+x^2} + \frac{2x}{1+x^2}y=\frac{2x}{1+x^2}(1+y)$$
which is a separable ODE.
$$\frac{dy}{1+y}=\frac{2x\:dx}{1+x^2}$$
Therefore we may integrate to find
$$\ln\left|y+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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FInding the exact value of a series I would like to see if anyone can find the EXACT value of
$$5(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+...)+7(\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+...)+9(\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+...)+...$$
I have tried to regroup the t... | Since all summands are positive, you can rearrange and get
$$
\sum_{n=2}^\infty (2n+1)\sum_{k=n}^\infty \frac1{k^4}=\sum_{n=2}^\infty\frac1{n^4}\sum_{k=2}^n (2k+1)
$$
The coefficient of $\frac1{n^4}$ is $5+7+\dots+(2n+1)=n^2+2n-3$, so by absolute convergence,
\begin{align*}
\sum_{n=2}^\infty\frac{n^2+2n-3}{n^4}
&=\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to solve $\lim _{x\to \infty}\dfrac{x^5}{2^x} $ without L'Hospital's Rule Considering that asymptotically, $2^x$ grows faster than $x^5$ (in the beginning, $x^5$ grows faster than $2^x$, but there will be a point where $2^x$ outgrows $x^5$) then $\dfrac{x^5}{2^x} \rightarrow 0$ as $x \rightarrow \infty$. Therefore,... | As an alternative way, by ratio test
$$\frac{\dfrac{(n+1)^5}{2^{n+1}}}{\dfrac{n^5}{2^n}}=\frac12\left(1+\frac1n\right)^5 \to \frac12 \implies \dfrac{n^5}{2^n} \to 0$$
and since $\forall x>0\quad \exists n$ such that $n\le x\le n+1$ we have
$$\dfrac{x^5}{2^x}\le \dfrac{(n+1)^5}{2^{n}}=2 \dfrac{(n+1)^5}{2^{n+1}} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
Find two complex numbers which satisfy the equation $\frac{3+2i}{z^2}=1$ There is one really obvious solution to the equation which I ignored because it's lazy $$z=\pm \sqrt{3+2i}$$
When trying to answer the question I tried many different approaches but never got an answer. I tried thinking of a general case where $z=... | Your quartic equations are not so ugly, because they are really just quadratic equations in $x^2$ and $y^2$ respectively. So for instance $x^4-3x^2-1$ has solutions
$$x^2=\frac32\pm\frac12\sqrt{13}$$
And $x$ is real, so we can discard the solution $x^2=\frac32-\frac12\sqrt{13}$. Therefore
$$x=\pm\sqrt{\frac32+\frac12\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergent limit $\lim_{n\to+\infty} \left(\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}}\right)$ which seems divergent by graph. Question $$\lim_{n\to+\infty} \left(\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}}\right)$$
My solution $$=\lim_{n\to+\infty} \left(\frac{9^n-2\cdot3^n+1-9^n}{\sqrt{9^n\cdot(1+\frac{n^9}{9^n})}}\right)$$
$$=\li... | Your work is fine, as an alternative way to check the result we can proceed as follows
*
*$(3^n-1)^2-9^n=((3^n-1)+3^n)((3^n-1)-3^n)\sim -2\cdot 3^n$
*$\sqrt{9^n+n^9}\sim \sqrt{9^n}=3^n$
that is
$$\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}} \sim \frac{ -2\cdot 3^n}{3^n} \to -2$$
As noticed, the problem with the graph is a n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Shifted Gaussian I am looking for a closed form expression for the following Gaussian integral, shifted by $N$ parameters $b_i$:
$$
\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N (x-b_i)^2}dx.
$$
The first few $i=1,2,3,\dots$ are easy to compute but I haven’t spotted the pattern yet. Even better than a... | $$\sum_{i=1}^N a_i(x-b_i)^2=\left(\sum_{i=1}^N a_i\right)x^2 -2\left(\sum_{i=1}^N a_ib_i\right)x+\left(\sum_{i=1}^N a_i(b_1)^2\right)$$
$$\begin{cases}
A=\sum_{i=1}^N a_i\\
B=\sum_{i=1}^N a_ib_i\\
C=\sum_{i=1}^N a_i(b_1)^2
\end{cases} $$
$$\sum_{i=1}^N a_i(x-b_i)^2=Ax^2-2Bx+C=A\left(x-\frac{B}{A}\right)^2+C-\frac{B^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve below somewhat symmetric equations Solve $x,y,z$ subject to
$$x^2+y^2 - xy = 3$$
$$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$
$$(x-z)^2+y^2 - y(x-z) = 1$$
$$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$
$x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z... | We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.
\begin{align*}
&[ (x-z)^2+(y-z)^2 - (y-z)(x-z) ]\\
- 4&[ (x-z)^2+y^2 - y(x-z) ]=0 \\
,\space &x = \dfrac{\sqrt{3}\sqrt{4 - y^2} + y}{2}\\
\\
\implies &y = \dfrac{14 - 3 \sqrt{3}}{13},
\space z = \pm\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.