Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to change the value of $a$ in a Taylor series yet still get the correct approximation? Set up
For example, if $f(x) = x^3 + 4x + 6$, then the second degree Taylor polynomial
is
$$
P_2(a) = (a^3 + 4a + 6) + (3a + 4)( x - a) + \left(\frac{6a}{2!}(x - a)^2\right)
$$
If I want to approximate $f(1.2)$ then I can use $... | I think you've gotten a little confused between your $x$ and your $a$. The Taylor series is a function of $x$, parameterised by $a$. So, for example, the 2nd-order Taylor polynomial of your cubic is:
$$T_{2;a}(x) = (a^3 + 4a + 6) + (x - a)(3a^2 + 4) + 6a\frac{(x - a)^2}{2}$$
If we evaluate this series at $a = 0$, we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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if $x+y+z=1$ then find the maximum of the form
Let $x>0$, $y>0$ and $z>0$ such that $x+y+z=1$. Find the maximal value of
$$x\sqrt{y}+y\sqrt{z}$$
I think $x=y=\dfrac{4}{9},z=\dfrac{1}{9}$ then the maximum $\dfrac{4}{9}$,but how to use AM-GM prove it?
| Without AM-GM, the Lagrange multiplier method will do:
$$L=x\sqrt{y}+y\sqrt{z}+t(1-x-y-z).$$
$$\begin{cases} L_x=\sqrt{y}-t=0 \\ L_y=\frac{x}{2\sqrt{y}}+\sqrt{z}-t=0 \\ L_z=\frac{y}{2\sqrt{z}}-t=0 \\ L_t=1-(x+y+z)=0 \end{cases} \Rightarrow x=y=\frac{4}{9}, z=\frac{1}{9}.$$
Note: To solve the system of equations, square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Mu... | Let $s=x+y$ and let $p=xy$. Then you know that $10=x^2+y^2=(x+y)^2-2xy=s^2-2p$ and that $\frac43=\frac1x+\frac1y=\frac sp$. So let us solve the system$$\left\{\begin{array}{l}s^2-2p=10\\p=\frac34s.\end{array}\right.$$Replacing $p$ by $\frac34s$ in the first equation gives $s^2-\frac32s=10$. This equation has two soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate root interval of $x^5+x^4+x^3+x^2+1$ So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$
I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible ... | Let $f(x)=x^5+x^4+x^3+x^2+1$.
Hence, $f'(x)=x(5x^3+4x^2+3x+2)$ and since $(5x^3+4x^2+3x+2)'=15x^2+8x+3>0$,
we see that the polynomial $5x^3+4x^2+3x+2$ has one real root $x_1$ and this root is negative.
Thus, $x_{min}=0$ and $x_{max}=x_1$.
But $f(0)>0$, which say that $f$ has an unique real root and since
$$f(-1.25)f(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Show that the following matrix has $4$ as an eigen value of multiplicity $3$ Show that the following matrix has $4$ as an eigen value of multiplicity $3$
$\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&0&1\\1&1&1&0&3&0\\1&1&1&1&0&4\end{bmatrix}$.
How should I try this?Isn't this equivalent to finding t... | Hint:$A=\begin{bmatrix}
4 & 0 & 0 &0&0&0\\
0&4&0&0&0&0\\
0&0&4&0&0&0\\
0&0&0&3&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&3
\end{bmatrix}+\begin{bmatrix}
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1
\end{bmatrix}$
$$\text{Both matrices commute, eigenvalues for the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I wa... | You're simply finding a common denominator to work with. If you have an expression $\frac{n_1}{d_1}+\frac{n_2}{d_2}+...+\frac{n_k}{d_k}$, then you want the denominator to be $\text{lcm}(d_1,d_2,...,d_k)$.
This now gives for your expression
$$\frac{\frac{n_1}{d_1}\text{lcm}(d_1,d_2,...,d_k)+\frac{n_2}{d_2}\text{lcm}(d_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$, for $a > 0$ I want to show that
$$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$$
for $a > 0$.
I try several methods
*
*Substitutions to rationalize
*The famous $U = \tan(\frac{\theta... |
\begin{align} \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a +
> \sin^2(\theta)} &= \frac{\pi}{2[a(a+1)]^\frac{1}{2}} \end{align}
\begin{align}
I=
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)}
&=
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{(\sqrt{a+1} + \cos\theta)
(\sqrt{a+1} - \cos\theta}
\\
&=\frac1{2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If... | we can use some common sense.
If $n$ is even $16|n^4$. But if $n$ is odd then $2\not \mid n^4$. However $n+1$ and $n-1$ are both even so $2^3|(n-1)(n+1)^2$. But if $n$ is odd then either $n + 1$ or $n -1$ is divisible by $4$ so $2^4|(n-1)(n+1)^2$. So $2^4$ will divide all such numbers.
Can any higher powers of $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Maximum value of $a^3 + b^3 + c^3$ If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers.
My Attempt
writing $a+b=5-c$ and $ \frac{1}{a} +\frac{1}{b} =\frac{1}{5} -\frac{1}{c}$ and after algebraic manipulation I obtain $(a+... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our conditions give $u=\frac{5}{3}$ and $w^3=15v^2$ and since
$$0\leq(a-b)^2(a-c)^2(b-c)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$
we obtain:
$$v^2(3v^2+25)^2\leq0$$ or $v^2\leq0$.
Thus, $$a^3+b^3+c^3=27u^3-27uv^2+3w^3=125+30v^2\leq125... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can one prove $-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$? How can one prove the following?
$$
-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$$
with $H(x)$ the Heaviside function.
I have the impres... | For $y>0$ you'd like to prove that
$$
\frac{\pi}{2}-\arctan\frac{x}{y}=2\arctan\frac{y}{\sqrt{x^2+y^2}+x}
$$
Note that the left-hand side is a number in the interval $[0,\pi]$ and the same for the right-hand side. The identity is true for $x=0$.
Suppose $x>0$, so both sides are in $(0,\pi/2)$. Then take the tangent of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why does rotation preserve angles and distance in the Euclidean plane? An angle in radians is a distance moved around a unit circle. A rotation of a point around a center of rotation, moves the point a distance around a circle around the center that goes through the point. The distance is given by the angle of the rota... | A linear algebra approach:
Any linear transformation of $\mathbb{R}^n$, such as a rotation about the origin, can be represented with an $n \times n$ matrix $A$, where $A\mathbf{v}$ is the output when the transformation acts on a vector $\mathbf{v}$. Any counterclockwise rotation of $\mathbb{R}^2$ about the origin look... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Examining the convergence with parameter $a$ For $a \in R$, let $x_1=a$ and $x_{n+1}=\frac{1}{4}(x_{n}^2+3)$ for all $n≥2$. Examine the convergence of the sequence ${x_n}$ for different values of $a$. Also, find $\lim x_n$, whenever it exists.
I am having problems on how to take $a$ as a parameter. I am unable to thin... | *
*Let $a>3$.
Hence, $$x_{n+1}-3=\frac{(x_n-3)(x_n+3)}{4}>0$$ and by induction we obtain here $x_n>3$ for all $n\in\mathbb N$.
In another hand, $$x_{n+1}-3=\frac{(x_{n-1}-3)(x_{n-1}+3)}{4}>$$
$$>\frac{3}{2}(x_{n-1}-3)>\left(\frac{3}{2}\right)^2(x_{n-2}-3)>...>\left(\frac{3}{2}\right)^{n-1}(x_1-3),$$
which says that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$? Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$?
Got this in a sample question paper for a contest but i have no idea how to solve it. I tr... | Interesting problem and elegant solution of @quasi.
Alternatively: Without loss of generality, assume $1\le a\le b$. Then:
$$a+b=\frac{a}{b}+\frac{b}{a}\le \frac{b}{b}+\frac{b}{1}=1+b\Rightarrow a\le 1 \Rightarrow a=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix to power 30 using eigen values issue
$$ \text{If matrix } A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{ then find } A^{30}.$$
I tried to approach through diagonalization using eigen values method.
I got eigen values as $-1, 1, 1$
As per diagonalization $ A = P*D*P^{-1}.$ So $ A^{... | You have computed the eigenvalues of $A$ to be $\{-1, 1, 1\}$. The repeated eigenvalue may be an obstacle to diagonalization. In this case, the geometric and algebraic multiplicities of the eigenvalue $1$ are different, so A is not diagonalizable. You continued as if $A$ were diagonalizable, so this is the mistake i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\be... | HINT: use the tan-half angle substution
$$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}$$
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
with $t=\tan(x/2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
An inequality with $\cos$ and triangle sides Here is the problem:
Let $ABC$ be a triangle with sides $a, b, c$. Show that $\dfrac{\cos A}{a^3}+\dfrac{\cos B}{b^3}+\dfrac{\cos C}{c^3}\geq\dfrac{3}{2abc}.$
Here's my attempt:
By the cosine formula, we have $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ etc, which the left hand sid... | You're almost done. If you factor out $\frac{1}{2abc}$ from the expression you obtained, you get
$$\frac{1}{2abc}\left(\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\right).$$
So you just need to prove $\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\ge 3$. Writing
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality - Maximize k such that I'm not sure how to solve this problem I came across online. Here it is:
Let $a, b, c$ be nonnegative reals. Maximize $k$ such that
$$(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) + \dfrac{k(ab + bc+ca)}{a^2+b^2+c^2} \geq 9 + k.$$
What I know is this:
$a, b, c$ must be pos... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$.
We know that $a$, $b$ and $c$ are positive roots of the following equation
$$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $m$, find three numbers $a,b,c$ such that $c-b=m(b-a)$ and $a+b,a+c,b+c$ are all squares I've worked with the next problem
*
*Given an integer $m$, find three integers $a,b,c$, with $b\neq a$, such that $c-b=m(b-a)$ and
$$a+b=x^2, \ \ a+c=y^2, \ \ b+c=z^2,$$
where $x,y, z \in \mathbb{Z}$.
I've not been able... | The condition requires $$y^2-x^2=m(z^2-y^2)$$ or $$x^2-(m+1)y^2=-mz^2.$$
You also require $x+y+z$ even, and then you get $$a=\frac{x^2+y^2-z^2}{2},b=\frac{x^2+z^2-y^2}{2},c=\frac{y^2+z^2-x^2}{2}.$$
But there is a general approach to finding all rational solutions to:
$$aX^2+bY^2=c$$
if you have one rational solution.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Identify if some special name is given to polynomials Is there some special name given to the two polynomials below:
$${n \choose 1}x - {n \choose 3}x^3 + \ldots \label{} \tag{1}$$ and $$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots \label{} \tag{2}$$.
| HInt:try $$(1+ix)^n$$ wher $i^2=-1$
$$(1+ix)^n=\\\left(\begin{array}{c}n\\ 0\end{array}\right)1^{n}(ix)^{0}+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}(ix)^{1}+
\left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}(ix)^{2}+
\left(\begin{array}{c}n\\ 3\end{array}\right)1^{n-3}(ix)^{3}+
\left(\begin{array}{c}n\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$ Is the Following Proof Correct?
Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true
$$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$
Proof. We construct the proof by recourse to Mathematical-Inductio... | Looks good to me :)
NB - those are Fermat numbers, not Fermat primes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqr... | Here is my, as Michael Rozenberg rightly dubs it, ugly proof of the last step in Sergic Primazon's proof that is $g(t)$ decreases for $t\in(0,1]$.
$$g'(t) = -3^{-\frac23}\Big(\frac1{t^2}+\frac1t-1\Big)^{-\frac23}\Big(\frac2{t^3}+\frac1{t^2}\Big)+\frac12\Big(\frac1t\Big)^{\frac12}\ .$$
We want to show $g'(t)<0$ or
$$3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Let $a_{1}>0,a_{2}>0$ and $a_{n}=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$.
My attempt:
\begin{alig... | Following in the footsteps of @martycohen elsewhere on this page, a generalized version of the sequence in the OP can be constructed. Thus, consider
$$g_n=\frac{ag_{n-1}g_{n-2}}{bg_{n-1}+cg_{n-2}}=\frac{a}{\frac{b}{g_{n-2}}+\frac{c}{g_{n-1}}}$$
The let $f_n=1/g_n$ to find that
$$f_n=\left(\frac{c}{a} \right)f_{n-1}+\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\o... | Let assume that for small $x$, $$\sin x = x + a x^2 + bx^3 + O(x^4)\tag{1}$$
In this trigonometric circle we can compute $BC$ in two ways:
First, considering the $ABC$ triangle:
$$BC= \sqrt{AB^2+AC^2}=\sqrt{\sin^2 x+\left(1-\sqrt{1-\sin^2 x}\right)^2}=\\= x+ax^2+\left(b+\frac18\right)x^3+O(x^4) \tag{2}$$
[*]
Second,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 0
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Solve Diophantine equation: $xy+5y=2x+(y+2)^2$
Solve Diophantine equation: $xy+5y=2x+(y+2)^2$ for positive integers $x,y$.
Rearranging the equation gives: $y^2=xy-2x+y-4$.So the RHS must be a perfect square,how can it be more restricted?
| It is equivalent to
$$x(y-2)=y^2-y+4\iff x=\frac{y^2-y+4}{y-2}=y+1+\frac6{y-2}$$
for $y\neq2$.
EDIT
As @lab bhattacharjee suggested, we may also arrange the equation with $y$ as the subject.
$$y^2-(1+x)y+(2x+4)=0\tag1$$
For integer $y$, we must have some integer $z$ such that
$$\Delta=z^2=(1+x)^2-4(2x+4)=x^2-6x-15=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Maclaurin series with zero denominator when evaluating derivative I have following function:
$\displaystyle f(x) = \frac{\ln(1+x^2) - x^2}{\sqrt{1+x^4} - 1}$
As you can see, when doing the quotient rule for the denominator in your head, the derivative of this function results to $0$ in the denominator when evaluating f... | Technically speaking, even the function itself is undefined at zero, as the denominator of $f(x)$ turns into zero when we try to plug in $x=0$. But this is a removable discontinuity, as the graph of $f(x)$ (available on the page that you linked to) shows. And moreover, with this discontinuity "removed" (or patched, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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First four nonzero terms of the McLaurin expansion of $\frac{xe^x}{\sin x}$ at $x_0=0$
Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series.
$$ \frac{xe^x}{\sin x}$$
Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x... | I didn't check the computations but, yes, you are doing it right.
As for the other question, note that the expression $\frac{xe^x}{\sin x}$ is undefined if $x=0$. In order to make $f$ continuous at $0$; you must define $f(0)$ as $\lim_{x\to0}\frac{xe^x}{\sin x}$, which is equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving the equation $2x^4-3x^2-9=0$ How would I go about solving $2x^4-3x^2-9=0$?
I started by taking $x^2$ out, thus getting $x^2(2x^2-3) = 9$ but I don't know if that did anything useful, usually if this equation is equal to $0$ it's easy but this time it's equal to $9$ so I'm lost.
| We can add $3x^2-3x^2$ to the left-hand side and factor by grouping.$$\begin{align*}2x^4-6x^2+3x^2-9=0 & \iff 2x^2(x^2-3)+3(x^2-3)=0\\ & \iff (x^2-3)(2x^3+3)=0\end{align*}$$However, if you did not see that sneaky trick, we can always make the substitution $X=x^2$ to get your polynomial into a quadratic.$$2X^2-3X-9=0\if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$
Prove that:
$$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$$
I computed the indefinite integral:
$$\int\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}(\ln 2)^2}+C$$
How can I continue from here?
| You are on the right track. Note that $2^x-x$ is increasing for $x\geq 1$. Therefore $f(x)=\frac{1}{2^{2^x-x}}$ is decreasing and
$$\sum_{n=1}^\infty\frac{2^n}{2^{2^n}}=\sum_{n=1}^\infty f(n)\geq \sum_{n=1}^\infty \int_{n}^{n+1}f(x) dx=\int_1^{\infty}f(x)dx=\left[-\frac{1}{2^{2^x}(\ln 2)^2}\right]_1^{\infty}=\frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13... | Numerical method.
The sum is equal to $1716$.
Now we want to find $x,y$ in:
$$\frac{x!}{y!(x-y)!}=1716=2^2\cdot 3\cdot 11\cdot 13=\frac{2^3\cdot 3\cdot 11\cdot 13}{2!}=\frac{2^3\cdot 9\cdot 11\cdot 13}{3!}=\frac{2^5\cdot 9\cdot 11\cdot 13}{4!}=$$
$$\frac{2^4\cdot 9\cdot 10\cdot 11\cdot 13}{5!}=\frac{8\cdot 9\cdot 10\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that: $\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$ Show that: $$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$$
I tried two approaches.
Approach 1: $$\sin ^2 (a \pm b)=\sin a\cos b \pm \sin b\cos a$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
I can't see where to go from here... | $$S=\sin ^2 (a+b)+\sin ^2 (a-b)=[\sin (a+b)+\sin (a-b)]^2-2\sin(a+b)\cdot\sin(a-b)$$
Now use that sum-product relations:
$$\sin (a+b)+\sin (a-b)=2\sin (a)\cos (b)$$
$$2\sin(a+b)\cdot\sin(a-b)=\cos(2b)-\cos(2a)=2(\cos^2(b)-\cos^2(a))$$
So,
$$S=4(1-\cos^2(a))\cos^2(b)+2\cos^2(a)-2\cos^2(b)$$
$$S=4(1-\cos^2(a))\cos^2(b)+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality : $ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right)\geq \frac{n}{n+1} $ Let $a_1, a_2, \ldots, a_n $ be positive real numbers such that $\displaystyle\sum^n_{i=1} a_i=1$. Prove that
$$ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2... | We consider two cases:
1) Let's first consider the case when $a_1,...,a_n$ fulfill $\sum a_ia_{i+1}\leq \frac{1}{n}.$
By CS we have
\begin{equation}
\left(\sum a_ia_{i+1} \right)\left(\sum\frac{a_i}{a_{i+1}^2+a_{i+1}} \right)\geq \left(\sum a_i \sqrt{\frac{a_{i+1}}{a_{i+1}^2+a_{i+1}}} \right)^2=\left(\sum a_i \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | $(\sqrt{5}-\sqrt{1})^2= 6-\sqrt{20}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that for any natural number n between $n^2$ and$(n+1)^2$ there exist 3 distinct natural numbers a, b, c, so that $a^2+b^2$ is divisible by c Show that for any natural number n ,one can find 3 distinct natural numbers a, b, c, between $n^2$ and$(n+1)^2$, so that $a^2+b^2$ is divisible by c.
It's easy to prove that... | Since $n^2+2n<(n+1)^2$ and since $n^2$ is included, one obvious answer is
$$\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
Extended exposition
Since $n^2+2n<(n+1)^2=n^2+2n+1$, let $a=(n^2+n)$, $b=(n^2+2n)$ and $c=n^2$ then
$$\frac{a^2+b^2}{c}=\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
$n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | We can simplify the algebra by noticing that the expressions are homogenous:
$
2(a^4+b^4+c^4)
\\\quad=2(a^4+b^4+(a+b)^4)
\\\quad=2b^4(x^4+1+(x+1)^4),
\quad x=a/b
$
$
2(x^4+1+(x+1)^4)
\\\quad=4 x^4 + 8 x^3 + 12 x^2 + 8 x + 4
\\\quad=4 (x^2 + x)^2 + 8 (x^2 + x) + 4
\\\quad=4(y^2+2y+1),
\quad y=x^2+x
\\\quad=4(y+1)^2
\\\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
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Reverse Logarithmic Inequality Given
$$\frac{1}{\log_{4}\left(\frac{x+1}{x+2}\right)}<\frac{1}{\log_{4}(x+3)}.$$
Then what is the range of values of $x$ for which this inequality is satisfied.
My Try On simplification, I got $x$ from $(-\infty,-2)$. That however is not the correct answer. Can someone tell me each and ... | The arguments of the logarithms should be positive. Hence $x+3>0$ and $\frac{x+1}{x+2}>0$, that is $x\in (-3,-2)\cup (-1,+\infty).$
If $x\in (-3,-2)$ then $0<x+3<1$
and $\frac{x+1}{x+2}>1$. Therefore
$$\log_4(x+3)<0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)>0$$
and the inequality does not hold.
If $x\in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Compute $\int_0^{\pi /2}\frac x {\tan x} \, dx$ using contour integration How can I calculate the integral $$\int_0^{\pi /2}\frac x {\tan x} \, dx$$ with complex integration? (Contours, residue theorem, etc.) I was thinking on using the fact that $\displaystyle \tan x=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}$, which im... | A complex-analytic solution. Here is a combination of calculus and a basic complex analysis: Perform integration by parts to sanitize the integrand:
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx
= \underbrace{ \left[ \vphantom{\int} x \log\sin x \right]_{0}^{\frac{\pi}{2}} }_{=0}
- \int_{0}^{\frac{\pi}{2}} \log \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{... | First rewrite the equation replacing $x$ by $X$: Thus we require $L=\lim_{X\rightarrow 0}\dfrac{e^{\sqrt{1+X^2}}-a-bX^2}{X^4}$
Now let $x:=\sqrt{1+X^2}$, then $X^2=x^2-1$ with $X^4=(x^2-1)^2$ and $x\rightarrow 1$ as $X\rightarrow 0$.
Now, we have $L=\lim_{x\rightarrow 1}\dfrac{e^x-a-b(x^2-1)}{(x^2-1)^2}=\dfrac{\lim_{x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Wrong result for a $1^\infty$ limit I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$
I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\righta... | Want
$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x
$.
Since,
for small $z$,
$\sqrt{1+z}
=1+\frac12 z + O(z^2)
$
and
$\ln(1+z)
=z+O(z^2)
$,
$\begin{array}\\
\sqrt{x^2+2x+4}
&=\sqrt{(x+1)^2+3}\\
&=(x+1)\sqrt{1+\frac{3}{(x+1)^2}}\\
&=(x+1)(1+\frac{3}{2(x+1)^2}+O(x^{-4}))\\
&=x+1+\frac{3}{2(x+1)}+O(x^{-3})\\
\text{so}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Volume made be a Revolving Pentagon I was looking to the volume created if I revolve the area of the pentagon having vertices
(1,0), (2,2), (0,4), (-2, 2), (-1,0) about the x-axis.
My work:
I used the Second theorem of Pappus, which states that the volume created is equal to the product
of circumference of the circle d... | It's
$$2\left(\frac{4\cdot\pi\cdot4^2}{3}-\frac{1\cdot\pi\cdot2^2}{3}-\frac{2\cdot\pi\cdot2^2}{3}\right)=\frac{104\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\... |
The graph of the function $g(x) = \lfloor\frac{2x-1}{3}\rfloor + \lfloor\frac{4x+1}{6}\rfloor - (5x-4)$
Once we know $0 < x < 1$, we have $\lfloor\frac{2x-1}{3} \rfloor = 0$ and $\lfloor \frac{4x+1}{6} \rfloor = 0$. Thus $x = \frac{4}{5}$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Review: Prove by induction Nicomachus's theorem Please help me out reviewing the way I wrote this proof:
Prove by induction: $1^3+2^3+3^3+...+n^3=\left(\frac{n(n+1)}{2}\right)^2$ with $n\geqslant1$
Proof:
Lets define the set, $S=\left \{n\in N:n\geqslant1, 1^3+2^3+3^3+...+n^3=\left(\frac{n(n+1)}{2}\right)^2 \right \}$... | Yes, the proof seems okay to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Expectation of minimum of two random numbers Let's assume we picked chose two numbers in the range $[1,n]$ where $n\in \mathbb N$ independently, such that each one of them is distributed randomly.
How can I find the expectation of the minimum of these two numbers?
I know that we have $2n-1$ possibilities of getting $1... | For $1 \le k \le n$, let $p_k$ be the probability that the minimum of the two numbers is $k$, and let $e$ be the expected minimum value.
\begin{align*}
\text{Then}\;\;p_k &=
2
\left(
{\small{\frac{1}{n}}}
\right)
\left(
{\small{\frac{n-k+1}{n}}}
\right)
-
{\small{\frac{1}{n^2}}}
\\[6pt]
&=\frac{2n-2k+1}{n^2}\\[8pt]
\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| You cannot simply keep the same coefficients but change the $10$'s to $9$'s. What you must do is something like the following:
$$\begin{align}397&=3\times \color{red}{10}^2+9\times\color{red} {10}^1+7\times \color{red}{10}^0\\&=3\times(9+1)^2+9\times(9+1)+7\times 1\\&=3\times\color{blue}9^2+(3\times18+3)+\color{blue}9^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| Here is a table of whole numbers $x$ so that $x^2+x\mid240$:
$$
\begin{array}{c|c}
x&1&2&3&4&5&15\\\hline
x^2+x&\color{#090}{2}&6&12&20&\color{#090}{30}&240
\end{array}
$$
For $x$ to work, we need $\frac{240}{x^2+x}+1=y^2$:
$\frac{240}{2}+1=11^2$
$\frac{240}{6}+1=41$
$\frac{240}{12}+1=21$
$\frac{240}{20}+1=13$
$\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the rest of the division of $(4^{103} + 2(5^{104}))^{102}$ by $13$ I already know some techniques to solve big exponents such as, Euler/Fermat Theorem,Euler/Carmichael,successive squaring.
But these problems seem to be more dificult as they involve operations insted of a single number.
How could I solve them ... | $$
(4^{103}+2 \cdot 5^{104})^{102}\overset{13}{\equiv}
(4^{3 \cdot 34+1}+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv}
(4^{3 \cdot 34}4^1+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv}
\\
\left((4^{3})^{34} \cdot 4^1+2 \cdot (5^{2})^{52}\right)^{102}\overset{13}{\equiv}
\left(64^{34} \cdot 4^1+2 \cdot 25^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+... | your inequality is equivalent to
$$ab(a-b)(a^2-b^2)+ac(a-c)(a^2-c^2)+bc(b-c)(b^2-c^2)\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
| If $ax^2+bx+c$ has two roots, then the sum of the roots is equal to $-b/a$ and the product of the roots is equal to $c/a$. In your case, if one root is $p$, then the other one would be $2p$; applying the formulas to:
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
$$\frac{3+2k}{k-1}=3p \quad\mbox{ and }\quad \frac{6+k}{k-1}=2p^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
asymptotic behavior of $\displaystyle{\sum_{r = 0}^k} { 2k - r \choose k} x^r $ for large $k$ Problem
This problem comes from the partition function $Z( n, r )$ in this question. I want to analyze the large $k$ behavior of
\begin{equation}
f(k) = \sum_{r = 0}^k { 2k - r \choose k} x^r
\end{equation}
For example,
\beg... | An upper bound, and a lower bound when $x>4$.
Assuming you mean $x^r$ rather than $x^{k}$, then you get:
$$\sum_{r=0}^{k}\binom{2k-r}{k}x^r = \sum_{s=0}^{k}\binom{k+s}{k}x^{k-s}=x^k\sum_{s=0}^{k}\binom{k+s}{k}x^{-s}$$
Now, $$\sum_{s=0}^{\infty}\binom{k+s}{k}y^s =\frac{1}{(1-y)^{k+1}}$$
for $|y|<1$.
So, when $x>1$, lett... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$?
How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$?
I realised that this looks a lot like the Pythagorean theorem -- it could be written as $^2 + (\sqrt{2}y)^2 = (2z)^2$ as well. Then wouldn't there be an infi... | Observe $x^2$ divisible by $2$, thus $x$ is divisible by $2$. Write
$$ \begin{align*} x = 2n &\implies 4n^2 + 2y^2 = 4z^2 \\
&\implies y^2+2n^2=2z^2 \\
&\implies y = 2m \\
&\implies 4m^2+2n^2=2z^2 \\
&\implies n^2+2m^2=z^2 \\
&\implies n = k(a^2-2b^2), \end{align*}$$
where $m = 2kab, z = k(a^2+2b^2), a,b,k \in \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Calculate the Limit of $f_n:= (\frac{2}{3}+\frac{i}{2})^n+\sqrt[n]{2n}-\frac{n^3}{2^n}$ $\lim_{n \rightarrow \infty}(\frac{2}{3}+\frac{i}{2})^n+\sqrt[n]{2n}-\frac{n^3}{2^n} = \lim_{n \rightarrow \infty}(\frac{2}{3}+\frac{i}{2})^n+1-0$
Now, how to evaluate $(\frac{2}{3}+\frac{i}{2})^n$?
| With $||.||_2$ we will denote the complex absolute value.
We see that: $$||\frac{2}{3}+\frac{i}{2}||_2=\sqrt{\frac{4}{9}+\frac{1}{4}}=\sqrt{\frac{25}{36}}=\frac{5}{6} <1$$
Thus $||\frac{2}{3}+\frac{i}{2}||_2^n=(\frac{5}{6})^n \to 0$
Now we have that: $$||f_n-1||_2=||(\frac{2}{3}+\frac{i}{2})^n+(\sqrt[n]{2n}-1)+\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $x^6 - x^4 +2 >0$ for all $x$
Show that $x^6 - x^4 +2 >0$ for all $x$.
I can prove this using calculus and I can prove it by factorising the expression.
Is there any way to prove it another way, perhaps by considering cases of $x$? I need a simpler way to explain to a student that cannot factorise degree $... | Use sum of squares! $x^6>x^4$ for $|x|>1$
Otherwise $|x|<1$ and we have $x^6-x^4+2=(x^3-\frac{1}{2}x)^2-\frac{1}{4}x^2+2>0$ since $2>\frac{1}{4}x^2$
Or, magically, we have
$x^6-x^4+2=(x^3-x)^2+(x^2-1)^2+x^2+1>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find $\iiint (x^2+y^2+z^2)~dV$, above the cone $z=\sqrt{3(x^2+y^2)}$, inside the sphere $x^2+y^2+z^2\leq a^2$.
Find
$$\iiint (x^2+y^2+z^2)~dV$$
above the cone $z=\sqrt{3(x^2+y^2)}$, and inside the sphere $x^2+y^2+z^2\leq a^2$.
I think that the solution for this problem is following:
Use cylindrical coordinates:
$... | Hint: Alternative approach is using the spherical coordinates for this problem:
$$\int_0^a \int_0^{2\pi} \int_{0}^{\frac{\pi}{3}} r^4\sin\varphi d\varphi d\theta dr=\color{blue}{\dfrac{\pi}{5}a^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
Evaluate
$$ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$
I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
| Alternatively:
$$\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}=\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}\cdot \frac{\sqrt{1-\cos(2x)}}{\sqrt{1-\cos(2x)}}\cdot \frac{\sqrt{\pi}+\sqrt{2x}}{\sqrt{\pi}+\sqrt{2x}}=$$
$$\lim_{x\to \pi/2} \frac{|\sin{(2x)|}\cdot(\sqrt{\pi}+\sqrt{2x})}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show that for $n\to \infty, u_n - \sqrt{n} - \frac{1}{2} \sim \frac{1}{8\sqrt{n}}$ $\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ with $u_1 = 1$.
I've already shown that $ u_n \sim \sqrt{n}$ and that $\underset{n\rightarrow \infty}{\lim} u_n - \sqrt{n} = \dfrac{1}{2}$
How to show that $n\rightarrow \infty, u_n - \s... | Write $y_n = u_n - \sqrt{n} - \frac{1}{2}$. Then the recurrence becomes
$$y_n = \sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} - \sqrt{n + \sqrt{n} + \frac{1}{4}} = \frac{\frac{1}{4} + y_{n-1}}{\sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} + \sqrt{n} + \frac{1}{2}}.$$
Now look at $8\sqrt{n}\cdot y_n$ and use that you alread... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\int x+1 \, dx$ using $u$-substition gives wrong answer. $\int (x+1) dx$
Substituting:
$u = x + 1$ and $du = dx$
$$\int u \,du = \frac{u^2}{2}$$
Which gives:
$$\int (x+1) \, dx = \frac{(x+1)^2}{2}$$
This is completely wrong as the correct answer is $\frac{x^2}{2}+x$
Why did this happen?
| You forgot the constant C. It is very important:
$$
\int u du = \frac{u^2}{2} + C = \frac{(x+1)^2}{2} + C = \frac{x^2}{2} + x + \frac{1}{2}+ C = \frac{x^2}{2} + x + C'
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
for $abc = 1$ and $a \le b \le c$ prove that $(a+1)(c+1)>3$ This inequality has been given to me by my teacher to keep me occupied and after hours of fumbling around with it, and later trying to google it. I found nothing at all.
For any three positive real numbers $a$, $b$ and $c$, where $abc = 1$ and $a\le b \le c$, ... | Our conditions give $c\geq1$ and $c\geq b$.
Thus, by AM-GM we obtain:
$$(a+1)(c+1)=ac+a+3\cdot\frac{c}{3}+1\geq5\sqrt[5]{ac\cdot a\cdot\left(\frac{c}{3}\right)^3}+1=$$
$$=\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2c^4}+1\geq\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2b^2c^2}+1=\frac{5}{\sqrt[5]{27}}+1>3.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\f... | Your inequality can be written
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<2\varepsilon\tag{*}
$$
If $x>0$, then $x^2+1>1$, so $\frac{1}{x^2+1}<1$. The condition that $x>0$ is satisfied as soon as $\delta<1$, when you take $0<|x-1|<\delta$.
In this case, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a... | You can consider $$(ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)\le a^2+b^2$$ so that $$(ax+by)^2\le a^2+b^2-(bx-ay)^2$$
Equality is attained when $x^2+y^2=1$ and $bx=ay$
How did I get that second square? Well I wanted to work with the square of the target expression and to get a term in $x^2+y^2$, so I needed to swap $a$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Show $k\mid 12$ with $2^k=1\bmod 13$ Let $k$ be the smallest positive integer such that $2^k=1\bmod 13$. Show that $k\mid 12$.
I'm not very good at proofs and I'm confused as how to prove this. I started by saying $2^k-1=13n$. But I don't know where to go from there.
| It seems quicker just to show directly that $k=12$.
Powers of $2 \bmod 13$:
$\begin{array}{c|rl|l}
k &2^k &\bmod 13 & \text{using -ve}\\ \hline
1 && 2 \\ 2 && 4 \\ 3 && 8 \\
4 & 16\equiv\! & 3 \\
5&& 6 \\ 6&& 12 &\equiv -1\\
7 & 24\equiv\! & 11 &\equiv -2 \leftarrow 2^6\cdot 2^1\\
8 & 22\equiv\! &9 &\equiv -4 \lefta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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For $x: (1 \to b)$ where $a$ & $b$ are co prime, why is $ax \bmod b$ distinct So, out of curiosity, I was wondering why $x (1 \to b)$ where $a$ & $b$ are co prime, why is $ax \bmod b$ distinct.
For example, let $a = 5$, and $b= 8$:
$\begin{array}{c|c|c}
x & 5x & 5x\bmod 8 \\ \hline
1 & 5 & 5 \\
2 & 10 & 2 \\
3 & 1... | Suppose $ax\equiv ay\pmod{b}$, with $1\le x\le b$ and $1\le y\le b$. Without loss of generality, we can assume $x\ge y$. Then
$$
a(x-y)\equiv0\pmod{b}
$$
that is, $b\mid a(x-y)$; since $a$ and $b$ are coprime, we conclude that
$$
b\mid (x-y)
$$
but $0\le x-y<b$, so we only have one possibility, namely $x-y=0$.
In a sli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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What is the remainder when $p!$ is divided by $p+1$?
Let $p$ be a prime larger than $7$. What is the remainder when $p!$ is divided by $p+1$?
I tried plugging in the next prime (11), which doesn't help with such big numbers. Then I tried writing $\frac{p!}{p+1} = p(p-1)(p-2).../(p+1)$. Dividing each of $p$, $(p-1)$ e... | Remember that if $p > 7$, then $p! = 1 \times 2 \times 3 \times \ldots \times (p - 1) \times p$.
The smallest prime factor of $p + 1$ is less than $\sqrt{p + 1}$ and clearly $\lceil \sqrt{p + 1} \rceil < p$. The largest prime factor of $p + 1$ is at most $\frac{p + 1}{2}$, which is also less than $p$.
This means that a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right tr... | $$\tan\left(\arctan\frac{1}{7}+\arctan\frac{3}{4}\right)=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\cdot\frac{3}{4}}=\frac{4+21}{28-3}=1$$
and since $0^{\circ}<\arctan\frac{1}{7}+\arctan\frac{3}{4}<90^{\circ}$, we get the answer:
$$\arctan\frac{1}{7}+\arctan\frac{3}{4}=45^{\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proving an equation about exponential of matrices
Let $A, B$ be $n\times n$ matrices over $\mathbb{C}$. Assume that
$$[A,[A,B]]=[B,[A,B]] = 0$$
Show
$$e^{\epsilon A}Be^{-\epsilon A} = B + \epsilon[A,B]$$
Then, I have to show the equation below. Here $[A, B]$ is equal to $AB-BA$ and $\epsilon$ is a real parameter.... | $\textbf{Proof}$. Since
$$
e^{\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(\varepsilon A)^k}{k!}
= 1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} +
\frac{\varepsilon^3 A^3}{3!} + \ldots,
$$
we have
$$
e^{-\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(-\varepsilon A)^k}{k!}
= 1 - \varepsilon A + \frac{\varepsil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Characteristic $2$ ring $A$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$. Show that $x^2 = x$.
Let $A$ be a ring of characteristic $2$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$ (i.e. $k$ is specific to $x$). Show that $x^2 = x$ for every $x \in A$.
I have been fiddling about for a while now but hav... | We start with a lemma: By Theorem 2 of Binomial coefficients modulo a prime (Fine, 1947), we have that if the binary expansion of the positive integer $N$ has $m$ ones on it, then the number of odd entries of the $N$-th row of Pascal's Triangle is $2^m$. In particular, if $N=2^k$ then $m=1$ and $2^m=2$, so the only odd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$ is an integer if and only if $n$ is equal to twice a triangular number. $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$
Prove that $m$ is an integer if and only if $n$ is equal to twice a triangular number.
Supposing that $n$ is twice a triangular number. Then I... | We have
\begin{eqnarray*}
m= \sqrt{ n+ \sqrt{n+ \sqrt{n+ \cdots}}} \\
\end{eqnarray*}
So $m= \sqrt{n+m}$. Square this and solve the quadratic
\begin{eqnarray*}
m= \frac{-1 + \sqrt{1+4n}}{2}
\end{eqnarray*}
So $1+4n$ will need to be a perfect square to deal with the square root; indeed a square of an odd value to deal ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find limits of sequences Prove that
$$a) \lim\limits_{n \to \infty} (\frac{1^p+2^p+...+n^p}{n^p} - \frac{n}{p+1})=\frac{1}{2},$$
$$b) \lim\limits_{n \to \infty} \frac{1^p+3^p+...+(2n-1)^p}{n^{p+1}}=\frac{2^p}{p+1},$$
where is $p \in \Bbb N $.
Thanks to Stolz–Cesàro theorem in $a)$ I went to $$\lim\limits_{n \to \infty... | Solution for (b):
$$\lim_n\frac{1^p+3^p+\cdots +(2n-1)^p}{n^{p+1}}=\lim_n \frac{1^p+2^p+\cdots +(2n)^p}{n^{p+1}}-\lim_n\frac{2^p+4^p+\cdots+(2n)^p}{n^{p+1}}$$
$$=2^{p+1}\lim_n\frac{1^p+2^p+\cdots +n^p}{n^{p+1}}-2^p\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}$$
$$=(2^{p+1}-2^p).\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=2^p.\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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The asymptotic expansion of $x^x$ at $0$. If we put $x^x$ in Wlofram Alpha, we get the following:
$$
x^x = 1 + x \log(x) + \frac12 x^2 \log^2(x) + \frac16 x^3 \log^3(x) + \frac1{24} x^4 \log^4(x) + \frac1{120} x^5 \log^5(x) + O(x^6)
$$
Is there a name of this form of expansion and is there a systematic way to find such... | We know that
$$x^x=e^{x\log x}$$
Also by Taylor's Series,
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$
So
$$\begin{align}
x^x&=e^{x\log x}\\
&=1+x\log x+\frac{(x\log x)^2}{2!}+\frac{(x\log x)^3}{3!}+\frac{(x\log x)^4}{4!}+\frac{(x\log x)^5}{5!}+\cdots\\
&=1+x\log x+\frac{1}{2}x^2\log^2 x+\frac{1}{6}x^3\log^3x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inverting a $2 \times 2$ matrix $\mod 26$ I am trying to invert the matrix (in mod 26)
\begin{bmatrix}
19 & 7\\
19 & 0 \\
\end{bmatrix}
I compute the determinant
((19*0)-(19*7)=23 (Reducing in mod 26)
\begin{bmatrix}
0 & -7\\
-19 & 19 \\
\end{bmatrix}
Switch a & d and negate b & c
Multiply by 23^-1 =... | Let $A$ be the given matrix. Then $AA^{-1}=I_2$. Hence, we must row reduce the augmented system:
$$\begin{bmatrix}19&7&|&1&0\\19&0&|&0&1\end{bmatrix}\to\begin{bmatrix}19&7&|&1&0\\0&-7&|&-1&1\end{bmatrix}\to \begin{bmatrix}19&0&|&0&1\\0&-7&|&-1&1\end{bmatrix}\to\begin{bmatrix}1&0&|&0&11\\0&1&|&-11&11\end{bmatrix}$$
usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Draw the function $y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ I'm not allowed to use differentiation for this. I know the domain and range of the arctan function. The function $\arctan:\mathbb{R}\rightarrow \left(-\frac{\pi}{2},\frac{\pi}{2}\right).$ Is my job to see what happens to the argum... | Note:If you find $f'$ when $f(x)=y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ , you will see
$f'(x)=0$ so $f(x)$ is a constant function
$$f(x)=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}=\\
\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^{-x}(e^x-4)}{\sqrt{3}}}=\\
\arcta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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In how many ways can $5$ students be chosen such that at least one student is chosen from each state? There are $3$ states and $3$ students representing each state. In how many ways can $5$ students be chosen such that at least one student is chosen from each state?
According to me, the answer should be $3C1 \cdot 3C1 ... | Your answer over counts.
Suppose the representatives of state $A$ are $a, b, c$, the representatives of state $B$ are $d, e, f$, and the representatives of state $C$ are $g, h, i$. You count the selection $\{a, b, c, d, g\}$ three times, once for each of the ways we could choose a representative of state $A$.
$a, d, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute $\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)$ I note that $\sqrt{x^2-6x+9}=|x-3|$. Splitting upp the limit into cases gives
*
*$x\geq 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(|x-3|+x-1)=2\lim_{x\rightarrow -\infty}(x-2)=-\infty.$$
*
*$... | Since $\lim_{x \rightarrow - \infty}$ it suffices to consider 'small' negative $x$.
$|x-3| = -x + 3$.
Example: $x= -7$:
$|x-3| = |-7-3| = -x +3.$
Hence:
$\lim_{x \rightarrow -\infty} (|x-3| + x-1) = $
$\lim_{x \rightarrow -\infty}( -x +3 +x -1 )= 2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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In how many ways can a person score a total of $130$ marks in an exam consisting of $3$ sections of $50$ marks each? In how many ways can a person score a total of $130$ marks in an exam consisting of $3$ sections of $50$ marks each?
What I did:-
$$a+b+c=130$$
So applying partition rule
$n=130$ and $130$ marks has to ... | We wish to solve the equation
$$a + b + c = 130 \tag{1}$$
in the nonnegative integers subject to the restrictions that $a, b, c \leq 50$.
As you observed, equation 1 has
$$\binom{130 + 2}{2} = \binom{132}{2}$$
solutions in the nonnegative integers.
From these, we must exclude those in which or more of the variables ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A median of a triangle is the geometric mean of the adjacent sides; find the cosine of one angle in terms of the others
$AD$ is a median of $\triangle ABC$. $|AD|$ is the geometric mean of $|AB|$ and $|AC|$.
Find $1+\cos A$ in terms of $\cos B$ and $\cos C$.
Edit
This is the second part of the question
Also prove th... | $$\frac{1}{2}\sqrt{2b^2+2c^2-a^2}=\sqrt{bc},$$ which gives
$$a^2=2b^2+2c^2-4bc.$$
Thus, $b\neq c$ and
$$\cos\beta\cos\gamma=\frac{a^2+c^2-b^2}{2ac}\cdot\frac{a^2+b^2-c^2}{2ab}=$$
$$=\frac{(2b^2+2c^2-4bc+c^2-b^2)(2b^2+2c^2-4bc+b^2-c^2)}{4(2b^2+2c^2-4bc)bc}=$$
$$=\frac{(3c^2-4bc+b^2)(3b^2-4bc+c^2)}{8(b-c)^2bc}=\frac{(b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding equation of circle with the given restrictions
Tangents are drawn to circle $x^2+y^2-6x-4y-11=0$ from point $P(1,8)$ touching circle at $A$ and $B$. Let there be a circle whose radius passes through point of intersection of circles $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+6=0$ and intersect the circumcircle of $... | So here's how I tried to solve it :
Since the circle is passing through the point of intersection of the two circles.We can write the family of circles passing through them by S1 + $\omega $ S2, where $\omega $ is a parameter which produces the infinite circle passing through the point of intersection of the two circle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answe... | Rename $a= \alpha $ and $b=\beta$
So this is $(x-{a\over b})(x-{b\over a})=0$ thus
$$x^2-{a^2+b^2\over ab}x+1=0$$
Since $ab = {6\over 2}=3$ and $a^2+b^2 = (a+b)^2-2ab = ({3\over 2})^2-6 = {-15\over 4} $ we get:
$$x^2+{5\over 4}x+1=0$$
and thus the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, alg... | I found an AM-GM proof$:$
$$\text{LHS}-\text{RHS}=\frac{1}{a+b+c} \sum \Big({\dfrac {{a}^{3}}{b}}+{\dfrac {{a}^{4}}{14{b}^{2}}}+{
\dfrac {{a}^{3}c}{{b}^{2}}}+{\dfrac {11{b}^{4}}{14{c}^{2}}
}+{\dfrac {{c}^{4}}{7{a}^{2}}}-3{a}^{2}\Big)\geqslant0,$$
which is obvious by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
The area of curve $y^3=x$ bounded by lines. Find the area of region on $xy$ plane shaded by curve $y^3=x$ and lines $y=1$ and $x=8$
My solution: The line $x=8$ intersect curve at point $y=2$ and line $y=1$ intersect at point x=1. So the intended area is $$=\left(1-\int \limits_{0}^{1}y^3dy\right)+7=8-\frac{1}{4}=7,75$$... | Referring to the graph:
1-method:
$$S=\int_{x=1}^{x=8} x^{1/3}dx-7\cdot 1=\left(\frac{3}{4} x^{4/3}\right) \bigg|_1^8-7=\left(12-\frac34\right)-7=4\frac14.$$
2-method:
$$S=8\cdot 1-\int_{y=1}^{y=2} y^3dy=8-\left(\frac{1}{4} y^4\right) \bigg|_1^2=8-\left(4-\frac14\right)=4\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sum_{k=1}^\infty \frac{2^k a^{2^k}}{1+a^{2^k}}=\frac{a}{1-a}$ for $-1How to prove $$\sum_{k=1}^\infty \frac{2^k a^{2^k}}{1+a^{2^k}}=\frac{a}{1-a}$$ for $-1<a<1$?
It gives a hint: $2^{k+1}-1=1+2+\cdots+2^k$. How can we use this hint to cancel some terms?
| Here is a brutal-force computation: Expand everything and group according to the exponent of $a$.
$$
\sum_{k=0}^{\infty} \frac{2^k a^{2^k}}{1+a^{2^k}}
= \sum_{k=0}^{\infty} 2^k \sum_{j=1}^{\infty} (-1)^{j-1} a^{j \cdot 2^k}
= \sum_{n=1}^{\infty} \left( \sum_{(j,k) \ : \ j \cdot 2^k = n} (-1)^{j-1} 2^k \right) a^{n}.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of Trailing Zeros of Binomial Coefficient If $x+2=18181818...$ $n$ $digits$, find the number zeros at the end of ${x \choose x/2}$. I have tried using Legendre's formula for factorials, but I have got nowhere because of the strange value of $x$.
| We want to find how many Trailing zeros for $\binom{2n}{n}$.
First Write the binomial as factorials $\frac{(2n)!}{(n!)^2}$
And apply Legendre formula for every term separately.
For $(2n)!$ there are $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor $ $2$'s and $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality for $x$: $\log_4 (x^2 − 2x + 1) < \log_2 3$ Solve the inequality for $x$: $$\log_4 (x^2 − 2x + 1) < \log_2 3$$
I got two answers and I'm not sure if I did it correctly.
1st ans: $(-2,1)\cup (1,4)$
2nd ans: $x \neq -2,4$
| \begin{array}{c}
\log_4 (x^2 − 2x + 1) < \log_2 3 \\
x^2 − 2x + 1 < 4^{\log_2 3} \\
0<(x-1)^2 < 2^{2\log_2 3} \\
0<(x-1)^2 < 2^{\log_2 3^2} \\
0<(x-1)^2 < 3^2 \\
\text{$-3 < x-1 < 3$ and $x \ne 1$} \\
\text{$-2 < x < 4$ and $x \ne 1$} \\
x \in (-2,1) \cup (1,4)
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the matrix $X$ in $AX$
Given that $A$ is a $3$ x $3$ matrix such that $$ A \begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix} = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $$ and $$ A \begin{pmatrix}
0 \\
0 \\
1 \\
\end{... | You can take$$X=\begin{pmatrix}1&0&4\\0&0&0\\0&0&3\end{pmatrix}.$$
And how do I know that? Note that$$X.\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\implies AX\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix},$$that$$X.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $\sin x - \cos x = \frac{1}{2}$ then determine: $\sin^4 x + \cos^4 x$ If $$\sin x - \cos x = \frac{1}{2}$$ then determine: $$\sin^4 x + \cos^4 x$$
I tried making it $(\sin^2 x)^2+(\cos^2 x)^2$ but then I get nothing that can help. What is the trick to this?
| I'm using the identity $2(u^2+v^2)=(u+v)^2+(u-v)^2$ repeatedly:
$$(\cos x+\sin x)^2=2(\cos^2 x+\sin^2 x)-(\cos x-\sin x)^2=2-\frac14=\frac74,$$ so
\begin{align}2(\cos^4 x+\sin^4 x)&=(\cos^2 x+\sin^2 x)^2+(\cos^2 x-\sin^2 x)^2\\&=(\cos^2 x+\sin^2 x)^2+(\cos x+\sin x)^2\,(\cos x-\sin x)^2\\&=1+\frac74\cdot\frac14=\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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showing $f$ to be a bijection but stuck trying to show that $f$ is surjective Recall that the closed interval $[a,b]$={$x \in \mathbb{R} \vert a\le x \le b$}
prove that $\vert [2,5] \vert = \vert [-2,3] \vert$
(cardinality of [a,b] not absolute value)
proof: define $f:[2,5] \rightarrow [-2,3]$ by $f(x)=c+ \frac{d-c}{b... | You have
$$f(x) = \frac{5}{3}(x - 2) - 2$$
Let $y = f(x)$, then solve for $x$.
\begin{align*}
y & = \frac{5}{3}(x - 2) - 2\\
y + 2 & = \frac{5}{3}(x - 2)\\
\frac{3}{5}(y + 2) & = x - 2\\
\frac{3}{5}(y + 2) + 2 & = x
\end{align*}
Substituting this value for $x$ in the function $f$ yields
\begin{align*}
f\left(\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$ Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$
$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$
$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$
$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$
$$1+c=(1-c)x^2+\color{red}{(... | $$
-(1-c)y^2-2cy = -(1-c)\left[y^2+\frac{2c}{1-c}y\right]=-(1-c)\left[\left(y+\frac{c}{1-c}\right)^2-\left(\frac{c}{1-c}\right)^2\right]
$$
Not what you have above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve: $4 \log_5 x- \log_5y =1 \quad \&\quad 5\log_5 x+ 3\log_5 y =14$ Find $x,y$ given:
$$4 \log_5 x- \log_5y =1 \quad \&\quad
5\log_5 x+ 3\log_5 y =14$$
How do I help my son this assignment? Not so good in maths myself
| $$
\begin{align}
4 \log_5 x- \log_5y &=1\tag 1 \\
5\log_5 x+ 3\log_5 y &=14\tag 2
\end{align}
$$
Multiplying $(1)$ by $3$ and adding $(1)+(2)$ we have
$$17\log_5 x=17\quad\Longrightarrow\quad \log_5 x=1\quad\Longrightarrow\quad x=5$$
Multiplying $(1)$ by $-5$ and $(2)$ by $4$ and adding $(1)+(2)$ we have
$$17\log_5 y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation $ \ T=\begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \ $
Answer:
From the given matrix , we can write as
$ T(x,y)=(x-y,x+y) ... | \begin{align}T(x,y) &= (x-y, x+y)\\
&=\left(\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \right)^T \\
&=\sqrt{2} \left(\begin{bmatrix} \cos \left( \frac{\pi}{4}\right) & -\sin \left( \frac{\pi}{4}\right) \\ \sin \left( \frac{\pi}{4}\right) & \cos \left( \frac{\pi}{4}\right)\end{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How did x become -sqrt(x^2)?
In computing the limit as $x \to -\infty$, we must remember that for $x<0$, we have $\sqrt{x^2} = |x| = -x$. So when we divide the numerator by $x<0$, we get
$$
\frac{\sqrt{2x^2+1}}{x}
= \frac{\sqrt{2x^2+1}}{-\sqrt{x^2}}
= - \sqrt{\frac{2x^2+1}{x^2}}
= - \sqrt{2 + \frac{1}{x^2}}
$$
H... | Because we are looking at the limit $x\rightarrow-\infty,$ our target is negative x from the limit's perspective. That means since the value $\sqrt{x^2}=|x|$ (you can confirm this by doing casework on positive and negative values - note that square roots are always positive!), $\sqrt{x^2}=|x|=-x$ since $|x|=-x$ for neg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\frac{\arctan x}{x^2}\,\mathrm dx$ $$\int \frac{\arctan
(x)}{x^2}\,\mathrm dx$$
(I have a constant in front of the integral, but I figure it just distributes to each of the terms in the end answer, and simplified it for now).
I figured it would be a chance for using integration by parts. I let $u=\arctan x,... | $$\int { \frac { \arctan { x } }{ { x }^{ 2 } } dx } =-\int { \arctan { x } d\left( \frac { 1 }{ x } \right) } =-\frac { \arctan { x } }{ x } +\int { \frac { dx }{ x\left( { x }^{ 2 }+1 \right) } } =\\ =-\frac { \arctan { x } }{ x } +\int { \left[ \frac { 1 }{ x } -\frac { x }{ { x }^{ 2 }+1 } \right] dx } =-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x... | A better way to solve this problem is to use polar coordinates
$$x=r\cos \phi$$
$$y=r\sin \phi$$
$$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3(\cos^3 \phi+\sin^3 \phi)}{r^2}=\lim_{r\to 0}\left[r(\cos^3 \phi+\sin^3 \phi)\right].$$
Because $|\cos^3 \phi + \sin^3\phi|\leq 1$ the limit exists (by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Need hint for solving the following problem
If $f(x^{500}-1)=5x^{1015}+3x^{244}+7x+10$. Find the sum of coefficients of $f(x^5+1)$.
Let $x^{500}-1=y$, then $f(y)=5(y+1)^\frac{1015}{500}+3(y+1)^\frac{244}{500}+7(y+1)^\frac{1}{500}+10$.
But, I don't think that it is a right approach as each term of $f(y)$ gets transfor... | The sum of the coefficients of $x$ in $g(x)=f\!\left(x^5+1\right)$ is $g(1)=f(2)$. However, given the formula for $f\!\left(x^{500}-1\right)$, we plug in $x=3^{1/500}$ to get
$$
f(2)=45\cdot3^{3/100}+3\cdot3^{61/125}+7\cdot3^{1/500}+10=68.651334833\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{alig... | Since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$, we obtain:
$$a-x+b\sqrt[3]{2}+c\sqrt[3]{4}=0$$ or
$$(a-x)^3+2b^3+4c^3-6(a-x)bc=0,$$
which is third degree.
I hope now it's clear.
By the way, we got your polynomial exactly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Third order convergence of an iteration scheme Consider the iteration scheme
$x_{n+1}=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a}{x_n^2})$
For third order convergence to $\sqrt 2$, the values of $\alpha$ and $\beta$ are ......
I tried it by plugging $x_{n+1}=x_n=\sqrt a$ as $n\rightarrow\infty$ and got $\alpha ... | Let $x_n = \sqrt{2}+\epsilon$ (and of course $a=2$). Then
$$
x_{n+1} = \alpha (\sqrt{2}+\epsilon)(3-\frac12(\sqrt{2}+\epsilon)^2
+\beta (\sqrt{2}+\epsilon) \left( 1+\frac{2}{(\sqrt{2}+\epsilon)^2}\right)
$$
Expanding in $\epsilon$, to one order more than we need in the first line,
$$
x_{n+1} = (\sqrt{2}+\epsilon) \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $21\mid a^2+b^2\Rightarrow 441 \mid a^2 + b^2$ How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.
| If $3\mid(a^2+b^2)$ then $3$ divides both $a$ and $b$, since $-1$ is not a quadratic residue $\!\!\pmod{3}$.
The same applies $\!\!\pmod{7}$. If $21\mid(a^2+b^2)$, from the CRT we get that $3$ and $7$ divide both $a$ and $b$, hence $3^2$ and $7^2$ divide both $a^2$ and $b^2$ and $3^2\cdot 7^2\mid (a^2+b^2)$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that if $r\overrightarrow {X}=\overrightarrow {0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$
Suppose $\overrightarrow{X}\in{\bf R}^2$ and $r\in{\bf R}$. Show that if $r\overrightarrow {X}=\overrightarrow{0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$.
[Attempt:]
Let $\o... | Your proof is correct. As long as you have shown that $\vec{X}=\vec{0}$ assuming $r\neq 0$, all the rest of your work are redundant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| $$\begin{array}{rrcl}
& (a^3+b^3)^2 &\le& (a^2+b^2)(a^4+b^4) \\
\iff& a^6 + 2a^3b^3 + b^6 &\le& a^6+a^2b^4+b^2a^4+b^6 \\
\iff& 2a^3b^3 &\le& a^2b^4+b^2a^4 \\
\iff& 2ab &\le& b^2+a^2 \\
\iff& 0 &\le& b^2-2ab+a^2 \\
\iff& 0 &\le& (b-a)^2 \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving for $k$: $\sqrt{k-\sqrt{k+x}}-x = 0$
$$\sqrt{k-\sqrt{k+x}}-x = 0$$
Solve for $k$ in terms of $x$
I got all the way to
$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$
but could not factor afterwards. My teacher mentioned that there was grouping involved
Thanks Guys!
Edit 1 : The exact problem was solve for $x$ given that $$\s... | In this kind of problem, you have to be very careful about the domain of definition. Squaring the equation an find an equivalent polynomial equations is not enough, you need to verify if the solutions found are effective solutions of the original equation.
First two remarks :
*
*$x=\sqrt{\cdots}\quad$ thus $x\ge 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Number of ways for a 6 digit license plate such that each digit is unique except for one The problem asks how many different six digit license plates can be made if there are three digits that occur once and one digit that occurs thrice?
My idea is that there can be 10 numbers for the reoccurring digits, and $6 \choose... |
Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits.
This is not about the number of ways of 'laying them out', but of picking digits for the three remaining digits.
So yes, you have indeed $6 \choose 3$ ways to lay out the recurring digit, and you have $10$ choices for what that digit is.
But... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$|\int_{0}^{a+bi}\cos(z^2)\,dz|\leq(a^2+b^2)^{1/2}\sinh(2ab)/(2ab)$ Assuming that $a>0$ and $b>0$, derive the estimate
$$\left|\int_{0}^{a+bi}\cos(z^2)\,dz\right|\leq \frac{(a^2+b^2)^{1/2}\sinh(2ab)}{2ab}$$
I know $|\int_{\gamma}f(z)\,dz|\leq \int_{\gamma}|f(z)|\,dz$, but I do not know how to apply it here, I have trie... | First note that \begin{align}
\int_0^{a+bi} \cos (z^2) dz&=\int_0^1\cos((a+bi)^2t^2)\cdot (a+bi)dt\quad (z=(a+bi)t,\, 0\le t\le 1)\\
&=(a+bi)\int_0^1\cos((a^2-b^2)t^2+2abt^2i)dt.
\end{align}
Using the inequality \begin{align}
|\cos (\alpha +\beta i)|&=\left|\frac{\exp(i(\alpha +\beta i))+\exp(-i(\alpha +\beta i))}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$ with epsilon-delta I am looking to show that $$f:(0,1)\rightarrow \mathbb{R} \lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$$ Normally for these types of problems I have been doing epsilon-delta proofs but I cannot figure out how to define my $\... | Consider that $$\frac{\sqrt{9-x}-3}{x}+\frac{1}{6} = \frac{x+6\sqrt{9-x}-18}{6x} = -\frac{\left(\sqrt{9-x}-3\right)^2}{6x}$$ If $\lim_{x\to 0^+} \frac{f(x)}{x} = c$, then for $\epsilon > 0$, there is a $\delta > 0$ such that $$\left\lvert \frac{f(x)}{x}-c\right\rvert = \left\lvert \frac{f(x)-cx}{x}\right\rvert < \epsil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \f... | By Holder
$$\sum_{cyc}\frac{1}{ab}=\sum_{cyc}\frac{1}{ab}\sum_{cyc}a\sum_{cyc}b\geq\left(\sum_{cyc}\sqrt[3]{\frac{1}{ab}\cdot a\cdot b}\right)^3=\left(\sum_{cyc}1\right)^3=27.$$
The equality occurs for $a=b=c=\frac{1}{3}$, which says that $27$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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