Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is wrong with this proof of: $2+2 = 5$ I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.
Prove that, $2+2=5$.
We know that, $2+2=4$
$$\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\
&=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{16... | No,its not possible
your second step is wrong
$4-4.5$ is negative number
So $\sqrt{(4-4.5)^2}$ is not possible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
List of Three Letters I'm having issues with my logic on this problem:
How many ways are there to form a list of three letters from the letters in the word COMBINATORICS if the letters cannot be used more often than they appear in COMBINATORICS?
I'm trying to think of this as a set with 13 elements (2 Cs, 2 Os, 2 Ms... | A general method for solving this kind of problem uses exponential generating functions. Let
$$
\begin{aligned}
G(x,C,O,I,M,\ldots)=&\left(1+\frac{Cx}{1!}+\frac{C^2x^2}{2!}\right)\left(1+\frac{Ox}{1!}+\frac{O^2x^2}{2!}\right)\left(1+\frac{Ix}{1!}+\frac{I^2x^2}{2!}\right)\left(1+\frac{Mx}{1!}\right)\\
&\times\left(1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| For a lot of ways to solve quadratic equations, see Pat Ballew's manuscript Solving Quadratic Equations.
Here's a way to get the quadratic formula if you partially remember it. (I don't know off-hand if this is one of the methods in Pat Ballew's list.)
We remember that the quadratic formula gives us results that can be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Solve this Diophantine Equation: $a^3-b^3 = ab+61$ "solve this diophantine equation: $a^3-b^3 = ab+61$"
For Diophantine Equations I have always had the question in the form $ax+by = c$ and solved it via Extended Euclidean Algorithm. However, this question is not formatted in such a way.
What I am thinking is that we m... | Finding all integer solutions to the equation $a^3-b^3=ab+61$ is equal to solving the diophantine equation $a^3-ab-b^3=61$, and placing $-b$ instead of $b$ gives $a^3+ab+b^3=61$.
Since the equation above is symmetric ( that is, the value doesn't changes even if we swap the position of the variables a and b), it can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
| Note that $4-2x^2 = 2(2-x^2)$.
Muliplying both sides by $\sqrt{2-x^2}$ shows that any solution of the equation also solves $2 = 2 (2-x^2)^{\frac{3}{2}}$, or equivalently $(2-x^2)^{\frac{3}{2}} = 1$.
This is equivalent to solving $2-x^2 =1$, which yields $x^2 = 1$, from which we get $x = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to determine whether critical points (of the lagrangian function) are minima or maxima? $f(x,y) = 2x+y$ subject to $g(x,y)=x^2+y^2-1=0$. The Lagrangian function is given by
$$
\mathcal{L}(x,y,\lambda)=2x+y+\lambda(x^2+y^2-1),
$$
with corresponding
$$
\nabla \mathcal{L}(x,y,\lambda)= \begin{bmatrix} 2 + 2\lambda x... | Fortunately $\{x^2+y^2-1=0\}$ is compact and $f$ continuous thus $f$ must have a min and max. Just compare the values of $f$ at the critical values. In this case the Hessian is not needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Math question from the GMATprep If $xy=1$ what is the value of: $2^{(x+y)^2}/2^{(x-y)^2}$
A 1
B 2
C 4
D 16
E 19
$(x+y)^2/(x-y)^2$ because $2$ just cancels out from numerator and denominator, right?
| Using exponentiation rules, it can be simplified.
$$ \frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2^{(x+y)^² - (x-y)^2} = 2^{x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)} = 2^{4xy} = \left(2^{4}\right)^{xy} = 16 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/519922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that the sum of squares of four consecutive natural numbers may never be a square.
Show that the sum of squares of four consecutive natural numbers may never be a square.
I know (and I have the proof) a theorem that says that every perfect square is congruent to $0, 1$ or $4$ $\pmod8$ and wanted to demonstrate t... | Let $n$, $n+1$, $n+2$, and $n+3$ be any four consecutive natural numbers. Then we have $$ n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = 4n^2 + 12n + 14 = (2n + 3)^2 + 5. $$ Now if this sum were to be the square of a natural number $N$, say, then we must have $$ N^2 - (2n + 3)^2 = 5.$$ However we know that for any two distinct na... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Spot my error in solving a linear system I almost always get the unit matrix if I try to get to an row reduced echelon form. I probably always make a mistake. Can you spot the error? What illegal operations could a beginner do while trying to solve a linear system?
\begin{array}{}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 &... | In the move from $1$ to $2$, you incorrectly obtained $6$ when you should have obtained 4: $4 \times R_1 + R_2 \to R_2 = 0\;\;1\;\; 4\;$
thus giving to identical rows, rows 2 & 3.
\begin{pmatrix}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 & -2 \\
1 & 2 & 7 \end{pmatrix}
2.
\begin{pmatrix}
1 & 1 & 3 \\
0 & 1 & 4 \\
0 & 1 & 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find this numbers $a,b$ Question
let function $f(x)=ax^2+b$, find all positive real numbers $(a,b)$,such for any real numbers,then we have
$$f(xy)+f(x+y)\ge f(x)f(y)$$
My try:
since $$f(xy)+f(x+y)\ge f(x)f(y)\Longrightarrow a(xy)^2+b+a(x+y)^2+b\ge (ax^2+b)(ay^2+b)$$
| $f(xy)+f(x+y)≥f(x)f(y)⟹a(xy)^2+b+a(x+y)^2+b≥(ax^2+b)(ay^2+b)$
$ax^2y^2+a(x^2+y^2)+2axy+2b\ge a^2x^2y^2+ab(x^2+y^2)+b^2$
$2b-b^2\ge (a^2-a)x^2y^2+(ab-a)(x^2+y^2)-2axy$....(*)
$\frac{2b-b^2}{a}\ge (a-1)x^2y^2+(b-1)(x^2+y^2)-2xy$
In the last part, we assumed that $a\neq0$. Since the left side is a constant, the right side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the units digit in the number $7^{9999}$. I have step by step instructions from a previous example to follow, so I figure I know how to get the answer, but I don't understand fully why it works the way it does...
By Euler's theorem, if $(a,m)=1$ then $a^{\varphi(m)}\equiv 1\mod{m}$. Here we have $(7,10)=1$ and $\... | We know that
$$ 7^2 \equiv -1 \pmod{10} $$
and that
$$ 7^{9999} = 7 \cdot 7^{9998} = 7 \cdot (7^2)^{4999}, $$
and that
$$a_1 \equiv b_1 \pmod c \qquad\text{and}\qquad a_2 \equiv b_2 \pmod c \Rightarrow a_1a_2 \equiv b_1b_2 \pmod c.$$
Therefore,
$$ 7^2 \equiv -1 \pmod{10} \Rightarrow 7^{9998} = (7^2)^{4999} \equiv (-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Asymptotics of sequence depending on Tricomi's function I'm dealing with the following sequence
$$ p_n = U(a, a - n, 1)$$
where $a > 0$ and $U$ is Tricomi's function.
I suspect that asymptotically when $n \to \infty$ its behaviour is a power law (evidence)
$$ p_n \sim 1/n^a $$
but I cannot arrive to anything with known... | There are some details in the following derivation which will be omitted to keep this answer at a reasonable length. What follows is a set of clotheslines on which a rigorous proof can hopefully be hung.
We will use the integral representation
$$
\begin{align}
U(a,a-n,1) &= \frac{1}{\Gamma(a)} \int_0^\infty e^{-t} t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)}$ $$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)} =[1]\lim_{x\to 0}\frac{\sin x/\cos x-\sin x}{x^3}\\ =[2]\lim_{x\to 0}\frac{1-\cos x}{x^2}\\ =\frac{1}{2} $$
My question is how [1]=[2]?
$\tan x-\sin x=\tan x(1-\cos x)=x(1-\cos x)$?
| We have $\frac{\sin x}{\cos x}-\sin x=\frac{\sin x}{\cos x}(1-\cos x)$.
Thus our function is equal to
$$\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}.$$
The limit of $\frac{1}{\cos x}$ is $1$. So is the limit of $\frac{\sin x}{x}$.
You did not ask about the limit of $\frac{1-\cos x}{x^2}$. That can be obtained ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer? What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?
The root mean square is defined as:
$$\sqrt{\left(\frac{a_1^2 + a_2^2 + a_3^... | Starting with $1+48k^2= m^2$ or $m^2-48k^2 = 1$, it makes sense to try convergents to $\sqrt{48}$, which are $\left\{6,7,\frac{90}{13},\frac{97}{14},\frac{1254}{181},\frac{1351}{195},\frac{17466}{2521}, ...\right\}$.
We spot solutions $(m, k) = (7, 1), (97, 14), (1351, 195), \;\dots $ corresponding to $n = 1, \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$ This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\s... | As a backup check, you could use $\sqrt{x+a} = \sqrt{a}+\frac{1}{2\sqrt{a}} x +o(x)$, or $\sqrt{x+a} - \sqrt{a} =\frac{1}{2\sqrt{a}} x +o(x)$.
Then $\frac{\sqrt{x^2+2}-\sqrt{2}}{ \sqrt{x^2+1}-\sqrt{1} } = \frac{ \frac{1}{2\sqrt{2}} x^2 +o(x^2) }{ \frac{1}{2\sqrt{1}} x^2 +o(x^2) } =
\frac{ \frac{1}{2\sqrt{2}} +\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral of quartic function in denominator I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I ... | $$
\int\frac{x}{x^4+x^2+1}dx=\int\frac{x}{x^4+2x^2+1-x^2}dx=\int\frac{x}{\left(x^2+1\right)-x^2}dx=
$$
$$
=\int\frac{x}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=\frac{1}{2}\int\frac{\left(x^2+x+1\right)-\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=
$$
$$
=\frac{1}{2}\left(\int\frac{dx}{x^2-x+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving this logarithm equation? How do I solve this equation using common logarithms?
$\log x = 1-\log(x-3)$
| Let $1 = \log 10$ then, $$\log x = \log 10 - \log (x - 3).$$ For all $x$ and $y$ and a constant $a$, one always has $$\log_a xy = \log_a x + \log_a y.\tag1$$ This is one of the most important logarithm rules. From this rule, we can express $\log\dfrac xy$. $$\log \frac xy = \log x\bigg(\frac 1y\bigg),$$ and from $(1)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove geometric sequence question Prove that $x+2x^2+3x^3+4x^4+...+nx^n = \frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$
I see that this can be written as $$\sum_{n=1}^n nx^n = n\sum_{n=1}^n x^n$$
$$\sum_{n=1}^n x^n = \frac{x(x^n-1)}{(x-1)}$$
$$\therefore n\sum_{n=1}^n x^n = n(\frac{x(x^n-1)}{(x-1)})$$
Following this I squar... | Multiply $x + 2x^2 + 3x^3 + 4x^4+...+nx^n$ with $x-1$. You'll get:
$$(x + 2x^2 + 3x^3 + 4x^4+...+nx^n)(x-1) = x(x + 2x^2 + 3x^3 + 4x^4+...+nx^n) - (x + 2x^2 + 3x^3 + 4x^4+...+nx^n)$$
Expand it and you'll obviously end up with sort of telescopic series. Every number in the first term, except the last one will be of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\... | Rewrite it as
$$\sum_{cyc} \frac{a}{\sqrt{b}} + (\sqrt{b} - 2\sqrt{a}) = \sum_{cyc} \frac{a}{\sqrt{b}} -\sum_{cyc} \sqrt{a} \ge 3 - \sum_{cyc} \sqrt{a} = \sqrt{3(a+b+c)} - \sum_{cyc} \sqrt{a}$$
LHS becomes
$$\sum_{cyc} \frac{(\sqrt{a} - \sqrt{b})^2}{\sqrt{b}}$$
RHS becomes
$$\frac{3(a+b+c) - (\sum_{cyc} \sqrt{a})^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Can't solve this problem $lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x}$(Without using L'Hospital's rule) I don't see the way to solve this limit.
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} $$
My attempt is
1) Divide the numerator by $3^{5x}$
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}... | Using L'Hospital's rule, we find
\begin{align*}
\lim_{x \to 0} \frac{3^{5x} - 2^{7x}}{\arcsin{2x} - x} &= \lim_{x \to 0} \frac{3^{5x} (5 \ln{3}) - 2^{7x}(7 \ln{2})}{\frac{2}{\sqrt{1 - (2x)^2}} - 1} \\
&= \frac{3^0 (5 \ln{3}) - 2^0 (7 \ln{2})}{2 - 1} \\
&= 5 \ln{3} - 7 \ln{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If a line makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes, then $\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma} = 2 $ The following is the question in my textbook:-
If a straight line makes angle $\alpha$, $\beta$, $\gamma$ with the $x, y, z$ axes respectively, then show that $\sin^2{\alpha} + \sin^2{\b... | Yes, the solution is correct. Turning $\sin^2$ into $1-\cos^2 $ reduces the problem to the sum of squared cosines being $1$, and the latter fact is essentially the distance formula. Indeed, move the line so that it passes through the origin and pick a point $(u,v,w)$ on the line at distance $1$ from the origin. Then $u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fermat's 2 Square-Like Results from Minkowski Lattice Proofs
Minkowski's Convex Body Theorem for lattices in the plane: Suppose $\mathfrak{L}$ is a lattice in $\mathbf{R}^2$ defined as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$, where $\vec{v_1}$ and $\vec{v_2}$ are linearly independent. Suppose $d$ is ... | Unfortunately I'm answering my own question, but I was able to prove that if $p\equiv 1\pmod{8}$ or $p\equiv 3\pmod{8}$, then $p$ is expressible as $x^2+2y^2$ for $x,y\in\mathbf{Z}$, and if $p\equiv 1\pmod{3}$, then $p$ is expressible as $x^2+3y^2$ for $x,y\in\mathbf{Z}$. For posterity, here it is.
Lemma: $p\equiv 1\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Intersection of two tangents on a parabola proof There are two tangent lines on a parabola $x^2$. The $x$ values of where the tangent lines intersect with the parabola are $a$ and $b$ respectively. The point where the two tangent lines intersect has an $x$ value of $c$.
Prove that $c=(a+b)/2$
I have tried taking the d... | Consider the curve $C$ with the equation $y=x^2$. When $x=a$, we have $y=a^2$. The derivative is given by $y'=2x$ and so the gradient at the is exactly $2a$. The equation of the tangent line is
$$y-a^2=2a(x-a)$$
Expanding this gives $y-2ax=-a^2$. Similarily, the equation of the tangent line to $C$ at the point with $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating Artin symbol Consider the field $K=\mathbb Q(\sqrt{2})$. Let $mp=\frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}$.
For the field extension, $K(-1+2\sqrt{2})/K$, and $p\equiv 5\bmod{6}$ how can one show
$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\equiv 1\pmod{-1+2\sqrt{2} }$
and hence, Artin symbol corresponding to $ K... | The Galois group of the extension $K(-1+2\sqrt2)/K$ is trivial, so all the Frobenius automorphisms are equal to the identity for lack of alternatives.
In the specific example from comments we can also check it as follows. You already showed that $(-1+2\sqrt2)\mid 7$. Therefore
$$2+\sqrt2\equiv2-6\sqrt2=-1-3(-1+2\sqrt2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ I am trying to prove that
$$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +G\log 2$$
where $G$ is the Catalan's Constant. Numerically, it's value is $-0.199739$.
Using the substitution $x... | A solution proposed by Cornel Ioan Valean
As also shown in this post, an immediate connection can be made with the integral in the current OP. So, integrating by parts, rearranging, and employing the needed results, we arrive at once at the desired closed form
$$\int_0^1 \frac{x \arctan(x) \log(1-x^2)}{1+x^2}\textrm{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 0
} |
How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$? Let $$0\le a\le 1,0\le b\le 1$$
Prove or disprove
$$\sin{\left(\dfrac{\pi}{2}ab\right)}\le\sin{\left(\dfrac{\pi}{2}a\right )}\sin{\left(\dfrac{\pi}{2}b\right)}$$
My try:
Since
$$\... | The proof is related to the concavity of the $\sin\left(\frac\pi2 x\right)$ curve.
Consider the following blue curve, which shows $y = \sin\left(\frac\pi2 x\right)$. The red dot $P$ is at $\left(A, \sin\left(\frac\pi2 A\right)\right)$. Now I fit (the red curve) a smaller version of $y = \sin\left(\frac\pi2 x\right)$ en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| \begin{align}
x^3+y^3+z^3-3xyz\\
&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\\
&= (x+y)^3+z^3-3xy(x+y+z)\\
&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\\
&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\\
&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/543991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 2
} |
True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
| EDIT: I found a much better solution:
WLOG assume that $b,c$ are on the same side of $1$. Bringing the RHS over to the LHS, it suffices to show that
$$(a-1)^2+2a(b-1)(c-1)+(b-c)^2\ge 0$$
But since $b,c$ are on the same side of $1$, this is true.
Old solution:
We prove the inequality for $a,b,c\ge 0$.
The inequality is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$? How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$?
What else? Well.
Well can you use long division?
| Use method add something and take something.
$$\frac{t}{t-1}= \frac{t-1+1}{t-1}$$
Then splits equation on 2 parts.
$$\frac{t-1}{t-1}+ \frac{1}{t-1} $$
Then reduce fraction and you get: $$1+\frac{1}{t-1}$$
If this is true
$$\frac{t}{t-1}= 1-\frac{1}{t-1}$$
Than you get:
$$\frac{t}{t-1}= \frac{t-1-1}{t-1} $$
Than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to efficiently compute $17^{23} (\mod 31)$ by hand? I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 3... | We can do away with manipulating the powers of $17$ as follows
Observe that $17\equiv-14\pmod{31}$
$\displaystyle\implies 17^{23}\equiv(-14)^{23}\equiv-14^{23}\pmod{31}$
Now, $7^2=49\equiv18\pmod{31}, 7^3\equiv18\cdot7\equiv2\pmod{31}$
$\displaystyle\implies14^{23}=2^{23}\cdot7^{23}=2^{23}(7^3)^7\cdot7^2\equiv2^{23}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Quadratic equation and trig If the quadratic equation $ax^2+bx+c=0$ has equal roots where $a, b$ and $c$ denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$
Just some h... | Let $s=\frac ab,t=\frac cb$. Then, we have
$$b^2-4ac=0\Rightarrow 4st=1.$$
So, we have
$$\begin{align}&a+b\gt c,\quad b+c\gt a,\quad c+a\gt b\\&\iff s+1\gt t,\quad 1+t\gt s,\quad t+s\gt 1\\&\iff s+1\gt \frac{1}{4s},\quad 1+\frac{1}{4s}\gt s,\quad \frac{1}{4s}+s\gt 1\\&\iff \frac{\sqrt 2-1}{2}\lt s\lt \frac 12\quad\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$
let $x,y,z$ be positive numbers, and such $x+y+z=1$
show that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
My try:
let
$$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$
so
$$a=y\ln{x}-x\... | Another solution.
Without loss of generality, we can assume that $0<x\le y\le z$ and so $x=az$, $y=bz$, $0<a\le b\le1$. By the AM-GM inequality, we find that
\begin{align}
\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}&\ge 3\sqrt[3]{x^{y-z}y^{z-x}z^{x-y}}=3\sqrt[3]{(az)^{(b-1)z}(bz)^{(1-a)z}z^{(a-b)z}}=3\sqrt[3]{a^{(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
limit $\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$ Can you help me with
$$\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$$
I've already tried to multiply at conjugate expression but I failed.
| $$ \lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2} \\$$
Let $ a= a=\sqrt[3]{x-6} $ and $ b= {2} $ . Then $ a^3+b^3 = (a+b)(a^2-ab+b^2)$ . So $ a^3+b^3 = x+2 $ and $ a+b = \sqrt[3]{x - 6}+2 $ . So $ \frac{\sqrt[3]{x - 6}+2}{x + 2} $ = $ \frac{a+b}{a^3+b^3}$ = $\frac{1}{ a^2 - ab +b^2} $ . Hope so this helps a lot .
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
evaluation of $\int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$
By Using Substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
I have Tried it without using the given substution.
$\bf{My\; Try}::$ $\displaystyle \int_{0}^{1}\frac{1-x^2... | Nicely done. So they suggest setting
$$
\cos \theta =\frac{\sqrt{1+x^4}}{1+x^2} \quad\Rightarrow\quad \frac{2x(1-x^2)}{\sqrt{1+x^4}(1+x^2)^2}dx=\sin \theta \,d\theta
$$
and, since $\sin\theta\geq 0$ here:
$$
\sin\theta =\sqrt{1-\cos^2\theta}=\frac{\sqrt{2} \;x}{1+x^2} \quad\Rightarrow \quad \frac{1-x^2}{(1+x^2)\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$ \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0$ implies that $ \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0$ $a ,b , c$ are real numbers such that $ \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0$ , then how do we prove
(without routine laborious manipulation) that $\dfrac a{(bc-a^2)^2} + \... | $$\frac 1 {b c - a^2 } + \frac 1 {c a - b^2} + \frac 1 {a b - c^2} =
\frac {
(b c + a c + a b) (bc+ac+ab-a^2-b^2-c^2)
}
{
(bc-a^2)(ac-b^2)(ab-c^2)
}
$$
$$ \frac a {(bc - a^2)^2} + \frac b {(ca - b^2)^2} + \frac c {(ab - c^2)^2} =
\frac{
(bc+ac+ab)(bc+ac+ab-a^2-b^2-c^2)
}
{
(bc-a^2)(ac-b^2)(ab-c^2)
}
\times
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ ... | The method we were taught was to set up an augmented matrix like so:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13\\
\end{array}\right]
$$
And then solve the matrix to the left of the vertical bar like a homogeneous system.
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a matrix $X$ given $X^4$ Find the matrix $X$ such that
$$X^4=\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
This problem I can't work,and I think let the matrix the eigenvalue is $\lambda$,then
$\lambda^4$ is $$\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
eigenvalue?Thank you for your help.
| Hint: Write the Jordan Normal Form., then take:
$$X = S.J^{1/4}.S^{-1}$$
So, we have:
$$\begin{bmatrix} 3&0&0\\ 0&3&1\\ 0&0&0 \end{bmatrix}$$
We can write this in Jordan Normal Form as:
$$S.J.S^{-1} = \begin{bmatrix} 0&0&1\\ -1&1&0\\ 3&0&0 \end{bmatrix}.\begin{bmatrix} 0&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}.\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $\int_0^\infty \frac{dx}{1+x^3}$ Problem
Compute $$\displaystyle \int_0^\infty \frac{dx}{1+x^3}.$$
Solution
I do partial fractions
$$\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.$$
But we could simplify the left one $$\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\... | Since
$$
\frac{1}{x^3+1}=\frac{1}{(x+1)(x^2-x+1)},
$$
we can find some $a,b,c \in \mathbb{R}$ such that
$$
\frac{1}{x^3+1}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}.
$$
A simple computation shows that
$$
a=-b=\frac{c}{2}=\frac13,
$$
i.e.
\begin{eqnarray}
\frac{1}{x^3+1}&=&\frac13\cdot\frac{1}{x+1}-\frac13\cdot\frac{x-2}{x^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate the limit $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}$ First I tried direct substitution, which resulted in the indeterminate form. Then because of the square roots, I tried rationalizing (both numerator and denominator), but still get the indeterminate form. I can't use L'Hopita... | $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}\cdot\frac{\sqrt{3x+4}+\sqrt{2x+4}}{\sqrt{3x+4}+\sqrt{2x+4}}\cdot\frac{\sqrt{x+1}+\sqrt{2x+1}}{\sqrt{x+1}+\sqrt{2x+1}}=\lim_{x\to0}\frac{-x(\sqrt{3x+4}+\sqrt{2x+4})}{x(\sqrt{x+1}+\sqrt{2x+1})}=-\frac{4}{2}=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the... | The following is a proof of the hypergeometric identity $$_2F_1\left(a,1-a;\frac{1}{2};\sin^2(x)\right)=\frac{\cos[(2a-1)x]}{\cos x} \, , \quad - \frac{\pi}{2} < x < \frac{\pi}{2} \tag{1}.$$
This identity is used in R. J. Mathar's answer.
Similar to my answer here, I will first use the generalized binomial theorem to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 1
} |
Proving combinatorial identity $\sum_s (-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0$ I need to prove following combinatorial identities:
$$
\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0
$$
$$
\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s-1}{2m-s}2^s=(-1)^m\binom{p+m-1}{m}
$$
given the fact t... | Suppose we seek to verify that
$$\sum_{q=0}^{2m} (-1)^q {p-1+q\choose q}
{2m+2p+q-1\choose 2m-q} 2^q =
(-1)^m {p-1+m\choose m}.$$
Introduce
$${2m+2p+q-1\choose 2m-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m-q+1}} (1+z)^{2m+2p+q-1} \; dz.$$
Observe that this controls the range being zero when $q\gt 2m$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Residue/Contour integration problem Supposedly,
$\displaystyle\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right)$, $a>0$.
Using Residues/Contour integrals, I have
$\displaystyle\int_{-\infty}^\infty \frac{e^{iax}}{x^4+1}dx=\int_{... | Here, because $a >0$, it is easiest to use a contour $C$ that is a semicircle of radius $R$ in the upper half-plane in the complex plane. Thus, consider
$$\oint_C dz \frac{e^{i a z}}{z^4+1} $$
This contour integral is equal to
$$\int_{-R}^{R} dx \frac{e^{i a x}}{x^4+1} + i R \int_0^{\pi} d\theta \, e^{i \theta} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why is $(2+\sqrt{3})^{50}$ so close to an integer? I just worked out $(2+\sqrt{3})^{50}$ on my computer and got the answer
$39571031999226139563162735373.999999999999999999999999999974728\cdots$
Why is this so close to an integer?
| Let $$x=2+\sqrt { 3 }, \\$$then$$ { x }^{ 2 }=7+4\sqrt { 3 }, \\ 4x=8+4\sqrt { 3 } ,$$ therefore $${ x }^{ 2 }+1=4x,\\ x+\frac { 1 }{ x } =4.$$ Take the square of both side and repeat the process $${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ 4 }^{ 2 }-2\\ { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } ={ \left( { 4 }^{ 2 }-2 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 1
} |
What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3... | The formula $$\sum_{k=1}^{n}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ is incorrect. The correct formula is $$\sum_{{\Large k=}{\Huge0}}^{\Huge n-1}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ or $$\sum_{k=1}^{n}ar^{k}=a{\Large{r}}\left(\frac{1-r^{n}}{1-r}\right)$$. With $a=1$ and $r=2$, you should have $\sum_{k=1}^{n+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{ali... | Note that we can simply add zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. Then
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2(x^2 + 4) - 8 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2}{x - 1} + \frac{5x - 5}{(x - 1)(x^2 + 4)}.$$
Do you think you can integrate
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)}$$
now?
Hint:
$$\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the limit of $\frac{1−\cos 5x}{x^2}$ as $x\to 0$? What is the right approach to calculate the Limit of $(1-\cos(5x))/x^2$ as $x \rightarrow 0$?
From Wolfram Alpha, I found that:
$$\lim_{x \to 0} \frac{1- \cos 5x}{x^2} = \frac{25}{2}.$$
How do I get that answer?
| $\cos(2\alpha) = 1-2\sin ^2 \alpha$
Therefore
$\frac{1-\cos (5x)}{x^2}=\frac{2 \sin ^2 \left( \frac{5x}{2}\right)}{x^2} = \frac{25}{2} \left( \frac{\sin \frac{5}{2}x}{\frac{5}{2}x} \right)^2 \rightarrow \frac{25}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/567263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
} |
Show that $x^3+y^3 = 3\mod 9$ has no solutions Not even sure how to go about this. I tried $x^3 = 3-y^3 \mod 9$, but not sure what that does.
| Modulo 9, we have:
$$0^3=0, 1^3=1, 2^3=-1, 3^3=0, 4^3=1, 5^3=-1, 6^3=0, 7^3=1, 8^3=-1$$
No pair of these sums to 3.
In fact, we can use this table to conclude the stronger statement that $$x^3+y^3+z^3=4\pmod{9}$$ has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/567724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.
| Here's an "elementary" proof that doesn't use any theorems such as l'Hôpital. Start off by writing $\frac{n^5}{3^n} = \frac{n^5}{2^n} \left(\frac{2}{3}\right)^n$. The term $\left(\frac{2}{3}\right)^n$ tends to zero as $n \rightarrow \infty$, so we would be done if we showed that the term $\frac{n^5}{2^n}$ is bounded as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Finding polynomal function with given zeros and one zero is a square root I've been having trouble with this problem: Find a polynomial function of minimum degree with $-1$ and $1-\sqrt{3}$ as zeros. Function must have integer coefficients.
When I tried it, I got this:
\begin{align}
(x+1)(x-(1-\sqrt{3}))=& x^2 - x(1-\s... | If you're looking for a polynomial that has those two roots and integer coefficients, you'll need to add another root.
Substitute $w = x-1$ so that $(x+1)(x-(1-\sqrt{3})) = (w+2)(w+\sqrt{3})$.
Now multiply by $(w-\sqrt{3})$ to get
$(w+2)(w+\sqrt{3})(w-\sqrt{3}) = (w+2)(w^2-3)$.
Then substitute back in $x$ and you have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_... | (The following is basically obsolete because already answered in
https://math.stackexchange.com/questions/571967/integral-inta2-b2-sec2x-sqrta2-b2-sec2xdx)
The first integral of Harry Peter is easy with the substitution $x^2=t$.
The second, attacked with partial integration, requires to integrate
$(a^2-x^2)/2$ which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
How to evaluate a limit with subtractions $\lim_{x \rightarrow -1}(\frac{3}{x^3+1}-\frac{1}{x+1})$? I'm having trouble thinking of a way to solve this.
$$\lim_{x \rightarrow -1}\left(\frac{3}{x^3+1}-\frac{1}{x+1}\right)$$
| Because$$x^3+1=(x+1)(x^2-x+1)$$we get
$$\frac{3}{x^3+1}-\frac{1}{x+1}=\frac{3}{(x+1)(x^2-x+1)}-\frac{1}{x+1}$$
$$=\frac{3-(x^2-x+1)}{(x+1)(x^2-x+1)}=\frac{x-x^2+2}{(x+1)(x^2-x+1)}=$$
$$=\frac{x+1-(x^2-1)}{(x+1)(x^2-x+1)}=\frac{(x+1)(2-x)}{(x+1)(x^2-x+1)}=\frac{2-x}{x^2-x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding exactly one real solution to the system
I dont know how to go about finding the one real solution to the system where k is a real number
Thank you
| You can write $n=\frac{6}{m}$. Substituting in the first equation, we get $m+\frac{6}{m}=k$, and then $m^2-km+6=0$.
This is a quadratic equation, which we can solve using the Quadratic Formula. We get
$$m=\frac{k\pm\sqrt{k^2-24}}{2}.$$
There is a single solution if $k^2-24=0$. There is no real solution if $k^2-24\lt 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
| Another approach:
Put $x=\sqrt[3]{2y-1}$ then $x^3=2y-1$. So, $x^3+1=2y$
Then we have the system of equation:
$\left\{\begin{matrix} x^3+1=2y\\y^3+1=2x \end{matrix}\right.$
Subtracting these two equations from each other, we have
$(x-y)(x^2+xy+y^2+2)=0$
which gives $x=y$ as ($x^2+xy+y^2+2>0 $ $\forall$ $x,y$)
Hence we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
I need to find the second derivative! I need to find the second derivative of
$$f(x)=\frac{\sqrt{x}}{x+4}.$$
I think the first derivative is
$$f'(x)=\frac{4-x} {2 \sqrt{x} (x+4)^2}.$$
EDIT: I got the second derivative to be:
$$f''(x)=-\frac{1}{4 x^{\frac{3}{2}} (x+4) }-\frac{1}{\sqrt{x}(x+4)^2 }+\frac{2 \sqrt{x}}{(x+... | Hint.
$$\begin{align}
\frac{d}{dx} \frac{4 - x}{2\sqrt{x}(x + 4)^2} ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - (4 - x)\frac{d}{dx}(2\sqrt{x}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2} \\
~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - 2(4 - x)((x + 4)^2\frac{d}{dx}\sqrt{x} + \sqrt{x}\frac{d}{dx}(x + 4)^2)}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \... | Note that, since you can change $a,b,c$ to $-a,-b,-c$, you can assume that at least two of the $a,b,c$ are nonnegative. So, first start by supposing that two of your numbers are nonnegative, and after that order them. Therefore, without loss of generality, assume that $a\leq b\leq c$, with $0\leq b\leq c$. Then, for th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| since
$$a^2+1\ge 2a$$
so
$$a^2+b^2+c^2+3\ge a+b+c+(a+b+c)\ge a+b+c+3\sqrt[3]{abc}=a+b+c+3$$
so
$$a^2+b^2+c^2\ge a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Finding value of equation without solving for a quadratic equation How do I go about solving this problem:
If $α$ and $β$ are the roots of $x^2+2x-3=0$, without solving the equation, find the values of $α^6 +β^6$.
In my thoughts: I commenced by expanding $(α +β)^6$, such that:
$$(α +β)^6 =α^6+6α^5β+15α^4β^2+20α^3β^... | HINT:
$$a^6+b^6=(a^3)^2+(b^3)^2=(a^3+b^3)^2-2(ab)^3\text{ and } a^3+b^3=(a+b)^3-3ab(a+b)$$
or
$$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)^3-3(ab)^2(a^2+b^2) \text{ and } a^2+b^2=(a+b)^2-2ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/582960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Integration with substitution I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
| Using Trigonometric substitution
put $x-3=\sqrt5\tan\theta$ as $\displaystyle x^2-6x+14=(x-3)^2+(\sqrt5)^2$
$$\int \frac{2x+1}{(x^2-6x+14)^3}dx=\int\frac{2(\sqrt5\tan\theta+3)+1}{(5\sec^2\theta)^3}\sqrt5\sec^2\theta d\theta$$
$$=\frac1{25\sqrt5}\int(2\sqrt5\tan\theta+7)\cos^4\theta d\theta=\cdots$$
$$=\frac2{25}\int\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Triangle - coordinate geometry problem Let ABC be a triangle. Let BE and CF be internal angle bisectors of B and C
respectively with E on AC and F on AB. Suppose X is a point on the segment CF
such that AX is perpendicular to CF; and Y is a point on the segment BE such that AY perpendicular BE. Prove
that XY = $\frac{b... | The following is a diagram. And $B$ is the origin and assume each point $P$ has coordinate $(p_1,p_2)$.
*
*Point $A,B,C$
From the length three edges. We can immediately write $B(0,0)$ and $C(a,0)$. Then solve $\displaystyle A(\frac{a^2-b^2+c^2}{2a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{2a})$ from $a_1^2+a_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How we got $z\cdot(x+y)=x\cdot y$ This is from "Test of math at 10+2 level":
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferr... | The problem is symmetric in $x$ and $y$. That means that at any step, you should be able to reverse:
1) $x$ and $y$
2) Wine and Water
3) vessel 1 and vessel 2
And the result is the exact same relation.
You can also check at each step that wine in vessel1 plus water in vessel1 is x, and that wine in vessel2 plus water ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Exponential map of 2x2 matrix I saw the following formula (in (7.25) of Lie Groups, Physics, and Geometry)
$$
\exp\left(
\begin{array}{c}
w & z\\
0 & 1\\
\end{array}
\right)
=
\left(
\begin{array}{ccc}
e^w & (e^w-1)z/w\\
0 & 1 \\
\end{array}
\right).
$$
Is this correct or some kind of typo? At l... | The formula is indeed incorrect, including the case when $z=0$.
In the simple case that $z=0$, you have a diagonal matrix, and the exponential of a diagonal matrix $\exp\begin{pmatrix}a&0\\0&b\end{pmatrix}$ is the diagonal of the exponentials $\begin{pmatrix}e^a&0\\0&e^b\end{pmatrix}$. In your case it would give
$$\ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding Length Of Segment
$AB=12$ , $BC=24$ , $CA=20$
$\dfrac{BF}{CG}=\dfrac{3}{5}$ , $\angle FAG=\angle CAG$
Find the length of $FG$.
It seems simple but is not easy for me.
| let $FG=x,CG=y,AF=z,\dfrac{z}{20}=\dfrac{x}{y}, z^2=20^2+(x+y)^2-2*20*(x+y)cosC <1>$
$z^2-20^2=20^2\dfrac{x^2-y^2}{y^2},cosC=\dfrac{20^2+24^2-12^2}{2*20*24}=\dfrac{13}{15}$
so $<1>$ can be simplify as:
$\dfrac{20^2(x-y)}{y^2}=(x+y)-\dfrac{2*20*13}{15} <2>$
note $BF=\dfrac{3CG}{5}=\dfrac{3y}{5}, BF+FG+CG=24 \implies \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$? I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the
nearest value of $y^2 - y... | You were on the right track. Put
$$x=\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}\implies x^2=5+\sqrt{5-\sqrt{5+\ldots}}\implies$$
$$x^4-10x^2+25=5-\sqrt{5+\sqrt{5-\ldots}}\implies$$
$$x^4-10x^2+20=-\sqrt{5+\sqrt{5-\ldots}}=-x\implies\;\ldots\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 4
} |
why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x ten... | The only way you can do is to make $\sqrt{x}-\sqrt{2}$ into $x-2$ by $$\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x}+\sqrt{2}}=\frac{x-2}{\sqrt{x}+\sqrt{2}}$$ So the main fraction can be changed to $$\sqrt{\frac{x-2}{x+2}}\times\frac{1}{\sqrt{x}+\sqrt{2}}+\frac{1}{\sqrt{x+2}},~~x\neq 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let $a;b;c>0$. Prove : $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ Let $a;b;c>0$. Prove :
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$
P/s : Only use AM-GM and Cauchy-Schwarz
In th... | I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
$$ \frac{a^2}{1-a} + \frac{b^2}{1-b} + \frac{c^2}{1-c}
\ge \frac{b^2}{1-a} + \frac{c^2}{1-b} + \frac{a^2}{1-c} $$
Since $x\mapsto x^2$ and $x\mapsto\frac1{1-x}$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $... | First note that $\lfloor x^2 \rfloor \ge \lfloor x \rfloor ^2$. Now, if we let $y = \lfloor x \rfloor$, then $y^2 -3y+2 \le 0$ and $y $ is an integer, so $y=1$, or $2$.
Then, let $x = y+f$, with $f \in (0,1)$. If $y=1$, then
$$\lfloor2f+f^2\rfloor = 0 \iff 2f+f^2 <1 \iff f < \sqrt{2}-1$$
If $y=2$, then
$$\lfloor4f+f^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$
Show that $2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}$
I tried, $2730=13\cdot5\cdot7\cdot3\cdot2$
We have $13\mid n^{13}-n$, by Fermat's Little Theorem.
We have $2\mid n^{13}-n$, by if $n$ even then $n^{13}-n$ too is even; if $n$ is odd $n^{13}-n$ is od... | One Approach
If $k\mid n$, then $x^{k+1}-x\mid x^{n+1}-x$. Therefore,
$$
\begin{array}{}
13&\mid&n^{13}-n\\
7&\mid&n^7-n&\mid&n^{13}-n&\text{since }6\mid12\\
5&\mid&n^5-n&\mid&n^{13}-n&\text{since }4\mid12\\
3&\mid&n^3-n&\mid&n^{13}-n&\text{since }2\mid12\\
2&\mid&n^2-n&\mid&n^{13}-n&\text{since }1\mid12\\
\end{array}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 1
} |
Divisibility induction proof: $8\mid 7^n+3^n-2$ I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$
And this is how far I've gotten:
$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$
Any help on where to go after this would be great!
| You did a mistake:
$$3\cdot3^n+7\cdot7^n-2=3(3^n+7^n-2)+7^n(7-3)+4$$
Bow by $P(n)$ you know that $8|3(3^n+7^n-2)$. Thus, to complete the problem, you have to prove that
$$8|7^n\cdot 4+4$$
you can do this either by induction or by simply observing that $7^n+1$ is even, thus $4(7^n+1)$ is a multiple of $8$.
Added $P(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How do I find the Jordan normal form of a matrix with complex eigenvalues? I'm trying to obtain the Jordan normal form and the transformation matrix for the following matrix:
$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$
I've calculated its characteristic and mi... | Well, the dimension of the eigenspace corresponding to $\lambda=1$ is two and therefore there are two Jordan blocks for eigenvalue 1. Further calculation is unnecessary as we know that distinct eigenvalues gives rise to distinct Jordan blocks. Therefore in this case the Jordan normal form is,
$$A = \begin{pmatrix} 1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
proving $ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq 20$ If $a,b,c,d,e>1$, Then prove that $\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq 20$
$\bf{My\; Try}::$ Using Cauchy- Schtwartz Inequality
$\displaystyle \frac{a^2}{c-1}+\... | Apply CS inequality to get
$$LHS > \frac{(a+b+c+d+e)^2}{a+b+c+d+e - 5} = M$$
Then let $y = a + b + c + d + e - 5$, then $y > 0$, and so
$$M = \frac{(y + 5)^2}{y}= 10 + y + \frac{25}{y} \ge 10 + 2\sqrt{25} \overset{\strut\text{AM-GM}}= 10 + 10 = 20$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/603722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
| $(\sin x- \cos x+1)\cos x = \sin x \cos x -\cos^2 x +\cos x$
$(\sin x+ \cos x-1)(\sin x +1) = \sin^2 x + \sin x \cos x +\cos x -1 = \sin x \cos x +\cos x +(\sin^2 x -1)= \sin x \cos x -\cos^2 x +\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Get the last two digits of $16^{100}$ and $17^{100}$ I try to get the last two digits of $16^{100}$ and $17^{100}$
I started with:
$6^{1} = 6$
$6^{2} = 36$
$6^{3} = 216$ means last digit for $6$ is always $6$
That is why the last digit for $16^{100}$ is $6$
And for $17^{100}$ i calculated:
$7^{1} = 7$
$7^{2} = 49$
$7^{... | As $(17,100)=1,$ and using Carmichael function $\displaystyle\lambda(100)=20$
$\displaystyle\implies17^{20}\equiv1\pmod{100}$
$16^{100}=(2^4)^{100}=2^{400}$
As $(2^{400},100)=4=2^2\ne1$ let us find $2^{400-2}\pmod{25}$
As $\displaystyle\lambda(25)=\phi(25)=20, 2^{20}\equiv1\pmod{25}$
and $\displaystyle400-2\equiv18\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $1 + \cot^2\theta = \csc^2\theta$ Prove the following identity:
$$1 + \cot^2\theta = \csc^2\theta$$
This question is asked because I am curious to know the different ways of proving this identity depending on different characterizations of cotangent and cosecant.
| Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\sin^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}$$
$$\Rightarrow \left(\frac{\sin\theta}{\sin\theta}\right)^2 + \left(\frac{\cos\theta}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Where did I make a mistake
In a class, 2/3 of the students are girls. 1/4 of the girls are asleep, and 2/5 of the boys are asleep. What fraction of the class is awake.
This is what I did:
Let $x%$ be the total number of students.
Thus the number of girls is $\frac{2}{3}x$ and the number of boys is $\frac{1}{3}x$.
S... | Number of girls, who are awake is not $x- \dfrac{x}6$. It is $\color{red}{\dfrac{2x}{3}} - \dfrac{x}6$.
Similarly, number of boys, who are awake is not $x- \dfrac{2x}{15}$. It is $\color{blue}{\dfrac{x}{3}} - \dfrac{2x}{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$ How can I find a closed form for the following integral
$$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$
| Here is a proof of Cleo's claim that
\begin{align}
\int^\frac{1}{2}_0(2x-1)^6\ln^2(2\sin(\pi x))\ {\rm d}x=\boxed{\displaystyle\frac{11\pi^2}{360}+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)}
\end{align}
It doesn't take much to show that the integral is equivalent to
$$\mathscr{J}=\frac{64}{\pi^7}\int^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 2
} |
Calculate the inverse of a complex matrix I am trying to calculate the inverse of a given matrix but somewhere I have an error in my calculation that I cannot find
$$\begin{array}{ccc}
&&
\left(
\begin{array}{ccc|ccc}
1-i & 2 & 1+i & 1 & 0 & 0 \\
0 & 1+3i & 1+4i & 0 & 1 & 0 \\
4+i&6+2i&2+i & 0 & 0 & 1
\end{array}
\righ... | Here is an error that I found. In the step where you multiply row two by $1-3i$, the second row should become $(0,\ 10,\ 13+i)$. You have it as $(0,\ 4,\ 13+i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
two short doubts about the inverse function in a point the function is $F(x,y,z)=(y^2+z^2, z^2+x^2, x^2+y^2)$ the point is (-1,1,-1)
task: find the local inverse of F in that point.
I have already proved that F is actually invertible there. then i solved the system:
$$
\begin{cases}
y^2+z^2=a\\
z^2+x^2=b\\
x^2+y^2=c
\e... | First doubt: you are right
Secon doubt: the first way you have written is good, the other does not make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality in triangle involving medians Let $ABC$ be a triangle and $M$ a point on $(BC)$, $N$ a point on $(CA)$ and $P$ a point on $(AB)$ such that the triangles $ABC$ and $MNP$ have the same centroid.
Does the following inequality hold: $$AM^2+BN^2+CP^2\ge AA'^2+BB'^2+CC'^2 ?$$ where $A',B'$ and $C'$ are the midpoin... | If $MNP$ has the same centroid than $ABC$ then $$\frac{M+N+P}3=\frac{A+B+C}3$$ (if seen each point as a vector in $\mathbb R$.)
Also, as $M$ is in $BC$ then $$M=mB+(1-m)C$$ for some $0\le m\le1$. Analogous:
$$N=nC+(1-n)A,$$
$$P=pA+(1-p)B.$$
This means that:
$$N+M+P=(1+p-n)A+(1+m-p)B+(1+n-m)C$$
Where $A+B+C=N+M+P$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}... | Answer
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}(\dfrac{1}{1/n^2 + 1} +\dfrac{2}{2/n^2+1}+ \ldots+\dfrac{n}{n/n^2+1})$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}\dfrac{n*(n+1)}{2}$$
$$\lim_{n \to \infty}\dfrac{1}{2}*(1+\dfrac{1}{n})$$
$$=\dfrac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I ca... | The cancellation make sense if instead of $x^2+16y^2$ we use $x^2-16y^2$
$$\frac{x^2-16y^2}{x}\frac{x^2+4xy}{x-4y}=\frac{x^2-(4y)^2}{x}\frac{x(x+4y)}{x-4y}=$$
$$=\frac{(x-4y)(x+4y)}{x}\frac{x(x+4y)}{x-4y}=(x+4y)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the ma... | HINT : Letting $k=\frac{4x-9y}{2}$, this represents a line $y=\frac{4}{9}x-\frac{2}{9}k.$
Then, find the max $k$ such that the line is tangent to the ellipse.
You can set $y=\frac{4}{9}x-\frac{2}{9}k$ in the equation of the ellipse, then you'll get an equation of $x$. Then, find the condition that the equation has only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Question on Sets Given a finite nonempty set $S$ of integers, let $p(S)$ denote the
product of all integers in $S$ (for example, if $S = \{3, 11, 61\}$,
then $p(S) = 2013$). Determine the least positive integer $n$ such
that in any sequence $A_1, A_2, \cdots A_n$ of $n$ nonempty subsets of
$\{1, 2, 3, 4, 5, 6, 7, 8, 9\... | This is not an answer. The following examples show that $n$ has to be larger than $48$.
$$\{1\},\{2\},\cdots,\{9\}$$
$$\{2,5\},\{2,6\},\{2,7\},\{2,8\},\{2,9\},\{3,7\},\{3,8\},\{3,9\},\{4,7\},\{4,8\},\{4,9\},\{5,8\},\{5,9\},\{6,8\},\{6,9\},\{7,8\},\{7,9\},\{8,9\}$$
$$\{2,8,9\},\{3,8,9\},\cdots,\{7,8,9\}$$
$$\{2,7,8,9\},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I ... | Bringing all to one side and clearing denominators, we wish to prove zero
$$f(c) = (a\!-\!b)(b\!-\!c)(c\!-\!a)+(a\!-\!b)(b\!+\!c)(c\!+\!a)+(b\!-\!c)(a\!+\!b)(c\!+\!a) + (c\!-\!a)(a\!+\!b)(b\!+\!c)$$
Note $\ f(a) = (a\!-\!b)(b\!+\!a)2a\!+\!(b\!-\!a)(a\!+\!b)2a = 0.\ $ By symmetry $f(b) = 0.$
The coefficent of $c^2$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$ Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$
I tried it to do using $AM \ge GM$ but don't know how to proceed.
Please help.
| There is a nice solution of this using convexity. Jensen's inequality states that if $f$ is a concave function and $p_1, ... ,p_n$ is a distribution, then for any numbers $x_1, ... ,x_n$ we have:
$$
f\left(\sum_{i=1}^n{p_i x_i}\right) \geq \sum_{i=1}^n{p_i f(x_i)} .
$$
We shall use this and the fact that $\log$ is a co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$ Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$.
I already proved that it converges but I can't find the sum, Although I think it should be $1+\frac{1}{\sqrt3 +1}$, from calculations.
| There's a very simple way to evaluate this sum. Let
$$a_n = \frac{2^n}{3^{2^{n-1}}-1}$$
Then
$$\begin{align}a_n-a_{n+1} &= \frac{2^n}{3^{2^{n-1}}-1} - \frac{2^{n+1}}{3^{2^{n}}-1} \\ &= \frac{2^n}{3^{2^{n-1}}-1} \left [1-\frac{2}{3^{2^{n-1}}+1} \right ]\\ &= \frac{2^n}{3^{2^{n-1}}-1} \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $ Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. I provide the answer below.
| We prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $.
Base Case: Let $ n=1 $. Then
$ 1^2 + 3^2 + ... + (2n-1)^2 = (2*1-1)^2 = 1 = \displaystyle \frac{4*1^3 - 1}{3} = \frac{4n^3 -n}{3} $.
Inductive Step: Assume that $ 1^2 + 3^2 + ... + (2k-1)^2 = \displaystyle \frac{4k^3 -k}{3} $ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it i... | Let $\omega = \exp\left(\frac{2\pi i}{5}\right) = \cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)$. Then $\omega$ is a fifth root of unity ($\omega^5 = 1$). Then $$1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.$$ By taking the real part of both sides (after applying De Moivre's Theorem), we obtain ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
Smallest Prime-Factor of $4^{52} + 52^{2013} + 2013^{52}$ I would like to find out the smallest prime factor of $4^{52} + 52^{2013} + 2013^{52}$ by hand.
Thanks in advance
| The given number is odd, so it cannot be divided by $2$.
In mod $3$, we have
$$\begin{align}4^{52}+52^{2013}+2013^{52}\equiv 1+1+0\equiv 2\end{align}$$
In mod $5$, we have
$$\begin{align}4^{52}+52^{2013}+2013^{52}&\equiv (-1)^{52}+2^{2013}+3^{52}\\&\equiv 1+2\cdot 2^{2012}+(-1)^{26}\\&\equiv 1+2\cdot {16}^{503}+1\\&\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/626832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How many integers less than $1000$ can be expressed in the form $\frac{(x + y + z)^2}{xyz}$? How many integers less than $1000$ can be expressed in the form $$\frac{(x + y + z)^2}{xyz}$$ where $x, y, z$ are positive integers?
| You need to check if the question restricted to only positive integers $x, y, z$, otherwise it is quite trivial to get any desired number.
For zero, take $x = 2, y =-1, z = -1$.
For any positive number $N$, take $x = -N, y = 1, z = -1$.
For any negative number $-N$, take $x = N, y = 1, z = -1$. Clearly all negative ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
compute limit (no l'Hospital rule) I need to compute
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$
I can not use the l'Hospital rule.
| As lcm $(2,3)=6, $ let $(\cos x)^\frac16=y$
$\displaystyle\implies (\cos x)^\frac12=y^3,(\cos x)^\frac13=y^2$ and $\displaystyle \cos x=y^6\implies\sin^2x=1-y^{12}$
$$\lim_{x\to0}\frac{(\cos x)^\frac12-(\cos x)^\frac13}{\sin^2x}=\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}\frac{-y^2(1-y)}{(1-y)(1+y+\cdots+y^{10}+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = ... | Let $$v=(1,1,1,1...1)^T$$
Your matrix is $$
I + v v^T$$
This has $n-1$ eigenvalues equal to $1$ and one with value $n+1$. Hence the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2... | We are given that $3n+1 = a^2 $.
We want to show that $n+1$ is the sum of 3 perfect squares.
Note that $a$ is not a multiple of 3.
If $ a \equiv 1 \pmod{3}$, then observe that $9n+9 = 3a^2 + 6 = (a-1)^2 + (a-1)^2 + (a+2)^2$, and hence
$$ n+1 = \left( \frac{a-1}{3} \right)^2 + \left( \frac{a-1}{3} \right)^2 + \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
How to prove this formula I want to know how to prove this:
$$
\sum_{k=1}^\infty \arctan\left( \frac{x}{k^2+a} \right)
= \pi \left\lfloor \frac{b}{\sqrt{2}} \right\rfloor
+ \arctan\left( \frac{x}{b^2} \right)
- \arctan\left[ \tanh\left( \frac{\pi}{\sqrt{2}} \frac{x}{b} \right)
\cot\left( \f... | Let $\;\displaystyle u + iv = \pi\sqrt{a+ix} = \frac{\pi}{\sqrt{2}}\left(\frac{x}{b} + ib\right)$ and follow nearly exactly the same step as in this answer, you get
$$\begin{align}
\sum_{k=1}\tan^{-1}\left(\frac{x}{k^2+a}\right)
&= \tan^{-1}\left(\frac{\tan v}{\tanh u}\right) - \tan^{-1}\left(\frac{v}{u}\right) + \pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/634470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1... | $$S=1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$$
$$=\sum_{0\le r<\infty} \left(\frac12\right)^r+\sum_{0\le r<\infty} \left(\frac13\right)^r$$
Using summation formula of Infinite Geometric Series
$$S=\frac1{1-\frac12}+\frac1{1-\frac13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/634558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
| The given equation is
$$\dfrac{1}{4}x^2y(x - 1)^3 - \dfrac{5}{4}xy(x - 1)^2$$
Since the GCF (greatest common factor) is $\frac{1}{4}xy(x - 1)^2$, write the expression as
$$\begin{aligned}
\underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot x(x - 1)}_{\frac{1}{4}x^2y(x - 1)^3} - \underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot 5}_{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
S... | You can approximate the sum by an integral to obtain the inequality
$$
\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\le1+\int_1^n\frac1{x^2}\mathrm dx=2-\frac1n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit with geometric sequence I computed
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^n+n^{n+1}+\cdots+n^{2n}} \cdot\left(1-\cos{\frac{3}{n}}\right)$$
$$=\lim_{n \rightarrow \infty } n^2 \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1} \cdot \left(1-\cos \frac{3}{n} \right) $$
$$=\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+ ... | We have these inequalities
$$n^2\le(n^n+\cdots+n^{2n})^{1/n}\le((n+1)n^{2n})^{1/n}=n^2(n+1)^{1/n}\sim_\infty n^2$$
and by Taylor series
$$1-\cos\left(\frac 3 n\right)\sim_\infty\frac 1 2\left(\frac{3}{ n}\right)^2=\frac{9}{2n^2}$$
hence
$$\lim_{n \rightarrow \infty } {\sqrt[n]{n^n+n^{n+1}+...+n^{2n}}\cdot(1-\cos{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as sayi... | You want to prove that
$$2^{-n} = 5^n \times 10^{-n}.$$
And
$$
5^n10^{-n} = \frac{5^n}{10^n} = \left(\frac{5}{10}\right)^n = \left(\frac{1}{2}\right)^n = \frac{1}{2^{n}} = 2^{-n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The beginning of a series in a limit using integrals Question:
evaluate:
$\lim \limits_{n \to \infty} n\left(\dfrac 1{(n+2)^2}+\dfrac 1{(n+3)^2}+1\ldots+\dfrac 1{(2n+1)^2}\right)$
What we did
We found that the limit (using the integral of $\displaystyle \int \limits_1^2\dfrac 1{x}\mathrm dx$) is $0.5$.
But that is true... | If you plot the graph of the function : $x \mapsto \frac{1}{x^2}$ you can easily be convinced by $$\int_{k}^{k+1} \frac{\text{d}x}{x^2} \leqslant \frac{1}{k^2} \leqslant \int_{k-1}^{k} \frac{\text{d}x}{x^2} $$ for all $k \geqslant 2$. Therefore, if you sum this from $k=n+2$ to $k=2n+1$ we got
$$ \int_{n+2}^{2n+2} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.