Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Number of natural numbers less than million with sum of the digits equal to $12$ Number of natural numbers less than million with sum of the digits equal to $12$
My Try: obviously we need 6 places to filled with digits $0$ to $9$ such that sum of the digits is $12$
so the required number is number of non negative integ... | We seek the number of solutions of the equation in the nonnegative integers
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 12 \tag{1}$$
subject to the restrictions that $x_k \leq 9$ for $1 \leq k \leq 6$.
As you determined, if there were no restrictions, equation 1 has
$$\binom{12 + 6 - 1}{6 - 1} = \binom{17}{5}$$
solutions i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Polynomial : $P(a)P(a^2)P(a^3)P(a^4)$ Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.
Find the value of $P(a)P(a^2)P(a^3)P(a^4)$
Is my answer correct ?
Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,
so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$... | Suppose $P(x)=x^3+2x^2+3x+4=(x-p)(x-q)(x-r)$ for some $p,q,r\in\mathbb{C}$. Then, $$\prod_{j=1}^4\,P\left(a^j\right)=Q(p)\,Q(q)\,Q(r)\,,$$
where $Q(x):=x^4+x^3+x^2+x+1$. Now, $$Q(x)=(x-1)\,P(x)+5\,.$$
Thus, $Q(p)=Q(q)=Q(r)=5$.
This is actually quite a nice technique. Let $P(x)$ and $Q(x)$ be two nonconstant polyn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Evaluating $\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$ I need some help with Complex Analysis:
To evaluate the integral
$$\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$$
Here is what I tried:
So, $f(z)=\frac{e^{\pi z}}{z^2(z^2+2z+2)}$ has 3 singularities, $z_0 = 0$, $z_0= -1+i ... | So checking the residues...
The pole at $z=0$ is double, so (in my opinion) it's easier to find the residue via the series expansion than with the explicit formula. Factoring the denominator, the integrand is $\frac{e^{\pi z}}{z^2(z+1-i)(z+1+i)}$. The series for $e^{\pi z}$ is just $1 + \pi z + \frac{\pi^2 z^2}{2!} + .... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all values of $x$ at which $P(x)=x^4-4x^3+22x^2-36x+18$ is a perfect square Find all values of positive integer $x$ at which the following expression is perfect square
$$P(x)=x^4-4x^3+22x^2-36x+18$$
I tried to assume $P= (x^2+ax+b)^2$ ; and comparing the cofactors , get that $a= -2 ; b= 9$
, but when ... | As Jyrki Lahtonen mentioned in comments, my bounds can be improved to
$$(x^2-2x-14)^2 < P(x) < (x^2-2x-13)^2,$$
or in expanded form:
$$196 + 56 x - 24 x^2 - 4 x^3 + x^4 < 18 - 36 x - 22 x^2 - 4 x^3 + x^4 < 169 + 52 x - 22 x^2 - 4 x^3 + x^4$$
for sufficiently large values of $x$ ($x \ge 48$ in first inequality, $x \ge -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a so... | Note that $\sin x$ is concave function when $0\leq x\leq \dfrac{\pi}{2}$. Then by Jensen's inequality on concave function, we have:
$$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\leq \sin \dfrac{A+B+C}{2\cdot 3}=\sin\dfrac{\pi}{6}=\frac{1}{2}$$
Now by AM-GM inequality $$\frac{\sin \dfrac{A}{2}+\sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of
$a+b+c=?$
for the $101$st such number?
There a... | Without loss of generality we can assume that $a > b > c$.
Let's consider the lexicographic ordering;
we will write $(a,b,c) > (a',b',c')$,
if one of the following occures:
*
*$a > a'$ ; or
*$a=a' $ and $b > b'$; or
*$a=a' $ and $b=b' $ and $c > c'$.
Remark:
Let's consider two ordered triples $(a,b,c) , \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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maybe $(2^m-1)(3^n-1)$ can never be a perfect square? Conjecture：
For any positive integer $m,n$, show that $(2^m-1)(3^n-1)$ is never a perfect square?
maybe this is old problem？ I find some $m,n$ the problem is right ,so How to prove it?
The conjecture was based on my solution to this following (different) proble... | Another partial solution, is to note that all squares of integers are 0 1 or 4 mod 8, $2^m-1$ is either 1,3 or 7 mod 8, and $3^n-1$ is either 0 or 2 mod 8.
$$1\cdot0\equiv0\bmod8;
1\cdot2\equiv2\bmod8;
3\cdot0\equiv0 \bmod8;
3\cdot2\equiv6\bmod8;
7\cdot0\equiv0\bmod8;
7\cdot2\equiv6\bmod8$$
None of these are 1 or 4 mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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${2^\left|(x+2)\right|}$ - $\left|2^{x+1} -1 \right| $=$ 2^{x+1}+1$ .What is the Minimum value of ${x}$? I try to use logarithm but I cannot be applied on the equation.
So how can I solve this equation.
Any help will be appreciate.
| If $x \in [-1 , \infty)$,
then we have $|x+2|=x+2$ and $|2^{x+1}-1|=2^{x+1}-1$;
so in this interval, the equation has this form:
$$
\ \ \ \ \
2^{x+2}-(2^{x+1}-1)=
2^{x+1}+1
\Longleftrightarrow
2.2^{x+1}-2^{x+1}+1=
2^{x+1}+1
\Longleftrightarrow
\\
(2-1).2^{x+1}+1=
2^{x+1}+1
\Longleftrightarrow
2^{x+1}+1=
2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove by Induction: $1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4} + \ldots + \frac{(-1)^n}{n}$ is always positive How do you prove by induction that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive.
The base case works out and the inductive step is n=k, so $1-\frac{1}{2}+\frac{1}{3}-\frac{... | Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive is equivalent to proving that:
$1+\frac{1}{3}+\dots +\frac{1}{2n-1} > \frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2n}$
For all $n\geq 1$
Clearly this is true when $n=1$. Assuming this is true for $n=k$, how could you show this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find $\int\frac{\tan^2(x)}{\sqrt{x}} \, dx$ I can not find the next antiderivative
$$\displaystyle\int\dfrac{\tan^2(x)}{\sqrt{x}}\,dx$$
I tried with integration by parts the second integral, please help me.
My try
$$
\begin{aligned}
\int \dfrac{\sec^2(x)-1}{\sqrt{x}}\,dx
=\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-\int\dfrac{... | $\int\dfrac{\tan^2x}{\sqrt x}~dx$
$=\int\dfrac{\sec^2x-1}{\sqrt x}~dx$
$=\int\dfrac{\sec^2x}{\sqrt x}~dx-\int\dfrac{1}{\sqrt x}~dx$
$=\int\dfrac{1}{\sqrt x}~d(\tan x)-2\sqrt x$
$=\dfrac{\tan x}{\sqrt x}-\int\tan x~d\left(\dfrac{1}{\sqrt x}\right)-2\sqrt x$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x-\int\tan u^2~d\left(\dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Complex transformation $ w = z + \frac{c}{z} $ where $ |z| = 1 $ I am to show that if $ w = z + \frac{c}{z} $ and $ |z| = 1 $, then $w$ is an ellipse, and I must find its equation.
Previously, I have solved transformation questions by finding the modulus of the transformation in either the form $ w = f(z) $ or $ z = f(... | $z$ is on the unit circle; let $z=e^{i \theta}$ so
\begin{eqnarray*}
w= (1+c) \cos( \theta) +i (1-c) \sin(\theta)
\end{eqnarray*}
which gives $ x= (1+c) \cos( \theta) , y= (1-c) \sin(\theta)$ and considered in cartesian coordinates
\begin{eqnarray*}
\frac{x^2}{ (1+c)^2} + \frac{y^2}{ (1-c)^2} =1 .
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that no perfect number of the form $3^m 5^n 7^k$ exists. For all perfect numbers $N$, $\sigma (N) = 2N$, where $\sigma$ is the divisor sigma function.
Let $s$ be a perfect number of the form $3^m 5^n 7^k$, where $m,n,k \geq 1$ are integers.
Then $\sigma (s)= \sigma (3^m 5^n 7^k)$
$ =\sigma (3^m) \sigma (5^n) \sig... | Note that the Question stipulates the exponents considered are $m,n,k\ge 1$. As @lulu points out, the cases where one of the exponents is zero can (if desired) be ruled out by this previous Question.
The following is a simplification of the proof that an odd perfect number cannot be divisible by $105$ found here, as p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Show that ${\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$
Prove that $\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ for each positive integer $n$.
My work. I think that i have to use induction, but i can't see how... What i did:
$$f(n)=\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \implies k(n)=f(n+1)-f... | Here there are two different proofs.
1) Note that since $1/(1+x)$ is decreasing and positive for $x\geq 0$ then
$$\sum^{n}_{k=1} \frac{1}{n+k}=\frac{1}{n}\sum^{n}_{k=1} \frac{1}{1+\frac{k}{n}}\leq
\sum^{n}_{k=1}\int_{(k-1)/n}^{k/n} \frac{dx}{1+x}
= \int_0^1 \frac{dx}{1+x}=\ln(2)<\frac{3}{4}.$$
2) For $n\geq 1$, show by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to find the one real root of $(x-1)(x-2)(x-3) + 1$, manually? The polynomial
$$ (x-1)(x-2)(x-3) + 1 $$
has one root (I have seen it by a plot in LaTeX), $0 < x_{root} < 1$. So I presume that the polynomial can be rewritten as :
$$ (x-1)(x-2)(x-3) + 1 = (x - a)^{3} $$
but this is not possible, since
$$ (x-1)(x-2)(... | We can solve $ (x-1)(x-2)(x-3) + 1 = 0 $ for $x \in \mathbb{R}$ by manipulating it in a quadratic equation with some convenient variable substitutions as follows:
Expand the polynomial
$$ x^3-6 x^2+11 x-5 = 0 $$
Substitute $ y = x - 2 $ to get rid of the quadratic term
$$ (y+2)^3 - 6(y+2)^2 + 11(y+2) - 5 = y^3 - y + 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use the AGM inequality to find the maximum Use the AGM inequality to find the maximum of $(5+\sqrt{x^4+1}) \cdot (9-\sqrt{x^4 + 1})$.
$$ab \le (\frac{a+b}{2})^2$$
I don't know how to relate this inequality to find the maximum of that.
| Take $a=5+\sqrt{x^4+1}, b=9-\sqrt{x^4+1}$. Then:
$$ab\le 49$$
Equality occurs when $x=\pm\sqrt[4]{3}$.
Because:
$$ab=(5+\sqrt{x^4+1})(9-\sqrt{x^4+1})\le \left(\frac{(5+\sqrt{x^4+1})+(9-\sqrt{x^4+1})}{2}\right)^2=7^2=49.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $ Prove that
$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$
My Try:
I started with $$I=\... | With substitution $x=e^u$:
\begin{align}
\int \frac{x^m}{\ln x}dx
&=\int\dfrac{1}{u}e^{u(m+1)}du \\
&=\int\dfrac{1}{u}\sum_{n=0}^\infty \dfrac{u^n(m+1)^n}{n!}du \\
&=\int\left(\dfrac{1}{u}+\sum_{n=1}^\infty \dfrac{u^{n-1}(m+1)^n}{n!}du\right) \\
&=\ln u+\sum_{n=1}^\infty \dfrac{u^n(m+1)^n}{n.n!} \\
&=\ln\ln x+\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$
Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$
(DO NOT use trigonometric methods)
| For $a=\frac{1}{\sqrt2}$, $b=\sqrt2$ and $c=\frac{1}{2\sqrt2}$ we'll get a value $\frac{10}{3}$.
We'll prove that it's a maximal value.
Indeed, the condition gives $c(ab+1)=b-a$.
If $ab=-1$ then $a=b$ and $a^2+1=0$, which is impossible.
Thus, $ab\neq-1$, $c=\frac{b-a}{ab+1}$ and we need to prove that
$$\frac{2}{1+a^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x\le y \le z$; $\text{ }x \cdot y \cdot z = 1 $; Prove that $(x+1) \cdot(z+1)>3$ Good morning,
Assuming that $x$,$y$,$z$ are positive real numbers and $x \le y \le z$ and $ x \cdot y \cdot z = 1$.
How can I prove that $(x+1) \cdot (z+1) > 3$?
I know that $x$ has to be less than $1$ and $z$ greater than $1$ and I have ... | Let $y < z$
Set $y_1 = z_1 = \sqrt{yz}$
$(x+1)(z+1)> (x+1)(z_1+1)$
So, you must prove you task only for $z = y$ => $z=\frac{1}{\sqrt{x}}$, $x \le 1$
$(x+1)(z+1) = 1 + x + \sqrt{x} + \frac{1}{\sqrt{x}} \ge 3+x$
(by $t + \frac{1}{t} \ge 2$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the center of the octagon, complex plane The problem given read as follows,
Two consecutive vertices of a regular convex octagon are $(1,2)$ and $(3,-2)$. Find the center of the octagon.
I would like to know if my answer is correct. Here's what I have thought:
I've drawn the complex plane and set the affixes giv... | $z_0=x+i y;z_1=1+2 i;z_2=3-2 i$
$z_1-z_0=e^{\frac{i \pi }{4}} (z_2-z_0)$
$-e^{\frac{i \pi }{4}} (-x-i y+(3-2 i))-x-i y+(1+2 i)=0$
$
\left\{
\begin{array}{c}
\frac{x}{\sqrt{2}}-x-\sqrt{2}-\frac{y}{\sqrt{2}}-\frac{3}{\sqrt{2}}+1=0 \\
\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}-y+\sqrt{2}-\frac{3}{\sqrt{2}}+2=0 \\
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$
Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing se... | Let $\sqrt{x}=\frac{t}{6}$.
Thus, we need to solve
$$18\cdot\frac{t^4}{1296}-18\cdot\frac{t^3}{216}-17\cdot\frac{t^2}{36}-8\cdot\frac{t}{6}-2=0$$ or
$$t^4-6t^3-34t^2-96t-144=0$$ or for all real $k$
$$(t^2-3t+k)^2-9t^2-k^2-2kt^2+6kt-34t^2-96t-144=0$$ or
$$(t^2-3t+k)^2-((2k+43)t^2-(6k-96)t+k^2+144)=0.$$
Now, we'll choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$ where $n\in\mathbb{N}-\{0\}$ with induction? So the question is as above is shown:
How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$?
$n=1: 8^2 - 5^2 = 39$ is a multiple of $13$.
I.H.: Suppose $8^{2^n} - 5^{2^n}$ is divisible by $13$ is true $\forall n \in \ma... | To complete the proof with the following identity $ A^2 - B^2 = (A + B) (A-B) $.
I get
$\begin{align}\\
\underline {\text{n= m+1:}} \\
8^{2^n} -5^{2^n} & = 8^{2^{m+1}} -5^{2^{m+1}} \\
& = (8^{2^m} + 5^{2^m})(8^{2^m} -5^{2^m})\\
& = (8^{2^m} + 5^{2^m})\cdot \text{some multiple of 13}
\end{align}$
Thus concluding that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
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Proving pointwise convergence I'm having trouble proving: $$f_n(x) = \frac{1-n^2x^2}{(1+n^2x^2)^2}$$
converges pointwise to zero for $x \in \,[-1,0)\cup(0,1]$.
My attempt for finding : $N(\epsilon,x)$
\begin{align*}
\left|\frac{1-n^2x^2}{(1+n^{2}x^{2})} - 0\right|<\epsilon &\iff \frac{|1-n^2x^2|}{(1+n^2x^2)^2} < \epsil... | HINT
Let $x \in [-1,0) \cup (0,1]$ and $\epsilon>0$
\begin{align*}
|f_n(x)| \leq \frac{|1-n^2x^2|}{n^4x^4}\leq\frac{1}{n^4x^4}+\frac{n^2x^2}{n^4x^4}=\frac{1}{n^4x^4}+\frac{1}{n^2x^2}
\end{align*}
Exists $n_1 \in \mathbb{N}$ such that for all n$\geq n_1$ we have $\frac{1}{n^4x^4} \leq \epsilon\Rightarrow n_1 \geq \fra... | {
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"url": "https://math.stackexchange.com/questions/2425951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equat... | On one hand, we have $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$, then one has $$-8 \leq x+ y +z \leq 6.$$
On other hand, we have $$(x+y+z)^2 \geq 3(xy+yz+zx) = 36,$$
then $$x+y+z \geq 6$$ or $$x+y+z \leq -6.$$
So, you have $x+y+z = 6$ or $-8 \leq x+y+z \leq -6$ (but x, y, z is non-negative!).
Thus, $x+y+z = 6$ and $x=y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1... | The system reduces to
$\left[\begin{matrix}
1 & 0 & 1 & 1+a\\
0 & -1 & 0 & 1 \\
0 & 0 & 1-a & 1+(1+a)^2
\end{matrix}\right]$ which means $z=\frac{1+(1+a)^2}{1-a}$
It implies for $a$ any value other than $1$ the system can have infinite solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
limit of $\frac{1-\sin x+\cos x }{x-\frac{\pi }{2}}$ how can i show that
$$\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=-1$$
$$\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=\left(\frac{1+\cos \:x -\sin x}{x-\frac{\pi }{2}}\right)$$
any help thanks
| $$\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=\\
\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:}{x-\frac{\pi }{2}}\right)+\lim _{x\to \frac{\pi }{2}}\left(\frac{\cos \:x\:}{x-\frac{\pi }{2}}\right)=\\
\lim _{x\to \frac{\pi }{2}}\left(\frac{\sin(\frac{\pi}{2})-\sin x}{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Defining a piecewise function using restricted operations Question
Can the piecewise function
$$f(x) = \begin{cases}
0 & \text{if $x > 0$} \\
1 & \text{if $x = 0$} \\
0 & \text{if $x < 0$} \\
\end{cases}$$
be defined using only the operations $+ , -, *, /, |\cdots|, \max$, and $\min$?
What I have tried
I c... | You asked how to define your function with an infinite number of operations
from your list.
Here it is defined by means of an infinite sum:
$$
f(x) =
\max(0,1-|x|)
+ \frac12\sum_{n=1}^\infty (\min(1,|2^n x+1|) + \min(1,|2^n x-1|) - 2).
$$
The first part of this, $\max(0,1-|x|),$ is zero except for $-1<x<1,$
where it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
substitution in $y' = \cos(y-x)$ equation I have one simple DE with separable variables to test substitutions skill:
$$\begin{align*}
y' &= \cos(y-x) \\
\frac{dy}{dx} &= \cos(y-x) \\
t &= y -x \\
\frac{dy}{dx} &= \cos t
\end{align*}$$
But what should I do next? I mean I do not know if there's an algorithm to handle sub... | $y-x=u$ derive $y'-1=u'\to y'=u'+1$
The equation becomes
$u'+1=\cos u$
$\dfrac{du}{dx}=\cos u-1$
$\dfrac{du}{\cos u-1}=dx$
$\int dx=x+C$
$\cos u=\dfrac{1-t^2}{1+t^2}$ substituting $t=\tan\frac{u}{2}$
$u=2\arctan t$ and $du=\dfrac{2dt}{1+t^2}$
$$\int \frac{1}{\cos u-1} \, du=\int \frac{2}{\left(\frac{1-t^2}{1+t^2}-1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$... | We provide a method for determining when a binomial coefficient is a multiple of a prime $p$.
We first prove the following:
Lemma: Let $p$ be a prime and let $m,n$ be natural numbers. Let $n = lp+t$ and $m = kp+s$, where $0 \leq t < p$ and $0 \leq s < p$. Then
\begin{align*}
\binom{n}{m} = \binom{l}{k}\binom{t}{s}\mod ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}... | Here I am using the results provided above by Markus Scheuer. Firstly I provide a closed form expression for the quantity $B_m(z)$. We have:
\begin{eqnarray}
B_m(z)&=& \sum\limits_{i=0}^\infty \binom{m\cdot i}{i} \cdot \underbrace{\frac{1}{(m-1) i+1}}_{\int\limits_0^1 \theta^{(m-1) i} d\theta} \cdot z^i\\
&=& \int\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | hint: A classic trick is for example $1^2 - 2^2 + 3^2 - 4^2 = (1^2 +2^2+3^2+4^2) - 2(2^2+4^2)= S_4 - 2^3(1^2+2^2)= S_4 - 8S_2$. Now you can generalize this with $4,2$ are replaced by $2n, n$. Can you finish it? If the last term is odd, then isolate it and use up to the preceeding term which is even. So it can always be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gi... | Put $\arctan 2 = x$ then $\tan x= 2$, so $\sin x = {2\over \sqrt{5}}$ and $\cos x = {1\over \sqrt{5}}$.
Let $\arcsin {1\over \sqrt{10}}= y$ so $\sin y = {1\over \sqrt{10}}$, thus $\cos y = {3\over \sqrt{10}}$.
Now:
$$\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})} = \sin (x-y) = \sin x \cos y- \cos x \sin y= {\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
Determinant of a symmetric zero-diagonal matrix I am trying to write an algorithm which computes determinants of $n\times n$ real-symmetric matrices of the form
$$\mathbf S = \begin{pmatrix}
0 & a_{1,1} & a_{1,2} & a_{1,3} & \cdots &a_{1,n-1}\\
a_{1,1} & 0 & a_{2,1} & a_{2,2} & \cdots& a_{2,n-2}\\
a_{1,2} & a_{2,1} &... | You won't be able to do anything that's asymptotically better than an algorithm for computing general determinants (which is $O(n^3)$ the easy way, though there are algorithms that are $O(n^k)$ for $k \approx 2.373$). The reason for this is that a special case of your problem is finding the determinant of a block matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)$ I am trying to evaluate, at least in the limit $|x|\to0$,
$$
F_m(r)\equiv\int \frac{d^3q}{(2\pi)^3}\left(\frac{1}{2\sqrt{\mathbf q^2-m^2}}-\frac{1}{2|\mathbf q|}\right) e^{i\mathbf q \cdot \mathbf x}\,.
$$
Going to spherical coor... | With $q=mt$ and $x=rm$,
\begin{align}
I&=\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\\
&=\frac{m}{r}\int_0^\infty \sin (xt) \left( \frac{t}{\sqrt{1+t^2}}-1 \right)\,dt
\end{align}
We integrate by parts,
\begin{align}
I&=\frac{m}{r}\left[\left.-\frac{\cos (xt)}{x}\left( \frac{t}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can I compute $\frac{dy}{dx}$ of $y=x\sqrt{3x+1}\sqrt{x+1}$ by taking $ln$ on both sides My teacher has suggested that to compute $\frac{dy}{dx}$ of $y=x\sqrt{3x+1}\sqrt{x+1}$, it's better to take $ln$ on both sides of the equation $y=x\sqrt{3x+1}\sqrt{x+1}$, and try taking derivative of logarithms on both sides, and t... | Multiply both sides of the equation by $-1$,
$$-y=-x\sqrt{3x+1}\sqrt{x+1}$$
Now that both sides are positive you can take the logarithm:
$$\ln(-y)=\ln(-x)+\frac{1}{2}\ln(3x+1)+\frac{1}{2}\ln(x+1)$$
Then take the derivative with respect to $x$ (note that $\ln(-x)'=\frac{-1}{-x}=\frac{1}{x}$)
$$\frac{y'}{y}=\frac{1}{x}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find values such that the curves $y=\frac{a}{x-1}$ and $y=x^2-2x+1$ intersect at right angles? Problem:
Find all values of $a$ such that the curves $y = \frac{a}{x-1}$ and $y = x^2-2x+1$ intersect at right angles.
My attempt:
First, I set the two curves equal to each other:
$ \frac{a}{x-1} = x^2 - 2x + 1 $
$ \fr... | Then,
$$\displaystyle a=\frac{X-1}2
a=(x-1)^3((\text{eq}
\frac{x-1}2=\frac{(x-1)^3}2
=\frac{(x-2)^2}2
2x^2-4x+2=1$$
So,
$$X=\frac{2+\sqrt 2}2, \frac{2-\sqrt 2}2$$
Put $x$ in eq $1$:
$$a= 2^{-\frac32}= \sqrt \frac 42$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $... | This problem seems a bit wonky to me because there are simpler expressions for these functions and easier ways to prove them, but let's roll with it.
Once you render $c^2=5\pm2\sqrt5$, take its square roots to get four roots
$c=a/b=\pm\sqrt{5\pm2\sqrt5}$
with the $\pm$ signs independent. Now from DeMoivre's Theorem we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Seeking the maximal parameter value s.t. two-variable inequality still holds Consider the expression
$$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.
The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-G... | Let $a=b-1$, so $\frac{1}{2}(\frac{a^2}{b}+\frac{b^2}{a}) =b+\frac{1}{2 (b-1)}+\frac{1}{2 b}-\frac{1}{2} < b-\frac{1}{3} $ true for $b>6.54138$.
Now $ \lim \limits_{x \to \infty} (\frac{a^x+b^x}{2})^{\frac{1}{x}} =(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} = ((b-1)^x +b^x)^{\frac{1}{x}} *0.5^{1/x} = (b^x(1+(1-1/b)^x))^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to prove closed form for a binomial sum? Let $l\ge 0$ be an integer.
In the process of solving A tough series related with a hypergeometric function with quarter integer parameters we have discovered the following identity:
\begin{eqnarray}
\frac{1}{4} \sum\limits_{j=0}^{l-1} \sum\limits_{\begin{array}{r} p=-l-1\\p... | This question is actually easier than I thought and it can be solved using elementary methods. First of all note that :
\begin{equation}
\binom{2j+1}{p+j+1}- \binom{2j+1}{p+j+2}= \binom{2j+1}{p+j+1} \left(1-\frac{j-p}{p+j+2}\right)= \binom{2j+1}{p+j+1}\cdot \frac{2p+2}{p+j+2}
\end{equation}
Therefore the sum over $p$ g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $a^2 \equiv b^2 $ mod $p$ implies that $a \equiv \pm b$ mod $p$ Prove that $a^2 \equiv b^2 $ (mod $p$) implies that $a \equiv \pm b$ (mod $p$). Where $p$ is a prime number.
So I know that $a^2 \equiv b^2 $ (mod $p$) implies $p|(a^2 -b^2)$ which implies that $a^2 -b^2 = mp$ for some $m \in \mathbb{Z}$. So $a^... | Just to be different (but essentially exactly the same):
Obviously if $a\equiv \pm b \mod p$ then $a^2 \equiv b^2 \mod p$. ($a = kp \pm b\implies a^2 = (k^2p \pm 2b)*p + b^2$.)
There are $p$ possible integers, $m$ so that $0\le m < p$ and for all integers $a$, $a \equiv m \mod p$ for one of these.
Let $0 \le n < m \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = ... | $$\begin{align}
\binom {n+3}4-3\binom {n+2}4+3\binom {n+1}4-\binom n4
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}4\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}{n+r-4}\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r (-1)^{n+r-4}\binom {-5}{n+r-4}\tag{*}\\
&=(-1)^{n-3}\sum_{r=0}^3\binom 3{3-r}\binom {-5}{n+r-4}\\
&=(-1)^{n-1}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
If we know range of a function , how can construct range of other function?
If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4 x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$
| We have
\begin{eqnarray*}
- \frac{1}{2} \leq \frac{x}{x^2+1} \leq \frac{1}{2}
\end{eqnarray*}
Let $x= \frac{3z}{5}$
\begin{eqnarray*}
- \frac{1}{2} \leq \frac{\frac{3z}{5}}{(\frac{3z}{5})^2+1} \leq \frac{1}{2} \\
- \frac{1}{2} \leq \frac{15z}{9z^2+25} \leq \frac{1}{2}
\end{eqnarray*}
Now multiply by $ \frac{4}{15}$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving sum and modulus of complex numbers Let $z_1,z_2,...,z_n$ be complex numbers, prove the inequality:
$$\frac{\vert\sum_{k=1}^n z_k\vert}{1+\vert\sum_{k=1}^n z_k\vert} \le \sum_{k=1}^n \frac{\vert z_k \vert}{1+\vert z_k \vert}$$
| For the base case $n=2$, let $a= \mid z_1 \mid $ and $b=\mid z_2 \mid $
\begin{eqnarray*}
&ab(a+b+2) & \geq & 0 \\
& ab(a+b)+b(2a+b) +b & \\ &+ab(a+b)+a(a+2b) +a & \geq & ab(a+b)+a(a+b)+b(a+b) +a+b \\
&a(1+b)(1+a+b)+b(1+a)(1+a+b) & \geq & (a+b)(1+a)(1+b) \\
& \frac{a}{1+a}+\frac{b}{1+b} \geq \frac{a+b}{1+a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please giv... | $$\sum_{cyc}\frac{1}{ab+c-1}=\frac{\sum\limits_{cyc}(ab+c-1)(ac+b-1)}{\prod\limits_{cyc}(ab+c-1)}=$$
$$=\frac{\sum\limits_{cyc}(a^2bc+a^2b+a^2c-ab-2a+1)}{a^2b^2c^2+abc-1+\sum\limits_{cyc}(a^3bc+a^2b^2-a^2bc-a^2b-a^2c+a)}=$$
$$=\frac{2\cdot4+2\cdot\frac{1}{2}-3\cdot4-\frac{1}{2}-2\cdot2+3}{4^2+4-1+4\cdot\left(2^2-2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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If $f(1)=1$, then is it true that $f(n)=n$ for all $n \in \mathbb{N}\cup\{0\}$. Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. The... | Let $N$ be the smallest number with $f(N)\neq N$.
Notice that $f(N^2+a^2)=f(N)^2 + a^2 \neq N^2 + a^2$ for all $a<N$.
On the other hand if we can write $N^2+a^2$ as $b^2 + c^2$ with $b,c<N$ then $f(b^2+c^2)=f(b)^2+f(c)^2=b^2+c^2$.
We now prove that for large $N$ we can always find $0\leq a,b,c < N$ such that $N^2+a^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\... | You are right to render
$\sqrt{x^2+3x}+x=\frac{3x}{\sqrt{x^2+3x}-x}$
The next step is to complete the square on $x^2+3x$ getting
$x^2+3x=(x+(3/2))^2-(9/4)$
Use this to show that for negative $x$ with $x<-3$ (why?):
$-x-(3/2)<\sqrt{x^2+3x}<-x$
and then
$-\frac{3x}{2x+(3/2)}<\frac{3x}{\sqrt{x^2+3x}-x}<-(3/2)$
Now get the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm... | Hint: Let $y=e^{\frac25t}$ and use $\sinh^2t+1=\cosh^2t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt... | $\displaystyle I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}\,dx$
Perform the change of variable $y=\sqrt{1-x^2}$,
$\begin{align}I&=\int_0^1 \frac{x}{(2-x^2)\sqrt{1-x}}\,dx\tag{1}\\
&=\Big[\sqrt{\sqrt{2}-1} \cdot\text{arctanh}\left(\sqrt{\sqrt{2}-1}\sqrt{1-x}-\sqrt{\sqrt{2}+1}\cdot\arctan\left(\sqrt{\sqrt{2}+1}\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $(abc)^2 = \frac{(c^6-a^6-b^6)}{3}$ We are given the length of the sides of a right triangle T, where $a \leq b \leq c$. We are asked to prove if $T$ is a right triangle. then $(abc)^2 = (c^6-a^6-b^6)/3$.
What I tried:
I tried substituting with the Pythagorean theorem $a^2 + b^2 = c^2$.
Where I am stuck:
I a... | with $$c^2=a^2+b^2$$ the left-hand side is given by
$$(ab\sqrt{a^2+b^2})^2$$ and the other side
$$\frac{(\sqrt{a^2+b^2})^6-a^6-b^6}{3}$$
can you compute this?
expanding the right-hand side we get $$a^4b^2+a^2b^4$$ and so is the left-hand side.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding integral solutions of $x+y=x^2-xy+y^2$
Find integral solutions of $$x+y=x^2-xy+y^2$$
I simplified the equation down to
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other met... | $x+y=x^2-xy+y^2
$.
Since this is
symmetrical in
$x$ and $y$,
we can assume that
$x \le y$.
If $x=y$,
this becomes
$2x = x^2$,
so
$x=0$ or $x=2$.
If $x < y$,
then
$2y
\gt x+y
=(x-y/2)^2+3y^2/4
\ge 3y^2/4
$.
There is no solution if
$y \le 0$.
If $ > 0$,
$2 \gt 3y/4
$
or
$y \lt 8/3$
so
$y \le 2$.
If $y=2$,
then
$x=0$ or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding equation of circle under the given geometric conditions
Finding the equation of the circle which touches the pair of lines
$7x^2 - 18xy +7 y^2 = 0$ and the circle $x^2 + y^2 - 8x -8y = 0$ and
contained in the given circle??
My attempt
The centre of required circle would lie on angle bisector of the pair o... | The angle bisector of the two lines $y=\dfrac{1}{7} \left(9 \pm4 \sqrt{2} \right)x$ is the line $y=x$
The wanted circle has then centre $H(k,k)$ and its radius is the distance from the given lines $\left(9+4 \sqrt{2}\right) x-7 y=0$
$$r_k=\frac{\left|\left(9+4 \sqrt{2}\right) k-7 k\right|}{\sqrt{\left(9+4 \sqrt{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the perfect numbers of the product of two primes, $2^p-1$ and $2^{p-1}$ A number $n\in N$ Show that if $p$ is a prime, such that $2^p - 1$ is also a prime, a Mersenne prime that is, then $n = 2^{p-1}(2^p-1)$ is a perfect number.
So I know that $n$ must be divisible by, $2^p-1$, $2^{p-1}$ and $1$ (the proper diviso... | Assuming $2^p-1$ is prime number then $2^{p-1} (2^p -1)$ is perfect number if
$\sigma(2^{p-1} (2^p -1)) = \sigma(2^{p-1}) \sigma(2^p-1) = 2^p \sigma(2^{p-1}) = 2^p (2^{p}-1) = 2 * 2^{p-1} (2^{p}-1) $ twice the number we start with so its perfect number.
Done.
| {
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Prove that there exists a $c$ such that $f(c-1/8)=f(c+1/8)$ Suppose that the function $f$ is continuous on $[0,1]$ and $f(0)=f(1)$, prove that there exists a number $c$ such that $$f(c-\frac{1}{8})=f(c+\frac{1}{8})$$
I am not sure how to prove its existence, but what I did was to find the possible range that $c$ is ins... | Let $g:[\frac{1}{8},\frac{7}{8}]\to \mathbb{R}$ be defined by
$\displaystyle{g(x) = f(x+{\small{\frac{1}{8}}}) - f(x-{\small{\frac{1}{8}}})}$.
The goal is to show that $g(c) = 0$, for some $c$.
If $g(\frac{k}{8})=0$, for some $k \in \{1,3,5,7\}$, then let $c=\frac{k}{8}$, and we're done.
Suppose then that $g(\frac{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac{x+1}{y+1}$ is equal to or less than $\frac{x-1}{y-1}$ I know that $\frac{x+1}{y+1}$ is equal to (when $x = y$) or less than $\frac{x-1}{y-1}$. Suppose we cross multiply, then
*
*$xy - x+ y -1$ is less than or equal to $xy + x - y -1$
So,
*$y- x$ is less than or equal to $x- y$.
But when $y$ is greate... | If $ 1 \leq y \leq x $ then $ y-x \leq x-y $. Now add $xy-1$ to both sides and factorise to get
\begin{eqnarray*}
\underbrace{xy+y-x-1}_{(x+1)(y-1)} \leq \underbrace{xy+x-y-1}_{ (x-1)(y+1)} \\
\frac{x+1}{y+1} \leq \frac{x-1}{y-1}.
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
I see that $n^2+3n+1 =n^2-2n+1+5n\equiv n^2-2n+1=(n-1)^2 \pmod{5}$
I also see that $n^2+3n+1=n^2+3n-10+11=(n-2)(n+5)+11\equiv(n-2)(n+5) \pmod{11}$
After that what?
| from what you deduced we have $n\equiv 1 \bmod 5$ and $n\equiv 2$ or $n\equiv 6 \bmod 11$.
This tells us $n\equiv 46$ or $n\equiv 6 \bmod 55$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Base case $T(1)=8$
$T(n)=T(n-1)+6n^{2}+2n$
$T(n-1)=T(n-2)+6(n-1)^{2}+2(n-1)............(1)$
$T(n-2)=T(n-3)+6(n-2)^{2}+2(n-2)...........(2)$
Substituting $(1)\,\, \text{and} \,\,(2) \text{in our question},$
$T(n)=T... | Since the equation contains a quadratic in $n$, it is legitimate to suppose that
$$T_n=a+bn+c n^2+d n^3$$ Just replace in the equation and group terms to end with
$$(3 d-6) n^2+ (2 c-3 d-2)n+(b-c+d)=0$$ Since this must hold for all $n$, then
$$3d-6=0 \qquad 2c-3d-2=0 \qquad b-c+d=0$$ that is to say $b=2$, $c=4$, $d=2$.... | {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to unde... | $$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=$$
$$=\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0.$$
Because
$$\sum_{cyc}(2a^2b^2-a^4)=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$
$$=(2ab)^2-(a^2+b^2-c^2)^2=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=$$
$$=(c^2-(a-b)^2)((a+b)^2-c^2)=(c-a+b)(c+a-b)(a+b-c)(a+b+c).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve identity: $\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$ $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
| Set $t=\tan \frac x2$ so that $\sin x = \cfrac {2t}{1+t^2}$ and $\cos x = \cfrac {1-t^2}{1+t^2}$ (Weierstrass substitution) or $$(1+t^2)\sin x =2t; (1+t^2)\cos x=1-t^2$$
Since $1+t^2\ge 1 \gt 0$ we can multiply the original fractions by $1=\cfrac {1+t^2}{1+t^2}$ to obtain $$\frac {\sin x}{1-\cos x}=\frac {2t}{2t^2}=\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving that: $ | a + b | + |a-b| \ge|a| + |b|$ I am trying to prove this for nearly an hour now:
$$
\tag{$\forall a,b \in \mathbb{R}$}| a + b | + |a-b| \ge|a| + |b|
$$
I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ?
Thanks in advance.
| Very Simple Trick: We have that
\begin{split} (|a|-|b|)^2 +2|a^2-b^2| \ge 0&\Longleftrightarrow & a^2+b^2 -2|a||b|+ 2|a +b||a-b| \ge 0\\
&\Longleftrightarrow & a^2+b^2 + 2|a +b||a-b|\ge 2|a||b|\\
&\Longleftrightarrow& \color{red}{2a^2+2b^2} + 2|a +b||a-b|\ge \color{red}{a^2+b^2}+2|a||b|\\
&\Longleftrightarrow& (|a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 8
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How much smaller is the set of ratios than the set of ordered pairs? For integers $a, b$ with $0 < a < N$ and $0 < b < N$ for some integer $N$, let $\{(a, b)\}$ be the set of all their ordered pairs and let $\{\frac{a}{b}\}$ be the set of all their ratios.
Clearly, $|\{\frac{a}{b}\}| < |\{(a, b)\}|$ (the number of poss... | I will use inclusive bounds, so $1 \leq a, b \leq n$.
$|\{\frac{a}{b}\}|$ is the number of coprime pairs $1 \leq a, b \leq n$. This is OEIS sequence A018805. For large $n$ it's noted that we have $\text{A018805(n)} \approx \dfrac{6}{\pi^2} n^2$.
$|\{a, b\}|$ is the number of pairs $1 \leq a, b \leq n$. This is simply ... | {
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What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. H... | $$\sum _{k=1}^{\infty } \frac{k}{3^k}=\frac34$$
To prove this I'll use generating functions
Consider $$f(x)=\sum _{k=1}^{\infty } \frac{ x^k}{3^k}=\sum _{k=1}^{\infty } \left(\frac{ x}{3}\right)^k=\frac{1}{1-x/3}=\frac{3}{3-x}\quad(*)$$
$$f'(x)=\frac{1}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}$$
To make... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving inequality.. Question:
$$
f(a,b)= \begin{cases} ja & \text{if } b \le 2 \\ jb & \text{if } a \le 2 \\ jab + f(d, \frac{b}{2}) + f(a-d, \frac{b}{2}) & \text{if } a,b >2\end{cases}
$$ where $\ j$ is a positive constant and $ \ 1 \le d<a$.
Prove using some form of induction that $f(a,b) = \Theta(ab)$ or $ f(a,b) ... | We assume that the domain of $f$ is a set $\Bbb R^2_+$ of pairs $(a,b)$ of non-negative real numbers and the third row of the definition means that there exists $d$ such that the claim holds. Remark that $f(a,b)=O(ab)$ fails, for instance for the sequence $a_n=\frac 1n$ and $b_n=n^2$, because then $a_nb_n=n$, whereas $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
| Suppose $b$ is in the range of the given function. Then equation
$$ \frac{8x(x^2-1)}{(x^2+1)^2} = b $$
$$ \frac{8x^2(x-{1\over x})}{x^2(x+{1\over x})^2} = b $$
Mark $t = x-{1\over x}$ and $t$ takes all real values. Then we have
$$ b= \frac{8t}{t^2+4} $$ If $t$ is positive then $b\leq 2$ since $4t\leq t^2+4$ or $(t-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483031",
"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$ How can I prove by induction that for every natural number $n$,
$$1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$$
| The inductive proof requires the Bernoulli inequality (using $3$ terms).
\begin{eqnarray*}
(2n+3)^{n+2}=(2n+1+2)^{n+2} \geq \\(2n+1)^{n+2} +2(n+2) (2n+1)^{n+1} + \frac{(n+2)(n+1)}{2} 2^2 (2n+1)^{n}=(2n+1)^n (10n^2+20n+9) > (3(n+1))^2 (2n+1)^n.
\end{eqnarray*}
So we have $(2n+3)^{n+2} \geq (3(n+1))^2 (2n+1)^n$ & the ... | {
"language": "en",
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How many solutions does the equation $a + b + c = 10$ if $(1, 4, 5)$ and $(1, 5, 4)$ are not considered distinct?
Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
| Since $10$ is not a multiple of $3$, it is not possible for all three numbers to be the same.
However, it is possible for two numbers of the same. Since each number is positive, the repeated number must be $1$, $2$, $3$, or $4$. There are $\binom{3}{2}$ ways to choose the locations of the repeated number. Choosing t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$ The equation is $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
I cannot find any way to simplify this, other than squaring repeatedly, which is, btw, again not simplifying. Also tried substitutions like $1-x = t^2$ and trigonometric substitutions. None seem to work.
By trial an... | $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
$$\sqrt{x-\sqrt{1-x}} = 1 -\sqrt{x}$$
$$x-\sqrt{1-x} = (1 -\sqrt{x})^2$$
$$x-\sqrt{1-x} = x- 2\sqrt{x} +1$$
$$\sqrt{1-x} = 2\sqrt{x} +1$$
$$1-x = (2\sqrt{x} +1)^2$$
$$1-x = 1 + 4 \sqrt{x} + 4 x$$
$$- 5x= 4 \sqrt{x} $$
$$- \frac{5}{4}x= \sqrt{x} $$
$$\left(- \frac{5}{4}x\right)^2= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$
Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$
I have no idea which test to use here because of this complicated expression.
| $n^{th}$ term of the series is
= $ (n^3+1)^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{n^3})^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{3n^3} + other terms)- n$ . . . . . . . .(other terms goes to 0 as n approaches $\infty$)
= $ \frac{1}{3n^2}$ + other terms
Now as $n\to\infty$ ,our $n^{th}$ term can be shown with following inequalit... | {
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"timestamp": "2023-03-29T00:00:00",
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Rationalize denominator: $\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$ $$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$
So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})$. In my head I see that this is the first part for the sum of cubes formula. I multipli... | Use the identity: $$\dfrac{1}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}} = \dfrac{\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}-\sqrt[3]{xy}-\sqrt[3]{yz}-\sqrt[3]{zx}}{x+y+z - 3\sqrt[3]{xyz}}$$
and then use the difference of cube formula. Pretty sure there is no easier way to do this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Number of sequences that have at least five consecutive positions in which the numbers are in increasing order? A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five conse... | Both myself and @N.F.Taussig used the following approach, although I'd like to see if it could be generalised to increasing runs of arbitrary length.
Define set $S_{i,j}$ as the set of permutations of $[7]$ with an increasing run between position $i$ and $j$ inclusive. Then by inclusion-exclusion the desired success co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Justify the recurrence relation $a_{n+2}=2a_{n+1}+a_n$ and find $a_n$ Let $a_n$ be the number of ways to color the squares of a $1$ x $n$ chessboard using the colors red, white, and blue, so that no red square is adjacent to a white square. Justify the relation $a_{n+2}=2a_{n+1}+a_n$ (for certain $n$), and then find $a... | I'm also learning recurrence relations as well so here's my take on it :) (please let me know if this makes sense)
I got a recurrence relation of $$a_n = a_{n-1} + 2a_{n-2} + 2\times 3^{n-2}$$
for $n\geq 2$ with $a_0 = 1, a_1 = 3$.
Here's how I derived it:
Case 1: The first block is blue. The remaining board is f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factoring a quartic expression
We have a function $$f(x)=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ and let line segment $AB$ be represented by $g(x)$ such that $$g(x)=x-4$$
Find the greatest positive real solution representing the intersection between the two lines.
The first step is obviously to let them equal... | You are very close, just haven't tried $a=8$ yet.
\begin{align}
f(x)&=x^4 -16x^3+81x^2-137x+8 \\
f(8) &= 8^4 - 16 \cdot8^3 + 81 \cdot8^2 - 137\cdot8+8 \\
&= 4096-8192+5184-1096+8\\
&= 0
\end{align}
Then we know $(x-8)$ is a factor of $f(x)$.
\begin{align}
f(x)&=(x-8)(x^3-8x^2+17x-1) \\
\end{align}
Used polynomial divis... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k... | HINT
If $4^k + 6k -10$ is divisible by $18$, then $4(4^k + 6k -10)$ is divisible by $18$ as well.
Subtract the resulting expression from the expression you get when plugging in $k+1$:
$$4^{k+1} +6(k+1)-10 - 4(4^k + 6k -10) = $$
$$4 \cdot 4^k +6k+6-10 - 4 \cdot 4^k -24k + 40 = $$
$$-18k+36$$
Since $-18k+36$ is clearly d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum of an expression involving radicals: $\sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}}$ The link below is a MSE question discussing the maximum value of the expression
$$ \sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}} $$
which is $3$.
Prove inequality $\sqrt{\frac{2... | For $b=\epsilon^2$ and $c=\epsilon$, where $\epsilon\rightarrow0^+$ our expression is closed to $\sqrt2$.
We'll prove that for positive variables it's an infimum.
Indeed, we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}\geq1,$$ which is true by Holder:
$$\left(\sum_{cyc}\sqrt{\frac{a}{a+b}}\right)^2\sum_{cyc}a^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$
| The expression equals
$$\frac{(3^x-3^0) - (5^x-5^0)}{(4^x - 4^0)-(10^x-10^0)}.$$
Divide top and bottom by $x=x-0.$ Then by definition of the derivative, the desired limit equals
$$ \frac{(3^x)'(0) - (5^x)'(0)}{(4^x)'(0)- (10^x)'(0)}= \frac{\ln 3 - \ln 5}{\ln 4 - \ln 10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Can someone explain the steps of this Partial fraction decomposition?
My thoughts:
$$\frac{(Ax + B)}{(x^2+1)} + \frac{(Cx + D)}{(x^2+4)} = \frac{x}{(x^2+1)(x^2+4)}$$
I combined the left terms
Set the numerator of the combined left term to "x" which is the numerator of the right term
I got
$$4Ax + Cx = x$$
I am not sur... | I’m not entirely sure what you did to get there, but here are the steps to decomposing your partial fraction. You can compare and see where went wrong.
Starting off with$$\frac x{(1+x^2)(4+x^2)}=\frac {Ax+B}{1+x^2}+\frac {Cx+D}{4+x^2}$$We get rid of the fractions to see$$x=(Ax+B)(4+x^2)+(Cx+D)(1+x^2)$$Now, we set $x^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that length of Randomly chosen chord of a circle Find Probability that length of Randomly chosen chord of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try: I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $... | The distance between $\alpha$ and $\beta$ cannot be negative.
$p = 2|\sin \frac {\beta - \alpha}{2}| = p = 2\sin \frac {|\beta - \alpha|}{2}$
Now, $\beta - \alpha$ is uniformly distributed. Let $\theta = \beta - \alpha.$ (In fact, we can fix $\alpha = 0$ and get the same result.) And rather than dealing with the absol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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What is the exact value of $\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$
I would like to get the exact value of the following integral.
$$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$
I was able to prove the convergence as well. But I don't how to compute its exact value. I tried with the Residue Theorem of the comple... | Performing the change of variables $2u = x^2$ together with two integration by parts, we get, $$ \int_0^\infty \cos(x^2)dx = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx\\=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Proving that $\int_0^\infty e^{-a^2x^2} \cos (bx) dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$ Any clue regarding this:
$$\int_0^\infty e^{-a^2x^2} \cos (bx) dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$$
| Let
$$I(b) = \int^\infty_0 e^{-a^2 x^2} \cos (b x) \, dx, \quad a > 0.$$
Integrating by parts gives
$$I(b) = \frac{2a^2}{b} \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx.$$
Also, differentiating $I(b)$ with respect to the parameter $b$ we have
$$I'(b) = - \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx = - \frac{b}{2a^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding exact solutions of an equation I have the question "Find the exact solutions of the following equation, giving your answer in terms of i and simplifying any surds (where possible):
2X^2 + 8X + 9 = 0"
Here is my attempt is this correct ?
| Looks correct to me. Here's how I would do it though:
$$
2x^2 + 8x + 9 = 0\implies\\
x^2 + 4x + \frac{9}{2} = 0\implies\\
x^2+2\cdot x\cdot 2+2^2-2^2 +\frac{9}{2} = 0\implies\\
(x+2)^2=\frac{8}{2}-\frac{9}{2}\implies\\
(x+2)^2=-\frac{1}{2}\implies\\
x_{1,2}+2=\pm\sqrt{-\frac{1}{2}}\implies\\
x_{1,2}=-2\pm\sqrt{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
estimate the maximum error of taylor approximation of $\ln(x)$ find the error bound using the Lagrange Remainder:
$$\ln(x)\approx (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}, |x-1|<\dfrac{1}{64}$$
My attempt:
\begin{align}
R_3(x)&=\dfrac{f^{(4)}(c)}{4!}(x-1)^4\\
f^{4}(c)&=\dfrac{-6}{(x-1)^4}\\
\implies R_3(x)&=\dfrac{-6}... | Note that,
\begin{equation}
f(x) = log(x) \implies f^{(4)}(x) = \frac{-6}{x^4}
\end{equation}
Therefore,
\begin{equation}
R_3(x) = -\frac{6}{4!c^4}(x-1)^4\quad; \quad \text{where } c \text{ is in between 1 and x}
\end{equation}
Since $c$ depends on $x$, first find an upper bound as;
\begin{equation}
|R_3(x)| \leq
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Elementary method to solve this equation I have to describe ''the sign of the root(s)''$$\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}=0$$ for k-11 students , who did not learned derivation.
I can solve the equation by taking $f(x)=\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}$ also by graphing https://www.desmos.c... | Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and since
$$3+\sqrt{x}=3-x=-6$$ is impossible,
our equation is equivalent to
$$3+\sqrt{x}+3-x-6+3\sqrt[3]{6(3+\sqrt{x})(3-x)}=0$$ or
$$(\sqrt{x}-x)^3+162(3+\sqrt{x})(3-x)=0,$$
which after substitution $\sqrt{x}=t$ gives
$$t^6-3t^5+3t^4+161t^3+486t^2-486t-1458=0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It s... | You want to study the function
$$
f(x)=x+\sqrt{a^2-x^2}
$$
defined over $[-a,a]$. We have $f(-a)=-a$, $f(a)=a$; moreover
$$
f'(x)=1-\frac{x}{\sqrt{a^2-x^2}}
$$
for $x\in(-a,a)$. The derivative can only vanish where
$$
x=\sqrt{a^2-x^2}
$$
so $x\ge0$ and $x^2=a^2-x^2$, that is, $x=a/\sqrt{2}$. Note that
$$
f(a/\sqrt{2})=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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When $f(x) = \frac{ax + b}{a -x + 1}$ is an integer Given that $a$ and $b$ are positive integers. and $$f(x) = \frac{ax + b}{a -x + 1}$$ is an integer, what integer values can x have?
If I could only somehow move $x$ from numerator to the denominator I would be able to solve this by factoring the numerator.
| Since
$$a-x+1\mid ax+b \;\;\;{\rm and} \;\;\;a-x+1\mid a(a-x+1)$$
we have $$a-x+1\mid (ax+b)+(a^2-ax+a)= a^2+a+b$$
So $x=a-d+1$ where $d$ divides $a^2+a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Inverse Laplace transform of $f(s)=\frac{1}{s\sinh^2(c\sqrt{s})}$ I want to calculate inverse Laplace transform of $f(s)=\frac{1}{s\sinh^2(c\sqrt{s})}$, where $s$ the is Laplacian variable.
The function has one pole at $s=0$, for which the residue can be easily found, and infinitely many poles at $c\sqrt{s} = n\pi i$. ... | $$f(s)=\frac{1}{s \sinh ^2\left(c \sqrt{s}\right)}$$
Integrate f(s) as a function a:
$$\text{Int}(f(a,s))=\int \frac{1}{s \sinh ^2\left(a c \sqrt{s}\right)} \, da=-\frac{\coth \left(a c \sqrt{s}\right)}{c
s^{3/2}}+\text{c1}
$$
Using this identity and substituting:
$$\coth (s)=\frac{1}{s}+\sum _{k=1}^{\infty } \frac... | {
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"url": "https://math.stackexchange.com/questions/2516670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem about the dimension of the intersection of two subspaces I'm trying to solve the problem below.
$U$ and $W$ are subspaces of polynomials over $\mathbb{R}$.
$U = Span(t^3 + 4t^2 - t + 3, t^3 + 5t^2 + 5, 3t^3 + 10t^2 -5t + 5)$
$W = Span(t^3 + 4t^2 + 6, t^3 + 2t^2 - t + 5, 2t^3 + 2t^2 -3t + 9)$
What is $dim(... | Consider a polynomial $p(t)=at^{3}+bt^{2}+ct+d$ belonging to $U$ and $V$ at the same time and solve two separated systems to find the conditions for $p(t)$ to be in $U$ and beside find the coditions for $p(t)$ to be in $V$ and once you find those two list of conditions consider a system on the unkowns $a,b,c$.
$$p(t)=... | {
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"url": "https://math.stackexchange.com/questions/2516783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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To find the values of a for $f$ has no critical number Determine the values of $a$ for which $$f(x)=(a^2+a-6)\cos{2x}+(a-2)x+\cos{1}$$has no critical number
My attempt:
A critical number of $f$ is a number $c$ in the domain of $f$ such that $f'(c)=0$ or $f'(c)$ does not exist
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)$$
So,
$$f... | We need
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)=0$$
Then
$$f'(x)=-2(a+3)(a-2)\sin{2x}+(a-2)$$
Because we want that $f$ has no critical points, we need that $f'(x)$ is never $0$. But $-2(a+3)(a-2)\sin{2x}$ can be $0$ (since $\sin{2x}=0$ an infinite number times in the domain).
So really, we need that $(a-2)$ is never $0$, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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If $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$?
Is it true that if $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$. Prove it, or prove the contrary.
I tried saying that if:
$$A= \begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}\quad\text{and}\quad ... | For example you may try with
$$A=B=\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the number of 5 digit numbers divisible by 3? What is the number of $5$ digit numbers divisible by $3$ using the digits
$0,1,2,3,4,6,7$ and repetition is not allowed?
| the rule of three says that a number is divisible by three the sum of the digits is divisible by three. SO if one five digit number is divisible by three then any arrangement of its digits is divisible by three.
So it's a matter of finding the ways of picking $5$ out of $7$ numbers that add to a multiple of $3$ and mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Divisibility theorem based proof for any square mod 4 being either 0 or 1 Kindly help me in understanding the below proof for given statement:
If $n$ is a square, then leaves a remainder $0$ or $1$ when divided by 4.
Proof: The divisibility theorem states that for two integers $a,b$ with $b>0$, then there is a unique p... | Any integer is of the form $2n$ or $2n-1$ (i.e odd or even)
Now sq. of any number will be in the form $$(2n)^2 =4p or (4n-1)^2 =4q +1$$
So we find when that if $n$ is a square, then leaves a remainder $0$ or $1$ when divided by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find value of $a_{2012}$ A sequence $\left\{a_n\right\}$ is defined as:
$a_1=1$, $a_2=2$ and
$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$
Find $a_{2012}$
My Try:
we have
$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$
$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$
Replacing $n$ with $n-1$ we get
$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \ta... | You already have $$a_n a_{n+1}-a_{n-1}a_n=2$$
So if you set the new sequence $b_k=a_{k+1}a_{k}$ then you get
$$b_k-b_{k-1}=2$$ so,
$b_k$ is an arithmetic sequence with $b_1=a_1a_2=2$ and step $2$. Then
$$b_k=2+2(k-1)=2k$$
then
$$a_{2}a_{1}=2$$
$$4=a_{3}a_{2}$$
$$a_{4}a_{3}=6$$
$$8=a_{5}a_{4}$$
$$...$$
$$2\cdot 2010=a_{... | {
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"url": "https://math.stackexchange.com/questions/2524929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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>Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$ Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t)\, dt$.
Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$
I try to show that $$\int_0^x \sin^2\left(t +\frac1n\right) \, dt = \frac{x}{2} - \frac{1}{4}\sin\l... | So $\dfrac{1}{2n}+\dfrac{1}{4}|\sin(2/n)|\leq\dfrac{1}{2n}+\dfrac{1}{4}\cdot\dfrac{2}{n}=\dfrac{1}{n}\rightarrow\infty$ uniformly in $x$, so the convergence is uniform and hence pointwise by fixing $x$, and then taking $n\rightarrow\infty$ to $|f_{n}(x)-f(x)|<\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of the $n$th root of the product How one can evaluate $$\lim_{n\to\infty} \frac{\left( \prod_{k=1}^n (n^2+k^2) \right)^{1/n}}{n^2}?$$ I was unable to found any trick to do the computation.
| $$\frac{\left( \prod_{k=1}^n (n^2+k^2) \right)^{1/n}}{n^2}=\frac{\left( \prod_{k=1}^n (n^2(1+\frac{k^2}{n^2}) \right)^{1/n}}{n^2}= \frac{n^2 \left( \prod_{k=1}^n (1+\frac{k^2}{n^2} \right)^{1/n}}{n^2}$$
$$\log\frac{n^2 \left( \prod_{k=1}^n (1+\frac{k^2}{n^2} \right)^{1/n}}{n^2}= \sum_{k=1}^n \frac{1}{n}\log (1+\frac{k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x+1}{x^2+2x}dx$ in two different ways I am trying to evaluate
$$\int \frac{x+1}{x^2+2x}dx$$
Sounds like $$\int \frac{x+1}{x^2+2x}dx=\int \frac{(x+1)}{(x+1)^2-1}dx\\u=x+1 \implies\int\frac{u}{u^2-1}du\\t=u^2 \implies \frac{1}{2}\int\frac{1}{t-1}dt $$ Which is also known as $$\frac{1}{2}\ln|t-1|=\fr... | You can use substitution to solve. I will be a little bit faster. In fact, let $u=x^2+2x$. Then $du=2(x+1)dx$ and so
$$ \int\frac{x+1}{x^2+2x}dx=\frac12\int\frac{du}{u}=\frac12\ln|u|+C=\frac12\ln|x^2+2x|+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Combinatorics and Matrices Find the number of $4\times4$ matrices such that $|a_{ij}| = 1 \forall i,j\in[1,4]$ , and sum of every row and column is zero.
I tried 'counting' the number of matrices that satisfy the above conditions, that is, elements are $1$ or $-1$ and sum of every row and column is zero.
In the attempt... | The first row has two $1$s and two $-1$s. There are $\binom{4}{2} = 6$ ways they can be arranged in the first row. We'll count the number of matrices with first row $(1,1,-1,-1)$, then multiply by $6$ to account for all the other arrangements of the first row.
The second row can be one of the three following things: $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Probability of rolling two dice with either showing a six We roll two dice, one red and one green. Under each assumption below, what is the probability that the roll is double sixes?
a) The red die shows six.
b) At least one of the dice shows a six.
I think this is pretty straight forward:
For a) it is $\dfrac{1}{6}$
... | I am suspicious of your second computation. I think it would be easier to think in terms of inclusion-exclusion and independence:
\begin{align}
P(\text{at least one 6})
&= P(\color{red}{\text{red 6}} \lor \color{green}{\text{green 6}})\\
&= P(\color{red}{\text{red 6}}) + P(\color{green}{\text{green 6}}) - P(\color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatri... | Let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix};$$its columns are the eigenvectors that you found. Then$$T^{-1}.A.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}.$$Therefore,$$T^{-1}.A^n.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}^n=\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}$$and so$$T^{-1}.e^A.T=\sum_{n=0}^\infty\frac1{n!}\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 ... | If $f(m)=5^m+2\cdot3^m-3,$
$$f(n+2)-f(n)=5^n(5^2-1)+2(3^2-1)3^n$$ which is clearly divisible by $8$
$\implies8|f(n+2)\iff8|f(n)$
Now establish the base cases $f(0),f(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$ So as the title states I've got the following problem:
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$
So I'd guess that this probably involves the formula for half-angles, but that is is a dead-end.
Any suggestions?
Thank you in advance!
| $$
\sin x = \overbrace{\sin\left( 2\cdot \frac x 2 \right) = 2\sin\left( \frac x 2 \right) \cos\left( \frac x 2 \right)}^{\large\text{the double-angle formula for the sine}}$$
If $z = \tan \frac x 2,$ then what are $\sin \frac x 2$ and $\cos \frac x 2\,$?
In the first quadrant, we can say $\tan = \dfrac \sin \cos$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factor $z^5 + z +1 =0$ The problem stated in the title is:
Factor $z^5 + z + 1 = 0$
(naturally without the use of computers or calculators)
How do I go about solving this? Is there a more systematic approach than simply guessing a root and then applying polynomial division etc. ? Thank you for any help!
| If $z^5 + z + 1 = f(z)g(z)$, then:
*
*$\deg f =1, \deg g=4$, or
*$\deg f =2, \deg g=3$
In both cases, you can try to determine the coefficients. Perhaps simplify things by first trying monic polynomials with independent term $1$.
The second case gives
$$
(z^2+a z+1)(z^3+bz^2+c z+ 1) =
z^5 + (a + b)z^4 + (a b + c +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many possibilities are there for two full houses to be dealt to two players in one game? Imagine dealing cards from a classic 52 card deck to two poker players.
How many possibilities are there for both of them to be dealt a full house(three cards in same rank and two cards of another rank) in same round?
As I kno... | There are two possibilities. Either both players receive two cards of the same rank or they do not.
Both players receive two cards of the same rank: There are $\binom{13}{1}$ ways to choose the rank from which player A receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank. That leaves $\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.
My work :
Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies | a_n - a_... | A slightly unorthodox approach to this problem (as I have nothing else to add to Robert's spot on answer) is to solve the recurrence using (for example) characteristic polynomials, which for
$a_{n+2}=\frac{a_{n+1}+a_n}{2}$ is:
$$2x^2-x-1=0$$
which has the following roots $x_1=1$ and $x_2=-\frac{1}{2}$ and the general ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.