Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the root of a complex number I have a complex number $8-6 i$ . I have to find the square root . I did all the steps and I got $\pm(-3+i)$. I also got $\pm(3-i)$. On squaring I am getting the same $8-6 i$, but what is the right square root . Please someone point out my mistake.
| $$\sqrt{8-6i}=\sqrt{\left|8-6i\right|e^{\arg\left(8-6i\right)i}}=$$
$$\sqrt{\sqrt{8^2+6^2}e^{-\tan^{-1}\left(\frac{6}{8}\right)i}}=\sqrt{\sqrt{100}e^{-\tan^{-1}\left(\frac{3}{4}\right)i}}=$$
$$\sqrt{10e^{-\tan^{-1}\left(\frac{3}{4}\right)i}}=\left(10e^{-\tan^{-1}\left(\frac{3}{4}\right)i}\right)^{\frac{1}{2}}=$$
$$\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4... | Hint: consider the term $$-4 \left(a^4 c^4+a^4 d^4-4 a^2 b^2 c^2 d^2+b^4 c^4+b^4 d^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$ I need to prove that there are no rational numbers $x, y$ that
$$\sqrt{5} = x \sqrt{7} + y$$
We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is ir... | If $x,y\in Q$, then $\sqrt 5= x\sqrt 7+y\implies \sqrt 5-y=x\sqrt 7\implies 5-2 y\sqrt 5 +y^2=7x^2\implies$ $$(-2 y)\sqrt 5=7x^2-5-y^2\in Q\implies y=0\implies$$ $$\sqrt 5=x\sqrt 7\implies x=\sqrt {5/7}\not \in Q$$contradicting $x\in Q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove inequality $\left| \frac{x+yz}{x^2+y^2} \right| \leq 1$ for $x^2+y^2-z^2=1$ Prove this:
If $x,y,z \in \mathbb R$ and $x^2+y^2-z^2=1$, then $$\left| \frac{x+yz}{x^2+y^2} \right| \leq 1 $$ holds.
Own ideas:
If we eliminate $z$ the inequality is equivalent to $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right|... | $(1+z^2)(x^2+y^2)\geq(x+yz)^2$ by Cauchy Inequality
Since $(1+z^2)=(x^2+y^2)$ the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
MyApproach
Since I cannot form pattern above
I simplified this as $21^3$ = $3^3$ $. $ $7^3$
Similarly I did $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)... | By exploiting $a^3+b^3=(a+b)(a^2-ab+b^2)$ we have:
$$ 21^3+23^3+25^3+27^3 = 48\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$$
hence $ 21^3+23^3+25^3+27^3$ is a multiple of $96$, since $\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$ is an even number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number Theory: Prove that $x^{p-2}+\dots+x^2+x+1\equiv 0\pmod{p}$ has exactly $p-2$ solutions I just completed this homework problem, but I was wondering if my proof was correct:
If $p$ is an odd prime, then prove that the congruence $x^{p-2}+\dots+x^2+x+1\equiv 0\pmod{p}$ has exactly $p-2$ incongruent solutions, and t... | If $a \in Z_p \setminus \{0,1\}$ and $p$ is prime, then:
$$1+a+a^2+\cdots +a^{p-2}=(a^{p-1} - 1)\cdot(a-1)^{-1} = 0,$$because $a^{p-1}=1$.
On the other hand, if $a=1$, then
$$1 + a + a^2 + \cdots + a^{p-2} = \underbrace{1+1+ \cdots +1}_{p-1} =(p-1) \ne 0.$$
Finally if $a = 0$,
$$1 + a + a^2 + \cdots + a^{p-2} = 1 \ne 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simple Division Problem I have the equation:
$$(1-\frac{1}{2^2})...(1-\frac{1}{n^2}) = \frac{n+1}{2n}$$ for n ≥ 2
Trying to prove by induction and I get the following equation.
$$\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} = \frac{k+2}{2(k+2)}$$
I can't to simplify it to the final answer.
I multiplied $$\frac{k+1}{2k}.{(k... | The statement is true for $n=2$. Suppose it holds for $n$; then
$$
\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{n^2}\right)
\left(1-\frac{1}{(n+1)^2}\right)=
\frac{n+1}{2n}\left(1-\frac{1}{(n+1)^2}\right)
$$
The final term becomes
$$
\frac{n+1}{2n}\frac{n(n+2)}{(n+1)^2}=\frac{(n+1)+1}{2(n+1)}
$$
which is exactly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\ri... | Hint: (for the first version, before multiplying by 2)
$$
\left(1+\frac1n\right)^n\leq\left(1+\frac1n+\frac1{n^2}\right)^n\leq\left(1+\frac{1+\epsilon}n\right)^n
$$
for large $n$. Hence all partial limits are $e$ and the final one — 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Limit of $\frac{(n+1)^{2n}}{(n^2+1)^n}$ as $n\to \infty$ So it is given to find $$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}$$
So what I did is
$$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}=\lim_{n\to \infty}\dfrac{(n^2+2n+1)^{n}}{(n^2+1)^n}=\lim_{n\to \infty}\left(1+\frac{2n}{n^2+1}\right)^n$$
Now the rightmo... | Your guess is right.
We have
$$
1+\frac{2n}{n^2+1} \le 1+\frac{2n}{n^2} = 1+\frac{2}{n}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n \le \left(1+\frac{2}{n}\right)^n \to e^2
$$
On the other hand,
$$
1+\frac{2n}{n^2+1} \ge 1+\frac{2n}{n^2+n} = 1+\frac{2}{n+1}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n
\ge \left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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LU decomposition on 5 by 3 matrix. This is a problem given in a quiz. Even after reading up a related question, I cannot figure it out.
Determine an LU-factorization of $$
C =
\begin{bmatrix}
3 & 1 & -4 \\
6 & -3 & 10 \\
-9 & 5 & -11 \\
-3 & 0 & -7 \\
6 & -4 & 2
\end{bmatrix}
$$
In the related question, I can acquir... | Because $C$ is not square, $U$ will not be in upper triangular form but rather in row echelon form.
In your case, proceed as you would for a square matrix, that is, reduce $C$ to row echelon form to get $U$, and keep track of the multipliers you used to get $L$:
Use the first row, $R_1$, to make the first entries of th... | {
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"url": "https://math.stackexchange.com/questions/1513252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\... | If $m+n\equiv 1 \bmod 2$, then $m-n\equiv 1 \bmod 2$, so $m^2-n^2\equiv 1 \bmod 2$, and $m^2+n^2=m^2-n^2+2n^2\equiv 1\bmod 2$.
If $d\mid m^2-n^2$ and $d\mid m^2+n^2$, then $d\mid 2m^2$ and $d\mid 2n^2$.
Since both $m^2+n^2$ and $m^2-n^2$ are odd, $d$ is not even, so $d\mid m^2$ and $d\mid n^2$, which implies $d\mid m$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+... | The condition for your answer is when $x_1=x_2=...=x_5=4$,so we can find that they are roots of $$(x-4)^5=0$$
Expand this equation of left side we have the coefficient of $x^4$ is $$C(5,4)*(-4)^1=-20$$
and coefficient of $x^3$ is $$C(5,3)*(-4)^2=160$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Evaluate $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ Let $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ for all $n\geq0$.
How to find the value of $\displaystyle\prod_{n=1}^\infty A_n$ ?
I don't have any idea.
Thank you.
| If we set $A_n = \cos^2(\theta_n)$, we get:
$$ A_{n+1} = \frac{1+\cos(\theta_n)}{2} = \cos^2\left(\frac{\theta_n}{2}\right),$$
and since $\theta_0=\frac{\pi}{6}$, induction gives:
$$ A_n = \cos^2\left(\frac{\pi}{6\cdot 2^n}\right)=\left(\frac{\sin\left(\frac{\pi}{6\cdot 2^{n-1}}\right)}{2\sin\left(\frac{\pi}{6\cdot 2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1... | $$0\leq(1-x^2)^2\iff 4x^2\leq(1+x^2)^2\iff\frac{x^2}{(1+x^2)^2}\leq\frac14
\iff\left|\frac{x}{1+x^2}\right|\leq\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluate $\sum_{n=0}^\infty\left(-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\right)$ I proved that the given series is convergent and to find the sum I tried to compute the partial sum and then I pass to the limit but I didn't find a closed form to this partial sum. How I should proceed?
| Set $a_n=n\ln(2n+1)$, and rewrite the general term of the series as \begin{align*}b_n&=-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\\&=-1+(n+1)\ln(2n+3)-n\ln(2n+1)-\ln(2n+1)=-1-\ln(2n+1)+a_{n+1}-a_n.\end{align*} Summing from $k=0$ to $n$, we obtain that $$\sum_{k=0}^nb_k=\sum_{k=0}^n\left(-1-\ln(2k+1)+a_{k+1}-a_k\right)=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The identity $\tan({\pi\over4}-{a\over2}) = \sec(a)-\tan(a)$ Today I was solving an exercise and while checking the solution on WolframAlpha, the website used the following identity:
$$\tan \left({\pi\over4}-{\alpha\over2} \right) = \sec(\alpha)-\tan(\alpha)$$
Since I've never seen that formula, I tried to calculate it... | Using the identity $\sin^2(a)+cos^2(a)=1$ and dividing by $\cos^2(a)$ gives you the identity $\tan^2(a)+1=\sec^2(a)$. So $\tan(a)=\sqrt{sec^2(a)-1}$. Using this we have:
$$
\begin{split} \tan\left(\frac{\pi}{4}-\frac{a}{2}\right)&=\sqrt{\sec^2\left(\frac{\pi}{4}-\frac{a}{2}\right)-1} \\ &=\sqrt{\frac{1}{\cos^2\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series
$$\sum\limits_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$
Second derivative is
$$\left(\frac{1}{1-x^2}\right)^{''}=\frac{1}{(1-x^2)^4}\cdot x(1+8x-2x^2-8x^3+x^4)$$
This gives
$$\frac{1}{(1-x^2)^4}=\sum\limits_{n=0}^... | Perhaps it could be easier expand $(1-x^2)^{-4}$ using the binomial series.
$\begin{align}
(1-x^2)^{-4} &= 1+4x^2 + \frac{(-4)(-4-1)}{2}(-x^2)^2+\frac{(-4)(-4-1)(-4-2)}{3!}(-x^2)^3+ \dots\\
&=1+4x^2+10x^4+20x^6+35x^8+\dots\\
&=\sum_{n=1}^{\infty} \binom{n+2}{3}x^{2n-2}\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:
if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac ... | It's not clear what the OP means by "adding" the numerator and denominator but there is a way using componendo and dividendo. If $\frac{x+y}{z} = \frac{x}{y}$ then $\frac{(x + y) - x}{z - y} = \frac{x}{y} \implies \frac{y}{z - y} = \frac{x}{y}$.
This implies that $\frac{y}{z - y} = \frac{y}{x - z} \implies z - y = x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Where did I mistake to integrate $I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx\; ?$ It was given to integrate $$I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx.$$
Attempt:
\begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\sin 2\alpha}\int\sqrt{\cot 2\alpha ... | HINTS:
We have $$\frac{\sin(x-\alpha)}{\sin(x+\alpha)}=\frac{\sin x\cos\alpha-\sin\alpha\cos x}{\sin x\cos\alpha+\sin\alpha\cos x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x\cos^2\alpha-\sin^2\alpha\cos^2x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x-\sin^2\alpha}$$
and $$u=(\sin^2x-\sin^2\alpha)^{1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\... | $\frac{n^2}{n^2+1}=1-\frac{1}{n^2+1}$
As $n \to \infty$,
$\because(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} > 1^\frac{n-1}{n+1} = 1$
$(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} < (1-\frac{1}{n^2+1})^\frac{n}{n}=1-\frac{1}{n^2+1}$
$\because \lim \limits_{n \to \infty} 1 = 1$
$\lim \limits_{n \to \infty} (1-\frac{1}{n^2+1}) = 1$
$\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mz-2=0$ is The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mz-2=0$ is
$(A)-2\hspace{1cm}(B)8\hspace{1cm}(C)-18\hspace{1cm}(D)11$
My Attempt:The plane $2x-y+mz-2=0$ is parall... | Note that if you write only $3x-2y+z+3=0$, then it represents a plane.
Here, we have
$$3x-2y+z+3=0=4x-3y+4z+1,$$
which represents a line.
We can write the line with the equation
$$3x-2y+z+3=0=4x-3y+4z+1$$
as
$$\frac{x-0}{1}=\frac{y-\frac{11}{5}}{\frac 85}=\frac{z-\frac 75}{\frac 15}$$
(you can get this by expressing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the formula for the sequence $a(n)$ Find the formula for the sequence $a(n)$ given by the following recurrence relations and prove that the formula is correct:
$a(0) = 1, ~a(1) = -2,~ a(n) = -2a(n-1)-a(n-2)$
How to solve it? I don't know how to start?
| Let $$f(x) = \sum_{n = 0}^\infty a_nx^n$$ be the generating function for the sequence $\left\{a_n\right\}_{0}^\infty$. Then
$$\begin{align}f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots \\
&= 1 -2x + (-2a_1 - a_0)x^2 + (-2a_2 - a_1)x^3 + (-2a_3 - a_2)x^4 + \dots\\
&= 1 - 2x + (-2x)(f(x) - a_0) + (-x^2)f(x)\\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When is EF longer than AC? (a generalization) ABC is an isosceles right triangle,
M is on AC,
and EMF is a straight line.
When is EF longer than AC?
]1
Note:
This is a generalization
of the following problem,
which has M in the
center of AC:
Prove that EF is longer than AC
In that case,
I showed that
EF is always l... | In my solution to
the original problem,
I used algebra and
analytic geometry.
I tried to do the same thing
for this generalization,
but the final simplification
does not occur here.
Here is what I've got.
Suppose
$AB = BC = 1$.
Then
$M
=(a, 1-a)
$,
where
$0 < a < 1$.
Let
$E = (0, 1+v)
$,
where $v > 0
$.
Then
line $EM... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Every sixth polynomial shares a factor of $(a^2-6)$ I currently looking at the polynomials you get from the series expansion of
$$
\frac{1-x^2}{1-ax+2x^2}=1+a x
+(a^2-3) x^2
+(a^3-5a) x^3
+\underbrace{(a^4-7a^2+6)}_{(a-1)(a+1)(a^2-6)} x^4+\dots
$$
W|A helped here...
What I found is that from $x^{4}$ onwards, every ... | Consider $\displaystyle f(a,x)=\frac{1-x^2}{1-a x+2x^2}+{1\over2}=\frac{3-ax}{2(1-a x+2x^2)}$, which is just your function augmented by $1/2$. What you ask amounts at proving that the series expansions of both $f(\sqrt6, x)$ and $f(-\sqrt6, x)$ around $x=0$ do not contain powers of $x$ with exponent ${4+6k}$.
That is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Find number of roots of the equation $e^x(x^4 + 4x^3 + 12x^2 + 24x + 24) + 1 = 0$ Find number of roots of the equation $e^x(x^4 + 4x^3 + 12x^2 + 24x + 24) + 1 = 0$
Using Descartes rule, number of positive roots is zero and there can be a maximum of 4 negative roots.
Also, for the function $P(x)=x^4 + 4x^3 + 12x^2 + 24... | Just observe that
\begin{align}
P(x) &=(x^4+4x^3+4x^2)+(8x^2+24x+18)+6\\
&=(x^2+2x)^2+2(2x+3)^2+6>0
\end{align}
Therefore
$$e^xP(x)+1>0$$ for all $x \in \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the ... | Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Form of a real orthogonal matrix Q. Let $A$ be a real orthogonal matrix of order $2$ with $|A|=1$. Show that $\exists\theta$ s.t. $$A=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}\tag{4 marks}$$
My answer:
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ where $a,b,c,d\in\Bbb R$. Now, since $A$... | The solution in the original post is correct.
Here we present another way forward, which seems fairly efficient. We note that
$$a^2+b^2=1\implies a=\cos \theta\,\,\text{and}\,\,b=\sin \theta$$
$$c^2+d^2=1\implies c=\cos \phi\,\,\text{and}\,\,d=\sin \phi$$
for some $\theta$ and some $\phi$.
Then, we have
$$ac+bd=0\im... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that if $A\ge B$ then $\left[ {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right]\ge0$. Let $A$ and $B$ be $n \times n$ matrices, i.e., $A, B \in M_n$. Also, $A \ge 0$, $B \ge 0$, and $A-B \ge 0$ which mean all these matrices are semi-positive-definite.
Why does
$\left[ {\begin{array}{*{20}{c}}
A... | $\left[ {\begin{array}{*{20}{c}}
A & B \\
B & A \\
\end{array}} \right] =\left[ {\begin{array}{*{20}{c}}
A-B & 0 \\
0 & A-B \\
\end{array}} \right] +\left[ {\begin{array}{*{20}{c}}
B & B \\
B & B \\
\end{array}} \right] \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Product of generating functions Let $f(x) = \sum_{i=0}^\infty a_ix^i$ and $g(x) = \sum_{i=0}^\infty b_ix^i$ where $a_n = 1$ and $b_n = 2^n$ for all natural numbers $n$
What are the first three terms of the sequence generated by $f(x)g(x)$
?
So I know that $f(x)$ will generate $\{1,1,1,1,1,.... \}$ and $g(x)$ will g... | $f(x)=\frac{1}{1-x}$, $g(x)=\frac{1}{1-2x}$, $f(x)g(x)=\frac{1}{(x-1)(2x-1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Computing $\sum{\frac{1}{m^2+n^2}}$ I want to prove that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=2\pi\log R+O(1)$ as $R\rightarrow\infty$.
For this, I'm trying to approximate the sum by using the integral $\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy=\int_{0}^{2\pi}\int_{1}^{R}{\frac{1}{r}}drd\theta=2\pi\log R$... | $$
\begin{align}
\sum_{m,n=1}^R\frac1{m^2+n^2}
&=2\sum_{m=1}^R\sum_{n=1}^m\frac1{m^2+n^2}-\sum_{n=1}^R\frac1{2n^2}\\
&=2\sum_{m=1}^R\frac1m\sum_{n=1}^m\frac{\frac1m}{1+\left(\frac nm\right)^2}-\sum_{n=1}^R\frac1{2n^2}\\
\end{align}
$$
Comparing the Riemann sum $\sum\limits_{n=1}^m\frac{\frac1m}{1+\left(\frac nm\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Define a bijection Okay so I know that I asked this already but I want to ask how much progress I have made. so the question is
Construct a bijection between $[1,2]$ and $[3,5)$
So I have:
\begin{equation}
f(x) = \left \{ \begin{array}{ll} 3 + 2^{1-n} & \textrm{ if } x = 1+2^{1-n} \textrm{ for } n \in \mathbb{N} \\
... | The function $f$ is onto:
Case 1
Let $y \in [3,5) $ be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = 1 + 2^{1-n}$ is such that $f(x) = y$. In particular, if $n$ is such that $3 + 2^{1-n} \in [3,5)$ then $n \geq 0$. This in turn implies that $1 \leq 1 + 2^{1-n} \leq 2$.
Case 2
Let $y \in [3,5) $ ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx$ Evaluation of $$\displaystyle \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx\;,$$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx = 2\int_{0}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx$$
Now Substiute $$\displaystyle \cos x ... | I am trying to continue from your method but before that let me post my method:
use substitution $$\sqrt{cosx-cosk}=t$$ so $$\frac{dx}{\sqrt{cosx-cosk}}=\frac{-2dt}{sinx}=\frac{-2dt}{\sqrt{1-(t^2+cosk)^2}}=\frac{-2dt}{\sqrt{1-(t^2+cosk)}\sqrt{1+(t^2+cosk)}}=\frac{-2dt}{\sqrt{2sin^2\frac{k}{2}-t^2}\sqrt{2cos^2\frac{k}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$ Let $X$ be as set containing $n$ elements.Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$
I hav... | First let's solve this simplier problem : What's the number $\mathcal{N}$ of sets of $X$ such that $|A| = a$, $|B| = b$ and $|C| = c$ with $n \geq a>b>c >0$?
There is $n\choose a$ way to choose $A$, then $a\choose b$ ways to choose $B$ inside $A$, then $b\choose c$ way to choose $C$ inside $B$. This way, we have all th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{dx}{3^x-8}$ I have a problem with solving this Integrate
$$\int \frac{dx}{3^x-8}$$
At first I use substitute $$3^x =t $$
$$\log_3{t} = x$$
$$\frac{1}{t\ln{3}}dt = dx$$
Next:
$$\int \frac{dt}{t^2\ln3} - \int\frac18dx$$
$$=\frac{1}{\ln3}\int \frac{dt}{t^2} - \int\frac18dx$$
$$=\frac{1}{\ln3} \left(\... | Let $3^x-8=u\implies 3^x\ln(3)\ dx=du$ or $dx=\frac{1}{(u+8)\ln (3)}\ du$
$$\int \frac{1}{3^x-8}\ dx=\int \frac{1}{u(u+8)\ln 3}\ du $$
$$=\frac{1}{\ln 3}\int \frac{1}{u(u+8)}\ du $$
$$=\frac{1}{8\ln 3}\int\left( \frac{1}{u}-\frac{1}{u+8}\right)\ du $$
$$=\frac{1}{8\ln 3}\ln\left| \frac{u}{u+8}\right| +C$$
$$=\frac{1}{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is my integration of $\sqrt{4z^2 - 4z + 2}$ correct? I'm trying to
$$
\int \sqrt{4z^2 - 4z + 2}\ dz
$$
the integrand I first rewrite to (completing the square)
$$
4\left(z^2 - z +\frac{1}{4} - \frac{1}{4}\right) + 2 = 4\left(z - \frac{1}{2}\right)^2 + 1
$$
I apply $\int \sqrt{a^2 + x^2} \ dx = \frac{x}{2} \sqrt{a^2+x^2... | Your integration methods are all right.
Few silly mistakes that I observed are:
*
*No need to keep it $4(z-\frac{1}{2})^2+1$. You could have made it $(2z-1)^2+1$ directly.
*You should change the $dz$ to $\frac{1}{2}d(2z-1)$,which you have not done. If you take this into account, your result will be according to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determinant of rank-$1$ update of multiple of identity matrix I've got to calculate determinant for such matrix:
$$ \begin{bmatrix}
a_1+b & a_2 & \cdots & a_n\\
a_1 & a_2+b & \cdots & a_n\\
\vdots & \vdots & \ddots & \vdots\\
a_1 & a_2 & \cdots & a_n+b\\
\end{bmatrix} $$
Please give me some tips how to calculate this.... | Set $A=\sum\limits_{i=1}^na_i$. By multilinearity,
\begin{align*}
&\begin{vmatrix}
a_1+b &a_2&\dots&a_n\\
a_1&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
a_1 a_2&\dots &a_n+b
\end{vmatrix}=
\begin{vmatrix}
A+b &a_2&\dots&a_n\\
A+b&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
A+b & a_2&\dots &b+a_n
\end{vmatrix}\\[1e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | In order to prove anything whatsoever about $\frac{0}{0}$, we need a definition of $\frac{a}{b}$ where $a$ and $b$ are, say, integers. The correct definition is $ \frac{a}{b} = ab^{-1}$, where the definition of $b^{-1}$ is the number (which is unique, when it exists, according to some basic properties of numbers) such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 5
} |
Prove that $\overrightarrow{GH}.\overrightarrow{IJ}=-2x^2+8x-2$ in a regular hexagon We know that the hexagon is regular and:
$\overline{AB}=1$;
$\overline{AG}=\overline{CI}=\overline{DH}=\overline{FJ}=x$;
How would you prove that $\overrightarrow{GH}.\overrightarrow{IJ}=-2x^2+8x-2$?
| $\qquad\qquad\qquad$
Consider $\triangle{GAK}$ where $K$ is the intersection point between $GH$ and $IJ$. ($K$ is the center of the hexagon.) Since $GA=x,AK=1,\angle{GAK}=60^\circ$, by the law of cosines, we have
$$GK=\sqrt{x^2+1^2-2\cdot x\cdot 1\cos(60^\circ)}=\sqrt{x^2-x+1}$$
Here, note that $GK=HK=IK=JK$.
Also, con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| Using
$$\log_{10}(x) = \frac{\ln(x)}{\ln(10)}$$
then
\begin{align}
D \left[ 10^{x} \, \log_{10}(x) \right] &= D \left[ e^{x \, \ln(10)} \, \frac{\ln(x)}{\ln(10)} \right] \\
&= \frac{1}{\ln(10)} \, \left[ \frac{10^{x}}{x} + \ln(10) \, 10^{x} \, \ln(x) \right] \\
&= 10^{x} \, \left[ \frac{1}{ x \, \ln(10)} + \ln(10) \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Find a recurrence relation and generating function of....
Model the amount of crab being caught per year based on the assumption that the # of crab caught in a year is the average of the # caught in the 3 preceding years.
a.) Find a recurrence relation for $C_n$, the # of crab caught in a year $n$.
For this I got:
$$... | Caveat: This is maybe not the best method. But it gets the job done.
Let the generating function of $f$ be given by $f(x)=\sum_{n=1}^\infty C_n x^n$. Then,
$$
\begin{align}
3f(x)&=3(C_1x+C_2x^2+C_3x^3)+\sum_{n=4}^\infty 3C_n x^n
= 3(x+2x^2+2x^3)+\sum_{n=4}^\infty (C_{n-1}+C_{n-2}+C_{n-3}) x^n \\
&= 3(x+2x^2+2x^3)+x\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find Matrix T of a Linear Transformation $T: M_{2\times 2} \to P_3$ Been staring at this practice problem and scribbling nonsense for quite a while now googled extensively and such but nothing triggered any spark of understanding.
All in all a helpful shove off the correct cliff would be appreciated.
Consider the linea... | For the $2\times 2$ matrices, the standard basis would be
$$
\begin{pmatrix} 1 & 0 \\ 0&0 \end{pmatrix}, \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, \begin{pmatrix} 0&0\\1&0 \end{pmatrix}, \begin{pmatrix} 0&0\\0&1 \end{pmatrix}
$$
and for the polynomials the standard basis is $\{1, x, x^2, x^3\}$. Now you want to see w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$ $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$
How do I solve this without using derivatives or integrals.
| Take a look at the following steps
$$\eqalign{
& L = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}} = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}}{{2 + {{(5x + 4)}^{0.5}}} \over {2 + {{(5x + 4)}^{0.5}}}}{{1 + {{(3x + 1)}^{0.5}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A nice double integral on the square $[0,1]$x$[0,1]$ It is recalled that the fractional part of a real $x$, noted {$x$}, is defined by
{$x$} $=x-\lfloor x\rfloor $ where $\lfloor x\rfloor $ is the floor function (largest integer not greater than $x$, many times noted $[x]$).
I don’t remember where I saw the double inte... | $$\int_0^1\int_0^1\{\frac{x}{y}\}dxdy \\
= \lim_{N\to\infty} \sum_{n=1}^{N}\int_{\frac{1}{n+1}}^{\frac{1}{n}}((\sum_{k=0}^{[\frac{1}{y}]-1}\int_{ky}^{(k+1)y} (\frac{x}{y}-[\frac{x}{y}]) dx) + \int_{ny}^1 (\frac{x}{y}-[\frac{x}{y}])dx) dy \\
= \lim_{N\to\infty} \sum_{n=1}^{N} \frac{1}{2}\ln(\frac{n+1}{n})-\int_{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567597",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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On proving $\lim \frac{n^2}{n^2+n+1} = 1$
Prove that $\lim \frac{n^2}{n^2+n+1} = 1$
Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$
But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n... | if $\forall n.a_n \gt 0$
$$
\lim_{n\to\infty} a_n = 1 \Leftrightarrow \lim_{n\to\infty} \frac1{a_n} = 1
$$
set
$$
a_n = \frac{n^2+n+1}{n^2} = 1+\frac1{n}+\frac1{n^2}
$$
so for $n \gt 1$
$$
0 \lt a_n-1 \lt \frac2n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Maximum remainder $(a-1)^n+(a+1)^n\mod a^2$ for $3\le a\le 1000$ Here's the problem:
Let $r$ be the remainder when $(a−1)^n + (a+1)^n$ is divided by $a^2$.
For example, if $a = 7$ and $n = 3$, then $r = 42$ since $63 + 83 = 728 \equiv 42 \pmod{49}$. And as $n$ varies, so too will $r$, but for $a = 7$ it turns out that... | As you said the remainder is
$$r\mod a^2 = \begin{cases}2&2|n\\2an&2\not|n\end{cases}.$$
Now observe that the maximum value of $2an \mod a^2$ could be $a(a-1)$. IT holds if and only if the following system of congruence
$$\begin{cases}n\equiv1&\mod 2\\
2an\equiv a(a-1)&\mod a^2
\end{cases}$$
has solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit involving exponentials of $\arcsin(x)$ and $\arctan(x)$ How can I calculate this limit without L'Hospital rule and Taylor series?
$$
\lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)}
$$
| Let $L=\lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{(e^{\arctan(x)} -1)- (e^{\arcsin(x)}-1)}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{\arctan (x)\frac{(e^{\arctan(x)} -1)}{\arctan(x)}- \arcsin (x)\frac{(e^{\arcsin(x)} -1)}{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The chart-problem; problem solving In how many ways can we construct a $6\times 6$ chart with only $1$ and $-1$ such that in every row and column, the product is always positive?
| There are $2^{25} $ combinations. Like Jack's answer fill the $5 \times 5$ subgrid. I will prove that there is a unique $6 \times 6$ grid with the desired property.
First we add the correct $1$'s or $0$'s in the first five places of the last row and column. We only have to determine the last digit in the lower left cor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
I would like to prove convergence of the following series: $\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)}$ I would like to prove the following series:
$$\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)} $$
is convergent (absolutely?) or divergent. I think $\arctan\left(\frac{n}... | To show that the terms are decreasing:
$\arctan\left(\frac{n}{1+n^2}\right)- \arctan\left(\frac{n+1}{1+(n+1)^2}\right)
=\arctan\left(\dfrac{\frac{n}{1+n^2}-\frac{n+1}{1+(n+1)^2}}{1+\frac{n}{1+n^2}\frac{n+1}{1+(n+1)^2}}\right)
$
and
$\dfrac{n}{1+n^2}-\dfrac{n+1}{1+(n+1)^2}
=\dfrac{n(1+(n+1)^2)-(n+1)(1+n^2)}{(1+n^2)(1+(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n})$ I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$
I know it is $1$ but I don't understand why this is wrong :
$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$... | $$\sqrt{n^2+n} = n \sqrt{1 + \frac{1}{n} } = n (1 + \frac{1}{2n} + O(1/n^2)) =$$
$$ = n + \frac{1}{2} + O(1/n) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$. Who can help me to prove this equation?
$$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$
| We have
$$\left(\sqrt{a-\sqrt{b}}+\sqrt{a+\sqrt{b}}\right)^2 = a-\sqrt{b}+a+\sqrt{b} +2\sqrt{a^2-b} = 2a+2\sqrt{a^2-b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\sec\theta+\tan\theta=p$ and $\sec\theta\tan\theta=q$. Eliminate $\theta$ to form a equation between $p$ and $q$. $\sec\theta+\tan\theta=p$ and $\sec\theta\tan\theta=q$. Eliminate $\theta$ to form a equation between $p$ and $q$.
$\sec\theta+\tan\theta=p$
$(\sec\theta+\tan\theta)^2=p^2$
$\sec^2\theta+\tan^2\theta+2\ta... | Now, $\sec^2\theta \tan^2\theta=q^2$.
So, $(1+\tan^2\theta)\tan^2\theta=q^2$.
You have got that $1+2\tan^2\theta+2q=p^2$.
So, $\tan^2\theta=\frac{p^2-1-2q}{2}$.
Now plug in this $\tan^2\theta$ in the equation got above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Calculating the value of $\lfloor(1+\sqrt{2})^{2n}\rfloor$ Problem: Calculate the value of $\lfloor(1+\sqrt{2})^{2n}\rfloor$ where $n$ is an arbitrary non-negative integer and $\lfloor x\rfloor$ indicates the largest integer not greater than $x$.
What I have tried:
Based on the Binomial Theorem:
$$
(1+\sqrt{2})^{2n} +... | You can get recursions, which might be easier to compute.
Define $a_n=1+\left\lfloor (1+\sqrt{2})^{2n}\right\rfloor$, for $n>0$, and get a recursion:
$$a_{n+1} = 6a_n - a_{n-1}; a_1=6; a_2=34$$
So if you need a bunch of values $a_{n}$, recursion will be faster. It still might be faster to use this formula than your ori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqr... | Hint
First let us give a name to the nested radicals
$$x=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}$$
Then you can observe that
$$((x^2-4)^2-4)^2=4-x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$ Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$=\int_{0}^{\frac{\pi}{12}}\frac{1}{3}\frac{\tan^2x-\frac{1}{3}+\frac{1}{3}-3}{\tan... | $$u=\tan(x)$$
$$du=(\tan^2x+1) dx$$
$$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx=\int_{0}^{\tan\frac{\pi}{12}}\frac{u^2-3}{(3u^2-1)(u^2+1)}du$$
$$=\int_{0}^{\tan\frac{\pi}{12}}(\frac1{u^2+1}-\frac2{3u^2-1})du$$
$$=\int_{0}^{\tan\frac{\pi}{12}}(\frac1{u^2+1}+\frac{\sqrt3}{3u+\sqrt3}-\frac{\sqrt3}{3u-\sqrt3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is there a closed form for $n^k$ in terms of $\Delta n^{k+1},\Delta n^k$, ...? Let $\Delta$ be a sort of difference operator on a function $f(n)$ such that
$$\Delta f(n)=f(n+1)-f(n)$$
Take the basic power function $f(n)=n^k$, $k\in\mathbb{N}\cup\{0\}$. Then we get
$$\begin{cases}\Delta 1=0\\[1ex]
\Delta n=1\\[1ex]
\Del... | We derive a formula for $n^k$ in terms of $\Delta n^k$ by use of binomial inverse pairs and their generating functions. We show
the following is valid
\begin{align*}
n^k=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}\qquad\qquad k\geq 0
\end{align*}
with $B_k$ the Bernoulli numbers.
$$ $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| First, show that this is true for $n=1$:
$\left(\sum\limits_{k=1}^{1}k\right)^2=\sum\limits_{k=1}^{1}k^3$
Second, assume that this is true for $n$:
$\left(\sum\limits_{k=1}^{n}k\right)^2=\sum\limits_{k=1}^{n}k^3$
Third, prove that this is true for $n+1$:
$\left(\sum\limits_{k=1}^{n+1}k\right)^2=$
$\left(\color\green{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
$3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$ I came across this problem:
Prove there arent't any $a$, $b$ integers that satisfy
equation $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$
Firstly, I've thought something like this:
$$(a^3 + b^3)-3b^3 + 3ab - 4a + 4b - 4 + b = 3$$
$$(a+b)(a^2-ab+b^2) - 3(b^3 - ab) -4(a-b+1)+b = 3$$
I thin... | $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$
Consider the equation mod $3$. Since $x^3\equiv x\pmod 3$ for all $x$, we have
$3ab+a-2b-4a+2b-7\equiv 0$ which is the same as $2\equiv 0\pmod 3$.
Contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find a continuous function $f$ such $\int_{a}^{a^2+1}f(x)dx=0\quad\forall a\in \Bbb R$ Find example : exsits a nonzero continuous function $f:\Bbb R\to \Bbb R$ satisfying
$$\int_{a}^{a^2+1}f(x)dx=0\quad\forall a\in \Bbb R$$
the book answer this: $f(x)=x^2(x-1)(x-\frac{3}{5}),0\le x\le 1$,
$f(x)=-f(-x),-1\le x\le 0$.
... | I think the answer you have is the seed of a full answer. You have a formula for $f$ on $[-1,1]$, but then you would need to use the defining relation to extend $f$ to larger and larger domains.
Take $$f(x)=\begin{cases}x^2(x-1)(x-\frac{3}{5})&0\le x\le 1
\end{cases}$$ where we know that with $a=0$, $$\int_0^1 x^2(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Since $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, you know that
$$
\tan A+\tan B=-p,
\qquad
\tan A\tan B=q
$$
so
$$
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=
\frac{-p}{1-q}=\frac{p}{q-1}
$$
provided $q\ne1$.
The relation to be proved can be written
$$
\sin^2C+p\sin C\cos C+q\cos^2C=q
$$
where $C=A+B$, or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is there a formula for computing the number of arrangements from a set of N elements groupped in K groups? Eg: N=3, K=2
There will be two groups in each solution. We need to calculate the number of such possible solutions. Consider the set S={1,2,3}. The possible solutions are:
{1} {2,3}
{1} {3,2}
{2} {1,3}
{2} {3,1}
{... | Consider an arrangement of a set $S = \{ a_1, a_2, a_3, \dots, a_n\}$.
We need to partition this set into $K$ groups such that each group contains more than one element such that the relative order of elements remain same. For example for $S = \{1,2,3\}$, the partition $\{1\},\{2,3\}$ will be valid but $\{2\},\{1,3\}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determinant $n\times n$ problem $$
D_n = \left|
\begin{matrix}
n & -1 & -3 & 0 & 0 & \cdots & 0 & 0 & 0 \\
n & 1 & 2 & -3 & 0 & \cdots & 0 & 0 & 0 \\
n & 0 & 1 & 2 & -3 & \cdots & 0 & 0 & 0 \\
n & 0 & 0 & 1 & 2 & \cdots & 0 & 0 & 0 \\
n & 0 & 0 & 0 & 1 & \cdots & 0 & 0 & 0 \\
\vdots&\... | Add the second up to the last rows to the first. Then the first row becomes $(n^2,0,\ldots,0)$. The new matrix is lower block-triangular and is in the form of $\pmatrix{n^2&0\\ n\mathbf1&A}$, where $A$ is by itself an upper triangular matrix with determinant 1. Hence $\det D_n=n^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| $$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1\\=x(x((x-1)(x+1)(x(x+2)-1)x^4+3)+6)+1$$
$\Longrightarrow f(\sqrt 2 -1)=\boxed{\color{red}{4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
| With limits going to infinity, usually factoring out the greatest term is the way to go, since then you'll have a lot of terms going to $0$. Here the term that grows faster is $x!$. So:
\begin{align}
\require{cancel}
\lim_{x \to +\infty} \frac{(x+2)(x+1)x!\left(1+\frac{4^x}{(x + 2)!}\right)}{((2x+1)^2+\ln x)x!} &= \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Using Lagrange's diagonalization on degenerate linear forms
Let $A=\begin{pmatrix}1 & 2 & 3\\
2 & 3 & 4\\
3 & 4 & 5
\end{pmatrix}$ be a real matrix.
Find an invertible matrix $P\in M_{3}(\mathbb{R})$ such that $P^TAP$ is diagonal whose elements on the diagonal are all in the set $\{-1, 0, 1\}$
I've been trying to s... | Denote the quadratic form by
$$ q \left( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right) = (x_1, x_2, x_3) \cdot A \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = (x_1 + 2x_2 + 3x_3)^2 - (x_2 + 2x_3)^2. $$
We see that if we make the change of variables
$$ u_1 = x_2 + 2x_2 + 3x_3, \,\, u_2 = x_2 + 2x_3, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx,\alpha\in(0,\pi)$ For the equation $x^2-2x\cos\alpha+1=0$ solutions are
$$x_1=\cos\alpha-\sqrt{\cos^2\alpha-1},x_2=\cos\alpha+\sqrt{\cos^2\alpha-1}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\int_{-1}^1 \frac{1}{(x-\cos\alp... | Since the denominator has no real roots you should complete the square and write it as $$(x-\cos\alpha)^2+\sin^2\alpha$$
Applying the standard integral $$\int\frac{1}{a^2+x^2}dx=\frac 1a \arctan(\frac xa),$$ we get
$$\frac{1}{\sin\alpha}\left[\arctan\frac{1-\cos\alpha}{\sin\alpha}-\arctan\frac{-1-\cos\alpha}{\sin\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose ... | New Method
assume this Matrics
\begin{bmatrix} 1&2&2&-3\\4&2&5&0\\1&3&4&2\\-3&1&-1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original matrix from 1234 to 1342
\begin{bmatrix} 1&2&-3&2\\4&5&0&2\\1&4&2&3\\-3&-1&1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 4
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\... | Notice, $$\int \frac{1}{x^4+x^2+1}\ dx$$ $$=\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\ dx$$
$$=\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx$$
$$=\frac{1}{2}\left(\int \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx-\int \frac{\left(1-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that $f(x,y)=g(x+y,xy)$. Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. Prove that $f(x,y)=g(x+y,xy)$.
My try.
$f\in \Bbb R[x,y]$ and $f(x,y)=f(y,x)$ .If I take for example $f(x,y)=a_{00}+a_{01}y+a_{11}xy+a_{10}x+a_{13}xy^3$ then only the coefficients which are symmetric w.r.t $x $ and $y$ will remain in th... | Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. To prove that $f(x,y)=g(x+y,xy)$.
Let $$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{1,0}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{2,0}x^2y^0+a_{2,1}x^2y^1+a_{2,2}x^2y^2$$
We then have
$$f(y,x)=a_{0,0}y^0x^0+a_{0,1}y^0x^1+a_{0,2}y^0x^2+a_{1,0}y^1x^0+a_{1,1}y^1x^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that the sequence $a_0 = 1$, $a_{n+1 }= \sqrt{2+a_n}$ is monotonically increasing Given the sequence $a_0 = 1$, $a_{n+1}= \sqrt{2+a_n}$, how can I show that it is monotonically increasing?
I need it to show, that the sequence converges. I already proved boundedness but I can't figure out this last part.
| Option 1: First, prove that $1 \le a_n < 2$ for all $n \in \mathbb{N}$. This can be done using induction.
After that, prove $\sqrt{2+x} > x$ for $1 \le x < 2$, and thus, $a_{n+1} = \sqrt{2+a_n} > a_n$ for all $n \in \mathbb{N}$.
Option 2: Define $\theta_n = \arccos\dfrac{a_n}{2}$. Then, $a_n = 2\cos\theta_n$ for each... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct
So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equati... | Just another way:
Inductive hypothesis: $3^{2n+1} = 7k - 2^{n+2}$
Inductive step:
$$3^{2n+3} + 2^{n+3} = 3^2 * 3^{2n+1} + 2 * 2^{n+2} $$
$=3^2(7k - 2^{n+2})+2*2^{n+2}$
$=7k(3^2) - 3^2*2^{n+2} + 2* 2^{n+2}$
$=7k(3^2) + 2^{n+2}(-3^2+2)$
$=7k(3^2) -7*(2^{n+2})$
Both terms are divisible by 7
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$ To solve the equation
$$x^2-5=0$$
There exitsts a iteration method
$$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$
I know that it is cubic convergence but I don't know how to prove it.
I have tried the following
$$
\begi... | Hint: Define $f(x):=\frac{x(x^2+15)}{3x^2+5}$, and Taylor expand it around $x=\sqrt{5}$.
If you want to continue with the method you're currently working with, then you'll want to replace then $x_n$ terms by $e_n+\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2... | 1) $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\implies\frac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1$
$\implies(x^2-5x+4)(x^2-5x+6)=(x^2+5x+4)(x^2+5x+6)$
We know that the even-indexed terms will cancel out because the LHS is just the RHS with $x\to-x$
$\implies-10x^3-50x=10x^3+50x\implies10x^3+50x=0\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)$ without L'Hospitals rule? $$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)=?$$
How to solve it without using L'Hospitals rule?
| Use equivalents and Taylor's formula at order $2$: we know that
*
*$\cos x=1-\dfrac{x^2}2+o(x^2)$,
*$\sqrt{1+u}=1+\dfrac u2+o(u)$,
*$\sqrt[3]{1+u}=1+\dfrac u3+o(u)$
Thus the numerator gives, by composition:
$$\sqrt[2]{\cos x}-\sqrt[3]{\cos x}=\Bigl(1-\frac{x^2}4\Bigr)-\Bigl(1-\frac{x^2}6\Bigr)+o(x^2)=-\frac{x^2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-... | $$
\begin{aligned}
\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2\left(1-x^2\right)}{\left(ln\left(x^2-3\sqrt{2}x+5\right)\right)^2}\right)
& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(1-\left(t+\sqrt{2}\right)^2\right)}{\left(ln\left(\left(t+\sqrt{2}\right)^2-3\sqrt{2}\left(t+\sqrt{2}\right)+5\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all pairs $(n,k)$ such that $n(n+1) \, \mid\,(k+1)! \,(1^k+2^k+3^k+\cdots+n^k)$ How can I solve (find all the solutions) the following problem?
Find all pairs of postive integers $(n,k)$ such that
$$n(n+1) \,\mid \,(k+1)!\, (1^k+2^k+3^k+\cdots+n^k).$$
I included here what I had done so far. If $k=1$ and $n=1$, the... | It is true for all $n,k \in \mathbb{N}, n,k \geq 1$.
Induction by $k$.
1) For $k=1$ we have $(k+1)! (1^k + \ldots + n^k = 2(1+\ldots+n) = 2 \frac{(n+1)n}{2}=(n+1)n $ , so this is obviously true.
2) For $k \geq 1$ let we assume that that this conditions it true for all previous $j \geq 1, j<k$.
Cosider following sum:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Example of normal linear maps Let $V$ be a real inner product space and $g:V\rightarrow V$ a normal linear map.
Can you give me an example of:
1- A map $g$ that is not self-adjoint?
2- A map $g$ that is not an isometry?
3- A map $g$ that is not orthodiagonalizable?
| Consider the matrix
$$ A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \in M_2(\mathbb{R}). $$
We have
$$ AA^T = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix}, \\
A^TA = \begin{pmatrix} a & b \\ -b & a \end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is g... | Your first step is good. Now you want to compute
$$
\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}
$$
because the factor $x+3$ at the denominator poses no problem. This should remind you of the definition of derivative and indeed it's the derivative of $f(x)=\sqrt{x+6}$ at $3$. Since
$$
f'(x)=\frac{1}{2\sqrt{x+6}}
$$
for $x>-6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculte indefinte integral of $\int \frac{dx}{\sqrt{(x+2)(3-x)}}$ I have to calculate $$\int \frac{dx}{\sqrt{(x+2)(3-x)}}$$.
I tried to use -
$\int u'v = uv - \int v'u$, but im pretty stuck.
Thanks.
| $$\int\frac{1}{\sqrt{(x+2)(3-x)}}\space\text{d}x=\int\frac{1}{\sqrt{-x^2+x+6}}\space\text{d}x=\int\frac{1}{\sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}}\space\text{d}x=$$
Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$:
$$\int\frac{1}{\sqrt{\frac{25}{4}-u^2}}\space\text{d}u=\int\frac{2}{5\sqrt{1-\frac{4u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find complex roots of $\frac{2i}{1+i}$
Find 6th roots of $$\frac{2i}{1+i}$$
$$\frac{2i}{1+i}=\frac{2e^{i\pi/2}}{\sqrt 2 e^{i \pi/4}}=\sqrt 2 e^{i \pi/4}$$
Now if I set $z^{1/6}=\sqrt 2 e^{i \pi/4}$ and knowing the fact that the roots are distibuted equality with an angle $k\pi/3$ for $k=1,2,3,4,5,6$ I get the answer ... | $$z^6=\frac{2i}{1+i}\Longleftrightarrow$$
$$z^6=\left|\frac{2i}{1+i}\right|e^{\arg\left(\frac{2i}{1+i}\right)i}\Longleftrightarrow$$
$$z^6=\frac{\left|2i\right|}{\left|1+i\right|}e^{\left(\arg(2i)-\arg(1+i)\right)i}\Longleftrightarrow$$
$$z^6=\frac{2}{\sqrt{2}}e^{\left(\frac{\pi}{2}-\arctan\left(\frac{1}{1}\right)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \... | $$\left(\frac{2}{\sin x}-3\sin x\right)^2\geq0$$
$$\iff\csc^2 x+9\sin^2 x\geq12$$ with equality achieved when $\sin^2x=2/3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is ... | As one of the comments stated, your first error lies in the first step where you multiply "everything" by $\sqrt{3}$. While you claim to do this, you did not multiply both terms in the denominator by $\sqrt{3}$. This was an error that let you to an incorrect expression.
That having been said, your approach is not corre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
... | $$\frac{i^4+3}{i-1}=\frac{iiii+3}{i-1}=\frac{i^2i^2+3}{i-1}=\frac{-1-1+3}{i-1}=\frac{1+3}{i-1}=\frac{4}{i-1}=$$
$$\frac{4(i+1)}{(i-1)(i+1)}=\frac{4i+4}{-1^2-1^2}=\frac{4+4i}{-2}=\frac{4}{-2}+\frac{4i}{-2}=-2-2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:
Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2... | $\sqrt{6}-\sqrt{2}$ is positive because $\sqrt{x}$ is minimal at $0$ and always increasing at $x>0$.
Because $x^2$ is $1$ at $x=1$ and always increasing at $x>0$, numbers greater than 1 remain greater than $1$ when squared and $8-\sqrt{48}$ results from squaring $\sqrt{6}-\sqrt{2}\,$.
Since $8-\sqrt{49}=1$ and $\sqrt{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Eigenvectors of $\begin{pmatrix}6&2\\-10&-1\end{pmatrix}$ (linear equations with complex numbers) I want to compute the eigenvectors of the matrix
$$\begin{pmatrix}6&2\\-10&-1\end{pmatrix}$$
and thus far I got the eigenvalues $\lambda_{1,2}=\frac{1}{2}(5\pm\sqrt{31}i)$. However solving for example the system
$$\begin{p... | We have:
$$\begin{pmatrix}6-\frac{1}{2}(5+\sqrt{31}i)&2\\-10&-1-\frac{1}{2}(5+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
Lets simplify:
$$\begin{pmatrix}\frac{1}{2}(7-\sqrt{31}i)&2\\-10&-\frac{1}{2}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
Divide the first row by $\frac{1}{2}(7-\sqrt{31}i)$ and the second row by $-10$:
$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find coordinate vector for an ordered none standart basis of polynomials? I know that it's trivial to know the coordinate vector with respect to the standart basis of polynomials.
The standart basis of $P_2[R]$ is:
$$\{1,t,t^2\}$$
A general vector spanned by the above set is of the form:
$$a+bt+ct^2$$
such that ... | method $1$:
Find the change of basis matrix from $E$ to $B$:
$1+t=1(1)+1(t)+1(t^2)$
$t^2\;\;\;\;\,=0(1)+0(t)+1(t^2)$
$t\;\;\;\;\;\;= 0(1)+1(t)+0(t^2)$
tranpose the coefficients and put as columns in a $3x3$ matrix, you get:
$$C = \left( {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right)$$
take $C$ and m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+sinx}-3}$? $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
I have tried with Taylor:
$$\lim _{x\to \:0^+}\frac{\left(1+x+\frac{x^2}{2}+1+2x+2x^2\right)^2-4}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}=\lim _{x\to \... | $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
$$=\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sin x}\cdot\lim _{x\to \:0^+}(\sqrt{9+\sin x}+3)$$
$$=\left(\lim _{x\to0^+}\frac{e^{2x}-1}x+\lim _{x\to0^+}\frac{e^{4x}-1}x+2\lim _{x\to0^+}\cdot\frac{e^{3x}-1}x\right)\cdot\dfrac1{\lim _... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that for every natural $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$ Prove that for every natural number $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$.
I've noticed that $(n^2 + n) = n(n+1)$ so these are two successive numbers hence one of them can be divided by two.
I suppose that I should prove that $(n^2 +... | HINT:
$$\begin{align} & (n^2+n)(n^2+2) \\
&=n(n+1)(n^2+2) \\
&=n(n+1)(n^2-1+3)\\
&=\color{red}{n(n+1)(n-1)} \cdot (n+1)+3\cdot\color{blue}{n(n+1)}\end{align}$$
Get the clue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solve limit with Lagrange theorem I tried to solve this limit:
$$ \lim_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) $$
Instead of solving it with Taylor series (using $u = 1/x$), I noticed that the difference within the parenthesis is the $\Delta f$ of the function $e^{\frac{1}{x}}$.
$\lim\limits_... | \begin{align}
& \lim_{x \rightarrow \infty} x^2 \left( e^{\frac{1}{x+1}} -e^{\frac{1}{x}} \right)\\
= & \lim_{x \rightarrow \infty} \left( 1+ \frac{1}{x+1} + \frac{1}{2 (x+1)^2} + O(x^{-3}) - 1 -\frac{1}{x} - \frac{1}{2 x^2} + O(x^{-3})\right) \\
= & \lim_{x \rightarrow \infty} \frac{-x^2}{(x+1)x} + \frac{-2x-1}{2(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
the minimum value of $a^2+b^2-a-\frac{2b}{3}$ using known standard inequalities If $a,b$ are real numbers satisfying $a+2b=3,$ then the minimum value of $a^2+b^2-a-\frac{2b}{3}$
Here $a+2b=3\implies a=3-2b$
$a^2+b^2-a-\frac{2b}{3}=(3-2b)^2+b^2-(3-2b)-\frac{2b}{3}$
$=9+4b^2-12b+b^2-3+2b-\frac{2b}{3}$
I diiferentiated i... | Hint : complete squares !can you see the answer?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is this relationship already known? I like math because it's a puzzle to me, but am really not very good at it. But I figured out the relationship below myself. Just curious, is this already pretty common knowledge? Kind of proud of myself for figuring it out, but my son who's getting math minor had never heard of it. ... | I would prove it this way:
\begin{align}
b^2&=a^2+a+b&\iff\\
b^2-b&=a^2+a&\iff\\
(b-1)(b)&=(a)(a+1)
\end{align}
which is clearly true if $b=a+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Inequality $2a^nb^nc^n+1\geq a^{2n}+b^{2n}+c^{2n}$ Let $a,b,c\in[-1,1]$ be such that $$2abc+1\geq a^2+b^2+c^2.$$ Prove that $$2a^nb^nc^n+1\geq a^{2n}+b^{2n}+c^{2n}$$ for any positive integer $n$.
The case $n=1$ is of course the same as the assumption. For $n=2$, squaring the assumption gives $$4a^2b^2c^2+4abc+1\geq a^4... | Here is a partial solution : $a>0,b>0,c>0$ case
$b=\cos{B},c=\cos{C}, 0 \le B,C \le \dfrac{\pi}{2}$
$a^2-2abc\le 1-b^2-c^2 \iff bc-\sqrt{(1-b^2)(1-c^2)} \le a \le bc+\sqrt{(1-b^2)(1-c^2)} \iff \cos{(B+C)} \le a \le \cos{(B-C)}$
now we need to prove:
$b^{n}c^{n}-\sqrt{(1-b^{2n})(1-c^{2n})} \le a^{n} \le b^nc^n+\sqrt{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic.
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$
is a root of the cubic.
My effort
Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$.
Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14)... | Since the term in $x^2$ is missing, the sum of the three roots is zero; so $a^2+2a-14$ is a root if and only if $-a-(a^2+2a-14)=-a^2-3a+14$ is also a root.
Since
$$
(x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax-a^4 - 5a^3 + 22a^2 + 70a - 196
$$
and the remainder of $-t^4 - 5t^3 + 22t^2 + 70t - 196$ divided by $t^3-21t+35$ is $t^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$
Prove that if $a,b,c$ are non-negative real numbers, then
$(ab + bc + ca)^2 \geq 3abc(a+b+c)$.
I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
| $a,b,c$ can be any real numbers.
Let $x=ab$, $y=bc$, $z=ca$. Then your inequality is $$(x+y+z)^2\ge 3(xy+yz+zx),$$
which is true, because $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
and the inequality is equivalent to $$x^2+y^2+z^2\ge xy+yz+zx,$$
which is true, because, as you said, it's equivalent to $$\frac{1}{2}\left((x-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for all integers n such that n ≥ 3, $ 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$ I am trying to prove this using mathematical induction, but I'm lost once I get to comparing the two sides of the equation.
Proposition:
For all integers n such that n ≥ 3,
$ 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$
Pro... | Using the induction hypothesis, the last line you wrote should be
$\frac{4(4^k - 16)}{3} + 4^{k + 1}$. Then:
\begin{align*}
\frac{4(4^k - 16)}{3} + 4^{k + 1} &= \frac{(4^{k+1} - 4\cdot16)}{3} + \frac {3\cdot4^{k + 1}}3
\\ &= \frac{4^{k+1} - 4\cdot16+3\cdot4^{k + 1}}3
\\ &= \frac{4\cdot 4^{k+1} - 4\cdot16}3
\\ &= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What am I counting wrong? EDIT: I made a mistake in the beginning, the second condition has changed. Sorry for this.
I'm asked to count the number of sets of 4 elements that satisfy the two following conditions:
1) Each element of the set is a two-digit number (from 10 to 99).
2) There are no repeated digits in the set... | Case $1$ - digit $0$ appears:$$\frac{4\cdot\binom97\cdot7!}{4!}$$
Case $2$ - digit $0$ does not appear: $$\frac{\binom98\cdot8!}{4!}$$
Total count:$$\frac{4\cdot\binom97\cdot7!+\binom98\cdot8!}{4!}=\frac{3\cdot9!}{4!}=45360=2^4\cdot3^4\cdot5\cdot7$$
As for the error in your counting, you seem to have mixed your cases.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatri... | Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
MY ATTEMPT:
Since it is an orthogonal matrix So modulus of It's eigenvalue is equal to 1
Also since $det(A)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Require assistance proving $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$ Theorem: $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$
Attempted Solution:
We use induction. Additionally, we prove the stronger inequality omitting the floor function. That is, ... | Another approach is to show that
$$\left(1+\frac1n\right)^{2n}=\left(\frac{n+1}{n}\right)^{2n}>2.$$
The inequality follows directly from the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On a hypothetical computer with a word length of three digits and truncation, compute the solution of a system of equations
On a hypothetical computer with a word length of three digits and truncation, compute the solution of
$$
\begin{matrix}
-3x & + & y & = & -2 \\
10x & - & 3y & = & 7 \\
... | You have arrived at the correct conclusion. It is however possible to argue that you have made mistakes and reduce your score during a test. This depends on how strict an instructor you have. You specifically state that $\frac{10}{3}$ has a finite decimal expansion and that the last digit is the digit $5$. Of course, y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the following congruence: $x(x+1)(x+2) \equiv 0 \pmod{221}$
Find the first five solutions for,
$$x(x+1)(x+2) \equiv 0 \pmod{221}$$
I am very confused. By CRT,
$x(x+1)(x+2) \equiv 0 \pmod{13}$ and $x(x+1)(x+2) \equiv 0 \pmod{17}$
But these two congruences are also reckless.
| $$x(x+1)(x+2)\equiv 0\pmod{221}\iff \begin{cases}x(x+1)(x+2)\equiv 0\pmod{13}\\x(x+1)(x+2)\equiv 0\pmod{17}\end{cases}$$
By Euclid's Lemma:
$$\iff \begin{cases}x\equiv \{0,-1,-2\}\pmod{13}\\x\equiv \{0,-1,-2\}\pmod{17}\end{cases}$$
By the Chinese Remainder Theorem (CRT) there are $9$ solutions mod $221$.
E.g., if $x\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Covariance of dice tosses that result in 1 or 2 (fake proof) Question:
Consider n independent tosses of a $k$-sided fair dice. Let $X_i$ be the number of tosses that result in $i$.
What is the covariance $\mathrm{cov}(X_1,X_2)$ of $X_1$ and $X_2$.
\begin{align}
\mathrm{cov}(X_1,X_2) = \mathbf{E}[X_1X_2] - \mathbf{E}[... | Unfortunately it turns out MIT was right and I was wrong.
Thanks to the help from JeanMarie I was able to come up with the following derivation.
\begin{align}
\mathbf{E}[{X_2}^2] &= \mathrm{var}(\mathbf{E}[{X_2}^2]) + \mathbf{E}[X_2]^2 \\
\mathrm{var}(\mathbf{E}[{X_2}^2]) &= \frac{n(k-1)}{k^2} \\
\mathbf{E}[X_2]^2 &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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.