Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the missing coordinates. Fill in the missing coordinates on the unit circle, represented by the letters.
Using sin and cos, we have a $\sin(45^\circ)$ of $\frac{\sqrt{2}}{2}$ and a $\cos(45^\circ)$ of $\frac{\sqrt{2}}{2}$, and a $\cos(60^\circ)$ of $\frac{1}{2}$, and a $\sin(60^\circ)$ of $\frac{\sqrt{3}}{2}$. The... | Point represented as (value of cos, value of sin)
Point A represent $45^\circ$
$sin45^\circ = \frac{1}{\sqrt2}$,
$cos45^\circ = \frac{1}{\sqrt2}$,
Point B represent $120^\circ$
$sin120^\circ = \frac{\sqrt3}{2}$,
$cos120^\circ = -\frac{1}{\sqrt2}$, [ because in second quadrant]
Point C represent $150^\circ$
$sin1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $ where $0 < \alpha <1$ Evaluate
$$P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $$
where $0 < \alpha <1$
Thm
Let $P$ and $Q$ be polynomials of degree $m$ and $n$,respectively, where $n \geq m+2$. If $Q(x)\neq 0$. for $Q$ has a zero of order at most... | We assume $0<\alpha<1$. We have
$$
P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx =\frac{\pi}{\sin(\alpha \pi)}.
$$
Hint. One may prove that
$$
\begin{align}
& \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx
\\\\&=\int^1_{0} \frac{x^\alpha }{x(x+1)} dx+\int^{\infty}_{1} \frac{x^\alpha }{x(x+1)} dx
\\\\&=\int^1_{0} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Functions have similar graphs I graphed the two functions $y=x^2\sin\left(\frac{x}{50}\right)$ and $y=\frac{1}{50}x^3$ and I noticed they have extremely similar graphs from $-50 < x <50$. What is the reason for this?
| Hint. By differentiation, one may prove that
$$
u-\frac{u^3}6\le\sin u\le u,\qquad u \in [0,1]
$$ giving
$$
\frac{x^3}{50}-\frac{x^5}{750000}\le x^2 \sin \left(\frac{x}{50}\right)\le \frac{x^3}{50}, \quad -1<\frac{x}{50}<1
$$ or
$$
\left| x^2 \sin \left(\frac{x}{50}\right)-\frac{x^3}{50}\right|\le \left|\frac{x^5}{750... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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continued fraction for $\sqrt{n^2 − 1}$ Question:
Find the continued fraction for $\sqrt{n^2 − 1}$, where $n \ge 2$ is
an integer.
My attempt:
$n - 1 = \sqrt{n^2} - 1 \lt \sqrt{n^2 − 1} \lt \sqrt{n^2}$
So far, $[n-1; ]$
$\sqrt{n^2 − 1} = n - 1 + \frac{1}{x}$
$\to \frac{1}{x} = \sqrt{n^2 - 1} - (n-1) \to \frac{1}{... | Use that $$x = \frac{1}{\sqrt{n^2-1} - (n-1)} = 1+ \frac{\sqrt{n^2-1} - (n-1)}{2(n-1)} = 1 + \frac{1}{2(n-1)x} $$
to conclude that $$\sqrt{n^2-1} = [n-1; \overline{1, 2(n-1)}]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$ Let $\{a,b,c,d\}\subset[-1,1]$ such that $a+b+c+d=0$. Prove that:
$$\sqrt{1+a+b^2}+\sqrt{1+b+c^2}+\sqrt{1+c+d^2}+\sqrt{1+d+a^2}\geq4$$
I tried Holder and more, but without success.
| Some thoughts
(For 3-variable problem, see: Prove $\sum_{\mathrm{cyc}} \sqrt{34x^2 + 28y^2 + 7z^2 - xy - 28yz + 41zx} \ge 9x + 9y + 9z$)
It suffices to prove the following:
Problem 1: Let $a, b, c, d \ge -1$ with $a+b+c+d = 0$. Prove that
$$\sqrt{1 + a + b^2} + \sqrt{1 + b + c^2} + \sqrt{1 + c + d^2} + \sqrt{1 + d + a^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove by induction that $n^3 < 3^n$ The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$.
The way I have been presented a solution is to consider:
$$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac{5}{4})^3 = \frac{125}{64} <2 < 3$$
Then using this $$(d+1)^3 = d^3 \times \frac{(d+1)^3}{d^... | This base case holds because $4^3 < 3^4$.
To show that the inductive step holds, we need to show that:
$(n + 1)^3 < 3^{n + 1}$ holds if $n^3 < 3^n$ holds.
Note that:
$3^{n + 1} = 3 * 3^n > 3n^3$(since $3^n > n^3$ by the inductive hypothesis) > $(n + 1)^3$.
By binomial expansion: $(n + 1)^3 = n^3 + 3n^2 + 3n + 1$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
| $\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2\right)=\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2-\frac{3}{2}(a-1)\right)$.
Let $f(x)=\ln(x^3+1)-\ln2-\frac{3}{2}(x-1)$.
Easy to see that $\min\limits_{[0,2]}f=0$.
Indeed, $f'(x)=\frac{3x^2}{x^3+1}-\frac{3}{2}=\frac{3(1-x)(x^2-x-1)}{2(x^3+1)}$ and since $x_{max}=\frac{1+\sqrt5}{2}$, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 ... | For the sake of an answer:
YES. You are thinking properly. What you did is nice and correct.
Similar Related questions:
What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?
what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by 47?
| {
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"url": "https://math.stackexchange.com/questions/2069717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
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To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$
Problem Statment:-
Prove:-$$\dfrac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\dfrac{4^m}{\sqrt{2m+1}}$$
My Attempt:-
We start with $\binom{2m}{m}$ (well that was obvious), to get
$$\binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}$$
Now, since $2... | Taking the product of the ratios of the terms gives
$$
\binom{2n}{n}=\prod_{k=1}^n4\frac{k-1/2}{k}\tag{1}
$$
Bernoulli's Inequality says
$$
\sqrt{\frac{k-1}k}\le\frac{k-1/2}{k}\le\sqrt{\frac{k-1/2}{k+1/2}}\tag{2}
$$
Applying $(2)$ to $(1)$, we get
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{2n+1}}\tag{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $abc+1=0$, verify that' If $abc+1=0$, prove that:
$\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}} +\frac {1}{1-c-a^{-1}}=1$.
My Attempt:
$abc+1=0$
$abc=-1$.
Now,
$$L.H.S=\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}}+\frac {1}{1-c-a^{-1}}$$
$$=\frac{b}{b-ab-1} + \frac {1}{1-b-c^{-1}} + \frac {c^{-1}}{c^{-1}-1-(ca)^{-1}}$$... | $$\dfrac1{1-a-b^{-1}}=\dfrac{bc}{bc-abc-c}=\dfrac{bc}{bc+1-c}$$
$$\dfrac1{1-b-c^{-1}}=\dfrac{-c}{-c+bc+1}$$
As $-abc=1,-a^{-1}=bc$
$$\dfrac1{1-c-a^{-1}}=\dfrac1{1-c+bc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Short technique of tackling or another method $\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$ Prove that,
$$I=\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$$
I try:
$x^4+2x^2+1=(x^2+1)^2$
$$\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx$$
Let $u=x^2+1$, $du=2xdx$
$$\int_{1}^{\inft... | Note that $$I=\int_{0}^{\infty}\frac{2x}{\left(x^{2}+1\right)\left(x^{2}+2\right)}dx=2\int_{0}^{\infty}\frac{1}{x}\left(\frac{x^{2}}{x^{2}+1}-\frac{\left(x/\sqrt{2}\right)^{2}}{\left(\frac{x}{\sqrt{2}}\right)^{2}+1}\right)dx
$$ hence by the Frullani's theorem we get $$I=-2\log\left(\frac{1}{\sqrt{2}}\right)=\color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The sum of $\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}+\cdots \cdots $
The sum of $$\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \cdots \cdots $$
$\bf{My\; Try::}$ We can write above sum as $$\sum^{n}_{k=0}(-1)^k\binom{n... | This also has a very simple algebraic proof. Suppose we seek to
evaluate
$$S_n = \sum_{k=0}^n {n\choose k} (-1)^k {3n-3k\choose n}.$$
Introduce
$${3n-3k\choose n} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-3k+1}} \frac{1}{(1-z)^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
... | {
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"url": "https://math.stackexchange.com/questions/2071532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots. Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots.. I know $\Delta \ge 0$ and $\sqrt{\Delta}$ must be rational.
| This is partial solution.
As Χpẘ suggested, we need to determine $a,b,c$ which makes $a^4+b^4+c^4+a^2b^2+a^2c^2+3b^2c^2$ perfect square. We know that $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2$ is a square.
We know that $(a,0,0), (0,b,0), (0,0,c)$ is solution. Therefore, let's assume that at most one of $a,b,... | {
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How to differentiate $y=\ln(x+\sqrt{1+x^2})$? I'm trying to differentiate the equation below but I fear there must have been an error made. I can't seem to reconcile to the correct answer. The problem comes from James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 25.
Please differentiate $y=\ln(... | You wrote:
Distribute the left term across the two right terms
That is the same mistake that we see when we consider simplifying $\dfrac 3 4\times \dfrac 5 3.$ One can multiply:
$$
\frac{3\times 5}{4\times 3} = \frac{15}{12}
$$
but that is not a good way to simplify. Usually one should cancel before multiplying:$\re... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Prove $10^{n+1}+3\cdot 10^n+5$ is divisible by $9$? How do I prove that an integer of the form $10^{n+1}+3\cdot 10^{n}+5$ is divisible by $9$ for $n\geq 1$?I tried proving it by induction and could prove it for the Base case n=1. But got stuck while proving the general case. Any help on this ? Thanks.
| Let $S(n)$ be the statement: $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$
Basis step: $S(1)$:
$\Rightarrow 10^{(1)+1}+3\cdot{10^{(1)}}+5=10^{2}+30+5$
$\hspace{45.5 mm}=100+35$
$\hspace{45.5 mm}=135$, which is divisible by $9$
Inductive step:
Assume $S(k)$ is true, i.e. assume that $10^{k+1}+3\cdot{10^{k}}+5$ is divi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 6
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Simplifying $9^{3/4}$, I get $3\sqrt[4]{9}$, but that's not the answer. Why? I am trying to simplify:
$9^\frac{3}{4}$
So this is what I did:
$9^\frac{3}{4} = \sqrt[4]{9^3}$
$\sqrt[4]{3*3*3*3*3*3}$
$3\sqrt[4]{3*3}$
$3\sqrt[4]{9}$
$3\sqrt[4]{3^2}$
I don't see how I can simplify this even more, however the answer I provid... | We know that $9=3^2$ .So, $$\sqrt [4]{9^3} =\sqrt [4]{(3^2)^3} =\sqrt [4]{3^2*3^2*3^2} =\sqrt {3*3*3*3*3*3} $$ After this you have proceeded correctly. You can simplify the last step as: $$ 3\sqrt [4]{3^2} =3\times 3^{2/4} =3\times 3^{1/2} =3\sqrt {3} $$ Hope it helps.
| {
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"url": "https://math.stackexchange.com/questions/2074362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$
The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$ for $a,b,c,d>0$
The inequality is equivalent to:
$$\displaystyle\frac{a^cb^d}{(a+b)^{c+d}}\le\frac{c^cd^d}{(c+d)^{c+d}}$$
and the right side should be at least $\left(\frac{1}{2}\right)^{c+d}$ if the rea... | Since $f(x)=x\ln x$ is a convex function, by Jensen we obtain:
$$\frac{a}{a+b}\left(\frac{c}{a}\ln\frac{c}{a}\right)+\frac{b}{a+b}\left(\frac{d}{b}\ln\frac{d}{b}\right)\geq\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)\ln\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$ Problem :
Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$
My approach :
Let $\tan^{-1}x =t $
$y = \sqrt{t+1}+\sqrt{1-t}$
Squaring both sides we get :
$y^2= t+1+1-t +2\sqrt{(t+1)(1-t)}$
$\Rightarrow y^2=... | Note that $f(x)$ is even, so we may find extrema in either $x\geq 0$ or $x \leq 0$.
Also, we can find the domain of the function to be $[\tan(1), \tan(1)]$.
Now, consider $f'(x)$ for x $\geq0$:
$$f'(x)=\left(\dfrac{-1}{\sqrt{1-\arctan(x)}}+\dfrac{1}{\sqrt{1+\arctan(x)}}\right)\times\dfrac{1}{2(1+x^2)}$$
$f'(x)< 0$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Another variant of vandermonde's identity: $\binom{r}{m+k} \binom{s}{n+k}$ It can be shown that $$\sum_{k}\binom{r}{m+k} \binom{s}{n+k} = \binom{r+s}{r-m+n}$$ using either lattice paths or manipulation of the binomial coefficients in the identity. How so? I've played with this for hours, and the only worthwhile offerin... | Observe that we certainly get all non-zero values when $k\ge -m-n$
so we may write
$$\sum_k {r\choose m+k} {s\choose n+k}
= \sum_{k\ge -m-n} {r\choose m+k} {s\choose n+k}
\\ = \sum_{k\ge 0} {r\choose k-n} {s\choose k-m}
= \sum_{k\ge 0} {r\choose r+n-k} {s\choose s+m-k}.$$
Now put
$${r\choose r+n-k} =
\int_{|z|=\epsilo... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls ... | The answer is given by:
$$
P = \frac{3}{9} \cdot \frac{2}{8}\cdot \frac{1}{7} = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{7} = \frac{1}{84}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have... | $\left(\frac{1}{an+b}\right)_{\substack{n\in\mathbb{N}\\an+b\neq0}}$ is not summable if $a\neq0$ (this has the same behavior as the harmonic series). Thus $\sum\frac{1}{an+b}$ diverges, and your $F(a,b)$ is not well-defined.
However you can sum up to some integer $N$, and use an asymptotic development of:
$$\sum_{n=1}^... | {
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"url": "https://math.stackexchange.com/questions/2075828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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System of $3$ nonlinear equations
Find positive solutions to the following:
\begin{align*}
x^2+y^2+xy=&1\\
y^2+z^2+yz=&3\\
z^2+x^2+xz=&4
\end{align*}
I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?
| Through the cosine theorem, the problem can be stated in the following way:
In a triangle $ABC$ with side lengths $1,\sqrt{3},2$, what are the
distances of the Fermat point $F$ from the vertices of $ABC$?
Well, such a triangle is a right triangle, since $1^2+\sqrt{3}^2=2^2$, and $FA+FB+FC$ is the length of the Stei... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help wo... | Method 1: Consider this denesting algorithm:
Denested square roots: Given a radical of the form $\sqrt{X\pm {Y}}$ with $X,Y\in\mathbb{R}$ and $X>Y$, we have a possible simplification as$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$
Using $(1)$ on $\sqrt{4+2\sqrt3}$, we ha... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
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Linear transformation matrix of vector space $\mathbb R^{2x2}$
Select a basis B of a vector space $\mathbb R^{2x2}$ and for linear transformation $f:\mathbb R^{2x2}→\mathbb R^{2x2}$ given by $f(X) = \begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T$ compute the matrix relative... | If it's indeed the case that
$$f\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} = 2 \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + 2 \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}$$
(I didn't check your calculations), then this would tell us that the first row of the matrix is $2, 0, 2, 0$, and so forth.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sum of series $\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms sum of series $\displaystyle \frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms
assuming $\displaystyle S_{n} =\frac{n}{1\cdot 2 \cdot 3}+\fra... | Hint:
$\dfrac{n-r}{(r+1)(r+2)(r+3)}=\dfrac{n+1-(r+1)}{(r+1)(r+2)(r+3)}$
$=(n+1)\cdot\dfrac1{(r+1)(r+2)(r+3)}-\dfrac1{(r+2)(r+3)}$
Now $\dfrac1{(r+2)(r+3)}=\dfrac{r+3-(r+2)}{(r+2)(r+3)}=?$
See Telescoping series
Again, $\dfrac2{(r+1)(r+2)(r+3)}=\dfrac{r+3-(r+1)}{(r+1)(r+2)(r+3)}=\dfrac1{(r+2)(r+1)}-\dfrac1{(r+2)(r+3)}$
| {
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Help find closed form for:$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)$ What is the closed form for
$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$
My try:
I have found a few values of $F(k)$, but was unable to find a closed form for it.
$F(0)=0$
$F(1... | The sum can also be expressed in terms of the Regularized Incomplete Gamma function ($Q(a,z)$).
In fact, premised that for a general function of a integer $f(k)$ we have
$$
\begin{gathered}
\frac{{\left( {i^{\,k} + \left( { - i} \right)^{\,k} } \right)}}
{2}f(k) = \left( {\frac{{e^{\,i\,k\frac{\pi }
{2}} + e^{\, - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find the sum of this infinite series
Sum the series $$\frac{1^2}{2^2}+\frac{1^2\cdot3^2}{2^2\cdot4^2}+\frac{1^2\cdot3^2\cdot5^2}{2^2\cdot4^2\cdot6^2}+\ldots$$
Someone please help me in finding sum of this infinite series. I have never found the sum of this kind of series.
| The function $${}_2 F_1(1/2,1/2;1;z) = 1 + \sum_{n=1}^{\infty} \frac{((1/2)(3/2)...(1/2 + n - 1))^2}{n!^2} z^{n} = 1 + \sum_{n=1}^{\infty} \Big( \frac{1 \cdot 3 \cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot ... \cdot (2n)} \Big)^2 z^{n}$$
can be expressed as an elliptic integral $$\frac{2}{\pi} \int_0^{\pi/2} \frac{1}{\sqrt{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large n... | Since
$$ \sec[..]^2 = 1+ \tan[..]^2 $$
we directly have identities
$$ \sec^{-1}x = \tan^{-1}\sqrt {x^2 - 1} $$
and
$$ \tan ^{-1} x= \sec^{-1}\sqrt {1 + x^2} $$
Also plot between $ \sec^{-1}...,\, \tan^{-1}. $
Note the $ \pi/4,\pi $ intercept on $y$, and period respy.Proper sign to be taken.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ Calculate the sum of series:
$$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$
I tried to spread this logarithm, but I'm not seeing any method for this exercise.
| In another, more straight, way:
$$
\begin{gathered}
\sum\limits_{1\, \leqslant \,n} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)} = \ln \left( {\prod\limits_{1\, \leqslant \,n} {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} } \right) = \hfill \\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 0
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Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exi... | $-1^2 + 0^2 = 0^2 + 1^2 = 1^2 + 0^2$ if you want a trivial example including negative and zero integers.
So $x = -2$ etc.
| {
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"source": "stackexchange",
"question_score": "6",
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Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equival... | Let's look at the equations modulo $8$. Notice that the only squares modulo $8$ are $0,1$ and $4$.
Case $(1)$: $x \equiv 0,4 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+3\\
z^2&\equiv w^2+5\\
w^2 &\equiv t^2+7\end{align*}
The first equation implies $y^2\equiv 4$ and $z^2\equiv 1$. This implies $w^2 \... | {
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Weird Integration problem: $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$ $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$
I am stuck at the first step and have tried replacing $5^x$ with $e^{\ln(5^x)}$ but nothing simplifies out in the end.
Any hints how I should proceed?
| $$I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx$$
Let $y=-x$, then:
$$I=\int_{-2}^{2} \frac{y^2}{1+5^{-y}}dy=\int_{-2}^{2} \frac{y^25^y}{1+5^y}dy=\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx$$
So:
$$I+I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx+\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx=\int_{-2}^2x^2\,dx=2\int_0^2x^2\,dx=2\left(\frac{2^3}{3}\right)=... | {
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"source": "stackexchange",
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Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$ Had this question popup while studying for a test but unsure how to go about solving it:
Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$
Initial thoug... | First, lets solve for $y$ in $y^2-x^2=\frac{1}{3}$. Since $y>0$, we find $$y=\sqrt{\frac{1+3x^2}{3}}.$$ Now define $g(x,y)=x^2-y^2+2x+2y-2xy-1$. As you noted, we can substitute to get $$g(x,y)=-\frac{1}{3}+2x+2y-2xy-1.$$ However, we can also get rid of $y$ with the previous condition: define $$f(x)=g(x,\sqrt{\frac{1+3x... | {
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"source": "stackexchange",
"question_score": "3",
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Hint in integration $\int\frac{x^{2}}{\left(x\cos x-\sin x \right )\left( x\sin x+\cos x \right )}\,\mathrm{d}x$ In the following integration
$$\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x$$
I tried alot. But does not get any proper start.
Can anybody provide me a hin... | Notice
$$x^{2}=x^2\left ( \sin^2x+\cos^2x \right )$$
then
\begin{align*}
\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x&=\int \frac{x^2\left ( \sin^2x+\cos^2x \right )}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x\\
&=\int \frac{x\cos x}{... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Calculate $A \cdot B$ and say if it's equal to $B \cdot A$ (matrixes)
Calculate $A \cdot B$ and $B \cdot A$ for $A= \begin{pmatrix} 1 & 1\\
1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1\\ 1 & 1
\end{pmatrix}$
Is $A \cdot B=B \cdot A$?
This is a task from an old exam and I'd like to know if I did it correctly... | Your calculation is correct.
Notice that if you were not required to calculate $B \cdot A$ then you would not have needed to calculate it entirely in order to get that $A \cdot B \neq B \cdot A$.
| {
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"source": "stackexchange",
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What must be the simplest proof of the sum of first $n$ natural numbers? I was studying sequence and series and used the formula many times $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$ I want its proof.
Thanks for any help.
| I'm not sure how simple this gets, but I still think it's worth noting that this can be applied to higher powers such as $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.
Solving by the use of Indeterminate Coefficients:
Assume the series$$1+2+3+4+5\ldots+n\tag1$$
Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn... | {
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Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+... | You can use the property given here:
$$\color{blue}{\int_a^b f(x)dx=\int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx \Rightarrow \\ \int_a^b f(f(x))dx=\int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\\
\int_{1/4}^{3/4} f(f(x))dx=\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[1]dx=\frac14,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1+8k$ is a quadratic residue modulo $2^n$
Prove that $1+8k$, for all nonnegative integers $k$, is a quadratic residue modulo $2^n$, where $n$ is a positive integer.
For $n = 5$, we have $0,1,9,17,25$, which are all quadratic residues modulo $2^5$. How do we prove this in general?
| We can assume $n\ge 4$, because otherwise it is trivial.
Fact 1: Two odd numbers have the same square modulo $2^n$ if and only if they are equal or opposite modulo $2^{n-1}$.
Proof: If $(2s+1)^2\equiv (2t+1)^2\pmod {2^n}$ then $s^2+s\equiv t^2+t\pmod {2^{n-2}}$ and
$$(s-t)(s+t+1)\equiv 0\pmod{2^{n-2}}$$
Only one of t... | {
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Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square. As written in title, I want to prove that
If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square.
I m struggling in making a start . Please help.
| Here's another answer which uses the Pell equation.
As mentioned above, in order for $2 + 2\sqrt{28y^2+1}$ to be an integer, it is necessary and sufficient that $28y^2 + 1$ is a perfect square.
Hence we must look at the solutions to Pell's equation $x^2 - 28y^2 = 1$.
All solutions to this equation are of the form $x_n+... | {
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"source": "stackexchange",
"question_score": "12",
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Prove conjecture $a_{n+1}>a_{n}$ if $a_{n+1}=a+\frac{n}{a_{n}}$ Let sequence $\{a_{n}\}$ such $a_{1}=a>0$,and
$$a_{n+1}=a+\dfrac{n}{a_{n}}$$
I used the software to find this following conjecture :
if $n>\dfrac{4}{a^3}$,we have
$$a_{n+1}>a_{n}$$
| Here is an answer which proves that the conjecture holds for $a > 1$. For the general case I couldn't show it but I present some ideas.
The proof follows by induction.
We want $a_{n+1} > a_n$. So we need (see the calculations by Han de Bruijn):
$$
a_n < \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n} = U(n)
$$
So ... | {
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"url": "https://math.stackexchange.com/questions/2093442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Prove or disprove $|x+y+z|\ge\sqrt{3}$ Let $x,y,z\in R$, such $|x|\neq|y|\neq|z|$,and $|x|,|y|,|z|>1$ and $xy+yz+xz=-1$
prove or disprove
$$|x+y+z|\ge\sqrt{3}$$
I try
$$(x-y)^2+(y-z)^2+(z-x)^2\ge 0$$
so $$|x+y+z|^2\ge 3(xy+yz+xz)=-3$$
why?
| Counterexample:
Suppose $x = 1.02, y = -1.0001, z = \frac {-1-xy}{x+y} = \frac {0.020102}{0.0199} \approx 1.01015$
$|x|,|y|,|z| > 1$
and
$xy + yz + xz = -1$
and $|x+y+z| = 1.03 < \sqrt 3$
Since $|x|, |y|, |z| > 1, x^2 + y^2 + z^2 > 3$
$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \ge 0\\
(x+y+z)^2 \ge 3 -2 \ge 0\\
(x+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the bes... | Also a Cauchy-Schwarz approach works:
\begin{align*}
1 = x + y \le \sqrt{(x^2+y^2)(1^2+1^2)} \implies x^2 + y^2 \ge \frac{1}{2} \\
\frac{1}{2} \le x^2 + y^2 \le \sqrt{(x^4 + y^4)(1^2 + 1^2)} \implies x^4 + y^4 \ge \frac{1}{8}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 10,
"answer_id": 1
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Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^... | It suffices to consider $x, y > 0$. Other case simply changes the sign of $x,y$ and of $\theta$. Thus: $\dfrac{|\text{Re(z)}|}{|z|} = \dfrac{x}{\sqrt{x^2+y^2}}= \sin \theta, \dfrac{\text{Im(z)}}{|z|}= \dfrac{y}{\sqrt{x^2+y^2}} = \cos \theta, \theta \in \left(0,\dfrac{\pi}{2}\right)$. Thus you prove: $\sin \theta + \co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Number of cubes We have $X$ cubes with $8000\le X\le10000$.
We have built columns with $2×2$ bases, leaving 2 cubes. We have also built columns with $3×3$ and $5×5$ bases, leaving 4 cubes in these cases.
How can we calculate the number of cubes?
I have created the equations
$$n\equiv2\bmod4$$
$$n\equiv4\bmod9$$
$$n\e... | The Chinese Remainder is used for calculating a number $n$ in this case that when divided by $4$ has remainder $2$, by $9$ remainder $4$ and by $25$ remainder $4$.
Then $lcm(4,9,25) = 900$ and we need $a+b+c \equiv r \pmod{900}$
Let's start calculating $a=2\cdot9\cdot25\cdot t$ where $t_1 \equiv (9\cdot25)^{-1} \pmod 4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
... | Let's take the decomposition you give and see how quick we can go
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
Multiplying by $x$ the two sides and making $x=0$ one gets $A=1$.
Multiplying by $(x^2+1)^2$ and making $x=i$ we get $-i=Di+E$ and this means $D=-1$ and $E=0$ . Now rewrite t... | {
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"source": "stackexchange",
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Find $\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )\mathrm{d}x$ How to prove
$$\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )dx=\frac{1}{2}\sqrt{\frac{\pi }{2}}\left ( \cos\frac{1}{4}-\sin\frac{1}{4} \right )$$
any hint ?Thanks!
| \begin{align*}
\int_{0}^{\infty }\cos x\sin x^2\, \mathrm{d}x&=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( x^{2}+x \right )+\sin\left ( x^{2}-x \right ) \right ]\mathrm{d}x\tag1\\
&=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( \left ( x+\frac{1}{2} \right )^{2}-\frac{1}{4} \right )+\sin\left ( \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How solve the given logarithmic problem Let
$$\log_{12}(18) = a$$
Then $$\log_{24}(16)$$
is equal to what in terms of a?
| We have: $12^a = 18 \implies 2^{2a}\cdot 3^a = 2\cdot 3^2\implies 2a\ln 2+ a\ln 3 = \ln 2+2\ln 3$. Thus $b = \log_{24}16\implies 24^b = 16 = 2^4\implies 2^{3b}\cdot 3^b = 2^4\implies 3b\ln 2+b\ln 3=4\ln2\implies 3b+b\dfrac{\ln3}{\ln2}=4$. You can divide by $\ln2$ the first equation, and solve for $\dfrac{\ln3}{\ln2}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | moving $$\frac{3}{7}$$ to the left and clearing the denominators we get
$$3\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-3\,abc-a{c}^{2}+3\,{b}^{3}-{b}^{
2}c-b{c}^{2}+3\,{c}^{3}
\geq 0$$ and this is equivalent to $$a^3+b^3+c^3-3abc+(a-b)(a^2-b^2)+(b^2-c^2)(b-c)+(a^2-c^2)(a-c)\geq 0$$
which is clear
(the first Terms are AM-GM)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
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Prove that the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than$2^{(a-1)n}$
Let $a,n$ be positive integers. Find all $a$ such that for some $n$ the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than $2^{(a-1)n}$.
Since I thought there were no such $a$, I thought about proving... | There are $(a-1)n$ factors in the original expression.
$\lfloor \frac {(a-1)n + 1}{2}\rfloor$ are divisible by $2.$
$\lfloor\frac {(a-1)n + 1}{4}\rfloor$ are divisible by $4$.
$\lfloor\frac {(a-1)n + 1}{2^i}\rfloor$ are divisible by $2^i$
The largest power of $2$ that divides $(n+1)\cdots(an) = 2^{\left(\lfloor \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $f(x)+2f(\frac{1}{x})=2x^2$ what is $f(\sqrt{2})$ Question
if $f(x)+2f(\frac{1}{x})=2x^2$ what is $f(\sqrt{2})$
My steps
I tried to plug in $\sqrt{2}$ into the equation but that didn't get me anywhere because then i would have $2f(\frac{1}{\sqrt{2}})$ in the way. I was wondering on how to solve this equation?
| If $f(x)+2f(\frac{1}{x})=2x^2$, then by substitution $f(\frac{1}{x})+2f(x)=\frac{2}{x^2}$ and so $f(\frac{1}{x})=\frac{2}{x^2}-2f(x)$. Substituting $f(\frac{1}{x})$ into the first equation yields $f(x)+2[\frac{2}{x^2}-2f(x)]=2x^2$. Solve for $f(x)$: $f(x)=\frac{2}{3}[\frac{2}{x^2}-x^2]$. So $f(\sqrt{2})=-\frac{2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n... | If $-1< x < 1$, we have, by differentiation and adding :
$$\begin{align} \sum_{n=0}^\infty x^n & = \frac 1{1-x} \implies & \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} \implies \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} \implies & \small \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} \implies & \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity. Factorise $z^3 - 1 $.
If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.
Factorising gives you :
$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.
... | You are done. The two possible values of $z^2+z+1$ are $0$ for $z$ a third root of unity different from $1$, as you have computed, and $1+1+1=3$ for $z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$ I want to find the following series.
$$\sum_{n\ge1} \frac{1}{4n^3-n}$$
So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$
Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-... | You can do this way. It is easier. Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{4n^3-n}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^\infty\frac{1}{4n^3-n}$ and
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{(2n-1)n}x^{2n}, f''(x)=2\sum_{n=1}^\infty\frac{1}{(2n-1)}x^{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac2{1-x^2}$$
and hence
$$ f(1)=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How many triples satisfy $ab + bc + ca = 2 + abc $
$a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc$
How many triples $(a,b,c)$ satisfies the statement? Here $a,b,c > 1$.
It is easy to simplify the statement to
$$ab + bc + ca = 2 + abc.$$
But now how to proceed I don't know. I think this is someh... | If $a$, $b$, and $c$ are required to only be positive integers and some of them is $1$, then we have a unique solution $(a,b,c)=(1,1,1)$. For solutions with $a,b,c>1$, note that
$$(a-1)(b-1)(c-1)=abc-bc-ca-ab+a+b+c-1=a+b+c-3\,.$$
Set $x:=a-1$, $y:=b-1$, and $z:=c-1$. Therefore,
$$xyz=x+y+z\,.$$
Without loss of general... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
| If $n= 1$ then we have $3^{n-1} + 5^{n-1} = 1+ 1 = 2|8 = 3^{n}+5^{n}$. So that's one solution.
Assume $n > 1$.
$3^n + 5^n = 3*3^{n-1} + 5*5^{n-1} = 3*3^{n-1} + 3*5^{n-1} + 2*5^{n-1} = 3(3^{n-1} + 5^{n-1}) + 2*5^{n-1}$.
So $3^{n-1} + 5^{n-1}|2*5^{n-1}$
If $p$ is prime and $p|3^{n-1} + 5^{n-1}$ then $p|2*5^{n-1}$ the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{... | Hint:
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}=\\
\lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}+\lim_{x\to1}\dfrac{2\sqrt{3x+1}-4}{\sqrt[4]{x}-1}\\
\lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}\times\dfrac{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}\dfrac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Field with 2 elements Let $ A $ be a ring with $ x^{2}+y^{2}+z^{2}=xy+yz+zx+xyz+1,\forall x,y,z\in A^{*} $.
Prove that $ A $ is a field with 2 elements.
If we put $ x=y=z=1 $ we obtain that $ 1+1=0 $.
If we put $y=z=1 $ we have that $ x^2=x $, which means $ A $ is a boolean ring.
That's all I did so far.
| $x^2 + x^2 + x^2 = x*x + x*x + x*x + x*x *x + 1 =x^2 + x^2 +x^2 + x^3 + 1$ so $x^3 + 1 = 0$ and $x^3 = -1$
$x^2 + x^2 + 1 = x*x + x*1 + 1*x + x*x *1 = x^2 + x + x + x^2 + 1$ so $x + x =0$ and$x = -x $ and therefore $x^3 = -1 = 1$
$x^2 = x^2 + ( 1-1) = x^2 + 1 + 1 = x^2 + 1^2 + 1^2 = x*1 + 1*1 + 1*x + x*1*1 + 1=x+x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... | You have to prove two things:
1) If $b = 0$ or $b = a$ then $(a-b)^2 = a^2 - b^2$.
And
2) If $(a-b)^2 = a^2 - b^2$ then either $b = 0 $ or $b = a$.
To prove 1: we do what you did correctly:
If $b = 0$ then $(a - b)^2 = (a-0)^2 = a^2$. An $a^2 - b^2 = a^2 - 0^2 = a^2 - 0 = a^2$. So $(a - b)^2= a^2 - b^2$.
If $b =a$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Approximation of a summation by an integral I am going to approximate $\sum_{i=0}^{n-1}(\frac{n}{n-i})^{\frac{1}{\beta -1}}$ by $\int_{0}^{n-1}(\frac{n}{n-x})^{\frac{1}{\beta -1}}dx$, such that $n$ is sufficiently large.
*
*Is the above approximation true?
*If the above approximation is true, by which theorem or me... | If we write $s = 1/(\beta-1)$, your sum and integral can be re-written as
$$ \sum_{i=0}^{n-1} \left( \frac{n}{n-i} \right)^s = n^s \sum_{k=1}^n \frac{1}{k^s}, \qquad \int_{0}^{n-1} \left( \frac{n}{n-x} \right)^s \, dx = n^s \int_{1}^{n} \frac{dx}{x^s}. $$
In this case, the Euler-Maclaurin formula provides a way of esti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface. Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface generated by the movement of a line and explicite its rectilinear generatrices.
I have tried to make its matrix, knowing that $a_{11}=1$, $a_{12}=3$...$a_{00}=8$.. But I don't think that'... | If you want to make a matrix do it like this.
$\mathbf x^T \begin {bmatrix}
1 &0 &0\\
0 &3 &2\\
0 &2 &0\\ \end{bmatrix}\mathbf x + \begin {bmatrix} -6&8&0\end{bmatrix} \mathbf x = 0$
Since that matrix is symmetric is is diagonalizable with ortho-normal basis.
$\mathbf x^t P^T D P \mathbf x + BP^TP\mathbf x = -8\\
\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that determinant of $\small\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$ is divisible by $19$ Using that the numbers
$228,
323$
and
$456$
are
divisible
by
$19$.
Show
that
the
determinant of matrix
$\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$
is
divisible
by
$19$.
| There's an alternative and maybe easier way to prove the result, with Gaussian elimination.
Consider the matrix with coefficients over $\mathbb{Z}/19\mathbb{Z}$; the inverse of $2$ is $10$, so
\begin{align}
\begin{bmatrix}
2 & 2 & 8 \\
3 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 4 \\
3 & 2 & 3 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$ Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!
| By using hint of @dxix we get
$$\lim _{ x\to 0 } \frac { 8^{ x }-7^{ x } }{ 6^{ x }-5^{ x } } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\left[ { \left( \frac { 8 }{ 7 } \right) }^{ x }-1 \right] }{ { 5 }^{ x }\left[ { \left( \frac { 6 }{ 5 } \right) }^{ x }-1 \right] } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\frac { \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to show that $\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$ How to show that
$$\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$$
| Here is an answer based upon Short Proofs of Saalschütz's and Dixon's theorems by Ira Gessel and Dennis Stanton.
The answer is provided in three steps
*
*Step 1: Relationship between coefficients of the constant term of a bivariate Laurent series and a transformed of it.
*Step 2: Application of this rela... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Is the following limit finite ....? I would like to see some clue for the following problem:
Let $a_1=1$ and $a_n=1+\frac{1}{a_1}+\cdots+\frac{1}{a_{n-1}}$, $n>1$. Find
$$
\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right).
$$
| Quick and dirty proof.
Assume for a moment that the sequence $\,a_n\,$ is a continuous
and differentiable function $\,a(n)\,$ of $\,n\,$ as is suggested in the picture below.
Then consider the following sequence. Fasten your seatbelts!
$$
a_{n+1} = a_n + \frac{1}{a_n} \\
\frac{a_{n+1} - a_n}{1} = \frac{1}{a_n} \\
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that...
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that: $AD$=5,$BD=3$,$CD=6$.
| Let $\angle{ADC}$ be $\alpha$. Then by the cosine theorem
$b^2=25+36-60\cos\alpha$,
$c^2=25+9-30\cos(180^{\circ}-\alpha)=34+30\cos\alpha$.
This gives
$b^2+2c^2=129$.
Since $b^2+c^2=9^2=81$, it gives $c^2=48.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute $\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$ $$\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Then the question ask me to change it into
$$\int_0^3 \frac{(x-3)^4}{x^4+(x-3)^4} \,dx$$
Then how to evaluate it
if let $$u=x-3$$
$$du=dx$$
$$x^4 =(u+3)^4$$
$$\int_3^0 \frac{u^4}{(u+3)^4+u^4} \,du$$
it is still the same
so what ... | Let,
$$I=\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Let $x=3-u$ and change dummy variable back to $x$.
$$I=\int_{0}^{3} \frac{(3-x)^4}{x^4+(x-3)^4} dx$$
$$=\int_{0}^{3} \frac{(x-3)^4}{x^4+(x-3)^4} dx$$
Add this to the first form of $I$.
$$2I=\int_{0}^{3} 1 dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of the combinatorial series $$\frac{\binom{b+i}{1}}{\binom{i}{1}} + \frac{\binom{b+i}{2}}{\binom{i}{2}} + \frac{\binom{b+i}{3}}{\binom{i}{3}} +... + \frac{\binom{b+i}{i}}{\binom{i}{i}} $$
where $b \in \mathbb{R}$ and $b\in [-1, 1]$ .
| Proof of $ \displaystyle\enspace \sum\limits_{j=1}^i \binom{b+i}{j}\binom{i}{j}^{-1} = \frac{b+i}{1-b}-\frac{b}{1-b} \binom{b+i}{i} \enspace$ by induction for $b\ne 1$.
$i:=1$ : $\displaystyle \enspace \sum\limits_{j=1}^1 \binom{b+1}{j}\binom{1}{j}^{-1}=\frac{b+1}{1-b}-\frac{b}{1-b} \binom{b+1}{1} \enspace $ which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
| Continue from your last step.
Put $(x + 1)^2 = z$
We have,
$z^2 + 36 = 13z$
$z^2 - 13z + 36 = 0$
$z^2 - 9z - 4z + 36 = 0$
$z(z - 9) - 4(z - 9) = 0$
$(z - 9)(z - 4) = 0$
When z = 9
$(x + 1)^2 = 9$
$(x + 1) = \pm 3$
x = 2 or -4
When z = 4
$(x + 1)^2 = 4$
$(x + 1) = \pm 2$
x = 1 or -3
x = -4, -3, 1, 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How does this simplify to 1? I am working on this differetiation problem:
$ \frac{d}{dx}x(1-\frac{2}{x})$
and I am currently stuck at this point:
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$
Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;
$1\cdot \left(1-\frac... | Do not get confused by fractions and exponents.
You should remember that $\frac {k}{n} = kn^{-1}$ and not $k^{-n} $ and that $\frac {k}{n} l = kln^{-1}$ and not $kl^{-n} $.
We have $$1\times (1-\frac {2}{x}) = 1-\frac {2}{x} \tag {1}$$ and then $$\frac {2}{x^2}x = \frac {2}{x^2} \times x = \frac {2}{x} \tag {2} $$
Wha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
| $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{3^n}\frac{1}{3^{k-n}}=\\
=\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}=\frac{3}{2}\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n =\\
= \frac{3}{2}\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{3}{2}\frac{1}{2}=\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Proving the inequality $(a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12$. This is a follow up question to my previous post "Inequalities of expressions completely symmetric in their variables". An answer provided a counterexample to me reasoning: under the constraints $a,b,c\in\Bbb{R}^+$ and $a+b+c=3$,
$$
(a^2-ab+b^2)(c^2-... | Let $a\geq b\geq c\geq0$.
Hence,
$$\prod_{cyc}(a^2-ab+b^2)\leq(a^2-ab+b^2)a^2b^2=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=$$
$$=12(3-ab)\cdot\frac{ab}{2}\cdot\frac{ab}{2}\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to do this question that talks about dependency of x Let $x > 0$. Prove that the value of the following expression doesn't depend on x
$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Attempt:
Left: f'(x) = $\frac{1}{1+x^2}$
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
... | Using the $u$ sub $t' = t^{-1}$, we see $dt'= -(t')^2dt$ that
\begin{align}
\int^{\frac{1}{x}}_0\frac{1}{1+t^2}\ dt = -\int^x_\infty \frac{1}{1+(t')^{-2}} \frac{dt'}{(t')^2}= \int^\infty_x \frac{1}{1+(t')^2}\ dt' = \frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'.
\end{align}
But we see that
\begin{align}
\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
| $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$
$$-3\sin(2x)+2\dfrac{1-\cos2x}{2}+3$$
$$-3\sin(2x)-\cos(2x)+4$$
now use
$$|a\sin\alpha+b\cos\alpha|\leq\sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | Your last integral, $\int \frac{-x}{u(2x-2)} du$, doesn't make me happy.
You have both $x$ and $u$ in the integral. The point of the substitution is to get rid of the $x$'s and leave the integral in terms of $u$.
I'll take it from the last integral in terms of $x$, everything looks fine before and it goes astray after... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Permute order of summation change the sum : $\sum_{n=1}^\infty \frac{(-1)^n}{n}$. In a course of analysis it's written that if $\sum_{k=1}^\infty x_k$ converge but it's not absolutely convergent, then changing any order of summation will change the sum.
For example, if I consider $$\sum_{n=1}^\infty \frac{(-1)^n}{n}=\... | You are right. Changing finite numbers wont change the sum.
Here is a nice example: Set $S: =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$. Then
\begin{align*}
\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n}\right)
\end{align*}
is a rearrangement of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ and since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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To prove that prove that $cos^8 \theta sec^6 \alpha , \frac{1}{2 } ,sin^8 \theta cosec^6 \alpha $ are in A.P If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P ,
then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6 \alpha $ are in A.P
Now i have reach... | You have done rightly. Now we can proceed as follows: $$\tan \theta = \pm \tan \alpha \Rightarrow \tan^2 \theta = \tan^2 \alpha \Rightarrow \sec^2 \theta -1 =\sec^2 \alpha -1 \Rightarrow \sec^2 \theta = \sec^2 \alpha \Leftrightarrow \sec^6 \theta = \sec^6 \alpha \tag{1}$$
Also, $$\tan^2 \theta = \tan^2 \alpha \Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to ... | Define $$f(x) = \frac{1-x^2}{1+3x^2+x^4}$$ so that $$f(x^{-1}) = -\frac{x^2(1-x^2)}{1+3x^2+x^4}.$$ Since $$\int_{x=0}^\infty f(x) \, dx = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx,$$ the transformation $$x = u^{-1}, \quad dx = -u^{-2} \, du$$ gives $$\int_{x=1}^\infty f(x) \, dx = \int_{u=1}^0 -\frac{u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Linear Equations system The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:
*
*has no solution
*has one solution
*has more than one solution
I started to do Gauss elimination on it, but i have no idea what i am looking fo... | Start with the matrix and data vector
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
1 & -1 & 3 \\
5 & 2 & -6 \\
2 & -1 & \alpha \\
\end{array}
\right], \ \alpha \in \mathbb{C}, \quad
%
b =
\left[
\begin{array}{r}
-5 \\
\alpha \\
-6
\end{array}
\right]
$$
The reduced row eschelon form is
$$
\mathbf{E}_{A} =
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving and proving inequalities? If $a,b$ and $c$ are positive real numbers, how do I prove that:
$$\frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
and when is equality?
Are there general techniques to solve these symmetric and or cyclic inequalities?
| Lemma 1: If $a,b$ and $c$ are positive real numbers, then $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c.$
Proof: Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ $\blacksquare$
Lemma 2: If $a,b$ and $c$ are positive real numbers, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\int_{-1}^{1}e^{-\frac{1}{1-x^2}}dx$, can it be computed? Is there a way to compute $\int^{1}_{-1} e^{-\frac{1}{1-x^2}}dx$ ?
I have tried a few change of variables and also to write down $\frac{1}{1-x^2} = \frac{1}{2} ( \frac{1}{1-x} + \frac{1}{1+x})$ But I didn't get anything so far.
Edit: changed $e^{\frac{1}{1-x^2}... | $$\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \int_{0}^{1}\exp\left(\frac{1}{x-1}\right)\frac{dx}{\sqrt{x}}$$
equals:
$$ \int_{0}^{1}\exp\left(-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x}} = \int_{1}^{+\infty}\frac{e^{-x}}{x^{3/2}\sqrt{x-1}}\,dx=\frac{1}{e}\int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know whe... | $\frac {3^{11} - 1}{2} = \frac {3^{11} - 1}{3-1}=$
$3^{10} + 3^9 + .... + 3^2 + 3 + 1=$
$9(3^8 + ..... + 1) + 4$
so the remainder is $4$.
..or...
$\frac {3^{11}-1}2 = 3^{11}*\frac 12 - \frac 12 \equiv k \mod 9$
$ 3^{11}-1 \equiv 2k \mod 9$
$2k + 1 \equiv 0 \equiv 9 \mod 9$
$2k \equiv 8 \mod 9$
$k \equiv 4 \mod 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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probability of discrete random variable In a nuclear reaction, a particle can either separate into two pieces,
or not separate, with respective probabilities $2/3$ and $1/3$. Knowing that
pieces behave like new independent particles, find the law of
probability and mean of the number of particles obtained after two rea... | You start with only one particle, let's call it $A$. The set of your particles at the beginning is $$\{A\}.$$
After the first reaction, the particle $A$ can produce another particle $B$, or not. Then we can have the following particle sets:
$$\begin{cases}
\{A\} & ~\text{with probability}~\frac{1}{3}\\
\{A,B\} & ~\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$
If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?
Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Le... | $$
\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0 \quad\&\quad \left|\,\cos{x}\,\right|\,\le1 \\[6mm]
-1\le\,\cos{x}=\frac12\left(\,-(1-a)\pm\sqrt{(1-a)^2+4a^2}\,\right)\,\le+1 \\
-1-a\,\le\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \\[6mm]
\begin{align}
&\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\ge\,-1-a \implies 5a^2-2a+1=a^2+2a+1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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For $a, b, c$ is the length of three sides of a triangle. Prove that $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$ For $a, b, c$ is the length of three sides of a triangle. Prove that $$\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$$
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ be positives and we need to prove that
$$(2x+y+z)(2y+x+z)(2z+x+y)\geq8\sum_{cyc}(y-x)(2x+y+z)(2y+x+z)$$ or
$$\sum_{cyc}\left(2x^3+15x^2y-x^2z+\frac{16}{3}xyz\right)\geq0,$$
which is obvious because $x^3+y^3+z^3\geq x^2z+y^2x+z^2y$ by Rearrangement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For $x,y,z>0$. Minimize $P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$ For $x,y,z>0$ and $xy+yz+xz=1$, minimize $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$$
My try: Let $xy=a; yz=b;zx=c \Rightarrow a+b+c=1$
$\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{... | You proved that $P \ge 1$.
And for $x=y=z=\frac{1}{\sqrt{3}}$ you have
$$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}= \frac{1}{3} + \frac{1}{3}+\frac{1}{3}=1.$$ Hence the minimum of $P$ for positive values of $x,y,z$ is one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating some integrals I need to evaluate two kind of similar integrals
The first one:
$$\lim _{n\to \infty }\int _0^{\frac{\pi }{3}}\:\frac{\sin ^n\left(x\right)}{\sin ^n\left(x\right)+\cos ^n\left(x\right)}dx$$
The second one:
$$\int _0^{2\pi }\:\frac{x\sin ^{100}\left(x\right)}{\sin ^{100}\left(x\right)+\cos ^{10... | For the first try substituting $x=\frac{2}{3}u$ and divide through by $\sin^n(\frac{2}{3}x)$ top and bottom.
So now we have,
$$\frac{2}{3} \lim_{n \to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^n (\frac{2}{3}x)} dx $$
Now for $\frac{2}{3}x \in (0,\frac{\pi}{4})$ we have $\cot \frac{2}{3} x>1$ so there the integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
| $x\geq1$ and $1$ is not root.
Hence we can rewrite our equation in the following form:
$$x-1=x\left(\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}\right)$$ or
$$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=1-\frac{1}{x}$$ and with the given we obtain
$$2\sqrt{x-\frac{1}{x}}=x+1-\frac{1}{x}$$ or
$$x-\frac{1}{x}-2\sqrt{x-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplifying Algebraic Expression Under A Square Root This is part of a problem for calculating the length of a curve. Unfortunately, I'm stuck on a pretty basic algebra concept.
Solutions for my problem say that:
$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} = \frac{\sqrt{t} + \frac{1}{\sqrt{t}}}{\sqrt{2}}$
I cannot underst... | Put over a common denominator: $$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} =\sqrt{\frac{1 + 2t + t^2}{2t}}$$
Factor the top term: $$\sqrt{\frac{(1+t)^2}{2t}}$$
Take the square root: $$\frac {1+t}{\sqrt{2t}}$$
Divide top and bottom by $\sqrt t$: $$\frac {\frac 1{\sqrt{t}} + \sqrt{t}}{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve derivative of $(2x^2+x+3)(5x+7)$ using product rule. I've to find the derivative of
$$(2x^2+x+3)(5x+7)$$
Using product rule I get
$$(2x^2+x+3)×5+5×(2x^2+x+3)\\20x^2+10x+30$$
Which is wrong. Please help, where I went wrong.
| $y=(2x^2+x+3)(5x+7) $ then
$$y'=(2x^2+x+3)'(5x+7)+(2x^2+x+3)(5x+7)'\\
=(4x+1)(5x+7)+5(2x^2+x+3) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$... | HINT:
$$\cos^4A=(\cos^2A)^2=\cdots=1-\cos^2A\iff\cos^4A+\cos^2A-1=0$$
Divide $a\cos^{12}A+b\cos^8A+c\cos^8A-1$ by $\cos^4A+\cos^2A-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How do I can simplify the below intersection? let $S$ be a sphere defined by this equation :$ \left(x-\frac{4}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(z-\frac{5}{3}\right)^2=\frac{25}{36}$ and $(p)$ the plane defined by the following equation :$2x+y-2z+4=0$ ,
My question here is : How do I can get the interse... | \begin{align*}
\text{Radius of the sphere} &= \frac{5}{6} \\
\text{Distance of the centre from the plane} &=
\frac{2\left( \frac{4}{3} \right)+
\left( \frac{1}{3} \right)-
2\left( \frac{5}{3} \right)+4}
{\sqrt{2^2+1^2+2^2}} \\
&= \frac{11}{3} \\
&> \frac{5}{6}
\end{align*}
Hence, no i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a complex number $w$ such that $w^2=-\sqrt{3} - i$ This is a problem in my undergrad foundations class.
\begin{equation}
w^2=-\sqrt{3} - i \\w=(-\sqrt{3}-i)^{\frac{1}{2}} \\w=\sqrt{2}\bigg(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\bigg)
\end{equation}
So I get to here and the next step is
\begin{equation}
\sqrt{2}... | $\omega^2 = $$2(-\frac {\sqrt 3}{2} + i \frac 12)\\
2(\cos \frac {7\pi}{6} + i \sin \frac {7\pi}{6})$
$\omega = \sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12})$
And here you are....
When we take the square root of the square of something, there are always two solutions. For example $x^2 = 9 \implies x = \pm 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $DEF+FEF=GHH$, $KLM+KLM=NKL$, $ABC+ABC+ABC=BBB$ She visits third class and is $8$ years old (you can imagine how ashamed I felt when I said so to her). I helped her with lots of maths stuff today already but this is very unknowable for me. Sorry it's in German but I have translated it :)
It's saying "Each lett... | For the 3rd sum:Let the carry from the 1st column to the 2nd column be $d$, and let the carry from the 2nd column to the 3rd column be $e$.
The second column implies that $3B+d $ is equal to $B$ plus a multiple of $10,$ so $2B+d$ is a multiple of $10$. So $d$ is even, so $d=0$ or $d=2$.
If $d=0$ then $2B=2B+d$ is a mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "73",
"answer_count": 13,
"answer_id": 4
} |
Is my proof consider to be correct? Problem: Let $a,b ∈ \mathbb N$, prove that at least one of the following $ab, a+b, a-b$ is evenly divisible by 3.
My solution:
Case 1: If $a_{mod}3=0 \text{ or } b_{mod}3=0$ then we can say that $a = 3k,\ ab=3kb$ which is evenly divisible by three
Case 2: If $a_{mod}3=b_{mod}3$ then ... | Seems correct, but if you're allowed to use Fermat's little theorem, you could go like this :
If $a$ or $b$ are divisible by $3$ then it is obvious that $ab$ is divisible by $3$.
If neither $a$ or $b$ are divisible by $3$, then let's take a look at $(a-b)$ and $(a+b)$. Take their product : $(a-b)(a+b) = a^2 - b^2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why is $ab=(a-b)(a+b)$ false when $a,b$ are coprime? I could not find this, and do not have a background in number theory so sorry if this sounds trivial.
If $a$ and $b$ are coprime integers, why is the statement
$$ab=(a-b)(a+b)$$
a contradiction?
| it becomes $$ a^2 - ab - b^2 = 0. $$ If $b \neq 0,$ divide through by $b^2$ and introduce $r = \frac{a}{b},$ this gives
$$ r^2 - r - 1 = 0 $$ So $r$ is either the Golden ratio
$$ \frac{1 + \sqrt 5}{2} $$ or negative its reciprocal. Put briefly, irrational, contradicting $a,b$ being integers.
It is also possible to u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 4
} |
Proving these binomial sums Need help proving with these two:
1)
$$\sum_{k=0}^m (-1)^k \binom{n}{k}= (-1)^m \binom{n - 1}{m}$$
2)
$$\sum_{k=0}^n \binom{n}{k} \binom{k}{m}= \binom{n}{m} 2^{n-m}$$
I tried using these two properties below, but got nowhere. $$\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$$
$$ \sum_{k=0}^n \binom{n}... | For the first equality, it can be easily proved by induction on $m$. When $m = 0$, the LHS is $1$ and the RHS is also $1$, thus the equality holds. For $m \geq 1$, we have
\begin{align}
\sum_{k=0}^m (-1)^k \binom{n}{k} &= \sum_{k=0}^{m-1}(-1)^k\binom{n}{k} + (-1)^m\binom{n}{m} \\
&= (-1)^{m-1}\binom{n-1}{m-1} + (-1)^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
| $$y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$$
$$\implies y=\sqrt{16-4-x^2+4x}-\sqrt{4-1-x^2+2x}$$
$$\implies y=\sqrt{16-(4+x^2-4x)}-\sqrt{4-(1+x^2-2x)}$$
$$\implies y=\sqrt{16-(2-x)^2}-\sqrt{4-(1-x)^2}$$
From the first radical, we have that $$-4 \le 2-x\le 4 \implies -2 \le x\le 6$$
And from the second radical, we have that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $b_n = b_{n+100}$
Let $b_n$ denote the units digit of $\displaystyle\sum_{a=1}^n a^a$. Prove that $b_n = b_{n+100}$.
I tried rewriting the sum, but didn't see how to prove the equality. For example, if $n = 178$ we have $$\displaystyle\sum_{a=1}^{178} a^a = (1^1+2^2+3^3+\cdots+78^{78})+(79^{79}+80^{80}+81^... | We can reduce the numbers to the residues modulo $10$, i.e:
$$79^{79} + 80^{80} + \cdots + 178^{178} \equiv 9^{79} + 0^{80} + \cdots + 8^{178} \pmod {10}$$
Now note that for each residues we have exactly $10$ numbers such that it appears in the base of the exponent. Group them and take one group. Now notice as their pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{... | Observe that checking for absolute convergence we get
$$\frac1{k(4k^2-1)}\le\frac1{k^2}\;\implies\;\text{since}\;\;\sum_{k=1}^\infty\frac1{k^2}$$
converges so does the series of the left side. What can you then deduce?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
solving a recurrence relation - finding the general solution Solve the recurrence relation:
$u_{n+2} = 2u_{n+1}-u_n$
$u_0 = 1 $ and $u_1 = 4$
My calculations:
I have calculated that the characteristic equation is: $t^2-2t+1 = 0$ so the roots are $r_1=1$ and $r_2=1$
here is where I am stuck. The answer says that the g... | $\begin{bmatrix}u_{n+2}\\u_{n+1}\end{bmatrix} = \begin{bmatrix} 2&-1\\1&0\end{bmatrix}\begin{bmatrix}u_{n+1}\\u_{n}\end{bmatrix}$
$\mathbf u_n = B^n \mathbf u_0$
Unfortunately B is not diagonalizable.
$\lambda^2 - 2\lambda + 1 = 0\\(\lambda-1)^2$
Choose $v_1$ such that
$(B-\lambda I)v_1 = 0\\
v_1 = \begin{bmatrix}1\\1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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.