Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence Prove that the sequence $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence for $n>2.$
We have to show that $f(n+1)<f(n)$ for all $n>2.$ We have
$$
f(n+1)-f(n)=((n+1)!)^{\frac{1}{n+1}}-\frac{n+2}{2}-\left((n!)^{\frac{1}{n}}-\frac{n+1}{2}\right)=((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}-\frac{1}{2} <0,
$$
and reduce the problem to the following inequality
$$
((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}<\frac{1}{2}.
$$
With Maple I calculate
$$
\lim\limits_{n \to \infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})=\frac{1}{e}<\frac{1}{2},
$$
but in is not enought to prove the inequality. Any ideas?
| We have
$$\begin{align}\frac{\left(\frac12+n!^{1/n}\right)^{n+1}}{(n+1)!}&=\frac{(n!^{1/n})^{n+1}+(n+1)\cdot\frac12(n!^{1/n})^{n}+{n+1\choose2}\cdot\frac14(n!^{1/n})^{n-1}+\ldots}{(n+1)!}
\\&>\frac{n!^{1/n}}{n+1}+\frac12+\frac n{8\cdot n!^{1/n}}\end{align}$$
Using $a x+\frac bx\ge 2\sqrt{ab}$ (arithmetic-geometric inequality), we find that for $n\ge 2$
$$\frac{\left(\frac12+n!^{1/n}\right)^{n+1}}{(n+1)!}>\frac12+\sqrt{\frac{n}{2(n+1)}} \ge \frac12+\sqrt{\frac{1}{3}}>1$$
so that
$$ \frac12+n!^{1/n}>(n+1)!^{1/(n+1)}$$
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ We have to evaluate $$\int\frac{d\theta}{1+x\sin^2(\theta)}$$
Here is all my steps:
$$\tan\frac{\theta}{2}=\omega\Rightarrow \sin(\theta)=\frac{2\omega}{1+\omega^2}\Rightarrow 1+x\sin^2{(\theta)}=1+\frac{x 4\omega^2}{(\omega^2+1)^2}$$
Therefore: $$\int\frac{d\theta}{1+x\sin^2(\theta)}=2\int\frac{\omega^2+1\:\ d \omega}{(\omega^2+1)^2+x(2\omega)^2}=2\left(\int\frac{\omega^2\:\ d \omega}{(\omega^2+1)^2+x(2\omega^2)}+\int\frac{d\omega}{(\omega^2+1)^2+x(2\omega)^2}\right)$$
I don't know if I'm wrong somewhere but I don't have ideea how can I continue...
| Hint: Let $$\sin \theta = \pm\frac{\omega}{\sqrt {1 +\omega^2}} \implies 1 +x\sin ^2 \theta = 1 + \frac{x\omega^2}{1 + \omega^2} = \frac{1 + (1 + x) \omega^2}{1 + \omega^2 }$$
where $\omega = \tan \theta$. Then
$$\int \frac{1 + \omega^2}{1 +(1 + x)\omega^2} d\theta = \int \frac{\sec^2 \theta}{1 + (1 +x)\tan^2 \theta} d\theta$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus differentiable gra Hey guys so stuck on a calculus question. So far all I know is that $d$ should equal $4$. I then got the derivative of the a b c d function to $3ax^2 + 2bx + c$ , subbed in $(0,4)$ to get $c$, which was also $4$. Just wondering if I'm on the right track and where I should go from here.
A section of road, represented by the line $y = x + 4$ when $x \leq 0$, is to
be smoothly connected to another section of road, represented by $y = 4
> –x$ when $x \geq 4$ , by means of a curved section of road, represented by a
cubic curve $y = ax^3+ bx^2+ cx + d$ . Find $a, b, c$ and $d$ such that the
function $f(x)$ is everywhere differentiable (and therefore everywhere
continuous), where
$$f(x) = \begin{cases}
x+4 &x\leq 0 \\
ax^3 +bx^2 +cx +d &0<x<4 \\
4-x &x\geq 4
\end{cases}$$
| ok, so
$y = $
$x + 4, x ≤0$
$ax^3+ bx^2+ cx + d, 0 < x < 4$
$4 – x, x ≥4$
and y has to be differentiable everywhere (and therefor continuous, too)
so, the derivative at x = 0 and at x = 4 have to be the same when approached from both sides, as do the values of y.
Let $P(x) = ax^3 + bx^2 + cx + d$
and
$DP(x) = 3ax^2 + 2bx + c$ (the derivative of P)
so, the limit as $x$ goes to $0$ of $P = f(0) = 4$, so $P(0) = 4$, so $d = 4$.
$DP(0) = 1$, and $DP(4) = -1$ (in order for it to be differentiable)
$DP(0) = c = 1 $
so $c=1$, $d=4$
$DP(4) = -1 = 3a*16 + 2b*4 + 1$
$2 + 48a + 8b = 0 = 1 + 24a + 4b$
and finally, $P(4) = 4 - 4 = 64a + 16b + 4(1) + 4$
$0 = 64a + 16b + 8$
$0 = 16a + 4b + 2$
so
$24a + 4b + 1 = 16a + 4b + 2$
$24a - 16a = 1$
$a = 1/8$
then
$0 = 16(1/8) + 4b + 2$
$0 = 2 + 2 + 4b$
$-4 = 4b$
$b = -1$
In conclusion:
$a=1/8$, $b=-1$, $c=1$, $d=4$
hope it helps. (sorry for the for the poor formatting.)
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the inequality $ (1+a+ab)(1+b+bc)(1+c+ca) \leq (1+a+a^2)(1+b+b^2)(1+c+c^2)?$ For $a,b,c>0$ prove the inequality
$$
(1+a+ab)(1+b+bc)(1+c+ca) \leq (1+a+a^2)(1+b+b^2)(1+c+c^2).
$$
I know that I should use the multiplicative rearrangement inequality but I am not sure how to choose the involved sequences correctly. Any ideas?
| HINT: use that $$x^2+y^2+z^2\geq xy+xz+yz$$ for all real $x,y,z$ is hold.
And your inequality can be written as
$$(ab)^2+(ac)^2+(bc)^2-a^2bc-ab^2c-abc^2+a^2b+ac^2+b^2c-3abc+a^2+b^2+c^2-ab-ac-bc\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine Positive Integer in a Series Let $a_{n} = 4 - 3n,$ $\forall$ Integers $n \geq 1.$
$$f(x, y) = x + \sum_{i=1}^{\infty}\left [\left(\frac{\prod_{j=1}^{i}a_{j}}{3^i\cdot i!} \right )x^{a_{i+1}}y^{i}\right]$$
$\forall$ Real $x$ and $y.$ How can I find a positive integer $k$ such that $f(19, k) = 20$.
| Consider the function $F(x, y) = \sqrt[3]{x^3 + y}.$ If we write the function in the form $(x^3 + y)^{1/3},$ we can use the generalized version of the binomial theorem to expand this. We get
$$(x^3 + y)^{1/3} = \binom{1/3}{0}x + \binom{1/3}{1}x^{-2}y + \binom{1/3}{2}x^{-5}y^2 \cdots$$
Also,
$$\begin{align*} \binom{1/3}{n} &= \frac{(1/3)(1/3 - 1)(1/3 - 2)\cdots(1/3 - n + 1)}{n!} \\ &= \frac{(1/3)(-2/3)(-5/3)\cdots(a_{n}/3)}{n!} \\ &= \frac{\prod_{i=1}^{n}a_{i}}{3^n \cdot n!}. \\ \end{align*}$$
Finally, notice that the exponent on $x$ for the $n$th term in our expanded expression, excluding the first $x$, actually equals $a_{n+1}$, and the corresponding exponent on $y$ is $n.$ Combining all this, we get
$$\binom{1/3}{0}x + \binom{1/3}{1}x^{-2}y + \binom{1/3}{2}x^{-5}y^2 \cdots = x + \sum_{n=1}^{\infty}\left [\left(\frac{\prod_{i=1}^{n}a_{n}}{3^n\cdot n!} \right )x^{a_{n+1}}y^{n}\right] = f(x, y).$$
Thus, $F(x, y) = f(x, y) = \sqrt[3]{x^3 + y}.$ To solve the problem, we need to find the integer $k$ such that $19^3 + k = 20^3.$ This value is $k = 20^3 - 19^3 = 20^2 + 20(19) + 19^2 = 1141.$
We still need to show that this value of $k$ will indeed make the series equal $20$ i.e. if the series converges or diverges. The binomial expansion of $(x+y)^n$, for a real number $n$ that is not a nonnegative integer, will converge if $\left | x \right | > \left | y \right |$ and diverge otherwise. In this case, since $\left | 19^3 \right | = 6859$ is clearly greater than $1141,$ our series will converge, so $f(19, 1141) = 20.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results that the integral is $0$ (as it should, since it is not $1-1$ function in this interval) I got down down the way of contour integration. Hence:
$$\begin{aligned}
\int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\
&= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\
&=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\
&= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\
&= \frac{4\pi}{1+ \sqrt{1-8ab}+2a}
\end{aligned}$$
However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for $a=6, b=3$ we have that:
$$\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}$$
My formula cannot derive the result because then radical would be negative. What am I doing wrong here?
| The issue I see with your formula is that scaling $a$ and $b$ by $\lambda$ fails to scale the result by $\lambda^{-1}$. Anyway, I do not know how you computed the residue(s), but you did that wrong somehow.
Also, the substitution is $x=i^{-1}\ln z$, not $x=i\ln z$. Anyway, we get
$$\frac{2}{i}\oint\limits_{|z|=1}\frac{dz}{bz^2+2zaz+b}=\frac{2}{ib}\oint\limits_{|z|=1}\frac{dz}{z^2+\frac{2a}{b}z+1}=\frac{2}{ib}\oint\limits_{|z|=1}\frac{dz}{(z+\frac{a}{b})^2-(\frac{a^2}{b^2}-1)} $$
$$=\frac{2}{ib}\oint\limits_{|z|=1}\frac{dz}{\left(z+c+\sqrt{c^2-1}\right)\left(z+c-\sqrt{c^2-1}\right)}\qquad {\rm where}~~c:=\frac{a}{b}>1. $$
The poles are $-c\pm\sqrt{c^2-1}$. Notice
$$\left(-c-\sqrt{c^2-1}\right) < -1 < \left(-c+\sqrt{c^2-1}\right) < 0 $$
so the relevant pole is $-c+\sqrt{c^2-1}$. Plug that into $(z+c+\sqrt{c^2-1})^{-1}$ to obtain
$$\frac{2}{ib}(2\pi i)\frac{1}{2\sqrt{c^2-1}}=\frac{2\pi}{\sqrt{a^2-b^2}} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $\Delta x^{(n)} = n x^{(n-1)}$ Define $\Delta f(x) = f(x+1) - f(x)$ (the difference operator). Define $x^{(n)} = x(x-1) \dots (x-n+1)$ (the falling factorial function). There's a rather simple theorem which shows that $\Delta x^{(n)} = n x^{(n-1)}$:
$\Delta x^{(n)} = (x+1)^{(n)} - x^{(n)}$
$= (x+1)(x+1-1)(x+1-2)\dots (x+1 - n+1) - x(x-1) \dots (x-n+2)(x-n+1)$
$= (x+1)(x)(x-1) \dots (x-n+2) - x(x-1) \dots (x-n+2)(x-n+1)$
$= [(x+1) - (x-n+1)][(x)(x-1) \dots (x-n+2)]$
$= n [(x)(x-1) \dots (x-n+2)]$
$= n x^{(n-1)}$
My (probably silly) question is about the step from the second line to the third line (the rest of the proof is straightforward). Shouldn't the third line contain the expression $(x+1)(x)(x-1) \dots (x-n)$, instead of $(x+1)(x)(x-1) \dots (x-n+2)$?
| That first term is the product of $n$ factors of the form $x+1-k$, where $k$ starts at $0$ and increases by $1$ from one factor to the next. The highest value of $k$ must therefore by $n-1$, and the last factor is therefore $x+1-(n-1)=x-n+2$.
If you went all the way to $x-n=(x+1)-(n+1)$, you’d have $k$ running from $0$ to $n+1$, for a total of $n+2$ factors.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can we simplify the expression $P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$? Is there a way to reduce the expression
$P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$?
| So our target expression is:
$$ P + \sqrt{P^2 + \sqrt{P^4...}}$$
An indirect approach that is quite simple is to consider
$$ P \left( 1 + \sqrt{1 + \sqrt{1 ...}} \right)$$
It follows that of we distribute the P we will have
$$ P +P \sqrt{1 + \sqrt{1 ...}} $$
Which yields
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
And again we can distribute under the next root for $P^4$ then $P^8$ etc...
So now the trick is to evaluate:
$$S = 1+\sqrt{1+\sqrt{...}}$$
It follows that
$$(S-1)^2 = 1+\sqrt{1+\sqrt{...}}$$
And therefore
$$ (S-1)^2 = S$$
Using this we can solve and find
$$S^2-3S + 1 = 0$$
Which has solutions given by the quadratic formula:
$$S = \frac{3 \pm \sqrt{5}}{2}$$
A simple test of evaluating the first few terms of
$$S = 1+\sqrt{1+\sqrt{...}}$$
Suggests that solution closer to $2$ is in order this we have
$$S = \frac{3 + \sqrt{5}}{2}$$
And this to evaluate:
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
Is just
$$P \left( \frac{3 + \sqrt{5}}{2} \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Finding the maximum of an expression in three variables I am trying to find the max of $\frac{(a)(a-1)(9-a)}{2} + \frac{(b)(b-1)(9-b)}{2} + \frac{(c)(c-1)(9-c)}{2}$ subject to $a+b+c=9$ over nonnegative integers. I originally tried to see if something could work with Jensen's or calculus, and I have a conjecture that I haven't been able to prove that the maximum occurs when $a=5, b=4, c=0$ by simply plugging in an initial guess $a=b=c=3$ and then trying $a=4, b=3, c=2$ and then eventually $a=4, b=4, c=1$. Then because the function $f(x) = \frac{(x)(x-1)(9-x)}{2}$ is increasing when $x<6$ and $f(0)=f(1)=0$ we know that $(5,4,0) > (4,4,1)$.
Is there a more rigorous approach to this?
| $\dfrac{(a)(a-1)(9-a)}{2} + \dfrac{(b)(b-1)(9-b)}{2} + \dfrac{(c)(c-1)(9-c)}{2}$
$=\dfrac{-\sum a^3+10\sum a^2-9\sum a}2$
$\sum a=9$
$\sum a^2=(\sum a)^2-2\sum ab=9^2-2\sum ab$
$\sum a^3=(\sum a)\{(\sum a)^2-3\sum ab\}=9[9^2-3\sum ab]$
Finally $2[(\sum a)^2-3\sum ab]=\sum(a-b)^2\ge0$
$\implies6\sum ab\le2(\sum a)^2=2\cdot9^2$
Fortunately, this gives integer solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328592",
"timestamp": "2023-03-29T00:00:00",
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Solving a three variable equation I have three given values, suppose a=1.86, b=2.6 and c=4.2. Now I have to figure out x,y,z such that
*
*$x\gt 0,y\gt 0$ and $z\gt 0$
*$x+y+z=1$
*$a*x\gt 1, b*y\gt 1$ and $cz\gt 1$
I need a generalized solution steps for this to implement in programming.
Thanks.
| $\left\{ \begin{array}{l}
x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + \varepsilon \\
y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + \varepsilon \\
z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + \varepsilon
\end{array} \right.$ where $0<\varepsilon$.
$\Rightarrow \frac{1}{a} + \varepsilon + \frac{1}{b} + \varepsilon + \frac{1}{c} + \varepsilon = 1 \Rightarrow \varepsilon = \frac{1}{3} - \frac{1}{{3a}} - \frac{1}{{3b}} - \frac{1}{{3c}} = {\rm{0}}{\rm{.1053}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Differential equation $y\cdot y'' - (y')^2 - 1 = 0$ I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$:
$$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {dy}{y} \implies \frac{1}{2} \ln \left|C_1(p^2 + 1) \right | = \ln y \implies y = C_1 \sqrt{p^2+1} \implies p = \sqrt {C_1y^2 - 1} $$
Next step:
$$ y' = \sqrt {C_1 y^2 - 1} \implies \frac{dy}{dx} = \sqrt {C_1 y^2 - 1} \implies \int \frac {dy}{C_1 \sqrt{ y^2 - \frac {1}{C_1^2} }} = \int dx \implies \\ C_1 \ln \left |y + \sqrt{y^2 - C^2_1} \right | = x + C_2 \implies y + \sqrt{y^2 - C_1^2} = \exp {\frac {x+C_2}{C_1}} $$
Here I don't know how to go on. The answer should be $ y = \frac {C_1}{2} \left ( \exp(\frac{x+C_2}{c_1}) + \exp(-\frac{x+C_2}{c_1}) \right )$
| You obtained :
$$y + \sqrt{y^2 - \frac{1}{C_1^2}} = \exp {\frac {x+C_2}{C_1}} $$
it remains to solve this equation for $y$
$$\sqrt{y^2 - \frac{1}{C_1^2}} = -y+\exp {\frac {x+C_2}{C_1}} $$
$$y^2 - \frac{1}{C_1^2} = \left( -y+\exp {\frac {x+C_2}{C_1}}\right)^2 $$
Then, simplify and express $y$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How exactly do we do Gauss elimination?
This is a matrix:
$$\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & k
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
3\\
6\\
4+k
\end{bmatrix}$$
Find $k$ so that it has no unique solution. Solve the equations for this value of $k$.
I found $k = 5$ by using $\text{determinant}=0$.
Then I tried Gauss elimination, from what I understand from my lecturer, I use the 3 rules of Gauss elimination to solve, I just randomly use whatever I need to achieve echelon form. I have come to this:
$$\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 0\\
\end{bmatrix}$$
The answer is
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\3\\0\end{bmatrix}+t\cdot\begin{bmatrix}1\\-2\\1\end{bmatrix}$$
Where does the $t$ come from? Does it involve eigenvalues?
| By Gauss-Jordan elimination I get the following:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 6 \\
1 & 3 & k & 4+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 2 &k-1& 1+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 &k-5& k-5
\end{array}\right)$$
In the next step I would like to divide the last row by $k-5$. Notice that this is valid only for $k-5\ne0$, i.e., $k\ne 5$.
If $\boxed{k\ne5}$ I get
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 &k-5& k-5
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 0 & 2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}\right)$$
I see that for $k\ne5$ I have the unique solution $(1,1,1)$.
I can check that this is indeed a solution by pluggin these values into the original system:
\begin{align*}
1+1+1&=3\\
1+2+3&=6\\
1+3+k&=4+k
\end{align*}
For $\boxed{k=5}$ I get
$$
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 &-1 & 0 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right)
$$
This equation has infinitely many solutions $(t,3-2t,t)$, $t\in\mathbb R$, as explained in Adriano's answer.
| {
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Sum of Square-Weights For positive reals $a,b,c$, prove that $$\frac{a^3+b^3+c^3}{3abc}+\sum_{\text{cyc}} \frac{a(b+c)}{b^2+c^2}\geq 4$$
I've heard of a lemma stating if a polynomial expression $f(a,b,c)$ satisfies both
$f(a,b,c)=f(b,a,c)$ and $f(a,a,c)=0$, then it is divisable by $(a-b)^2$. I tried splitting up the expression into different polynomials where the lemma could be applied but had no success.
I would appreciate any help.
Thanks in advance.
| $\sum_{\text{cyc}} \dfrac{a(b+c)}{b^2+c^2} -3=\sum_{\text{cyc}}(\dfrac{a(b+c)}{b^2+c^2} -1 )=\sum_{\text{cyc}}\dfrac{b(a-b)+c(a-c)}{b^2+c^2}=\sum_{\text{cyc}}(\dfrac{b(a-b)+}{b^2+c^2}+\dfrac{a(b-a)}{a^2+c^2})=\sum_{\text{cyc}}(\dfrac{(a-b)^2(ab-c^2)}{(b^2+c^2)(a^2+c^2)})=\sum \dfrac{(a-b)^2(ab)}{(b^2+c^2)(a^2+c^2)}-\sum \dfrac{(a-b)^2c^2}{(b^2+c^2)(a^2+c^2)}\ge \sum \dfrac{(a-b)^2(ab)}{(b^2+c^2)(a^2+c^2)}-\sum \dfrac{(a-b)^2c}{4abc}$
$\dfrac{a^3+b^3+c^3}{3abc}-1 =\sum \dfrac{(a+b+c)(a-b)^2}{6abc}$
$\sum \dfrac{(a+b+c)(a-b)^2}{6abc} -\sum \dfrac{(a-b)^2c}{4abc}=\dfrac{1}{12abc}\sum (a-b)^2(2(a+b+c)-3c)=\dfrac{1}{12abc}\sum (a-b)^2(a+b)+\dfrac{1}{12abc}\sum (a-b)^2(a+b-c)$
$\sum (a-b)^2(a+b-c)\ge 0 \iff a^3+b^3+c^3+3abc \ge \sum a^2b$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The number of ways to write $10$ as the sum of five natural numbers not equal to $3$ How many answers are there for the equation
$$x_1+x_2+x_3+x_4+x_5=10$$
given that $x_1,x_2\dots x_5\in\Bbb{Z^{0+}}\setminus\{3\}$.
| The number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$
is the number of ways four addition signs can be inserted into a row of ten ones, which is $\binom{10 + 4}{4} = \binom{14}{4}$. From these, we must exclude those solutions in which one or more of the numbers is equal to $3$. Since $3 \cdot 3 = 9 < 10 < 12 = 4 \cdot 3$, at most three of the addends are equal to $3$.
If one of the addends is equal to $3$, then the sum of the four remaining addends is equal to $10 - 3 = 7$. The number of solutions of the equation
$$y_1 + y_2 + y_3 + y_4 = 7$$
in the nonnegative integers is $\binom{7 + 3}{3} = \binom{10}{3}$. Since there are five ways in which one of the five addends could equal $3$, the number of solutions in which at least one addend is equal to $3$ is $\binom{5}{1}\binom{10}{3}$.
If two of the addends are equal to $3$, then the sum of the three remaining addends is equal to $10 - 2 \cdot 3 = 4$. The number of solutions of the equation
$$z_1 + z_2 + z_3 = 4$$
in the nonnegative integers is $\binom{4 + 2}{2} = \binom{6}{2}$. Since there are $\binom{5}{2}$ ways in which two of the five addends could equal $3$, the number of solutions in which at least two addends are equal to $3$ is $\binom{5}{2}\binom{6}{2}$.
If three of the addends are equal to $3$, then the sum of the two remaining addends is equal to $10 - 3 \cdot 3 = 1$. The number of solutions of the equation
$$w_1 + w_2 = 1$$
in the nonnegative integers is $\binom{1 + 1}{1} = \binom{2}{1}$. Since there are $\binom{5}{3}$ ways in which three of the five addends could equal $3$, the number of solutions in which three addends are equal to $3$ is $\binom{5}{3}\binom{2}{1}$.
By the Inclusion-Exclusion Principle, the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$
in which none of the addends is equal to $3$ is
$$\binom{14}{4} - \binom{5}{1}\binom{10}{3} + \binom{5}{2}\binom{6}{2} - \binom{5}{3}\binom{2}{1} = 531$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known:
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
(2.) $\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
(3.) In third step I multiplied these two series, but I am not sure whether it is correct:
$\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
EDIT:
How did you get this sum $\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$ ?
And, in case I have to determine the product of the (some other) series, which one of these should I write?
(a) $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
or
(b) $ = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
| Do not multiply, tackle it directly by looking at the Taylor recipe...
Just a short answer for your check:
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$ Let $\varphi=\frac{1+\sqrt5}2$ (the golden ratio).
How can I simplify the following expression?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
| The answer of @Vladimir Reshetnikov: is hard to improve on, so we'll try to answer some of the questions posed in the comments.
If $\phi=\frac{1+ \sqrt{5}}{2}$ is the positive root of $x^2 - x-1=0$ then
\begin{eqnarray}
\tan(2 \arctan(\phi^n)) = \frac{2\phi^n}{ 1- \phi^{2n}} = \begin{cases} -\frac{2}{L_{n}} \text{ if $n$ odd} \\
-\frac{2}{F_{n}\sqrt{5}} \text{ if $n$ even }
\end{cases}
\end{eqnarray}
where $(F_n)$ is the Fibonacci sequence and $(L_n)$ is the Lucas sequence.
Indeed, for every $v$ we have
$$(1+ iv)^2 = 1- v^2 + 2 i v= -v\cdot ( v -\frac{1}{v} - 2 i)$$
Now consider $v = \phi^n$. We get
$$(1+ i \phi^n)^2 = - \phi^n\cdot ( \phi^n - \phi^{-n} - 2 i)$$
Note that $\phi$, $-\frac{1}{\phi}$ are the roots of $x^2 - x -1=0$. Therefore, if $n$ odd then $\phi^n - \phi^{-n}= \phi^n + \left(-\frac{1}{\phi}\right)^n=L_n$, while if $n$ even then $\phi^n - \phi^{-n} = \phi^n - \left(-\frac{1}{\phi}\right)^n= F_n \sqrt{5}$ .
Assume $n$ is odd. In order to write $2\arctan(\phi^n)$ as a combination of several $\arctan$'s of rational numbers, we decompose the number $L_n - 2 i$ in the Gaussian integers. It appears that for $n >5$ the number $L_n - 2 i$ keeps involving some distinct Gaussian primes from the previous ones.
Assume now $n$ is even. We need to decompose the number $2 + F_{n} \sqrt{-5}$ into some products of elements. Note that in the ring $\mathbb{Z}[\sqrt{-5}]$ the decomposition into irreducibles is not unique, so we have some problems.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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Irrational numbers, induction I have $\sqrt[3]{2}^{2^n}$.
Can I prove that this number is irrational by showing that $3$ does not divide $2^n$?
| If $\dfrac{p}{q} = 2^{\frac{2^n}{3}}, (p,q) = 1 \Rightarrow \dfrac{p^3}{q^3} = 2^{2^n}\Rightarrow p^3=2^{2^n}q^3\Rightarrow 2\mid p^3\Rightarrow 2\mid p$. Now if $p = 2^k$, then $p^3 = 2^{3k} \Rightarrow q^3 = 2^{2^n-3k}$. Observe $3\nmid 2^k \Rightarrow 2^n - 3k \geq 1 \Rightarrow 2\mid q^3 \Rightarrow 2\mid q \Rightarrow (p,q) \geq 2 \neq 1$, contradiction. Thus if $p = (2r+1)2^k \Rightarrow p^3 = (2k+1)^3\cdot 2^{3k} = 2^{2^n}q^3$. Now if $2^{3k} > 2^{2n} \Rightarrow q$ is even. Thus $(p,q) \geq 2 \neq 1$, contradiction. And if $2^{3k} < 2^{2^n}$, then $(2k+1)^3$ is even, contradiction, since $2k+1$ is odd, so is its cube power. Thus $\sqrt[3]{2^{2^n}}$ is irrational.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
| $$x-\frac1x=3.$$
Then
$$\left(x-\frac1x\right)^2+3=x^2+1+\frac1{x^2}=12.$$
Then
$$\left(x^2+1+\frac1{x^2}\right)\left(x-\frac1x\right)=x^3-\frac1{x^3}=36.$$
Finally,
$$\left(x^3+\frac1{x^3}\right)^2=\left(x^3-\frac1{x^3}\right)^2+4=1300.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341237",
"timestamp": "2023-03-29T00:00:00",
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Long polynomial expansion with 34 roots
This is a very tricky problem, I just need a few hints. I think the $(-x^{17})$ is also there for a specific trick. In the end if it is $ax^{17}$, I see that $a = 17 - 1 + 1 = 17$.
Also, another possible approach is:
$$(1 + x + \cdots + x^{17})^2 = x^{17}$$
$$1 + x + \cdots + x^{17} = x^{17/2}$$
But that doesn't do much. Only hints please!
UPDATE:
$$P(x)=0\implies x\ne 1.$$
By the geometric series formula this changes to:
$$\left(\sum_{n=0}^{17} x^n \right)^2 = x^{17} \text{ where } |x| < 1.$$
$$ \left( \frac{1 - x^{18}}{1-x} \right)^2 = x^{17}.$$
$$(1 - x^{18})^2 = (1-x)^2(x^{17}) = x^{19} - 2x^{18} + x^{17}.$$
$$x^{36} - 2x^{18} + 1 = x^{19} - 2x^{18} + x^{17}.$$
$$x^{36} - x^{19} - x^{17} + 1 = 0.$$
$$x^{19}(x^{17} - 1) - (x^{17} - 1) = 0.$$
$$(x^{19} - 1)(x^{17} - 1) = 0.$$
With zero prod. property, we have to use roots of unity.
$$x^{19} = 1 = e^{2\pi i*k}.$$
$$1\ne x = e^{2\pi i \cdot k/19}.$$
$$1\ne x = e^{2\pi i \cdot k/17} \space \text{for the other case}.$$
the smallest root obviously is $a_1 = 1/19, a_2 = 1/17, a_3 = 2/19, a_4 = 2/17, a_5 = 3/19$.
$$\sum a_k = \frac{6}{19} + \frac{3}{17} = \frac{102 + 57}{323} = \frac{159}{323} = \frac{m}{n}$$
$m + n = 482$.
| From the update to the question:
The value of $|z_k| < 1$, because $(1 + x + ... + x^{17})^2 > x^{17}$.
By the geometric series formula this changes to:
$$\left(\sum_{n=0}^{17} x^n \right)^2 = x^{17} \space \text{where} \space |x| < 1$$
$$ \left( \frac{1 - x^{18}}{1-x} \right)^2 = x^{17}$$
$$(1 - x^{18})^2 = (1-x)^2(x^{17}) = x^{19} - 2x^{18} + x^{17}$$
$$x^{36} - 2x^{18} + 1 = x^{19} - 2x^{18} + x^{17}$$
$$x^{36} - x^{19} - x^{17} + 1 = 0$$
$$x^{19}(x^{17} - 1) - (x^{17} - 1) = 0$$
$$(x^{19} - 1)(x^{17} - 1) = 0$$
With zero prod. property, we have to use roots of unity.
$$x^{19} = 1 = e^{2\pi i*k}$$
$$x = e^{2\pi i *k/19}$$
$$x = e^{2\pi i *k/17} \space \text{for the other case}$$
the smallest root obviously is $a_1 = 1/19, a_2 = 1/17, a_3 = 2/19, a_4 = 2/17, a_5 = 3/19$.
$$\sum a_k = \frac{6}{19} + \frac{3}{17} = \frac{102 + 57}{323} = \frac{159}{323} = \frac{m}{n}$$
$m + n = 482$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
| There is more simple way (for me). You have
$$a_n=\frac{2n-1}{n(n+1)(n+2)}=-\frac52\frac{1}{n+2} + \frac{3}{n+1} - \frac{1}{2n};$$
hence
$$S_N = \sum_{n=1}^N a_n = -\frac52\left(H_{N+2}-1-\frac12\right) + 3(H_{N+1}-1) - \frac12 H_N,$$
where $H_N$ is $N$-th harmonic number. Simplify it:
$$S_N = -\frac52\left(H_N + \frac{1}{N+1} + \frac{1}{N+2}-\frac32\right) + 3\left(H_N+\frac{1}{N+1}-1\right) - \frac12H_N=\\=
\frac34 + \frac{1}{2(N+1)}-\frac{5}{2(N+2)}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| The relation "go modulo 4" is very nice. It respects addition and multiplication.
So
$$
6 \cdot 7^{32} + 7 \cdot 9^{45}
$$
is the same as
$$
6 \cdot 1 + 7 \cdot 1
$$
by what you have already computed!
| {
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Intersection of a cone and a plane In $\mathbb{R}^3$, given the cone $K$ and the plane $E_c$ with the equations $4x^2=y^2+z^2$ and $z=c(1-x)$.
How do I find out which different geometric objects I get for all $c\geq 0$ if I intersect both $K$ and $E_c$?
| Expanding on the above comment, if we substitute $z = c(1-x)$ in the equation of $K$ we get $4x^2 = y^2 + c^2(1-x)^2$, i.e.
$$
(4-c^2)x^2 - y^2 + 2c^2 x = c^2 \label{eq:1} \tag{1}
$$
Now all we have to do is distinguish a few cases.
If $c = 0$ then \eqref{eq:1} becomes
$$
0 = 4x^2 - y^2 = (2x + y)(2x - y)
$$
so $E_0 \cap K$ is a degenerate conic, the union of the lines $y = \pm 2x$.
If $0 < c < 2$ then \eqref{eq:1} becomes the equation of a hyperbola, because the terms in $x^2$ and $y^2$ have opposite signs. In particular, for $c = \sqrt{3}$ we have a rectangular hyperbola.
If $c = 2$ then \eqref{eq:1} becomes
$$
y^2 = 8 x - 4
$$
so $E_2 \cap K$ is a parabola.
If $c > 2$ then \eqref{eq:1} becomes the equation of an ellipse, because the terms in $x^2$ and $y^2$ have the same sign.
In particular, for $c = \sqrt{5}$ we have a circle.
| {
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How do I evaluate this : $\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ dx $ for $a > 0$? How do I evaluate this integral if I suppose that $a > 0$
$$\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ \mathrm{d}x .$$
For $a=2$ I got $2\pi$ I think the result will be $a\pi$.
| According to an answer by rae306 on Art of Problem Solving:
Using integration by parts:
$$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\right)\right|_0^{\infty}}_{L}+\underbrace{\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx}_{I}$$
$$I=\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx=\left.2a^2\cdot\frac{1}{a}\arctan\left(\frac{x}{a}\right)\right\vert_0^\infty=2a\cdot\frac{\pi}{2}=\pi a$$
We now have to show that $L=0$:
$$\lim_{x\to\infty}x\cdot\ln\left(1+\frac{a^2}{x^2}\right)=\lim_{x\to\infty}x\left(\frac{a^2}{x^2}-\frac{\left(\frac{a^2}{x^2}\right)^2}{2}+\frac{\left(\frac{a^2}{x^2}\right)^3}{3}-\frac{\left(\frac{a^2}{x^2}\right)^4}{4}+\ldots\right)\\=\lim_{x\to\infty}\frac{a^2}{x}-\frac{a^4}{2x^3}+\frac{a^6}{3x^5}-\frac{a^8}{4x^7}+\ldots=0$$ by using the series expansion.
$$\lim_{x\to0}x\cdot\ln\left(1+\frac{a^2}{x^2}\right)=\lim_{x\to 0} \frac{\ln\left(1+\frac{a^2}{x^2}\right)}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{\frac{2a^2}{x}}{x^2+a^2}}{\frac{1}{x^2}}=\lim_{x\to0}\frac{2a^2 x}{x^2+a^2}=0$$ using L'Hôpital's Rule.
Therefore $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx= a\pi$$
$\square$
| {
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If $f(x+y^3)=f(x)+[f(y)]^3$ and $f'(0)\ge0$, what can $f(10)$ be? A real valued function satisfies the condition: $f(x+y^3)=f(x)+[f(y)]^3$ for all real $x$, $y$.
If $f'(0)\ge0$ how to find $f(10)$ ?
| Here is my attempt:
Let $(x,y)=(0,t)$:$$
f(0+t^3) = f(t^3) = f(0) + [f(t)]^3
$$We can use this to find $f(0)$:$$
f(0^3)=f(0)+[f(0)]^3 \implies f(0) - f(0) = [f(0)]^3 \implies f(0) = 0
$$This means our function simplifies to:$$
f(t^3) = [f(t)]^3
$$Next let $(x,y)=(t,c)$ where $c$ is some arbitrary constant with respect to $t$:$$
f(t+c^3) = f(t) + [f(c)]^3
$$Differentiate both sides with respect to $t$:$$
f'(t+c^3) = f'(t) \implies f'(c^3) = f'(0) \geq 0
$$Since $c$ is any arbitrary constant, this means that $f'$ is always non-negative and so $f$ is always increasing or constant. So $0<1 \implies f(0)\leq f(1) \implies 0 \leq f(1)$.
Next let $(x,y)=(0,1)$:$$
f(1^3)=f(1) = [f(1)]^3 \implies f(1) = 1 \lor f(1) = 0
$$The fact that $0 \leq f(1)$ rules out $f(1)=-1$.
Next let $(x,y)=(1,1)$:$$
f(1+1^3) = f(1) + [f(1)]^3 \implies f(2) = 1 + 1^3 = 2 \lor f(2) = 0
$$Next let $(x,y)=(2,2)$:$$
f(2+2^3) = f(2) + [f(2)]^3 \implies f(10) = 2 + 2^3 = 10 \lor f(10) = 0
$$
It would appear that $f(x)=x$ satisfies both conditions of the provided function.$$
f'(0) = 1 \geq 0 \quad \land\quad f(x+y^3) = x+y^3 = f(x) + [f(y)]^3
$$
It would appear that $f(x)=0$ also satisfies both conditions.$$
f'(0) = 0 \geq 0 \quad \land\quad f(x+y^3) = 0 = 0 + [0]^3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integrating $\int \frac{x^2}{1+x^5} \, dx $ I just encountered the following integral
$$\int \frac{x^2}{1+x^5} \, dx $$
At first it appeared to be simple, but I don't know how to solve it. Please share any ideas.
| In this problem, the hard part is the algebra.
\begin{align}
x^5 + 1 & = (x+1)(x^4-x^3+x^2-x+1) \\[10pt]
& = (x+1)\left(x^2 - 2x\cos\frac\pi5 + 1\right)\left(x^2 - 2x\cos\frac{3\pi}5 + 1\right)
\end{align}
These two quadratic factors are irreducible, as can be seen by the fact that their discriminants are negative.
Next, proceed to partial fractions. Completing the square, you get
\begin{align}
x^2 -2x\cos\frac\pi 5 +1 & =\left( x^2 - 2x\cos\frac\pi 5 + \cos^2\frac\pi5\right) + \sin^2 \frac\pi5 \\[10pt]
& = \left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5.
\end{align}
If you have $\dfrac{Ax+B}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5} \, dx$, you can write it as
\begin{align}
& \frac{A\left(x-\cos\frac\pi5\right)}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx + \frac{B + A\cos\frac\pi 5}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx \\[10pt]
= {} & \frac{\frac 1 2\,du}{u} + \frac{B + A\cos\frac\pi 5}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx
\end{align}
The first term yields a logarithm and the second an arctangent.
Moral: In this probelm, the hard part is the algebra.
So how did I get $\pi/5$ and $3\pi/5$?
The point is that $x^5+1=0$ iff $x^5 = -1$, and the $5$th roots of $-1$ are $\cos\dfrac\pi5 + i\sin\dfrac\pi 5$ and other points on the circle differing from that by a fifth of a circle, i.e. $2\pi/5$ radians. One of those points is $-1$, and that's where $(x+1)$ came from. Two of those points are $\cos\frac\pi5 \pm i\sin\frac\pi5$, and two are $\cos\frac {3\pi}5\pm i\sin\frac{3\pi}5$. So
$$
\left(x - \cos\frac\pi 5 - i \sin\frac\pi5\right)\left(x - \cos\frac\pi 5 + i \sin\frac\pi5\right) = \left(x^2 - 2x\cos\frac\pi5+1\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Trigonometry equation maximum Given the equation: $\cos x + \sqrt3 \sin x = a^2$ find the maximum value for $a$ for which the equation has solutions and for this case solve the equation, $a \in \mathbb{R}$.
I'm guessing I need to find the maximum for the function and for this I have to differentiate it and solve it when it's derivative is $0$? At a first glance I'd say the maximum for $a$ is $\sqrt2$ when $x=\frac{\pi}{3}$, but how do I go around proving this?
| More generally,
and with no originality:
If
$f(x)
=a \sin x + b \cos x
$,
let
$c = \sqrt{a^2+b^2}$.
Then
$f(x)
=c(\frac{a}{c} \sin x + \frac{b}{c} \cos x)
=c(A \sin x + B \cos x)
$.
Since
$A^2+B^2 = 1$,
there is an angle
$\theta$
such that
$\cos \theta = A$
and
$\sin \theta = B$.
Therefore
$f(x)
=c(\cos \theta \sin x + \sin \theta \cos x)
=c \sin(x+ \theta)
$.
Therefore,
$|f(x)|
\le c
$
and this value is attained
for $x = -\theta$.
In this case,
$a = 1$
and $b = \sqrt{3}$,
so the max is
$\sqrt{1+3}
= 2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Differentiate the Function: $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
$g(y)= ln(2y+1)^5-ln\sqrt{y^2-1}$
g'(y)=$\frac{5(2y+1)^4\cdot (2)}{(2y+1)^5}-\frac{2(y^2+1)(2y)}{(y^2+1)^2}$
At this point I can cancel the (2y+1) from the numerator and denominator of the first rational equation and from the second rational equation $(y^2+1)$. However, how would I get $\frac{y}{y^2+1}$ my answer is
$g'(y)=\frac{10}{(2y+1)}-\frac{4y}{y^2+1)^2}$
| Hint
$$g=\log\frac{(2y+1)^5}{\sqrt{y^2+1}}=5\log(2y+1)-\frac 12 \log(y^2+1)$$ $$g'=5 \frac 2{2y+1}-\frac 12 \frac {2y}{y^2+1}=\frac {10}{2y+1}-\frac {y}{y^2+1}=\frac{8 y^2-y+10}{(2y+1)(y^2+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do I show that:if$p$ is prime $>5$ then $p^4-20p^2+19$ is always divisible by $180$.? Is there someone who can show me How do i show that :If $p$ is a prime number greater than $5$ then : $$p^4-20p^2+19$$ is always divisible by $180$.
Note : i think should factor $p^4-20p^2+19=$ as:$p^4-20p^2+19 = (p^2-1)(p^2-19)$ but how i do continue this Idea ?
Thank you for any help .
| Let $p^4-20p^2+19 = (p^2-1)(p^2-19)$, and $180 = 6^2\cdot 5$. But $p\equiv \pm 1\pmod{6}$, so $p^2\equiv 1\pmod{6}$, thus $p^2-1 \equiv 1-1 = 0 \pmod{6}$ and $p^2-19 \equiv 1-19 = -18 \equiv 0 \pmod{6}$, therefore $6^2 \mid p^4-20p^2+19$. On the other hand $p\not \equiv 0\pmod{5}$, so $p^2\equiv \pm 1\pmod{5}$ and $p^4\equiv 1\pmod{5}$, thus $p^4-20p^2 + 19 \equiv 1 + 19 = 20 \equiv 0 \pmod{5}$, therefore $5 \mid p^4-20p^2+19$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
| We can solve this problem completely, without looking at anything larger than a quartic polynomial.
Let $f(x)=\sqrt{4+\sqrt{4-x}}$. The equation we are trying to solve is $f(f(x))=x$.
Observe that $f(x)$ is strictly decreasing from $f(0)=\sqrt6\gt0$ to $f(4)=2\lt4$, which means the equation $f(x)=x$ has a unique solution, which is relatively easy to find:
$$\begin{align}
x=\sqrt{4+\sqrt{4-x}}
&\implies x^2=4+\sqrt{4-x}\\
&\implies (x^2-4)^2=4-x\\
&\implies x^4-8x^2+x-12=0\\
&\implies(x^2-x-3)(x^2+x-4)=0
\end{align}$$
There are two roots that satisfy $0\le x\le4$:
$$x={1+\sqrt{13}\over2}\quad\text{and}\quad x={\sqrt{17}-1\over2}$$
Only the first of these satisfies the equation $x=\sqrt{4+\sqrt{4-x}}$; the other one, $x=(\sqrt{17}-1)/2$, satisfies the equation $x=\sqrt{4-\sqrt{4-x}}$.
Since $x=(1+\sqrt{13})/2$ satisfies $f(x)=x$, it also satisfies $f(f(x))=x$, so we have found $a$ solution of our equation. To see that there are no other solutions, we can appeal to calculus.
From $f(x)=\sqrt{4+\sqrt{4-x}}$ we have
$$f'(x)={-1\over4\sqrt{4+\sqrt{4-x}}\sqrt{4-x}}$$
from which it will turn out to be convenient to note that, for $x\le3$, $|f'(x)|\le|f'(3))|={1\over4\sqrt5}\lt1$
Meanwhile, since $f(x)$ is decreasing on $0\le x\le4$, the function $F(x)=f(f(x))$ is increasing, hence
$$2\lt\sqrt{4+\sqrt{4-\sqrt6}}=f(f(0))\le f(f(x))\le f(f(4))=\sqrt{4+\sqrt2}\lt3$$
(The crude outer bounds, $2$ and $3$, are just there to simplify some of the arithmetic we're about to do.) Thus any solution to $x=f(f(x))$ satisfies $2\lt x\lt3$, which means $f(x)\lt f(2)=\sqrt{4+\sqrt2}\lt3$.
Now suppose the equation $f(f(x))=x$ had two solutions, say $x=a$ and $x=b$ with $a\lt b$. Then by the Mean Value Theorem applied to $F(x)=f(f(x))$, $F'(c)=1$ for some $c\in(a,b)$. But $2\lt a\lt c\lt b\lt 3$ implies $f(c)\lt3$, and we thus see that
$$|F'(c)|=|f'(f(c))f'(c)|\lt|f'(3)|^2\lt1$$
which is a contradiction. We thus conclude that $x={1+\sqrt{13}\over2}$ is the only solution to the equation $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Summation Theorem how to get formula for exponent greater than 3 I'm studying in the summer for calculus 2 in the fall and I'm reading about summation. I'm given these formulas:
\begin{align*}
\sum_{i=1}^n 1 &= n, \\
\sum_{i=1}^n i &= \frac{n(n+1)}{2},\\
\sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6},\\
\sum_{i=1}^n i^3 &=\left[ \frac{n(n+1)}{2}\right]^2.
\end{align*}
But how do I come up with the right formula for exponents greater than 3?
| For any polynomial of $p$th powers $S_p(n) = \sum_{k=1}^n k^p$ $(p \in \Bbb N)$ there is a formula of higher degree $S_p(n)=\sum_{k=1}^{p+1} a_kn^k$
The question is how to find the unknown $a_k$ coefficients which are real numbers.
So use the following formulas to find them:
$a_k= \frac{1}{k!} \cdot \sum_{j=1}^{p+1-k} (-1)^{j+1} \frac {a_{k+j} \cdot (k+j)!}{(j+1)!}$ for $(0 \lt k \lt p+1)$
$a_k=\frac {1}{p+1}$ for $(k=p+1)$
Example:
$S_3(n) = \sum_{k=1}^n k^3$
Solution:
$p=3$
$S_3(n)=\sum_{k=1}^{4} a_kn^3 = a_1n+a_2n^2+a_3n^3+a_4n^4$
Now using the above two formulas to find the unknown coefficients $a_1,a_2,a_3,a_4$
$a_4 = \frac{1}{3+1}=\frac{1}{4}$
$a_3=\frac {1}{3!} \cdot \sum_{j=1}^{1} (-1)^{j+1} \frac {a_{3+j} \cdot (3+j)!}{(j+1)!} =\frac {1}{3!} \cdot \frac {a_4 \cdot 4!}{2!} = \frac {\frac{1}{4} \cdot 4!}{3! \cdot 2!}=\frac {1}{2}$
$a_2=\frac {1}{2!} \cdot \sum_{j=1}^{2} (-1)^{j+1} \frac {a_{2+j} \cdot (2+j)!}{(j+1)!} = \frac {1}{2!} \cdot (\frac {a_3 \cdot 3!}{2!} - \frac {a_4 \cdot 4!}{3!}) = \frac {1}{2} \cdot (3a_3-4a_4)= \frac {1}{2} \cdot (3 \cdot \frac {1}{2}-4 \cdot \frac {1}{4})= \frac {1}{4}$
$a_1=\frac {1}{1!} \cdot \sum_{j=1}^{3} (-1)^{j+1} \frac {a_{1+j} \cdot (1+j)!}{(j+1)!} = \sum_{j=1}^{3} (-1)^{j+1} a_{1+j}=a_2-a_3+a_4 =\frac {1}{4}-\frac {1}{2}+\frac {1}{4}=0$
Finally
$S_3(n)=0*n+\frac {1}{4}*n^2+\frac {1}{2}*n^3+\frac {1}{4}*n^4 =\frac {n^2+2n^3+n^4}{4}=\left[ \frac{n(n+1)}{2}\right]^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 9,
"answer_id": 5
} |
How can we solve a multi-variable recurrence relation in closed form, when the number of terms is also variable? Consider the formula
$f(x, y) = f(x, y-1) + 2 \sum\limits_{i=1}^{x-1} f(i, y-1) $
The factor '2' makes this not expressible cleanly as $f(x, y) = f(x, y-1) + f(x-1, y)$, which is solved here using generating functions.
In this case, the edge conditions are
$f(x, 1) = 2x - 1$
$f(1, y) = 1$
How can this sort of formula be expressed in closed form?
| Argh... better write this as:
$$
f(k + 1, n + 1) = f(k + 1, n) + 2 \sum_{1 \le i \le n} f(i, n)
$$
Define $F(x, y) = \sum_{k, n \ge 1} f(k, n) x^k y^n$, multiply the recurrence by $x^k y^n$, sum over $k \ge 1$ and $n \ge 1$. Recognize the resulting sums
$$
\frac{F(x, y) - \sum_{k \ge 1} f(k, 1) x^k - \sum_{n \ge 1} f(1, n) y^n + f(1, 1)}{x y}
= \frac{F(x, y) - \sum_{n \ge 1} f(1, n) y^n}{x}
+ 2 \frac{F(x, y)}{1 - y}
$$
You also have the boundary conditions:
$\begin{align}
\sum_{k \ge 1} f(k, 1) x^k
&= \sum_{k \ge 1} (2 k - 1) x^k \\
&= \frac{2 x}{(1 - x)^2} - \frac{1}{1 - x} \\
&= \frac{3 x - 1}{(1 - x)^2} \\
\sum_{n \ge 1} f(1, n) y^n
&= \frac{1}{1 - y} \\
f(1, 1)
&= 1
\end{align}$
Plugging all this in and solving for $F(x, y)$:
$$
F(x, y)
= \frac{(1 - 3 x)(1 - y)}
{(1 - x)^2 ((1 - y)^2 - 2 x y)}
$$
Next step would be to split into partial fractions by $x$, and the results by $y$, to read off the coefficients. It looks doable, but messy. I'll leave it here.
| {
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"source": "stackexchange",
"question_score": "1",
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quadratic approximation of $e^{x^3}$ at $x=0$ we know that
$$P(x)=f(a)+f'(a)x+\frac{f''(a)}{2}x^2 $$
therefore
$f(0)=1$
if $x^3=u$, so $$f'(e^u)=e^{x^3}3x^2$$ and $$f''(e^u) = 4(e^u)'3x^2+e^u (3x^2)'=e^{x^3}3x^23x^2+e^{x^3}6x=e^{x^3}(9x^4+6x)$$
$$P(x)=1+e^{x^3}3x^3+\frac{e^{x^3}(9x^4+6x)}{2}x^2 $$
is it correct?
can you please point out my mistake.
cause the system marked my answer as a wrong one.
and what is the magnitude of the error?
| We can just kick it and do: $$e^x = \sum_{n \geq 0}\frac{x^n}{n!} \implies e^{x^3} = \sum_{n \geq 0}\frac{x^{3n}}{n!},$$so that: $$e^{x^3} = 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6}+\cdots.$$We have no $x$ and $x^2$ terms.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.
Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$
I want to show the last step, that is, the inductive step. Assume that this equation is true for some $n=k$. Note that
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\prod_{i=1}^{n}\left [ 2-\frac{1}{i} \right ].$$
For the case $n=k+1$ would be
$$\prod_{i=1}^{k+1}\left [ 2-\frac{1}{i} \right ]=\prod_{i=1}^{k}\left [ 2-\frac{1}{i} \right ]\left [ 2-\frac{1}{k+1} \right ]\leq 2^{k}\left [ 2-\frac{1}{k+1} \right ]=2^{k+1}-\frac{2^{k}}{k+1}.$$
I have to show that the last term is $\leq 2^{k+1}$. But it's not possible to show that $2^{k}/(k+1)\geq 0$.
| For the step from $n=k$ to $n=k+1$, we multiply by $\frac{2k+1}{k+1}$, which is less than $2$.
| {
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Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Starting with the base case $N = 1$:
$$\displaystyle \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} = \frac{1}{1} + \frac{-1}{2} = \frac{1}{2} = \frac{1}{1+1} = \sum^{1}_{n=1} \frac{1}{N+n}.$$
My problem is the inductive step for $N+1$, starting with
$$\displaystyle \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N+1}_{n=1} \frac{1}{(N+1)+n}.$$
And now my problem:
\begin{align}
\sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} &\Leftrightarrow \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \\[1em]
&\underset{\Leftrightarrow}{\text{ind. hyp.}} \sum^{N}_{n=1} \frac{1}{N + n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n}
\end{align}
Is this the correct start, and if so, how do I continue?
| It's easy to prove it using integration,
$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$
$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag1$$
$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2r}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k=1}^{2r}(-x)^{k-1}\ dx$$
$$=\int_0^1\frac{1-x^{2r}}{1+x}\ dx=\ln2-\int_0^1\frac{x^{2r}}{1+x}\ dx=H_{2r}-H_r\tag2$$
where the last result, follows from $\int_0^1\frac{x^{2r}}{1+x}dx=\ln2+H_r-H_{2r}$.
Hence by $(1)$ and $(2)$ , we conclude that
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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How many four digit numbers are perfect square whose first and last two digits are same? I tried it by assuming the number as $\sqrt{1100a+11b}$ and than tried to find figure out perfect square but I am unable to approach further.
| It's easy to solve the problem with brute force, but here is a handmade solution.
So, we are trying to find integers $a,b,x$ so that $1100a+11b=x^2$, $x>0$, $1\leq a\leq9$ and $0\leq b\leq9$.
(In fact, we must have $30<x<100$.)
Considering the equation modulo 11, we get $x^2\equiv0\pmod{11}$.
It is easy to check (there are only 11 cases, or one could use the fact that 11 is square free) that this implies $x\equiv0\pmod{11}$.
Therefore $x=11y$ for some $y\geq1$.
Plugging this into the equation gives $100a+b=11y^2$.
Since we must have $x<100$, we must have $y\leq9$.
Now there are only nine cases to check, and it should be easy enough to find all solutions by hand.
Even this amount of brute force can be avoided.
From our new equation $100a+b=11y^2$ we get $b\equiv 11y^2\pmod{100}$.
Therefore $b\equiv y^2\pmod{10}$ and the only squares modulo 10 are easy to find, so $b\in\{0,1,4,5,6,9\}$.
We also get $b\equiv -y^2\pmod{4}$, and the only squares modulo 10 are 0 and 1, so $b\in\{0,3,4,7,8\}$.
Combining these two conditions gives $b\in\{0,4\}$.
If $b=0$, then from the condition $1100a=x^2$ we get that $x=10z$ for some integer $z$ which must satisfy $z^2=11$.
There is no such integer $z$, so necessarily $b=4$.
Since $b=4$, $y$ must be even.
Therefore $y=2w$ and we get $25a+1=w^2$ and for an integer $w$ with $0<w<5$.
Since $w^2\equiv1\pmod5$, we have $w=1$ or $w=4$.
If $w=1$, then $x=22$, which is too small, so $w=4$ and $x=88$ is the only option.
And a little calculation shows that $x=88$ indeed gives a solution: $88^2=7744$.
(There are many ways to go about this problem. For example, I never used the fact that $100a+b=11y^2$ implies $a+b\equiv0\pmod{11}$ and therefore $a+b=11$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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About $(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16$ Studying polynomials I got the follows:
$$
(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16
$$
I can't understand from where we got this $-8x^3$.
I got to simplify this polynomial just to:
$$
2x^5 + 16
$$
Can someone help me understand from where we got the expression $-8x^3$ ?
| When expanding $(x^3-4)^2$, you get $x^6-4x^3-4x^3+16=x^6-8x^3+16.$ Then, the $x^6$ is subtracted off, and you are left with $2x^5-8x^3+16$, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the matrix of a linear transformation with respect to given (non-standard) ordered bases I am working on the following question from a study guide for my exam on Friday.
Let $L: ℝ^3 → ℝ^3$ be a linear transformation, represented in standard basis by: $$ L(e_1) = \begin{pmatrix}-3 \\ -1 \\ -2 \end{pmatrix}, L(e_2) = \begin{pmatrix}1 \\ 4 \\ 2 \end{pmatrix}, L(e_3) = \begin{pmatrix}-5 \\ -6 \\ -5 \end{pmatrix}$$
Represent the same transformation from $B$ to $B'$ basis where:
$$B = \begin{bmatrix}\begin{pmatrix}0 \\ -1 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \begin{pmatrix}3 \\ -4 \\ 6 \end{pmatrix}\end{bmatrix}$$
$$ B' = \begin{bmatrix}\begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \begin{pmatrix}8 \\ 3 \\ -9 \end{pmatrix}\end{bmatrix}$$
The first thing I thought was: $$L(e_1) = L\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, L(e_2) = L\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}, L(e_3) = L\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$So then I was thinking, that means that I can just apply it to $B$, as $L(B)$, for each vector in $B$. But then I got lost. I am now thinking that this is not what I should do, because apparently there needs to be a linear transformation from $B → B'$. But I don't know what to do now. How should I go about solving this?
| You should apply $L$ to each column of $B$, then write them in linear combination of $B'$. The coefficients are the columns of the transformation. For example,
$$LB_1=\begin{pmatrix}-3&1&-5\\-1&4&-6\\-2&2&-5\end{pmatrix}\begin{pmatrix}0\\-1\\2\end{pmatrix}=\begin{pmatrix}-11\\-16\\-12\end{pmatrix}=-25\begin{pmatrix}1\\1\\0\end{pmatrix}-15\begin{pmatrix}-2\\-1\\2\end{pmatrix}-2\begin{pmatrix}8\\3\\-9\end{pmatrix}$$
So $\begin{pmatrix}-25\\-15\\-2\end{pmatrix}$ is the first column of your transformation. I believe you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abcd}=1000a+100b+10c+d\end{equation}
I see that $4d+1=\overline{..a},\ d\neq0,\ a\neq 0,\ a<d$ but that gives too many pairs to start with $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$
| I'm late to the game here, but I find it interesting that no one used the mod 3 identity for digit sums in base 10, which lets you go from the inside digits out, instead of the outside in.
\begin{align}
\overline{abcd} + \overline{bcd} + \overline{cd} + d + 1 &= \overline{dcba}&(\textrm{mod 3}) \\
(a + b + c + d) + (b + c + d) + (c + d) + d + 1 &= (a + b + c + d) &\quad(\textrm{mod 3}) \\
a + 2b + 3c + 4d + 1 &= a + b + c + d &(\textrm{mod 3}) \\
a + 2b + d + 1 + 3(c+d) &= a + b + c + d &(\textrm{mod 3}) \\
a + 2b + d + 1 &= a + b + c + d &(\textrm{mod 3}) \\
b + 1 &= c &(\textrm{mod 3}) \\
\end{align}
Since $a \le d$, we know that the $4d+1$ in the ones place produces a carry of at least 1, so for some carry digits $k$ and $l$:
\begin{align}
l &\in \{1, 2, 3\} \\
3c + l &= \overline{kb} \\
b + 1 &= c &(\textrm{mod 3}) \\
2b + k &= c &(\textrm{mod 10})
\end{align}
You can go through each of the 9 possible values of $c$, and for any given $c$, you only need to pick one of the three possible values of $l$ to uniquely determine $k$ and $b$. You can usually tell pretty quickly which $l$ value (if any) will give a $b$ with the right value mod 3, so this is only a little tedious by hand. This leads to only one solution, $(b, c, k, l) = (7, 5, 1, 2)$.
From this you know that $5 \le d \le 7$ (to get the carry digit $l=2$), and $d = a+1$ (because $2b+l=15$, carrying a 1). Since $a$ must be odd, $d$ must be even, so the answer is $5756$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the relation between $a,b $ and $c$ in quadratic equation.
If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0,\ \ \{a,b,c,x\}\in \mathbb{R}$ are equal, then $a,b,c$ are in
Options
$a.)\ AP\\
b.)\ GP\\
\color{green}{c.)\ HP}\\
d.)\ \text{cannot be determined}\\$
by using discriminant property
$[b(c-a)]^2-4ac(b-c)(a-b)=0$
I cannot reach any conclusion and is also cumbersome , though I don't know how wolfram
reached this
$[b(c-a)]^2-4ac(b-c)(a-b)=0\ \Longleftrightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$
I look for a short and simple way .
I have studied maths up to $12$th grade.
| $$\begin{align}&b^2(c-a)^2=4ac(b-c)(a-b)\\&\Rightarrow b^2(c^2-2ac+a^2)=4ac(ab-b^2-ca+bc)\\&\Rightarrow b^2c^2-2ab^2c+a^2b^2=4a^2bc-4ab^2c-4a^2c^2+4abc^2\\&\Rightarrow b^2c^2+a^2b^2=4a^2bc-2ab^2c-4a^2c^2+4abc^2\end{align}$$Dividing the both sides by $a^2b^2c^2$, we have$$\begin{align}&\Rightarrow \frac{1}{a^2}+\frac{1}{c^2}=\frac{4}{bc}-\frac{2}{ac}-\frac{4}{b^2}+\frac{4}{ab}\\&\Rightarrow \frac{1}{a^2}-\frac{4}{ab}+\frac{4}{b^2}+\frac{1}{c^2}+\frac{2}{ac}-\frac{4}{bc}=0\\&\Rightarrow \left(\frac 1a-\frac 2b\right)^2+\frac 2c\left(\frac 1a-\frac 2b\right)+\frac{1}{c^2}=0\\&\Rightarrow \left(\frac 1a-\frac 2b+\frac 1c\right)^2=0\\&\Rightarrow \frac 1a-\frac 2b+\frac 1c=0\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1365523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomials: Linking equations via their roots The equation $ax^3+bx^2+cx+d=0$ has solutions $1,p$ and $q$. The equation $x^3+sx^2+tx+r=0$ has solutions $1, 1/p$ and $1/q$. Show that $r=a/d$ , and find an expression for $s$ in terms of $c$ and $d$ , simplifying your answer.
| By the fundamental theorem of algebra:
\begin{align}
ax^3+bx^2+cx+d&=a(x-1)(x-p)(x-q)\\&=a\left(x^3-(1+p+q)x^2+(p+q+pq)x-pq\right)\end{align}
and
\begin{align}
x^3+sx^2+tx+r&=(x-1)\left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\\& = x^3-\left(1+\frac{1}{p}+\frac{1}{q}\right)x^2+\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{pq}\right)-\frac{1}{pq}
\end{align}
Thus $d=-apq$ and $r=-\frac{1}{pq}=\frac{a}{d}$. Also note that $c=a(pq+p+q)$ and $pq=-\frac{d}{a}$, so
$$s=-\left(1+\frac{1}{p}+\frac{1}{q}\right)=-\frac{pq+p+q}{pq}=-\frac{\frac{c}{a}}{-\frac{d}{a}}=\boxed{\frac{c}{d}}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$ How can one prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$?
I know that if i put: $x=y=0$, i will get: $(0-a)^2+(0-b)=a^2+b^2=a^2+b^2$
But that's not a proof but checking.
Thanks.
| Notice, the equation $(x-a)^2+(y-b)^2=a^2+b^2$ represents a circle with a radius $\sqrt{a^2+b^2}$ & center at the point $(a, b)$.
Now, the distance of the given point $(0, 0)$ from the center $(a, b)$ of the circle $$=\sqrt{(a-0)^2+(b-0)^2}$$ $$=\sqrt{a^2+b^2}$$ $$=\text{radius of circle}$$
But the above result is true only when the point $(0, 0)$ lies on the circumference of the circle i.e. the circle $(x-a)^2+(y-b)^2=a^2+b^2$ passes through the point $(0, 0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a-b = 6$
$7c-d = 2$
$a^2+bc = a$
$ab+bd = b$
$ac + cd = c$
$bc + d^2 = d$ from trying that method. What should I do?
| Since
$A^2= A, \tag{1}$
we have
$A(I - A) = (I - A)A = A - A^2 = 0, \tag{2}$
whence, for any vector $x$,
$A(I - A)x = (I - A)Ax = 0. \tag{3}$
(3) indicates that vectors $Ax \ne 0$ in the image of $A$ are in the kernel of $I - A$, i.e., are eigenvectors of $A$ with eigenvalue $1$, and likewise that vectors $0 \ne (I - A)x \in \text{Im}(I - A)$ are eigenvectors corresponding to eigenvalue $0$.
Since we are given that
$A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}, \tag{4}$
we find via the above remarks that
$A \begin{pmatrix} 6 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}; \tag{5}$
also, using (4),
$(I - A) \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix} - \begin{pmatrix} 6 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}; \tag{5}$
we thus also conclude that
$A \begin{pmatrix} 1 \\ -3 \end{pmatrix} = 0. \tag{6}$
We note that the vectors $(6, 2)^T$, $(1, -3)^T$ are linearly independent, being eigenvectors of different eigenvalues; this may also be seen by evaluating the determinant
$\det(\begin{pmatrix} 6 & 1 \\ 2 & -3 \end{pmatrix}) = 6 \cdot -3 - 1 \cdot 2 = -20 \ne 0. \tag{7}$
Being linearly independent, $(6, 2)^T$ and $(1, -3)^T$ form a basis; knowing the action of $A$ on a basis allows the computation of $a, b, c, d$ to proceed; from (6),
$a - 3b = 0, \tag{8}$
$c - 3d = 0; \tag{9}$
so,
$b = \dfrac{a}{3}; d = \dfrac{c}{3}; \tag{10}$
from (5),
$6a + 2b = 6, \tag{11}$
$6c + 2d = 2; \tag{12}$
combining (10)-(12):
$\dfrac{20}{3} a = 6; \dfrac{20}{3} c = 2, \tag{13}$
or
$a = \dfrac{18}{20} = \dfrac{9}{10}; c = \dfrac{6}{20} = \dfrac{3}{10}; \tag{14}$
now from (10),
$b = \dfrac{3}{10}; d = \dfrac{1}{10}, \tag{15}$
so we finally see that
$A = \begin{bmatrix} \dfrac{9}{10} & \dfrac{3}{10} \\ \dfrac{3}{10} & \dfrac{1}{10} \end{bmatrix}. \tag{16}$
It is easily checked that (4), (5), and (6) all bind, and that $A^2 = A$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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IMC 2011 Day 1 Problem 2 - Linear Algebra I have been reading the solutions of a past IMC paper (from 2011, Day 1)
and I did not understand the solution to Problem 2 completely.
Problem 2: Does there exist a real $3$x$3$ matrix $A$, such that $tr(A)=0$ and $A^2+A^t=I$?
Official solution:
"The Anwser is NO. Suppose that $tr(A)=0$ and $A^2+A^t=I$. Taking the transpose, we have $$A=I-(A^2)^t = I - (A^t)^2 = I - (I-A^2)^2= 2A^2 - A^4,$$
$$A^4-2A^2+A=0.$$
The roots of the polynomial $x^4-2x^2+x=x(x-1)(x^2+x-1)$ are $0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})$ so these numbers can be the eigenvalues of $A$; the eigenvalues of $A^2$ can be $0,1,(\frac{1+\sqrt 5}{2}),(\frac{1-\sqrt 5}{2})$.
By $tr(A)=$, the sum of the eigenvalues of $A$ is $0$, and by $tr(A^2)=tr(I-A^t)=3 the sum of squares of the eigenvalues is 3. It is easy to check, that this two conditions cannot be satisfied simultaneously."
It says, that if the only possible eigenvalues for a real $3$x$3$ matrix $A$ are $0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})$, then the only possible eigenvalues for $A^2$ are $0,1,(\frac{1+\sqrt 5}{2}),(\frac{1-\sqrt 5}{2})$.
But why is this so? Shouldn´t it be, that if $\lambda_{i}$ are the eigenvalues for $A$, then $\lambda_{i}^2$ are the eigenvalues for $A^2$? But if the latter is true and they made an mistake, the only possible eigenvalues for $A^2$ are $0,1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})$ and their argument fails, since you can take the eigenvalues to be $0,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})$ so that $tr(A^2)=3$ and there is no contradiction.
Or am I missing something (is there a theorem, etc. that supports their reasoning)?
Please help
| So first of all, you're right, they made a mistake, in the stated setting, we have for the eigenvalues of $A$ the possible set of
$$\left\{0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})\right \}
$$
which gives us, since for $\lambda$ being an eigenvalue to the eigenvector $v$ of $A$
$$
A(Av)=A\lambda v=\lambda Av=\lambda^2v
$$
the following set of possible eigenvalues for $A^2$
$$\left\{0,1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})\right \}
$$
Now we have the restrictions $\operatorname{tr}(A)=0 \tag 1$ and $\operatorname{tr}(A^2)=3 \tag 2$
which arises from $A^2+A^t=I$. Assume now we only have $0$ eigenvalues, then $(1)$ holds but $(2)$ doesn't. So we can exclude this.
Assume now we take $(\frac{-1+\sqrt 5}{2})$, then we also have to take $(\frac{-1-\sqrt 5}{2}),1$. Because otherwise we never fulfill $(1)$. If we take any of $\left\{1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})\right \}$, we always have to take all three of them.
This means, that the eigenvalues of $A^2$ look in this case like
$$\left\{1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})\right \}$$
but because of $(2)$ we need to satisfy
$$
\operatorname{tr}(A^2)=3 \text{ but } \operatorname{tr}(A^2)=1+\frac{3+\sqrt 5}{2}+\frac{3-\sqrt 5}{2}\equiv4
$$
and therefore $(2)$ can never hold, which makes it impossible to have a real $3\times3$ matrix $A$, such that $\operatorname{tr}(A)=0$ and $A^2+A^t=I$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction Proof using factorials Recall that for $n \in N$, $n! = 1 \cdot 2 \cdots n$.
Prove the following for each $n \in N$:
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$$
I understand how to do the proof, but in the inductive step I am facing some difficulty proving the left-hand side is equivalent to the right-hand side.To be direct I am facing some difficulty with the algebra required to make LHS = RHS.
Here is what I have done so far:
1) Base Case
$n = 1$
LHS:
$1/2$
and RHS is $1/2$ $\checkmark$
2) Inductive Step
For $k \geq 1$, Assume $n = k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
$$n = k + 1$$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(n+1)!}$$
$$\implies 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(k+2)!}$$
Here is where i do not know how to make the LHS = RHS.
| $$1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}$$
Note that to simplify this we need a common denominator. Let it be $(k+2)!$. Recall that $(k+1)! = (k+1)(k)(k-1)(k-2) \cdots$ So to get a $(k+2)!$ in the denominator of the fraction we must multiply the numerator and denominator by $k+2$ and get:
\begin{align*}
1-\frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!} &=1-(\frac{k+2}{(k+2)!} - \frac{k+1}{(k+2)!}) \\
&=1-(\frac{k+2-k-1}{(k+2)!}) \\
&= 1-\frac{1}{(k+2)!} \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.
Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$
$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$
We can view it as the sequence(on $n$) of partials sums of
$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$
Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.
By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.
But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.
Is my proof correct? Any alternative solutions are appreciated.
| Here is a general result:
Let $s$ be an infinite sum over a sequence $(a_n)_{n\in\mathbb N}$, where $a_n$ are a permutation of $\frac{(-1)^{k+1}}{k}$, but where the sub sequences $(a_{n_k})_{k\in\mathbb N},\space a_{n_k}=-\frac{1}{2k}$ and $(a_{m_k})_{k\in\mathbb N},\space a_{m_k}=\frac{1}{2k-1}$ exist with $n_k<n_{k+1}$ and $m_k<m_{k+1}$. So
$$
s=1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac1{11}-\frac{1}{4}+...
$$
is such a sum, but
$$
s=1-\frac14+\frac13-\frac12+...
$$
isn't.
Now, for such a sequence, define $p_n$ as the number of plus signs up to $a_n$, $q_n$ as the number of minus signs up to $a_n$ and $l=\lim_{n\to\infty}\frac{p_n}{q_n}$.
We have:
$$
\sum_{k=1}^{n} a_k=\sum_{k=1}^{p_n}\frac{1}{2k-1}-\sum_{k=1}^{q_n}\frac{1}{2k}
$$
But also:
$$
\sum_{k=1}^{p_n}\frac{1}{2k-1}=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k}
$$
And therefore, if $H_n$ is the $n$-th harmonic number and using the well known result $H_n=\ln(n)+\gamma+\epsilon_n$ where $\lim_{n\to\infty}\epsilon_n=0$, we obtain:
$$
\sum_{k=1}^{n} a_k=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k}-\sum_{k=1}^{q_n}\frac{1}{2k}=\\
\sum_{k=1}^{2p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{q_n}\frac{1}{k}=\\
H_{2p_n}-\frac{1}{2}H_{p_n}-\frac{1}{2}H_{q_n}=\\
\ln{(2p_n)}+\gamma+\epsilon_{2p_n}-\frac{1}{2}\ln{(p_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\ln{(q_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{q_n}=\\
\ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n}
$$
And thus:
$$
\lim_{n\to\infty}\sum_{k=1}^{n} a_k=\lim_{n\to\infty}\left(\ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n}\right)=\ln(2)+\frac{1}{2}\ln(l)
$$
In the original sum, we have $l=2$ and therefore $s=\frac{3}{2}\ln(2)$
| {
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Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqrt{x}\ln 10}+\log_{10}\sqrt{x}\cdot 2$
Answer in book is $y'= \frac{1}{\ln10}+\log_{10}x$
| Firstly, note that $$\log_{10} \sqrt{x} = \dfrac{1}{2} \log_{10} x$$
Then we have that $$ \begin{aligned} f(x) = \log_{10} (x) & \iff 10^{f(x)} = x \\ & \implies f(x) \log 10 = \log x \\ & \implies \dfrac{\text{d}}{\text{d}x} f(x) \log 10 = \dfrac{\text{d}}{\text{d}x} \\ & \implies f'(x) \log 10 = \dfrac{1}{x} \\ & \implies f'(x) = \dfrac{1}{x \log 10} \end{aligned} $$
Use the product rule and you should get your answer.
(Note that the above method for finding the derivative applies to any arbitrary base $0 < a \in \mathbb{R}$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
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} |
Is there a way to show the following equality without induction I wanted to show the following equality without using induction:
$$
\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}
$$
Any hint on how to do it?
| Hint: Partial fraction decomposition yields $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$$ which makes your series a special little beast called a telescoping series, try writing out the first few terms and see if you can spot a pattern that allows cancelling.
In particular, you get $$\sum_{k=2}^n \frac{1}{k(k-1)} = \sum_{k=2}^n \left(\frac{1}{k-1} - \frac{1}{k}\right)$$ Which, when written out term by term looks something like $$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-2} - \frac{1}{n-1}\right) + \left(\frac{1}{n-1} - \frac{1}{n}\right)$$
Can you spot what cancels out?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Diagonalize a symmetric matrix let $$A = \left(\begin{array}{cccc} 1&2&3\\2&3&4\\3&4&5 \end{array}\right)$$
I need to find an invertible matrix $P$ such that $P^tAP$ is a diagonal matrix and it's main diagonal may have only the terms from the set $\{ 1,-1,0 \}$
I'd be glad if you could explain to me how to solve this. I haven't found the right theorem/algorithm.
Thanks.
| This trick is due to Hermite. It is especially useful when you have a symmetric matrix of integers. First, we write a certain function in three variables,
$$ f(x,y,z) = x^2 + 3 y^2 + 5 z^2 + 8 yz+ 6 zx +4xy, $$ because this is exactly the result of calculating $v^t A v,$ with
$$
v =
\left(
\begin{array}{c}
x \\
y \\
z
\end{array}
\right)
$$
In order to clear the terms with $x,$ we write
$$ (x + 2 y + 3 z)^2 = x^2 + 4 y^2 + 9 z^2 + 12 yz+ 6 zx +4xy. $$
So far,
$$ f(x,y,z) - (x + 2 y + 3 z)^2 = -y^2 - 4 z^2 - 4 y z. $$
Next we clear all $y,$
$$ (y + 2 z)^2 = y^2 + 4 z^2 + 4 y z, $$ and
$$ f(x,y,z) - (x + 2 y + 3 z)^2 + (y + 2 z)^2 = 0, $$
$$ \color{red}{ f(x,y,z) = (x + 2 y + 3 z)^2 - (y + 2 z)^2 }. $$
The matrix multiplication that this shows is
$$
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{array}
\right) =
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}
\right)
$$
That is actually the right way to do it.
However, I see that someone asked for invertible $P,$ despite non-full rank. Also can be done, and easily:
$$
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}
\right)
$$
The effect of this is to add $0$ to the function in the shape of $0z^2.$
ADDED: looking at the question again, we do need the extra $1$ in the lower right, because the matrix $P$ requested is actually the inverse of the on I display above. Life goes on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$ - basic question
Find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$
Ok so using the product rule on the LHS and the quotient rule on the RHS I differentiated both sides of the equation and got:
$2xy(x\frac{dy}{dx}+y)=\frac{(x+1)\frac{dy}{dx}-y-1}{(x+1)^2}$
Now the textbook gives the answer as $-\frac{y(y+1)(3x+2)}{x(x+1)(y+2)}$ and I'm wondering how to get that answer. I think I'm going about this the wrong way so any hints/guidance would be much appreciated.
| Notice, $$x^2y^2=\frac{y+1}{x+1}$$ $$x^2(x+1)y^2=y+1$$ $$(x^3+x^2)y^2=y+1$$ Now, differentiating both the sides w.r.t. $x$ by applying chain-rule as follows $$\frac{d}{dx}(x^3
+x^2)y^2=\frac{d}{dx}(y+1)$$ $$(3x^2+2x)y^2+2(x^3+x^2)y\frac{dy}{dx}=\frac{dy}{dx}$$ $$2(x^3+x^2)y\frac{dy}{dx}-\frac{dy}{dx}=-(3x^2+2x)y^2$$ $$\frac{dy}{dx}=\frac{-(3x^2+2x)y^2}{2(x^3+x^2)y-1}$$ $$\frac{dy}{dx}=\frac{-(3x+2)xy^2}{2(x+1)x^2y-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| As I'd suggested like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
Using Generalized Binomial Expansion, $$(1+y)^n=1+ny+\frac{n(n-1)}{2!}y^2+\frac{n(n-1)(n-2)}{3!}y^3+\cdots$$ given the converge holds
Comparing with given Series
$ny=2x\ \ \ \ (1)$
$\dfrac{n(n-1)}{2!}y^2=3x^2\ \ \ \ (2)$
$(1)\implies y=\dfrac{2x}n$
From $(2),3x^2=\dfrac{n(n-1)}2\left(\dfrac{2x}n\right)^2\iff n=-2$ as $x\ne0$ for non-trivial cases
$(1)\implies y=\dfrac{2x}n=\dfrac{2x}{-2}=-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 3
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Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^2+y^2+z^2=12$. This was my solution:
$(x.1+y.1+z.1)^2 \le (x^2+y^2+z^2)(1^2+1^2+1^2)$
$\implies (x+y+z)^2 \le 12*3$
$\implies (x+y+z)^2 \le 12*3 \le 36$
Taking square root on both sides, I got
$(x+y+z) \le 6$
Now I am wondering whether I detected a typo in the problem ($x^2+y^2+z^2=12$, not $x^2+y^2+z^2=1$) or is there a real way to reach $(x+y+z) \le 6$ starting from $x^2+y^2+z^2=1$ based on Cauchy-Schwarz principles!
The main reason for this doubt is I am stuck on an extension of this problem in which I need to prove $x^3+y^3+z^3 \ge 24$
Any help is appreciated.
| We have $x^2\le1$, and thus $x\le1$. Similarly, we have $x+y+z\le3<6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$$
However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance!
| $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx\;,$ Now Let
Let $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt\;,$ Then Integral Convert into
$\displaystyle = -\int \frac{t^3}{(1+t^4)^{\frac{3}{4}}}dt\;,$ Now Let $(1+t^4) = u\;, $ Then $\displaystyle t^3dt = \frac{1}{4}du$
So Integral $\displaystyle = -\frac{1}{4}\int t^{-\frac{3}{4}}dt = -u^{\frac{1}{4}}+\mathcal{C} = -\left(1+t^4\right)^{\frac{1}{4}}+\mathcal{C}$
So Integral $\displaystyle \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx = - \left(\frac{1+x^4}{x^4}\right)^{\frac{1}{4}}+\mathcal{C.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
given $2f(x) + f(1-x) = x^2$ find $f(-5)$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
A function $f$ has property that $2f(x)+ f(1-x) = x^2$ for any number of x. What is the number for $f(-5)$?
(A) $\frac {15}4$ (B) $\frac {21}5$ (C) $\frac {25}3$ (D) $\frac {14}3$ (E) $\frac {34}3$
It doesn't give us what $f(x)$ is equal to.
Tried to do comparison
\begin{align}
2f(-5) + f(1-(-5)) &= (-5)^2 \\
2f(-5) + f(6) &= 25 \quad(1) \\
2f(6) + f(1-(6)) &= (6)^2 \\
2f(6) + f(-5) &= 36 \quad(2) \\
\end{align}
so we have the $f(-5)$ and $f(6)$ in both functions, we can add them together
\begin{align}
(1)+(2) \\
[2f(-5) + f(6)] + [2f(6) + f(-5)] &= 25+36 \\
2f(-5) + f(-5) + f(6) + 2f(6) &= 61 \\
3f(-5) + 3f(6) &= 61 \\
f(6) &= \frac {61}3 - f(-5) \quad(3) \\
\end{align}
replacing $f(6)$, at equation $\quad(3)$ to the first equation, $\quad(1)$
\begin{align}
2f(-5) + [\frac{61}{3} - f(-5)] &= 25 \\
2f(-5) - f(-5) &= 25 - \frac {61}3 \\
f(-5) &= \frac{14}{3}
\end{align}
EDIT: Well, I am blind, saw + to minus
| We have, $$2f(x)-f(1-x)=x^2$$ $$\implies 2f(6)-f(1-6)=6^2$$ $$\implies 2f(6)-f(-5)=36\tag 1$$ $$ 2f(-5)-f(1-(-5))=(-5)^2$$ $$\implies 2f(-5)-f(6)=25$$ $$\implies 4f(-5)-2f(6)=50\tag 2$$ Now, adding (1) & (2), we get $$3f(-5)=36+50=86$$ $$\implies f(-5)=\frac{86}{3}$$ Your answer is correct. There may be some printing mistake in the options provided in your book.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of divisors of the form $(4n+1)$
Find the number of divisors of $$2^2\cdot3^3\cdot5^3\cdot7^5$$ which are of the form $(4n+1)$
I know how to find the total number of divisors. But, to find the number of divisors of the form $(4n+1)$, I'm thinking of listing down the divisors and then finding, but that'd be very tedious. Is there any elegant way to do this?
Any help will be appreciated.
Thanks.
| Number of divisors of $$N= 2^2\times3^3\times5^3\times7^5,$$which are of the form $4n+1$ exculuding
$$\begin{align}1 &=\{ \text{number of terms in product}\}\\
&=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+\{\text{number of terms in product}\}\\
&= ( 3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1\\
&= 2\times4\times3+2\times3\times4-1\\
&= 47
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$ The problem is the following (Velleman's exercise 3.2.10):
Suppose that $x$ and $y$ are real numbers. Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$.
My approach so far: Suppose $x \neq 0$. Suppose $ y = \frac{3x^2+2y}{x^2+2}$. Suppose $y \neq 3$. Suppose $y=4$. Then $4x^2=3x^2 \iff x=0$. Contradiction to $x \neq 0$.
Is it correct? I am a noob so feel free to bash it as good as you can. Thanks in advance.
| Givens
*
*$x\neq 0$
*
*$y = \frac{3x^2+2y}{x^2+2}$
Goal
*
*$y=3$
Proof sketch:
*
*Suppose $x\neq 0$.
*Suppose $y = \frac{3x^2+2y}{x^2+2}$. This means that $y (x^2+2)=3x^2+2y \Rightarrow x^2y+2y=3x^2+2y $.
$\qquad \quad$ [Proof of $y=3$ goes here.]
$\qquad$ 3. Therefore $y=3$
*Thus, if $x\neq 0$, then if $y = \frac{3x^2+2y}{x^2+2}$, then $y=3$
The key is the algebraic manipulation to get $x^2y+2y=3x^2+2y$. Now can you get rid of the $2y$ and $x^2$ on both sides? (for the latter, recall that $x\neq 0$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)={n(4n^2+6n-1) \over 3}$ by induction for $n\geq 1$
Prove using mathematical induction that
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)= {n(4n^2+6n-1) \over 3}.$$
Step 1: If we assume that the equation is true for a natural number, $n=k$, then we get
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)= {k(4k^2+6k-1) \over 3}$$
Step 2: When a statement is true for a natural number $n = k$,
then it will also be true for its successor, $n=k+1$. Hence, we have to prove that it is also true for $n=k+1$.
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) = {k(4k+1^2+6k+1-1) \over 3}$$
I replace the LHS by step 1.
$${k(4k^2+6k-1) \over 3} + 2(k+1-1)(2k+1+1)={k(4k+1^2+6k+1-1) \over 3}$$
Now I need to make LHS equal to RHS.
| From the formula for the sum of the square of the first $n$ natural numbers we have
\begin{align*}
\sum_{k=1}^n(2k-1)(2k+1)&=\sum_{k=1}^n(4k^2-1)\\
&=4\sum_{k=1}^nk^2-\sum_{k=1}^n1\\
&=\frac{4n(n+1)(2n+1)}{6}-n\\
&=\frac{n\left[4(n+1)(2n+1)-6\right]}{6}\\
&=\frac{n(4n^2+6n-1)}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality - Cauchy Schwarz Let $a, b, c, d > 0 \in \mathbb{R}$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Show that:
$S = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq 4$
My approach:
I used the Cauchy-Schwarz inequality to show that $S \geq a + b + c + d$ but that is useless as $a + b + c + d \leq 4$. How would you approach this problem? (Only hints desired)
| By Holder's Inequality,
$$S^2\sum_{cyc} a^2b^2\ge (a^2+b^2+c^2+d^2)^3$$
So it remains to note that by AM-GM
$$\sum_{cyc}a^2b^2 = (a^2+c^2)(b^2+d^2) \le \frac{\left((a^2+c^2)+(b^2+d^2)\right)^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2}$$
$$m_2=\frac{-2x+6}{2x-2}$$
Is this right?
I don't know if the tag is right, so please don't be based on it.
| $$(m^2-m-2)x=m^2+4m+3$$
$$(m-2)(m+1)x=(m+1)(m+3)$$
$$m=-1\,\,\,\,\,\text{or}\,\,\,\,\,(m-2)x=(m+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the area of this pentagon $4-\sqrt 5$?
Consider a regular pentagon with vertices (in clockwise order) $A, B, C, D, E$, let $A'$ be the point of intersection of $BD$ and $CE$, let $B'$ be the point of intersection of $CE$ and $DA$, and so on. If $\triangle AC'D'$ has area 1, what is the area of the pentagon $A'B'C'D'E'$?
I tried first to compute the are of $\triangle A'C'D'$ then by using golden triangle, I compute the are of $\triangle C'B'A'$ and $\triangle E'D'A'$, thus getting the area of the smaller pentagon $A'B'C'D'E'$ being $4-\sqrt 5$.
| Let $\color{red}{C'D'=a}$ in isosceles $\triangle AC'D'$ then the angle of vertex A is given as $$\angle C'AD'=\frac{180^\circ}{5}=36^\circ\implies \angle C'AM=\frac{\angle C'AD'}{2}=18^\circ$$ Now, drop a perpendicular say $AM$ from vertex $A$ to the side $C'D'$ in $\triangle AC'D'$, we get $$\tan\angle C'AM=\frac{\frac{C'D'}{2}}{AM}$$ $$\implies \color{blue}{AM}=\frac{C'D'}{a\tan 18^\circ}=\color{blue}{\frac{a}{2\tan 18^\circ}}$$ Hence, the area of isosceles $\triangle AC'D'$ $$=\frac{1}{2}(C'D')(AM)=1\ \text{(given in the question)}$$ $$\implies \frac{1}{2}(a)\left(\frac{a}{2\tan 18^\circ}\right)=1\implies \color{blue}{a^2=4\tan 18^\circ}$$ Hence the area of regular pentagon $A'B'C'D'E'$, having each side, $C'D'=a$, $$=\frac{1}{4}(5)(a^2)\cot\left(\frac{180^\circ}{5}\right)$$ substituting the value of $a^2$, we get area of $A'B'C'D'E'$ $$=\frac{5}{4}(4\tan 18^\circ)(\cot 36^\circ)$$ $$=5\left(\sqrt{\frac{5-2\sqrt{5}}{5}}\right)\left(\sqrt{\frac{5+2\sqrt{5}}{5}}\right)=\frac{5\sqrt{25-20}}{5}$$$$=\sqrt 5$$ That is the correct answer.
Edit Area of any regular n-gon (polygon) having each side, $a$ is given as $$\frac{1}{4}na^2\cot\left(\frac{\pi}{n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| \begin{align*} a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
6+\sqrt{x} + 6-\sqrt{x} + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
12 + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= -9 \end{align*}
Divide by $3\sqrt[3]{ab}$:
$$\sqrt[3]{b} + \sqrt[3]{a} = \frac{-9}{3\sqrt[3]{ab}}$$
Using the original equation we see that the right handed side equals $\sqrt[3]{3}$. We use that. Then substituting $a=6+\sqrt{x}$, $b=6-\sqrt{x}$ will yield a simple lineair equation in $x$ which we can solve.
\begin{align*} \frac{-9}{3\sqrt[3]{ab}} &= \sqrt[3]{3} \\
\frac{-729}{27ab}&=3 \\
ab &=-9 \\ \\ (6+\sqrt{x})(6-\sqrt{x}) &=-9 \\
36-x &= -9 \\
x &= 45
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How should I go about solving this definite integral? The integral is:
$$\int_{-1}^1\sqrt{4-x^2}dx$$
I'm having difficulty figuring out how to go about this. I attempted to use u-substitution, both by substituting $u$ for $\sqrt{4-x^2}$ entirely, and then just $x^2$, but I quickly realized that neither of those work. I've just been introduced to integrals, so I'm not the best at figuring out what to do with them, yet.
| Let $x = 2 \sin \theta$. Then $dx = 2\cos \theta \; d\theta$. When $x = -1, \theta = -\frac{\pi}{6}$ and when $x = 1, \theta = \frac{\pi}{6}$. Therefore
$$\int_{-1}^1 \sqrt{4-x^2} \; dx = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{4 - 4 \sin^2 \theta} \; 2\cos \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 4\cos^2 \theta \; d\theta.$$
Now you can use, say the double angle formula,
$$2\cos^2\theta - 1 = \cos 2\theta$$
to obtain
$$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 4\cos^2 \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 2(1+\cos 2\theta) \; d\theta = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} (1+\cos \theta) \; d\theta.$$
I suppose the remaining steps are easy.
Added: When you see $a^2-x^2$ and $a^2+x^2$, you can always substitute $x = a\sin \theta$ and $x = \tan \theta$ respectively to see if it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $x$ - Logarithm Equation $\ln x+\ln(x+1)=\ln 2$ My attempt:
$\ln x(x+1)=\ln 2$
$e^{\ln x(x+1)}=e^{\ln 2}$
$x(x+1)=2$
$x^2+x-2=0$
$(x-1)(x+2)=0$
therefore $x=1, -2$
| $$\ln { x+\ln { \left( x+1 \right) =\ln { 2 } } } \\ \ln { x\left( x+1 \right) =\ln { 2 } } \\ x\left( x+1 \right) =2\\ x^{ 2 }+x-2=0\\ \left( x-1 \right) \left( x+2 \right) =0\\ { x }_{ 1 }=-2,{ x }_{ 2 }=1\\ $$
x should be $x>0$ hence ${ x }_{ 2 }=1$ is root
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377748",
"timestamp": "2023-03-29T00:00:00",
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Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
| $B=A^2-2A+2I$. Now multiplying the sides respectively by $A$ and $A^2$, and using the assumption $A^3=2I$ yields:
$$AB=BA=-2A^2+2A+2I$$
$$A^2B=BA^2=2A^2+2A-4I$$
Therefore
$$AB+A^2B=4A-2I$$
$$AB+2B=-2A+6I$$
By straightforward algebraic manipulation, it comes that:
$$A^2B+3AB+4B=BA^2+3BA+4B=10I$$
$$(A^2+3A+4I)B=B(A^2+3A+4I)=10I$$
Therefore
$$B^{-1}=(A^2+3A+4I)/10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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This expression is always a perfect square How to show that for $x,y\in \Bbb R$, the expression $xy+\left(\frac{x-y}{2} \right)^2$ is always a perfect square?
For example $x=7, y=3$, $7\times 3+\left(\frac{7-3}{2} \right)^2=25=5^2$
| $$xy+\left(\frac{x-y}{2} \right)^2\\
=\frac{4xy+x^2+y^2-2xy}{4}\\
=\left(\frac{x+y}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $D,E,F$ be (respectively) the projections of $O$ on $BC,CA,AB$. Prove that $\cot{\angle ADB} + \cot{\angle BEC} + \cot{\angle CFA} =0$
Let $O$ be an arbitrary point located inside the triangle $ABC$. Let $D, E, F$ be (respectively) the projections of $O$ on $BC, CA, AB$. Prove that $$\cot{\measuredangle ADB} + \cot{\measuredangle BEC} + \cot{\measuredangle CFA} =0$$
| This is a nice question. I am assuming all the angles are directed. (Directed angles modulo $180^\circ$ are sufficient here, since $\cot$ is periodic with period $180^\circ$.)
We direct our three sidelines $BC$, $CA$ and $AB$. All distances along these lines will thus be understood to mean directed distances. Let $X$, $Y$, $Z$ be the feet of the altitudes of triangle $ABC$ emanating from $A$, $B$, $C$. Let $U$, $V$ and $W$ be the midpoints of the segments $BC$, $CA$, $AB$.
1. Pythagoras in the right-angled triangles $ODB$ and $ODC$ yields $BO^2 = BD^2 + OD^2$ and $CO^2 = CD^2 + OD^2$. Subtracting these two equalities yields $BO^2 - CO^2 = \left(BD^2 + OD^2\right) - \left(CD^2 + OD^2\right) = BD^2 - CD^2$. Similarly, $CO^2 - AO^2 = CE^2 - AE^2$ and $AO^2 - BO^2 = AF^2 - BF^2$. Summing the last three equalities, we obtain
$\left(BO^2 - CO^2\right) + \left(CO^2 - AO^2\right) + \left(AO^2 - BO^2\right) = \left(BD^2 - CD^2\right) + \left(CE^2 - AE^2\right) + \left(AF^2 - BF^2\right)$.
The left hand side of this equality simplifies to $0$, and so it becomes
(1) $0 = \left(BD^2 - CD^2\right) + \left(CE^2 - AE^2\right) + \left(AF^2 - BF^2\right)$.
This is nothing new so far (just a well-known equality, I think due to Carnot, which can be regarded as an additive analogue of Ceva's theorem, or at least a direction of it).
2. Since $U$ is the midpoint of $BC$, we have $BU = - CU$ (remember that our lengths are directed), so that $BU + CU = 0$. Now, adding the equalities $BD = BU + UD$ and $CD = CU + UD$, we obtain $BD + CD = \left(BU + UD\right) + \left(CU + UD\right) = \underbrace{BU + CU}_{=0} + 2 \cdot UD = 2 \cdot UD$. Now,
$BD^2 - CD^2 = \underbrace{\left(BD + CD\right)}_{=2 \cdot UD} \underbrace{\left(BD - CD\right)}_{= BC} = 2 \cdot UD \cdot BC$.
Similarly, $CE^2 - AE^2 = 2 \cdot VE \cdot CA$ and $AF^2 - BF^2 = 2 \cdot WF \cdot AB$. Using the last three equalities, we can rewrite (1) as
$0 = 2 \cdot UD \cdot BC + 2 \cdot VE \cdot CA + 2 \cdot WF \cdot AB$.
Dividing this by $2$, we obtain
(2) $0 = UD \cdot BC + VE \cdot CA + WF \cdot AB$.
3. Let $H$ be the orthocenter of triangle $ABC$. Then, $X$, $Y$, $Z$ are the feet of the perpendiculars from $H$ to $BC$, $CA$, $AB$. Thus, we can apply (2) to $H$, $X$, $Y$ and $Z$ instead of $O$, $D$, $E$ and $F$, respectively. We thus obtain
$0 = UX \cdot BC + VY \cdot CA + WZ \cdot AB$.
Subtracting this equality from (2), we obtain
$0 = \left(UD \cdot BC + VE \cdot CA + WF \cdot AB\right) - \left(UX \cdot BC + VY \cdot CA + WZ \cdot AB\right)$
$= \underbrace{\left(UD - UX\right)}_{= XD} \cdot BC + \underbrace{\left(VE - VY\right)}_{= YE} \cdot CA + \underbrace{\left(WF - WZ\right)}_{= ZF} \cdot AB$
(3) $= XD \cdot BC + YE \cdot CA + ZF \cdot AB$.
4. Let us direct the line $AX$ in such a way that rotating a positive-length vector on the line $BC$ by $90^\circ$ clockwise produces a positive-length vector on the line $AX$. Let us direct the lines $BY$ and $CZ$ similarly. Let $\Delta$ be the signed area of triangle $ABC$. Recall that the area of a triangle equals half times an altitude times the corresponding sidelength. Hence, $\Delta = \dfrac{1}{2} \cdot AX \cdot BC = \dfrac{1}{2} \cdot BY \cdot CA = \dfrac{1}{2} \cdot CZ \cdot AB$. (This is a honest equality, not just an equality that holds up to sign; and the reason for this is how we directed the lines $AX$, $BY$ and $CZ$.) From $\Delta = \dfrac{1}{2} \cdot AX \cdot BC$, we obtain $BC = \dfrac{2 \Delta}{AX}$. Similarly, $CA = \dfrac{2 \Delta}{BY}$ and $AB = \dfrac{2 \Delta}{CZ}$. Using the last three equalities, we can rewrite (3) as
$0 = XD \cdot \dfrac{2 \Delta}{AX} + YE \cdot \dfrac{2 \Delta}{BY} + ZF \cdot \dfrac{2 \Delta}{CZ}$.
Dividing this equality by $2 \Delta$ and simplifying, we obtain
(4) $0 = \dfrac{XD}{AX} + \dfrac{YE}{BY} + \dfrac{ZF}{CZ}$.
Now, the right-angled triangle $AXD$ yields $\dfrac{XD}{AX} = \cot \measuredangle ADX = \cot \measuredangle ADB$. (We are using the interplay between directed lengths and directed angles here -- again, the way how we directed the line $AX$ is important here.) Similarly, $\dfrac{YE}{BY} = \cot \measuredangle BEC$ and $\dfrac{ZF}{CZ} = \cot \measuredangle CFA$. Using the last three equalities, we can rewrite (4) as
$0 = \cot \measuredangle ADB + \cot \measuredangle BEC + \cot \measuredangle CFA$.
Done!
| {
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"url": "https://math.stackexchange.com/questions/1380945",
"timestamp": "2023-03-29T00:00:00",
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Number Theory Problem involving fractional part of a number If
$x = ( 9 + 4 \sqrt {5} )^{48}$
where $x = [x] + f$, where $[x]$ is he integral part of $x$ , and $x$ is its fractional part
How do I go about finding the value of $x(1-f)$ ?
Thanks!
| Imagine expanding $A=(9+4\sqrt{5})^{48}+(9-4\sqrt{5})^{48}$ using the Binomial Theorem. The terms that involve odd powers of $\sqrt{5}$ cancel, so $A$ is an integer.
Note that $(9-4\sqrt{5})^{48}$ is positive and close to $0$. It follows that the integer part of $(9+4\sqrt{5})^{48}$ is $A-1$, so the fractional part $f$ is $1-(9-4\sqrt{5})^{48}$. Now we have a nice expression for $1-f$. Multiply by $x$, using the fact that $(9+4\sqrt{5})(9-4\sqrt{5})=1$.
Remark: The numbers $9+4\sqrt{5}$ and $9-4\sqrt{5}$ are conjugates. Whenever $9+4\sqrt{5}$ has a problem, $9-4\sqrt{5}$ can be reliably counted on for help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382081",
"timestamp": "2023-03-29T00:00:00",
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Expressing the integral in terms of the original variable In evaluating the integral:
$$ \int{dx\over(a^2-x^2)^{3/2}} $$ or $$ \int{dx\over(a^2-x^2)^{1/2}\ (a^2-x^2)}$$
Let $ x=a\sin\theta $ and $ dx=a\cos\theta\ d\theta $. Then
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\int {{d\theta}\over{1-\sin^2\theta}} = {1\over a^2}\int {{d\theta}\over{\cos^2\theta}} = {1\over a^2}\int \sec^2\theta\ d\theta $$
$$ ={{\tan \theta} \over {a^2}}$$
My question has to do with the method of expressing the transformed variable $\theta$ in terms of the original variable $x$. Taking the initial substitution $x=a\sin\theta$, I am able to derive the following:
$$ \theta=\arcsin{x \over a}$$
And substitute into the answer in the following manner:
$$ \tan(\arcsin {x\over a})\over a^2$$
However, the book expresses the integral as such:
$$ {1 \over a^2}{{x} \over {\sqrt {a^2-x^2}}} + C$$
Although the term $1 \over a^2$ remains the same, how do I express the trigonometric functions in this particular form?
| Solution without trigonometric functions.
Given
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }.
$$
Write this as
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{a^2 - x^2 + x^2}{ \sqrt{ a^2 - x^2 }^3 } dx
$$
or
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{dx}{ \sqrt{ a^2 - x^2 } }
+ \color{blue}{ \frac{1}{a^2} \int \frac{x^2}{ \sqrt{ a^2 - x^2 }^3 } dx }.
$$
By parts on the blue part gives
$$
\begin{eqnarray}
\int \frac{d x}{ \sqrt{ a^2 - x^2 }^3 }
&=& \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } }
+ \color{blue}{ \frac{1}{a^2} \int x \frac{x}{ \sqrt{ a^2 - x^2 }^3 } dx }\\
&=& \color{red} { \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } } }
+ \bbox[16px,border:2px solid #800000] { \frac{1}{a^2} \frac{x}{ \sqrt{ a^2 - x^2 } } }
- \color{red} { \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } } }.
\end{eqnarray}
$$
So the final result is
$$
\int \frac{d x}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \frac{x}{ \sqrt{ a^2 - x^2 } }.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An infinite sum of lengths. Let both the base and height of the following triangle have the length 6m. If the lines drawn inside the triangle bisect each right angle formed while proceeding. Then find the length of all these lines. The reason i could not do it is that i did not find any series representation of the lengths of the lines.
| First length element is
$$
\frac{1}{2} \sqrt{2} a
$$
Second length element is
$$
\frac{1}{2} a
$$
Now you repeat this for a smaller triangle with
$$
\frac{1}{2} a
$$
Then
$$
\Bigg( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \Bigg)
\Big( 1 + \sqrt{2} \Big) a
$$
Result
$$
= \Big( 1 + \sqrt{2} \Big) a
$$
$$
S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots
$$
So
$$
2 S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots
$$
Difference gives
$$
S = 2S - S = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
| $$2(x^2+y^2)-(x+y)^2=(x-y)^2\ge0\iff(x+y)^2\le2(x^2+y^2)=?$$
| {
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"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
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A puzzling step in a solution for "Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$" A textbook problem:
Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$
The solution from the textbook:
Let's divide each term of this equation by $\sqrt{a^2+b^2}$:
$$\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
Since the sum of squares of $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ equals 1, then there will always exist an angle, let's call it $\phi$, for which
$$\sin(\phi)=\frac{a}{\sqrt{a^2+b^2}};$$
$$\cos(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
Using this, let's transform our equation to
$$\sin(\phi)sin(x)+cos(\phi)cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
This will bring us to
$$\cos(x-\phi)=\frac{c}{\sqrt{a^2+b^2}}$$
This I understand. But the next step is:
It is evident from this that
$$\sin(x-\phi)=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{a^2+b^2}}$$
Could you give me a hint regarding this last transformation?
| $$\sin^2(x-\phi)+\cos^2(x-\phi)=1$$
$$\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}$$
But $\cos(x-\phi)=\frac{c}{\sqrt{a^2+b^2}}$
Therefore $\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}=\pm\sqrt{1-(\frac{c}{\sqrt{a^2+b^2}})^2}=\pm\sqrt{1-\frac{c^2}{a^2+b^2}}$
This gives: $\pm\sqrt{\frac{a^2+b^2}{a^2+b^2}-\frac{c^2}{a^2+b^2}}=\pm\sqrt{\frac{a^2+b^2-c^2}{a^2+b^2}}$
I believe this is explanatory enough
| {
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Proving $(1 + \frac{1}{n})^n < n$ for natural numbers with $n \geq 3$. Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.
Attempt at proof: Basic step. This can be verified easily.
Induction step. Suppose the assertion holds for $n >3$, then we now prove it for $n+1$. We want to prove that \begin{align*} \big( 1+ \frac{1}{n+1})^{n+1} < n+1. \end{align*} So we have \begin{align*} \big( 1+ \frac{1}{n+1}\big)^{n+1} &= \big( 1+ \frac{1}{n+1} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & < \big (1 + \frac{1}{n} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & = n \cdot \big( 1 + \frac{1}{n+1} \big) \qquad \text{(Induction hypothesis)} \\ &= n \cdot \big( \frac{n+1+1}{n+1} \big) \\ &= \frac{n^2 + 2n}{n+1} \\ &= \frac{(n+1)^2 -1}{(n+1)}
\end{align*}
And now I'm stuck. I don't know how to get $n+1$ on the RHS. Please help!
| HINT:
$$
\frac{(n+1)^2-1}{n+1}<\frac{(n+1)^2}{n+1}.
$$
| {
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For positive integers $x, y, k$, prove that $4^x (4^y+1)=k(k+1)$ implies $x = y$ In the proof that I read, even $k$ implies $4^x=k$ and $4^y+1=k+1$. I am wondering why we don't need to factorize $4^y+1$ into $pq$, such that $p, q > 1$, and solve for $4^x p=k$, $q=k+1$.
| Note that $k^2 < k(k+1) = 4^{x+y}+4^x < (2^{x+y}+2^x)^2$, i.e. $k < 2^{x+y}+2^x$,
and that $(2^{x+y})^2 < 4^{x+y}+4^x = k(k+1) < (k+1)^2$, i.e. $2^{x+y}-1 < k$.
Since $2^x$ divides $4^x(4^y+1)$, either $2^x$ divides $k$, or $2^x$ divides $k+1$.
If $k$ is a multiple of $2^x$, then since $2^{x+y}-1 < k < 2^{x+y}+2^x$, we must have $k = 2^{x+y}$.
Then, $4^{x+y}+4^x = k(k+1) = 2^{x+y}(2^{x+y}+1) = 4^{x+y}+2^{x+y}$. Solving yields $x = y$.
If $k+1$ is a multiple of $2^x$, then since $2^{x+y} < k+1 < 2^{x+y}+2^x+1$, we must have $k+1 = 2^{x+y}+2^x$.
So, $4^{x+y}+4^x = k(k+1) = (2^{x+y}+2^x)(2^{x+y}+2^x+1) = 4^{x+y}+4^x+2^{2x+y+1}+2^{x+y}+2^x$, which is a contradiction since $2^{2x+y+1}+2^{x+y}+2^x > 0$.
Therefore, if $x,y,k$ are positive integers such that $k(k+1) = 4^x(4^y+1)$, then $x = y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this inequality $\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx$ Let $f(x)\in \mathcal C^2([a,b]),f(\frac{a+b}{2})=0$.
Show that$$\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$
I try to use Taylor's Theorem with Integral form of the Remainder,then I have $$f(x)=f(\frac{a+b}{2})+f'(\frac{a+b}{2})(x-\frac{a+b}{2})+\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt.$$ Since $f(\frac{a+b}{2})=\int_{a}^{b}f'(\frac{a+b}{2})(x-\frac{a+b}{2})dx=0, $ we get$\int_{a}^{b}f(x)dx=\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx.$
I think the point of this problem is to prove:
$$\left|\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx\right|\leq\frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$The coeffient of $\frac{1}{8}$ is strange ,and how find it ?
| Changing the order of integration, we obtain
\begin{align}
\int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx
&= \int_{\frac{a+b}{2}}^b \int_t^b (x-t)\,dx\, f''(t)\,dt\\
&= \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2}f''(t)\,dt.
\end{align}
Now we can take the absolute value to get
\begin{align}
\biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert
&\leqslant \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2} \lvert f''(t)\rvert\,dt\\
&\leqslant \frac{(b-a)^2}{8} \int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt,
\end{align}
since $0 \leqslant b-t \leqslant b - \frac{a+b}{2} = \frac{b-a}{2}$ for $\frac{a+b}{2} \leqslant t \leqslant b$.
The analogous estimate for the integral over the first half of the interval gives the desired conclusion
\begin{align}
\biggl\lvert \int_{a}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx\biggr\rvert
&\leqslant \biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert
+ \biggl\lvert \int_a^{\frac{a+b}{2}} \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert\\
&\leqslant \frac{(b-a)^2}{8} \Biggl(\int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt + \int_a^{\frac{a+b}{2}}\lvert f''(t)\rvert\,dt\Biggr).
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$
But I've only been able to get till $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}.\sqrt{\frac{1-y^2}{1-x^2}}.$$
How can the two terms be cancelled off? Thanks.
| Solve explicitly for $y$:
$y = \frac{a^2 x-2 a \sqrt{1-x^2}-x}{a^2+1}$.
Then take the derivative:
${d y \over d x} = \frac{a^2+\frac{2 a x}{\sqrt{1-x^2}}-1}{a^2+1}$
or
$-\frac{2 a \sqrt{1-x^2}}{a^2+1}-\frac{2 x}{a^2+1}+x$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of non-real roots of equation? What is the sum of all non-real, complex roots of this equation -
$$x^5 = 1024$$
Also, please provide explanation about how to find sum all of non real, complex roots of any $n$ degree polynomial. Is there any way to determine number of real and non-real roots of an equation?
Please not that I'm a high school freshman (grade 9). So please provide simple explanation. Thanks in advance!
| $\bf{My\; Solution::}$ Given $$x^5 = 1024 = 2^{10} = 4^5\Rightarrow x^5-4^5 = 0$$
So $$(x^5-4^5) = (x-5)\cdot (x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4) = 0$$
Above we use the formula $\displaystyle x^n-a^n = (x-a)\cdot (x^{n-1}+x^{n-2}\cdot a+x^{n-3}\cdot a^2+.......+a^{n-1})$
So We Get $x=4(\bf{Real \; Root})$ and $$x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4=0$$
So Sum of Roots in $\bf{Second}$ Equation is $\displaystyle = -\frac{4}{1} = -4$
Above we use the Formula $\displaystyle \bf{Sum\; of \; Roots} = -\frac{\bf{coeff.\; of \; x^3}}{\bf{coeff.\; of \; x^4}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x}{a}$ but my friend said it should be something like $arcsin?$
| Notice, we have $$\int\frac{adx}{x\sqrt{x^2-a^2}}$$ Let $x=a\sec \theta \implies a\sec\theta\tan \theta d\theta=dt$
$$\int\frac{a^2\sec\theta\tan \theta d\theta}{a\sec\theta\sqrt{a^2 \sec^2\theta-a^2}}$$
$$=\int\frac{a^2\sec\theta\tan \theta d\theta}{a^2\sec\theta\tan \theta}$$ $$=\int d\theta=\theta+c_1$$ Now, setting the value of $\theta$ $$=\sec^{-1}\left(\frac{x}{a}\right)+c_1$$ Since, $\sec^{-1}\left(\frac{x}{a}\right)=\cos^{-1}\left(\frac{a}{x}\right)$ $$=\cos^{-1}\left(\frac{a}{x}\right)+c_1$$ $$=\frac{\pi}{2}-\sin^{-1}\left(\frac{a}{x}\right)+c_1$$ $$=-\sin^{-1}\left(\frac{a}{x}\right)+c_2$$
Hence, we have $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int\frac{a}{x\sqrt{x^2-a^2}}dx=\sec^{-1}\left(\frac{x}{a}\right)+c_1=-\sin^{-1}\left(\frac{a}{x}\right)+c_2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| Hint: By multiplying the numerator and denominator by $(\sqrt{x+25}+5)(\sqrt{x+16}+4)$, and then using the difference of squares identity $(a-b)(a+b) = a^2-b^2$, we have:
$\dfrac{\sqrt{x+25}-5}{\sqrt{x+16}-4}$ $= \dfrac{(\sqrt{x+25}-5)(\sqrt{x+25}+5)(\sqrt{x+16}+4)}{(\sqrt{x+16}-4)(\sqrt{x+16}+4)(\sqrt{x+25}+5)}$ $= \dfrac{x(\sqrt{x+16}+4)}{x(\sqrt{x+25}+5)}$.
Can you take the limit of this expression as $x \to 0$?
| {
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"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \dfrac{\pi}{2}+6x+2k\pi$$
$$ -3x > \dfrac{\pi}{2}+2k\pi$$
$$ x < -\dfrac{\pi}{6}+ \dfrac{2k\pi}{3}$$
and
$$ 3x < -\dfrac{\pi}{2} - 6x + 2k\pi$$
$$ 9x < -\dfrac{\pi}{2} + 2k\pi$$
$$ x < - \dfrac{\pi}{18} + \dfrac{2k\pi}{9} $$
as you can see the solution is not correct
| $$ \sin (x) \sin (2x) - \cos (x) \cos (2x) > \sin (6x) \Longleftrightarrow$$
$$ -\cos(3x) > \sin (6x) $$
So we find $4$ solutions:
$$x=\frac{2\pi n +\pi}{3}$$
$$\frac{2\pi n -\pi}{3}<x<\frac{12\pi n -5\pi}{18}$$
$$\frac{4\pi n -\pi}{6}<x<\frac{12\pi n -\pi}{18}$$
$$\frac{4\pi n +\pi}{6}<x<\frac{2\pi n +\pi}{3}$$
With $n \in \mathbb{Z}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the integral $PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$ Using the Cauchy Integral Principal to evaluate:
$$PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$$
I know this integral has a pole at $x=i$, $x = -1-i$ and $x = -1+i$. Can someone please show me how to use Cauchy Principal value to evaluate this integral ?
| Let $C_R$ denote the closed semi-circle of radius $R$ centered at $z=0$ in the lower-half plane oriented in anti-clockwise fashion.
$$
\int_{C_R}{dz\over(z^2+1)(z^2+2z+2)}=\\\int_{\pi}^{2\pi}{iRe^{it}dt\over(R^2e^{2it}+1)(R^2e^{2it}+2Re^{it}+2)}+\int_{R}^{-R}{dx\over(x^2+1)(x^2+2x+1)}
$$
Now $|R^2e^{2it}+1|=R^2|e^{2it}+R^{-2}|>{R^2\over2}$ for large enough $R$.
Simliarly,
$|R^2e^{2it}+2Re^{it}+2|>{R^2\over2}$ for large enough $R$.
So for large enough $R$, the absolute value of the integrand in the first integral is less than ${4\over R^3}$.
$$
\therefore\lim_{R\to\infty}\int_{C_R}{dz\over(z^2+1)(z^2+2z+2)}=\int_{\infty}^{-\infty}{dx\over(x^2+1)(x^2+2x+1)}
$$
As Winther notes Cauchy principal value is redundant here. Using residue theorem we have
$$
\int_{-\infty}^{\infty}{dx\over(x^2+1)(x^2+2x+1)}=-2\pi i\left[\text{Res}_{-1-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)+\text{Res}_{-i}\left({1\over(z^2+1)(z^2+2z+2)}\right)\right]=-2\pi i{1\over(z^2+1)(z+1-i)}\Big{|}_{z=-1-i}-2\pi i{1\over(z-i)(z^2+2z+2)}\Big{|}_{z=-i}\\
={2\pi\over5}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can someone show me HOW to do this, I don't just want the answer $4n$ to the power of $3$ over $2 = 8$ to the power of negative $1$ over $3$
Written Differently for Clarity:
$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$
EDIT
Actually, the problem should be solving $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$. Another user edited this question for clarity, but they edited it incorrectly to add parentheses around the right hand side, as can be seen above.
| $${ \left( 4n \right) }^{ \frac { 3 }{ 2 } }={ \left( 8 \right) }^{ -\frac { 1 }{ 3 } }\\ \left( { \left( 4n \right) }^{ \frac { 3 }{ 2 } } \right) ^{ 2/3 }=\left( { \left( { 2 }^{ 3 } \right) }^{ -\frac { 1 }{ 3 } } \right) ^{ 2/3 }\\ 4n=2^{ -\frac { 2 }{ 3 } }\\ n=\frac { 2^{ -\frac { 2 }{ 3 } } }{ 4 } =\frac { 1 }{ 4\sqrt [ 3 ]{ 4 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of Interesting Quadruples
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many interesting ordered quadruples are there?
This is a bit of trouble here actually, I am to use $a + d \gt b + c$ as a constraint.
Without any restrictions (no $a + d \gt b + c$) there are: $\binom{10}{4} = 210$ possible values for $a, b, c, d$.
We could have three cases: $a + d \gt b + c$ or $a + d < b + c$ or $a + d = b + c$.
We need to take out $a + d = b + c$ cases first: It is possible that:
$a + b = \{1 + \sum_{k=4}^{10}k, 2 + \sum_{k=5}^{10}, 3 + \sum_{k=6}^{10}, 4 + \sum_{k=7}^{10}k, ..., 7 + 10 \}$
Total (incl. overcounting): $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$ possible.
But I think I have messed the whole problem.
Hints Please!
| For each set $a,b,c,d$ with $a+d \neq b+c$, either $a+d \gt b+c$ or $(10-d)+(10-a) \gt (10-b)+(10-c)$. Exactly half the selections with $a+d \neq b+c$ will meet your restriction.
I haven't found a neat way to count the cases $a+d=b+c$. We can note that there are two ways for two numbers to sum to $5$, two ways to sum to $6$, three ways to sum to $7$ up to five ways to sum to $10$, then decreasing to two ways to sum to $17$. If you pick two ways out of any of these, you get a case $a+d=b+c$, so there are $4{2\choose 2}+4{3 \choose 2}+4{4 \choose 2}+{5 \choose 2}=4(1+3+6)+10=50$ ways to have $a+d=b+c$ and $(210-50)/2=80$ ways to have $a+d \gt b+c$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{7}{5} \equiv 11 \pmod{12}$. Why is it $11$? I know this equation is true but I don't get the reason $\frac{7}{5}$ is congruent to $11$ here.
The quotient is supposed to be $1.4$ in non modular arithmetic and I don't get where $11$ has come from.
| $\frac{7}{5}$ is not an integer, the congruence relation $\bmod n$ is defined over the integers only.
On the other hand there is such thing as an inverse when working with modular arithmetic. Just like when working in the real numbers we have an inverse for $a$, we also have inverses in modular arithmetic. The inverse of the integer $a\bmod n$ is an integer $b$ so that $ab\equiv 1 \bmod n$ (just like in real numbers $a\cdot\frac{1}{a}=1$), we shall denote the inverse of $a$ as $a^{-1}$. It is also interesting to note not every integer has an inverse $\bmod n$ , in general $a$ has an inverse $\bmod n$ if and only if $a$ and $n$ are relatively prime.
So what is the inverse of $5\bmod 12$? it is the number $k$ so that $ak\equiv 1\bmod 12$. We can find it by trying:
$1\cdot 5=5$ no
$2\cdot 5=10$ no
$3\cdot 5=3$ no
$4\cdot 5=8$ no
$5\cdot 5=1$ yes
$6\cdot 5=6$ no
$7\cdot 5=11$ no
$8\cdot 5=4$ no
$9\cdot 5=9$ no
$10\cdot 5=2$ no
$11\cdot 5=7$ no
Hence the inverse of $5$ is $5$ itself (this rarely happens).
With this in mind we can know calculate $7\cdot5^{-1}=7\cdot5=35\equiv 11\bmod 12$
There are much better ways to obtain the inverse of a number, such as logarithmic exponentiation along with fermat or euler's theorem or via the euclidean algorithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integer values for $m$ such that $x_1,x_2\in\mathbb{Z}$ We have $(m+1)x^2-(2m+1)x-2m=0$ where $m\in\mathbb{R}\backslash\left\{-1\right\}$. We need to find all integer values for $m$ such that both roots of the equation are integers.
Here is all my steps:
*
*$x_{1,2}=\frac{(2m+1)\pm\sqrt{12m^2+12m+1}}{2(m+1)}\in\mathbb{Z}$
$\Rightarrow 2(m+1)\:|\:(2m+1)\pm\sqrt{12m^2+12m+1}$
$\Rightarrow 2(2m+2)+1\pm\sqrt{12(2m+2)^2+12(2m+2)+1}=0$
*$x_1=2(2m+2)+1+\sqrt{12(2m+2)^2+12(2m+2)+1}=0$
$\Rightarrow \sqrt{12(2m+2)^2+12(2m+2)+1}=-4m-5$
$\Rightarrow 12(2m+2)^2+12(2m+2)+1=(-4m-5)^2$
$\Rightarrow 12(4m^2+8m+4)+24m+25=16m^2-40m+25$
$\Rightarrow 48m^2+96m+48+24m+25=16m^2-40m+25$
$\Rightarrow 32m^2+160m+48=0$|:16
$\Rightarrow 2m^2+10m+3=0$ but $\sqrt{\Delta}\notin\mathbb{N}$
Now we have to verify for $x_2$ like above:
*
*$x_2=48m^2+96m+48+24m+25=16m^2+40m+25$
$\Rightarrow 32m^2+80m+48=0|:16$
$\Rightarrow 2m^2+5m+3=0 \Rightarrow \sqrt{\Delta}=1$
$\Rightarrow m_1=-1\in\mathbb{Z}, m_2\notin\mathbb{Z}$
But as the author says $m\in\mathbb{R}\backslash\left\{-1\right\}$
| Let's express $x$:
$$
-m = \frac{x^2-x}{x^2-2x-2}.
$$
So,
$$
-m - 1 = \frac{x+2}{x^2-2x-2}.
$$
Denominator never equal to $0$ for integers $x$. If $|x+2|<|x^2-2x-2|$, $m$ cannot be integer (unless $x=-2$; we'll check this case later). Solve this inequality:
$$
|x+2|<|x^2-2x-2|\Leftrightarrow (x^2-2x-2)^2 - (x+2)^2 > 0 \Leftrightarrow x^4 - 4x^3 + 4x - x^2 > 0
$$
But
$$
x^4 - 4x^3 + 4x - x^2 = x(x-1)(x+1)(x-4).
$$
Hence, if $x>4$ or $x<-1$, $m$ cannot be integer. Now we just should check integers $-2\le x\le 4$ (don't forget about $x=-2$).
*
*$x=-2$, $m=-1$
*$x=-1$, $m=-2$
*$x=0$, $m=0$
*$x=1$, $m=0$
*$x=2$, $m=1$
*$x=3$, $m=-6$
*$x=4$, $m=-2$
Excluding $m=-1$, possible $m$ are
$$
-6, -2, 0, 1.
$$
But both roots must be integers; so, $\boxed{m=-2}$ or $\boxed{m=0}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?
A differentiable function f is satisfying the relation $$f(x+y) = f(x) + f(y) + 2xy(x+y) - \dfrac{1}{3} $$ $ \forall $ $ x , y $ belongs to $\Re$ and $$lim_{h \to 0} \dfrac{3f(h)-1}{6h} = \dfrac{2}{3}. $$Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?
I think $g(x+y)=g(x)+g(y)$ then we use the limit condition, I need help, and If someone can complete my solution I'll appreciated
| Using the formula $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}...............(1)$$
Now Given $$\displaystyle f(x+y) = f(x)+f(y)+2xy(x+y)-\frac{1}{3}$$
So Put $x=h\;$ We get $$\displaystyle f(x+h) = f(x)+f(h)+2xh(x+h)-\frac{1}{3}$$
So $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+2xh(x+h)-\frac{1}{3}-f(x)}{h}$$
$$\displaystyle f'(x) = \lim_{h\rightarrow 0}2x(x+h)+\lim_{h\rightarrow 0}\frac{3f(h)-1}{3h} = 2x^2+\frac{4}{3}$$
So $$\displaystyle \int f'(x)dx = \int 2 x^2+\frac{4}{3}\int 1dx$$
So $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\mathcal{C}$$
Now Put $x=y=0$ in original functional equation, we get $$\displaystyle f(0) =\frac{1}{3}$$
So Put $x=0$ in $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\mathcal{C}$$
We get value of $\mathcal{C} = \frac{1}{3}$
So we get $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\frac{1}{3}$$
Put $x=2\;,$ We get $\displaystyle f(2) = \frac{25}{3}$
So we get $\lfloor f(x)\rfloor = 8$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution of System of linear equations I have three equations:
$$
\begin{cases}
4y + z = 2\\
2x + 6y - 2z = 3\\
4x + 8y - 5z = 4
\end{cases}
$$
Applying Gauss elimination I get:
$$ \left[
\begin{array}{ccc|c}
1&0&-\frac{7}{4} & 0\\
0 & 1 & \frac{1}{4} & \frac{1}{2} \\
0 & 0 & 0 & 0
\end{array}
\right] $$
Now how can I get the solution as there one row is eliminated leaving 2 equations with 3 unknowns?
| You can arbitrarily choose $z$. Then, $x$ and $y$ can be computed from that:
$$
x - \frac{7}{4}z = 0 \;\Rightarrow\; x = \frac{7}{4} z\\
y + \frac{1}{4}z = \frac{1}{2} \;\Rightarrow\; y=\frac{1}{2} - \frac{1}{4}z
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The sum of the lengths of the hypotenuse and another side of a right angled triangle is given. The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)\frac{\pi}{3}\hspace{1cm}(D)\frac{5\pi}{12}$
Let $a,b$ are sides of a right triangle other than hypotenuse.Then given that $\sqrt{a^2+b^2}+a$=constant=$k$
Area of triangle=$\frac{1}{2}ab=\frac{1}{2}a\sqrt{(k-a)^2-a^2}$
and then?
| The area of the triangle can be expressed as
$$\frac 12ab=\frac 12a\sqrt{(k-a)^2-a^2}=\frac{a}{2}\sqrt{k^2-2ak}=\frac{\sqrt k}{2}\sqrt{a^2k-2a^3}$$
Here, let $f(a)=a^2k-2a^3$. Then, we have
$$f'(a)=2ak-6a^2=2a(k-3a).$$
So, since we know that the maximum of $f(a)$ is $f(k/3)$, it follows that the maximum of the area of the trangle is attained when $a=k/3$.
Then, the angle $\theta$ satisfies
$$\cos\theta=\frac{a}{\sqrt{a^2+b^2}}=\frac{k/3}{k-(k/3)}=\frac 12\Rightarrow \theta=\frac{\pi}{3}.$$
| {
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Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$ Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from
using CAS, do you see any very easy way of calculating it?
$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \sqrt{1+\sqrt{2+x+x^2+x^3}}} \, dx$$
EDIT: It's a part from the generalization
$$\int \frac{1+2x +3 x^2+\cdots n x^{n-1}}{\left(2+x+x^2+\cdots+ x^n\right) \sqrt{1\pm\sqrt{2+x+x^2+\cdots +x^n}}} \, dx$$
Supplementary question: How would you calculate the following integral using the generalization above? Would you prefer another way?
$$\int_0^{1/2} \frac{1}{\left(x^2-3 x+2\right)\sqrt{\sqrt{\frac{x-2}{x-1}}+1} } \, dx$$
As a note, the generalization like the one you see above and slightly modified versions can be wisely used for calculating very hard integrals.
| HINT:
As $\dfrac{d(2+x+x^2+x^3)}{dx}=1+2x+3x^2,$
let $\sqrt{1+\sqrt{2+x+x^2+x^3}}=u\implies 2+x+x^2+x^3=(u^2-1)^2$
and $(1+2x+3x^2)dx=(u^2-1)2u\ du$
Now use Partial Fraction Decomposition,
$\dfrac1{(u^2-1)^2}=\dfrac A{u-1}+\dfrac B{(u-1)^2}+\dfrac C{u+1}+\dfrac D{(u+1)^2}$
| {
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selection of balls of three colors with restrictions I have asked a similar question here and answers were very helpful. I tried doing similar questions and could solve them comfortably. However, I myself came up with a question like this and wondering how to solve this.
Suppose there are 20 black balls, 10 yellow balls and 5 brown balls. Balls of same color are identical. Order of selection does not matter.
*
*How many ways can a selection of 15 balls be made?
*How many ways can a selection of 15 balls be made if 2 brown must be selected always?
My approach
(1) number of solutions of $x_1+x_2+x_3 = 15$ with $0\le x_1\le 20; 0 \le x_2 \le 10; 0 \le x_3 \le5$ is the solution. Is this right? if so, is there any formula for this?
(2) Select the 2 brown balls in (only 1 way) and then find number of solutions of $x_1+x_2+x_3 = 13$ with $0\le x_1\le 20; 0 \le x_2 \le 10; 0 \le x_3 \le3$
Please give directions on how to approach similar problems. thanks.
| Your approaches to both problems are correct.
Since balls of the same color are indistinguishable, the number of ways $15$ balls can be selected from $20$ black balls, $10$ yellow balls, and $5$ brown balls is the number of nonnegative integer solutions of the equation
$$x_1 + x_2 + x_3 = 15 \tag{1}$$
subject to the restrictions $x_2 \leq 10$ and $x_3 \leq 5$. Note that since $20 > 15$, we do not need a restriction on the number of black balls.
If there were no restrictions, the number of solutions of the equation
$x_1 + x_2 + x_3 = 15$ would be equal to the number of ways of inserting two addition signs in a row of $15$ ones, which is $\binom{15 + 2}{2} = \binom{17}{2}$ since we must select which two of the seventeen symbols ($15$ ones and two addition signs) are to be addition signs.
From these, we must remove those solutions in which $x_2 > 10$ or $x_3 > 5$. Note that it is not possible for both of these conditions to hold simultaneously since $11 + 6 = 17 > 15$.
Suppose $x_2 > 10$. Let $y_2 = x_2 - 11$. Then $y_2$ is a nonnegative integer. Moreover, substituting $y_2 + 11$ for $x_2$ in equation 1 yields
\begin{align*}
x_1 + y_2 + 11 + x_3 & = 15\\
x_1 + y_2 + x_3 & = 4
\end{align*}
Thus, the number of solutions of equation 1 in which $x_2 > 10$ is equal to the number of ways two addition signs can be placed in a row of four ones, which is $\binom{4 + 2}{2} = \binom{6}{2}$.
Now, suppose $x_3 > 5$. Let $z_3 = x_3 - 6$. Then $z_3$ is a nonnegative integer. Moreover, substituting $z_3 + 6$ for $x_3$ in equation 1 yields
\begin{align*}
x_1 + x_2 + z_3 + 6 & = 15\\
x_1 + x_2 + z_3 & = 9
\end{align*}
Thus, the number of solutions of equation 1 in which $x_3 > 5$ is the number of ways in which two addition signs can be placed in a row of nine ones, which is $\binom{9 + 2}{2} = \binom{11}{2}$.
Hence, the number of ways $15$ balls can be selected from $20$ indistinguishable black balls, $10$ indistinguishable yellow balls, and $5$ indistinguishable brown balls is
$$\binom{17}{2} - \binom{6}{2} - \binom{11}{2}$$
which agrees with the answer ml0105 obtained using generating functions.
You can apply the same method to solve the second problem you posed.
| {
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Find $\lim_{n\to\infty}\left(2n\int_{0}^{1}\frac{x^n}{1+x^2}dx\right)^n$ Find this limits
$$\lim_{n\to\infty}\left(2n\int_{0}^{1}\dfrac{x^n}{1+x^2}dx\right)^n\tag{1}$$
since following links post this question have solve it.$$\lim_{n\to\infty}2n\int_{0}^{\frac{\pi}{4}}\tan^n{x}dx=\dfrac{1}{2}$$
Solving $\lim_{n\to\infty}(n\int_0^{\pi/4}(\tan x)^ndx)$?
Could you suggest a helpful idea with $(1)$ ?
| Integration by parts gives:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = \left.x^n\frac{2x}{1+x^2}\right|_{0}^{1}-2\int_{0}^{1}x^{n}\frac{(1-x^2)}{(1+x^2)^2}\,dx$$
so performing it twice:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = 1-2\left.\frac{x^{n+1}}{n+1}\frac{(1-x)^2}{(1+x^2)^2}\right|_{0}^{1}+2\int_{0}^{1}\frac{x^{n+1}}{n+1}\cdot\frac{2x(3-x^2)}{(1+x^2)^3}\,dx$$
but since $\frac{2x(3-x^2)}{(1+x^2)^3}$ is non-negative and bounded on $(0,1)$, it follows that:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = 1+O\left(\frac{1}{n^2}\right),$$
so the wanted limit is just $\color{red}{1}$.
To be more accurate, we may evaluate $I_n = 2n\int_{0}^{1}\frac{x^n}{1+x^2}\,dx$ in terms of the digamma function:
$$ I_n = \frac{n}{2}\left(\psi\left(\frac{n+3}{4}\right)-\psi\left(\frac{n+1}{4}\right)\right)=\sum_{m\geq 1}\frac{4n}{(4m+n-1)(4m+n-3)}$$
or through the simple series representation:
$$ I_n = 2\sum_{m\geq 1}\left(\frac{4m-1}{n+4m-1}-\frac{4m-3}{n+4m-3}\right)$$
to get:
$$ I_n \approx 1-\frac{1}{n^2}$$
for large $n$s.
| {
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.
| The substitution $t=1/x$ works. Then, $dx=-\frac{1}{t^2}dt$ and
$$\begin{align}
I&=\int_0^{\infty}\frac{1}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\frac{1}{x^2}\,dx\\\\
&=\frac12\int_0^\infty \frac{1+\frac{1}{x^2}}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\frac12\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\,d\left(x-\frac1x\right)\\\\
&=\left.\frac1{2a} \arctan\left(\frac{x-\frac1x}{a}\right)\right|_{0}^{\infty}\\\\
&\bbox[5px,border:2px solid #C0A000]{=\frac{\pi}{2a}\text{sgn}(a)}
\end{align}$$
If $a>2$, then setting $\frac{\pi}{2a}=\frac{\pi}{5050}$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{a=2525}$$
| {
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Angle chase:In $\Delta ABC, AB=AC $ and $\angle BAC=20°.$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$. In $\triangle ABC, AB=AC $ and $\angle BAC=20^\circ$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.
| Notice, we have $$\angle A+\angle B+\angle C=180^\circ$$ But $AB=AC\iff \angle B=\angle C$ Hence, we get $$\angle A+\angle C+\angle C=180^\circ$$ $$20^\circ+2\angle C=180^\circ$$ $$\angle C=\frac{180^\circ-20^\circ}{2}=80^\circ$$
Let the unknown angle $\angle ADC=x$
Applying sine rule in $\triangle ACD$ as follows $$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle CAD}{CD}$$ Setting the corresponding values, we get
$$\frac{\sin(180^\circ-(x+20^\circ))}{AD}=\frac{\sin 20^\circ}{CD}$$
$$\frac{\sin(x+20^\circ)}{AD}=\frac{\sin 20^\circ}{CD}\tag 1$$
Similarly, applying sine rule in $\triangle BCD$ as follows $$\frac{\sin\angle BCD}{BD}=\frac{\sin\angle CBD}{CD}$$ Setting $BD=AD$ & the corresponding values, we get
$$\frac{\sin(x-80^\circ)}{AD}=\frac{\sin 80^\circ}{CD}\tag 2$$
Now, diving (2) by (1) we get
$$\frac{\frac{\sin(x-80^\circ)}{AD}}{\frac{\sin(x+20^\circ)}{AD}}=\frac{\frac{\sin 80^\circ}{CD}}{\frac{\sin 20^\circ}{CD}}$$
$$\frac{\sin(x-80^\circ)}{\sin(x+20^\circ)}=\frac{\sin 80^\circ}{\sin 20^\circ}$$
$$\frac{\sin x\cos 80^\circ-\cos x\sin 80^\circ}{\sin x\cos 20^\circ+\cos x\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$
$$\frac{\tan x\sin 10^\circ-\cos 10^\circ}{\tan x\cos 20^\circ+\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$ $$(\cos 20^\circ\cos 10^\circ-\sin 20^\circ\sin 10^\circ)\tan x=-2\sin 20^\circ\cos 10^\circ$$
$$\cos 30^\circ\tan x=-2\sin 20^\circ\cos 10^\circ$$
$$\tan x=\frac{-2\sin 20^\circ\cos 10^\circ}{\cos 30^\circ}=\frac{-2\sin 20^\circ\cos 10^\circ}{\frac{\sqrt 3}{2}}$$
$$\tan x=\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}$$$$\iff x=\tan^{-1}\left(\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$
$$=180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$
Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\angle ADC}=\color{blue}{180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)\approx 142.12^\circ}}$$
| {
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Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\Rightarrow 2\leq x^2+2\leq6\Rightarrow\frac{1}{6}\leq\frac{1}{x^2+2}\leq\frac{1}{2}\Rightarrow\frac{1}{6}\int_{0}^{2}1dx\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{1}{2}\int_{0}^{2}1dx$$\Rightarrow\frac{1}{3}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq1$
In the second part,$0\leq x\leq2\Rightarrow -2<x^2-x<4\Rightarrow e^{-2}<e^{x^2-x}<e^4\Rightarrow2e^{-2}<\int_{0}^{2}e^{x^2-x}dx<2e^4$
Where have i gone wrong?What is the correct method to solve it.
| Both answers are fine, I would just point out than since $f(x)=\frac{\sqrt{x}}{2+x}$ is a concave function on $[0,4]$ and $e^{x^2-x}$ is a convex function on $[0,2]$, both inequalities can be improved through the Hermite-Hadamard inequality, i.e. by using the usual rectangle/trapezoid method for approximating an integral.
About the first integral, another chance is given by the identity $\frac{1}{2+x^2}=\frac{1}{2}-\frac{x}{6}+\frac{x(x-1)(x-2)}{6(2+x^2)}$, from which:
$$ I=\int_{0}^{2}\frac{dx}{2+x^2}=\frac{2}{3}+\frac{1}{6}\int_{0}^{2}\frac{x(x-1)(x-2)}{2+x^2}\,dx, $$
and since $x(x-1)(x-2)$ over $[0,2]$ is between $-\frac{2}{3\sqrt{3}}$ and $\frac{2}{3\sqrt{3}}$,
$$ I \leq \frac{2}{3}+\frac{1}{9\sqrt{3}}\,I,\qquad I\geq \frac{2}{3}-\frac{1}{9\sqrt{3}}\,I $$
lead to:
$$ \color{red}{\frac{5}{8}}<0.626478\ldots=\frac{2}{3+\frac{1}{3\sqrt{3}}}\leq \color{red}{I} \leq \frac{2}{3-\frac{1}{3\sqrt{3}}}=0.712364\ldots<\color{red}{\frac{5}{7}}. $$
The second integral is clearly concentrated around $x=2$:
$$ J=\int_{0}^{2}e^{x^2-x}\,dx = e^2\int_{0}^{2}e^{-3x+x^2}\,dx\geq e^2\int_{0}^{2}e^{-3x}\,dx= \frac{e^2-e^{-4}}{3},$$
$$ J \leq e^2\int_{0}^{2}e^{-x}\,dx = e^2-1,$$
but we may improve the last inequality up to:
$$ J\leq e^{2}\int_{0}^{2}e^{-x}\left(1-\left(1-\frac{1}{e}\right)x(2-x)\right)\,dx = e^2-5+\frac{4}{e} $$
and the first inequality can be improved up to:
$$ J\geq e^2\int_{0}^{2}e^{-3x}(1+x^2)\,dx = \frac{11 e^2-59 e^{-4}}{27} $$
hence $J$ is between $2.97$ and $3.87$.
| {
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Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$
This is a problem on Mathlinks.ro and it have had no solution. So I hope it would have a nice answer and as simply as possible.
Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$$
| We can trade off the cyclic symmetry to reduce the number of variables to consider. The homogeneous form is:
$$4^4(a^3b+b^3c+c^3d+d^3a+23abcd) \le 27(a+b+c+d)^4$$
WLOG, let $a$ be the minimum among $a, b, c, d$; and so we have non-negatives $x, y, z$ s.t. $b = a+x, \, c = a+y, \, d = a+z$. Hence we get an inequality of form
$$f(x, y, z)\cdot a^2+g(x, y, z)\cdot a+h(x, y, z) \ge 0$$
It is easy to check $f = 32\cdot23[(x-y)^2+(y-z)^2+(z-x)^2] + 32\cdot11(x^2+y^2+z^2) \ge 0$
$g = 16 (11x^3 + 33x^2y + 81x^2z + 81xy^2 - 206xyz + 81xz^2 + 11y^3 + 33y^2z + 81yz^2 + 11z^3)$ which is non-negative as AM-GM gives:
$$11x^3 + 33x^2y + 81x^2z + 81xy^2 + 81xz^2 + 11y^3 + 33y^2z + 81yz^2 + 11z^3 \\ \ge 9\cdot9\cdot11^{5/9} \cdot xyz \ge 9\cdot9\cdot3\cdot xyz \ge 206xyz$$
It only remains to show $h \ge 0$, which is the same as setting one variable to zero, or showing $4^4(a^3b+b^3c) \le 27(a+b+c)^4$.
In fact it is not hard to prove the stronger cyclic version $4^4(a^3b+b^3c+c^3a) \le 27(a+b+c)^4$ or the tougher version $a+b+c=4 \implies a^3b+b^3c+c^3a+\frac{473}{64}abc \le 27$ using exactly the same technique to reduce variables from $3 \to 2$!
| {
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,c\in N$ and $\frac{b}{c}$is in its lowest form.Find the value of $a+b+c+abc$.
I tried to solve this question.Let $\left[\frac{3}{x}\right]=t$,therefore $\left[\frac{4}{x}\right]=5-t$.
$t\leq \frac{3}{x}<t+1$......(i) and $5-t\leq \frac{4}{x}<6-t$.....(ii)
Add (i) and (ii),we get $5\leq\frac{3}{x}+\frac{4}{x}<7$
$5\leq\frac{7}{x}<7$
Then i solved $5\leq\frac{7}{x}$ and $\frac{7}{x}<7$
and took the intersection of the two solution sets and got $x\in (1,\frac{7}{5}]$,so my $a+b+c+abc=48$ but the answer is given to be 20.What is wrong in my method.Please help me.
| So first of all, for $x=4$ you have $1$, for $x>4$ you have $0$. Next, your function has jumps whenever $x=3/n$ or $x=4/m$ for some integers $n,m$. Unless both of these happen at once (for instance, at $x=1$), these jumps are by one unit at a time. Can you find the closest four jumps to the left of $x=4$? (Hint: the first one is at $x=3$.)
Your actual mistake was that you essentially simplified $\left \lfloor \frac{3}{x} \right \rfloor + \left \lfloor \frac{4}{x} \right \rfloor$ into $\left \lfloor \frac{7}{x} \right \rfloor$, which is not correct.
| {
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Simplifying a complication max operation I have dervied an inequality and have arrived to the following
$$\max\{1, \frac{b}{2}+1\} \leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}$$
I am trying to simplify further and arrived to the following conclusions
$$a \geq 1$$
$$ a \geq \frac{b}{2}+1$$
$$ a\geq 2 $$
$$\frac{b}{2}+ \frac{a}{2} \geq 2 $$
How can I proceed and further simplify, are these inequalities redundant?
Thanks
| Starting with
$$\max\{2, \frac{b}{2}+1\}
\leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}
$$
Let's look at the left side.
Case 1:
$2 > \frac{b}{2}+1
$.
Then
$b < 2$.
This then becomes
$2
\le \max\{a, \frac{b}{2}+ \frac{a}{2}\}
< \max\{a, \frac{1}{2}+ \frac{a}{2}\}
= \max\{a, \frac{a+1}{2}\}
$.
If
$a > \frac{a+1}{2}
$,
then
$a < 1
$
so that
this becomes
$2 < 1$
which is false.
Therefore,
$a \le \frac{a+1}{2}
$,
so that
$a \ge 1
$
so that
this becomes
$2 < \frac{a+1}{2}$
or
$a > 1
$
which we already know.
Case 2:
$2 \le \frac{b}{2}+1
$.
Then
$b \ge 2$.
This then becomes
$\frac{b}{2}+1
\le \max\{a, \frac{b}{2}+ \frac{a}{2}\}
=\frac{a}{2}+ \max\{\frac{a}{2}, \frac{b}{2}\}
$.
If
$a \ge b
$,
this becomes
$\frac{b}{2}+1
\le a
$.
But since
$b \ge 2$,
$b \ge \frac{b}{2}+1 $
which is true.
If $a < b$,
this becomes
$\frac{b}{2}+1
< \frac{a}{2}+\frac{b}{2}
$
or
$a > 2$.
In all cases here,
$a \ge 2$.
Therefore,
the solutions are
$b < 2, a > 1$;
$b \ge 2, a \ge 2$.
| {
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} |
convergence proof without finding 'N' I tried to prove $\sqrt{n^2 +n}-n$ converges to $\frac{1}{2}$ I am not sure what I have proved is correct.
$$\left|\frac{n}{\sqrt{n^2+n} +n} - \frac{1}{2} \right| = \left| \frac{2n - 2(\sqrt{n^2+n}+n)}{2(\sqrt{n^2+n}+n)} \right|< \left|\frac{2n - 2\cdot 2n}{2\cdot 2 n} \right|<\left|\frac{2n - 2n}{ 2 n} \right| =0<\epsilon\\\forall \epsilon>0$$
Can I do this way?
| Let $\epsilon>0$ be given. Then, we have
$$\begin{align}
\left|\frac{n}{\sqrt{n^2+n} +n} - \frac{1}{2} \right| &= \left| \frac{2n - \left(\sqrt{n^2+n}+n\right)}{2\left(\sqrt{n^2+n}+n\right)} \right|\\\\
&= \left| \frac{n -\sqrt{n^2+n}}{2\left(n+\sqrt{n^2+n}\right)} \right|\\\\
&= \left| \frac{n -\sqrt{n^2+n}}{2\left(n+\sqrt{n^2+n}\right)}\frac{n +\sqrt{n^2+n}}{n+\sqrt{n^2+n}} \right|\\\\
&= \left| \frac{n^2 -(n^2+n)}{2\left(n+\sqrt{n^2+n}\right)^2} \right|\\\\
&=\left| \frac{-n }{2\left(n+\sqrt{n^2+n}\right)^2} \right|\\\\
&\le\frac{n}{8n^2}\\\\
&=\frac{1}{8n}\\\\
&<\epsilon
\end{align}$$
whenever $n>N=\lceil\frac{8}{\epsilon}\rceil$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
computing an integral (fraction of real powers) I want compute the following integral depending on the real parameters $\alpha, \beta > 0$ and $C >0$
$$ \int_0^1 \frac{u^{2\beta}}{C+u^{2(\alpha+\beta)}} du$$
Thanks a lot for any clue !
| Consider
\begin{align}
I(\alpha, \beta) &= \int_{0}^{1} \frac{u^{2\beta}}{c+u^{2(\alpha + \beta)}} \, du \\
&= \frac{1}{c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \int_{0}^{1} u^{2\beta + 2 n (\alpha + \beta)} \, du \\
&= \frac{1}{2c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \frac{1}{ n \alpha + (n+1) \beta} \\
&= \frac{1}{2c} \, \sum_{n=0}^{\infty} \frac{1}{\beta + (\alpha + \beta) \, n} \, \frac{1}{c^{n}}
\end{align}
This expression is better used for $c$ being large.
For $c$ being small the integral can be done in the following way.
\begin{align}
I &= \int_{0}^{1} u^{1 - 2\alpha} \, \partial_{u}\left( \frac{\ln(c+u^{2(\alpha + \beta)})}{2(\alpha + \beta)} \right) \, du \\
&= \ln c + \frac{2 \alpha -1}{2(\alpha + \beta)} \, \int_{0}^{1} u^{-2 \alpha} \, \ln(c + u^{2(\alpha + \beta)}) \, du \\
&= \ln c + \frac{2 \alpha -1}{2(\alpha + \beta)} \, \int_{0}^{1} u^{-2 \alpha} \, \left[ \ln(c) + \ln\left(1 + \frac{1}{c} \, u^{2(\alpha + \beta)} \right) \right] \, du \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
My attempt:
Let $z = x + iy$
$\displaystyle |z+1|=2 \iff | (x + iy)+1|=2 \iff |(x+1) +iy|=2 \iff (x+1)^2 + y^2 = 4$
Let $w = u + iv$
Now let $\displaystyle w = \frac{1}{z}$ hence we have that \begin{align}z &= \frac{1}{w} \\ &= \frac{1}{u + iv} \\ &= \frac{u-iv}{u^2 + v^2} \\ &= \frac{u}{u^2 + v^2} + i \big( - \frac{v}{u^2 + v^2} \big)\end{align}
From which we can deduce that $\displaystyle x = \frac{u}{u^2 + v^2}$ and $\displaystyle y = - \frac{v}{u^2 + v^2}$
and thus $$\displaystyle \bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4$$
This is where I am stuck. I keep on messing up the simplification. Can someone please show me how to simplify this?
| Proceeding with your result,
$$
\bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ we get
$$
\bigg(\frac{u}{u^2 + v^2}\bigg)^2+2\frac{u}{u^2 + v^2}+1 + \bigg(\frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ or
$$
\frac{u^2}{(u^2 + v^2)^2}+ \frac{v^2}{(u^2 + v^2)^2}+\frac{2u}{u^2 + v^2} = 3
\\$$$$
2u+1 = 3u^2+3v^2
$$ or, with a little algebra,
$$
\left(u-\frac13 \right)^2+v^2=\frac49
$$
which is an easy set to identify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \ge 0
\end{cases}
$$
$$3(\sin^2(x)+\cos^2(x))+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x)$$
$$2\cos^2(x)+\frac{\sin^2(x)}{3}-3\sin^2(x)+\frac{6}{\sqrt 3}\sin(x)\cos(x)=0$$
I multiply by 3 and divide by $\cos^2(x)$:
$$8\tan^2(x)-6\sqrt{3}\tan(x)-6=0$$
Let $t=\tan(x)$, then
$$4t^2-3\sqrt{3}t-3=0$$
$$t_1=\frac{7\sqrt 3}{2}$$
$$t_2=\frac{11\sqrt 3}{4}$$
The solutions for $x$ would be arc-tangents of these values.
But the textbook's answer is
$$\color{green}{x_1=\frac{\pi}{3}+2\pi n; x_2=-\arctan\left(\frac{\sqrt 3}{4}\right)+2\pi n}$$
Where did I make a mistake?
P.S. From the textbook
| The solution of the quadratic equation $4t^2 -3\sqrt3 t -3 = 0$ is wrong. Here, $D = b^2 -4ac$ i.e. D=75.
Hence, $t = \frac{3√3 + or - √D}{2*4}$
$t = \frac{3√3 + 5√3}{8}$
And $t = \frac{3√3 -5√3}{8}$
Hence, $ t = √3, -\frac{\sqrt3}{4}$
Now you will get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integration of the square root of a quadratic I am in the tricky situation of trying to integrate the following.
$$\sqrt{4 a^2 (y-b)^2+c^4}$$
$a, b$ and $c$ are all known constants.
Can anybody provide insight as to how to do this?
I have tried to rearrange to fit the form:
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
But do not seem able to do so. Maybe excessive toiling has hidden an obvious answer from my eyes.
Thanks for the help.
| Before applying Inverse Trigonometric Substitution, I prefer to preprocess the expression as follows
$$\begin{array}{lll}
\int\sqrt{4a^2(y-b)^2+c^4}dy &=& \displaystyle\int\sqrt{\frac{4a^2c^4(y-b)^2}{c^4}+c^4}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{4a^2(y-b)^2}{c^4}+1}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{(2ay-2ab)^2}{c^4}+1}dy\\
&=& \displaystyle \int c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}dy\\
\end{array}$$
Next, making the substitution
$$u=\frac{2ay-2ab}{c^2}\implies u'(y) = \frac{2a}{c^2}$$
we have
$$\int\frac{c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}du(y)}{\frac{2a}{c^2}} = \frac{c^4}{2a}\int \sqrt{u^2+1}du=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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.