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HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$ As the L.c.m contains $2^7$ as a factor and G.c.d contains $2^4$ as a factor , we can assure that one element contains $2^7$ and the other one contains $2^4$. And $3$ and $5^3$ will be divided between two numbers in a manner so that the L.C.M and G.C.D remain as given.
So the possible pairs are $(2^7 .3 . 5^3 , 2^4)$ , $(2^7 .3 , 5^3.2^4)$ , $(2^7 , 3.5^3.2^4)$ , $(2^7.5^3 , 3.2^4)$.
So I think the number of elements in the set $S$ is $4$. Have I gone wrong anywhere? Can anyone please help me ?
| WLOG let $\dfrac xX=\dfrac yY=16$ so that $(X,Y)=1$
We have $48000\cdot16=xy=16^2XY\iff XY=3000=3\cdot2^3\cdot5^3$
So, the possible values of $X$ can be
take none of the factors $$1$$
take one of the factors $$3;2^3;5^3$$
take two of the factors $$2^3\cdot5^3;3\cdot5^3;2^3\cdot3$$
take all three of the factors $$3\cdot2^3\cdot5^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the second derivative of y in $y^2 = x^2 + 2x$? I have a problem to solve:
use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^2y}{dx^2}$. Write the solutions in terms of x and y only
It means that I need to differentiate the equation one time to find $y'$ and then once more to find $y''$.
The correct answer from the textbook is $y' = \frac{x + 1}{y}$ and $y'' = \frac{x^2 + 2x}{y^3}$. I got the first derivative right, but I can't understand how did they get the second one, or is it a typo (unlikely), since I have $y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}$
I did this:
$$
y^2 = x^2 + 2x\\
2yy' = 2x + 2\\
yy' = x + 1\\
y' = \frac{x + 1}{y}\\
$$
I tried to get to the second derivative from both $yy' = x + 1$, $y' = \frac{x + 1}{y}$ and $2yy' = 2x + 2$. But every time I had that dangling constant (1 or 2), which lead to the dangling $\frac{1}{y}$ in my answer.
Like here:
$$
yy' = x + 1\\
y'y' + yy'' = 1\\
yy'' = 1 - (y')^2\\
y'' = \frac{1 - (y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(\frac{x + 1}{y})^2}{y}\\
y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}
$$
I don't see any way to get from my answer to the textbook's one with a transformation, no way to get rid from y in the numerator. And the correct answer doesn't have a "y" there.
Could someone either point to an error in my solution, or corroborate the suspicion that it indeed may be a typo.
| From $y'y'+yy''=1$ multiply by $y^2$.
Then $(yy')^2+y^3y''=(x+1)^2+y^3y''=y^2=x^2+2x \iff y^3y''=x^2+2x-x^2-2x-1=-1$
If we continue your calculation
$y''=\dfrac 1y-\dfrac{(x+1)^2}{y^3}=\dfrac{y^2-(x+1)^2}{y^3}=\dfrac{(x^2+2x)-(x^2+2x+1)}{y^3}=\dfrac{-1}{y^3}$
Gives the same result, so I guess the textbook result is erroneous (i.e. it gives $yy''=1$ which does not agree with derivatives of $\pm\sqrt{x^2+2x}$)
| {
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
| As advised by Maximilian Janisch, you should use the $\tan x$ formula rather $\tan^{-1}x$:
$$\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)\right]=\tan\left[\dfrac{\pi}{4}\right] \iff \\
\frac{\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]-\frac1{239}}{1+\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]\cdot \frac1{239}}=1 \iff \\
\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]=\frac{120}{119} \iff \\
\frac{2\tan\left[2\tan^{-1} \left(\dfrac{1}{5}\right)\right]}{1-\tan^2\left[2\tan^{-1} \left(\dfrac{1}{5}\right)\right]}=\frac{120}{119} \iff \\
\frac{2\cdot \frac{2\cdot \frac15}{1-\frac1{5^2}}}{1-\left[\frac{2\cdot \frac15}{1-\frac1{5^2}}\right]^2}=\frac{120}{119} \iff \\
\frac{\frac5{6}}{1-\frac{25}{144}}=\frac{120}{119} \ \checkmark$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
| Let $a_n = 5^{2n+1} - 3^{2n+1} - 2^{2n+1} = 5 \cdot 25^n - 3 \cdot 9^n - 2 \cdot 4^n$. Then $a_{n+3} = 38 a_{n+2} - 361 a_{n+1} + 900 a_n$ (*). Therefore, you only need to check the claim for $n=0,1,2$, which is immediate.
(*) Because $(x-25)(x-9)(x-4) = x^3 - 38 x^2 + 361 x - 900$. The coefficients are not important here. The important point is that $a_n$ satisfies a linear recurrence with integer coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$ $a,b,c,d > 0$, then find the solution for
$$a+b+c+d = 4$$
and
$$ \left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$$
Attempt:
We know that $a=b=c=d=1$ is a solution. Are there any other solutions? By AM-GM,
$$ \frac{4}{4} \ge (abcd)^{1/4} \implies abcd \le 1 $$
Also by HM-GM we have
$$ \frac{4}{ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} } \le \left( a^{12} b^{12} c^{12} d^{12} \right)^{1/4} $$
$$ \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} $$
then
$$ 4 \le \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} $$
and of course $1 + 3abcd \le 4$.
Next,
$$ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} =
\frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}} $$
so we have
$$\left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right)
= \frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}}
+ \frac{ 3(bcd)^{12} + 3(acd)^{12} + 3(abd)^{12} + 3(abc)^{12} }{(abcd)^{11}}
= 16 $$
| Note that
\begin{align*}
\left(\frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}\right)(1 + 3abcd)
&\geq \frac{4}{(abcd)^3}(1 + 3abcd) \\
&= \frac{4}{(abcd)^3} + \frac{12}{(abcd)^2} \\
&\geq 4 + 12 \\
&= 16
\end{align*}
where the first inequality holds by AM-GM, and the last inequality holds since $abcd \leq 1$ (again established by AM-GM). But by the equality conditions of AM-GM, equality is attained in each of these inequalities only when $a, b, c, d$ are equal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality:
$$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$
and I calculated partial derivatives:
$$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$
$$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$
I noticed that:
$$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$
I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).
| Since $x^3$ is convex for $x\ge 0$, we have that $x^3 \ge 1 + 3(x-1)$, using the tangent at $x=1$.(*) With this observation, it is enough to prove that
$$
\sum_{cyc} (3x-2)(1+x)\geq \frac{3}{4}(1+x)(1+y)(1+z)
$$
Expanding the terms, and using $x y z = 1$, gives the equivalent
$$
-30 + \sum_{cyc} x + 12 \sum_{cyc} x^2 - 3 \sum_{cyc} xy \ge 0
$$
Now note
$$
3 \sum_{cyc} x^2 - 3 \sum_{cyc} xy = 3(\frac12 \sum_{cyc} x^2 +\frac12 \sum_{cyc} y^2 - \sum_{cyc} xy ) = \frac{3}{2}(\sum_{cyc} (x-y)^2) \ge 0
$$
So it is enough to show
$$
-30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge 0
$$
For the two sums in here, we have by AM-GM:
$\sum_{cyc} x \ge 3 \sqrt[3]{xyz} = 3$
and by Titu's Lemma (Cauchy - Schwarz)
$\sum_{cyc} x^2 \ge \frac13 (\sum_{cyc} x )^2 \ge \frac93= 3$
So we have
$$
-30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge -30 + 3 +9 \cdot 3 = 0
$$
which proves the claim. $\qquad \Box$
(*) Alternatively, if you do not want calculus arguments, $x^3 \ge 1 + 3(x-1)$ can be shown elementary. Let $y = x-1$. Then we have $x^3 = (1+y)^3 = 1 + 3y + y^2(3+y)$
and this is $\ge 1 + 3y$ as long as $y \ge -3$ or $x \ge -2$, which is given since we are interested in $x \ge 0$.
| {
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"url": "https://math.stackexchange.com/questions/3444270",
"timestamp": "2023-03-29T00:00:00",
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$x^4 + 1020x^3 - 4x^2 + 2039x+1$ is divisible by 1019. Give all integers x in the range [1,2018] such that $x^4 + 1020x^3 - 4x^2 + 2039x+1$
is divisible by 1019.
Applying mod 1019 to the coefficients, we have
$x^4 + x^3 - 4x^2 + x + 1$ which is equal to $(x-1)^2(x^2+3x+1)$
I tried making 1 of the factors equal to a multiple of 1019, but it looks tedious and i have not found a solution other than 1020.
answer given : 1, 492, 524, 1010, 1511, 1543
| As $1019$ is prime, one of the factors must be $\equiv 0\pmod{1019}$. For $(x-1)$ this obviously means that $x=1$ or $x=1020$.
For the factor $x^2+3x+1$, we'd expect two solutions $x=\frac{-3\pm\sqrt{5}}2$ - but what is $\sqrt 5$ in modular arithmetic? Any $y$ with $y^2\equiv 5\pmod{1019}$. Fortunately, just playing with $1019+5=1024$, we see that $32^2\equiv5\pmod{1019}$. Hence $\frac{-3+32}{2}\equiv\frac{1016+32}{2}=524$ and $\frac{-3-32}{2}\equiv\frac{1016-32}{2}=492$ are remainders mod $1019$ that solve too (i.e., $492$, $524$, $1511$,$1543$)
| {
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| By Minkowski (triangle inequality) we obtain:
$$\sum_{cyc}\sqrt{2a^2+b+1}=\sum_{cyc}\sqrt{2a^2+\frac{b(a+b)}{2}+\frac{(a+b)^2}{4}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{9a^2+4ab+3b^2}=\frac{1}{2}\sum_{cyc}\sqrt{7a^2+b^2+8}\geq$$
$$\geq\frac{1}{2}\sqrt{7(a+b)^2+(b+a)^2+8(1+1)^2}=4.$$
We see that our inequality is true for all reals $a$ and $b$ such that $a+b=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at $(a,b)$ Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at (a,b) when $a<b$
$a-b = ...$
Find gradient of the line, df(x) / dx
$\frac{(4x+3)3 - (3x-6)(4)} {(3x-6)^2}$
Which the same as $ -11/3$
$(12x+9 - 12x + 24 ) 3= -11(3x-6)^2$
If i solve x, literally i solve $a $ right?
Is there less complicated way to solve it?
| Rewrite
$$f(x)=\frac43+\frac{11}{3x-6}.$$
Now
$$f'(x)=-\frac{11}{(3x-6)^2}\cdot3$$
and $f'(x)=-11/3$ gives $(3x-6)=\pm1$, that is $x=7/3$ and $x=5/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve complex equation $\left(\frac{8}{z^3}\right) - i = 0$ from $$(a^3 + b^3) = (a+b)(a^2-ab+b^2)$$
I have $${(2/z)}^3 + i^3 =0$$
I have $$\left(\frac{2}{z} + i\right)\left(\left(\frac{2}{z}\right)^2-(2/z)(i)-1)\right) = 0$$
i.e.$ \left(\frac{2}{z}\right)+i = 0 $ or $ \left(\left(\frac{2}{z}\right)^2-(\frac{2}{z})(i)-1\right) = 0$
......
but I'm not sure that is the correct answer;
help me please.
Thank you.
| Since $i^{3} = -i$
Then $(\frac{2}{z})^{3} + i^{3} = 0 \iff (\frac{2}{z})^{3} - i = 0 \iff z^{3} = -2^{3}i = -8i $
Then the solutions are $z_{1} = 2i, z_{2} = 2e^{i\frac{2\pi }{3}}i, z_{3} = 2e^{i \frac{4\pi }{3}}i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450012",
"timestamp": "2023-03-29T00:00:00",
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Proving an interesting inequality with square roots Let $a,b,c>0$ be real numbers such that $c \geq a \geq b$ and $a^2 \geq bc$. Show that
$$\frac{\sqrt{a^2b+b^2c}}{a+c}+\frac{\sqrt{b^2c+c^2a}}{b+a}+\frac{\sqrt{c^2a+a^2b}}{c+b} \geq \frac{\sqrt a +\sqrt b +\sqrt c}{2}.$$
I tried to augment each term on the left using what we have in the hypothesis:
$$\frac{\sqrt{a^2b+b^2c}}{a+c} \geq \frac{\sqrt{a^2b+b^2c}}{2c} \geq \frac{\sqrt{a^2b+b^2a}}{2c}=\frac{\sqrt{ab(a+b)}}{2c}$$
but this kind of inequality doesn't get me anywhere. Could you give me a hint?
| Even the following inequality is true for any positives $a$, $b$ and $c$:
$$\sum_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}.$$
Indeed, let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, by Holder
$$\sum_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{\sqrt{a^2b+b^2c}}{a+c}\right)^2\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}{\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}}\geq$$
$$\geq \sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}\frac{b^2(a+c)^2}{a^2+bc}}}=\sqrt{\frac{(x^2+y^2+z^2)^3}{\sum\limits_{cyc}\frac{y^4(x^2+z^2)^2}{x^4+y^2z^2}}}$$ and it's enough to prove that
$$2(x^2+y^2+z^2)^3\geq(x+y+z)^2\sum\limits_{cyc}\frac{y^4(x^2+z^2)^2}{x^4+y^2z^2},$$
which is obviously true by BW:
https://www.wolframalpha.com/input/?i=2%28x%5E2%2By%5E2%2Bz%5E2%29%5E3-%28x%2By%2Bz%29%5E2%28y%5E4%28z%5E2%2Bx%5E2%29%5E2%2F%28x%5E4%2By%5E2z%5E2%29%2Bz%5E4%28x%5E2%2By%5E2%29%5E2%2F%28y%5E4%2Bx%5E2z%5E2%29%2Bx%5E4%28y%5E2%2Bz%5E2%29%5E2%2F%28z%5E4%2Bx%5E2y%5E2%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
| {
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finding a constant $c$ so that $ \hat{f} (m)=0 $ Let be $ c \in (0, 2 \pi) $
and
$$f_c(x):= \begin{cases} \frac{x}{c} , 0 \leq x \leq c \\ \frac{2 \pi -x}{2 \pi -c}, c < x \leq 2 \pi \end{cases} $$
I want to determine $c$,
so that
$\hat{f} (m)=0 $
for $ m \in 7 \mathbb{Z} \backslash \{ 0\} $
it is
$ \hat{f}(m)= \frac{1}{2 \pi} \int_0^{2 \pi} f(x) e^{-imx} dx $
$= \frac{1}{2 \pi} [\int_0^c \frac{x}{c} e^{-imx} dx + \int_c^{2 \pi} \frac{2 \pi -x}{2 \pi -c} e^{-imx} dx] $
=$\frac{1}{2 \pi}[ [ \frac{(imx+1) e^{-imx}}{cm^2} ]_0^c +[ \frac{(im(x- 1 \pi)+1)e^{-imx}}{(c-2 \pi )m^2}]_c^{2 \pi} $
=$ \frac{1}{2 \pi} [ \frac{(cm-i)sin(cm)+(icm+1)cos(cm)-1}{cm^2}] + \frac{1}{2 \pi} [- \frac{((c- 2 \pi)m-i)sin(cm)+((ic-2i \pi )m+1)cos(cm)+isin(2 \pi m)-cos(2 \pi m)}{(c-2\pi)m^2}] $
i was trying to reshape the equation $ \hat{f} (m) =! 0 $
but i get stuck at determining a $c$.
And what will it mean for $ m \in 7 \mathbb{Z} \backslash \{0 \} $
do you see a mistake here? Or is there an other approach?
Would be very thankful for any help!
| We have
\begin{align}
\hat{f}(m) &= \frac{1}{2\pi}\int_0^c \frac{x}{c}\mathrm{e}^{-\mathrm{i}mx} dx
+ \frac{1}{2\pi}\int_c^{2\pi} \frac{2\pi - x}{2\pi - c}\mathrm{e}^{-\mathrm{i}mx} dx\\
&= \frac{1}{2\pi (2\pi - c) m^2}\left(1 - \mathrm{e}^{-\mathrm{i}2m\pi}\right)
- \frac{1}{(2\pi - c) cm^2}\left(1 - \mathrm{e}^{-\mathrm{i}cm}\right).
\end{align}
Remark: I used Maple to simplify the expressions.
For $m \in 7 \mathbb{Z} \backslash \{ 0\}$, we have
$$\hat{f}(m) = -\frac{1}{(2\pi - c) cm^2}\left(1 - \mathrm{e}^{-\mathrm{i}cm}\right).$$
Since $\hat{f}(m) = 0$ for all $m \in 7 \mathbb{Z} \backslash \{ 0\}$,
we have $c = \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}, \frac{8\pi}{7}, \frac{10\pi}{7}, \frac{12\pi}{7}$.
| {
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Solve the equation exponential radical Solve the equation$$31+8\sqrt{15}=(4+\sqrt{15})^x$$for $x$.
I think you could set up a recursion from the coefficients of $(4+\sqrt{15})^{x}$ to the coefficients of $(4+\sqrt{15})^{x+1}$ then find the general formula using the characteristic equation?
It should look like $a_{n+1}=4a_{n}+15b_{n}$ and $b_{n+1}=4b_{n}+a_{n}$, cancel out one of variables, and then take the characteristic equation.
| If $x = \log_{4+\sqrt {15}} (31+8\sqrt {15})$ is not an acceptable solution (and they should specify that it is not; it satisfies all the requirements of a solution; It exists and it is a unique value and it solves) then I'm not really sure there is anything to do but guess.
We can note for any positive integer $k$ that $(4+\sqrt{15})^k = \sum_{j=0}^k (\sqrt{15})^2 \sqrt{15}^j*4^{k-j}C_{k,j}=$
$\sum_{j=0;j\text{ even}}^k 15^{\frac j2}*4^{k-j}C_{k,j} + \sqrt{15}\sum_{j=0;j\text{ odd}}15^{\frac{j-1}2}4^{k-j}C_{k,j}$
So if we can get $31 = 4^{2m} + 15*4^{2m-2}C_{2m\{+1\},2} + .... + \{4;1\}\{2m;1\}15^m$.
And $8 = 4^{2m\pm 1} + 15*4^{2m-3;-1}C_{whatever} + .....$ we'd .... have something.
Now
$31 = 15 + 16 = (\sqrt {15})^2 + 4^2$ and $8\sqrt{15} = 2*4*\sqrt{15}$.
So $31+8\sqrt{15} = 4^2 + 2*4*\sqrt{15} + \sqrt{15}^2 = (4+\sqrt{15})^2$
So $x=\log_{4+\sqrt{15}} (4+\sqrt{15})^2 = 2$.
.....
I don't really like this because it is basically guessing.
But it does seem reasonable that if $x$ is not an integer then $(4+\sqrt {15})^x = \text{a mess}$. An if $x$ is an integer we must have $31 =$ some power of $4 + $ some power of $15$ + several combinations thereof. Well $8\sqrt{15} =\sqrt{15}($ combination of powers of $4$ and powers of $15$).
And given that $8\sqrt{15} = 2*4\sqrt{15}$ and $31 = 4^2 + 15$ and $x=2$ is a good.
Now, confession.... did I see it right away. Not really. I first tried factoring and got $31+ 8\sqrt{15} = 8(4+\sqrt {15}) -1$ which was odd. Then I figure $31 + 8\sqrt{15} = 31 + 2*4\sqrt{15} = 15 + 16+2*4\sqrt{15} + 16+2*4\sqrt{15} + \sqrt{15}^2$>
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
What's wrong in my calculation of $\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \frac{k}{n^2}$ I have the following limit to find:
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2}$$
This is what I did:
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2} = \lim\limits_{n \to \infty} \bigg ( \arcsin \dfrac{1}{n^2} + \arcsin \dfrac{2}{n^2} + ... + \arcsin \dfrac{n}{n^2} \bigg )$$
$$ \hspace{.8cm} = \arcsin 0 + \arcsin 0 + ... + \arcsin 0 $$
$$= 0 + 0 + ... + 0 \hspace{2.9cm}$$
$$=0 \hspace{5.2cm}$$
However, my textbook claims that the actual answer is in fact $\dfrac{1}{2}$. I don't see how I could reach this answer.
| As noted by others, there are infinite many summands, one cannot simply distribute the limit operator to them.
The following might be over-killed, but I think it is somehow interesting:
We know that
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\sin^{-1}x}{x}=1,
\end{align*}
given $\epsilon\in(0,1)$, there is an $N$ such that
\begin{align*}
1-\epsilon<\dfrac{\sin^{-1}x}{x}<1+\epsilon
\end{align*}
for all $n\geq N$ and $0<x<1/n$.
Note that
\begin{align*}
\sum_{k=1}^{n}\sin^{-1}\left(\dfrac{k}{n^{2}}\right)&=\sum_{k=1}^{n}\dfrac{\sin^{-1}\left(\dfrac{k}{n^{2}}\right)}{\dfrac{k}{n^{2}}}\cdot\dfrac{k}{n^{2}}\\
&=\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{\sin^{-1}\left(\dfrac{k}{n^{2}}\right)}{\dfrac{k}{n^{2}}}\cdot\dfrac{k}{n},
\end{align*}
pluggint to the $\epsilon$-inequality for large $n$, we have
\begin{align*}
(1-\epsilon)\cdot\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k}{n}<\sum_{k=1}^{n}\sin^{-1}\left(\dfrac{k}{n^{2}}\right)<(1+\epsilon)\cdot\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k}{n}.
\end{align*}
Taking $n\rightarrow\infty$, the sum $\dfrac{1}{n}\displaystyle\sum_{k=1}^{n}\dfrac{k}{n}$ is simply the Riemann sum of $\displaystyle\int_{0}^{1}xdx=\dfrac{1}{2}$.
The arbitrariness of $\epsilon\in(0,1)$ gives the limit as $\dfrac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1.
2^5 ends with 2.
3^5 ends with 3.
And so on.
Do I simply add the ending digits to get my answer?
| $1^5+99^5=(1+99)(1-99+99^2-99^3+99^4)=100(\text{positive number})$ what can you see??
Also
$2^5+98^5=(100)(2^4-2^3(98)+2^2(98)^2-2(98)^3+98^4)$
My solution is a special case of lab's
| {
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What is the Basis of the Kernel and the Image $R^4 \to R^3$ where $ f(x,y,z,w) = \left[\begin{array}{ccc}2x + z -w\\x +w\\x +z-2w\end{array}\right]$
I started with matrix $ A = $$\left[\begin{array}{ccc}2 & 0 & 1 & -1\\1 & 0 & 0 & 1\\1 & 0 & 1 &-2\end{array}\right]$
and set it to 0 and the result $\left[\begin{array}{ccc}1 & 0 & 0 & 1 & 0\\0 & 0 & 1 & -3 & 0\\0 & 0 & 0 &0 &0\end{array}\right]$
The basis of the kernel or $\ker(F)$ is $\Biggl\{\left(\begin{array}{c}0\\1\\0\\0\end{array}\right),\left(\begin{array}{c}-1\\0\\3\\1\end{array}\right)\Biggl\} $
With dim = 2
and the basis of the image is $\Biggl\{\left(\begin{array}{c}2\\1\\1\end{array}\right),\left(\begin{array}{c}1\\0\\1\end{array}\right)\Biggl\} $
With Dim = 2
Did I miss up somewhere ?
| You are right indeed we can check that
$$\left[\begin{array}{ccc}2 & 0 & 1 & -1\\1 & 0 & 0 & 1\\1 & 0 & 1 &-2\end{array}\right]\left[\begin{array}{c}0\\1\\0\\0\end{array}\right]=\vec 0$$
$$\left[\begin{array}{ccc}2 & 0 & 1 & -1\\1 & 0 & 0 & 1\\1 & 0 & 1 &-2\end{array}\right]\left[\begin{array}{c}-1\\0\\3\\1\end{array}\right]=\vec 0$$
and by RREF first and third colum of the original matrix are linearly independent.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Prove by mathematical induction that $(3n+1)7^n -1$ is divisible by $9$ for integral $n>0$ $7^n(3n+1)-1=9m$
$S_k = 7^k(3k+1)-1=9P$
$\Rightarrow 7^k(3k+1) = 9P+1$
$S_{k+1} = 7\cdot7^k(3(k+1)+1)-1$
$= 7\cdot7^k(3k+1+3)-1$
$= 7\cdot7^k(3k+1) +21\cdot7^k -1$
$= 7(9P+1)+21\cdot7^k -1$
$= 63P+7+21\cdot7^k -1$
$= 63P+6+21\cdot7^k$
$= 9(7P +2/3+21\cdot7^k/9)$
therefore it is divisible by $9$
So I believe I have done this right but I've ended up with non-integers in the answer which im pretty sure isn't right.
Where have I gone wrong?
Thanks
| You're almost finished.
You assume that
$S_k = 7^k(3k+1)-1$ is divisible by 9, and therefore divisible by 3.
$S_k = 7^k(3k)+ 7^k-1$ is divisible by 3.
Therefore $7^k-1$ is divisible by 3.
let $3x = 7^k-1$.
$7^k=3x+1$
Then in your final steps:
$=63P+6+21\cdot7^k$
$=63P+6+21(3x+1)$
$=63P+63x+27$
$=9(7P+7x+3)$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Derivative of $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ with respect to $\tan^{-1}x$
Derivative of $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ with respect to $\tan^{-1}x$
Method 1
$$
w=\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}\quad\&\quad z=\tan^{-1}x
$$
Put $\theta=\tan^{-1}x\implies\tan\theta=x$
$$
w=\tan^{-1}\frac{|\sec\theta|-1}{x}=\tan^{-1}\frac{|\sec\theta|-1}{\tan\theta}=\tan^{-1}\frac{\dfrac{1}{|\cos\theta|}-1}{\dfrac{\sin\theta}{\cos\theta}}=\tan^{-1}\frac{1-|\cos\theta|}{|\cos\theta|}.\frac{\cos\theta}{\sin\theta}\\
=\tan^{-1}\frac{1\mp\cos\theta}{\pm\cos\theta}.\frac{\cos\theta}{\sin\theta}=\pm\tan^{-1}\frac{1\mp\cos\theta}{\sin\theta}\\
w=\begin{cases}\tan^{-1}\dfrac{1-\cos\theta}{\sin\theta}=\tan^{-1}\tan(\theta/2)\quad;\quad\theta\in(-\pi/2,\pi/2)\\
-\tan^{-1}\dfrac{1+\cos\theta}{\sin\theta}=-\cot^{-1}\cot(\theta/2)\quad\&\quad\text{elsewhere}
\end{cases}\\
=\begin{cases}n\pi+\dfrac{\tan^{-1}x}{2}\quad;\quad\theta\in(-\pi/2,\pi/2)\\
-n\pi-\dfrac{\tan^{-1}x}{2}\quad\&\quad\text{elsewhere}
\end{cases}\\
\frac{dw}{dx}=\dfrac{\pm1}{2(1+x^2)}\\
\frac{dz}{dx}=\frac{1}{1+x^2}\implies\boxed{\frac{dw}{dz}=\pm\frac{1}{2}}
$$
Method 2
$$
\frac{dw}{dx}=\frac{1}{1+\Big(\frac{\sqrt{1+x^2}-1}{x}\Big)^2}.\frac{x.\dfrac{x}{\sqrt{1+x^2}}-(\sqrt{1+x^2}-1)}{x^2}\\
=\frac{x^2}{2(x^2+1-\sqrt{1+x^2})}.\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}}=\frac{1}{2(1+x^2)}\\
\frac{dz}{dx}=\frac{1}{1+x^2}\implies\boxed{\frac{dw}{dz}=\frac{1}{2}}
$$
Why do I seem to get slightly different solutions in methods 1 and 2 ?
| In the first method for $\theta\in(-\pi/2,\pi/2)$ we have $|\cos\theta|=\cos\theta$ therefore only plus sign holds.
Second method seems more effective and clear to me.
| {
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$(a+bi)$ of $\frac{3+4i}{5+6i}$ and $i^{2006}$ and $\frac{1}{\frac{1}{1+i}-1}$ How can one get the Cartesian coordinate form $(a+bi)$ of the following complex numbers?
$$\frac{3+4i}{5+6i}$$
$$i^{2006}$$
$$\frac{1}{\frac{1}{1+i}-1}$$
Regarding $\frac{3+4i}{5+6i}$ I tried expanding it with $\frac{5+6i}{5+6i}$ and got $\frac{-9+38i}{-9+60i}$, but that doesn't get me anywhere.
Regarding $i^{2006}$ I have $i^{2006} = -1$ (because $2006$ is an even number and $i^2 = -1$). The Cartesian coordinate form would then be $-1 + 0i$. So the real part is $-1$ and the imaginary is $0i$ is that correct?
Regarding $\frac{1}{\frac{1}{1+i}-1}$ I tried expanding the denominator with $\frac{1+i}{1+i}$ and got $\frac{1}{\frac{1}{1+i}-\frac{1+i}{1+i}} = \frac{\frac{1}{1}}{\frac{1+i}{i^2}} = \frac{-1}{1+i}$, but how do I continue from there?
| For $\frac{3+4i}{5+6i}$ multiply the numerator and denominator by the conjugate of the denominator, $5 - 6i$, what do you observe ?
For $i^{2006}$. If $i^2 = -1$, what can you say about $i^{2006} = (i^2)^{1003} = (-1)^{1003} $ ?
For $\frac{1}{\frac{1}{1+i}-1}$, multiply numerator and denominator by $1+i$, you get $\frac{1+i}{-i} = -\frac{1}{i} - 1 = -1 +i$
| {
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Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
| There are indeed other ways to do this. Your solution is great, but if you were just curious about another method, here is one:
$$
\begin{align}
\sqrt{7} + \sqrt{10} \quad &? \quad \sqrt{3} + \sqrt{19} \\
\sqrt{10} - \sqrt{3} \quad &? \quad \sqrt{19} - \sqrt{7}
\end{align}
$$
Note that instead of comparing $a$ and $b$ directly, we can just compare these values.
Define the function
$$
f(x) = \sqrt{9x+10} - \sqrt{4x+3}
$$
We do this because $f(0) = \sqrt{10} - \sqrt{3}$ and $f(1) = \sqrt{19} - \sqrt{7}$.
The magic step is now figuring out that for all positive $x$, this function is increasing, which tells us that $f(1) > f(0)$.
Of course, seeing that this function is increasing is not exactly obvious, but it is not a difficult task if you have a calculus background.
Perhaps there is another step we can take or a different function we can use that would make the fact it is increasing more obvious?
| {
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Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y\cos y + \sin y + \cos y + 1)$
$= 2(\sin y\cos y + \sin y + \cos y + \sin^2 y + \cos^2 y) (\sin^2 y + \cos^2 y = 1)$
$= 2(\sin^2 y + \sin y\cos y + \sin y + \cos^2 y + \cos y)$
I got stuck here. I do not know what to do from here.
I have tried and tried several days even contacted friends but all to no avail.
| Expanding the RHS,
$$\color{blue}{(\sin y + \cos y + 1 )^2} = \sin^2 y + \cos^2 y + 1 +2\cos y \sin y + 2\cos y + 2\sin y$$
$$= 1+ 1 +2\cos y \sin y + 2\cos y + 2\sin y=2(1+\cos y \sin y + \cos y + \sin y) = 2 \left[(1+\cos y) + (\sin y + \cos y \sin y ) \right] = 2 \left[(1+\cos y) + \sin y(1 + \cos y) \right]=\color{blue}{2(1+\cos y)(1+\sin y )}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Evaluating the limit $\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$ $$\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$$
This is a compound question from me.
*
*I don't know how to begin evaluating this limit. My guess would be that I would have to find the value of this Riemann's integral and then plug the result into the limit. Is this the right direction to head?
Which brings me to...
*I am also stuck trying to resolve the integral. I tried integrating by substitution, trying with both $u = t^2$ and $u = \sin(t^2)$, but both have lead me to finding that $t$ or $dt$ popping back into the equation sooner or later and I'm not quite sure how to handle that. Any hints as to how I can integrate that function?
Thank you.
| Without L'Hopital:
\begin{align*}
\dfrac{1}{x^{3}}\int_{0}^{x}\sin(\sin t^{2})dt=\dfrac{1}{x^{3}}\left(x\sin(\sin x^{2})-\int_{0}^{x}t\cos(\cos t^{2})2tdt\right).
\end{align*}
Note that
\begin{align*}
\dfrac{1}{x^{3}}(x\sin(\sin x^{2}))=\dfrac{\sin(\sin x^{2})}{\sin x^{2}}\dfrac{\sin x^{2}}{x^{2}}\rightarrow 1.
\end{align*}
On the other hand,
\begin{align*}
\int_{0}^{x}t\cos(\cos t^{2})2tdt=\dfrac{2}{3}x^{3}\cos(\cos x^{2})-\dfrac{2}{3}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt.
\end{align*}
And we have
\begin{align*}
-\dfrac{\dfrac{2}{3}x^{3}\cos(\cos x^{2})}{x^{3}}=-\dfrac{2}{3}\cos(\cos x^{2})\rightarrow-\dfrac{2}{3},
\end{align*}
whereas fot the integral, by the change of variable $u=t^{4}$, we obtain that
\begin{align*}
\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt=\dfrac{1}{2}\int_{0}^{x^{4}}u\sin(\sin u^{1/2})\dfrac{du}{u^{3/4}}=\dfrac{1}{2}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du,
\end{align*}
and that
\begin{align*}
\dfrac{1}{x^{3}}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt&=\dfrac{1}{2}\cdot x\cdot\dfrac{1}{x^{4}}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du\\
&=\dfrac{1}{2}\cdot x\cdot\eta_{x}^{1/4}\sin(\sin\eta_{x}^{1/2})\\
&\rightarrow 0,
\end{align*}
where $\eta_{x}\in[0,x]$ is chosen by Integral Mean Value Theorem, therefore the whole limit is $1-2/3=1/3$.
| {
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
| You're asking to prove, for integers $m$ and $n$ (with the assumption $n \neq 0$), that
$$\left|\sqrt{3}-\frac{m}{n}\right| \le \frac{1}{5n^2} \tag{1}\label{eq1A}$$
Note if $m = 0$, \eqref{eq1A} obviously holds. Otherwise, as this other answer states, WLOG, we may assume both $m$ and $n$ are positive since if they have opposite signs, the result is trivial, and if they are both negative, the result is the same as if they were both their absolute value equivalents instead.
As you've shown by rationalizing the numerator and stating it must be at least $1$ is that you have
$$\left|\sqrt{3}-\frac{m}{n}\right| = \left|\frac{3n^2-m^2}{\sqrt{3}n^2 + mn}\right| \ge \frac{1}{\sqrt{3}n^2 + mn} \tag{2}\label{eq2A}$$
If the denominator on the right side is $\le 5n^2$, then you get
$$\begin{equation}\begin{aligned}
\sqrt{3}n^2 + mn & \le 5n^2 \\
\frac{1}{\sqrt{3}n^2 + mn} & \ge \frac{1}{5n^2}
\end{aligned}\end{equation}\tag{1}\label{eq3A}$$
so combined with \eqref{eq2A}, this shows \eqref{eq1A} will be true.
Consider instead that the denominator is $\gt 5n^2$ to get
$$\begin{equation}\begin{aligned}
\sqrt{3}n^2 + mn & \gt 5n^2 \\
mn & \gt (5 - \sqrt{3})n^2 \\
m & \gt (5 - \sqrt{3})n \\
\frac{m}{n} & \gt 5 - \sqrt{3} \\
-\frac{m}{n} & \lt - 5 + \sqrt{3} \\
\sqrt{3} -\frac{m}{n} & \lt - 5 + 2\sqrt{3} \lt -1.5 \\
\left|\sqrt{3}-\frac{m}{n}\right| & \gt 1.5 \gt \frac{1}{5n^2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
As such, \eqref{eq1A} will still hold in this case as well. Since all possibilities have been covered, it proves \eqref{eq1A} is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$
I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?
| We know that
$$\begin{align}
\cos(2x) &= \cos^2(x) + \sin^2(x)\\
\sin(x) &= \dfrac{\tan(x)}{\sec(x)}\\
\sec^2(x) &= 1 + \tan^2(x)
\end{align}$$
So,
$$\int\dfrac{\cos(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm dx \equiv \int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx$$
Let $u = \tan(x)$. So, $\dfrac{\mathrm du}{\mathrm dx} = \sec^2(x)\to\mathrm dx = \dfrac{\mathrm du}{\sec^2(x)}$.
$$\implies\int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx\equiv-\int\dfrac{(u - 1)(u + 1)}{u^4 + 1}\mathrm du$$
Now, you should be able to factor the denominator and use partial fractions to get the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Uniform Continuity of a function with the simplest way I'm trying to show that
$f(x) =
\begin{cases}
x \sin\left(\frac1x\right),\quad\text{if $x \in (0,1]$ }\\[2ex]
0, \quad \quad \quad \quad \ \text{if $x=0$}
\end{cases}$
is uniformly continuous on $[0,1]$
Let $\epsilon \gt 0$ and let $x, y \in (0,1)$.
Then
$$
\left|x\sin\frac{1}{x} - y \sin\frac{1}{y} \right|=\left| x\sin\frac{1}{x} - y\sin\frac{1}{x} + y\sin\frac{1}{x} - y \sin\frac{1}{y} \right|
= \left| (x-y)\sin\frac{1}{x} + y \left(\sin\frac{1}{x} - \sin\frac1y\right) \right| $$
and by triangle inequality
\begin{equation}
\label{eq: star}
\left | x\sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x - y| + y \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |
\end{equation}
$$\left | \sin\frac{1}{x} -\sin\frac{1}{y} \right|= \left | 2 \cos \left ( \frac{1}{2} \left ( \frac{1}{x} + \frac{1}{y} \right ) \right ) \sin \left ( \frac{1}{2} \left ( \frac{1}{x} - \frac{1}{y} \right ) \right ) \right |$$
I could not continue from there. Is there any basic way for this? I've seen other answers on this question but I'm looking for really simple one for formal proof.
Thanks for any help.
| Here is an ad-hoc proof. Let $\epsilon>0$ be fixed. We can and do adjust it to $1$ if it is bigger.
The function $|f'|$ is bounded and continuous on $[\epsilon/3, \; 1]$, let $M> 3$ be an upper bound.
We set $\delta = \epsilon/M<\epsilon/3$.
Let $x,y$ be two points in $[0,1]$ at distance $<\delta$, and we can and do assume $0\le x\le y\le x+\delta$. Two cases:
*
*If $x$ is in the interval $[0,\epsilon/3]$ then we have:
$$|f(x)-f(y)|\le |f(x)|+|f(y)|\le x+y\le 2x+(y-x)<
\frac 23\epsilon +\delta<\epsilon\ .
$$
*
*If $x>\epsilon/3$ then both $x,y$ live in the interval $[\epsilon/3,\; 1]$ where we can find an intermediate point $\xi$ with
$$|f(x)-f(y)|=|f'(\xi)|\;|x-y|<M|x-y|<M\delta<\epsilon\ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $\int\limits_0^1\frac{\sqrt{x}}{(x+3)\sqrt{x+3}}dx$ without trigonometric substitution? I have the following integral to solve:
$$\int_0^1 \dfrac{\sqrt{x}}{(x+3)\sqrt{x+3}}dx$$
without using trigonometric substitution. My textbook gives me the following hint:
$$t = \sqrt{\dfrac{x}{x+3}}$$
But I don't see how this would help me. If I differentiate that, I get:
$$dt = \dfrac{1}{2} \cdot \sqrt{\dfrac{x+3}{x}} \cdot \dfrac{3}{(x+3)^2} dx$$
$$dt = \dfrac{3}{2(x+3)^2} \cdot \dfrac{1}{t} dx$$
And I'm stuck. If I substitute in the original integral, I'll have terms with both $x$ and $t$. So how can I use the given hint?
| Solve for $x$:
$$t^2 = \frac{x}{x+3} = 1 - \frac{3}{x+3} \implies x = \frac{3}{1-t^2} - 3$$
then we have
$$dx = \frac{6t}{(1-t^2)^2}dt$$
and plugging in to the integral gets us
$$ \int_0^{\frac{1}{2}} \left(\frac{1-t^2}{3}\right)\cdot (t) \cdot \left(\frac{6t}{(1-t^2)^2}\right)dt = 2\int_0^{\frac{1}{2}} \frac{t^2}{1-t^2}dt = \int_0^{\frac{1}{2}} \frac{2}{1-t^2}-2dt$$
$$ = \int_0^{\frac{1}{2}} \frac{1}{1+t}+\frac{1}{1-t}-2dt = \log\left(\frac{1+t}{1-t}\right)-2t\Biggr|_0^{\frac{1}{2}} = \log(3)-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find result of system $x+y^2-4=0$ and $y+x^2-4=0$ without using the quartic formula How do I get the values of $x$ and $y$, without using the quartic formula?
\begin{align*}
\begin{cases} x+y^2-4=0 \\ y+x^2-4=0 \end{cases}
\end{align*}
| Subtract the first equation from the second to obtain:
$$x^2-y^2+y-x=0\Leftrightarrow \\
(x-y)(x+y-1)=0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the integrand $\frac{x^3}{\sqrt{x^2+10x+16}}$ I just wanted to make sure that what I did to integrate $\frac{x^3}{\sqrt{x^2+10x+16}}$ is correct.
I assumed that it is classified as a trigonometric substitution problem. And so, what I first did is to apply "completing the square":
$\int \frac{x^3 dx}{\sqrt{x^2+10x+16}} = \int \frac{x^3 dx}{\sqrt{x^2+10x+25+16-25}} = \int \frac{x^3 dx}{\sqrt{(x+5)^2-9}}
$
After that, I assigned values into some variables:
let $a = 3 $
$ x + 5 = 3 \sec \Theta \rightarrow \sec \Theta = \frac{x+5}{3} $
$ dx = 3 \sec \Theta \tan \Theta\, d \Theta$
Next, I have substituted the value of (x+5) to $\sqrt{(x+5)^2-9}$ which leads to $3 \tan \Theta$. And, $\tan \Theta$ is equal to $\frac{\sqrt{x^2+10x+16}}{3}$.
Afterwards, I have replaced all of the variables to the values that I assigned to them:
$\int \frac{x^3}{\sqrt{x^2+10x+16}} \rightarrow \int \frac{(3 \sec \Theta)^3 (3 \sec \Theta \tan \Theta)\, d\Theta}{3 \tan \Theta} \rightarrow \int (3 \sec\Theta - 5)^3 \sec\Theta\, d \Theta$
Expanding the trinomial, distributing $\sec \theta$ to each term and applying the constant theorem will result to:
$27\int \sec^4\Theta d\Theta - 135\int sec^3\Theta\, d\Theta + 225 \int sec^2\Theta d\Theta - 125 \int sec \Theta d \Theta$
Now, by making $\sec^2 \Theta$ into $(1 + \tan^2 \Theta)$ and applying u-substitution: $27 \int \sec^4 \Theta \,d\Theta \rightarrow 27 \tan \Theta + 9 \tan^3 \Theta + C$
By applying integration by parts, $\int \sec^3 \Theta\, d\Theta$ would become $\frac{\sec\Theta tan\Theta + \ln\left | \sec\Theta + \tan\Theta \right |}{2}$.
Finally, $\int \sec^2 \Theta\, d\Theta$ would simply become $\tan \Theta$ and $\int \sec \Theta$ would be $ln\left | \sec\Theta + \tan\Theta \right |$ .
Since $\sec \Theta$ is equal to $\frac{x+5}{3}$ and $\tan\Theta$ is equal to $\frac{\sqrt{x^2+10x+16}}{3}$, the whole integral would be (I had added all like terms before this):
$9\left [ \frac{\sqrt{x^2+10x+16}}{3} \right ]^3 + 252\left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) -\frac{135}{2} \left ( \frac{x+5}{3}\right ) \left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) - \frac{385}{2} ln \left | \frac{x+5}{3} + \frac{\sqrt{x^2+10x+16}}{3}\right | + C$
Lastly, I simplified the integral:
$\frac{1}{3} \left ( x^2 + 10x+16 \right )^\frac{3}{2} + \frac{(93-15x)\sqrt{x^2+10x+16}}{2} - \frac{385}{2} ln \left | \frac{x+5+\sqrt{x^2+10x+16}}{3}\right | + C$
Have I integrated the integrand appropriately? Thanks in advanced.
| It is correct. Here is another method without trigonometric substitution:
$$\int\frac{x^3}{\sqrt{x^2+10x+16}}dx=\int\frac{x^3+10x^2+16x-10x^2-100x-160+84x+160}{\sqrt{x^2+10x+16}}dx$$
$$ = \int\frac{(x-10)(x^2+10x+16)+84x+160}{\sqrt{x^2+10x+16}}dx$$
$$ = \int(x-10)\sqrt{x^2+10x+16}\ dx+\int\frac{84x+420}{\sqrt{x^2+10x+16}}dx-260\int\frac{1}{\sqrt{x^2+10x+16}}dx$$
The last integral can be directly solved using formula, second last one will follow if you substitute $t=x^2+10x+16$, remains to solve the first integral.
$$\int(x-10)\sqrt{x^2+10x+16}\ dx = \frac{1}{2}\int(2x+10)\sqrt{x^2+10x+16}\ dx-15\int\sqrt{x^2+10x+16}\ dx$$
Last integral follows directly from formula and second last is solved by substituting $x^2+10x+16$.
Note that this is a general method. In case there is any polynomial in numerator and any quadratic (with or without square root) in denominator, divide the numerator by denominator to get a linear in numerator, then reduce the numerator to a constant times the derivative of quadratic plus another constant, which can be solved easily using substitution and existing formulas.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An interesting contest math problem: find the maximum value of $f(a_1,a_2,...,a_n)$
Suppose the sequence $a_1,a_2,...,a_n$ is a permutation of the sequene $1+2^1,2+2^2,...,n+2^n$. Find the maximum value of $f(a_1,a_2,...,a_n)=\vert a_1-a_2\vert+\vert a_2-a_3\vert+\cdots+\vert a_{n-1}-a_n\vert$.
This is an interesting problem found in a math contest. I used to be a contest math lover when I was a high school student. I know I may solve it by finding the possible regular formula when $n=2,3,4,...$, but the method is not always effective so I hope someone can give me some hints.
By the way, as a student majored in math now, if there exists an advanced method behind it, can anyone share it with me or at least tell me what it may related with. Thanks for your answer.
| Inspired by John Omielan's answer,here is my attempt.
Our goal is to find the maximum value of \begin{equation}\begin{aligned}
f(a_1,a_2,...,a_n) & = \vert a_1-a_2\vert+\vert a_2-a_3\vert+...+\vert a_{n-1}-a_n\vert \\
& = \sum_{i=1}^{n-1}\vert a_{i} - a_{i+1} \vert
\end{aligned}\end{equation}
We can regard $f(a_1,a_2,...,a_n)$ as the foumula with coefficients$$f(a_1,a_2,...,a_n)=x_1a_1+x_2a_2+...+x_na_n$$
An important observation is that the sum of the coefficients $x_1+x_2+...+x_n$ has to be $0$.On the other hand, $a_1$ and $a_n$ appear only once and the other terms $a_i,(0\leqslant i \leqslant n-1)$ appear twice, so we get $x_1,x_n \in \{-1,1\}$, and $x_i\in\{-2,0,2\}$, $(0\leqslant i \leqslant n-1)$.
Now the problem seems to get clear.It's time to discuss $n$ is even or odd now.We first let $b_i=i+2^i,1\leqslant i \leqslant n$
$(1)$ When $n$ is even:
The coefficients $2$ appear ${n\over 2}$ times,the coefficients $-2$ appear ${n\over 2}$ times either.And the coefficients $1$ and $-1$ both appear once.
Suppose $n=2k,k\geqslant 2$
\begin{align}
f(a_1,a_2,...,a_n)&=2(b_{2k}+...+b_{k+2})+b_{k+1}-b_k-2(b_{k-1}+...+b_1)\\
&=2[(2k)+...+(k+1)-k-...-1]+k-(k+1)\\
&+2[2^{2k}+...+2^{2k-1}-2^k-...-2]+2^k-2^{k+1}\\
&=2k^2-1+2^{2k+2}-2^{k+3}-2^k+4\\
&=\frac 12 n^2+4\cdot 2^{n}-9\cdot 2^{\frac n2}+3
\end{align}
$(2)$ When $n$ is odd:
The situation is a bit different but actually both $[1]$ and $[2]$ are the same.
$[1]$ The coefficients $2$ may appear ${n-1\over 2}$ times,while $-2$ appear ${n-3\over 2}$ times,$1$ appear twice and no $-1$.
$[2]$ The coefficients $2$ appear ${n-3\over 2}$ times,while $-2$ appear ${n-1\over 2}$ times,$-1$ appear twice and no $1$.
Suppose $n=2k+1,k\geqslant 1$
\begin{align}
f(a_1,a_2,...,a_n)&=2(b_{2k+1}+...+b_{k+2})-b_{k+1}-b_k-2(b_{k-1}+...+b_1)\\
&=2[(2k+1)+...+(k+2)-(k+1)-k-...-1]+(k+1)+k\\
&+2[2^{2k+1}+...+2^{k+2}-2^{k+1}-...-2]+2^k+2^{k+1}\\
&=2k^2+2k-1+2^{2k+3}-13\cdot 2^{k+1}+4\\
&=\frac 12 n^2+4\cdot2^{n}-13\cdot 2^{n-1 \over 2}+\frac52
\end{align}
So we find the maximal value of it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solution of an algebraic equation having unknown within modulus The question is this:
find the number of real values of $x$ for which $|x-3|+(x-3)^2+\sqrt{x-3}+|x+3|=0.$
Here is my attempt to answer:
Since it's a real equation and $\sqrt{x-3}$ is present here, therefore we must have $x>3$. In this case this equation becomes
$(x-3)+(x-3)^2+\sqrt{x-3}+(x+3)=0$
$\Rightarrow 2x+(x-3)^2+\sqrt{x-3}=0$
$\Rightarrow 2(x-3)+(x-3)^2+\sqrt{x-3}+6=0$
$\Rightarrow (x-3+1)^2+\sqrt{x-3}+5=0$
$\Rightarrow \{(x-2)^2+5\}^2=x-3$
$\Rightarrow (x-2)^4+10(x-2)^2-(x-2)+26=0,$ which is a biquadratic equation.
I want to know whether I am going in the right direction.
| Hint (easier direction):
None of the summands is negative.
Could they all be zero?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$ Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$
My Attempts
$$x^2(xdx+ydy)+2y(xdy-ydx)=0$$
$$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Put $x=r\cos (\theta)$ and $y=r\sin (\theta)$
So, $r^2=x^2+y^2$ and $\tan (\theta)=\dfrac {y}{x}$
Now,
$$x^2+y^2=r^2$$
Differentiating both sides,
$$2xdx+2ydy=2rdr$$
$$xdx+ydy=rdr$$
| $$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Duvide both sides by $({x^2+y^2})$:
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( \dfrac {xdy-ydx}{x^2+y^2} \right )$$
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( d(\arctan (\frac {y}{x})) \right )$$
Susbtitute $\dfrac {y}{x}=z$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2zd(\arctan z)$$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-\dfrac {2zdz}{z^2+1}$$
Integrate.
Edit I didn't pay attention that the right side of the DE was wrong so:
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2\frac {y}{\color{red}{x^2}}d(\arctan \frac {y}{x})$$
$$\frac 12\dfrac {dr^2}{r^2}=-2\frac {r\sin \theta}{r^2 \cos^2 \theta}d\theta$$
More simply:
$${dr}=-2\frac {\sin \theta}{\cos^2 \theta}d\theta$$
Integrate.
| {
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"url": "https://math.stackexchange.com/questions/3488027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int\limits_0^a x^b (1 - c x)^d\ \cosh x\,\mathrm dx$ How does one calculate
$$\frac{2\pi^{\frac{m-1}{2}}}{\Gamma \left(\frac{m-1}{2} \right)} \left(\frac{\alpha}{\kappa} \right)^{\frac{1}{\alpha}-1}\int\limits_0^{\kappa/\alpha}x^{m+\frac{1}{\alpha}-2}\cosh(x)\left(1-\left(\frac{\alpha}{\kappa}x\right)\right)^{\frac{2}{\alpha}}dx ?$$
| Doing basically the same as @Eric Towers, for the integral
$$I=\frac{2\pi^{\frac{m-1}{2}}}{\Gamma \left(\frac{m-1}{2} \right)} \left(\frac{\alpha}{\kappa} \right)^{\frac{1}{\alpha}-1}\int\limits_0^{\kappa/\alpha}x^{m+\frac{1}{\alpha}-2}\cosh(x)\left(1-\frac{\alpha}{\kappa}x\right)^{\frac{2}{\alpha}}\,dx $$
we have
$$I=K\, _2F_3\left(\frac{1}{2 \alpha }+\frac{m-1}{2},\frac{1}{2 \alpha
}+\frac{m}{2};\frac{1}{2},\frac{3}{2 \alpha }+\frac{m}{2},\frac{3}{2 \alpha
}+\frac{m+1}{2};\frac{\kappa ^2}{4 \alpha ^2}\right)$$
where
$$K=\frac 2 {\sqrt \pi}\left(\frac { \kappa\sqrt \pi}{\alpha}\right)^m \,\,\frac{ \Gamma \left(1+\frac{2}{\alpha }\right)\,
\Gamma \left(m+\frac{1}{\alpha
}-1\right)}{\Gamma \left(\frac{m-1}{2}\right)\, \Gamma
\left(m+\frac{3}{\alpha }\right)}$$
provided $\qquad \alpha \kappa >0\land \Re\left(m+\frac{1}{\alpha }\right)>1\land
\Re\left(\frac{1}{\alpha }\right)>-\frac{1}{2}$
| {
"language": "en",
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"source": "stackexchange",
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Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron whose vertices are the given points:
$$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$
Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
| Note that the given volume is a cone with the height 2 and a right isosceles triangle of side 2 as the base. Thus, its volume can be calculated as
$$\frac13 Area_{base} \cdot Height = \frac13 (\frac12 \cdot 2\cdot 2)2=\frac43$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 1
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Hidden random walk in shallow sums? Prove that $\sum_{k=0}^n k \cdot \binom{2n-k}n 2^k = (2n+1) \cdot \binom{2n}n - 4^n $ Motivation: I was perusing through Laurent's solution to a recent online puzzle here.
Basically he managed to prove that, from the expected value of the probability distribution (in that question), we can obtain $\displaystyle \sum_{k=0}^n \binom{2n-k}n 2^k = 4^n $.
Now suppose that I want to find the variance of the same distribution, then I'm left to prove that $\displaystyle \sum_{k=0}^n k \binom{2n-k}n 2^k = (2n+1) \binom{2n}n - 4^n $.
Plugging in the values of $n=1,2,3,4,5,\ldots$ outputs this Random walk sequence.
Which got me curious: Is there a combinatorial proof for this sum? (Actually, I'm interested in any proof)
Further notes: If my claim is correct, then $$\text{Var}[T] = 4n + 2 - \frac{4^n}{\binom{2n}n} - \frac{16^n}{\binom{2n}n ^2} \to n(4-\pi) $$
which agrees with Laurent's claim that the probability distribution of $T$ asymptotically follows a Rayleigh distribution.
I also want to point out that this question is related. That is (upon division by $4^n$),
$$ \frac1{4^n} \sum_{k=0}^n k \binom {2n-k}n 2^k = \frac{2n+1}{4^n} \binom{2n}n - 1 $$
is the expected number of returns in a symmetric random walk of $2n$ moves.
| In evaluating
$$\sum_{k=0}^n {2n-k\choose n} k 2^k$$
we write
$$\sum_{k=0}^n {2n-k\choose n-k} k 2^k
= \sum_{k=0}^n k 2^k [z^{n-k}] (1+z)^{2n-k}
\\ = [z^n] (1+z)^{2n} \sum_{k=0}^n k 2^k z^k (1+z)^{-k}.$$
Now we may extend the sum in $k$ beyond $n$ because there is no
contribution to the coefficient extractor $[z^n]$ in that case:
$$[z^n] (1+z)^{2n} \sum_{k\ge 0} k 2^k z^k (1+z)^{-k}.$$
We also have
$$\sum_{k\ge 0} k w^k = \frac{w}{(1-w)^2}$$
which yields for the sum
$$[z^n] (1+z)^{2n} \frac{2z/(1+z)}{(1-2z/(1+z))^2}
= 2 [z^n] (1+z)^{2n+1} \frac{z}{(1-z)^2}$$
This is
$$2 \sum_{q=0}^n (n-q) {2n+1\choose q}
= 2n \sum_{q=0}^n {2n+1\choose q}
- 2 \sum_{q=0}^n q {2n+1\choose q}
\\ = 2n \frac{1}{2} 2^{2n+1}
- 2 \sum_{q=1}^n q {2n+1\choose q}
\\ = n 2^{2n+1}
- 2 (2n+1) \sum_{q=1}^n {2n\choose q-1}
\\ = n 2^{2n+1}
- 2 (2n+1) \sum_{q=0}^{n-1} {2n\choose q}
\\ = n 2^{2n+1}
- 2 (2n+1) \frac{1}{2} \left(2^{2n}-{2n\choose n}\right)
\\ = (2n+1) {2n\choose n} + 2 n 2^{2n} - (2n+1) 2^{2n}
= (2n+1) {2n\choose n} - 4^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that $\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=3\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx$
Prove without evaluating the integrals that:$$2\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\label{*}\tag{*}$$
Or equivalently:
$$\boxed{\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=3\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx}$$
In contrast we have:
$$\boxed{\int_0^\pi\frac{\ln(1-\sin x)}{\sin x}dx=2\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx}$$
This is of course easily provable by splitting the integral as $\int_0^\frac{\pi}{2}+\int_\frac{\pi}{2}^\pi$ and letting $x\to \pi-x$ in the second part, unfortunately this method doesn't work for the other one.
I am already aware how to evaluate the integrals as we have:
$$\mathcal I= \int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx\overset{\tan \frac{x}{2}\to x}=-2\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-\frac{\pi^3}{8}$$
And the latter integral is evaluated in many ways here, so if you have other approaches please add them there.
Here's how I came up with $\eqref{*}$:
I knew from here that:
$$I\left(\frac{3\pi}{2}\right)=\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx=-\frac{3\pi^2}{8}$$
And since this result is very similar to the one from above, I tried to show that $\mathcal I=\frac{\pi}{3} I\left(\frac{3\pi}{2}\right)$, equivalent to:
$$\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$$
I also noticed that we have:
$$\mathcal J=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\overset{x\to \pi-x}=\int_0^\frac{\pi}{2}\frac{(\pi-x)\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)-\mathcal I$$
$$\Rightarrow \mathcal I+\mathcal J=\int_0^\pi \frac{x\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)=-\frac{3\pi^3}{8}$$
Of course now it's trivial to deduce that $2\mathcal I=\mathcal J$ as we know the result for $\mathcal I$, but I'm interested to show that relationship without making use of the result or by calculating any of the integrals. If possible showing $\eqref{*}$ using only integral manipulation (elementary tools such as substitution/integration by parts etc).
I hope there's a nice slick way to do it as it will give an easy evaluation of the main integral.
| It suffices to show the vanishing integral below
\begin{align}I=& \int^\frac{\pi}{2}_0\frac{(3x-\pi)\ln(1-\sin x)}{\sin x}dx\\
=& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 \frac{(\pi-3x)\cos y}{1-\sin y \sin x}dy\>dx\\
=& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 (\pi-3x)\frac{d}{dx}
\left(2\tan^{-1}\frac{\sin\frac{x-y}2}{\cos\frac{x+y}2} \right)
\overset{ibp}{dx}\> dy\\
=& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 6\tan^{-1}\frac{\sin\frac{x-y}2}{\cos\frac{x+y}2}\>\overset{x \leftrightarrows y}{dxdy}=-I=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 1,
"answer_id": 0
} |
Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method with convex function but no result:
Since $f(x)=\frac{1}{x}$ is a convex function, by Jensen we obtain:
$$\frac{1}{x+y+z}\sum_{cyc}\frac{x^3}{x^2+y^2}=\sum_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{1}{\frac{x^2+y^2}{x^2}}\right)\geq$$
$$\geq\frac{1}{\sum\limits_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{x^2+y^2}{x^2}\right)}=\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}.$$
Thus, it's enough to prove that
$$\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}\geq\frac{1}{2}$$ or
$$x+y+z\geq\sum_{cyc}\frac{y^2}{x},$$ which is wrong.
thanks
| Hint: We have $$ \frac {x^3}{x^2 + y^2}=x-\frac{xy^2}{x^2+y^2}\ge x-\frac{y}{2}$$ because by AM-GM $x^2+y^2\geq 2xy$ so that $$\frac{xy}{x^2+y^2}\le\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\binom{a_1}{2} + \binom{a_2}{2} + \cdots + \binom{a_n}{2} \ge r\binom{k+1}{2} + \left(n-r\right)\binom{k}{2}$
If $a_1,a_2,\cdots,a_n$ are positive integers, and $a_1+a_2+\cdots +a_n=nk+r$, where $k$ and $r$ are integers such that $0\le r<n$, prove that $$\dbinom{a_1}{2} + \dbinom{a_2}{2} + \cdots + \dbinom{a_n}{2} \ge r\dbinom{k+1}{2} + \left(n-r\right)\dbinom{k}{2}$$
Here is what I do :
Consider the function $f(x) = \dbinom{x}{2}$. Since it is convex, then by applying the Jensen's inequality, we have :
$$\frac{\dbinom{a_1}{2}+\dbinom{a_2}{2}+\cdots+\dbinom{a_n}{2}}{n}\ge\dbinom{\frac{a_1+a_2+\cdots+a_n}{n}}{2}$$ $$\Rightarrow \dbinom{a_1}{2}+\dbinom{a_2}{2}+\cdots+\dbinom{a_n}{2}\ge \frac{n}{2}\left(\frac{nk+r}{n}\right)\left(\frac{nk+r}{n}-1\right)$$ $$\Rightarrow \dbinom{a_1}{2}+\dbinom{a_2}{2}+\cdots+\dbinom{a_n}{2}\ge \frac{1}{2}\left(r(k+1)+(n-r)k\right)\left(\frac{nk+r}{n}-1\right)$$
But I am stuck till here, I don't know how to get the form $r\dbinom{k+1}{2} + \left(n-r\right)\dbinom{k}{2}$.
Any help is surely appreciated, Thanks!
|
We obtain with OP's problem setting and applying Jensen's inequality
\begin{align*}
\color{blue}{\sum_{j=1}^n\binom{a_j}{2}}&\geq n\binom{\frac{1}{n}\sum_{j=1}^n a_j}{2}\\
&=n\binom{\frac{1}{n}(nk+r)}{2}\\
&=\frac{n}{2}\left(\frac{nk+r}{n}\right)\left(\frac{nk+r}{n}-1\right)\tag{1}\\
&=\frac{1}{2}\left(nk+r\right)\left(k+\frac{r}{n}-1\right)\\
&\,\,\color{blue}{=\frac{1}{2}\left(nk(k-1)+2rk-\frac{r}{n}(n-r)\right)}\tag{2}
\end{align*}
The expression (1) is stated by OP and can be rearranged to (2).
On the other hand the right side of OPs inequality can be written as
\begin{align*}
&r\binom{k+1}{2}+(n-r)\binom{k}{2}\\
&\qquad=\frac{1}{2}\left(r(k+1)k+(n-r)k(k-1)\right)\\
&\qquad\,\,\color{blue}{=\frac{1}{2}\left(nk(k-1)+2rk\right)}\tag{3}
\end{align*}
Since $0\leq r<n$ we see the expression in (2) is less than (3) by $\frac{r}{2n}(n-r)$ whenever $0<r<n$. We conclude Jensen's inequality is not strong enough to prove the claim.
Note, the reference given in the comment by @MartinSleziak provides a nice solution (which also makes plausible that Jensen's inequality does not work).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Is $x^2 \geq \alpha(\alpha-1)$?
If $\alpha$ is a nonnegative real and $x$ is a real satisfying $(x+1)^2\geq \alpha(\alpha+1),$ is $x^2 \geq \alpha(\alpha-1)$?
The answer is yes. Consider two cases: $1) \, x < -1$ and $2)\, x\geq -1.$
In case $1,$ taking the square root of both sides of the inequality gives $-(x+1) \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x \leq -\sqrt{\alpha(\alpha+1)}-1.$ Hence $x^2\geq \alpha(\alpha+1)+2\sqrt{\alpha(\alpha+1)}+1.$ Since $2\sqrt{\alpha(\alpha+1)}+1\geq 2\alpha+1 > -2\alpha, x^2>\alpha(\alpha+1)-2\alpha=\alpha(\alpha-1).$
Now in case $2,$ if $\alpha = 0, x^2 \geq 0,$ so we are done. Suppose $\alpha > 0$. Taking the square root of both sides gives $x+1 \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x\geq \sqrt{\alpha(\alpha+1)}-1\Rightarrow x^2\geq \alpha(\alpha+1)-2\sqrt{\alpha(\alpha+1)}+1.$ $2\alpha-2\sqrt{\alpha(\alpha+1)}+1 =2\alpha\left(1-\sqrt{1+\frac{1}{\alpha}}\right)+1 =\dfrac{\sqrt{1+\frac{1}{\alpha}}-1}{1+\sqrt{1+\frac{1}{\alpha}}}>0.$ Hence $-2\sqrt{\alpha(\alpha+1)}+1>-2\alpha\Rightarrow x^2 > \alpha(\alpha+1)-2\alpha = \alpha(\alpha-1)$. I think this argument works, but I was wondering if there was a faster method?
| Here is one way by contraposition that has some advantage in not requiring cases. We can assume $\alpha > 1$ otherwise $\alpha(\alpha-1) \leq 0 \leq x^2$. (This is so we can take the square root.)
Suppose $x^2 < \alpha(\alpha-1)$, so
$$ (x + 1)^2 = x^2 + 2x + 1 < \alpha(\alpha - 1) + 2\sqrt{\alpha(\alpha-1)} + 1. $$
It suffices to show this is less than $\alpha(\alpha + 1)$. Expanding, this amounts to showing
$$ -\alpha + 2\sqrt{\alpha(\alpha-1)} + 1 < \alpha. $$
This is easy, because you can rearrange to get $2\sqrt{\alpha(\alpha-1)} < 2\alpha - 1$ and then square (as both LHS and RHS are positive) to get $4\alpha(\alpha-1) < (2\alpha-1)^2 = 4\alpha(\alpha - 1) + 1$. That is, $x^2 < \alpha(\alpha-1)$ implies $(x + 1)^2 < \alpha(\alpha + 1)$.
EDIT: I forgot to add that when we write $2x < 2\sqrt{\alpha(\alpha-1)}$ in the first step, this is because $2x \leq 2|x| < 2\sqrt{\alpha(\alpha-1)}$, i.e. we can assume $x$ is nonnegative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Prove that for all acute triangles $\triangle ABC$, $r_a + r_b + r_c \ge m_a + m_b + m_c$.
Let $r_b$ and $m_b$ respectively be the exradius of the excircle opposite $B$ and the median drawn from the midpoint of side $CA$ of acute triangles $\triangle ABC$. Prove that $$\large r_a + r_b + r_c \ge m_a + m_b + m_c$$
We have that $$[ABC] = \sqrt{\frac{r_a + r_b + r_c}{2} \cdot \prod_{cyc}\frac{r_a - r_b + r_c}{2}} = \frac{4}{3}\sqrt{\frac{m_a + m_b + m_c}{2} \cdot \prod_{cyc}\frac{m_a - m_b + m_c}{2}}$$
Let $r_a - r_b + r_c = r_b'$, $m_a - m_b + m_c = m_b'$ and so on, we have that $$\sum_{cyc}r_b' \cdot \prod_{cyc}r_b' \ge \frac{16}{9} \cdot \sum_{cyc}m_b' \cdot \prod_{cyc}m_b'$$
In order to prove that $r_a + r_b + r_c \ge m_a + m_b + m_c$, which could be rewritten as $$r_a' + r_b' + r_c' \ge m_a' + m_b' + m_c'$$, we need to prove that $r_a' \cdot r_b' \cdot r_c' \le \dfrac{16}{9} \cdot m_a' \cdot m_b' \cdot m_c'$.
You could do that preferentially... or let $p - a = a'$, $p - b = b'$, $p - c = c'$, we need to prove that $$\sqrt{(a' + b' + c') \cdot (a'b'c')} \cdot \left(\frac{1}{a'} + \frac{1}{b'} + \frac{1}{c'}\right)\ge \sum_{cyc}\sqrt{b'(a' + b' + c') + \frac{(c' - a')^2}{4}}$$
$\left(p = \dfrac{a + b + c}{2}\right)$ in which I don't know what to do next.
| Your inequality is true for any triangle!
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and in the standard notation we need to prove that:
$$\sum_{cyc}\frac{2S}{b+c-a}\geq\frac{1}{2}\sum_{cyc}\sqrt{2b^2+2c^2-a^2}$$ or
$$\sum_{cyc}\frac{2\sqrt{xyz(x+y+z)}}{2x}\geq\frac{1}{2}\sum_{cyc}\sqrt{4x(x+y+z)+(y-z)^2}$$ or
$$\frac{2(xy+xz+yz)\sqrt{x+y+z}}{\sqrt{xyz}}\geq\sum_{cyc}\sqrt{4x(x+y+z)+(y-z)^2}$$ or
$$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}\geq\sum_{cyc}(4x(x+y+z)+(y-z)^2)+$$
$$+2\sum_{cyc}\sqrt{(4x(x+y+z)+(y-z)^2)(4y(x+y+z)+(x-z)^2)}$$ or
$$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-3\sum_{cyc}(4x(x+y+z)+(y-z)^2)+$$
$$+\sum_{cyc}\left(\sqrt{4x(x+y+z)+(y-z)^2}-\sqrt{4y(x+y+z)+(x-z)^2}\right)^2\geq0$$ or
$$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-18\sum_{cyc}(x^2+xy)+$$
$$+\sum_{cyc}\frac{(x-y)^2(4(x+y+z)-(x+y-2z))^2}{\left(\sqrt{4x(x+y+z)+(y-z)^2}+\sqrt{4y(x+y+z)+(x-z)^2}\right)^2}\geq0,$$ for which it's enough to prove that:
$$\frac{4(xy+xz+yz)^2(x+y+z)}{xyz}-18\sum_{cyc}(x^2+xy)+$$
$$+\sum_{cyc}\frac{9(x-y)^2(x+y+2z)^2}{\frac{1}{2}\left(4x(x+y+z)+(y-z)^2+4y(x+y+z)+(x-z)^2\right)}\geq0$$ or
$$\frac{2(xy+xz+yz)^2(x+y+z)}{xyz}-9\sum_{cyc}(x^2+xy)+$$
$$+\sum_{cyc}\frac{9(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or
$$\sum_{cyc}(2x^3y^2+2x^3z^2-5x^3yz+x^2y^2z)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or
$$\sum_{cyc}(2x^3y^2+2x^3z^2-4x^3yz-x^3yz+x^2y^2z)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or
$$\sum_{cyc}(x-y)^2\left(2z^3-\frac{1}{2}xyz\right)+\sum_{cyc}\tfrac{9xyz(x-y)^2(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\geq0$$ or
$$\sum_{cyc}(x-y)^2\left(4z^3-xyz+\tfrac{18xyz(x+y+2z)^2}{4(x+y)(x+y+z)+(y-z)^2+(x-z)^2}\right)\geq0,$$
For which it's enough to prove that:
$$18(x+y+2z)^2\geq4(x+y)(x+y+z)+(y-z)^2+(x-z)^2$$ or
$$13x^2+28xy+13y^2+70(x+y+z)z\geq0$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/3497183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ABCD$ is a trapezium, $MN=\dfrac{AB-CD}{2}$ where $M,N$ are the midpoints of $AB,CD$, respectively; show $\angle BAD+\angle ABC=90^\circ$
$ABCD$ is a trapezium in which $AB$ is parallel to $CD$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$. If $MN=\dfrac{AB-CD}{2}$, I should show that $\angle BAD+\angle ABC=90^\circ$.
Let $KP$ be the midsegment of $ABCD: K\in AD$ and $P\in BC$. We know $KP=\dfrac{AB+CD}{2}$. If $MN$ intersects $KP$ at $X$, $X$ is the midpoint of $MN$ and $KP$. How can I continue? Does this help?
Edit: I have another idea. Let $NN_2||AD$ and $NN_1||BC$. Can we show $\angle N_1NN_2=90^\circ$?
|
Extend $AD$ and $BC$ to meet at $P$. We are given that $m=x-y$. Since $CD\parallel AB$, $\triangle PDN\sim\triangle PAM$ and thus
$$\frac yn=\frac x{n+m}=\frac x{n+x-y}\\nx=y(n+x-y)=yn+xy-y^2\\nx-xy=yn-y^2\\x(n-y)=y(n-y)\\x=y\quad\text{or}\quad n-y=0$$
Since generally $x\neq y$, we conclude that $n-y=0$. Thus $y=n$ and $m+n=(x-y)+y=x$.
Therefore, $2PM=AB$, so $M$ is the circumcenter of $\triangle PAB$. This is only possible in a right triangle, so $\angle P=90^\circ$ and thus $\angle A+\angle B=90^\circ$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{1}4x^3\left(\frac{d^2}{dx^2}(1-x^2)x^5\right)dx$ $$\int_{0}^{1}4x^3\left(\dfrac{d^2}{dx^2}(1-x^2)x^5\right)dx$$
$$\int_{0}^{1}4x^3\left(\dfrac{d}{dx}(5x^4-7x^6)\right)dx$$
$$\int_{0}^{1}4x^3\left(20x^3-42x^5\right)dx$$
$$8\int_{0}^{1}10x^6-21x^8 dx$$
$$8\left(\dfrac{10x^7}{7}-\dfrac{21x^9}{9}\right)_{x=1}$$
$$8\left(\dfrac{10}{7}-\dfrac{21}{9}\right)=8\cdot\dfrac{90-147}{63}=-\dfrac{152}{21}$$
But actual answer is $2$. What am I missing here?
| Although you've gotten to the point of knowing that the "actual answer" you've been given is wrong, you might want to consider a different approach to the problem -- integration by parts. You have an integral of the form
$$
\int_0^1 f(x) g'(x) ~ dx
$$
(where $g(x) = ((1-x^2)x^5)'$), so you can convert it, with "parts", as follows:
\begin{align}
I &= \int_{0}^{1}4x^3\left(\dfrac{d^2}{dx^2}(1-x^2)x^5\right)dx\\
&= \int_{0}^{1}4x^3\left(\dfrac{d}{dx}\left(\frac{d}{dx}[(1-x^2)x^5]\right)\right)dx \\
&= \left. 4x^3 \left(\frac{d}{dx}[(1-x^2)x^5]\right)\right|_0^1 - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
\end{align}
In the left hand term, when $x = 0$ the $4x^3$ factor is zero; when $x = 1$, the "derivative" factor turns out to be $5 - 7 = -2$, so we have
\begin{align}
I
&= 4 \cdot (-2) - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
&= -8 - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
\end{align}
and we can repeat the integration by parts, to get
\begin{align}
I &= -8 - \left.\left(12x^2 \cdot (1-x^2)x^5\right) \right|_0^1 + \int_{0}^{1}24x\left((1-x^2)x^5\right)dx \\
\end{align}
In this case, the middle term is zero at both $x = 0$ and $x = 1$, so we have
\begin{align}
I
&= -8 + \int_{0}^{1}24x\left((1-x^2)x^5\right)dx \\
&= -8 + \int_{0}^{1}24x(x^5-x^7)dx \\
&= -8 + 24\int_{0}^{1}(x^6-x^8)dx \\
&= -8 + 24 \left(\left. \frac{x^7}{7}-\frac{x^9}{9}\right|_{0}^{1}\right)\\
&= -8 + 24 \left( \frac{1}{7}-\frac{1}{9}\right)\\
&= -8 + 24 \left( \frac{2}{63}\right)\\
&= -8 + \frac{16}{21}\\
\end{align}
Is this really the best way to go? Probably not -- I had to multiply stuff out in the middle to do the evaluations from $0$ to $1$, etc. -- but the general notion that whenever you see $fg'$ inside an integral, it's not a bad idea to try integration by parts, is still worth knowing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a^{\cos(b)^2}-b^{\cos(a)^2}\leq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$ and the reverse inequality . It'a problem of my own :
Let $a\geq b>0$ such that $a+b=\frac{\pi}{2}$ then we have :
$$a^{\cos(b)^2}-b^{\cos(a)^2}\leq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$$
Let $b\geq a>0$ such that $a+b=\frac{\pi}{2}$ then we have :
$$a^{\cos(b)^2}-b^{\cos(a)^2}\geq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$$
For the first the equality case is when $b=0$ or $b=\frac{\pi}{4}$ for the second when $b=\frac{\pi}{4}$ or $b=\frac{\pi}{2}$
The main remark is : the line $f(x)=\frac{4}{\pi}(1-\frac{\pi}{2})x+\frac{\pi}{2}-1$ is a chord of the curve defines by the function $g(x)=\Big(\frac{\pi}{2}-x\Big)^{\cos^2(x)}-\Big(x\Big)^{\sin^2(x)}$
So my idea was to derivate the function $g(x)$ we obtain :
$$g'(x)= (\frac{\pi}{2} - x)^{\cos^2(x)} \Big(-\frac{\cos^2(x)}{(\frac{\pi}{2} - x)} - 2 \log\Big(\frac{\pi}{2} - x\Big) \sin(x) \cos(x)\Big) - x^{\sin^2(x)} \Big(\frac{\sin^2(x)}{x} + 2 \log(x) \sin(x) \cos(x)\Big)$$
See that $g'(x)<0$ on the interval $[0,\frac{\pi}{2}]$ and apply the mean value theorem but I'm stuck after this.
If you have nice idea it would be cool.
Ps: Is there a symmetry with respect to the line ($f(x)$)?If yes we can cut in two the problem.
Thanks for sharing your time and knowledge .
| put $x=\dfrac{\pi}{4}+a$
$h(a)=(\frac{\pi}{4}-a)^{\frac{1}{4}(\cos{a}-\sin{a})}+(\frac{\pi}{4}+a)^{\frac{1}{4}(\sin{a}+\cos{a})}$
$=i(a)+i(-a)$
$i''(x)$ is increasing in the area $(-\frac{\pi}{4},0)$ and $i'(0)\geq-1$, h(x) is the shape like which add two parabola.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $E(\frac{1}{2018})+E(\frac{2}{2018})+\dots+E(\frac{2017}{2018})$ Let $E(n) = \dfrac{4^n}{4^n+2}$.
If the value of $E(\frac{1}{2018})+E(\frac{2}{2018})+\dots+E(\frac{2017}{2018})=\frac{a}{b}$ (in lowest terms), find $b$.
I tried solving this question using telescopic sum but was unable to solve
| Let $\displaystyle f(x)=\frac{4^x}{4^x+2},$ Then $\displaystyle f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{2}{4^x+2}$
So $$f(x)+f(1-x)=\frac{4^x+2}{4^x+2}=1.$$
So Put $\displaystyle x=\frac{1}{2018},\frac{2}{2018},\frac{3}{2018},\cdots \cdots ,\frac{1008}{2018}$
So $$f\bigg(\frac{1}{2018}\bigg)+f\bigg(\frac{2}{2018}\bigg)+f\bigg(\frac{3}{2018}\bigg)+\cdots +\cdots +f\bigg(\frac{1009}{2018}\bigg)+f\bigg(\frac{1010}{2018}\bigg)\cdots +f\bigg(\frac{2017}{2018}\bigg)=1008+\frac{1}{2}$$
For calculation of $\displaystyle f\bigg(\frac{1009}{2018}\bigg),$
Put $\displaystyle x=\frac{1009}{2018}$ in $f(x)+f(1-x)=1,$
Getting $$\displaystyle f\bigg(\frac{1009}{2018}\bigg)=\frac{1}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of Independent Normal Random Variables There are two Random Variables X and Y. Both of them are Normally distributed with 0 mean and variance a and b respectively. X and Y are independent of each other.
What is the probability of
What should be the easiest way to do this?
| The joint probability distribution of $X$ and $Y$ is $\frac{1}{2\pi \sqrt{ab}}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)}$.
Hence, we are looking for the value of the following integral.
\begin{align*}
\frac{1}{2 \pi \sqrt{ab}} \iint_{x+y>0, y>0}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)} dxdy
\end{align*}
We make the substitution $s = \frac{x}{a}$ and $t = \frac{y}{b}$, and this becomes
\begin{align*}
\frac{\sqrt{ab}}{2 \pi} \iint_{as+bt>0, t>0}e^{\frac{-1}{2} \left(s^2 + t^2 \right)} dsdt
\end{align*}
Now, we want to switch to polar coordinates, but we have to be careful about the domain over which we integrate. You might want to draw a picture to find that it is the sector $0 < \theta < \pi + \arctan\left( \frac{-a}{b} \right)$. This gives the following expression.
\begin{align*}
\frac{\sqrt{ab}}{2 \pi} \int_{0}^{\pi + \arctan\left( \frac{-a}{b} \right)}\int_0^{\infty} e^{\frac{-r^2}{2}} rdrd\theta =& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi} \int_0^{\infty}e^{\frac{-r^2}{2}}d\left( \frac{r^2}{2} \right) \\
=& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi}
\end{align*}
which is your solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality Solution Correctness Let $a,b,c \in \Re $ and $ 0 < a < 1 , 0 < b < 1 , 0 < c < 1 $ & $ \sum_{cyc} a = 2$
Prove that
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
My solution
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
or
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$
From $ \sum_{cyc} a = 2$ we can conclude $\prod_{cyc}a \le \frac{8}{27} $ , thus getting a maximum bound on $\prod_{cyc}a$ .
Going back to
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$
We can say $$\prod_{cyc}(1-a) \le \left(\frac{\sum_{cyc}(1-a)}{3}\right )^3=\left(\frac{1}{3}\right )^3=\frac{1}{27}.$$
Which is true from $AM-GM$ inequality .
Is this solution correct ?
| As shown by Michael Rozenberg, the proof given is not correct. Here is another AGM approach, using
$$
\sum_{k=1}^n\frac1{p_k}=1\implies\sum_{k=1}^nx_k\ge\prod_{k=1}^n(p_kx_k)^{1/p_k}\quad\text{where}\quad x_k,p_k\gt0\tag1
$$
Suppose $x+y+z=1$, then $(1)$ says that
$$
\begin{align}
&(1-x)(1-y)(1-z)\\
&=1-(x+y+z)+(xy+yz+zx)-xyz\\
&=(x+y+z)(xy+yz+zx)-xyz\\
&=x^2y+xy^2+2xyz+y^2z+yz^2+zx^2+z^2x\\
&\ge\left(8x^2y\right)^{1/8}\left(8xy^2\right)^{1/8}\left(8xyz\right)^{1/4}\left(8y^2z\right)^{1/8}\left(8yz^2\right)^{1/8}\left(8zx^2\right)^{1/8}\left(8z^2x\right)^{1/8}\\
&=8xyz\tag2
\end{align}
$$
Let $a=1-x$, $b=1-y$, $c=1-z$, then we get that if $a+b+c=2$, then $(2)$ is equivalent to
$$
abc\ge8(1-a)(1-b)(1-c)\tag3
$$
Thus, if $a+b+c=2$, then
$$
\frac{a}{1-a}\frac{b}{1-b}\frac{c}{1-c}\ge8\tag4
$$
and $a=b=c=\frac23$ shows that $(4)$ is sharp.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite complex nested radical and it's complex conjugate. Today I want to play with $i$ the imaginary unit I have this :
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4\sqrt{\cdots}}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{\cdots}}}}}$$
Main remarks we have :
$$\sqrt{1+i\sqrt{1+i^2}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}}}=1$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}}}}$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}}}}}$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4\sqrt{1+i^5}}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{1+\frac{1}{i^5}}}}}}$$
And so on...
My question :
What happend in the infinite case ?
Have we a closed form for this special nested radical ?
Edit : It was in fact the complex conjugate.
Thanks a lot for sharing your time and knowledge .
| Just a comment on this. Well first and foremost, I see that you're doing a lot of radicals of this form & also much trivially we can get;
$$\sqrt{1+\frac{1}{x}\sqrt{1+\frac{1}{x^2}\sqrt{1+\frac{1}{x^3}\sqrt{1+\frac{1}{x^4}\sqrt{\cdots}}}}}=\frac{1}{x^{2}}\sqrt{x^{4}+\sqrt{x^{6}+\sqrt{x^{8}+\sqrt{x^{10}+\sqrt{x^{12}+...}}}}}$$
So In a way;
$$\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{\cdots}}}}}=\frac{1}{i^{2}}\sqrt{i^{4}+\sqrt{i^{6}+\sqrt{i^{8}+\sqrt{i^{10}+\sqrt{i^{12}+...}}}}}$$
$$=-1$$
One can make a sense out of this, and I still don't personally like nested imaginary units myself. But the solution above is most likely wrong, because the identity claimed at first is by my observation only applicable for positive real quantities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim\limits_{n \to \infty} \int\limits_0^n \frac1{1 + n^2 \cos^2 x} dx$. I have to find the limit:
$$\lim\limits_{n \to \infty} \displaystyle\int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx$$
How should I approach this?
I kept looking for some appropriate bounds (for the Squeeze Theorem) that I could use to determine the the limit, but I didn't come up with anything useful.
| Here is a direct approach. Let $f(x)$ be the integrand, and note that $f(x)$ has a period $\pi$. Let $k$ be the largest positive integer such that $(2k+1)\frac{\pi}{2}<n$. Then:
$$\begin{align}
\int_0^nf(x)\,dx&=\int_0^{\pi/2}f(x)\,dx+\int_{\pi/2}^{3\pi/2}f(x)\,dx+\dots+\int_{(2k+1)\pi/2}^nf(x)\,dx \\
&=\int_0^{\pi/2}f(x)\,dx+k\int_{\pi/2}^{3\pi/2}f(x)\,dx+\int_{(2k+1)\pi/2}^nf(x)\,dx
\end{align} $$
Each of those can be evaluated with the substitution $t=\tan(x) \Rightarrow \cos^2(x)=\frac{1}{1+\tan^2(x)}=\frac{1}{1+t^2}$ and $dx=\frac{dt}{1+t^2}$
$$\begin{align}
\int\frac{1}{1+n^2\cos^2(x)}\,dx&=\int\frac{1}{1+n^2\frac{1}{1+t^2}}\frac{1}{1+t^2}\,dt\\
&=\int\frac{1}{t^2+n^2+1}\,dt\\
&=\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)+C
\end{align} $$
Then:
$$\begin{align}
\int_0^{\pi/2}f(x)\,dx&=\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_0^\infty \\
&=\frac{\pi}{2\sqrt{n^2+1}}\overset{n\to\infty}{\to} 0
\end{align}$$
and
$$\begin{align}
\int_{(2k+1)\pi/2}^nf(x)\,dx &= \frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_{-\infty}^{\tan(n)} \\
&=\frac{1}{\sqrt{n^2+1}}\left(\tan^{-1}\left(\frac{\tan(n)}{\sqrt{n^2+1}}\right)+\frac{\pi}{2}\right)\\
&\overset{n\to\infty}{\to} 0
\end{align}$$
because $1/\sqrt{n^2+1}\to 0$ and the expression in the parentheses is bounded. Finally,
$$\begin{align}
k\int_{\pi/2}^{3\pi/2}f(x)\,dx &= k\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_{-\infty}^{\infty} \\
&= \frac{\pi k}{\sqrt{n^2+1}}
\end{align}$$
So you just have to find
$$\lim_{n\to\infty}\frac{\pi k}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{\pi k}{n} $$
Note that the choice of $k$ implies $(2k+1)\frac{\pi}{2}<n<(2k+3)\frac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}} \, dx$
Evaluate $$\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx$$
$$
\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}} \, dx = \int_0^1 \bigg[\sqrt{\frac{1+x}{1-x}}+\frac{1}{(1-x)^{3/2}\sqrt{1+x}}\bigg] \, dx=\int_0^1[f(x)+f'(x)]\,dx
$$
Does making it into the above final form helps somehow solve the given definite integral ?
Note: The solution given is $\pi/2$
| Let $x=\sin t$. Then,
$$\int _0^1 \frac{2-x^2}{\left(1+x\right)\sqrt{1-x^2}}dx
=\int _0^1\left(\frac{1}{\left(1+x\right)\sqrt{1-x^2}}
+\frac{\sqrt{1-x^2}}{1+x}\right)dx$$
$$=\int_0^{\pi/2} \left( \frac{dt}{1+\sin t}+1-\sin t\right)dt$$
$$=\left(-\frac{\cos t}{1+\sin t} + t + \cos t\right)_0^{\pi/2}=\frac\pi2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear transformation with respect to basis problem Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $T(\begin{bmatrix} 3 \\ 1 \end{bmatrix}) = \begin{bmatrix} 1 \\2 \end{bmatrix}$ and $T(\begin{bmatrix} -1 \\ 0 \end{bmatrix}) = \begin{bmatrix} 1 \\1 \end{bmatrix}$. Find the matrix $A$ representing $T$.
I understand that to approach this problem, I have to view $\begin{bmatrix} 3 \\1 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ as a basis, $B = \{ v_1, v_2 \}$, where $[v1 v2]$ is the transition matrix from $[x]B$ to $x$. How do I use $\begin{bmatrix} 1\\2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\1 \end{bmatrix}$? I'm unclear on their connection to the basis vectors $v_1$ and $v_2$.
| Relative to the bases $B$ and the standard basis, the matrix is:. $\begin{pmatrix}1&1\\2&1\end{pmatrix}$.
The change of basis matrix is: $\begin{pmatrix}3&-1\\1&0\end{pmatrix}$.
The latter changes basis from $B$ to the standard basis.
Thus you want: $\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}3&-1\\1&0\end{pmatrix}^{-1}$.
I will leave this calculation for you.
$\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}0&1\\-1&3\end{pmatrix}=\begin{pmatrix}-1&4\\-1&5\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508294",
"timestamp": "2023-03-29T00:00:00",
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
| By Cauchy-Schwarz,
$$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2$$
Then by rearrangement inequality,
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b} \\ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}$$ Sum up and get $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}$$
So from the first inequality, we get:
$$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \dfrac{9}{4} \\ \dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2} \ge \dfrac{9}{4(a+b+c)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{Q}^{2\times2}, f=X^2+\alpha \in \mathbb{Q}[X].$
Consider $$f(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}+\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\ \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} + \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\\begin{pmatrix} a^2+bc + \alpha & ab+bd \\ ac+cd & bc+d^2 + \alpha \end{pmatrix}.$$
How do I get to the claim from here on?
Thanks in advance!
| HINT: What happens if $a=-d$ and $bc=-a^2-\alpha$?
| {
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What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
Then $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left(\frac{1}{2^n}-\frac{1}{4^n}\right)=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{4^n}$$
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}$$
$ii$) We found our series can be written as the difference of two geometric series with $|r|<1$, and the solution is
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}=\frac{\frac{1}{4}}{1-\frac{1}{2}}-\frac{\frac{1}{16}}{1-\frac{1}{4}}=\frac{5}{12}$$
I have given this series to an online series calculator, and it claims the result is not $\frac{5}{12}$ but $\frac{2}{3}$, and yet I can not find the mistake in my procedure. Can anyone point it out to me?
EDIT: Please remember, I'm asking more than "what is the right way to solve this problem"; I'm asking precisely what was the mistake I made. Correlated, but not the same.
| $\sum_\limits{n=1}^{\infty} k^n = \frac {k}{1-k}\\
\sum_\limits{n=1}^{\infty} (\frac {1}{2})^n = \frac {\frac 12}{1-\frac 12} = 1\\
\sum_\limits{n=1}^{\infty} (\frac {1}{4})^n = \frac {\frac 14}{1-\frac 14} = \frac {1}{3}\\
\sum_\limits{n=1}^{\infty} (\frac {1}{2})^n - \sum_\limits{n=1}^{\infty} (\frac {1}{4})^n = \frac 23$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Sum of combinations formula Is there an explicit formula for the sum $0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+n\dbinom{n}{n} = \sum_{k=0}^nk\dbinom{n}{k}$?
| Actually, I think I got it because I realized I forgot to show what work I have already done in the question but as I was trying to refine my work I realized what I need to do:
$$0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+n\dbinom{n}{n}=$$
$$1\dbinom{n}{1}+2\dbinom{n}{2}+\dots+n\dbinom{n}{n}=$$
$$1\cdot\dfrac{n!}{1!\cdot (n-1)!}+2\cdot\dfrac{n!}{2!\cdot (n-2)!}+\dots+n\cdot\dfrac{n!}{n!\cdot 0!}=$$
$$n\left(1\cdot\dfrac{(n-1)!}{1!\cdot (n-1)!}+2\cdot\dfrac{(n-1)!}{2!\cdot (n-2)!}+\dots+n\cdot\dfrac{(n-1)!}{n!\cdot 0!}\right)=$$
$$n\left(\dfrac{(n-1)!}{0!\cdot (n-1)!}+\dfrac{(n-1)!}{1!\cdot (n-2)!}+\dots+\dfrac{(n-1)!}{(n-1)!\cdot 0!}\right)=$$
$$n\left(\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dots+\dbinom{n-1}{n-1}\right)=$$
$$n\left(2^{n-1}\right)=\boxed{n \cdot 2^{n-1}}$$
Based on Lucas Henrique and Lucas De Souza's answers, this is what I came up with:
$$f(x)=(x+1)^n=\dbinom{n}{0}x^n+\dbinom{n}{1}x^{n-1}+\dots+\dbinom{n}{n}x^0$$
Taking the derivative of $f(x)$, we get
$$f'(x)=n\cdot(x+1)^{n-1}=n\dbinom{n}{0}x^{n-1}+(n-1)\dbinom{n}{1}x^{n-2}+\dots+0\dbinom{n}{n}$$
This can also be written as
$$f'(x)=n\dbinom{n}{n}x^{n-1}+(n-1)\dbinom{n}{n-1}x^{n-2}+\dots+0\dbinom{n}{0}$$
because $\binom{n}{k}=\binom{n}{n-k}$. Plugging in $1$ for $x$, we get
$$f'(1)=n\dbinom{n}{n}+(n-1)\dbinom{n}{n-1}+\dots+0\dbinom{n}{0}$$
as desired. Thus, our sum can be represented by $f'(1)$, which can be calculated as $n\cdot(1+1)^{n-1}=\boxed{n\cdot2^{n-1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Using Beta Gamma function, show that :$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta).\sin^4 (3\theta) d\theta$ Using Beta Gamma function, show that :$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta)\cdot\sin^4 (3\theta) d\theta$
My Attempt
$$\int_0^{\frac {\pi}{6}} \cos^2 (6\theta) \cdot \sin^4 (3\theta) d\theta$$
Put $3\theta=t$
$$3d\theta=dt$$
$$d\theta=\frac {dt}{3}$$
When $\theta=0$, $t=0$
When $\theta=\frac {\pi}{6}$, $t=\frac {\pi}{2}$
Now,
$$=\int_0^{\frac {\pi}{2}} \cos^2 (2t)\cdot\sin^4 (t) \frac {dt}{3}$$
| Hint:
Use $\cos2t=1-2\sin^2t$
$$\cos^2(2t)=(1-2\sin^2t)^2=?$$
$$\beta\left(m,n\right)=2\int_0^{\pi/2}\sin^{2m-1}t\cos^{2n-1}t\ dt$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find $\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)}{x-8}$ by using the conjugate rule? I need to find: $\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)}{x-8}$
I cannot solve this by substitution because that would cause the denominator to equal 0.
Normally, I would simply use the conjugate trick, however I am uncertain how I would rationalize the numerator.
$$\frac{(\sqrt[3]{x} -2)}{x-8}\times\frac{\sqrt[3]{x}+2}{\sqrt[3]{x}+2}$$
However, clearly this won't help me with anything, as I won't be able to factor anything.
$$\frac{(\sqrt[3]{x^2} -4)}{(x-8)(\sqrt[3]{x^2}+2)}$$
I am unsure about how to continue from here. Perhaps I am on the wrong track entirely. Any form of guidance would be welcome. Thank you.
| Take the steps below$$\lim_{x \to 8} \frac{\sqrt[3]{x} -2}{x-8}
=\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}$$
$$=\lim_{x \to 8}\frac{x-8}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}
=\lim_{x \to 8}\frac{1}{(\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4}
=\frac1{4+4+4}=\frac1{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can any polynomials in the rational field be decomposed like this I' ve learned that the following examples can be used to decompose a
factor in this way:
x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) +
1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 -
3/16 (5 a^2 - 12 a) + 1/16 (12 a^3 - 5 a^4) +
1/8 (12 a^2 - 5 a^3) - (9 a)/4) x^2 + ((5 a^4)/16 + a^3/4 + (
5 a^2)/16 + 1/16 (12 a - 5 a^2) + 1/16 (12 a^3 - 5 a^4) + a/2 +
1/4 (12 - 5 a) - 3) x + a x^3 + a/4 + 1/4 (5 a - 12) + x^4 - 2)
Can any polynomials f(x) in the field of rational numbers be factorized into the form of (x - a) g (x, a)?
Besides a, other coefficients of g (x, a) should also be in the rational number field.
In the case of $x^5-5 x+12$, we can know the algebraic relations of his five roots (The letter a is a root of equation $x^5−5x+12$):
$x^5-5 x+12=(x-a) \left(x-\frac{1}{8} \left(-a^4-a^3-a^2-2 \sqrt{2}
\sqrt{3 a^3-2 a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8}
\left(-a^4-a^3-a^2+2 \sqrt{2} \sqrt{3 a^3-2
a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8} \left(a^4+a^3+a^2-2
\sqrt{2} \sqrt{a^4+a^2+6 a-8}-3 a-4\right)\right)
\left(x-\frac{1}{8} \left(a^4+a^3+a^2+2 \sqrt{2} \sqrt{a^4+a^2+6
a-8}-3 a-4\right)\right)$
I used the function in this link to find the relationship between a polynomial Galois group and a root set.
| It seems to me that you are asking two separate questions. One is whether every $f$ can be factored as $(x-\alpha)g(x)$ where the coefficients of $g$ are simple expressions in (rationals and) $\alpha$. The other is whether every $f$ can be factored as $(x-\alpha)(x-p_1(\alpha))\cdots(x-p_{n-1}(\alpha))$ where each $p_i$ is a radical expression in $\alpha$.
The answer to the first question is Yes. Let $K={\bf Q}(\alpha)$. Then $f$ has a zero in $K$, so it factors over $K$, and one factor is $x-\alpha$, and the other is a polynomial $g(x)$ with coefficients in $K$, so the coefficients of $g$ are polynomials in $\alpha$ (of degree less than the degree of $\alpha$).
The answer to the second question is Yes if the degree of $f$ is five (or less), but in general it's No if the degree of $f$ is larger. If the degree of $f$ is five, then $f(x)=(x-\alpha)g(x)$, where $g$ is a polynomial of degree four, whose coefficients are polynomials of degree at most four in $\alpha$. Since $g$ has degree four, its roots can be expressed in radicals in its coefficients, which is to say, radicals of polynomials in $\alpha$. That's where stuff like $\sqrt{\alpha^4+\alpha^2+6\alpha-8}$ comes from in your example. But if the degree of $f$ exceeds five, then the degree of $g$ is at least five, and (in general) there will be no radical expression for the roots of $g$ in terms of radicals of polynomials in $\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume between surfaces
Find $V(T),$ where $T$ is the region bounded by the surfaces $y = kx^2+kz^2$ and $z=kx^2+ky^2,$ where $k\in\mathbb{R}, k > 0.$
I tried solving for the area over which these curves intersect, which gave me $y-kz^2 = z-ky^2.$ Solving gives $y+ky^2 - (z+kz^2) =0\Rightarrow (y-z)(1 + k(y-z)) = 0.$ Since $y$ and $z$ are nonnegative, this implies $y=z.$ We thus obtain $y = kx^2 + ky^2\Rightarrow ky^2 - y +kx^2 = 0\Rightarrow (y -\dfrac{1}{2k})^2 +x^2 = \dfrac{1}{4k^2},$ which is a circle centered at $(0,\dfrac{1}{2k})$ with radius $\dfrac{1}{2k}.$
I am able to solve for $A(x),$ the area in terms of $x$ at a given value of $x,$ and I know $x$ ranges from $-\dfrac{1}{2k}$ to $\dfrac{1}{2k},$ but the resulting integral I have to evaluate is absolutely disgusting!
Is there a "cleaner" integral I can use?
| Observe that the enclosed volume by $y = kx^2+kz^2$ and $z=kx^2+ky^2$ are symmetric with respect to the plane $y=z$. So, the total volume is twice the volume between the surfaces $y=z$ and $z=kx^2+ky^2$.
Recognize that the integration region in $xy$-coordinates is the circle given by
$$x^2+ \left(y -\frac{1}{2k}\right)^2 = \dfrac{1}{4k^2}$$
and re-center the circle with the variable changes $x=u$ and $v = y - \frac{1}{2k}$. Then, the integration region becomes $u^2+ v^2 = \frac{1}{4k^2}$ and the two enclosing surfaces are respectively,
$$z_1 = v+ \frac 1{2k},\>\>\>\>\> z_2=ku^2+k\left(v+\frac1{2k}\right)^2$$
As a result, the volume integral can be expressed as,
$$V= 2\int_{u^2+v^2\le \frac1{4k^2}} (z_1-z_2)dudv
= 2\int_{u^2+v^2\le \frac1{4k^2}} (\frac1{4k}-ku^2-kv^2)dudv$$
Then, integrate in polar coordinates to obtain,
$$V=2\int_0^{2\pi} \int_0^{\frac1{2k}}(\frac1{4k}-kr^2)rdrd\theta
=4\pi\int_0^{\frac1{2k}}(\frac1{4k}-kr^2)rdr =\frac\pi{16k^3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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When does $2n-1$ divide $16(n^2-n-1)^2$? Find all integers $n$ such that
$\dfrac{16(n^2-n-1)^2}{2n-1}$
is an integer.
| Starting off with the initial expression:
$$\frac{16(n^2-n-1)^2}{2n-1}$$
Completing the square:
$$=\frac{4^2(n^2-n-1)^2}{2n-1}$$
$$=\frac{(4n^2-4n-4)^2}{2n-1}$$
$$=\frac{((2n-1)^2-5)^2}{2n-1}$$
Some manipulation:
$$=(2n-1)^3(1-\frac{5}{(2n-1)^2})^2$$
Expanding the expression:
$$=(2n-1)^3-10(2n-1)+\frac{25}{2n-1}$$
Since the first two terms will always be integers,
If this expression is to be an integer, then:
$$(2n-1)= factor\ of\ 25$$
$$n=\frac{1+factor\ of\ 25}{2}$$
The factors of 5 are $\pm 1, \pm 5, \pm 25$
thus,
n=$1\pm1\over 2$
or
n=$1\pm 5 \over 2$
or
n=$1\pm 25 \over 2$
n=1,0,-2,3,-12,13
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic function with roots in $[0,1]$. Prove that $f(0) \geq \frac49$ or $f(1) \geq \frac49$ Let $a,b$ in $[0,1]$ be such that the polynomial $f(x) = (x-a)(x-b)$ satisfies $f(\tfrac12) \geq \frac1{36}$.
I have found a quite complicated proof of the following inequality using calculus:
$$f(0) \geq \frac49 \quad\text{or}\quad f(1) \geq \frac49.$$
Can you find a simple argument? (with geometric flavour if possible)
| I'm gonna prove this algebraically. First some observations:
$$f\left(\frac{1}{2}\right) \geq \frac{1}{36}\Leftrightarrow 2ab-(a+b)+\frac{4}{9} \geq 0 \ \ \ \ \ \ \ (1)$$
$$f(0) \geq \frac{4}{9} \Leftrightarrow ab \geq \frac{4}{9}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
$$f(1) \geq \frac{4}{9} \Leftrightarrow ab-a-b + \frac{5}{9} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
Assume for the sake of contradiction that both $(2)$ and $(3)$ are false, that means:
$$ab \leq \frac{4}{9}\text{ and } ab+\frac{5}{9} \leq a+b$$
Notice that from $(1)$ and AM-GM, we have:
$$0 \leq 2ab-(a+b)+\frac{4}{9} \leq 2ab-2\sqrt{ab}+\frac{4}{9}=2\left(\sqrt{ab}-\frac{1}{3}\right)\left(\sqrt{ab}-\frac{2}{3}\right)$$
However, since $ab \leq \dfrac{4}{9}$, we have $\sqrt{ab} \leq \dfrac{2}{3}$. This, together with the above inequality, implies that $\sqrt{ab} \leq \dfrac{1}{3}$. But combining the negation of $(3)$ and $(1)$, we get:
$$ab+\frac{5}{9}\leq a+b \leq 2ab+\frac{4}{9}\Rightarrow ab \geq \frac{1}{9}\Rightarrow \sqrt{ab} \geq \frac{1}{3}$$
a contradiction. Therefore, at least one of $(2)$ or $(3)$ is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$ I have to solve the following integral:
$$\int\frac{1}{(x^4+1)^2}dx$$
I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex roots and got stuck. Is there a trick for this kind of integral or some kind of helpful substitution? Thanks.
EDIT:
I did the following:
Let $x^2=\tan\theta$, then $x = \sqrt{\tan\theta}$ and $dx=\frac{\sec^2\theta}{2x}d\theta$
Then:
$$I=\int\frac{1}{(x^4+1)^2}dx = \int\frac{1}{(\tan^2\theta+1)^2} \frac{\sec^2\theta}{2x}d\theta=\int\frac{1}{\sec^4\theta} \frac{\sec^2\theta}{2x}d\theta$$
$$I=\frac{1}{2}\int{\frac{1}{\sec^2\theta \sqrt{\tan\theta}}}d\theta$$.
After this I don't know how to proceed.
| Hints:
$$\frac1{(x^4+1)^2}=\frac{x^4+1-x^4}{(x^4+1)^2}=\frac1{x^4+1}-\frac{x^4}{(x^4+1)^2}$$ and by parts
$$4\int\frac{x^3x}{(x^4+1)^2}dx=-\frac x{x^4+1}+\int\frac{dx}{x^4+1}.$$
This way we can get rid of the square at the denominator, and we are left with
$$\frac1{x^4+1}.$$
Now using the factorization of the quartic binomial,
$$\frac{\sqrt8}{x^4+1}=\frac{x+\sqrt2}{x^2+\sqrt2x+1}-\frac{x-\sqrt2}{x^2-\sqrt2x+1}.$$
Here, by completing the square, we can handle the terms $\sqrt2x$ in the denominators, and solve with terms $\log(x^2\pm\sqrt2x+1)$ and $\arctan(\sqrt2x\pm1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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determination coefficients of quadratic equation using sum and product of its solutions? let $x_1$ and $x_2$ are real solutions of the quadratic equation $ f(x)=ax^2+bx+c=0$ with $a, b, c$ are real such that its solutions satisfy the following system :
$$\begin{cases}x_1+x_2=16, \\ x_1 x_2=55.\end{cases}$$
Now my question here is : what are $a, b, c$ ?
I have got $a=1,b=-16, c=55$ but someone said me that I am wrong , $a$ should be arbitrary as a reason the system have infinitely many solutions
| Let's assume the solutions are $x_1$ and $x_2$. Now, we can write:
\begin{align*}
f(x) &= ax^2 + bx + c \qquad (1)\\
f(x) &= (x - x_1)(x - x_2) = x^2 - x(x_1 + x_2) + x_1 x_2 \qquad (2)
\end{align*}
Since the first term of $(1)$ has a highest power of $1$, we scale equation $(2)$ into:
$$
f(x) = a(x - x_1)(x - x_2) = ax^2 - xa(x_1 + x_2) + ax_1 x_2 \qquad (3)
$$
We can use the relations $x_1 + x_2 = 16$, $x_1x_2 = 55$ in equation $(3)$ to arrive at:
$$
f(x) = ax^2 -16ax + 55a
$$
which means we have an (infinite) family of polynomials $f(x)$ for different choices of $a$.
EDIT: showing that the roots of $f(x)$ are the same as asked:
Note that we can find the roots of $x^2 - 16x + 55$ using (say) the quadratic equation:
$$
x_{1, 2} = \frac{16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 55}}{2 \cdot 1} = \frac{16 \pm 6}{2} = 8\pm3 = \{ 11, 5\}
$$
Hence, $$f(x) = a(x^2 - 16x + 55) = a(x - 11)(x - 5)$$
which means that the roots are $\{11, 5\}$
EDIT: An attempt to clarify the confusion between roots and solutions
We have two "solutions" floating around in this question.
*
*The collection of polynomials of the form $f_a(x) = a(x - 11)(x - 5)$
*The roots of each $f_{a_0}(x)$, for each $a$. which are $x_1 = 11, x_2 = 5$.
Given a fixed $a = a_0$, the solution of $f_{a_0}(x)$ is $x_1 = 11, x_2 = 5$.
Given that the roots obey the equation $x_1 + x_2 = 16, x_1 x_2 = 55$. Equivalently, given that the roots are $x_1 = 11, x_2 = 5$, .there are an infinite number of polynomials $f_a(x)$ which have the roots, one for each choice of $a$.
So the fact that, for example, $f_1(x) = (x - 11)(x - 5)$ and $f_2(x) = 2(x- 11)(x - 5)$ both have the same roots ($x_1 = 11, x_2 = 5$) should not contradict the fact that there are two such polynomials, one called $f_1(x)$ and one called $f_2(x)$.
Indeed, this is analogous to this situation which might be easier to think about:
$$
a(x) \equiv x - 10 \quad
b(x) \equiv 2x - 20 \quad
c(x) \equiv 3x - 30
$$
All of $a(x), b(x), c(x)$ have the root $(x = 10)$, but $a(x) \neq b(x) \neq c(x)$.
Similarly, all of the $f_a(x)$ have roots $(x_1 = 11, x_2 = 5$), but they are different polynomials, one for each choice of $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which arose from the deduction that $\frac{1}{4}\sqrt{\frac{256\sin^{4}40^{\circ}-80\sin^{2}40^{\circ}+12-\ 8\sqrt{3}\sin40^{\circ}}{\left(16\sin^{4}40^{\circ}-4\sin^{2}40^{\circ}+1\right)}}=\cos50^{\circ}$. Despite the apparent simplicity of the relationship, it seems quite tricky to prove. I managed to prove it by solving the equation as a quadratic in $(\sin\frac{\pi}{9})$ and then using the identity $\sqrt{\sec^2 x-1}=|\tan x|$, the double angle formulae and finally that $\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x$ can be written in the form $\sin\left(x+\frac{2\pi}{3}\right)$.
But it seems like quite a neat problem. So, does anyone have a better way of proving it?
| We need to prove that:
$$4\sin^220^{\circ}-4\sin60^{\circ}\sin20^{\circ}+1=\frac{1}{4\cos^220^{\circ}}$$ or
$$4\sin^240^{\circ}-8\sin60^{\circ}\sin40^{\circ}\cos20^{\circ}+4\cos^220^{\circ}=1$$ or
$$2-2\cos80^{\circ}-4\sin60^{\circ}(\sin60^{\circ}+\sin20^{\circ})+2+2\cos40^{\circ}=1$$ or
$$\cos40^{\circ}-\cos80^{\circ}-2\sin60^{\circ}\sin20^{\circ}=0,$$ which is obvious.
| {
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"url": "https://math.stackexchange.com/questions/3539618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving recursive function with floor The recursive function is this:
$$ T(n) =
\begin{cases}
2 & \text{ for }n=1;\\
T \left( \lfloor \frac{n}{2} \rfloor \right) + 7 &,\text{ otherwise}
\end{cases}
$$
Based on the definition of the function, the right side becomes:
$T(n) = T( \lfloor \frac{n}{2^i} \rfloor) + 7 * i;$
The procedure stops when $\lfloor \frac{n}{2^i} \rfloor == 1$
The problem is how do I continue it from there?
| We list first some terms:
$$\underbrace{2}_{1}, \underbrace{9, 9}_{2}, \underbrace{16, 16, 16, 16}_{4},
\underbrace{23, 23, 23, 23, 23, 23, 23, 23}_{8}, 30, 30, \cdots$$
Conjecture:
$$T(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$$
We need to prove it. To this end, let
$$S(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$$
Let us prove that $S(n) = T(n)$ for $n = 1, 2, 3, \cdots$.
We use mathematical induction.
First, $S(1) = T(1) = 2$, and $S(2) = T(2) = 9$.
Assume that $S(k) = T(k)$ for $k = 1, 2, \cdots, n$ ($n\ge 2$).
We need to prove that $S(n+1) = T(n+1)$.
There exist integer $m\ge 1$ and integer $r$ with $0\le r < 2^m$
such that $n = 2^m + r$. We split into two cases:
1) $r = 2^m - 1$: We have $S(n+1) = 2 + 7(m+1)$
and $S(\lfloor \frac{n+1}{2}\rfloor ) = 2 + 7m$
which results in
$S(n+1) = S(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(n+1)$.
2) $r < 2^m - 1$: We have $S(n+1) = 2 + 7m$
and $S(\lfloor \frac{n+1}{2}\rfloor ) = 2 + 7(m-1)$
which results in
$S(n+1) = S(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(n+1)$. $\quad$ Q.E.D.
Thus, $T(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?
| Let $n=2m+1$. Observe that $$z^n+1=(z+1)(z^{2m}-z^{2m-1}+\cdots+z^2-z+1)$$ has simple roots (consisting of roots of unity). Let $f(z)$ be the second factor of the right hand side. Then $$f(z)=(z-1)(z^{2m-1}+z^{2m-3}+\cdots+z)+1=0$$ if $z$ is a root of $z^n+1$ and $z\neq -1$.
Applying similar argument as yours, one has $$z-1=\frac{-1}{z^{2m-1}+z^{2m-3}+\cdots+z}$$ $$\Rightarrow |z-1|\geq \frac 1m=\frac 2{n-1},$$ as required. QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Determine the value of the sum to infinity of $(u_{r+1} + \frac{1}{2^r})$ Determine the value of $$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$.
In earlier parts of the question, it is given $\displaystyle u_r= \frac{2}{(r+1)(r+3)}$, which when expressed in partial fractions is $\displaystyle\frac{1}{r+1}-\frac{1}{r+3}$.
Also previously, the question asks to show that $$\sum_{r=1}^n u_r = \frac{5}{6}-\frac {1}{n+2} -\frac{1}{n+3}$$ which I managed to.
I have also determined the values of $$\sum_{r=1}^\infty u_r$$ which is $$\sum_{r=1}^\infty u_r= \frac{5}{6}$$
But I don’t know how to solve for
$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
My attempts so far:
$\displaystyle u_{r+1} =\frac{2}{(r+1+1)(r+1+3)}$
$=\displaystyle\frac{2}{(r+2)(r+4)}$
Let $\displaystyle \frac{2}{(r+2)(r+4)}\equiv \frac{A}{(r+2)(r+4)}$
$2\equiv A(r+4)+B(r+2)$
Let r $=-4,B=-1$
Let $r=-2,A=1$
So, is $\displaystyle u_{r+1} \equiv \frac {1}{r+2}- \frac{1}{r+4}$?
Assuming that it is,$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
= $$\sum_{r=1}^\infty\Bigl(\frac{1}{r+2}+\frac{1}{r+4} + \frac{1}{2^r}\Bigr)$$
$=\frac{1}{3}-\require{cancel}\cancel{\frac{1}{5}}+\frac{1}{2}$
$\frac{1}{4}-\require{cancel}\cancel{\frac{1}{6}}+\frac{1}{4}$
$\require{cancel}\cancel{\frac{1}{5}}-\cancel{\frac{1}{7}}+\frac{1}{8}$
$\require{cancel}\cancel{\frac{1}{6}}-\cancel{\frac{1}{8}}+\frac{1}{16}$
.....
I don't know how to cancel the middle terms, how should this question be solved?
| Write $$U_{r+1}=\frac{2}{(r+2)(r+4)}= 2 \left[\left(\frac{1}{r+2}-\frac{1}{r+3}\right)+\left(\frac{1}{r+3}-\frac{1}{r+4}\right) \right]= [(F_r-F_{r+1})+(F_{r+1}-F_{r+2})], F_r=\frac{1}{r+2}.$$
Now telescoping summation can be done easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. Define $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right).$$ Find $\lim\limits_{n\to\infty} P_n$.
My approach: $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right)$$ $$=\left\{\frac{(a_1+1)(a_2+1)...(a_n+1)}{a_1.a_2...a_n}\right\}$$ $$=\left\{\frac{1}{a_1}\left(\frac{a_1+1}{a_2}\right)\left(\frac{a_2+1}{a_3}\right)...\left(\frac{a_{n-1}+1}{a_n}\right)(1+a_n)\right\}$$ $$=\left\{\frac{1}{a_1}.\frac{1}{2}.\frac{1}{3}.....\frac{1}{n}(1+a_n)\right\}$$ $$=\frac{1+a_n}{n!}, \forall n\in\mathbb{N}.$$
Thus, we need to find $$\lim_{n\to\infty} \frac{1+a_n}{n!}.$$ How to proceed after this?
| $$\begin{align}
a_{n}&=na_{n-1}+n\\
&=n[(n-1)a_{n-2}+(n-1)]+n\\
&=n(n-1)[(n-2)a_{n-3} + (n-2)]+n(n-1)+n\\
& \phantom{a bit of space here would be nice}\vdots\\
&=\left[n(n-1)(n-2)\dots 3 \cdot 2\cdot 1\right]\left( 1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!} \right)\\
&=n!\left(1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!}\right)\rightarrow n!e, \quad (n \to \infty)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| Just to give yet another approach, let $u=x^2+3$. Then, for $x\ge0$,
$$x^2(x^2-x\sqrt{x^2+6}+3)=(u-3)\left(u-\sqrt{u^2-9}\right)={9u\over u+\sqrt{u^2-9}}-{27\over u+\sqrt{u^2-9}}\to{9\over1+1}-0={9\over2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
| Let $y=x+1$ then
$$\begin{align}\sum\sin&=\sin\left(\frac1{x+2}\right)-2\sin\left(\frac1{x+1}\right)+\sin\left(\frac1x\right)\\
&=\sin\left(\frac1y-\frac1{y^2}+\frac1{y^2(y+1)}\right)-2\sin\left(\frac1y\right)+\sin\left(\frac1y+\frac1{y^2}+\frac1{y^2(y-1)}\right)\\
&=\sin\left(\frac1{y^2(y+1)}\right)\cos\left(\frac1y-\frac1{y^2}\right)+\left(1-2\sin^2\left(\frac1{2y^2(y+1)}\right)\right)\\
&\quad\times\left(\sin\left(\frac1y\right)\left(1-2\sin^2\left(\frac1{2y^2}\right)\right)-\cos\left(\frac1y\right)\sin\left(\frac1{y^2}\right)\right)-2\sin\left(\frac1y\right)\\
&\quad+\sin\left(\frac1{y^2(y-1)}\right)\cos\left(\frac1y+\frac1{y^2}\right)+\left(1-2\sin^2\left(\frac1{2y^2(y-1)}\right)\right)\\
&\quad\times\left(\sin\left(\frac1y\right)\left(1-2\sin^2\left(\frac1{2y^2}\right)\right)+\cos\left(\frac1y\right)\sin\left(\frac1{y^2}\right)\right)\\
&=\sin\left(\frac1{y^2(y+1)}\right)\cos\left(\frac1y-\frac1{y^2}\right)-4\sin\left(\frac1y\right)\sin^2\left(\frac1{2y^2}\right)\\
&\quad-2\sin^2\left(\frac1{2y^2(y+1)}\right)\sin\left(\frac1y-\frac1{y^2}\right)-2\sin^2\left(\frac1{2y^2(y-1)}\right)\sin\left(\frac1y+\frac1{y^2}\right)\\
&\quad+\sin\left(\frac1{y^2(y-1)}\right)\cos\left(\frac1y+\frac1{y^2}\right)\end{align}$$
So
$$\begin{align}\lim_{x\rightarrow\infty}x^3\sum\sin&=\lim_{y\rightarrow\infty}\left\{\left(1+\frac1y\right)^2\frac{\sin\left(\frac1{y^2(y+1)}\right)}{\frac1{y^2(y+1)}}\cos\left(\frac1y-\frac1{y^2}\right)\right.\\
&\quad-\frac1{y^2}\left(1+\frac1y\right)^3\frac{\sin\left(\frac1y\right)}{\frac1y}\frac{\sin^2\left(\frac1{2y^2}\right)}{\left(\frac1{2y^2}\right)^2}\\
&\quad-\frac{1-\frac1{y^2}}{2y^4}\frac{\sin^2\left(\frac1{2y^2(y+1)}\right)}{\left(\frac1{2y^2(y+1)}\right)^2}\frac{\sin\left(\frac1y-\frac1{y^2}\right)}{\frac1y-\frac1{y^2}}\\
&\quad-\frac{\left(1+\frac1y\right)^4}{2y^4\left(1-\frac1y\right)^2}\frac{\sin^2\left(\frac1{2y^2(y-1)}\right)}{\left(\frac1{2y^2(y-1)}\right)^2}\frac{\sin\left(\frac1y+\frac1{y^2}\right)}{\frac1y+\frac1{y^2}}\\
&\quad\left.+\left(1+\frac1y\right)^2\frac{\sin\left(\frac1{y^2(y-1)}\right)}{\frac1{y^2(y+1)}}\cos\left(\frac1y+\frac1{y^2}\right)\right\}\\
&=1-0-0-0+1=2\end{align}$$
I just wanted to see how this looked in brute force trigonometric identities...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Check if $\lim_{n\rightarrow\infty}\sum_{k=1}^n\ln\Big(1-\frac{1}{k}+\frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)$ converges. How can I check if the following expression converges as $n \rightarrow\infty$? I am confused because $n$ appears twice...
$$\lim_{n\rightarrow\infty} \sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)$$
My initial thoughts are that $\cos(\frac{\theta}{\sqrt{\ln n}}) \rightarrow 0$ and so we are essentially summing $\ln(1) = 0$ infinitely many times, so it converges to 0. But is this correct?
| Since $\cos$ is even, you may assume WLOG that $\theta \geq 0$.
Note that
$\displaystyle \sum_{k=1}^n\frac{-\theta^2}{2k\ln n} = \frac{-\theta^2}{2}\frac{1}{\ln n}\sum_{k=1}^n \frac 1k \xrightarrow[n\to \infty]{}\frac{-\theta^2}{2}$ and
$$\left|\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big) - \sum_{k=1}^n\frac{-\theta^2}{2k\ln n} \right|
\leq A_n + B_n
$$
where $\displaystyle A_n = \left|\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big) - \left(- \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big) \right) \right| $
and
$\displaystyle B_n = \left|\sum_{k=1}^n - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big) - \frac{-\theta^2}{2k\ln n} \right|$
Since $\ln(1+x) = x+O(x^2)$, there is some $K>0$ and $\epsilon$ such that $$|x|\leq \epsilon \implies |\ln(1+x)-x|\leq Kx^2$$
Since $\cos(x) = 1-\frac{x^2}2+O(x^4)$, there is some $K'>0$ and $\epsilon'$ such that $$|x|\leq \epsilon' \implies |\cos(x)-1+\frac{x^2}2|\leq K'x^4$$
For $n\geq \max\left(\exp\left(\frac{\theta}{\arccos(1-\epsilon)} \right), \exp\left(\frac{\theta}{\epsilon'} \right) \right)$,
$$\begin{align}A_n+B_n&\leq K\left(\cos(\frac{\theta}{\sqrt{\ln n}})-1 \right)^2\sum_{k=1}^n \frac 1{k^2} + K' \frac{\theta^4}{\ln^2 n} \sum_{k=1}^n \frac 1k \\&= O\left( \left(\cos(\frac{\theta}{\sqrt{\ln n}})-1 \right)^2 \right) + O\left(\frac{1}{\ln n} \right)
\\ &=o(1)
\end{align}$$
Finally,
$$\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)\xrightarrow[n\to \infty]{}\frac{-\theta^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find general term of recursive sequences $ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$ Please help to solve:
*
*$ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$
*$x_{n+1}= \frac{2}{3-x_n}, x_1=1/2$
I know answers, but can't figure out the solution.
The first one is obvious if you calculate first 3-5 terms by hand. But how can I get the result not by guessing, but mathematically?
Answers are:
*
*$x_n = \frac{n}{n+1}$
*$x_n = \frac{3\cdot2^{n-1}-2}{3\cdot2^{n-1}-1}$
| If we recursively apply the recursive relation we get
$$x_{n+1} = \frac{1}{2-x_n} = \frac{2-x_{n-2}}{3-2x_{n-2}} = \frac{3-2x_{n-3}}{4-3x_{n-3}}$$
and in general
$$x_{n+1} = \frac{k-(k-1)x_{n-k}}{k+1-kx_{n-k}}$$
Setting $k=n-1$ we get
$$x_{n+1} = \frac{n-1-(n-2)\frac{1}{2}}{n-(n-1)\frac{1}{2}} = \frac{n}{n+1}$$
I haven't checked the second one but I believe the same method should produce the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Isolate Variable in Fraction I can approximate u with a calculator by guessing or using excel but I want to isolate it.
$100 = \dfrac{1 + \dfrac{1}{(1+u)^6}}{u}$
Can not seem to do it by hand myself. Is it possible using only simple algebra?
| This looks very much as a finance problem.
Let us rewrite it as
$$\frac{u}{1+\frac{1}{(1+u)^6}}=a$$
Develop the lhs as a Taylor series to get
$$\text{lhs}=\frac{1}{2}u+\frac{3 }{2}u^2-\frac{3 }{4}u^3-4 u^4+\frac{51 }{8}u^5+O\left(u^6\right)$$ and use series reversion to get
$$u=2 a-12 a^2+156 a^3-2392 a^4+40560 a^5+O\left(a^6\right)$$ Making $a=\frac 1 {100}$ would give
$$u=\frac{2367017}{125000000}=0.0189361$$ while the exact solution is $\approx 0.0189355$
Edit
We could make the problem more general considering
$$\frac{u}{1+\frac{1}{(1+u)^n}}=a$$
$$\text{lhs}=\frac{1}{2}u+\frac{n }{4}u^2-\frac{n }{8}u^3-\frac{n
\left(n^2-4\right)}{48} u^4+\frac{n \left(n^2-2\right)}{32} u^5+O\left(u^6\right)$$ from which
$$u=2 a-2 na^2+2 n (2 n+1)a^3-\frac{2}{3} n (2 n+1) (7 n+4) a^4+$$ $$2 n (2 n+1)^2 (3 n+2)a^5+O\left(a^6\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it.
A hint or a detailed answer would be much appreciated
| $$ \left( x^{2} + x + 1 \right) \left( \frac{ - x - 7 }{ 31 } \right) - \left( x^{2} + 2 x - 4 \right) \left( \frac{ - x - 6 }{ 31 } \right) = \left( -1 \right) $$
is all you need. Cleaning up,
$$ (x+7) \left( x^{2} + x + 1 \right) - (x+6) \left( x^{2} + 2 x - 4 \right) = 31 $$
There is no guarantee of integer coefficients, even when both polynomials have integer coefficients. The units in $\mathbb Q[x]$ are nonzero rational constants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$ How can I simplify $\sqrt{1+\sqrt{i}}$?
I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$
But I'm not really sure
| $$(1+i^\frac{1}{2})^\frac{1}{2}=(1+(e^{i\frac\pi2+i2\pi n})^\frac12 )^\frac12
=(1+e^{i\frac\pi4+i\pi n} )^\frac12$$
$$=\left(1+\cos(\frac\pi4 +\pi n)+ i \sin(\frac\pi4 +\pi n) \right)^\frac12
=(re^{i\theta})^\frac12\tag 1$$
where,
$$r=\sqrt{\left(1+\cos(\frac\pi4 +\pi n)\right)^2+\sin^2(\frac\pi4 +\pi n) }
=\sqrt{2+2\cos(\frac\pi4 +\pi n) }=2| \cos(\frac\pi8 +\frac{\pi n}2) |$$
$$ \tan\theta = \frac{\sin(\frac\pi4 +\pi n)}{1+\cos(\frac\pi4 +\pi n)}
= \tan (\frac\pi8 +\frac{\pi n}2)\implies \theta =\frac\pi8 +\frac{\pi n}2+k\pi $$
Substitute $r$ and $\theta$ into (1),
$$(1+i^\frac{1}{2})^\frac{1}{2}
=r^\frac12 e^{i\frac{\theta}2} = \sqrt{2| \cos(\frac\pi8 +\frac{\pi n}2) |}
\>e^{i\pi(\frac1{16} +\frac{ n}4+\frac{k}2)}$$
which assumes multiple values. For the special case $n=k=0$,
$$(1+i^\frac{1}{2})^\frac{1}{2}
= \sqrt{2\cos\frac\pi8}\>e^{\frac{i\pi}{16}}
=\sqrt{2\cos\frac\pi8}\>(\cos\frac{\pi}{16} + i\sin\frac{\pi}{16})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I prove that if $x^2+bx+c$ is factorable, then $x^2-bx+c$ is also factorable? I want to prove that if $x^2+bx+c$ is factorable, then $x^2-bx+c$ is also factorable(by factorable I mean that it can be expressed with the product of $2$ binomials $(x+y)(x+z)$, where $y,z\in\mathbb Z$). Also, $b,c\in\mathbb Z$. It seems to be true in all the quadratic expressions I have tested, but I'm not too sure how to prove such a thing. Can someone help me prove this or provide a counterexample?
MY ATTEMPT:
$x^2+bx+c$ is factorable, so I can write it as $(x+y)(x+z)$ for some integer values $y$ and $z$. $y+z=b$, and $yz=c$.
I will initially assume that $x^2-bx+c$ is factorable and then see what I get:
$x^2-bx+c$ is factorable, so I can write it as $(x+p)(x+q)$ for $p,q\in\mathbb Z$. $p+q=-b$, and $pq=c.$
I have to somehow prove that $p$ and $q$ are integers but I'm not too sure how. Any advice would be greatly appreciated.
| If $x^2 + bx + c$ factors then by quadratic equation it must factor to
$(x - \frac {-b+ \sqrt{b^2 - 4c}}2) (x - \frac {-b- \sqrt{b^2 - 4c}}2)$ and this factors if and only if $b^2 - 4c$ is a perfect square (note: if $b$ is even/odd then $b^2 -4c$ is even/odd so $-b\pm \sqrt{b^2-4c}$ will be even so either $\frac {-b \pm \sqrt{b^2 -4c}}2$ is an integer or $x^2 + bx + c$ is not factorable.
But if $b^2-4c$ is a perfect square then
$x^2 -bx +c = (x -\frac {b+\sqrt{(-b)^2-4c}}2)(x-\frac {b+\sqrt{(-b)^2 -4c}}2)$ is also factorable.
Although a much easier idea is by user744868 in the comments.
If $f(x) = x^2 + bx + c = (x+r)(x+s)$ then $f(-x) = x^2 - bx + c= (-x+r)(-x+s)=(x-r)(x-s)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the residual theorem:
I started with $\gamma: [0,\pi] \to \mathbb{C}, t \to e^{2it}$. Since $\cos(t) = \frac{1}{2}\left(e^{it}+e^{-it}\right)$ and $\gamma'(t) = 2ie^{2it}$ we find that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(e^{it}+e^{-it}\right)} \cdot \frac{2ie^{2it}}{2ie^{2it}}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t
\end{align}
I did this with the aim to use
\begin{align}
\int_{\gamma} f(z) \mathrm{d}z = \int_{a}^{b} (f\circ \gamma)(t)\gamma'(t) \mathrm{d}t,
\end{align}
so we find
\begin{align}
\int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t = \int_{\gamma} \frac{-i}{6z+2z\sqrt{z}+2\sqrt{z}} \mathrm{d}z
\end{align}
At this point I don't know how to continue. Can anyone help?
| Using function transformations, compress the integral by a factor of $2$ in the $x$-axis, then multiply by $2$ to get:
$$2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{3+2\cos(2t)} = 2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{4 \cos^2 t+1} = 2 \int_{0}^{\pi/2} \frac{\mathrm{\sec^2 t\ d}t}{4 + \tan^2 t+1}$$
and substituting $u = \tan t$, $\mathrm{d} u = \sec^2 t \ \mathrm{d}t$:
$$2 \int_{0}^{\infty} \frac{\mathrm{d} u}{(\sqrt5)^2 + u^2} = \lim_{a \to \infty}2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{u}{\sqrt5} \right]_0^{a} = 2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{\tan t}{\sqrt5} \right]_0^{\pi/2} $$
$$=\frac{2}{\sqrt5} \left( \tan^{-1} ( \tan \pi/2) - \tan^{-1} (\tan 0)\right)= \frac{2}{\sqrt5} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt5} $$
since $\tan^{-1} (\tan x)= x, x \in [-\pi/2, \pi/2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $p(x)=x^4-x+\frac{1}{2}$ has no real roots.
What is the simplest way to prove that the polynomial
$p(x)=x^4-x+\frac{1}{2}$ has no real roots?
I did with Sturm's theorem:
$p_0(x)=x^4-x+\frac{1}{2}$
$p_1(x)=4x^3-1$
$p_2(x)=\frac34x-1$
$p_3(x)=-\frac{229}{27}$
The signs for $-\infty$ are $+,-,-,-$ and for $\infty$ are $+,+,+,-$. In the end $1-1=0$ real roots.
Can it be done faster?
| $f(x)=x^4$ is a convex function, hence its graph lies above the graph of the tangent line at $x=\frac{1}{2^{2/3}}$, whose equation is $g(x)=x-\frac{1}{2^{2/3}}+\frac{1}{2^{8/3}}$. $f(x)\geq g(x)$ implies
$$ x^4-x+\frac{1}{2}\geq -\frac{1}{2^{2/3}}+\frac{1}{2^{8/3}}+\frac{1}{2}=\frac{4-3\sqrt[3]{2}}{8} $$
but $64>27\cdot 2$, so the RHS is positive and the LHS has no real zeroes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A linear transformation such that $T(AB)=T(BA)$ The question goes as follows:
Let $V$ be a vector space and let $T: M_{2 \times 2} (R) —> V$ such that $T(AB)=T(BA)$ for all $A, B \in M_{2 \times 2}$. Show that $T(A) = 1/2(trA)T(I2)$ for all $A \in M_{2 \times 2}$.
I have no clue how to approach this. I’ve tried everything but I keep going in circles. Please help me.
| First,we know that $T$ is a linear transfromation.
Then,we just need to considering a basis of $M_{2\times 2}$.There we choose a basis as following.
$$\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right),\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right),\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right),\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
We observe that
$$\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
$$\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)$$
so,$T\left(\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)\right)=\mathbf{0}$.
similarly,$T\left(\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)\right)=\mathbf{0}$.
One can observe that
$$\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\left(\begin{array}{c}0 & 1\\1& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 1\\1& 0\\\end{array}\right)\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
Hence,$$T\left(\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\right)=\dfrac{1}{2}T\left(\left(\begin{array}{c}1 & 0\\0& 1\\\end{array}\right)\right)$$.
we conclude that $$T(A)=T\left(\left(\begin{array}{c}a & b\\c& d\\\end{array}\right)\right)=aT\left(\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\right)+bT\left(\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\right)+cT\left(\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right)\right)+dT\left(\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\right).$$
$$=\dfrac{1}{2}tr(A)T(I_2)$$
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi - c.$$ This follows that \begin{align*} \cos(a+b)=\cos(\pi - c) & \implies \cos a \cos b - \sin a \sin b = - \cos c \\ & \implies xy-\sqrt{1-x^2} \sqrt{1-y^2}=z.\end{align*} Now i am not able to proceed from here. Please help me to solve this.
| Another way to solve the same could be $\cos^{-1}x=A$, $\cos^{-1}y=B$ and $\cos^{-1}z=C$
and thus $A+B+C=\pi$ and the condition becomes $\cos A+\cos B+\cos C=\frac{3}{2}$,
which can be simplified to $2\cos\frac{A+B}{2}\frac{A-B}{2}+1-2\sin^2{\frac{c}{2}}=\frac{3}{2}$, and we know $\cos\frac{A+B}{2}=\cos\frac{\pi-C}{2}=\sin\frac{C}{2}$
Rearranging as quadratic in $\sin\frac{c}{2}$ we can write the equation as $2\sin^2\frac{c}{2}-2\cos\frac{A-B}{2}\sin\frac{c}{2}+\frac{1}{2}=0$
For real roots $D\ge0$, so we get $4\cos^2\frac{A-B}{2}-4\ge0$
i.e. $\sin^2\frac{A-B}{2}\le0$ which is only possible when $\sin^2\frac{A-B}{2}=0$
Therefore, $A=B=C=\frac{\pi}{3}\Rightarrow x=y=z=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Any shortcuts for integrating $\frac{x^6}{(x-2)^2(1-x)^5}$ by partial fractions? Is there a faster way to get the partial fraction decomposition of this $\frac{x^6}{(x-2)^2(1-x)^5}$?
$\frac{x^6}{(x-2)^2(1-x)^5} = \frac{A_1}{x-2} + \frac{A_2}{(x-2)^2} + \frac{B_1}{1-x} + \frac{B_2}{(1-x)^2} + \frac{B_3}{(1-x)^3} + \frac{B_4}{(1-x)^4} + \frac{B_5}{(1-x)^5}$
It's fairly easy to get $A_2$ and $B_5$, just by "covering up" $x-2$ and substituting $x=2$ and "covering up" $1-x$ and substituting $x=1$. So $A_2 = 64$ and $B_5=1$. But to proceed from there is there no other faster way other than to mulitply both sides of the equation by $(x-2)^2(1-x)^5$ and compare coefficients, which would result in a system of 6 equations with 6 unknowns?
| The "cover-up" rule gives correct answers but it suffers greatly from lack of rigour, in fact no rigour at all. Here's a more mathematically proper way to find the coefficients.First multiply through by $$(x-2)^2(1-x)^5$$ Your new equation is
$$x^6=A_1(x-2)(1-x)^5+A_2(1-x)^5+B_1(1-x)^4(x-2)^2+B_2(1-x)^3(x-2)^2+B_3(1-x)^2(x-2)^2+B_4(1-x)(x-2)^2+B_5(x-2)^2$$
Note that the original equation is true for all $x$ except 2 or 1 iff your new equatiion is true for all $x$ except possibly 2 or 1 iff your new equation is true for all x.[Two polynomials in one variable are equal for all $x$ iff they are equal for infinitely many $x$.There are infinitely many $x$ other than 1 or 2.]In this new equation put $x=2$ to find $A_1$ and then $x=1$ to find $B_5$Now take the derivative. $$6x^5=A_1(1-x)^5-5(1-x)^4(A_1(x-2)+A_2)+(-4B_1(1-x)^3-3B_2(1-x)^2-2B_3(1-x)-B_4)(x-2)^2+2(B_1(1-x)^4+B_2(1-x)^3+B_3(1-x)^2+B_4(1-x)+B_5)(x-2)$$ Put $x=2$ in the derived equation. Since you already know $A_2$ you can find $A_1$ Put $x=1$ in the derived equation. Since you already know $B_5$ you can find $B_4$. Another derivative will give you $B_3$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$. Question: Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$.
My approach: After trying some examples I can conjecture that, the minimum value of $x^2+y^2$ is attained at $$x=\left\lceil \frac{k}{2}\right\rceil \text{and } y=\left\lfloor\frac{k}{2}\right\rfloor \text{and equivalently at } x=\left\lfloor\frac{k}{2}\right\rfloor \text{and }y=\left\lceil \frac{k}{2}\right\rceil.$$ This also implies that the minimum value of $x^2+y^2=\left\lceil \frac{k}{2}\right\rceil^2+\left\lfloor\frac{k}{2}\right\rfloor^2.$
But, how to prove the same?
Also, since $x+y=k$, this implies that $(x+y)^2=k^2\implies x^2+y^2=k^2-2xy.$ Therefore, we need to maximize $xy$ in order to minimize $x^2+y^2$.
But, again this is leading me nowhere.
| Try this:
$$
\begin{aligned}
x^{2}+y^{2}&=\frac{(x+y)^{2}+(x-y)^{2}}{2}
\end{aligned}
$$
Since $(x+y)$ is fixed, we minimize $x^{2}+y^{2}$ by minimizing $|x-y|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing moderate decrease property in the real line for $f(z)=\frac{a}{a^2+z^2}$. Let $f(z) = \frac{a}{a^2 + z^2}$ for $a>0$. Then $f$ is holomorphic in the horizontal strip
$|\Im(z)| < a$. I would like to show that in if $|\Im (z)| < a/2$, we have some constant $A>0$ such that
$$|f(x+iy)| \le \frac{A}{1+x^2}$$ for all $x \in \mathbb{R}$ and $|y|<a/2$.
To get this bound, first note that $f(x+iy) = \frac{a}{a^2 + x^2 -y^2 + 2xy i}$.
So when we take the absolute value, we need a bound like $|a^2+x^2 - y^2+2xy i | \ge 1+x^2$.
I can see that we have $|a^2+x^2-y^2 + 2xyi| \ge |a^2+x^2-y^2| \ge (a^2+x^2) - y^2 > \frac{3a^2}{4} + x^2$.
So we have $|f(x+iy)| \le \frac{a}{\frac{3a^2}{4} + x^2}$. But how can I bound this right fraction by some $\frac{A}{1+x^2}$?
| All you need is the following: If $c>0,$ then
$$\tag 1 \frac{1+x^2}{c+x^2}\, \text{is bounded on }\mathbb R.$$
Can you prove this?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$ Let be $a,b,c,x,y,z>0$ such that $ax\ge \sqrt{(b^2+c^2)(y^2+z^2)}$. Prove that
$$(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$$
I tried to expand
$$a^2(y^2+z^2)+x^2(b^2+c^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
Here my idea was to use the condition after the means inequality:
$$a^2(y^2+z^2)+x^2(b^2+c^2) \ge 2ax\sqrt{(b^2+c^2)(y^2+z^2)}$$
$$\ge 2(b^2+c^2)(y^2+z^2)$$
but it's not good enough to prove the question
$$2(b^2+c^2)(y^2+z^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
is not true when $a$ and $x$ can be very big.
Thank you for your help.
| Using Cauchy-Schwarz:
$$
\begin{aligned} \left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[(ay-bx)^2+(cx-az)^2\right]&\geq \left[\frac{c}{a}(ay-bx)+\frac{b}{a}(cx-az)\right]^2\\
&=(cy-bz)^2\\
\end{aligned}
$$
and similarly
$$
\begin{aligned} \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]\cdot \left[(bx-ay)^2+(az-cx)^2\right]&\geq (bz-cy)^2\\
\end{aligned}
$$
Multiplying the two inequalities:
$$\left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]\cdot \left[(ay-bx)^2+(cx-az)^2\right]^2 \geq (bz-cy)^4$$
and notice that using the condition:
$$\left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]=\frac{(b^2+c^2)(y^2+z^2)}{a^2x^2}\leq 1$$
It follows that:
$$\left[(ay-bx)^2+(cx-az)^2\right]^2 \geq (bz-cy)^4$$
which is equivalent with the inequality to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^6-6x^4+12x^2-11$ is irreducible over $\mathbb{Q}$ Extracted from Pinter's Abstract Algebra, Chapter 27, Exercise B1:
Let $p(x) = x^6-6x^4+12x^2-11$, which we can transform into a polynomial in $\Bbb{Z}_3[x]$:
\begin{align*}
x^6+1
\end{align*}
Since none of the three elements $0,1,2$ in $\Bbb{Z}_3$ is a root of the polynomial, the polynomial has no factor of degree 1 in $\Bbb{Z}_3[x]$. So the only possible factorings into non constant polynomials are
\begin{align*}
x^6+1 &= (x^3+ax^2+bx+c)(x^3+dx^2+ex+f)
\end{align*}
or
\begin{align*}
x^6+1 &= (x^4+ax^3+bx^2+cx+d)(x^2+ex+f)
\end{align*}
From the first equation, since corresponding coefficients are equal, we have
\begin{align}
x^0:\qquad & cf &= 1 \tag{1} \\
x^1:\qquad & bf + ce &= 0 \tag{2} \\
x^2:\qquad & af + be + cd &= 0 \tag{3} \\
x^3:\qquad & c + f + bd + ae &= 0 \tag{4} \\
x^5:\qquad & a + d &= 0 \tag{5} \\
\end{align}
From (1), $c = f = \pm1$, and from (5), $a + d = 0$. Consequently, $af + cd = c(a + d) = 0$, and by (3), $eb = 0$. But from (2) (since $c = f$), $b + e = 0$, and therefore $b = e = 0$. It follows from (4) that $c + f = 0$, which is impossible since $c = f = \pm1$. We have just shown that $x^6 + 1$ cannot be factored into two polynomials each of degree 3.
For the second equation, however, $x^6+1=(x^2+1)^3$ in $\Bbb{Z}_3[x]$. So we cannot say $p(x)$ is irreducible over $\Bbb{Q}$ because $x^6+1$ is irreducible over $\Bbb{Z}_3$. What am I missing here?
| Update: The answer is wrong but see my comment!
I think the reasoning should be like the following. As $p(x)$ has integer coefficients and is monic every zero of $p$ that lies in $\mathbb{Q}$ is also integer. But every integer zero of $p$ must divide the absolute term which is 11. Therefore, it could only be $\pm1$ or $\pm11$. Neither is a solution. The explanation above shows that $p$ cant be the product of two cubic polynomials. So if $p$ were reducible it had an irreducible quadratic monic polynomial with as factor which would have itself two zeros of the form $\pm\sqrt{q}+r$ whose square is in $\mathbb{Q}$. But then $p$ has only even powers of $x$ so substituting $x^2\rightarrow{}y$ gives a cubic polynomial which is either irreducible or has a linear factor. But then the same reasoning as above applies and since neither $\pm1$ nor $\pm11$ are zeros of that polynomial we are done.
| {
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"url": "https://math.stackexchange.com/questions/3579410",
"timestamp": "2023-03-29T00:00:00",
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Proving a function(with two variables) is continuous I am having a difficulty solving this problem. First of all, I'm sorry that the problem isn't well written but I am not very good with typing out math problems, due to the fact that I am new to this, so I hope it's at least understandable. Next, I want to say that I've tried solving this using polar coordinates and also by changing y with $y=kx$, $y=kx^2$. But it didn't work. Anyway, the problem says: Find parameter a so that a function is continuous. (I tried to translate it correctly in English.) I hope someone can help me solve this problem, I would be so grateful.
$$
f(x,y) =
\begin{cases}
\dfrac{5 - \sqrt{25-x^2-y^2}}{7 - \sqrt{49-x^2-y^2}} & (x,y)\neq (0,0) \\
\\
a & (x,y)=(0,0)
\end{cases}
$$
| In these problems with roots, a typical strategy is the one of “rationalize” the fraction:
If you write:
$$\frac{5-\sqrt{25-x^2-y^2}}{7-\sqrt{49-x^2-y^2}}\cdot \frac{5+\sqrt{25-x^2-y^2}}{5+\sqrt{25-x^2-y^2}}\cdot \frac{7+\sqrt{49-x^2-y^2}}{7+\sqrt{49-x^2-y^2}}= \frac{x^2+y^2}{x^2+y^2}\cdot\frac {7+\sqrt{49-x^2-y^2}} {5+\sqrt{25-x^2-y^2}}=\frac {7+\sqrt{49-x^2-y^2}} {5+\sqrt{25-x^2-y^2}}$$
So your limit is $\frac 75$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Two Polynomials with a Common Quadratic Factor Let $f(x)=x^3-ax^2-bx-3a$ and $g(x)=x^3+(a-2)x^2-bx-3b$. If they have a common quadratic factor, then find the value of $a$ and $b$.
My Attempt
Let $h(x)$ be the common quadratic factor. Then $h(x)$ also the factor of $g(x)-f(x)$, that is
$$(2a-2)x^2+(3a-3b)$$
Since $h(x)$ a quadratic factor, then
$$(2a-2)x^2+(3a-3b)=k \cdot h(x)$$
Where $k$ is a constant.
But, i don't know how to continue, because there are many possible values for $k$.
Any advice?
| Since $(2a-2)x^2+(3a-3b)$ is a common factor and the common factor is said to be a quadratic, the (monic) common factor must look like $h(x) = x^2+\dfrac 32 \cdot \dfrac{a-b}{a-1}$.
This is ugly. So let's try and avoid the direction that this is taking us. We can assume that $h(x) = x^2 - \alpha$ where $\alpha = -\dfrac 32 \cdot \dfrac{a-b}{a-1}$ and, for some $u$ and for some $v$
\begin{align}
f(x) &= (x-u)(x^2-\alpha) \\
x^3-ax^2-bx-3a &= x^3 - ux^2 - \alpha x +\alpha u \\
u &= a \\
\alpha &= b \\
\alpha u &= -3a
\end{align}
\begin{align}
g(x) &= (x-v)(x^2-\alpha) \\
x^3+(a-2)x^2-bx-3b &= x^3 - vx^2 - \alpha x +\alpha v \\
v &= 2-a \\
\alpha &= b \\
\alpha v &= -3b
\end{align}
Since $\alpha = b$, then $\alpha v = -3b
\implies b v = -3b
\implies b(v+3)=0$.
So, either $b=0$ or $v=-3$
If $b=0$, then we must also have $a=0$.
In which case,
\begin{align}
f(x) &=x^3 \\
g(x) &=x^3-2x^2 \\
h(x) &=x^2
\end{align}
If $v=-3$, then $a=5$, $u=5$, $\alpha = -3$, and $b=-3$.
In which case
\begin{align}
f(x) &= x^3-5x^2+3x-15 \\
g(x) &= x^3+3x^2+3x+9 \\
h(x) &= x^2+3
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Surprising fact about a certain number-theoretic function Ante suggested the following function :
For natural number $n$ we can observe the $n$ remainders $b_1,...,b_n$ by writing $n$ as $n=a_k \cdot k+b_k$ for $1 \leq k \leq n$
Because of the familiar division-with-remainder-theorem we have $0 \leq b_k <n$
Now we can study the sum $$r(n)=\sum_{k=1}^{\lfloor \frac{n-1}{2} \rfloor}b_k$$
After playing around with some values with support of Haran and Ante, we noticed that $$r(b)=r(b+1)$$ seems to hold if and only if $b+1$ is a power of $2$ or $3$ , including $2$ and $3$
There is no counterexmple upto $10^4$
? r={(p)->su=0;for(j=2,(p-1)/2,su=su+lift(Mod(p,j)));su}
%94 = (p)->su=0;for(j=2,(p-1)/2,su=su+lift(Mod(p,j)));su
? for(j=1,10^4,if(r(j)==r(j+1),print(j," ",factor(j+1))))
1 Mat([2, 1])
2 Mat([3, 1])
3 Mat([2, 2])
7 Mat([2, 3])
8 Mat([3, 2])
15 Mat([2, 4])
26 Mat([3, 3])
31 Mat([2, 5])
63 Mat([2, 6])
80 Mat([3, 4])
127 Mat([2, 7])
242 Mat([3, 5])
255 Mat([2, 8])
511 Mat([2, 9])
728 Mat([3, 6])
1023 Mat([2, 10])
2047 Mat([2, 11])
2186 Mat([3, 7])
4095 Mat([2, 12])
6560 Mat([3, 8])
8191 Mat([2, 13])
?
Is this in fact true, and if yes, why ?
| We have:
$$r(b)=r(b+1)$$
$$\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k}) =\sum_{k=1}^{\lfloor \frac{b}{2} \rfloor} ((b+1) \bmod{k}) $$
since $n \equiv b_k \pmod{k}$. Now, we take two cases:
Case $1$ : When $b$ is odd
We have:
$$\sum_{k=1}^{\frac{b-1}{2}} (b \bmod{k}) =\sum_{k=1}^{\frac{b-1}{2}} ((b+1) \bmod{k})$$
$$\sum_{k=1}^{\frac{b-1}{2}} \bigg((b \bmod{k})-((b+1) \bmod{k})\bigg)=0$$
We can see that:
$$ (b \bmod{k})-((b+1) \bmod{k}) =
\begin{cases}
-1 & \text{if $k \nmid (b+1)$} \\
k-1 & \text{if $k \mid (b+1)$}
\end{cases}$$
Essentially, we are to substitute $-1$ for all the values $1 \leqslant k \leqslant \frac{b-1}{2}$ and add $k$ when it is a divisor of $b+1$. $k$ will be all the divisors of $b+1$ excpet $b+1$ and $\frac{b+1}{2}$ since $k \leqslant \frac{b-1}{2}$. We have:
$$\sum_{k=1}^{\frac{b-1}{2}} \bigg((b \bmod{k})-((b+1) \bmod{k})\bigg)=\bigg( \sum_{k=1}^{\frac{b-1}{2}} (-1) \bigg) + \sigma(b+1)-(b+1)-\bigg(\frac{b+1}{2}\bigg)$$
$$ \implies -\bigg(\frac{b-1}{2}\bigg)+\sigma(b+1)-(b+1)-\bigg(\frac{b+1}{2}\bigg)=0 \implies \sigma(b+1)=2b+1$$
Since $b+1$ is even, we have $b+1=x$ for all even $x$ satisfying:
$$\sigma(x)=2x-1$$
Clearly, $x=2^k$ works for all non-negative integers $k$. Such $x$ are known as 'almost perfect numbers'. It is unknown whether powers of $2$ are the only such solutions. For your claim to be true, we need all even almost perfect numbers to be powers of $2$.
Case $2$ : When $b$ is even
This works similarly:
$$\sum_{k=1}^{\frac{b}{2}-1} (b \bmod{k}) =\sum_{k=1}^{\frac{b}{2}} ((b+1) \bmod{k})$$
We can easily see that since $\frac{b}{2} \mid b$, we have $\sum_{k=1}^{\frac{b}{2}-1} (b \bmod{k})=\sum_{k=1}^{\frac{b}{2}} (b \bmod{k})$. Now, we have:
$$\sum_{k=1}^{\frac{b}{2}} (b \bmod{k}) =\sum_{k=1}^{\frac{b}{2}} ((b+1) \bmod{k})$$
$$\sum_{k=1}^{\frac{b}{2}} \bigg((b \bmod{k})-((b+1) \bmod{k})\bigg)=0$$
Using the same logic as in the first case (but we are only to exclude the divisor $b+1$):
$$\sum_{k=1}^{\frac{b}{2}} \bigg((b \bmod{k})-((b+1) \bmod{k})\bigg) = \bigg( \sum_{k=1}^{\frac{b}{2}} (-1) \bigg)+\sigma(b+1)-(b+1)=0$$
$$\sigma(b+1)=\frac{3b+2}{2}$$
We have $b+1=x$ for all odd $x$ satisfying:
$$\sigma(x)=\frac{3x-1}{2}$$
It is clear that $x=3^k$ works for all powers of $3$. However, I doubt it is possible to prove whether these are the only odd solutions, with elementary methods. I wasn't able to find any literature on this subject.
Link:
Almost Perfect Numbers : https://mathworld.wolfram.com/AlmostPerfectNumber.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| What you have done is absolutely correct, except where you forgot to mention that $\theta$ is in $(0, \pi)$, but you can simplify your answer further.
The book's answer might be something like $-\frac{1}{3} \sqrt{9-x^2} (x^2+18)$, which you can get by factoring out a factor of $\sqrt{9-x^2}$:
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$= \sqrt{9-x^2} \left(-9 + \frac{1}{3}(9-x^2) \right)+ C$$
and you can surely continue from here.
| {
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"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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For fixed hypotenuse, can the number of primitive Pythagorean triples exceed the number of non-primitive ones? For the equation,
$$a^2+b^2=c^2$$
if $c$ is fixed and the number of natural solutions for $a, b$ is greater than $1$, then can the number of primitive solutions (solutions in which $a, b, c$ are coprime) exceed the number of non-primitive ones?
After testing a large number of cases I believe that the number of non-primitive solutions will always exceed the number of primitive ones although I have no proof for this.
If false, what is the smallest counter-example for $c$?
| For some hypotenuse $c$, let the number of primitive solutions be greater than the number of non-primitive solutions.
Assume that $p \mid c$ for some prime $p$. Clearly, there is atleast one primitive solution $(a,b,c)$. Then, we have:
$$p^2 \mid c^2 \implies p^2 \mid (a^2+b^2)$$
It is easy to see that since $p \nmid a$, we must have $p \neq 2$ and $p \not\equiv 3 \pmod{4}$. Thus, any prime $p \mid c$ satisfies $p \equiv 1 \pmod{4}$.
Now, let the prime factorization of $c$ be:
$$c=\prod_{k=1}^n p_k^{x_k}$$
where all $p_k$ are $1 \bmod{4}$. Let $p_k=(a_k+b_ki)(a_k-b_ki)$ where $(a_k+b_ki)$ and $(a_k-b_ki)$ are Gaussian primes (since all primes in naturals which are $1 \bmod{4}$ are products of two Gaussian primes). We clearly have $\gcd(a_k,b_k)=1$. Then:
$$c=\prod_{k=1}^n (a_k+b_ki)^{x_k}(a_k-b_ki)^{x_k}$$
Now, we have $c^2=a^2+b^2=(a+bi)(a-bi)$. We will write $a+bi$ as the product of some of these Gaussian Primes and $a-bi$ as the product of the rest.
For all solutions to $c^2=a^2+b^2$ (including negative), we have to split the Gaussian primes equally, i.e. since we need to maintain the fact that $a+bi$ and $a-bi$ are conjugates, whenever we write $a_k+b_ki$ in the product of $a+bi$, we are to write $a_k-b_ki$ in the product of $a-bi$ and vice versa.
For each $p_k$, we have $2x_k+1$ choices for this process since $p_k^2 \mid (a+bi)(a-bi)$ and we have to divide $(a_k+b_ki)^{2x_k}(a_k-b_ki)^{2x_k}$ (so $a+bi$ can have $a_k+b_ki$ for $t$ number of times for $0 \leqslant t \leqslant 2x_k$). Finally, we can multiply by units $i,-i,1,-1$ which is $4$ choices. Thus, the number of solutions is:
$$4\prod_{k=1}^n (2x_k+1) \geqslant 4 \cdot 3^n$$
Since we need to remove $(c,0),(0,c),(-c,0),(0,-c)$, we reduce $4$. Furthermore, we divide by $4$ since both $a$ and $b$ are positive and divide by $2$ since $(a,b)$ is the same as $(b,a)$, giving:
$$T_{\text{all}} \geqslant \frac{4\cdot3^n-4}{8} = \frac{3^n-1}{2}$$
For primitive solutions alone, we need to segregate either all of the $a_k+b_ki$ or all of the $a_k-b_ki$ to $a+bi$ since $\gcd(a,b)=1$. This only gives $2$ choices per $p_k$. Multiplying by units, total choices are $4 \cdot 2^n$.
Again, we are to do the necessary removal. $(c,0)$ won't work as primitive, so we are to only divide by $8$. Thus:
$$T_{\text{primitive}} = \frac{4 \cdot 2^n}{8}= 2^{n-1}$$
We need to have:
$$2T_{\text{primitive}} > T_{\text{all}}$$
$$2^n>\frac{3^n-1}{2} \implies 2^{n+1}>3^n-1$$
Clearly, we have $n=1$. Thus:
$$T_{\text{all}}=\frac{(2x+1)-1}{2}=x$$
$$T_{\text{primitve}}=2^{n-1}=1$$
Since we have $2T_{\text{primitive}} > T_{\text{all}}$, we have $x=1$ showing $c=p$ is prime.
Thus, only all odd prime hypotenuse of the form $4k+1$ are exceptions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Upper Bound on a en exponential function I am trying to upper bound the following function and find its growth rate:
\begin{equation}
\psi(y) \stackrel{\triangle}{=} \int_{0}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx,~y>0,
\end{equation}
where $f(x)>0$ satisfies $\int_{0}^{+\infty}f(x)\,dx = 1$ (it is a pdf of the random variable $X$). I would like to find the growth rate of $\psi(y)$ for large values of $y$, i.e., would like to know if $\psi(y)$ grows like $O(y^{-k})$, for some $k>1$.
Here are my attempts so far:
It is straightforward to show that the function $\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)\in[0,1]$ is increasing in $x$ for $y>x>0$ and is decreasing in $x$ for $0<y<x$.
Then one can break the bounds of the integral to $[0,y/2], \, [y/2,3y/2], \, [3y/2, y^{3/2}],\,[y^{3/2},\infty)$ and then evaluate each integral. Doing so will give the following:
\begin{align}
\int_{0}^{y/2}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1\\
\int_{y/2}^{3y/2}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &\stackrel{?}{=}O(y^{-k}),\, k>1\\
\int_{3y/2}^{y^{3/2}}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1\\
\int_{y^{3/2}}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1
\end{align}
Therefore, if we can prove that the second integral also grows like $O(y^{-k}),\,k>1$, then we are done. However, I have not been able to show this and am stuck. Any help would be highly appreciated!
| First, let us see an example in which $\lim_{y\to \infty} y\psi(y) = \infty$.
Let (log-Cauchy distribution)
$$f(x) = \frac{1}{\pi x (1 + (\ln x)^2)}, \ x > 0.$$
We have, for sufficiently large $y$,
\begin{align}
\psi(y) &= \int_{0}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)
\frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x\\
&\ge \int_{y - \sqrt[4]{y}}^{y + \sqrt[4]{y}}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)
\frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\
&\ge \int_{y - \sqrt[4]{y}}^{y + \sqrt[4]{y}}\exp\left(-\frac{\sqrt{y}}{2(\sigma_0^2 + \sigma_1^2 y/2)}\right)
\frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\
&\ge \int_{y - \sqrt[4]{y}}^{y + \sqrt[4]{y}} \frac{1}{2}\cdot
\frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\
&= \frac{\arctan(\ln(y + \sqrt[4]{y})) - \arctan(\ln(y - \sqrt[4]{y}))}{2\pi}\\
&= \frac{1}{2\pi}
\arctan \frac{\ln(y + \sqrt[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{y})}
\end{align}
where we have used $\tan (x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$.
Thus, we have
\begin{align}
\lim_{y\to \infty} y\psi(y) &= \lim_{y\to \infty} \frac{1}{2\pi} y
\arctan \frac{\ln(y + \sqrt[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{y})}\\
&= \lim_{y\to \infty} \frac{1}{2\pi} y \frac{\ln(y + \sqrt[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{y})}\\
&= \infty
\end{align}
where we have used $\lim_{z \to 0}\frac{\arctan z}{z} = 1$.
Second, if $\mathbb{E}[X]$ and $\mathbb{E}[X^2]$ are both finite, one can prove that $\psi(y) = O(y^{-2})$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many times must I apply L’Hopital? I have this limit:
$$\lim _{x\to 0}\left(\frac{e^{x^2}+2\cos \left(x\right)-3}{x\sin \left(x^3\right)}\right)=\left(\frac 00\right)=\lim _{x\to 0}\frac{\frac{d}{dx}\left(e^{x^2}+2\cos \left(x\right)-3\right)}{\frac{d}{dx}\left(x\sin \left(x^3\right)\right)}$$
$$\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\frac{d}{dx}\left(x\right)\sin \left(x^3\right)+\frac{d}{dx}\left(\sin \left(x^3\right)\right)x}=\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\sin \left(x^3\right)+3x^3\cos \left(x^3\right)+\sin \left(x^3\right)}$$
but yet we have $(0/0)$. If I apply L’Hopital again, I obtain
$$=\lim_{x\to0}\frac{2\left(2e^{x^2}x^2+e^{x^2}\right)-2\cos \left(x\right)}{15x^2\cos \left(x^3\right)-9x^5\sin \left(x^3\right)}$$
again giving $(0/0)$. But if I apply L’Hopital a thousand times I'll go on tilt. What is the best solution in these cases? With the main limits or applying upper bonds?
| Let's first attack the numerator alone, repetitively differentiating until we no longer get zero. Let $N = \mathrm{e}^{x^2} + 2 \cos x - 3$.
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} N &= 2 x \mathrm{e}^{x^2} - 2 \sin x \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^2}{\mathrm{d}x^2} N &= (4 x^2 +2)\mathrm{e}^{x^2} - 2 \cos x \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^3}{\mathrm{d}x^3} N &= (8 x^3 +12 x)\mathrm{e}^{x^2} + 2 \sin x \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^4}{\mathrm{d}x^4} N &= (16 x^4 +48x^2+12)\mathrm{e}^{x^2} + 2 \cos x \xrightarrow{x \rightarrow 0} 14 \text{.}
\end{align*}
Now let $D = x \sin x^3$. If any of its first three derivatives are nonzero, our limit is zero. Otherwise, the fourth derivative will resolve the value of the limit.
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} D &= 3 x^3 \cos x^3 + \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^2}{\mathrm{d}x^2} D &= 12 x^2 \cos x^3 - 9x^5 \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^3}{\mathrm{d}x^3} D &= (24x - 27 x^7) \cos x^3 - 81 x^4 \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\
\frac{\mathrm{d}^4}{\mathrm{d}x^4} D &= (24 - 432 x^6) \cos x^3 + (-396 x^3 + 81 x^9) \sin x^3 \xrightarrow{x \rightarrow 0} 24 \text{.}
\end{align*}
So the limit is $\frac{14}{24} = \frac{7}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\sin(x) - \sqrt3 \sin(3x) + \sin(5x) < 0$ for $0My attempt at solving this:
$\sin(x) - \sqrt3\sin(3x) + \sin(5x) < 0$
$2\sin \left(\frac{5x+x}2\right) + \cos\left(\frac{5x-x}2\right) - \sqrt 3\sin(3x) < 0$
I divide everything with 2:
$\sin(3x) + \frac12\cos(2x) - \frac {\sqrt 3}2\sin(3x) < 0$
I think I have gone the wrong way at solving this problem. Please advise.
| $$\sin (x)+\sin (5x) - \sqrt3\sin(3x) <0\Rightarrow 2\sin(3x)\cos (2x) - \sqrt{3}\sin(3x) < 0 \\\Rightarrow \sin(3x)\left(\cos(2x) -\frac{\sqrt{3}}{2}\right)<0 $$
Case $1$:
$$ \sin(3x) < 0 \text{ and } \cos (2x) > \frac{\sqrt3}2 \implies \frac{\pi}{12}<x<\frac\pi3$$
Case $2$:
$$ \sin (3x) > 0 \text{ and } \cos (2x) < \frac{\sqrt3}2 \implies \frac{2\pi}{3} <x< \frac{11\pi}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Easy way to find the partial fraction I always have trouble trying to find the partial fraction, especially for complicated ones.
For example, this is what I will do to find the partial fraction of $\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$
*
*$\displaystyle \frac{A}{x}+\frac{B}{2x+3}+\frac{C}{(2x+3)^2}+\frac{D}{(2x+3)^3}$
*$\displaystyle A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $
*When x = $0$ -> $\displaystyle 27A = 27, A=1$
*$\displaystyle (2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27$
After what I can do is to expand all the elements and group them based on their exponent, and solve the system of equation.
However, I remember seeing there exists an easier solution. Also given that there will be no calculator available on the exam, doing this way will take a long time and results in possible errors.
Does anyone have an easier way to solve this question or similar ones?
Thanks!
| A suggestion may be in form of $$\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$$
look for $(2x+3)^3=8x^3+36x^2+54x+27$ so ,we can rewrite $$\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}=\\\frac{(2x+3)^3-x^2-12x}{x(2x+3)^3}=\\
\frac{(2x+3)^3}{x(2x+3)^3}-\frac{x(x+12)}{x(2x+3)^3}=\\
\frac{1}{x}-\frac{(x+12)}{(2x+3)^3}=\\$$ can you take over ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the system of inequalities $3x^2+2x-1<0$ and $(3a-2)x-a^2x+2<0$ possesses a solution, find the least natural number $a$ If the system of equations $3x^2+2x-1<0$ and $(3a-2)x-a^2x+2<0$ possesses a solution, find the least natural number $a$
My attempt is as follows:-
$$3x^2+3x-x-1<0$$
$$3x(x+1)-1(x+1)<0$$
$$x\in\left(-1,\dfrac{1}{3}\right)$$
$$(-a^2+3a-2)x<-2$$
$$(a^2-3a+2)x>2$$
Case $1$: $a^2-3a+2<0$
$$a\in(1,2)$$
In this interval there is no natural number, hence no need to proceed further.
Case $2$: $a^2-3a+2\ge0$
$$a\in(-\infty,1]\cup[2,\infty)$$
Checking for $a=1$:
$0>2$, which is not possible
Checking for $a=2$
$(4-6+2)x>2$
$0>2$, which is not possible
$$x>\dfrac{2}{a^2-3a+2}$$
$$x\in\left(\dfrac{2}{a^2-3a+2},\infty\right)$$
If system of equations possess a solution, then $\dfrac{2}{a^2-3a+2}<\dfrac{1}{3}$
$$6<a^2-3a+2$$
$$a^2-3a-4>0$$
$$a^2-4a+a-4>0$$
$$a(a-4)+(a-4)>0$$
$$(a+1)(a-4)>0$$
$$a\in(-\infty,-1)\cup(4,\infty)$$
So $a=5$ should be the answer, but actual answer is $2$
| Inequalities can be sumized in this system:
$x<-1$ and $x>\frac{1}{3}$
$a<=\frac{3\sqrt{x}-\sqrt{x+8}}{2\sqrt{x}}$
$a>=\frac{3\sqrt{x}+\sqrt{x+8}}{2\sqrt{x}}$
The least natural number is $a=4$ for $x=\frac{1}{3}$.
Other values are:
$a=5$ for $x=\frac{1}{6}$;
$a=6$ for $x=\frac{1}{10}$;
$a=7$ for $x=\frac{1}{15}$;
$a=8$ for $x=\frac{1}{21}$;
$a=9$ for $x=\frac{1}{28}$;
$a=10$ for $x=\frac{1}{36}$,
etc…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle \sum_{k=1}^{n+1}\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
which can be summarised to:
$ \displaystyle \sum_{k=1}^{n}\frac{1}{(5k + 1) (5k + 6)} + \frac{1}{(5(n+1) + 1) (5(n+1) + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
Using the Induction Hyptohesesis this turns to:
$ ( \frac{1}{30} - \frac{1}{5(5n + 6)} ) + \frac{1}{(5n + 6 ) (5n+6+5)} $
when we let s = 5n +6 we get:
$ \frac{1}{30} - \frac{1}{5s} + \frac{1}{(s) (s+5)} = \frac{1}{30} - \frac{1}{5(s + 5)} $
substracting both sides by $\frac{1}{30}$ yields:
$- \frac{1}{5s} + \frac{1}{(s) (s+5)} = - \frac{1}{5(s + 5)}$ and this is where I'm stuck. I'm working on this for more than 6h but my mathematical foundation is too weak to get a viable solution. Please help me.
| Let $ n $ be a positive integer.
Observe that : $ \left(\forall k\in\mathbb{N}\right),\ \frac{1}{\left(5k+1\right)\left(5k+6\right)}=\frac{1}{5}\left(\frac{1}{5k+1}-\frac{1}{5k+6}\right) \cdot $
Thus, \begin{aligned} \sum\limits_{k=1}^{n}{\frac{1}{\left(5k+1\right)\left(5k+6\right)}}&=\frac{1}{5}\left(\sum\limits_{k=1}^{n}{\frac{1}{5k+1}}-\sum\limits_{k=1}^{n}{\frac{1}{5k+6}}\right)\\ &=\frac{1}{5}\left(\sum\limits_{k=0}^{n-1}{\frac{1}{5k+6}}-\sum\limits_{k=1}^{n}{\frac{1}{5k+6}}\right)\\ &=\frac{1}{5}\left(\frac{1}{6}-\frac{1}{5n+6}\right)\\ \sum\limits_{k=1}^{n}{\frac{1}{\left(5k+1\right)\left(5k+6\right)}}&=\frac{1}{30}-\frac{1}{5\left(5n+6\right)} \end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that if $(b^n-1)/(b-1)$ is a power of prime numbers, where $b,n>1$ are positive integers, then $n$ must be a prime number.
Show that if $(b^n-1)/(b-1)$ is the power of a prime number, where $b,n>1$ are positive integers, then $n$ must be a prime number.
My solution:
If $n$ is composite, then let $n=mk$, $m,k>1$,
\begin{align*}
\frac{b^n-1}{b-1} &= 1+b+\cdots+b^{n-1} \\
&=(1+b+\cdots+b^{k-1} )+(b^k+b^{k+1}+\cdots+b^{2k-1}) \\ &\quad\,+\cdots+(b^{(m-1)k}+b^{(m-1)k+1}+\cdots+b^{mk-1}) \\
&=(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k})
\end{align*}
Which is composite and distinct, thus, for $(b^n-1)/(b-1)$ to be a power of primes, $n$ is not composite, thus it must be prime.
However, $(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k})$ might be equal to $p^x \times p^y$, where $p$ is prime.
Is there any better solution?
| Let $(b^n-1)/(b-1)=p^x$ where $p$ is a prime and $x> 0$. If $n$ is composite, there are two cases.
*
*There exists a prime $q$ such that $n=q^m$ for some $m>1$. Note
$$p^x=\frac{b^n-1}{b-1}=\frac{b^{q^m}-1}{b^{q^{m-1}}-1}\cdot \frac{b^{q^{m-1}}-1}{b-1},$$
we can assume
$$\frac{b^{q^{m-1}}-1}{b-1}=p^y$$
for some $0< y< x$. Then we have
\begin{align}
1+q(b-1)p^y+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^i&=\left((b-1)p^y+1\right)^q\\
&=\left(b^{q^{m-1}}\right)^q\\
&=b^{q^m}\\
&=(b-1)p^x+1,
\end{align}
i.e.,
$$q+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^{i-1}=p^{x-y}.$$
Hence, $p\mid q$. Recall that $p$ and $q$ are both primes, so $p=q$, we further have
$$1+\binom{p}{2}(b-1)p^{y-1}+\sum_{i=3}^p\binom{p}{i}(b-1)^{i-1}p^{y(i-2)}=p^{x-y-1}.$$
Note the left hand side is no less than 2, so both sides are divisible by $p$, i.e., the term $\binom{p}{2}(b-1)p^{y-1}$ cannot be divisible by $p$, thus $p=2$ and $y=1$. We further have $b=p^{x-2}$, i.e., $p\mid b$ (recall that $b>1$). However, note
$$p^x=\frac{b^n-1}{b-1}=1+b+\cdots+b^{n-1},$$
it is impossible that $p\mid b$.
*There exist two co-prime numbers $s,t>1$ such that $n=st$. In this case, we have
$$p^x=\frac{b^n-1}{b-1}=\frac{b^{st}-1}{b^s-1}\cdot\frac{b^s-1}{b-1},$$
which means $(b^s-1)/(b-1)$ is divisible by $p$. Similarly, $(b^t-1)/(b-1)$ is also divisible by $p$. Since $s$ and $t$ are co-prime, there exists integers $w_s,w_t$ such that $w_ss+w_tt=1$. Without loss of generality, we assume $w_s>0$ and $w_t<0$. Then we have
$$\frac{b^{w_ss}-1}{b^s-1}\cdot\frac{b^s-1}{b-1}-b\cdot\frac{b^{-w_tt}-1}{b^t-1}\cdot\frac{b^t-1}{b-1}$$
is also divisible by $p$. Note the expression above is exactly
$$\left(1+b+\cdots+b^{w_ss}\right)-b\left(1+b+\cdots+b^{-w_tt}\right)=1,$$
which is impossible.
As a conclusion, $n$ must be a prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why to simplify $\sin x - 1 =\cos x$ you have to multiply both side by $(\frac{\sqrt 2}{2}) $? Why do you have to multiply both side of $$\sin x - 1 = \cos x$$
by:
$$(\frac{\sqrt 2}{2}) $$ to simplify it?
I am doing Shaum's pre-calculus and the final answer to this question is: π but I have no clue where the idea of multiplying by $(\frac{\sqrt 2}{2})$ came from. Like what is the taught process behind it?
| It is because $\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}$. Then we get
$$\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x=\cos\dfrac{\pi}{4}\sin x-\sin\dfrac{\pi}{4}\cos x=\sin(x-\dfrac{\pi}{4}).$$
Generally, for expressions like $$a\cos x+b\sin x=c$$
we do $$\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x=\frac{c}{\sqrt{a^2+b^2}}$$
Now considering $$\alpha=\sin^{-1}\frac{a}{\sqrt{a^2+b^2}}=\cos^{-1}\frac{b}{\sqrt{a^2+b^2}}$$
we reach to
$$\sin\alpha\cos x+\cos\alpha \sin x =\frac{c}{\sqrt{a^2+b^2}}$$
or $$\sin(\alpha+x)=\frac{c}{\sqrt{a^2+b^2}}$$
which can be solved easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find an analytical function if initial value and the imaginary part is given. So I must find $f(z) = u(x,y) + v(x,y), \ \ f(0)=1$ if $v(x,y) = 2x^3 - 6xy^2-4xy+2x$
$$\frac{\partial v}{\partial x} = 6x^2-6y^2-4y +2 \\ \frac{\partial v}{ \partial y} = -12xy -4x \\ f'(z) = -12xy-4x + i(6x^2-6y^2-4y+2) \\ \int-4z+i(6z^2+2) dz = -2z^2+i(2z^3+2z) + C \Rightarrow \\ f(0) = 1 \Rightarrow C =1 \\ f(z) = (-2x^2+3x^2y-2y^2) +i(-4xy+x^3+xy^2) $$
It is visible that the imaginary part is not equal to the given $v(x,y)$ function. Where is my mistake?
| You got the right answer: $f(z)$ is $-2z^2+2iz^3+2iz+1$ indeed. And\begin{multline}2(x+yi)^2+2i(x+yi)^3+2i(x+yi)+1=\\=-6 x^2 y-2 x^2+2 y^3+2 y^2-2 y+1+i(2x^3-6xy^2-4xy+2x).\end{multline}So, again, yes, you got the right answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
| You've already got adequate answers, but this is just another way to do it -- especially fitted for this particular curve, as we'll find out.
The problem, reformulated, is to find the smallest circle intersecting the curve so that no point on the curve lies within the circle; on the other hand we need the largest circle intersecting the curve so that no point on the curve lies outside it.
*
*To find this least circle, let its radius be $r.$ Then we seek the least such $r$ satisfying the equations $x^2+y^2=r^2$ and $x^4+y^4+3xy=2.$ Squaring the first and substituting gives $$2-3xy+2(xy)^2=r^4,$$ a quadratic in $xy.$ Thus we easily find that the minimum value of $r^4,$ occurring at $xy=3/4,$ is given by $r^4=7/8.$ Hence our points lie on $xy=3/4$ and $x^2+y^2=\sqrt{7/8}.$ Thus we find that $x^2$ and $y^2$ are the roots of the quadratic $$m^2-\sqrt{7/8}m+9/16=0.$$ From here I believe you may proceed.
*To find the largest circle, we instead use the complementary circle of the hyperbola $x^2-y^2=s^2,$ whose vertices define the circle. Hence the radius of this circle is $s.$ Proceeding as above, we obtain that $$2-3xy-2(xy)^2=s^4,$$ which gives the maximum of $s^4$ as $25/8,$ occurring when $xy=-3/4.$ Thus our farthest points lie on the curves $xy=-3/4$ and $x^2-y^2=5/\sqrt 8.$ To solve this note that $(x^2+y^2)^2-(x^2-y^2)^2=4(xy)^2,$ which gives us that $x^2+y^2=\sqrt{43/8}.$ Hence $x^2$ and $y^2$ are the roots of the quadratic $$n^2-\sqrt{43/8}n+9/16=0,$$ from where you may proceed.
| {
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"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$ Show that
$$\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$$
My Attempt:
Let $$I=\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
Using $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx$ we can write:
$$I=\int_{0}^{\pi} \frac {(\pi - x)dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} $$
$$I=\int_{0}^{\pi} \frac {\pi dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - \int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$
$$I=2\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - I$$
$$I=\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
| Continue with the substitution $t=\tan x$,
$$\begin{align}
& \pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} \\
& = \pi \int_0^\infty \frac{1+t^2}{(b^2+a^2t^2)^2}dt \\
&=\frac{\pi(a^2-b^2)}{2a^2b^2} \frac t {b^2+a^2t^2}\bigg|_ 0^\infty
+ \frac{\pi(a^2+b^2)}{2a^2b^2} \int_0^\infty \frac {dt}{b^2+a^2t^2} \\
& =0+\frac{\pi(a^2+b^2)}{2a^3b^3} \tan^{-1}\frac {at}b\bigg|_0^\infty \\
&=\frac{\pi^2(a^2+b^2)}{4a^3b^3}\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a polynomial of degree 5 that is irreducible over $\mathbb{Z}_3$. Can someone please let me know if this looks ok? Thanks in advance!
An irreducible degree $5$ polynomial over $\mathbb{Z}_3$is one such that $f(0)\neq0,f(1)\neq0,f(2)\neq0$.
Take e.g. $p(x)=x^5+x^4+x^3+x^2+1$
$p(0)=1$
$p(1)=2$
$p(2)=1$
And, $$x^2,x^2+1, x^2+x+1\ \nmid\ x^5+x^4+x^3+x^2+1.$$
$\therefore x^5+x^4+x^3+x^2+1$ is an irreducible polynomial over $\mathbb{Z}_3$.
| Your problem is to find an irreducible polynomial of degree $5$ over $\mathbb Z_3$. You haven't found one yet, because
$$x^5+x^4+x^3+x^2+1=(x^2+x+2)(x^3+2x+2).$$
Note that the irreducible (monic) polynomials of degree $2$ are $x^2+1$ and $x^2+x+2$ and $x^2+2x+2$. You need to check all three of them as possible factors.
Hint. Try $x^5+2x+1$.
| {
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