Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Equation of circle with minimum radius that can be drawn through the points $A$ and $B.$ The line $y=mx+c$ cuts the given circle $x^2+y^2=a^2$ at two distinct points $A$ and $B.$ Equation of circle having minimum radius that can be drawn through the points $A$ and $B.$
I have approached the problem this way:
Considering it intersects at two points $A$ and $B.$
Now solving equation of circle and line.
So we get an equation in terms of $x$ and $y$
$$x^2(1 + m^2) + c^2 + 2mcx = a^2$$
and
$$y^2(1 + m^2) -2cy + c^2 - a^2m^2 = 0$$
These are the equations of $x$ coordinate and $y$ coordinate of two points $A$ and $B.$
Now since minimum radius circle can be drawn only when $AB$ is the diameter of the circle, the centre of the circle is the midpoint of $AB,$ which comes out to be $\left(\frac{-mc}{1+m^2} , \frac{c}{1+m^2}\right)$
and the radius is half of the distance $|AB|.$
But this way equation of circle would be difficult to obtain.
So I am interested in any other way of approaching.
| A circle passing through $A$ and $B$ would have an equation $x^2+y^2-a^2+k(mx-y+c)=0$ (where $k$ is a constant) and its centre is located at $\displaystyle \left(\frac{-km}{2},\frac{k}{2}\right)$. The circle with the minimum radius will have $AB$ as a diameter and its centre lies on $AB$.
So, we have $\displaystyle m\left(\frac{-km}{2}\right)-\left(\frac{k}{2}\right)+c=0$. This implies that $\displaystyle k=\frac{2c}{m^2+1}$.
The equation of the required circle is $\displaystyle x^2+y^2-a^2+\frac{2c}{m^2+1}(mx-y+c)=0$, i.e. $\displaystyle x^2+y^2+\frac{2mc}{m^2+1}x-\frac{2c}{m^2+1}y-a^2+\frac{2c^2}{m^2+1}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear system of equations that arises from recurrence relation. So when solving a third order homogenous recurrence relation I end upp with the task of determening the constants $a,b$ and $c$. The initial conditions give rise to the following system of equations:
$$\left\{
\begin{array}{rcr}
2a+(1+\sqrt{6})b+(1-\sqrt{6})c & = & 0\\
4a+(1+\sqrt{6})^2b+(1-\sqrt{6})^2c & = & 0\\
8a+(1+\sqrt{6})^3b+(1-\sqrt{6})^3c & = & 1
\end{array}
\right.$$
I've verified that this system is correct but now I need to solve it. I find brute force Gaussian elimination quite tedious to do (or maybe there is some trick to make it simple?). Is there any other easier method to solve this without having to spend 3-4 pages of hand written scribble?
The answer is
$$a=-\frac{1}{10}, \ b=\frac{1}{10\sqrt{6}}, \ c=-\frac{1}{10\sqrt{6}}.$$
| Very basic (done by hand in less than a page with pen and paper).
Consider
$$2 a+A b+B c = 0\tag 1$$
$$ 4 a+A^2 b+B^2 c =0\tag 2$$
$$ 8 a+A^3 b+B^3 c =1\tag 3$$
Use $(1)$ and $(2)$ and solve for $(a,b)$ to get
$$a=\frac{B (B-A)}{2 (A-2)} c\qquad \text{and} \qquad b=\frac{(2-B) B }{(A-2) A}c$$ Plug in $(3)$ and solve for $c$ to get
$$c=\frac{1}{(2-B) B (A-B)}\implies a=\frac{1}{2 (A-2) (B-2)}\qquad \text{and} \qquad b=\frac{1}{(A-2) A (A-B)}$$
Now, make $A=1+\sqrt 6$ and $B=1-\sqrt 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all $n$ $\in$ $\Bbb Z^+$ such that: $\lfloor\frac{n}{2}\rfloor \cdot \lfloor \frac{n}{3} \rfloor \cdot \lfloor \frac{n}{4} \rfloor = n^2$ Find all the numbers $n$ $\in$ $\Bbb Z^+$ such that:
$$\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2$$
I never worked before with floor function so i'm not completely sure how to solve this. I think (only because $n$ $\in$ $\Bbb Z^+$), i can just multiply (skipping the floor function) and get the answer of $n=24$, but this is floor function so i don't know if there are more solutions.
Any hints?
| We have
$$
\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 =0
$$
Note that:
$$
x > \left\lfloor x \right\rfloor
$$
Thus:
$$
\frac{n}{2}\cdot \frac{n}{3}\cdot \frac{n}{4} - n^2 >\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 = 0
$$
Thus:
$$
\frac{n^3}{24}-n^2= n^2\left( \frac{n}{24}-1\right) >0
$$
Hence we are looking for a number larger than 24.
At the same time:
$$
\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 > \left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2
$$
Therefore:
$$
0>\left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2
$$
This function has only one real root around 32.216 I'm lazy, sorry about that.
Hence we are looking for $n$ in the interval $[24,32]$. I'll check it in Python:
from math import floor
for n in range(24, 33):
if (floor(n/2)*floor(n/3)*floor(n/4)-n*n) == 0:
print(n)
And the only answer is 24.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$,
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
| 1) Observe: $$x= \frac 12 \times \left ( (2x + 1) -1\right).$$
2) Use this to obtain $$\frac{x}{(2x+1)^3} = \frac{1}{2} \times \frac{1}{(2x+1)^2} - \frac 12 \times \frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ \int \frac{x}{(2x+1)^3} dx = -\frac{1}{4} (2x+1)^{-1} + \frac{1}{8} (2x+1)^{-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Relation between roots and coefficients - manipulation of identities
The polynomial $x^3+3x^2-2x+1$ has roots $\alpha, \beta, \gamma$ . Find $$\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
| $a,b,c$ are the three roots.
$$
\begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\\
={}&(a+b+c)(ab+ac+bc)-3abc\\
={}&(-3)*(-2)-3*(-1)\\
={}&6-(-3)\\
={}&9
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $ Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can simplify this..
$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$
| $$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2}=$$
$$=\lim _{x\rightarrow0}\frac{1-\cos2x+\cos2x\left(1-\cos\left(\frac{(1-\cos x)(2\cos^2x+2\cos{x}+1)}{x^2}-\frac {5}{2}\right)\right)}{x^2}=$$
$$=\lim _{x\rightarrow0}\frac{2\sin^2x+\cos2x\left(1-\cos\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)\right)}{x^2}=$$
$$=2+\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)}{x^2}=$$
$$=2+\lim _{x\rightarrow0}\frac{2\sin^2\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)}{x^2}=$$
$$=2+2\lim _{x\rightarrow0}\frac{\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)^2}{x^2}=2$$
because $$\lim _{x\rightarrow0}\frac{\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{\left(\frac{x}{2}\right)^2}-5}{x^2}=$$
$$=\lim _{x\rightarrow0}\frac{\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1-5)}{\left(\frac{x}{2}\right)^2}+5\left(\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2-1\right)}{x^2}=$$
$$=\lim_{x\rightarrow0}\frac{2(\cos^2x+\cos{x}-2)}{x^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2-1}{\frac{x^2}{4}}=$$
$$=-\lim_{x\rightarrow0}\frac{2(1-\cos{x})(2+\cos{x})}{x^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin{x}}{x}\right)^2-1}{x^2}=$$
$$=-3\lim_{x\rightarrow0}\frac{4\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin{x}}{x}-1\right)\left(\frac{\sin{x}}{x}+1\right)}{x^2}=$$
$$=-3+\frac{5}{2}\lim_{x\rightarrow0}\frac{\sin{x}-x}{x^3}=-3-\frac{5}{6}\lim_{x\rightarrow0}\frac{1-\cos{x}}{x^2}=-3-\frac{5}{12}=-\frac{41}{12}$$ and
$$\lim _{x\rightarrow0}\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Geometry Question Involving Side Lengths I can't figure out how to find the minimum value of $y$ here. I know why its maximum value is 8, and the textbook says the minimum value is $12-\sqrt{5}$. Thank you!
Edit: I forgot to add that $x=4$ and I believe $y$ is minimized when segment $CD$ is a diameter of the circle.
| Let $t=CD$ and $\angle BAF = \phi$. By the PoP with respect to $E$ we have $$EF\cdot EA = ED\cdot EC\implies x(x+12) = y(y+t)$$
By the PoP with respect to $B$ we have $$BA^2 = BD\cdot BC\implies \boxed{64 = y(y+t)}$$
so we have $x^2+12x-64 = 0\implies x=4$. By the law of cosine we have:
$$ (2y+t)^2 = 16^2+8^2-2\cdot 16\cdot 8\cos \phi$$ so
using boxed equation we have:
$$ 4\cdot 64 +t^2 = 4y^2+4yt+t^2 = 16^2+8^2-2\cdot 16\cdot 8\cos \phi$$
so $$t^2 = 8^2-2\cdot 16\cdot 8\cos \phi <8^2\cdot 5$$
When $\phi \to -\pi $ we get $ t= 8\sqrt{5}$. From boxed equation we also get $$ y ={128\over \sqrt{t^2+256}+t}> {128\over 24+8\sqrt{5}}=12-4\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum _{r=1}^n r^4$ by considering $\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)$
By considering $$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)$$ Show that $$\sum _{r=1}^n r^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2 + 3n-1)$$
So I'm a little unsure on how to really start on this so I've expanded the given expression an found that
$$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right) = \sum_{r=1} ^n 5r^4 +10r^3 + 10r^2 +5r+1$$
I know I want to get this into a form which I can then solve for $\sum _{r=1}^n r^4 $ using the fact I know the sums of $\sum _{r=1}^n r^3$ and $\sum _{r=1}^n r^2$. If anyone could helping me get started with this question that would be great.
| OK, i will also type the "same answer", but am not that quick...
We use below a natural $n$. Let $S(n,p)$ be the sum of the numbers $1^p, 2^p,\dots, n^p$. Then:
$$
\begin{aligned}
(n+1)^5-1^5
&=\sum_{1\le r\le n}((r+1)^5-r^5)\\
&=5S(n,4)+10S(n,3)3+10S(n,2)+5S(n,1)+S(n,0)\\
&=5S(n,4)
+\frac {10}4\cdot n^2(n+1)^2
+\frac {10}6\cdot n(n+1)(2n+1)
+\frac 52\cdot n(n+1)
+n
\\[2mm]
&\qquad\text{ so}
\\[2mm]
30S(n,4)
&=
6((n+1)^5-1)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+5)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
6n(n+1)(n^3+4n^2+6n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
n(n+1)\Big(\ 6(n^3+4n^2+6n+4)-15(n^2+n)-10(2n+1)-15\ \Big)
\\
&=
n(n+1)\Big(\ 3(n^2 + n + 3)(2n + 1)-10(2n+1)\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 9-10\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 1\ \Big)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Simplify an expression with a root $ \frac{\sqrt{-x+2}}{\sqrt{x-1}}$ Simplify an expression with a root
I don't understand a transformation.
For this, $f_n= \frac{n}{2}l_n\sqrt{1-l_n^2/4}$
And now I'd like to show that
$\frac{1}{1-l_{2n}^2/4}f_{2n}=\frac{2}{1+\sqrt{1-l_n^2/4}}\cdot\frac{n}{2}l_n$
holds
Can anyone help me?
According to my calculations
$f_{2n}=\frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^{2}/4}$ and then
$\frac{1}{1-l_{2n}^2/4}\frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^2/4}=n\frac{l_{2n}\sqrt{1-l_{2n}^2/4}}{1-l_{2n}^2/4}=n\cdot l_{2n}\cdot \frac{1}{\sqrt{1-l_{2n}^2/4}}=\frac{n}{2}\cdot l_{2n}\cdot \frac{2}{\sqrt{1-l_{2n}^2/4}}$
Also we have
$l_{2n}^2=l_n^2/4+\big(1-\sqrt{1-l_n^2/4}\big)^2=2\big(1-\sqrt{1-l_n^2/4}\big)$
So I substituted the expression $l_{2n}$, but I can't prove the desired formula.
After putting in $n\cdot l_{2n}\cdot \frac{1}{\sqrt{1-l_{2n}^2/4}}$ I got
$n\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)}\cdot \frac{1}{\sqrt{1-(2\big(1-\sqrt{1-l_n^2/4}\big))/4}}=n\frac{\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)}}=n\frac{\sqrt{2\big(1-\sqrt{1-l_n^2/4}\big)\sqrt{1-l_n^2/4}\big)}}{\sqrt{2\sqrt{1-l_n^2/4}-1}}=n\frac{\sqrt{-2\sqrt{1-l_n^2/4}+2}}{\sqrt{2\sqrt{1-l_n^2/4}-1}}$
Now I don't know how I can further simplify the Expression.
| Nothing will be changed if you bring both the numerator and denominator under a single radical and bring the twos inside the radical to obtain
$$ n\sqrt{\frac{2-\sqrt{4-l_n^2}}{\sqrt{{4-l_n^2}}-1}} $$
Then the fraction under the radical can be rationalized.
$$ n\sqrt{\frac{2-\sqrt{4-l_n^2}}{\sqrt{{4-l_n^2}}-1}\cdot\frac{\sqrt{{4-l_n^2}}+1}{\sqrt{{4-l_n^2}}+1}}=n\sqrt{\frac{l_n^2-2+\sqrt{4-l_n^2}}{3-l_n^2}} $$
Although you could rationalize the denominator, I would not take it any further.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
| Your approach is a good one, and in fact can lead to the general formula mentioned in the other answers.
You just made a few mistakes in your calculation.
For example, you say that $\frac{\partial}{\partial x} \left[ (x-8)^2+y^2+(-x-y+12)^2 \right]$ is $2(x-8)+2(-x-y+12)$, when really it is $2(x-8)-2(-x-y-12)$. You forgot the chain rule.
You make a similar chain rule mistake when calculating the partial derivative with respect to $y$.
I think if you fix these, you should be able to finish your calculation the way you started it.
Note to other posters: I think it is generally of greater pedagogical value to help a student realize how to fix their own reasoning, rather than supplying them with a totally different method. Our primary aim should be to empower students to realize that they have the power to think through these things logically and creatively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does this Sum follow a general trend? In my personal works I came across a sum that looks like $$f(x,k)= \sum_{n=0}^\infty \frac{x^{k(2n+1)}}{(k(2n+1))!}. $$
We have:
$$f(x,1)= \sinh(x)$$ and
$$f(x,2)= \frac{1}{2} (\cosh(x) + \cos(x)).$$
Rr-writing these in their exponential forms we get $$f(x,1)= \frac{e^x}{2} + \frac{e^{-x}}{2}$$
And again, $$f(x,2) = \frac{e^x}{4} + \frac{e^{-x}}{4} - \frac{e^{ix}}{4} - \frac{e^{-ix}}{4}$$
My question is, is there a general trend that can be found for any natural number, $k$?
How would one evaluate $f(x,3)$ or $f(x,4)$, for example?
| For $k>2$, I think that you can only face hypergeometric functions.
To show you the patterns
$$f(x,3)=\frac{x^3}{3!} \,
_0F_5\left(;\frac{4}{6},\frac{5}{6},\frac{7}{6},\frac{8}{6},\frac{9}{6};\frac{x^
6}{6^6}\right)$$
$$f(x,4)=\frac{x^4}{4!} \,
_0F_7\left(;\frac{5}{8},\frac{6}{8},\frac{7}{8},\frac{9}{8},\frac{10}{8},\frac{11
}{8},\frac{12}{8};\frac{x^8}{8^8}\right)$$
$$f(x,5)=\frac{x^5 }{5!} \,
_0F_9\left(;\frac{6}{10},\frac{7}{10},\frac{8}{10},\frac{9}{10},\frac{11}{10},
\frac{12}{10},\frac{13}{10},\frac{14}{10},\frac{15}{10};\frac{x^{10}}{10^{10}}\right)$$
$$f(x,6)=\frac{x^6}{6!} \,
_0F_{11}\left(;\frac{7}{12},\frac{8}{12},\frac{9}{12},\frac{10}{12},\frac{11}{12},
\frac{13}{12},\frac{14}{12},\frac{15}{12},\frac{16}{12},\frac{17}{12},\frac{18}{12};\frac{
x^{12}}{12^{12}}\right)$$
$$\color{blue}{f(x,k)=\frac{x^k}{k!}\,
_0F_{2k-1}\left(;\frac{k+1}{2k},\frac{k+2}{2k},\cdots, \frac{k+2k}{2k};\frac{
x^{2k}}{(2k)^{2k}}\right)}$$
$$f(1,k)=\frac 1 {k!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Condition for $x^4-18x^2+4dx+9=0$ has four real roots Prove that if $x^4-18x^2+4dx+9=0$ has four real roots, Then $d^4 \le 1728$
My try:
obviously the equation should have two unequal negative roots and two repeated positive roots OR vice versa.
Choosing first case here is the rough sketch:
Let the two negative roots are $-p_1, -p_2$ and repeated positive root be $p_3$
where $p_1,p_2,p_3 \gt 0$
Thus we have the polynomial identity as:
$$(x+p_1)(x+p_2)(x-p_3)^2=x^4-18x^2+4dx+9$$
Comparing coefficients we get:
$$p_1+p_2=2p_3 \tag{1}$$
$$p_1p_2=3p_3^2-18\tag{2}$$
$$p_1p_2p_3^2=9\tag{3}$$
$$p_3^2(p_1+p_2)-2p_1p_2p_3=4d\tag{4}$$
From $(2)$ and$ (3)$ we get:
$$\frac{9}{p_3^2}=3p_3^2-18$$
we get:
$$p_3^4-6p_3^2-3=0$$
Since $p_3 \gt 0$ we get
$$p_3=\sqrt{3+2\sqrt{3}}$$
Using $(4)$ we get:
$$2p_3^3-\frac{18}{p_3}=4d$$
From this we get a unique value of $d$
But how proof is to be done, what went wrong in my analysis?
| Here is an alternative approach to the original question. As you have reasoned, it is helpful to solve for the $d$ that make $f(x)=x^4-18x^2+4dx+9$ have a repeated root. If $f$ has a repeated root, then the gcd of $f$ and $f'$ also shares that root.
We have $f'(x)=4x^3-36x+4d=4\left(x^3-9x+d\right)$.
We look for a linear combination of $f$ and $f'$ that is just a linear polynomial. Starting with $$\left(-x^2+Ax+B\right)\frac{f'(x)}{4}+\left(x+C\right)f(x)=mx+b$$
we slowly deduce values for $A,B,C$ and find
$$\left(-x^2-\frac{d}{3}x+9\right)\left(x^3-9x+d\right)+\left(x+\frac{d}{3}\right)\left(x^4-18x^2+4dx+9\right)=\left(d^2-72\right)x+12d$$
Since we are assuming there is a repeated root, it must be the root of this linear polynomial, $\frac{12d}{72-d^2}$.
So $\frac{1}{4}f'\mathopen{}\left(\frac{12d}{72-d^2}\right)\mathclose=0$.
$$\begin{align}
\left(\frac{12d}{72-d^2}\right)^3-9\left(\frac{12d}{72-d^2}\right)+d&=0\\
(12d)^3-108d\left(72-d^2\right)^2+d\left(72-d^2\right)^3&=0\\
-d\left(d^6-108d^4-1728d^2+186624\right)&=0\\
-d\left(d^2-108\right)\left(d^4-1728\right)&=0
\end{align}$$
So (assuming we are searching for the positive $d$) either $d=\sqrt{108}$ or $d=\sqrt[4]{1728}$. The first would mean $-\sqrt{12}$ is the common root of $f$ and $f'$. But try this for $x$ in $f(x)$ (with $d=\sqrt{108}$) and it does not give $0$.
Therefore $d=\sqrt[4]{1728}$. From here you already have the reasoning for why $\lvert d\rvert\leq\sqrt[4]{1728}$ is the general condition for four real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Triangle problem, $AC=3$ , $AB=5$ ,..., then $PH=?$ In $\triangle ABC$ :
$AC=3$ , $AB=5$ , $\angle ACB= 90 ^ \circ$,
$P$ is a point inside $\triangle ABC$ such that $PA+BC=PB+AC=PC+AB$,
$H$ is a point on the line $AB$ such that $\angle PHB=90^\circ$
then, $PH=?$
My Approach:
In Right-angled $\triangle ABC$
$$AB^2=AC^2 + BC^2 \implies BC=4$$
Let $PC=x$
then,
$$PA+4=PB+3=PC+5 $$
$$[\because PA+BC=PB+AC=PC+AB]$$
$$\implies PA=x+1 \, , \, PB= x+2$$
so,
My thoughts: First we will find the value of $x$ by equating area of triangle $ABC$,
i.e,
$$Ar[\triangle ABC]=Ar[\triangle APB]+Ar[\triangle BPC]+Ar[\triangle APC] \dots (i)$$
$Ar[\triangle ABC]$ can be simply found as: $\frac{1}{2}*3*4=6$ sq. units
For $Ar[\triangle APB]$ , we will use Heron's formula as follows:
$$s=\frac{1}{2} * [(x+1)+(x+2)+5] = x+4 $$
$$\therefore Ar[\triangle APB]=\sqrt{(x+4)(x-1)(2)(3)}=\sqrt{6(x+4)(x-1)}$$
Similarly, $Ar[\triangle BPC]=\sqrt{3(x+3)(x-1)}$
Similarly, $Ar[\triangle APC]=\sqrt{2(x+2)(x-1)}$
So, from eq.(i) , we get:
$$6=\sqrt{6(x+4)(x-1)} \, + \, \sqrt{3(x+3)(x-1)} \, + \, \sqrt{2(x+2)(x-1)} \dots (ii)$$
and then we will find $PH$ by equating area of triangle $APB$,
i.e,
$$\frac{1}{2}*5*PH=\sqrt{6(x+4)(x-1)} \dots (iii)$$
So, how to solve eq.(ii) to find the value of $x$ ? Please help...
| Hint: Let $x$ the perpendicular on $AC$,$y$ the perpendicular on $BC$
then we get the following equations:
$$3\cdot 4=3x+4y+5PH$$ Now let $$PA=t-BC,PB=t-AC,PC=t-AB$$ then we obtain
$$(t-5)^2=x^2+y^2$$
$$(t-3)^2=(4-x)^2+y^2$$
$$(t-4)^2=x^2+(3-y)^2$$
Can you finish?
I got $$PH=\frac{132}{115},x=\frac{20}{23},y=\frac{21}{23},t=\frac{144}{23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation:
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it:
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1 \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1 \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid.
New attempt.
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1 \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers.
| Assume, without loss of generality, that $x\leq y\leq z$. We see from the original equation that $x>1$ (since $x = 1$ means $\frac1x + \cdots > 1$). At the same time, we must have $x<4$, as otherwise the sum is clearly less than $1$.
So, where does $x = 2$ actually take us? We insert and get
$$
\frac12 + \frac1y + \frac1z + \frac1{2y} + \frac1{2z} + \frac1{yz} + \frac1{2yz} = 1\\
\frac3{2y} + \frac3{2z} + \frac3{2yz} = \frac12\\
3z + 3y + 3 = yz\\
12 = yz - 3y - 3z + 9 = (y-3)(z-3)
$$
Since $y$ and $z$ are integers, this is an easy solve.
What about $x = 3$? We get
$$
\frac13 + \frac1y + \frac1z + \frac1{3y} + \frac1{3z} + \frac1{yz} + \frac1{3yz} = 1\\
\frac4{3y} + \frac4{3z} + \frac4{3yz} = \frac23\\
4z + 4y + 4 = 2yz\\
2z + 2y + 2 = yz\\
6 = yz - 2y-2z + 4 = (y-2)(z-2)
$$
which, again, is an easy solve using the fact that $y, z$ are integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Is $f(x)=\frac{1}{x+1} \cos x^2$ uniformly continuous?
Let $f:[0,\infty)\to \Bbb{R}$, $f(x)=\frac{1}{x+1} \cos x^2$, Is $f$ uniformly continuous?
My attempt:
Let $x,y\in [0,\infty),$ then
\begin{align*}
\Bigg|\frac{\cos x^2}{x+1}-\frac{\cos y^2}{y+1}\Bigg|
& =\Bigg|\frac{(y+1)\cos x^2-(x+1)\cos y^2}{(x+1)(y+1)}\Bigg| \\
& =\Bigg|\frac{(y \cos x^2 - x \cos y^2)+(\cos x^2 - \cos y^2)}{(x+1)(y+1)}\Bigg| \\
& \le \Bigg|\frac{y \cos x^2 - x \cos y^2}{(x+1)(y+1)}|+|\frac{\cos x^2 - \cos y^2}{(x+1)(y+1)}\Bigg| \\
& \le \Big|y \cos x^2 - x \cos y^2| + |\cos x^2 - \cos y^2\Big| \\
\end{align*}
How can I complete, please? Thanks.
| For a “direct” proof one can proceed as follows. Without loss of generality we can assume that $y \le x$. Then
$$
\frac{\cos x^2}{x+1}-\frac{\cos y^2}{y+1} = \frac{(y+1)\cos x^2-(x+1)\cos y^2}{(x+1)(y+1)} \\
= \frac{\cos x^2 - \cos y^2}{x+1} - \frac{(x-y)\cos y^2}{(x+1)(y+1)}
$$
(The idea here is to split the difference into one term containing the difference $\cos x^2 - \cos y^2$, and another term containing the difference $x-y$.)
The second term is easy to estimate:
$$
\left|\frac{(x-y)\cos y^2}{(x+1)(y+1)} \right| \le |x-y| \, .
$$
For the first term we need the trigonometric identity
$$
\cos a - \cos b = -2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}
$$
and get
$$
\left | \frac{\cos x^2 - \cos y^2}{x+1} \right | =
\left | \frac{2 \sin \frac{x^2+y^2}{2}\sin \frac{x^2-y^2}{2}}{x+1}\right | \\
\le \frac{2 |\sin \frac{x^2-y^2}{2}|}{x+1} \le \frac{|x^2-y^2|}{x+1}
= \frac{x+y}{x+1} |x-y| \le 2|x-y| \, .
$$
Combining these estimates we get
$$
|f(x) - f(x) | \le 3 |x-y|
$$
so that $f$ is Lipschitz continuous (and therefore uniformly continuous) on $[0, \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$ Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$
So I did the following:
Let $f(x) = \sin(x)\tan(x) - x^2$.
Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) > 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.
I was thinking of approximating it using the following inequality $\sin(x) < x < \tan(x)$.
Is that a good direction?
| If you can use Taylor series, then you may proceed as follows:
*
*$\sin(x)\tan(x) > x^2 \Leftrightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > \cos x}$ on $(0,\frac{\pi}{2})$
Using Taylor you get on $(0,\frac{\pi}{2})$:
*
*$\frac{\sin x}{x} > 1-\frac{x^2}{6} \Rightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > 1-\frac{x^2}{3} + \frac{x^4}{36}}$
*$1-\frac{x^2}{2} + \frac{x^4}{24} > \cos x$
Remains to show $1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24}$ on $(0,\frac{\pi}{2})$:
\begin{eqnarray*}
& 1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24} &\\
& \stackrel{x > 0}{\Leftrightarrow} & \\
& |x| < 2\sqrt{3} &
\end{eqnarray*}
Since $\frac{\pi}{2} < 2\sqrt{3}$, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
calcuating residue of a complex function I need to calculate the residue of the function $\frac{(z^6+1)^2}{(z^5)(z^2-2)(z^2-\frac{1}{2})}$ at $z$=0.
z=0 is a pole of order 5 so I tried using the general formula to calculate the residue but the calculation becomes very tedious since it involves finding the fourth derivative of the complex function. I even tried writing the laurent series but that too got me nowhere. Could anyone please tell how to proceed with this problem?
| Hints:
You may proceed as follows:
*
*$\frac{(z^6+1)^2}{(z^5)(z^2-2)(z^2-\frac{1}{2})} = \frac{z^{12} + 2z^6 + 1}{(z^5)(z^2-2)(z^2-\frac{1}{2})}
= $$ $$ \frac{z^{7}}{(z^5)(z^2-2)(z^2-\frac{1}{2})} + \frac{2z}{(z^2-2)(z^2-\frac{1}{2})} + \frac{1}{(z^5)(z^2-2)(z^2-\frac{1}{2})}$
Since the first summands are holomorphic at $z=0$ you only have to calculate the residue for the last term.
Now, use the formula:
*
*$Res_{z=0}f(z) = \frac{1}{4!}\lim_{z\to 0}\frac{d^4}{dz^4}\frac{z^5}{(z^5)(z^2-2)(z^2-\frac{1}{2})} = \frac{1}{4!}\lim_{z\to 0}\frac{d^4}{dz^4}\frac{1}{(z^2-2)(z^2-\frac{1}{2})}$
Before differentiating split up the fraction into its partial fractions:
$$\frac{1}{(z^2-2)(z^2-\frac{1}{2})} = \frac{2}{3(z^2-2)} - \frac{4}{3(z^2-\frac{1}{2})}$$ $$= \frac{1}{3\sqrt{2}}\left( \frac{1}{z - \sqrt{2}} + \frac{1}{z + \sqrt{2}}\right) - \frac{\sqrt{2}}{3}\left(\frac{1}{z - \frac{1}{\sqrt{2}}} + \frac{1}{z + \frac{1}{\sqrt{2}}} \right)$$
Now, differentiating is straight forward and plug in $z= 0$ at the end. (Do not forget to multiply by $\frac{1}{4!}$ at the end.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$
I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?
| A long version:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}=
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{n\cdot\frac{1}{n^4}\cdot\frac{(1-\cos^2(1/n^2))}{\frac{1}{n^4}}}\cdot(1+\cos(1/n^2))=\\
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}\cdot \color{red}{\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}}}\cdot(1+\cos(1/n^2))=\\
2\cdot \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}}=
2\cdot \lim_{n \rightarrow +\infty} n^3\ln\left(\frac{1+n+n^3}{n^3}\right)=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{n^3}=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}\cdot (n+1)} = \\
2\cdot \lim_{n \rightarrow +\infty} (n+1)\cdot\ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}} =
2\cdot \ln{e} \cdot \lim_{n \rightarrow +\infty} (n+1)\rightarrow +\infty$$
On the 2nd line
$$\lim_{n \rightarrow +\infty}\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}=
\lim_{n \rightarrow +\infty}\frac{\sin(1/n^2)}{\frac{1}{n^2}}\cdot \frac{\sin(1/n^2)}{\frac{1}{n^2}}=1$$ from
$$\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to find a closed form for $\sum_{i=0}^n \binom{a+i}{b+i}i$ Wolframalpha tells me it's $$\frac{b (b + 1) \binom{a + 1}{ b + 1} - (b + n + 1) (b (n + 1) - (a + 1) n) \binom{a + n + 1}{ b + n + 1}}{(a - b + 1) (a - b + 2)}$$ but how to come up with or at least prove that?
| This is not quite the same expression, but this is how I would go about finding the closed formula. We can first rewrite the summation as
$$\sum_{i=0}^n \binom{a+i}{b+i}i = \sum_{k=1}^n \sum_{i=k}^n \binom{a+i}{b+i}$$
Then we can use the hockey stick formula to get that the inner sum is
$$\sum_{k=1}^n \binom{a+n+1}{b+n} - \binom{a + k}{b+k-1}$$
The first term is the same every time, so we can rewrite this as
$$n\binom{a+n+1}{b+n} - \sum_{k=1}^n \binom{a+k}{b+k-1}$$
And use the hockey stick formula again to get that this is
$$n\binom{a+n+1}{b+n} - \binom{a+n+1}{b+n-1} + \binom{a+1}{b-1}$$
And there we have a closed formula!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| The probability is $\frac{1}{2}$ because the last flip determines it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 10
} |
Show that $19^{31}>13^{33}$
How can i prove that $19^{31}>13^{33}$?
What I tried
$$\bigg(\frac{19}{13}\bigg)^2=\frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $\displaystyle 19^{30}>13^{30}$
How do I show it. Help me please.
| $19^{30}=(19^3)^{10}=6859^{10}\gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859\gt 3\times 2197=6591$
$6859^{10}\gt (3\times 2197)^{10}=3^{10}\times 2197^{10}=3^3\times 3^3\times 3^3\times 3\times 2197^{10}=27^3\times 3\times 2197^{10}$
$27^3\times 3\times 2197^{10}\gt 13^3\times 2197^{10}$ already without even multiplying by $19.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + 2Dx^2 + Dx + Ex^3 + 2Ex^2 + E$
= $ \int (A + D + E)x^3 + (3A + B + C + 2D + 2E)x^2 + (2A + 2C + 4B + D)x + (4B + E)$
Turn into matrix, find reduced row echelon form to solve system of equations:
$\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
3 & 1 & 1 & 2 & 2 & 1 \\
2 & 4 & 2 & 1 & 0 & 1 \\
0 & 4 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$
This is where things go wrong, apparently this doesn't reduce down properly :( and I have been relying on RREF to solve systems of equations everytime up until now.
I am also confused where the $x^2 + x + 1$ is supposed to go exactly. I know I put it into the system of equations later but until then, I feel like I kinda just ignored it and left it out of all my work up until then (is that okay?).
Other note: I recognize that this is a proper rational fraction so no long division is nesecary and that this has irreducible factors that are repeated so that is why I split them up into the partial fractions up above in that manner. Did I miss any intermittent steps that made the RREF turn out wrong? I am not sure where I went wrong thus far either
| Another way without having to solve equations:
The function
$$f(x) = \frac{x^2 + x+1}{(x+1)^2 (x+2)}$$
has a simple pole at $x = -2$ and a second order pole at $x = -1$. Let's denote the the singular parts of the expansions around $x = -2$ and $x = -1$ by $P_1(x)$ and $P_2(x)$, respectively. The partial fraction expansion can then be written as:
$$f(x) = P_1(x) + P_2(x)$$
This follows from the fact that $g(x) = f(x) - \left[P_1(x) + P_2(x)\right]$ does not have any singularities, as all the singularities in $f(x)$ have been subtracted by subtracting the ones contained in the singular parts of the Laurent expansions around each pole. This means that $g(x)$ is a rational function without any singularities, therefore $g(x)$ is a polynomial (after removing the removable singularities). However, we can also see that $g(x)$ tends to zero at infinity, therefore $g(x)$ must be identical to zero.
We can easily find $P_1(x)$ by substituting $x=-2$ in the factor multiplying $\frac{1}{x+2}$ in $f(x)$. This yields:
$$P_1(x) = \frac{3}{x+2}$$
We can find $P_2(x)$ without doing all the work needed to expand around $x=-1$ by observing that for large $x$ we have:
$$f(x)=\frac{1}{x} + \mathcal{O}\left(x^{-2}\right)$$
Therefore we must have:
$$P_2(x) = -\frac{2}{(x+1)} +\mathcal{O}(x+1)^{-2}$$
The leading term of the expansion around $x = -1$ of $f(x)$ is easy to find by substituting $x =-1 $ in the term multiplying $\frac{1}{(x+1)^2}$, this is given by $\frac{1}{(x+1)^2}$, we therefore have:
$$P_2(x) =\frac{1}{(x+1)^2} -\frac{2}{(x+1)}$$
It thus follows that:
$$f(x) = \frac{1}{(x+1)^2} -\frac{2}{(x+1)} + \frac{3}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Order of a sixth root in group G This question originates from Chapter 10, H3 of the 2nd edition of A Book of Abstract Algebra by Charles C. Pinter.
Let $a$ denote an element of a group $G$. Let $a$ have order 10. If $a$ has a sixth root in $G$, say $a=b^6$, what is the order of $b$?
Here is what I think:
Given $\operatorname{ord}(a) = 10$ and $a=b^6$,
$\qquad a^{10} = e = (b^{6})^{10} = b^{60}$
Let $\operatorname{ord}(b) = x, x$ must divide $60 \implies x\in \{1,2,3,4,5,6,10,12,15,20,30,60\}.$
*
*$x=1: b = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=2: b^2 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=3: b^3 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=4: b^4 = e = b^{12} = a^2$ but $\operatorname{ord}(a) \ne 2.$
*$x=5: b^5 = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=6: b^6 = e = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=10: b^{10} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=12: b^{12} = e = a^2$ but $\operatorname{ord}(a) \ne 2.$
*$x=15: b^{15} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=20: b^{20} = e = b^{60} = a^{10}$.
*$x=30: b^{30} = e = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=60: b^{60} = e = a^{10}.$
Hence $x \in \{20,60\}$.
So it seems $x=20$ as $20 < 60$.
Or is there a way to rule out $20$?
| $$\begin{align} o(b^6) & = \frac {o(b)} {\text {gcd}\ (o(b),6)}. \\ \implies o(b) & = 10 \cdot \text {gcd}\ (o(b),6). \\ \implies o(b) & = 10 \cdot \text {gcd}\ (10 \cdot \text {gcd}\ (o(b),6),6). \\ \implies o(b) & = 20 \cdot \text {gcd} (5 \cdot \text {gcd}\ (o(b),6) , 3). \\ \end{align}$$
Now if $3 \nmid o(b)$ then we have $o(b)=20.$ Otherwise if $3 \mid o(b)$ we have $o(b) = 60.$ So $o(b) = 20\ \text {or}\ 60.$
Observe that both the options are equally valid. Because for $n=20,60$ consider a generator $b$ of $\Bbb Z_n.$ Observe that $a=b^6$ has the same property as mentioned in the question. But for the first case we have $o(b)=20$ and for the second case we have $o(b)=60.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\int_0^\infty \log\left (1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$
Prove $$\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$$where $\theta\in[0,\pi]$.
I've met another similar problem,
$$ \int_0^{2\pi} \log(1-2r\cos \theta +r^2) d\theta=2\pi \log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} =\left(\frac{1}{x}-e^{i\theta}\right)\left(\frac{1}{x}+e^{i\theta}\right)\left(\frac{1}{x}-e^{-i\theta}\right)\left(\frac{1}{x}+e^{-i\theta}\right)$$
But I couldn't move on.
Any hints? Thanks in advance.
| For $\theta \in [0;\pi]$,
\begin{align}
J(\theta)&=\int_0^\infty \ln\left(1-\frac{2\cos(2\theta)}{x^2}+\frac{1}{x^4}\right) \,dx
\end{align}
Perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}
J(\theta)&=\int_0^\infty \frac{\ln\left(1-2\cos(2\theta)x^2+x^4\right)}{x^2} \,dx
\end{align}
For $a\geq -1$, define the function $F$ by,
\begin{align}F(a)&=\int_0^\infty \frac{\ln\left(1+2ax^2+x^4\right)}{x^2} \,dx\\
&=\left[-\frac{\ln\left(1+2ax^2+x^4\right)}{x}\right]_0^\infty+\int_0^\infty \frac{4\left(x^2+a\right)}{1+2ax^2+x^4}\,dx\\
&=\int_0^\infty \frac{4\left(x^2+a\right)}{1+2ax^2+x^4}\,dx\\
\end{align}
Perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}F(a)&=\int_0^\infty \frac{4\left( \frac{1}{x^2}+a\right) }{x^2\left(1+\frac{2a}{x^2}+\frac{1}{x^4}\right) } \,dx\\
&=\int_0^\infty \frac{4\left( 1+ax^2\right) }{x^4+2ax^2+1 } \,dx\\
\end{align}
Therefore,
\begin{align}F(a)&=\int_0^\infty \frac{2(a+1)\left( 1+x^2\right) }{x^4+2ax^2+1 } \,dx\\
&=2(a+1)\int_0^\infty \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+2a } \,dx\\
&=2(a+1)\int_0^\infty \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+2(a+1) } \,dx\\
\end{align}
Perform the change of variable $y=x-\dfrac{1}{x}$,
\begin{align}F(a)&= 2(a+1)\int_{-\infty}^{+\infty}\frac{1}{x^2+2(a+1)}\,dx\\
&=4(a+1)\int_{0}^{+\infty}\frac{1}{x^2+2(a+1)}\,dx\\
&=\left[2\sqrt{2(a+1)}\arctan\left( \frac{x}{\sqrt{2(a+1)}} \right)\right]_0^\infty\\
&=\boxed{\pi\sqrt{2(1+a)}}
\end{align}
Observe that, $J(\theta)=F\big(-\cos(2\theta)\big)$.
\begin{align} 2(1-\cos(2\theta))&=2(1-\cos^2(\theta)+\sin^2 (\theta))\\
&=2\times 2\sin^2 (\theta)\\
&=4\times \sin^2 (\theta)\\
\end{align}
Since, for $\theta \in [0;\pi],\sin(\theta)\geq 0$ then $\sqrt{2(1-\cos(2\theta))}=2\sin(\theta)$
Therefore,
\begin{align}\boxed{J(\theta)=2\pi \sin(\theta)}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin series for $\frac{\cos2x-1}{x^2}$ using Maclaurin series for $\cos2x$ I am having some trouble finding the first three nonzero terms for the maclaurin series of $\frac{\cos2x-1}{x^2}$ using the maclaurin series for $\cos2x$.
So far I have the maclaurin series for $\cos2x$ being:
$$\cos2x=1-\frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!}\dots$$
However I am not sure how to proceed in finding the macluarin series for $\frac{\cos2x-1}{x^2}$ now.
| Since \begin{eqnarray}
cos(2x) = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} ...
\end{eqnarray}
$$\cos (2x) - 1 = \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} ...$$
$$\frac{\cos(2x)}{x^2} = \frac{4}{2!} + \frac{16x^2}{4!} - \frac{64x^4}{6!} ...$$
$$=2 + \frac{16x^2}{4!} - \frac{64x^4}{6!} ...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a second order nonlinear differential equation
Solve $$y’’ + 2y = 4 -\frac{1}{\sqrt{y}}$$ with initial conditions $y(0) = 1$ and $y’(0) = 2$.
I am trying different substitutions, but none seem to work.
| $$y'' + 2y - 4 +\frac{1}{\sqrt{y}}=0$$
$$2y''y'+4yy'-8y'+\frac{2y'}{\sqrt{y}}=0$$
$$(y')^2+2y^2-8y+4\sqrt{y}=c_1$$
$y(0)=1$ and $y'(0)=2 \quad;\quad 4+2-8+4=c_1=2$
$$y'=\pm\sqrt{-2y^2+8y-4\sqrt{y}+2}$$
$$dx=\pm\frac{dy}{\sqrt{-2y^2+8y-4\sqrt{y}+2}}$$
$$x=\pm\int \frac{dy}{\sqrt{-2y^2+8y-4\sqrt{y}+2}}+c_2$$
$y(0)=1$
$$x=\pm\int_1^y \frac{d\xi}{\sqrt{-2\xi^2+8\xi-4\sqrt\xi{}+2}}$$
$$x=\pm\int_1^{\sqrt{y}} \frac{2\zeta\:d\zeta}{\sqrt{-2\zeta^4+8\zeta^2-4\zeta+2}}$$
This is the solution for the inverse function $x(y)$.
Further calculus appears arduous, involving elliptic integral and inverse.
Better use numerical calculus directly to solve the original ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$
If
$$\sin(x) - \cos(x) = \frac{1}{3}$$
then determine
$$\sin(x)\cos(x)$$
I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
| To save typing, let $s = \sin x$ and $c = \cos x$. Then we have two equations in two variables:
$$
\begin{align}
s^2 + c^2 &= 1\\
s - c &= \frac13.
\end{align}
$$
A widely applicable approach in such a situation is to solve one equation for one of the variables, and use that result to substitute into the other. Here, the second equation can easily be solved for $s$, yielding $s = c + \frac13$. Substituting for $s$ in the first equation gives a quadratic equation in $c$:
$$
2c^2 + \frac23 c - \frac89 = 0.
$$
The roots of this quadratic are
$$
c = \frac{-1+\sqrt{17}}{6}, \frac{-1-\sqrt{17}}{6}.
$$
Since $s = c + \frac13$, the corresponding values for $s$ are
$$
s = \frac{1+\sqrt{17}}{6}, \frac{1-\sqrt{17}}{6}.
$$
Multiplying either pair of corresponding values gives the result $sc = \frac49$.
This method is much more laborious than the original solution, but requires no cleverness or flash of insight.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What does it mean to solve the equation of an ellipse as a quadratic? An illustration required me to find out the area of the curve $$5x^2 + 6xy + 2y^2 + 7x + +6y + 6 = 0 $$
They proceeded to solve the equation as a quadratic obtaining $y_1$ and $y_2$ as the two branches.
Why are $y_1$ and $y_2$ as depicted in the diagram?
if it isn't too broad what does it mean to solve a quadratic equation for any conic section (parabola, ellipse, hyperbola, circle.
| The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$:
\begin{align*} 0 &= 5x^2+6xy+2y^2+7x+6y+6\\
0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\\
0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)\end{align*}
Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations:
\begin{align*}y &= \frac{-(6x+6)\pm\sqrt{(6x+6)^2-4\cdot 2\cdot (5x^2+7x+6)}}{2\cdot 2}\\
&= \frac{-6(x+1)\pm\sqrt{36x^2+72x+36-40x^2-56x-48}}{4}\\
&= \frac{-6(x+1)\pm\sqrt{-4x^2+16x-12}}{4}\\
y &= \frac{-3(x+1)\pm\sqrt{(x-1)(3-x)}}{2}\end{align*}
There it is. That quadratic equation has a $\pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.
Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
| For this question I proved that for all real $t, e^t \ge 1 + t $ and that equality holds only for $t=0.$
From this it follows that, for $t>0,$
$$\exp\left(\frac {-1}{1+t}\right)\gt 1-\frac 1 {1+t}=\frac t {1+t},$$
which implies (taking reciprocal of both sides, reversing the order) $$\exp\left(\frac {1}{1+t}\right)\lt \frac {1+t} {t}$$ so (setting $x=1+t$) $$\exp\left(\frac {1}{x}\right)\lt \frac {x} {x-1} = 1 + \frac 1 {x-1}$$ for $x>1$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Differentiate $e^{y/x} = 20x-y$ I am trying to use implicit differentiation to differentiate $e^{y/x} = 20x-y$. I get $\frac{20}{2 e^{y/x} \cdot \frac{x-y}{x^2}}$, but according to the math website I'm using, "WebWork", this is wrong.
I'm not sure how to handle it when an equation has two $\frac{dy}{dx}$ floating around, but I'm not sure this is the problem.
Here are my steps:
$$\frac{d}{dx}(e^{y/x}) = \frac{d}{dx}(20x-y)$$
Factoring both sides independently:
$$\frac{d}{dx}(e^{y/x}) = e^{y/x}\cdot\frac{x-y}{x^2} \frac{dy}{dx}$$
$$\frac{d}{dx}(20x-y)=20-\frac{dy}{dx}$$
Finding $\frac{dy}{dx}$
$$e^{y/x} \cdot \frac{x-y}{x^2} \frac{dy}{dx} = 20 - \frac{dy}{dx}$$
$$\frac{dy}{dx}+\frac{x-y}{x^2} \frac{dy}{dx} = \frac{20}{e^{y/x}}$$
$$\frac{dy}{dx} + \frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$2\frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$\frac{dy}{dx} = \frac{20}{2e^{y/x}\cdot\frac{x-y}{x^2}}$$
| You get:
$$(e^{y/x})'_x=(2x-y)'_x \Rightarrow \\
e^{y/x}\cdot (y/x)'_x=2-y'\Rightarrow \\
(2x-y)\cdot \frac{y'x-y}{x^2}=2-y' \Rightarrow \\
(2x^2-xy)y'-(2x-y)y=2x^2-x^2y'\Rightarrow \\
y'=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
If you are familiar with multivariable calculus, consider: $F(x,y)=e^{y/x}-2x+y=0$. Then:
$$y'=-\frac{F_x}{F_y}=-\frac{e^{y/x}\cdot (-y/x^2)-2}{e^{y/x}\cdot (1/x)+1}=\frac{(2x-y)y+2x^2}{(2x-y)x+x^2}=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all integers of the form $\frac{(x-1)^2(x+2)}{2x+1}$ I made a python program and I have that
\begin{array}{|c|c|}
\hline x & \frac{(x-1)^2(x+2)}{2x+1} \\\hline
-14 & 100 \\\hline
-5& 12 \\\hline
-2&0 \\\hline
-1&4 \\\hline
0&2 \\\hline
1&0 \\\hline
4&6 \\\hline
13&80 \\\hline
\end{array}
And taking very large intervals of integers this seems like the only integers of this form.
However I don't know how can i proved.
Any ideas?
| Note that $\gcd(x,2x+1)=1$, then:
$$\frac{(x-1)^2(x+2)}{2x+1}=\frac{x^3-3x+2}{2x+1}=\frac{x^3-7x}{2x+1}+2=\frac{x(x^2-7)}{2x+1}+2\iff \\
\frac{x^2-7}{2x+1}=\frac{x^2+14x}{2x+1}-7=\frac{x(x+14)}{2x+1}-7 \iff \\
\frac{x+14}{2x+1}=14-\frac{27x}{2x+1} \iff \\
\frac{27}{2x+1} \iff \\
x\in \{-14,-5,-2,-1,0,1,4,13\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$
My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
| $$\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}$$
$$=\begin{vmatrix}1&1-1\\c^2+cb+b^2&a^2+ab+b^2-(c^2+cb+b^2)\end{vmatrix}$$
$$=\begin{vmatrix}1&0\\c^2+cb+b^2&a^2-c^2+b(a-c)\end{vmatrix}$$
$$=a^2-c^2+b(a-c)$$
$$=(a-c)(a+c+b)$$
| {
"language": "en",
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"answer_count": 6,
"answer_id": 2
} |
Math competition function with parameter problem Given a function $$f(x) = \frac{ax + 2}{3x - \frac{1}{a}}$$ find every possible value of the parameter $a$ such that for all real values $x$ for which $f(x)$ is defined it is true that $f(f(x))$ is also defined and $f(f(x))=x$
After manipulating the equation $$\frac{a\left(\frac{ax+2}{3x-\frac{1}{a}}\right)+2}{3\left(\frac{ax+2}{3x-\frac{1}{a}}\right)-\frac{1}{a}}$$ I got $$x\left(\left(a-\frac{1}{a}\right)x + \frac{1}{a^2} - a^2\right) = 0$$. This expression is zero if $x$ is zero or $\left(a-\frac{1}{a}\right)x + \frac{1}{a^2} - a^2$ is zero. Since the second expression must be zero for all $x$, $$a - \frac{1}{a}=0$$ and $$-a^2 + \frac{1}{a^2}=0$$.
Adding these two equations we get $$a^4 - a^3 - a + 1 = 0$$
Factoring this we get $$(a-1)(a+1)(a^2-a+1)=0$$
so $a$ can be 1 or -1 since $a^2-a+1$ doesn't have any real solutions.
| From $$x\left(\left(a-\frac{1}{a}\right)x + \frac{1}{a^2} - a^2\right) = 0$$
you can ignore the possibility $x=0$ because this is supposed to work for all $x$, so we need $$\left(a-\frac{1}{a}\right)x + \frac{1}{a^2} - a^2 = 0$$
Now note that $\frac 1{a^2}-a^2-=(a-\frac1a)(a+\frac 1a) $ so our equation becomes either $$a-\frac 1a=0\\a=\pm 1$$
or $$x-a+\frac 1a=0$$
but this depends on $x$, so is not true for all values of $x$. Our solution is $$a=\pm 1$$
| {
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Is there a closed form for the recurrence $V_{n+1}={V_n+\Delta V\over 1+{V_n\cdot \Delta V/C^2}}$, for constants $\Delta V$ and $C$? I was wondering if the following recurrence formula has a closed form:
$$V_{n+1}={V_n+\Delta V\over 1+{V_n\cdot \Delta V\over C^2}}$$
where $\Delta V$ and $C$ are positive constants, $V_n$ is the velocity of the $n$-the inertial frame and the primary velocity $V_0$ is given (take it $0$ if needed).
Attempt
This sequence obviously tends to $C$ (the speed-of-light supremum of speeds of observations), so I naturally tried to crack it using $$e_n=V_n-C$$but I failed. Any idea is appreciated.
Note
The above rule determines the Relativistic Velocity-addition Formula where $V_n$ is supposed to be the velocity of the inertial frame $2$ that is moving with respect to us (inertial frame $1$) and $\Delta V$ is an increase in the speed of the moving object (or we can assume it as the relative speed of object in the inertial frame 2). My work basis is the Lorentz Transformation.
| As I have noted in the comment, Wolfram Alpha gives the closed form, with $k = \Delta V$:
$$V_n = \frac{C_1 c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + C_1 (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n}$$
Taking $V_0 = 0$, we can see
$$ 0 = \frac{C_1 c(c+k) - c(c-k)}{(c-k) + C_1 (c+k) }$$
which is undefined if $C_1 = -\frac{c-k}{c+k}$, so let $C_1 \ne -\frac{c-k}{c+k}$, then
$$0 = C_1 c(c+k)-c(c-k)\Rightarrow 0 = C_1(c+k)-c+k \Rightarrow C_1=\frac{c-k}{c+k}$$
Hence $C_1$ is undefined when $k = \Delta V = -C$, so depending on the physical context of special realtivity (which I don't know enough of), this may never happen
Substituting this into the original closed form:
\begin{align}V_n &= \frac{(\frac{c-k}{c+k}) c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + (\frac{c-k}{c+k}) (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n} \\
&=\frac{c(c-k)\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{(c-k)\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\\
&= c \frac{\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\end{align}
Hence in terms of the way you formulated you have
$$ \boxed{V_n=C \frac{\left[ \left( \frac{1}{\Delta V-C} + \frac{1}{C}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{1}{\Delta V-C}+\frac{1}{C}\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$
if one wishes a preferable altenartive form without negatives inside the power terms:
\begin{align}V_n&=C \frac{\left[ \left( \frac{\Delta V}{C(\Delta V - C)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{\Delta V}{C(\Delta V - C)}\right)^n\right]} \\
&=C \frac{\left[ \left( -\frac{\Delta V}{C(C-\Delta V)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(-\frac{\Delta V}{C(C- \Delta V)}\right)^n\right]} \\
&=C \frac{\left[ (-1)^n \left(\frac{\Delta V}{C}\right)^n\left( \frac{1}{(C-\Delta V)}\right)^n - (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C+\Delta V)}\right)^n\right]}{\left[(-1)^n \left(\frac{\Delta V}{C}\right)^n\left(\frac{1}{(C+\Delta V)}\right)^n + (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C- \Delta V)}\right)^n\right]} \end{align}
which is written
\begin{align}V_n&=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \\
&=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \frac{(C+\Delta V)^n (C-\Delta V)^n}{(C+\Delta V)^n (C-\Delta V)^n}\end{align}
Finally,
$$\boxed{V_n=C \frac{\left[ \left( C+\Delta V\right)^n - \left(C- \Delta V\right)^n\right]}{\left[\left( C+ \Delta V\right)^n + \left(C- \Delta V\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$
which is in the form:
$$V_n = C \frac{a^n-b^n}{a^n+b^n}$$
which has limit $C$ as $n \to \infty $ (see this if you don't know how to show that), matching the limit in the OP's question.
If anyone knows how to derive the solution Wolfram gets, that would be preferable to this
| {
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Find the cluster points and one convergent subsequence of $a_n = \sqrt{n+1}-\sqrt{n} -\frac{(-1)^n}{n}, n \in \Bbb N$ Find the cluster points and one convergent subsequence of $a_n = \sqrt{n+1}-\sqrt{n} -\frac{(-1)^n}{n}, n \in \Bbb N$
For large $n$ would this not converge to $0$ and the only cluster point be also $0$?
Thus, any subsequence of $\{a_n\}$ would also converge to $0$?
Is this correct or have I missed something here?
| $$\sqrt{n+1} - \sqrt{n} - \frac{(-1)^n}{n} = \sqrt{n} \left( \sqrt{1+ \frac{1}{n}} - 1\right) - \frac{(-1)^n}{n}$$
$$= \sqrt{n} \left( \frac{1}{2n} + o \left( \frac{1}{n}\right)\right) - \frac{(-1)^n}{n}= \frac{1}{2 \sqrt{n}} - \frac{(-1)^n}{n} + o \left( \frac{1}{\sqrt{n}}\right)$$
So the sequence tends to $0$, so there is only one cluster point ($0$), and all the subsequences tend to $0$ as well.
| {
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Solving $4\sin^2{x}\cos^2{x}-\cos^2{x}=0$
Does anyone know why these answers are wrong?
$$4\sin^2{x}\cos^2{x}-\cos^2{x}=0$$
$(4\sin^2x)(\cos^2x)=\cos^2x$
$4\sin^2x=1$
$\sin x=1/2$
$x=30° , 150°$
Thank you
| $$4\sin^2 x \cos^2x - \cos^2 x= \cos^2 x(4\sin^2 x -1)=0$$
so $\cos^2 x = 0$ or $4\sin^2 x -1 = 0$; this gives $\cos x =0$ or $\sin^2 x = \frac{1}{4}$ or
$$\cos x = 0 \lor \sin x = \frac12 \lor \sin x = -\frac12$$
Now apply standard angles.
| {
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Find the number of different residues mod $p$ of $(x^2+y^2)^2$ where $(x,p)=(y,p)=1$ Let $p=4k+3$ be a prime number. Find the number of different residues mod $p$ of $$(x^2+y^2)^2$$ where $(x,p)=(y,p)=1$
This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?
I think the answer is $\dfrac{p-1}{2}$.But How to prove it?
| Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 \not\equiv 0 \pmod p$. As such, the number of possible residues of
$$(x^2+y^2)^2 \tag{1}\label{eq1}$$
is $\le \frac{p - 1}{2}$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 \le y \le \frac{p - 1}{2}$. In those cases, \eqref{eq1} becomes $4y^4$. To confirm they're all unique, let some $1 \le z \le \frac{p - 1}{2}$, where $z \neq y$, be such that
$$4y^4 \equiv 4z^4 \pmod p \; \Rightarrow \left(y^2 - z^2\right)\left(y^2 + z^2\right) \equiv 0 \pmod p \tag{2}\label{eq2}$$
Since $y^2 + z^2 \not\equiv 0 \pmod p$, this means that $y^2 - z^2 \equiv 0 \pmod p$. However, since the $\frac{p-1}{2}$ residues of values between $1$ and $\frac{p-1}{2}$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $\frac{p-1}{2}$.
| {
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Irrational integral $\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$ Would anyone be able to verify if this integral is calculated correctly?
$$\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$$
My attempt:
substitute:$\left(t = \sqrt{x}, t^2 = x, 2tdt=dx \right)$
$$
\begin{split}
\int \frac{1}{t} \sqrt{\frac{t-2}{t+2}}\,2t\,dt
&= 2\int\sqrt{\frac{t-2}{t+2}}dt
= 2\int \frac{\sqrt{t-2}}{\sqrt{t+2}}dt \\
&= 2\int \frac{\sqrt{(t-2)(t-2)}}{\sqrt{(t+2)(t-2)}}dt \\
&= 2\int \frac{t-2}{\sqrt{t^2-4}}dt \\
&= \int{\frac{2t-4}{\sqrt{t^2-4}}}dt \\
&= \int \frac{2t}{\sqrt{t^2-4}}dt - 4\int\frac{dt}{\sqrt{t^2-4}} \\
&= \sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C
\end{split}
$$
Substitute back $t = \sqrt{x}$:
result: $\Longrightarrow \sqrt{x-4} - 4\ln{|\sqrt{x} + \sqrt{x-4}|} + C$
| I got $2\sqrt{x-4}-8\operatorname{arcsinh}\left(\frac{\sqrt{\sqrt{x}-2}}{2}\right)+C$
| {
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Find the length of the parametric curve $x(t)=5+6t^4, \quad y(t)=5+4t^6\ , \quad0 ≤ t ≤ 2$ Find the length of the following parametric curve.
$$x(t)=5+6t^4\ ,\quad y(t)=5+4t^6\ ,\qquad 0 ≤ t ≤ 2.$$
I used the formula
$$\int_0^2\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$
And I found
$$\frac23\cdot 17^{3/2}+4-\frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$\left(\frac{dx}{dt}\right) = 24\cdot t^3 $$
$$\left(\frac{dy}{dt}\right) = 24\cdot t^5 $$
$$\int_0^2\sqrt{\left(24\cdot t^3\right)^2+\left(24\cdot t^5\right)^2}dt$$
$$\int_0^2\sqrt{\left(576\cdot t^6\right)+\left(576\cdot t^10\right)}dt$$
$$\int_0^2\sqrt{\left(576\cdot t^6\right) \cdot \left(1+t^4\right)}dt$$
$$24+\int_0^2\sqrt{\left(t^6\right) \cdot \left(1+t^4\right)}dt$$
$$\frac23\cdot 17^{3/2}+4-\frac23$$
| Line 4 should read $$\int_{t=0}^2 \sqrt{576 t^6 + 576 t^{10}} \, dt.$$ This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 \int_{t=0}^2 \sqrt{t^6 (1+t^4)} \, dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$\begin{align*}
24 \int_{t=0}^2 \sqrt{t^6(1+t^4)} \, dt
&= 24 \int_{t=0}^2 t^3 \sqrt{1+t^4} \, dt \qquad (u = 1+t^4; \; du = 4t^3 \, dt) \\
&= 6 \int_{u=1}^{17} \sqrt{u} \, du \\
&= 6 \left[\frac{2u^{3/2}}{3} \right]_{u=0}^{17} \\
&= 4 (17^{3/2} - 1) \\
&= 68 \sqrt{17} - 4.
\end{align*}$$
| {
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$\int \cos^4{x}dx$ unsolvable with $t = \tan{x}$? I have been told that "universal substitution always works", so I wanted to give it a try on this specific integral.
$\int \cos^4{x}dx$
For some reason it does not work. Please note that I solved this integral in a normal way, just don't understand why the universal substitution does not work.
$\Bigg(t=\tan{x}, \cos^2{x} = \frac{1}{1+t^2},dx=\frac{dt}{1+t^2} \Bigg)$
$\int \cos^4xdx = \int (\cos^2x)^2dx = \int \Big(\frac{1}{1+t^2} \Big)^2 \frac{dt}{1+t^2} = \int \frac{dt}{(1+t^2)(1+t^2)(1+t^2)}$
$\frac{1}{(1+t^2)^3} = \frac{At +B}{1+t^2} + \frac{Ct+D}{(1+t^2)^2} + \frac{Et+F}{(1+t^2)^3}$
$1 = (At+B)(t^4+2t^2+1)+(Ct+D)(1+t^2)+Et+F$
$1 = At^5 + 2At^3 + At + Bt^4 + 2Bt^2 + B + Ct^3 + Ct + Dt^2 + D + Et + F$
Now this boils down to six linear equations:
$0 = A$
$0 = B$
$0 = 2A + C$
$0 = 2B + D$
$0 = A + C + E$
$1 = B + D + F$
Which results in: $A = 0, B = 0, C = 0, D = 0, E = 0, F = 0$ making it unsolvable for me.
Any ideas what went wrong?
| Another way using reduction formula
$$\dfrac{d(\cos^nx\sin x}{dx}=\cos^{n+1}x-n\cos^{n-1}x(\cos^2x-1)=-(n-1)\cos^{n+1}x+n\cos^{n-1}x$$
Integrate both sides $$\cos^nx\sin x+K=-(n-1)I_{n+1}+nI_{n-1}$$
where $I_m=\int\cos^mx\ dx$ and $k$ is an arbitrary constant
Set $n=3,1$
| {
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An integral with logarithm and rational function Can someone solve:
$ \int_0^\infty dx \frac{\log (x)}{(x-1)(x+a)}\;,$
for $a$ say a positive real (or presumably any non real-negative complex number). Note there is no pole at $x=1$.
I can easily find the solution on mathematica, but I can't find a `proof'. My instinct is to use some nice contour, or bring the integral limits to [0,1] and expand the log, but I can't make either work!
Edit: Sorry I should have added the following. I know that the above integral is easily evaluated via partial fractions and using properties of dilogarithms, but I am wondering out of curiosity whether a simpler proof is possible. Certainly the result (see below) looks as if there should be an extremely elementary argument.
| Let $a>0$,
\begin{align}
J=\int_0^\infty \frac{\ln x}{(x-1)(x+a)} \,dx
\end{align}
Perform the change of variable $y=\dfrac{x-1}{x+a}$,
\begin{align}
J&=\frac{1}{1+a} \int_{-\frac{1}{a}}^1 \frac{\ln\left( \frac{1+ax}{1-x} \right)}{x}dx\\
&=\frac{1}{1+a}\int_{-\frac{1}{a}}^1 \frac{\ln\left( 1+ax\right)}{x}\,dx-\frac{1}{1+a}\int_{-\frac{1}{a}}^1 \frac{\ln\left( 1-x\right)}{x}dx
\end{align}
In the first integral perform the change of variable $y=-ax$,
\begin{align}
J&=-\frac{1}{1+a}\int_{-a}^1 \frac{\ln\left( 1-x\right)}{x}dx-\frac{1}{1+a}\int_{-\frac{1}{a}}^1 \frac{\ln\left( 1-x\right)}{x}dx\\
&=-\frac{2}{1+a}\int_0^1 \frac{\ln(1-x)}{x}dx-\frac{1}{1+a}\int_{-a}^0 \frac{\ln\left( 1-x\right)}{x}dx-\frac{1}{1+a}\int_{-\frac{1}{a}}^0 \frac{\ln\left( 1-x\right)}{x}dx
\end{align}
In the last two integrals perform the change of variable $y=-x$,
\begin{align}J&=-\frac{2}{1+a}\int_0^1 \frac{\ln(1-x)}{x}dx+\frac{1}{1+a}\int_{0}^a \frac{\ln\left( 1+x\right)}{x}dx+\frac{1}{1+a}\int_0^{\frac{1}{a}} \frac{\ln\left( 1+x\right)}{x}dx\\
&=\frac{2}{1+a}\int_0^1 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x}dx+\frac{1}{1+a}\int_{1}^a \frac{\ln\left( 1+x\right)}{x}dx+\frac{1}{1+a}\int_1^{\frac{1}{a}} \frac{\ln\left( 1+x\right)}{x}dx
\end{align}
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}J&=\frac{2}{1+a}\int_0^1 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x}dx+\frac{1}{1+a}\int_{1}^a \frac{\ln\left( 1+x\right)}{x}dx-\frac{1}{1+a}\int_1^{a} \frac{\ln\left( \frac{1+x}{x}\right)}{x}dx\\
&=\frac{2}{1+a}\int_0^1 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x}dx+\frac{1}{1+a}\int_{1}^a \frac{\ln x}{x}dx\\
&=\frac{2}{1+a}\int_0^1 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x}dx+\frac{\ln^2 a}{2(1+a)}
\end{align}
In the first integral perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}
J&=-\frac{4}{1+a}\int_0^1 \frac{\ln x}{1-x^2}dx+\frac{\ln^2 a}{2(1+a)}\\
&=\frac{4}{1+a}\int_0^1 \frac{x\ln x}{1-x^2}dx-\frac{4}{1+a}\int_0^1 \frac{\ln x}{1-x}dx+\frac{\ln^2 a}{2(1+a)}
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{4}{1+a}\left( \frac{1}{4}-1\right) \int_0^1 \frac{\ln x}{1-x}dx+\frac{\ln^2 a}{2(1+a)}\\
&=\frac{\ln^2 a}{2(1+a)}-\frac{3}{1+a}\int_0^1 \frac{\ln x}{1-x}dx\\
&=\boxed{\frac{\ln^2 a}{2(1+a)}+\frac{\pi^2}{2(1+a)}}
\end{align}
NB: I assume,
\begin{align}\int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}\end{align}
| {
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Find the values of $a,b \in \Bbb R$ (if exists) such that $-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$ for all $x \in \Bbb R$ Find the values of $a,b \in \Bbb R$ (if exists) such that $$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$$ for all $x \in \Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)\le x^2+ax+b $ $\space$ $\land$ $\space$ $4(x^2+2x+3) \ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
| Hint: $$x^2+2x+3=(x+1)^2+2\ge2$$
| {
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Convergent value of $\sum\limits^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$ The sum in question is:
$$\sum^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$$
It passes the ratio test:
\begin{align}
&\lim_{n\rightarrow \infty}\frac{(n+1)\left(\frac{5}{6}\right)^{n}}{n\left(\frac{5}{6}\right)^{n-1}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\frac{\left(\frac{5}{6}\right)^{n}}{\left(\frac{5}{6}\right)^{n}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}(1+ \frac{1}{n})\\
=\frac{5}{6} &< 1\Rightarrow \text{convergent}
\end{align}
But now I do not know how to find the convergent value.
| Hmm so write $f(x) = \sum_{n=1}^{\infty}x^n$. We know how to write this as a geometric series sum, and know the answer is $\frac{x}{1-x}$ (whenever $|x|<1$, which is the case with $x=5/6$ fortunately.
Now we will differentiate $f(x)$ and set $x=5/6$. We are allowed to do this by Taylor's theorem and term by term differentiation, because our series is just a Taylor series for $f(x)$.
If we differentiate the expression $\frac{x}{1-x}$ in $x$ and set $x=5/6$ we will get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}$
Evaluate: $$I=\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}.$$
Attempt. Working on substitution $x=\sqrt{1+y^2}\,\sinh u$ we get
$$I=\int\limits_{0}^{1}\frac{x}{(1+y^2)\sqrt{x^2+y^2+1}}\Bigg|_{0}^{1}\,dy=\int\limits_{0}^{1}\frac{dy}{(1+y^2)\sqrt{2+y^2}}$$ and then working on $y=\sqrt{2}\,\tan u$ we get:
$$I=\arctan\left(\frac{y}{\sqrt{y^2+2}}\right)\bigg|_{0}^{1}=\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}.$$
On the other hand, polar coordinates give:
$$I=\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{1}{\cos \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi+\int\limits_{\pi/4}^{\pi/2}\int\limits_{0}^{\frac{1}{\sin \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi$$
which gets us to:
$$I=\int\limits_{0}^{\pi/4}\left(1-\frac{\cos\phi}{\sqrt{1+\cos^2\phi}}\right)d\phi+\int\limits_{\pi/4}^{\pi/2}\left(1-\frac{\sin\phi}{\sqrt{1+\sin^2\phi}}\right)d\phi$$
and the computations seem difficult to handle.
Is there an easier way to approach the calculation of this integral?
Thank you in advance.
| First, note that the two integrals in the last line are equal. Just substitute $\phi=\frac{\pi}{2}-\varphi$ and see. Therefore
$$I=2\int_{0}^{\frac{\pi}{4}}\left(1-\frac{\cos\phi}{\sqrt{2-\sin^{2}\phi}}\right){\rm d\phi}=\left[\begin{matrix}\sin\phi=\sqrt{2}\sin\theta\\\cos\phi{\rm d}\phi=\sqrt{2}\cos\theta{\rm d}\theta\end{matrix}\right]=$$
$$=\frac{\pi}{2}-2\int_{0}^{\frac{\pi}{6}}\frac{\cos\theta}{\sqrt{1-\sin^{2}\theta}}{\rm d}\theta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I'm Puzzled inverting a function This is probably a very silly question, but I got puzzled inverted a function. The function to consider is the following
So, I want to solve for $u$ in terms of $x$ in the following equation
$$xu^{2}=x-u$$
I can get the right answer by the following procedure:
$\rightarrow4x^{2}u^{2}=4x^{2}-4ux$
(Multiplying by $4x$
)
$\rightarrow4u^{2}(1+4x^{2})=4x^{2}-4ux+u^{2}$
(Adding $u^{2}$
to both sides)
$\rightarrow u^{2}(1+4x^{2})=\left(2x-u\right)^{2}$
(factorizing the RHS)
$\rightarrow u\sqrt{1+4x^{2}}=\left(2x-u\right)$
(Taking the positive square root)
$\rightarrow u=\frac{2x}{1+\sqrt{1+4x^{2}}}$
==========================
The question is: Why using the quadratic formula to solve $xu^{2}=x-u$ for $u$ fails to give the correct answer ?
| $xu^2+u-x=0$
$a=x$
$b=1$
$c=-x$
So $u=\frac{-1 \pm \sqrt{1-4(x)(-x)}}{2x}$
Multiply top and bottom by -1 giving:
$u=\frac{1 \mp \sqrt{1+4x^2}}{-2x}$
Rationalize the numerator
$u=\frac{1 \mp \sqrt{1+4x^2}}{-2x} \cdot \frac{1 \pm \sqrt{1+4x^2}}{1 \pm \sqrt{1+4x^2}}$
$u=\frac{1-(1+4x^2)}{-2x(1 \pm \sqrt{1+4x^2}}$
$u=\frac{-4x^2}{-2x(1 \pm \sqrt{1+4x^2})}$
$u=\frac{2x}{1 \pm \sqrt{1+4x^2}}$
Now you must consider domain restrictions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.
Method $1$: (multiplicity of $y$)
Let $y=kx$ for some $k\in\Bbb R$. Then $$3x+2y=36\implies x(2k+3)=36\implies x=\frac{36}{2k+3}\\5x+4y=64\implies x(4k+5)=64\implies x=\frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)\implies (144-128)k=(192-180)\implies k=\frac34.$$ Now $$x=\frac{64}{4k+5}=\frac{64}{4\cdot\frac34+5}=8\implies y=kx=\frac34\cdot8=6.\quad\square$$
Method $2$: (use this if you really like quadratic equations :P)
How about we try squaring the equations? We get $$3x+2y=36\implies 9x^2+12xy+4y^2=1296\\5x+4y=64\implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28\implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
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Finding a better bound in an inequality Consider points $(x,y)$ on the curve $\sqrt{x^2-3x}+\sqrt{y^2-3y}=1$.
Prove that for all such pairs:
$$x^2+y^2\lt2(x+y)+8.$$
NOTE.- This problem was proposed by two mathematicians, from Romania and Spain, to a math blog in Madrid with the number $15$ on the $RHS$. In my solution I lowered this number to $8$.
By the way the number $8$ can also be improved, it is not the best bound. The task at hand is to find the tightest bound.
| We'll prove that $$\max_{\sqrt{x^2-3x}+\sqrt{y^2-3y}=1}(x^2+y^2-2(x+y))=\frac{11+\sqrt{13}}{2}.$$
Indeed, let $y\leq0.$
Thus, $$x^2-3x\leq1,$$ which gives $$x\leq\frac{3+\sqrt{13}}{2}$$ and
$$x^2+y^2-2(x+y)=x^2-3x+y^2-3y+x+y\leq$$
$$\leq\left(\sqrt{x^2-3x}+\sqrt{y^2-3y}\right)^2+x\leq1+\frac{3+\sqrt{13}}{2}<\frac{11+\sqrt{13}}{2}.$$
Id est, it's enough to prove our inequality for $x\geq3$ and $y\geq3.$
Now, let $\sqrt{x^2-3x}=a$ and $\sqrt{y^2-3y}=b$.
Thus, $a+b=1$, $x=\frac{3+\sqrt{9+4a^2}}{2}$, $y=\frac{3+\sqrt{9+4b^2}}{2}$ and we need to prove that
$$\left(\tfrac{3+\sqrt{9+4a^2}}{2}\right)^2+\left(\tfrac{3+\sqrt{9+4b^2}}{2}\right)^2-2\left(\tfrac{3+\sqrt{9+4a^2}}{2}+\tfrac{3+\sqrt{9+4b^2}}{2}\right)\leq\tfrac{11+\sqrt{13}}{2}$$ or
$$2(a^2+b^2)+\sqrt{9+4a^2}+\sqrt{9+4b^2}\leq5+\sqrt{13}.$$
Now, let $f(x)=\sqrt{9+4x^2}.$
Thus, $$f''(x)=\frac{36}{\sqrt{(9+4x^2)^3}}>0,$$
which says that $f$ is a convex function.
Thus, since for $a\geq b$ we have $$(a+b,0)\succ(a,b),$$ by Karamata $$f(a)+f(b)\leq f(a+b)+f(0)$$ or
$$\sqrt{9+4a^2}+\sqrt{9+4b^2}\leq\sqrt{9+4(a+b)^2}+\sqrt{9}=3+\sqrt{13}.$$
Id est, it's enough to prove that
$$2(a^2+b^2)+3+\sqrt{13}\leq5+\sqrt{13}$$ or $$2(a^2+b^2)\leq2(a+b)^2,$$ which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I am getting undefined as the answer of this integral problem $\int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)}$. Am I doing something wrong?
Find $$\int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$\begin{align}\frac{1}{(n-2)(3-n)}&=\frac{A}{n-2}+\frac{B}{3-n} \\ &= \frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\\ \Rightarrow 1 &= A(3-n)+B(n-2) \\ &= 3A - An+Bn-2B \\ &= n(B-A)+3A-2B.\end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 \qquad and \qquad 3A-2B=1.$$ $$\therefore A=B=1.$$ $$$$ $$\begin{align}\therefore \int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)} &= \int\limits_{2}^{3}\bigg(\frac{1}{n-2}+\frac{1}{3-n}\bigg)\,\mathrm dn \\ &= \int\limits_{2}^{3}\frac{1}{n-2}\,\mathrm dn+\int\limits_{2}^{3}\frac{1}{3-n}\,\mathrm dn \\ &=\left[\log(n-2)\right]_{\small 2}^{\small 3}+\left[\log(3-n)\right]_{\small 2}^{\small 3} \\ &= \left[\log(3-2)-\log(2-2)\right]+\left[\log(3-3)+\log(3-2)\right] \\ &= \left[\log 1-\log 0\right]+\left[\log 0 - \log 1\right]\end{align}$$ but $\log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
| No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I take the limit of this trigonometric function I was wondering how I could take this limit:
$\lim_{a→1}\frac{\sin(a^4 - 1)}{a^3-1}$
My idea was that if I can get the denominator and the inside of the sin to be the same I can use the sandwich theorem or maybe cancel out the a-1 in the denominator somehow. To go about this I did the following operations:
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a-1)(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a-1)(a+1)(a^2+1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a^4 - 1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)} *\lim_{a\to 1}\frac{(a^2 + 1)(a + 1)}{(a^2 + a + 1)}$$
At this step I tried to use the sandwich theorem to find $\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)}$
However I got $(-\infty,\infty)$ Like so:
$$\lim_{a\to 1} -1 < \sin(a^4-1) < 1$$
$$\lim_{a\to 1} \frac{-1}{a^4 - 1} < \frac{\sin(a^4-1)}{a^4-1} < \frac{1}{a^4 - 1}$$
$$= -\infty < \frac{\sin(a^4-1)}{a^4-1} < \infty$$
So I'm at a loss as to what I can do from here on. Or is what I've done just completely wrong?
| Here is a sexy trick:
$$\frac{\sin(a^4-1)}{a^3-1}=\frac{\sin(a^4-1)}{a^4-1}\times\frac{a^4-1}{a^3-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding real matrices of order $2\times 2$ in matrix equation
Finding all real matrices $X$ of order $2\times 2$ which satisfy the equation $X^2=\begin{pmatrix}
1 & 2 \\
3 & 7
\end{pmatrix}$
My Try: Let $\displaystyle X=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$. Then $\displaystyle X^2=\begin{pmatrix}a^2+bc & b(a+d)\\
c(a+d) & bc+d^2\end{pmatrix}=\begin{pmatrix}1 & 2\\ 3 & 7\end{pmatrix}$
So $a^2+bc=1\cdots (1)$ and $b(a+d)=2\cdots (2)$
And $c(a+d)=3\cdots (3)$ and $bc+d^2=7\cdots (4)$
But this is very complex method
Could some help me to solve it some easy way .Thanks
| You've already got
$$a^2+bc=1\tag1$$
$$b(a+d)=2\tag2$$
$$c(a+d)=3\tag3$$
$$bc+d^2=7\tag4$$
Now, $3\times (2)-2\times (3)$ gives
$$(3b-2c)(a+d)=0$$
Since $a+d\not=0$, we have
$$3b-2c=0,$$
i.e.
$$c=\frac 32b\tag5$$
Also, $(4)-(1)$ gives
$$(d-a)(d+a)=6$$
Multiplying the both sides by $b$ and using $(2)$ give
$$2(d-a)=6b\implies d=a+3b\tag6$$
So, from $(6)$, we have
$$(2)\implies 2ab=2-3b^2\tag7$$
Also, from $(5)$, we have
$$(1)\implies a^2+\frac 32b^2=1\implies 2a^2=2-3b^2\tag8$$
From $(7)(8)$, we get
$$2ab=2a^2\implies a(a-b)=0\implies a=0\quad\text{or}\quad a=b$$
Therefore, it is necessary that
$$(a,b,c,d)=\left(0,\pm\frac{\sqrt 6}{3},\pm\frac{\sqrt 6}{2},\pm\sqrt 6\right),\left(\pm\frac{\sqrt{10}}{5},\pm\frac{\sqrt{10}}{5},\pm\frac{3}{10}\sqrt{10},\pm\frac 45\sqrt{10}\right)$$
which is sufficient.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The breakdown of the solution to the ODE $y' = \sqrt{4-2y}$. Suppose $y=y(x)$ be a real valued function such that
$$
y(x) = \int_0^x \sqrt{4-2y(t)} dt.
$$
By differentiating both side with respect to $x$, we get the ODE
$$
y' = \sqrt{4-2y}, \quad y(0)=0.
$$
This is a separable equation which has
$$
y_1(x) = -\frac 12 (2-x)^2 + 2
$$
as the solution. By inspection,
$$\begin{align}
y_1'(x) &=2-x, \\
\sqrt{4-2y_1(x)} &= |2-x|
\end{align}$$
so this solution works for $x\le 2$.
Is it possible to continue to solution pass $x=2$? I know one trivial answer which is
$$
y(x) = \begin{cases} -\frac 12 (2-x)^2 + 2 &; x\le 2 \\
2 &; x>2.
\end{cases}
$$
Using the notation from the Picard-Lindelof theorem, we have
$$
y' = f(y)
$$
where $f(y) = \sqrt{4-2y}$ so $f$ is not Lipschitz at $y=2$ so the breakdown at $x=2$ (which causes $y(2)=2$) is to be expected. However, I don't know much about ODE apart from the general local existence result so I'd be very glad if someone could tell me if there is any other possible extension of the solution pass $x=2$.
| $$
\begin{align}
y'(x) &= \sqrt{4-2y(x)}\\
y'(x) &= \sqrt{2}\sqrt{2-y(x)}\\
\dfrac{y'(x)}{\sqrt{2-y(x)}} &= \sqrt{2}\\
\int\dfrac{y'(x)}{\sqrt{2-y(x)}} dx&= \int\sqrt{2}dx\\
-2\sqrt{-y(x)+2}&=\sqrt{2}x+K_1\\
y(x)&=-\dfrac{1}{4}(\sqrt{2}x+K_1)^2-2\\
\end{align}
$$
For, $y(0)=0$:
$$K_1=\{-2;2\}.$$
Therefore,
$$
y(x) = \begin{cases}
-\dfrac{1}{2}x(x-4), & \mbox{if } K_1 = -2 \mbox{, } x\geq 2 \\
-\dfrac{1}{2}x(x+4), & \mbox{if } K_1 = 2
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Substitute $g(x)=Ae^{-\beta x^2}$ into $\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} g(\eta)\exp(-(x-\eta)^2/4\alpha t) \ d\eta$
I am trying to show that by directly substituting $g(x)=Ae^{-\beta x^2}$ into $$\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} g(\eta)\exp(-(x-\eta)^2/4\alpha t) \ d\eta,$$ we obtain $$\theta(x,t)=\frac{A}{\sqrt{1+4\alpha\beta t}}\exp(-\beta x^2/(1+4\alpha\beta t).$$
My attempt:
Let $g(\eta)=Ae^{-\beta\eta^2}$. Then $$\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} \exp(-\beta\eta^2-(x-\eta)^2/4\alpha t) \ d\eta.$$ However, this doesn't give the desired result. Is someone able to please give me a hint on the correct path that I should take?
| Let us first compute the expression in exponential
\begin{align}
-\beta\eta^2 - \frac{1}{4\alpha t}(x-\eta)^2 &= -\beta\eta^2 - \frac{1}{4\alpha t}(x^2-2x\eta+\eta^2) \\
&=-\frac{x^2 }{4\alpha t}-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta^2 - 2\frac{x}{4\alpha t\beta+1}\eta\right) \\
&=-\frac{x^2 }{4\alpha t}-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2 -\frac{x^2}{(4\alpha t\beta+1)4\alpha t} \\
&=-\frac{\beta x^2}{(4\alpha t\beta+1)} -\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2
\end{align}
Replacing the previous expression in the initial calculation:
\begin{align*}
\frac{A}{2\sqrt{\pi\alpha t}}\exp\left(-\frac{\beta x^2}{(4\alpha t\beta+1)}\right)\int_{R}\exp\left(-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2\right)d\eta = \\ \frac{A}{\sqrt{4\alpha t\beta+1}}\exp\left(-\frac{\beta x^2}{(4\alpha t\beta+1)}\right)
\end{align*}
In the last equation I used the Gauss integral computation : $\int_{R}\exp(-\mu(x-\gamma)^2)dx = \sqrt{\pi/\mu}$
| {
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"url": "https://math.stackexchange.com/questions/3194158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Taylor series of $\sqrt{x}$ centered at $x=4$ and the order 3 Find Taylor series of $\sqrt x$, about $x=4$ and the order 3
I've tried a few timesm but I keep getting a result that does not comply with the answer. Following are the steps I've taken, hopefully I can get a pointer to the factor I'm missing in some of the denominators. I hope all the steps are clear.
Find derivatives of $\sqrt x$
If $f(x) = \sqrt x$, then:
*
*$f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$
*$f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}$
*$f'''(x) = \frac{3}{8}x^{-\frac{5}{2}}$
Write out the Taylor polynomial
$P_3(x) = \sqrt 4 + \frac{1}{2} 4^{-\frac{1}{2}}(x - 4) - \frac{1}{4} 4^{-\frac{3}{2}}(x - 4)^2 + \frac{3}{8} 4^{-\frac{5}{2}}(x - 4)^3$
$P_3(x) = 2 + (\frac{1}{2}) (\frac{1}{\sqrt 4})(x - 4) - (\frac{1}{4}) (\frac{1}{4^{\frac{3}{2}}})(x - 4)^2 + (\frac{3}{8})(\frac{1}{x ^{\frac{5}{2}}})(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{4} \frac{1}{4^{2/2} 4^{1/2}}(x - 4)^2 + \frac{3}{8}\frac{1}{4^{4/2}} \frac{1}{4^{1/2}}(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32}(x - 4)^2 + \frac{3}{256}(x - 4)^3$
But according to the book the series develops like:
$$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2 + \frac{3}{1536}(x - 4)^3$$
So I'm missing some (increasing) factor, I just can't seem to find it. Any hints?
| You should simplify your powers of $4$,
I.e. $4^{-3/2} = 2^{-3} = \frac{1}{8}$. Furthermore you should divide by $k!$ where $k$ is the power: $(x-4)^k$
| {
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"question_score": "1",
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"answer_id": 2
} |
Combinatorics problem, right solution? We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.
For the remaining two places, I could have $2$ more people of a single profession. This is $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}$ possibilities.
I could also have two people of different professions; a doctor and a laywer, $\binom{5}{1}\binom{3}{1}$; a doctor and an engineer, $\binom{6}{1}\binom{3}{1}$; or an engineer and a laywer $\binom{6}{1}\binom{5}{1}$.
This adds up to $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}+\binom{5}{1}\binom{3}{1}+\binom{6}{1}\binom{3}{1}+\binom{6}{1}\binom{5}{1}=91$ possible committees.
I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.
Thanks in advance!
| There are 6 types of possible committees:
\begin{array} {|r|r|r|}
\hline
3&1&1 \\
\hline
1&3&1 \\
\hline
1&1&3 \\
\hline
2&2&1 \\
\hline
2&1&2 \\
\hline
1&2&2 \\
\hline
\end{array}
For each type of committee, it is necessary to calculate the different groups of people that compose it:
\begin{array} {|c|c|c|c|}
\hline
3&1&1& {6\choose3}{7\choose1}{4\choose1}= 20\cdot7\cdot4 = 560 \\
\hline
1&3&1& {6\choose1}{7\choose3}{4\choose1}= 6\cdot35\cdot4 = 840 \\
\hline
1&1&3& {6\choose1}{7\choose1}{4\choose3}= 6\cdot7\cdot4 = 168 \\
\hline
2&2&1& {6\choose2}{7\choose2}{4\choose1}= 15\cdot21\cdot4 = 1260 \\
\hline
2&1&2& {6\choose2}{7\choose1}{4\choose2}= 15\cdot7\cdot6 = 630 \\
\hline
1&2&2& {6\choose1}{7\choose2}{4\choose2}= 6\cdot21\cdot6 = 756 \\
\hline
\end{array}
Adding 6 gives a total of different committees:
$$ 560+840+168+1260+630+756 = 4214 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity I know that one can simplify $\cos(2\arcsin(x))$ using $\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$:
\begin{alignat}{1}
\cos(2\arcsin(x))&=\cos^2(\arcsin(x))-\sin^2(\arcsin(x))
\\&=1-2\sin^2(\arcsin(x))
\\&=1-2x^2
\end{alignat}
I tried to make this simplification using only $\sin^2(x)+\cos^2(x)=1$:
\begin{alignat}{1}
\cos^2(2\arcsin(x))&=1-\sin^2(2\arcsin(x))
\\ \left|\cos(2\arcsin(x))\right|&=\sqrt{1-\sin^2(2\arcsin(x))}
\\ &=\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}
\\ &=\sqrt{1-4x^2 |1-x^2|}
\\ &=\sqrt{1-4x^2(1-x^2)}
\\ &=\sqrt{1-4x^2+4x^4}
\\ &=\sqrt{\left(2x^2-1\right)^2}
\\ &=|2x^2-1|
\end{alignat}
And then I could not figure out how to proceed. So, how to get rid of $|\cdot|$ in:
$$\left|\cos(2\arcsin(x))\right|=|2x^2-1|$$
and get $1-2x^2$ ?
Note: this is a part of my attempt to solve the integral $\int x^2\cdot\sqrt{1-x^2}\,\,\mathrm dx$ by trigonometric substitution.
| $\cos(2\arcsin x)$ will be $\ge0,$ if
$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2$
$\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$
In that case $$|2x^2-1|=-(2x^2-1)$$
Check if $x>\dfrac1{\sqrt2}$
or $x<-\dfrac1{\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Square root property to solve quadratic equation: $3(x-4)^2=15$ I get $\sqrt{21}$ but solution is $4+-\sqrt{5}$ I am to solve for x using square root property:
$3(x-4)^2=15$
The textbook solution is $4+-\sqrt{5}$ and I am unable to arrive at this. I arrived at $\sqrt{21}$
Here is my working:
$3(x-4)^2=15$
$(x-4)^2=5$ # divide both sides by factor 3
$x^2-16=5$ # multiply out $(x-4)^2$
$x^2 = 21$
$x=\sqrt{21}$
How can I arrive at $x=4+-\sqrt{5}$ # not sure syntax for 'plus or minus' symbol here, just used +-
| We have$(x-4)^2 = 5$
Taking sqrt on both sides,
$$x-4 = \pm\sqrt5$$
$$x = 4 \pm\sqrt5$$
That's it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Largest value of determinant
If $\alpha,\beta,\gamma \in [-3,10].$ Then largest value of the determinant
$$\begin{vmatrix}3\alpha^2&\beta^2+\alpha\beta+\alpha^2&\gamma^2+\alpha\gamma+\alpha^2\\\\
\alpha^2+\alpha\beta+\beta^2& 3\beta^2&\gamma^2+\beta\gamma+\beta^2\\\\
\alpha^2+\alpha\gamma+\gamma^2& \beta^2+\beta\gamma+\gamma^2&3\gamma^2\end{vmatrix}$$
Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.
| Being a $3 \times 3$ determinant, it is too small to be attacked with sophisticated techniques, but also too involved to be attacked by brute force. We shall compute it with successive simplifications of its rows and columns. In the following, $R_i$ and $C_j$ will mean the $i$-th row and the $j$-th column, and $[R_i \to aR_i + bR_j]$ means to replace the $i$-th row with the linear combination $aR_i + bR_j$ where $a$ and $b$ are numbers. The computation, then, goes as follows:
$$\begin{align*}
& \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha^2 + \alpha \beta + \beta^2 & 3\beta^2 & \gamma^2 + \beta \gamma + \beta^2 \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = [R_2 \to R_2 - R_3] = \\[10pt]
& \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
(\beta - \gamma) (\alpha + \beta + \gamma) & (\beta - \gamma) (2\beta + \gamma) & (\beta - \gamma) (\beta + 2\gamma) \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = \\[10pt]
(\beta - \gamma) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
\alpha^2 + \alpha \gamma + \gamma^2 & \beta^2 + \beta \gamma + \gamma^2 & 3\gamma^2\end{vmatrix} = [R_3 \to R_3 - R_1] = \\[10pt]
(\beta - \gamma) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
(\gamma - \alpha) (\gamma + 2\alpha) & (\gamma - \alpha) (\alpha + \beta + \gamma) & (\gamma - \alpha) (2\gamma + \alpha) \end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta^2 + \alpha \beta + \alpha^2 & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 2\beta + \gamma & \beta + 2\gamma \\
\gamma + 2\alpha & \alpha + \beta + \gamma & 2\gamma + \alpha \end{vmatrix} = [C_2 \to C_2 - C_1] = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) & \begin{vmatrix}
3\alpha^2 & (\beta - \alpha) (\beta + 2\alpha) & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & \beta - \alpha & \beta + 2\gamma \\
\gamma + 2\alpha & \beta - \alpha & 2\gamma + \alpha \end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha) (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma^2 + \alpha \gamma + \alpha^2 \\
\alpha + \beta + \gamma & 1 & \beta + 2\gamma \\
\gamma + 2\alpha & 1 & 2\gamma + \alpha
\end{vmatrix} = [C_3 \to C_3 - C_1] \\[10pt]
(\beta - \gamma) (\gamma - \alpha) (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & (\gamma - \alpha) (\gamma + 2\alpha) \\
\alpha + \beta + \gamma & 1 & \gamma - \alpha \\
\gamma + 2\alpha & 1 & \gamma - \alpha
\end{vmatrix} = \\[10pt]
(\beta - \gamma) (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma + 2\alpha \\
\alpha + \beta + \gamma & 1 & 1 \\
\gamma + 2\alpha & 1 & 1
\end{vmatrix} = [C_3 \to C_3 - C_2] \\[10pt]
(\beta - \gamma) (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & \gamma - \beta \\
\alpha + \beta + \gamma & 1 & 0 \\
\gamma + 2\alpha & 1 & 0
\end{vmatrix} = \\[10pt]
- (\beta - \gamma)^2 (\gamma - \alpha)^2 (\beta - \alpha) & \begin{vmatrix}
3\alpha^2 & \beta + 2\alpha & 1 \\
\alpha + \beta + \gamma & 1 & 0 \\
\gamma + 2\alpha & 1 & 0
\end{vmatrix} = \\[10pt]
- (\beta - \gamma)^2 (\gamma - \alpha)^2 (\beta - \alpha)^2
\end{align*}
$$
where the last $3 \times 3$ determinant has been expanded along the $3$rd column.
Since the determinant is the negative of the square of a product of real numbers, its maximum will be $0$, reached whenever any two of $\alpha, \beta, \gamma \in [-3,10]$ are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve in $C$ : $P(z)=z^4+2z^3+5z^2+4z+1=0$ where $P(i-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$ Question solve in $C$ :
$P(z)=z^4+2z^3+5z^2+4z+1=0$
where $P(i+-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$
My attempt :
Let $\lambda=i-\frac{1}{2}+i\frac{\sqrt 3}{2}$
$(2\lambda+1)^2=(i(2+\sqrt 3))^2$
$(2\lambda+1)^2+7)^2=(-4\sqrt 3)^2$
$({\lambda}^2+\lambda+2)^2=3$
So we find :
$\lambda^2++2\lambda^3+5\lambda^2+4\lambda+1=0$
But which step !? can be find all root of P(z) !!
| If $$z=-\frac{1}{2}-i-\frac{1}{2}\sqrt{3}i$$ is one solution then is also $$z=-\frac{1}{2}+i+\frac{1}{2}\sqrt{3}i$$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
An interesting list ordering result based on skipping indexes by value entered This is something interesting I've found while attempting to shuffle $n$ values in a list is a deterministic way intended to look "random-ish". The idea is that you sort a number of values $r$ from $0...n-1$ into a list by starting with an index value $i=0$, then calculate the index for the new item to be inserted using the formula $i_1 = (i_0 + r)\mod n$. What I've found is that in most cases, there are several collisions where a number is placed over another except when $n = 2^k, k\in Z$.
I wrote a Python program to do this and display the output.
>>> for n in range(5,11):
print('_____')
num_list = ['' for i in range(n)]
i = 0
for r in range(n):
i = (i + r) % n
num_list[i] += str(r) + ','
for x in range(n):
print(str(x)+'|'+num_list[x])
_____
0|0,4,
1|1,3,
2|
3|2,
4|
_____
0|0,3,
1|1,
2|
3|2,5,
4|4,
5|
_____
0|0,6,
1|1,5,
2|
3|2,4,
4|
5|
6|3,
_____
0|0,
1|1,
2|4,
3|2,
4|7,
5|6,
6|3,
7|5,
_____
0|0,8,
1|1,4,7,
2|
3|2,6,
4|
5|
6|3,5,
7|
8|
_____
0|0,4,
1|1,6,
2|
3|2,
4|
5|5,9,
6|3,8,
7|
8|7,
9|
>>>
As you can see, multiple numbers are assigned to the same value for all cases except where n is a power of 2. Can anyone explain this?
| In other words, there are $n$ buckets $B_0, \ldots, B_{n - 1}$ and we throw $r$ into a bucket $B_{0 + 1 + \ldots + r\pmod{n}}$. The question then is for which $n$ all the buckets contain exactly one element. This is true if and only if $0 + 1 + \ldots + r\pmod{n}$, $r = 0, \ldots, n - 1$ are all different. In turn, the latter is true if and only if for all $r_1, r_2\in\{0, 1, \ldots, n - 1\}$, $r_1 < r_2$ it holds that $(0 + 1 + \ldots + r_2) - (0 + 1 + \ldots + r_1)$ is not divisible by $n$. And I will show that indeed this is true if and only if $n$ is a power of $2$.
Re-write $(0 + 1 + \ldots + r_2) - (0 + 1 + \ldots + r_1)$ as follows:
$$(0 + 1 + \ldots + r_2) - (0 + 1 + \ldots + r_1) = \frac{r_2(r_2 + 1)}{2} - \frac{r_1(r_1 + 1)}{2} = \frac{(r_2 - r_1) (r_2 + r_1 + 1)}{2}.$$
First consider the case $n = 2^k$. Assume that the last expression is divisible by $n = 2^k$. This means that $(r_2 - r_1)(r_2 + r_1 + 1)$ is divisble by $2^{k + 1} = 2n$. Note that $(r_2 - r_1)$ and $(r_2 + r_1 + 1)$ have different parity. This means that one of these two numbers is divisible by $2n$. Both if these numbers are positive. Hence one number should be at least $2n$. This is impossible since $r_1, r_2 \le n - 1$.
Now consider the case $n = 2^k \cdot u$ for some $k\ge 0$ and some odd $u \ge 3$. Set
\begin{align*}
r_1 &= \begin{cases} \frac{2^{k + 1} - u - 1}{2} & \mbox{ if } u < 2^{k + 1}, \\ \frac{u - 2^{k + 1} - 1}{2} & \mbox{ if } u > 2^{k + 1},\end{cases} \\
r_2 &= \frac{2^{k + 1} + u - 1}{2}.
\end{align*}
You see that there are two subcases, $u < 2^{k + 1}$ and $u > 2^{k + 1}$ (equality is impossible because $u$ is odd). In the first case we have $r_2 - r_1 = u$ and $(r_2 + r_1 + 1) = 2^{k + 1}$. In the second case we have $r_2 - r_1 = 2^{k + 1}$ and $r_1 + r_2 + 1 = u$. In either case we obtain
$$\frac{(r_2 - r_1) (r_2 + r_1 + 1)}{2} = 2^k\cdot u = n,$$
which means that $(0 + 1 + \ldots + r_1) \equiv (0 + 1 + \ldots + r_2)\pmod{n}$. We should also check that $r_1, r_2\in\{0, 1, \ldots, n - 1\}$ and $r_1 < r_2$. This all is obvious with one exception: it is not clear why $r_2 \le n - 1$. Actually, it is the only place where we use the fact that $n$ is not a power of $2$. So our goal is to verify that
$$\frac{2^{k + 1} + u - 1}{2} \le n - 1 = 2^k \cdot u - 1.$$
This is equivalent to the following:
$$2^{k + 1} + u \le 2^{k + 1} \cdot u - 1 = 2^k\cdot u + 2^k \cdot u - 1. $$
So why is this true? Because $u\le 2^k \cdot u$ and $2^{k + 1} = 2\cdot 2^k < u \cdot 2^k$ (the latter is due to the fact that $u\ge 3$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $7x^2+3x=0$ by completing the square? I am to solve for x: $7x^2+3x=0$
I'm aware that there are multiple approaches to solving a quadratic. In this case, since there is no constant term I decided to go the completing the square route. I know from my textbooks answer that the solutions are $x=0$ and $x=-\frac{3}{7}$.
Here is how far I got:
$7x^2+3x=0$ # want to have leading coefficient 1 not 7
$x^2 + \frac{3}{7}x=0$
take 1/2 of the linear coefficient and then square it:
$\frac{1}{2}*\frac{3}{7}=\frac{3}{14}$
Then square it:
$(\frac{3}{14})^2$ = $\frac{9}{196}$
Add this term to both sides of my equation:
$x^2 + \frac{3}{7}x + \frac{9}{196}=\frac{9}{196}$
This is where I get stuck. Apparently I should be able to factor as a perfect square the left hand side of the equation. Perhaps because I'm working with fractions I cannot see how to do that? How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?
| $$7x^2+3x=0 \iff 7(x^2+(3/7)x)=0\iff$$ $$\iff 7 (\;(x+(1/2)(3/7)\,)^2 -(\,(1/2)(3/7)\,)^2\;)=0\iff $$ $$\iff (x+(1/2)(3/7)\,)^2-(\,(1/2)(3/7)\,)^2=0 \iff$$ $$\iff (x+(1/2)(3/7)\,)^2=(\,(1/2)(3/7)\,)^2\iff$$ $$\iff x+(1/2)(3/7)=\pm (1/2)(3/7)\iff$$ $$\iff (\;(x+(1/2)(3/7)=(1/2)(3/7)\; \lor \;x+(1/2)(3/7)=-(1/2)(3/7)\;) \iff$$ $$\iff (\,x=0\,\lor \,x=-3/7\,).$$
Of course if $A\ne 0$ then $Ax^2+Bx=0 \iff A(x)(x+B/A)=0 \iff (x=0 \lor x+B/A=0)\iff (x=0\lor x=-B/A).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If p arithmetic means are inserted If $p$ arithmetic means $A_1$, $A_2$, ..., $A_p$, are inserted between $5$ and $41$ so that the relation is satisfied:
$$
\frac{A_3}{A_{(p-1)}}=\frac{2}{5}
$$
Find the value of $p/11$
I've tried using the formula that number of means $(n) = a+n[(b-a) /n+1]$ but my value of $p$ keeps coming $-12/13$ which is wrong. The correct ans that $p/11$ is 1
Pls help me out.
| When we insert p terms, the $n^{th}$ mean term is,
$A_n = a + nd$
Here,
$ d= \frac{last term -first term}{no. of terms - 1} = \frac{41 - 5}{p+1} = \frac{36}{p+1}$
$a = 5$,$a_n = 41$
$\frac{A_3}{A_{p-1}} = \frac{5 + 3\frac{36}{p+1}}{5 + (p-1)\frac{36}{p+1}} = \frac{2}{5}$
Taking LCM and solving,
$\frac{5p+5 + 108}{5p + 5 + 36p - 36} = \frac{5p+113}{41p - 31}=\frac{2}{5}$
Cross multiplying,
$25p + 565 = 82p - 62$
$627 = 57p$
$\frac{627}{57} = 11 = p$
Or,$$ \frac{p}{11} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
primitives of $f(x)=\frac{1}{2\sqrt{x-x^{2}}}$ I have this function $f:(0,1)\rightarrow \mathbb{R},f(x)=\frac{1}{2\sqrt{x-x^{2}}}$ and I need to find the primitives of $f(x)$
So I calculated $\int \frac{1}{2\sqrt{x-x^{2}}}\,dx$ and I got $\frac{1}{2}\arcsin(2x-1)+C$ but the right answer is $\arcsin\sqrt{x}+C$ .Where's my mistake?How to start?
| If $\theta=\arcsin \sqrt{x}$, then $\sin\theta=\sqrt{x}$, so $\sin^2\theta=x$.
Since $\sin^2\theta=\frac{1-\cos(2\theta)}{2}$, we get
$$\frac{1-\cos(2\theta)}{2}=x$$
We deduce $-\cos(2\theta)=2x-1$.
Then, $\sin(2\theta-\pi/2)=2x-1$.
So, (maybe up to a constant) we have $\arcsin(2x-1)=2\theta-\pi/2$.
Which means that $\theta=\frac{1}{2}\arcsin(2x-1)+\pi/4$. So you have that $\arcsin \sqrt{x}$ and $\frac{1}{2}\arcsin(2x-1)$ differ by a constant. So, both results are okay.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Combination Summation problem
I obtained $np/2$ which has a similar form as q but I'm unable to prove that it is equal to $q$.
| As a first step, we have better to simplify and pass from the double sums to simple ones.
Let's put
$$
s(n) = \sum\limits_{1\, \le \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c} n \\ k\\
\end{array} \right)}}} = \frac{1}{{n!}}\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n - k} \right)!}
$$
Then
$$
\begin{array}{l}
\sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {\frac{1}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}} + \frac{1}{{\left( \begin{array}{c} n \\ j \\
\end{array} \right)}}} + \sum\limits_{1\, \le \,j\, < \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}} + \frac{1}{{\left( \begin{array}{c} n \\ j \\
\end{array} \right)}}} + 2\sum\limits_{1\, \le \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}}} = \\
= 2p(n) + 2s(n) = \\
= \sum\limits_{\begin{array}{*{20}c} {1\, \le \,k\, \le \,n} \\ {1\, \le \,j\, \le \,n} \\
\end{array}} {\left( {\frac{1}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}} + \frac{1}{{\left( \begin{array}{c} n \\ j \\
\end{array} \right)}}} \right)}
= \sum\limits_{1\, \le \,k\, \le \,n} {\sum\limits_{1\, \le \,j\, \le \,n} {\left( {\frac{1}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}} + \frac{1}{{\left( \begin{array}{c} n \\ j \\
\end{array} \right)}}} \right)} } = \\
= \sum\limits_{1\, \le \,k\, \le \,n} {\left( {\frac{n}{{\left( \begin{array}{c} n \\ k \\
\end{array} \right)}} + s(n)} \right)} = 2n\,s(n)\quad \Rightarrow \\
\Rightarrow \quad p(n) = \left( {n - 1} \right)\,s(n) \\ \end{array}
$$
Similarly, putting
$$
r(n) = \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {\binom{n}{k}}}}
$$
we get
$$
\eqalign{
& \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {
{k \over {\binom {n}{k}}} + {j \over {\binom {n}{j}}}}
+ \sum\limits_{1\, \le \,j\, < \,k\, \le \,n} {{k \over {\binom {n}{k}}}
+ {j \over {\binom {n}{j}}}}
+ 2\sum\limits_{1\, \le \,k\, \le \,n} {{k \over {\binom {n}{k}}}} = \cr
& = 2q(n) + 2r(n) = \cr
& = \sum\limits_{\matrix{ {1\, \le \,k\, \le \,n} \cr
{1\, \le \,j\, \le \,n} \cr } } {\left( {{k \over {\binom {n}{k}}}
+ {j \over {\binom {n}{j}}}} \right)}
= \sum\limits_{1\, \le \,k\, \le \,n} {\sum\limits_{1\, \le \,j\, \le \,n} {
\left( {{k \over {\binom {n}{k}}}
+ {j \over {\binom {n}{j}}}} \right)} } = \cr
& = \sum\limits_{1\, \le \,k\, \le \,n} {\left( {n{k \over {\binom {n}{k}}} + q(n)} \right)} = 2n\,r(n)
\quad \Rightarrow \cr
& \Rightarrow \quad q(n) = \left( {n - 1} \right)\,r(n) \cr}
$$
We can then work on $s(n),\, r(n)$ to obtain
$$
\eqalign{
& r(n) + s(n) = \sum\limits_{1\, \le \,k\, \le \,n} {{{k + 1} \over {\binom{n}{k}}}} = \cr
& = {1 \over {n!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n - k} \right)!} = \cr
& = \left( {n + 1} \right)\left( {{1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) = \cr
& = \left( {n + 1} \right)\left( {{1 \over {\left( {n + 1} \right)!}}\sum\limits_{2\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!} } \right) = \cr
& = \left( {n + 1} \right)\left( {s(n + 1) - {{n!} \over {\left( {n + 1} \right)!}}} \right) = \left( {n + 1} \right)s(n + 1) - 1 \cr}
$$
and
$$
\eqalign{
& s(n + 1) = \sum\limits_{1\, \le \,k\, \le \,n + 1} {{1 \over {\left( \matrix{
n + 1 \cr
k \cr} \right)}}} = \cr
& = {1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!} = \cr
& = {1 \over {\left( {n + 1} \right)!}}\left( {\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n + 1 - k} \right)!} + \left( {n + 1} \right)!} \right) = \cr
& = 1 + {1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {n + 1 - k} \right)k!\left( {n - k} \right)!} = \cr
& = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n - k} \right)!}
- \sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) = \cr
& = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n)
- \sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) = \cr
& = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n) - \sum\limits_{2\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!} } \right) = \cr
& = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n) - \left( {n + 1} \right)!s(n + 1) + n!} \right) = \cr
& = 1 + {{n + 2} \over {n + 1}}s(n) - s(n + 1) + {1 \over {n + 1}}\quad \Rightarrow \cr
& \Rightarrow \quad s(n + 1) = {{n + 2} \over {2\left( {n + 1} \right)}}\left( {1 + s(n)} \right) \cr}
$$
So, the final result is
$$ \bbox[lightyellow] {
\left\{ \matrix{
p(n) = \left( {n - 1} \right)\,s(n) \hfill \cr
q(n) = \left( {n - 1} \right)\,r(n) \hfill \cr
r(n) + s(n) = \left( {n + 1} \right)s(n + 1) - 1 \hfill \cr
s(n + 1) = {{n + 2} \over {2\left( {n + 1} \right)}}\left( {1 + s(n)} \right) \hfill \cr}
\right.\quad \Rightarrow \quad \left\{ \matrix{
r(n) = {n \over 2}\left( {s(n) + 1} \right) \hfill \cr
p(n) = \left( {n - 1} \right)\,s(n) \hfill \cr
q(n) = {n \over 2}\left( {p(n) + \left( {n - 1} \right)} \right) \hfill \cr} \right.
}$$
which is not as expected.
In fact, you should define the sums to start from zero
$$ \bbox[lightyellow] {
\eqalign{
& \matrix{
\matrix{
p^ * (n) = \sum\limits_{0\, \le \,k\, < \,j\, \le \,n} {{1 \over { \binom{n}{k} }}
+ {1 \over { \binom{n}{j} }}} = \hfill \cr
= \sum\limits_{0\, < \,j\, \le \,n} {\left( {1 + {1 \over {\binom{n}{j}}}} \right)}
+ \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k} }}
+ {j \over {\binom{n}{j}}}} = \hfill \cr
= n + s(n) + p(n) \hfill \cr} & {\;\;\;} & \matrix{
q^ * (n) = \sum\limits_{0\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k} }}
+ {j \over {\binom{n}{j}}}} = \hfill \cr
= \sum\limits_{0\, < \,j\, \le \,n} {\left( {{j \over {\binom{n}{j}}}} \right)}
+ \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k}}}
+ {j \over {\binom{n}{j}}}} = \hfill \cr
= r(n) + q(n) \hfill \cr} \cr
} \cr
& \cr}
}$$
to obtain
$$ \bbox[lightyellow] {
q^ * (n) = {n \over 2}p^ * (n)
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $x +y$ If $a=\frac{x}{x^2+y^2}$ and $b=\frac{y}{x^2+y^2}$ then find $x+y$
I find that $x+y/y=\frac{a+b}{b}$ but the ans in the form of a and B only.
| $a^2+b^2={x^2\over{(x^2+y^2)^2}}+{y^2\over{(x^2+y^2)^2}}= {1\over{x^2+y^2}}$, $x=a(x^2+y^2), y=b(x^2+y^2)$ implies that $x+y={{a+b}\over{a^2+b^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from the question, I tried:
Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.
But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
| A somewhat systematic (but laborious) approach: Assume
$$N=\sqrt 3+\sqrt 7+\sqrt{21} $$
is rational. Then also
$$N^2=3+7+21+2(\sqrt{21}+3\sqrt 7+7\sqrt 3)= 31+2\sqrt{21}+3\sqrt 7+7\sqrt 3$$
is rational, as well as
$$(N^2-31)^2 =4\cdot 21+9\cdot 7+49\cdot 3+2(42\sqrt 3+42\sqrt 7+21\sqrt{21}).$$
Thus also
$$(N^2-31)^2- (4\cdot 21+9\cdot 7+49\cdot 3)-84N=-42\sqrt{21}$$
is rational.
I guess you can see how this could be similarly applied to all specific sums of square roots ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Polynomial equation for $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}$ I have $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}, k=1,2,3,4$.
I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$
The right answer is $x^4+x^3+x^2+x+1=0$.I tried to replace k with 1,2,3,4 to find the roots and then to use $a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but I didn't get too far.
| Your "high school" approach is valid:
$(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)$$(x-\cos\dfrac{6\pi}5-i\sin\dfrac{6\pi}5)(x-\cos\dfrac{8\pi}5-i\sin\dfrac{8\pi}5)$
$=(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)$
$(x-\cos\dfrac{4\pi}5+i\sin\dfrac{4\pi}5)(x-\cos\dfrac{2\pi}5+i\sin\dfrac{2\pi}5)$
Combine conjugate pairs:
$$(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{2\pi}5+i\sin\dfrac{2\pi}5)$$
$$=x^2+\cos^2\dfrac{2\pi}5+\sin^2\dfrac{2\pi}5-2x\cos\dfrac{2\pi}5=x^2-2x\cos\dfrac{2\pi}5+1$$
and
$$(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)(x-\cos\dfrac{4\pi}5+i\sin\dfrac{4\pi}5)$$
$$=x^2+\cos^2\dfrac{4\pi}5+\sin^2\dfrac{4\pi}5-2x\cos\dfrac{4\pi}5=x^2-2x\cos\dfrac{4\pi}5+1.$$
Now multiply to get $$ (x^2-2x\cos\dfrac{2\pi}5+1)(x^2-2x\cos\dfrac{4\pi}5+1)$$
$$=(x^2+1)^2+4x^2\cos\dfrac{2\pi}5\cos\dfrac{4\pi}5-2x(x^2+1)(\cos\dfrac{2\pi}5+\cos\dfrac{4\pi}5)$$
$$=x^4+2x^2+1-x^2+x(x^2+1)=x^4+x^3+x^2+x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Suppose that $a=\binom70+\binom73+\binom76,b=\binom71+\binom74+\binom77,c=\binom72+\binom75$. How to algebraically compute $a^3+b^3+c^3-3abc$?
$$
\begin{align}
a = {7 \choose 0}+{7 \choose 3}+{7 \choose 6}\\
b = {7 \choose 1}+{7 \choose 4}+{7 \choose 7}\\
c = {7 \choose 2}+{7 \choose 5}
\end{align}
$$
then $a^3+b^3+c^3-3abc$ is equal to _____.
I tried to write $a^3+b^3+c^3-3abc$ in terms of $a+b+c$ and failed.
$$
\begin{align}
a^3+b^3+c^3-3abc & = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
& = (2^7)((a+b+c)^2-3(ab+bc+ca))\\
& = (2^7)((2^7)^2-3(ab+bc+ca))
\end{align}
$$
I think the expression should be written in terms of another binomial series which I can not think of.
| More generally, let $N\in \Bbb N$ and let
$$a=\sum_{0\le 3k\le N}\binom N{3k},$$
$$b=\sum_{0\le 3k+1\le N}\binom N{3k+1},$$
$$c=\sum_{0\le 3k+2\le N}\binom N{3k+2}.$$
(Yours is the case $N=7$.)
Let $\omega=\exp(2\pi i/3)=\frac12(-1+i\sqrt3)$. Then $\omega^3=1$.
Also, by the binomial theorem
$$a+b+c=(1+1)^N=\sum_{j=0}^N\binom Nj=2^N,$$
$$a+b\omega+c\omega^2=\sum_{j=0}^N\binom Nj\omega^j=(1+\omega)^N,$$
$$a+b\omega^2+c\omega=\sum_{j=0}^N\binom Nj\omega^{2j}=(1+\omega^2)^N.$$
Then
\begin{align}
a^3+b^3+c^3-3abc&=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)\\
&=[2(1+\omega)(1+\omega^2)]^N=2^N
\end{align}
since $(1+\omega)(1+\omega^2)=1$.
ADDED IN EDIT
Here is essentially the same argument, but avoiding complex numbers.
Let $A=\pmatrix{0&1&0\\0&0&1\\1&0&0}$. Then $A^2=\pmatrix{0&0&1\\1&0&0\\0&1&0}$ and $A^3=I$ etc. Then
$$(I+A)^N=\sum_{j=0}^N\binom Nj A^j=\pmatrix{a&b&c\\c&a&b\\b&c&a}.$$
As $\det(I+A)=2$, then taking determinants gives
$$2^N=a^3+b^3+c^3-3abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Recursion-like sequences which are hard to relate recursively Consider the sequence
\begin{align*}
a_1&=1\\
a_2&=2+\sqrt1\\
a_3&=3+\sqrt{2+\sqrt1}\\
&\kern5.5pt\vdots\\
a_n&=n+\sqrt{n-1+\sqrt{\cdots+\sqrt{1}}}.
\end{align*}
Something like this is easy to work with inductively, since we can simply relate $a_n = n+\sqrt{a_{n-1}}$, and prove things that way. But now consider something which instead unfolds "on the inside", such as
\begin{align*}
b_1&=1\\
b_2&=1+\sqrt2\\
b_3&=1+\sqrt{2+\sqrt{3}}\\
&\kern5.5pt\vdots\\
b_n&=1+\sqrt{2+\sqrt{\cdots+\sqrt{n}}},
\end{align*}
where it is not easy to relate $b_n$ to $b_{n-1}$. How does one work with such sequences, where we are typically interested in the same sorts of questions? Or another example, consider the sequence
\begin{align*}
c_1&=1\\
c_2&=1(1+2)\\
c_3&=1(1+2(1+3))\\
c_4&=1(1+2(1+3(1+4)))\\
&\kern5.5pt\vdots\\
c_n&=1(1+2(1+3(1+\cdots+(n-1)(1+n)\cdots))).
\end{align*}
Is it possible to prove inductively that $c_n = 1!+2!+\cdots+n!$, even though there is no obvious way to relate $c_n$ to $c_{n-1}$?
| You can add $x$.
Let
$$\begin{align*}
c_1'&=1x\\
c_2'&=1(1+2x)\\
c_3'&=1(1+2(1+3x))\\
c_4'&=1(1+2(1+3(1+4x)))\\
&\kern5.5pt\vdots\\
c_n'&=1(1+2(1+3(1+\cdots+(n-1)(1+nx)\cdots))).
\end{align*}$$
So we know that $c_n'$ is $c_{n-1}'$ put $x$ as $(1+nx)$. Let $c_n'=A_n+B_nx$. So we have $A_n=A_{n-1}+B_{n-1}$ and $B_{n}=nB_{n-1}$. So you may need the recurrance relation like this.
To prove $c_n' = 1!+2!+\cdots+n!$ recursively, we can use induction to prove $A_{n}=1!+2!+\cdots+(n-1)!$ and $B_n=n!$. If this holds for $n-1$, we have $A_n=A_{n-1}+B_{n-1}=(1!+2!+\cdots+(n-2)!)+(n-1)!=1!+2!+\cdots+(n-1)!$, $B_n=nB_{n-1}=n\times (n-1)!=n!$. So $c_n$ is $c'_n$ to make $x=1$, $c_n=A_n+B_n=1!+2!+\cdots+(n-1)!+n!$.
For $b_n$, you can do it similarly (although there is no closed form). You can see $b'_n$ as functions: $b'_n(x)=1+\sqrt{2+\sqrt{\cdots+\sqrt{n+x}}}$, so we have $b'_{n+1}=b_n'(\sqrt{n+1+x})$. However, there may be no yield...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Solving $3$x $6$-sided dice of different colours to find the probability of events An experiment consists of $3$ fair, different coloured dice being rolled. The dice are $6$-sided and the sides show numbers $1,\dots,6$. Let $A$ be the event that none of the dice shows numbers $1$ and $2$, and let $B$ be the event that all dice show an odd number.
A) What is the probability of $A$?
B) What is the probability of $B$?
C) What is the probability of $A$ intersecting $B$?
I've solved this question by finding the total number of possible outcomes:
$|S| = 6^3 = 216$
The results were way too long for the marks given which makes me question the method I used for these solutions.
I ended up with:
A) $P(A) = \frac{64}{216} = \frac{8}{27} $
B) $P(B) = \frac{26}{216} = \frac{13}{108}$
C) $P(A \cap B) = \frac{8}{216} = \frac{1}{27}$
| A) The number of ways $3$ dice can not show a $1$ or $2$ is $4\cdot4\cdot4 = 64$. The total number of outcomes is $6\cdot6\cdot6 = 216$
The probability is therefore $\frac{64}{216} = \frac{8}{27}$.
B) The number of ways $3$ dice can show an odd number is $3\cdot3\cdot3 = 27$. Like before, the total number of outcomes is $216$.
The probability is therefore $\frac{27}{216} = \frac{1}{8}$
C) The number of ways $3$ dice cannot be a 1 or 2 or an even number is $2\cdot2\cdot2 = 8$.
The probability is therefore $\frac{8}{216} = \frac{1}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Conditional probability - sum of dice is even given that at least one is a five Question:
Calculate the conditional probability that the sum of two dice tosses is even given that at least one of the tosses gives a five.
I'm a bit confused by this. Shouldn't the probability just be 1/2, since we know that at least one of the dice tosses gave us a five, thus the other must give us an odd number?
| If the dies were colored, say red and blue, we asked the question, what's chance that sum is even, given that the red die rolls a five, that's $1/2$. But, you see, either red or the blue die could roll a five. So, your sample space is slightly larger.
Fix $X_{i}$ to be the random variable representing the number on die $i$, $i=1,2$. You could draw a nice tree to think of the possible outcomes of this experiment and count them.
(1) When the first die is rolled, there are three possibilities - $X_{1}=5$, $X_{1}=1 \cup X_{1}=3$, $X_{1}\text{ is even}$.
$P(X_{1}=5)=1/6$.
$P(X_{1}=1 \cup X_{1}=3)=1/3$.
$P(X_{1} \text{ is even})=1/2$.
(2) The second die is now rolled, again having three possibilities each - $X_{2}=5$, $X_{2}=1 \cup X_{2}=3$, $X_{2}\text{ is even}$. So, we have $3 \times 3=9$ branches.
$P(X_{1}=5 \cap X_{2}=5)=1/6 \times 1/6=1/36$.
$P(X_{1}=5 \cap (X_{2}=1 \cup X_{2}=3))=1/6 \times 1/3=2/36$.
$P(X_{1}=5 \cap X_{2}\text{ is even})=1/6 \times 1/2=3/36$.
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}=5)=1/3 \times 1/6 = 2/36$.
$P((X_{1}=1 \cup X_{1}=3) \cap (X_{2}=1 \cup X_{2}=3))=1/3 \times 1/3 = 4/36$.
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}\text{ is even})=1/3 \times 1/2 = 6/36$.
$P(X_{1} \text{ is even} \cap X_{2}=5)=1/2 \times 1/6 = 3/36$.
$P(X_{1} \text{ is even} \cap (X_{2}=1 \cup X_{2}=3))=1/2 \times 1/3 = 6/36$.
$P(X_{1} \text{ is even} \cap X_{2}\text{ is even})=1/2 \times 1/2 = 9/36$.
Counting the outcomes where the sum is even and atleast one roll is 5, the numerator $1/36+2/36+2/36=5/36$.
Restricting our attention to events, where atleast one roll is 5, the denominator = $1/36+2/36+3/36+2/36+3/36=11/36$.
So, the chance is $5/11 < 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
How do I show $( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $? How do I show that the left hand side of the below equation is equal to the right hand side?
$$( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $$
| There is
$$( \csc\theta- \cot\theta)^2 =\left( \frac{1}{\sin \theta} - \frac {\cos \theta}{\sin \theta}\right)^2 =\left( \frac{1-\cos \theta}{\sin \theta} \right)^2$$
Since $$ \sin^2 \theta = 1- \cos^2 \theta$$ the expression becomes:
$$( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $$
and the identity is verified
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Points outside a circle mapped inside a circle by $1/z$ Prove or disprove the following conjecture: Let $C$ be a circle in the complex plane that passes by the points $z_1=-1$ and $z_2=a$, with $a$ real and greater that one (there are obviously infinitely many circles with that property). Then every point outside $C$ gets mapped by $1/z$ to a point inside $C$.
Note that the converse is not true. There are points inside $C$ that get mapped by $1/z$ to points also inside $C$. For example, all points in the real axis between $1$ and $a$ are inside $C$ and are mapped to points in the real axis between $0$ and $1$, which are also inside $C$, as the whole $(-1,a)$ interval in the real axis is inside the circle $C$.
For the record, I believe the conjecture to be true.
| Let $C_1$ be the image of $C$ under the Möbius transformation $T(z) = \frac 1z$. Then $C_1$ is a circle, and – since $T$ maps the real axis onto itself – $C$ and $C_1$ intersect the real axis at the same angle at $z=-1$, so that the two circles are tangent to each other at that point.
It follows that $C_1$ is either “inside” $C$ or “outside” $C$. Since $\frac 1a \in C_1$ is in the interior of $C$, only the first option is possible.
Finally, $T$ maps the exterior of $C$ to the interior of $C_1$, which is contained in the interior of $C$.
Alternative solution: Let $c$ be the center of $C$. The condition $a>1$ implies that $\operatorname{Re}(c) > 0$. The radius is $r = |c+1|$ and
$z$ is in the exterior of $C$ iff
$$
|z-c|^2 > |c+1|^2 \\
\iff z \overline z - \overline c z - c \overline z - (c + \overline c + 1) > 0
$$
Then $w = \frac 1z$ satisfies
$$
1 - \overline c \, \overline w - cw - (c + \overline c + 1)w \overline w > 0 \\
\iff w \overline w + \frac{1}{c + \overline c + 1} w + \frac{\overline c}{c + \overline c + 1} \overline w - \frac{1}{c + \overline c + 1} < 0 \\
\iff \left | w + \frac{\overline c}{c + \overline c + 1} \right|^2 < \frac{|c+1|^2}{(c+\overline c + 1)^2}
$$
so that $w$ is inside the circle $C_1$ with center $c_1 = -\frac{\overline c}{c + \overline c + 1}$ and radius $r_1 = \frac{|c+1|}{(c+\overline c + 1)} $ . In particular,
$$
|w - c| \le |w - c_1| + |c_1 - c| < r_1 + |c_1 - c| \\
= \frac{|c+1|}{(c+\overline c + 1)} + \frac{|c+1|(c+\overline c)}{c+\overline c + 1} = |c+1| = r
$$
so that $w$ is inside the circle $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the ratio $p:q:r$, if $p,q,r$ are in H.P, and their squares are in A.P.
If three unequal numbers $p,q,r$ are in H.P and their squares are in A.P, then find the ratio $p:q:r$
.
Attempt
A.P(1): $\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$
$$
\dfrac{1}{q}-\dfrac{1}{p}=\dfrac{1}{r}-\dfrac{1}{q}\implies\dfrac{p+r}{pr}=\dfrac{2}{q}\\
$$
A.P(2): $p^2,q^2,r^2$
$$ q^2-p^2=r^2-q^2\implies p^2+r^2=2q^2\\
$$
As this was asked as a multiple choice question i was wondering what is the easiest way to solve this?
| Start with your
$\dfrac{1}{q}-\dfrac{1}{p}=\dfrac{1}{r}-\dfrac{1}{q}\implies\dfrac{p+r}{pr}=\dfrac{2}{q}\\
q^2-p^2=r^2-q^2\implies p^2+r^2=2q^2\\
$
Then
$q= \dfrac{2pr}{p+r}$.
Putting this to the second equation,
$p^2+r^2
= 2(\dfrac{2pr}{p+r})^2
=\dfrac{8p^2r^2}{(p+r)^2}
$
or
$(p^2+r^2)(p+r)^2
=8p^2r^2
$.
Dividing by $r^4$
and letting
$x = p/r$,
$(1+x^2)(1+x)^2
=8x^2$.
With the help of Wolfy,
this is
$0 = (1-x)^2(x^2+4x+1)
$
so
$x = 1$
or
$x
=\dfrac{-4\pm\sqrt{12}}{2}
=-2\pm \sqrt{3}
$.
If $x=1$ then
$p = r = q$.
If $x = -2+\sqrt{3}$,
$p = r(-2+\sqrt{3})$
so
$\begin{array}\\
q
&=\dfrac{2r^2(-2+\sqrt{3})}{r+r(-2+\sqrt{3})}\\
&=r\dfrac{2(-2+\sqrt{3})}{-1+\sqrt{3}}\\
&=r(1+\sqrt{3})(-2+\sqrt{3})
\qquad\text{since }(-1+\sqrt{3})(1+\sqrt{3})=2\\
&=r(1-\sqrt{3})\\
\end{array}
$
If $x = -2-\sqrt{3}$,
$p = r(-2-\sqrt{3})$
so
$\begin{array}\\
q
&=\dfrac{2r^2(-2-\sqrt{3})}{r+r(-2-\sqrt{3})}\\
&=r\dfrac{2(-2-\sqrt{3})}{-1-\sqrt{3}}\\
&=-r\dfrac{2(-2-\sqrt{3})}{1+\sqrt{3}}\\
&=-r(-1+\sqrt{3})(-2-\sqrt{3})
\qquad\text{since }(-1+\sqrt{3})(1+\sqrt{3})=2\\
&=r(-1+\sqrt{3})(2+\sqrt{3})\\
&=r(1+\sqrt{3})\\
\end{array}
$
Check (with $r = 1$)
with $x = -2+\sqrt{3}$.
$p+r = -1+\sqrt{3}
$
and
$pr =-2+\sqrt{3}
$
so
$2pr =-4+2\sqrt{3}
$
and
$q(p+r)
=(1-\sqrt{3})(-1+\sqrt{3})
=-4+2\sqrt{3}
$
and
$q^2 = 4-2\sqrt{3}$
and
$p^2+r^2
=1+(-2+\sqrt{3})^2
=1+7-4\sqrt{3}
=8-4\sqrt{3}
$.
It also checks for
$x = -2-\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Jordan Decomposition of $\left(\begin{smallmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{smallmatrix}\right)$ over $\mathbb{F}_5$
Find the Jordan decomposition of
$$
A := \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}
\in M_3(\mathbb{F}_5),
$$
where $\mathbb{F}_5$ is the field modulo 5.
What I've done so far
The characteristic polynomial is
\begin{equation}
P_A(t) = (4 - t)(1-t)(3-t) - (1-t)
= -t^3 + 8t^2-18t+1
\equiv 4t^3 + 3t^2 + 2t + 1 \mod5.
\end{equation}
Therefore, $\lambda = 1$ is a zero of $P_A$, since $4+3+2+1 = 10 \equiv 0 \mod 5$.
By polynomial division one obtains
$$
P_A(t)
= (t + 4)(4t^2 + 2t + 4)
= (t + 4)(t + 4) (4t + 1)
\equiv 4 (t + 4)^3
$$
Therefore $\lambda = 1$ is the only eigenvalue of $A$.
To find the eigenspace we calculate the kernel of $A + 4 E_3$ and obtain
$$
\text{span}\left( \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right)
$$
Since $(A + 4 E_3)^2 = 0$, the kernel of $(A + 4 E_3)^2$ is the whole space.
Now, I choose $v := (1, 0, 0) \in \text{ker}(A + 4 E_3)^2$ such that $v \not\in \text{ker}(A + 4 E_3)$.
We calculate $(A + 4E)v = (3, 0, 1)$ and then
$$
(A + 4E)
\begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},
$$
but the zero vector can't be a basis vector of our Jordan decomposition.
Have I made a mistake in my calculations?
| Since $(A-I)(1,0,0)^T=(3,0,1)^T=3(1,0,2)^T$, a desired ordered basis is given by $\{(1,0,2)^T,\frac13(1,0,0)^T,(1,1,2)^T\}=\{(1,0,2)^T,(2,0,0)^T,(1,1,2)^T\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n \to \infty}\left(\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}\right).$ Problem
Let $a_1=3,a_{n+1}=a_n^2+a_n(n=1,2,\cdots)$. Evaluate
$$\lim_{n \to \infty}\left(\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}\right).$$
Attempt
Notice $$\frac{1}{1+a_n}=\frac{a_n}{a_n+a_n^2}=\frac{a_n}{a_{n+1}}$$
Then
$$\sum_{k=1}^{n}\frac{1}{1+a_k}=\sum_{k=1}^n \frac{a_k}{a_{k+1}}.$$
This will help?
| Prove that
$$
\sum_{k=1}^{n} \frac{1}{1 + a_{k}} = \frac{1}{3} - \frac{1}{a_{n+1}}
$$
by induction on $n$. Then the limit is $1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove that $5x^2−2xy−8x+ 2y^2−2y+ 5 \ge 0$ for all $x, y\in\mathbb R$. When does equality occur? I tried grouping the $x$'s and the $y$'s but that didn't get me anywhere. I know that $5x^2, 2y^2$, and $5$ are always positive. I am not sure what to try next.
| By inspection, equality occurs for $x=y=1$.
So the obvious choice is to group around these values. This also gives a somewhat more detailed explanation how you arrive at these magic squares:
$$
5x^2−2xy−8x+ 2y^2−2y+ 5 = \\
5((x-1) +1)^2−2((x-1)+1)((y-1) + 1)−8(x-1)+ 2((y-1)+1)^2−2(y-1) -5\\
= 5(x-1)^2 - 2(x-1)(y-1) +2(y-1)^2 =\\
= 4(x-1)^2 + ((x-1)-(y-1))^2 +(y-1)^2 = \\
= 4(x-1)^2 + (x-y)^2 +(y-1)^2 \\
\geq 0
$$
which is clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| Note that A,B,C are are positive acute angles, so both LHS and RHS are positive. Square both LHS and RHS, Then do LHS - RHS we get.
$\cos2A + \cos2B + \cos2C + 2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ note that $2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ is always -ve for acute triangle (all these angles fall in second quadrant).
Check $\cos2A + \cos2B + \cos2C$, If all are greater than 45 deg. then this becomes negative. Suppose $A= 45 -\theta$ (in deg.), writing
$\cos2(45 -\theta) + \cos2(45 +\theta + a) + \cos2(45 +\theta + b)$ where $0\leq \theta < 45$ , $0< a + \theta < 45$, $0< b + \theta < 45$
This part becomes
$\sin2\theta - \sin(2a+2\theta) - \sin(2b+2\theta)$, all these angles fall in first quadrant, so this is also negative. So
$LHS < RHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Solve the following system of equations - (7).
Solve the following system of equations (over the reals). $$\large \left\{ \begin{aligned} (x + y)^2 &= xy + 3y - 1\\ x + y &= \frac{x^2 + y + 1}{x^2 + 1}\end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{aligned} (x + y)^2 - 1 &= y(x + 3)\\ x + y - 1&= \frac{y}{x^2 + 1}\end{aligned} \right. \implies \frac{y}{x^2 + 1}\cdot (x + y + 1) = y(x + 3)$$
$$\left[ \begin{aligned} y &= 0\\ x + y + 1 &= (x^2 + 1)(x + 3) \end{aligned} \right.$$
Plugging in $y = 0$ into the first equation, we have that $x^2 = -1$, which is incorrect for $\forall x \in \mathbb R$.
Then $x + y + 1 = (x^2 + 1)(x + 3) \implies y = x^3 + 3x^2 + 2$.
And I am done with my life.
| Direct substitution method!
From the second:
$$x + y = \frac{x^2 + y + 1}{x^2 + 1} \Rightarrow y=\frac{(1-x)(x^2+1)}{x^2}$$
Subbing to the first:
$$\left(\frac{x^2-x+1}{x^2}\right)^2=\frac{(x+3)(1-x)(x^2+1)}{x^2}-1 \Rightarrow \\
x^6+2x^5-2x+1=0 \Rightarrow \\
(x^2+1)(x^2+x-1)^2=0 \Rightarrow \\
x^2+x-1=0 \Rightarrow \\
x=\frac{-1\pm \sqrt{5}}{2}.$$
Note: $x^6+2x^5-2x+1=0 \Rightarrow (x^2)^3+1+2x(x^4-1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$
Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$
I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain
$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$
I am stuck here. Any hints on solving the R.H.S will be appreciated.
| $$\frac{\sqrt{3} +1}{2\sqrt{2}}=\frac{\sqrt{3}\sqrt{2} +\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\sin{45^\circ}\cos{30^\circ}+\cos{45^\circ}\sin{30^\circ}=\sin{(45^\circ+30^\circ)}=\sin{75^\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\sum\limits_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}{n\choose k}(x^k-1)$ In my answer here, I reduce the problem of evaluating
$$J=\int_0^{\pi/6}\frac{x\cos x}{1+2\cos x}dx$$
to the evaluation of $S(8-4\sqrt3)$, were
$$S(q)=\sum_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}\sum_{k=1}^{n}\frac{(-1)^k}{k}{n\choose k}(q^k-1)\qquad q>0.$$
I am now interested in the evaluation of $S(q)$ in terms of special functions.
I have not gotten very far with this series. I have shown that
$$\sum_{k=1}^{n}\frac{(-1)^k}{k}{n\choose k}(q^k-1)=\int_1^q\frac{(1-x)^n-1}{x}dx$$
and consequently that
$$S(q)=\int_0^{1-q}\frac1{1-x}\left[\frac{\tan^{-1}\sqrt{x}}{\sqrt{x}}-\frac{\pi}{2\sqrt3}\right]dx$$
Which 'reduces' to
$$S(q)=\frac{\pi}{2\sqrt3}\ln q+2\int_0^{\sqrt{1-q}}\frac{\tan^{-1}x}{1-x^2}dx$$
But I do not know how to proceed. Could I have some help? Thanks.
Edit:
I've found out I messed up a little earlier on. I found that for $0\leq q\leq 1$ we actually have that
$$S(q)=-\frac{\pi}{2\sqrt3}\ln q+\sqrt{3}\int_1^q \frac{\tan^{-1}\left[\sqrt{\frac{1-t}{3}}\right]}{t\sqrt{1-t}}dt.$$
It is really quite similar to the previous integral.
| New Integral. First note that it's not good to have $\sqrt {1-x}$ when the integrating region is greater than $1$.
$$f=\int_1^{8-4\sqrt 3} \frac{\arctan\left(\sqrt{\frac{1-t}{3}}\right)}{t\sqrt{1-t}}dt=\int_1^{8-4\sqrt 3} \frac{\text{arctanh}\left(\sqrt{\frac{t-1}{3}}\right)}{t\sqrt{t-1}}dt$$
Change the variable $\frac{t-1}{3}=u^2$ and then $\frac{1-u}{1+u}=x$
$$f=\sqrt 3 \int_0^{\frac{2}{\sqrt 3}-1} \frac{\ln\left(\frac{1+u}{1-u}\right)}{1+3u^2}du=-\frac{\sqrt 3}{2}\int_{\sqrt 3-1}^1\frac{\ln x}{x^2-x+1}dx$$
The latter integral can be expressed in terms of Dilogarithms with $\phi =\frac{\sqrt 5+1}{2}$.
$$\int_{\sqrt 3-1}^1\frac{\ln x}{x^2-x+1}dx=\frac{1}{\sqrt 5}\left(\int_{\sqrt 3-1}^1 \frac{\ln x}{x-\phi}dx-\int_{\sqrt 3-1}^1 \frac{\ln x}{x+\phi^{-1}}dx\right)$$
Previous Integral. Integrate by parts first.$$2\int_0^{\sqrt{1-q}}\frac{\tan^{-1}x}{1-x^2}dx=\arctan(\sqrt{1-q})\ln\frac{1+\sqrt{1-q}}{1-\sqrt{1-q}}+\int_0^\sqrt{1-q}\frac{\ln\frac{1-x}{1+x}}{1+x^2}dx$$
For the last integral let $\frac{1-x}{1+x}=u$ followed by $u=\tan t$, also set $k=\frac{1-\sqrt{1-q}}{1+\sqrt{1-q}}$.
$$\int_0^\sqrt{1-q}\frac{\ln\frac{1-x}{1+x}}{1+x^2}dx=\int_k^1 \frac{\ln u}{1+u^2}du=\int_0^\frac{\pi}{4}\ln(\tan t) -\int_0^{\arctan k}\ln (\tan t)dt$$
$$=-C+\frac12 \text{Cl}_2(\pi-\arctan k)+\frac12 \text{Cl}_2(\pi-2\arctan k)$$
$C$ is Catalan's constant and there is https://en.wikipedia.org/wiki/Clausen_function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof technique regarding divisibility by $n^3$ I read in a note that-
Let, $z \geq3, n$ is even, and
$$(18n^2 + 1)^x + (7n^2 − 1)^y = (5n)^z$$ By taking modulo $n^3$
implies that
$$1 + 18n^2x − 1 + 7n^2y ≡ 0 \pmod{ n^{3}}$$
I couldn't get how $1 + 18n^2x − 1 + 7n^2y ≡ 0 \pmod{ n^{3}}$ is derived, how it can be proved?
| This is just use of the binomial theorem: $$(a+b)^x={x\choose 0}a^x+{x\choose 1}a^{x-1}b+{x\choose 2}a^{x-2}b^2...+ {x\choose x-1}b^{x-1}a +{x\choose x}b^x$$
where $a=1$ and $b = 18n^2$
$$(1+18n^2)^x=1+ x18n^2+{x(x-1)\over 2}18^2n^4+...+18^xn^{2x}$$
$$=1+ x18n^2+n^3\Big({x(x-1)\over 2}18^2n+...+18^xn^{2x-3}\Big)$$
$$ \equiv 1+x18n^2 \pmod{n^3}$$
and similary for the second bracket.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Check if $\frac{x^3+y^3}{|x|+|y|}$ is differentiable at $(0,0)$ We have the function $$f(x,y) = \begin{cases} \frac{x^3+y^3}{|x|+|y|},& (x, y) \ne (0,0) \\
0, & (x,y) = (0,0)
\end{cases}$$
My work so far:
I tried checking if $f$ is continuous at $(0,0)$:
When $x >0, y>0:$
$$\lim\limits_{(x,y)\rightarrow(0,0)} f(x) = \frac{(x+y)(x^2-xy+y^2)}{x+y} = 0$$
When $x >0, y<0:$ we can rewrite $y$ as $-y$, with $y>0$
$\lim\limits_{(x,y)\rightarrow(0,0)} f(x) = \frac{x^3-y^3}{x+y}$
Which I can not solve.
I calculated the partial derivatives:
\begin{align*}\frac{\partial{f}}{\partial{x}}(0,0)&=\lim\limits_{x\to 0} \frac{f(x,0)-f(0,0)}{x} =\lim\limits_{x\to 0}\frac{x^3}{|x|}=\lim\limits_{x\to 0} x^2 \mathrm{sgn}(x)= 0\\
\frac{\partial{f}}{\partial{x}}(x,y)&=\frac{3x^2(|x|+|y|)-(x^3+y^3)\mathrm{sgn}(x)}{(|x|+|y|)^2}
\end{align*}
and analogously for $y$.
I tried checking if the partial derivatives are continuous and failed.
I tried with the definition:
$$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{f(x,y)-f(0,0)-T(0,0)}{||(x,y)||} = 0,$$
where $T(x,y) = 0$.
we get $$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{(|x|+|y|)\sqrt{x^2+y^2}}$$
And I got stuck here.
| Yes, the partial derivatives exist at $(0,0)$ and they are both equal to $0$. Therefore either $f$ is not differentiable at $(0,0)$, or it is differentibale there and its derivative there is the null function.
And what actually occurs is that $f'(0,0)$ is the null function. In fact, if $\varphi\colon\mathbb R^2\longrightarrow\mathbb R$ is the null function, then\begin{align}f'(0,0)=\varphi&\iff\lim_{(x,y)\to(0,0)}\frac{\bigl\lvert f(x,y)-f(0,0)-\varphi(x-0,y-0)\bigr\rvert}{\bigl\lVert(x,y)\bigr\rVert}=0\\&\iff\lim_{(x,y)\to(0,0)}\frac{\lvert x^3+y^3\rvert}{\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}}=0.\end{align}But, if $x=r\cos\theta$ and $y=r\sin\theta$, then$$\lvert x^3+y^3\rvert\leqslant2r^3$$and$$\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}\geqslant r^2,$$since $\lvert\cos\theta\rvert+\lvert\sin\theta\rvert\geqslant\cos^2\theta+\sin^2\theta=1$. Therefore,$$\frac{\lvert x^3+y^3\rvert}{\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}}\leqslant r\to_{r\to0}0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two challenging sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}$ where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$
again, my goal of posting these two challenging sums is to use them as a reference.
I will provide my solutions soon.
I would like to mention that these two sums can also be found in Cornel's book " almost impossible integrals, sums, and series".
| Using the fact that $\displaystyle \sum_{n=1}^\infty x^nH_n^{(2)}=\frac{\operatorname{Li}_2(x)}{1-x}$
Replace $x$ with $-x$ then multiply both sides by $\ln^2x$ and integrate, we get
\begin{align}
S&=\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1x^{n}\ln^2x\ dx=2\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\underbrace{\int_0^1\frac{\ln^2x\operatorname{Li}_2(-x)}{1+x}\ dx}_{IBP}\\
&=\int_0^1\frac{\ln^2x \ln^2(1+x)}{x}\ dx-2\int_0^1\frac{\ln x\ln(1+x)\operatorname{Li}_2(-x)}{x}\ dx\\
&=I_1-2I_2
\end{align}
Lets evaluate the first integral and using $\quad \ln^2(1+x)=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n,\quad $ we get
\begin{align}
I_1&=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx\\
&=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{2}{n^3}\right)\\
&=4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-4\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\\
&=4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}{4}\zeta(5)
\end{align}
to evaluate the second integral, apply IBP , we get
\begin{align}
I_2&=\left.-\frac12\operatorname{Li}_2^2(-x)\ln x\right|_0^1+\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\
&=\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\
\end{align}
I proved here $\quad \displaystyle \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx=\frac58\zeta(2\zeta(3)+\frac78\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$
Collecting these two integrals and using $\quad \displaystyle \sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3),\quad$ we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\frac9{16}\zeta(5)+\frac18\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$
but $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}-\frac{15}{16}\zeta(5)$$
Thus $$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}=\frac32\zeta(5)+\frac18\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$
Plugging $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ gives
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}=\frac58\zeta(2)\zeta(3)-\frac{11}{32}\zeta(5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
| The substitution
$$x=-\left( 3+\frac{2}{u} \right)$$
reduces the integral to
$$\begin{align}
& =\frac{1}{8}\int{\frac{{{u}^{4}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du} \\
& =\frac{1}{8}\int{u-5+\frac{17{{u}^{2}}+36u+20}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du} \\
& =\frac{1}{8}\int{u-5+\frac{1}{u+1}+\frac{16}{u+2}-\frac{16}{{{\left( u+2 \right)}^{2}}}du} \\
\end{align}$$
and you can see that partial fraction decomposition becomes much easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find all positive integers x,y satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$ Find all positive integers $x$,$y$ satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$
$$ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}\\
\frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{2\sqrt{5}}$$
By hit and trial I found one pair value of $x$ & $y$ i.e $(80, 80)$
But is there any other way to solve this tricky question and find all possible value of $x$ and $y$.
| By symmetry, we can assume $x \ge y$ and then duplicate the solutions. A simple way to to note that $20 \lt x \le 80$ to make the sum large enough and not too large. That is only $60$ numbers to try. A spreadsheet with copy down will make it easy, finding the pair (45,180).
Another way to find it is to note $\frac 1{\sqrt{20}}=\frac 1{2\sqrt 5}$ and remember that $\frac 13+\frac 16=\frac 12$, again giving $(45,180)$ and $\frac 14+\frac 14=\frac 12$ giving $(80,80)$
The algebraic approach is a mess.
$$\frac 1{\sqrt x}+\frac 1{\sqrt y}=\frac 1{\sqrt {20}}\\
\frac 1{\sqrt x}=\frac 1{\sqrt {20}}-\frac 1{\sqrt y}\\
\frac 1x=\frac 1{20}-\frac 2{\sqrt{20y}}+\frac 1y\\
\frac 1x-\frac 1y-\frac 1{20}=\frac 1{\sqrt{20y}}\\
\frac 1{x^2}+\frac 1{y^2}+\frac 1{400}-\frac 2{xy}-\frac 1{10x}+\frac 1{10y}=\frac 1{20y}\\
400y^2+400x^2+x^2y^2-800xy-40xy^2+40x^2y=20x^2y$$
but at least we can note that $x^2y^2$ must have a factor of $20$, so either one has a factor $10$ or one has a factor $2$ and the other a factor $5$. We can view this as a quadratic in one of the variables and feed it to the quadratic equation, but that will lead to testing cases in a more complicated way.
| {
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"url": "https://math.stackexchange.com/questions/3257026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Tough substitution inequality Prove that if $x, y, z >0$ and $xyz=x+y+z+2$, then
$$
\sqrt{x}+\sqrt{y}+\sqrt{z} \leq \frac{3}{2}\sqrt{xyz}.
$$
By the way, the first equation implies the existence of positive $a, b, c$ such that $x=\frac{b+c}{a}, y=\frac{c+a}{b}, z=\frac{a+b}{c}$.
| Yes, your idea works.
Let $x=\frac{b+c}{a}$ and $y=\frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{a+b}{c}$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{b+c}{a}}\leq\frac{3}{2}\sqrt{\frac{\prod\limits_{cyc}(a+b)}{abc}}$$ or
$$\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\frac{3}{2},$$ which is true by AM-GM:
$$\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+c}+\frac{c}{c+a}\right)=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this proof of the $\lim_{(x,y)\to (0,0))} \frac{x^6y}{x^8+y^4}$ correct? Let $\varepsilon>0$ be arbitrary. Note that $||(x,y)-(0,0)||<\delta \implies \sqrt{x^2+y^2}<\delta$, which in turn yields $|x|<\delta$ and $|y|<\delta$.
Now, let $\delta=\varepsilon$ and assume that $||(x,y)-(0,0)||<\delta$. From what we just proved, $|x|<\varepsilon$ and $|y|<\varepsilon$. Finally, we have:
$$\bigg|\frac{x^6y}{x^8+y^4}\bigg|=\frac{|x|^6|y|}{|x|^8+|y|^4}<\frac{\varepsilon^7}{\varepsilon^8+\varepsilon^4}<\frac{\varepsilon^7}{\varepsilon^4}<\varepsilon^3$$
Since $\varepsilon^3\to 0$ as $\varepsilon\to0$, the limit is $0$.
| NO. $(|x|^6|y|<e^7$ and $|x|^8+|y|^4<e^8+e^4)$ does NOT imply that $\frac {|x|^6|y|}{|x|^8+|y|^4}<\frac {e^7}{e^8+e^4}$ for the same reason that $(2<3$ and $1<8)$ does not imply that $\frac {2}{1}<\frac {3}{8}.$
There is no limit. If $0\ne y=x^2$ then $\frac {x^6y}{x^8+y^4}=\frac {1}{2}$ but if $x=0\ne y$ then $\frac {x^6y}{x^8+y^4}=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove limit doesn't exist $\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$ I need to show that limit doesn't exist:
$\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
How can I show it?
| First, not that for $xy\ne 0$, we have
$$\begin{align}
\frac{1-\cos(x^2+y^2)}{x^2y^2(x^2+y^2)}&=\frac{2\sin^2(x^2+y^2)}{x^2y^2(x^2+y^2)}\\\\
&=2\left(\frac{\sin(x^2+y^2)}{x^2+y^2}\right)^2\times \left(\frac{x^2+y^2}{x^2y^2}\right)\tag1
\end{align}$$
The first term on the right-hand side of $(1)$, $\left(\frac{\sin(x^2+y^2)}{x^2+y^2}\right)^2$, tends to $1$ as $(x,y)\to(0,0)$ However, the second term, $\frac{x^2+y^2}{x^2y^2}$ tends to $+\infty$ since
$$\frac{x^2+y^2}{x^2+y^2}\ge \frac{1}{x^2}\to \infty$$
Therefore, we see that
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{x^2y^2(x^2+y^2)}=\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the probability that the circumferecnce of triangle CDE is smaller than $1+ \sqrt \frac{11}{2}$. Let page of square ABCD have length 1. On the page AB, point E is randomly selected. Calculate the probability that the circumferecnce of triangle CDE is smaller than $1+ \sqrt \frac{11}{2}$.
I have used Pythagorean theorem for pages ED and EC, and then observed case where circumference is equal to $1+ \sqrt \frac{11}{2}$, but I got to a polynomial equation of $4$th degree ( $64x^4 - 128x^3 -112x^2-16x-175=0$, where $x$ is distance between A and E) which doesn't have "easy" readable solution. Is there any better way to solve this?
| If you denote the distance $AE=t$ it follows that $EB=1-t$. Moreover the question can be reformulate as $CE+DE<\sqrt{\frac{11}{2}}$. $CE=\sqrt{1+t^2}$ and $DE=\sqrt{1+(1-t)^2}=\sqrt{2-2t+t^2}.$
Consider equality:
$$\sqrt{2-2t+t^2}+\sqrt{1+t^2}=\sqrt{\frac{11}{2}}$$
Then, we take the square:
$$2-2t+t^2+1+t^2+2\sqrt{2-2t+t^2}\sqrt{1+t^2}=\frac{11}{2}$$
$$2\sqrt{2-2t+t^2}\sqrt{1+t^2}=-2t^2+2t+\frac{5}{2}$$
$$4(2-2t+t^2)(1+t^2)=4t^4+4t^2+\frac{25}{4}-8t^3+10t-10t^2$$
$$4t^4-8t^3+8t^2+8-8t+4t^2=4t^4+4t^2+\frac{25}{4}-8t^3+10t-10t^2$$
$$18t^2-18t+\frac{7}{4}=0$$
The equation has two symmetric solution $t_1=\frac{1}{2}-\frac{\sqrt{\frac{11}{2}}}{6}$ and $t_2=\frac{1}{2}+\frac{\sqrt{\frac{11}{2}}}{6}$. The set of point that satisfy the request has Lebesgue measure $t_2-t_1=\frac{\sqrt{\frac{11}{2}}}{3}$
If we suppose a uniform probability, the answer is $\frac{\sqrt{\frac{11}{2}}}{3}$.
| {
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"url": "https://math.stackexchange.com/questions/3259773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$ . Prove
$$x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
We have
$$(\,x+ y+ z\,)^{\,2}+ (\,-\,x+ y+ z\,)^{\,2}+ (\,x- y+ z\,)^{\,2}+ (\,x+ y- z\,)^{\,2}= 4(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= 36$$
Or
$$\left ( z+ (\,y- x\,) \right )^{\,2}+ \left ( z- (\,y- x\,) \right )^{\,2}+ (\,3- 2\,z\,)^{\,2}= 27$$
Or
$$3\,z^{\,2}- 6\,z+ (\,y- x\,)^{\,2}= 9$$
Or
$$(\,y- x\,)^{\,2}= -\,3(\,z- 1\,)^{\,2}+ 12\leqq 12\,\therefore\,y- x\leqq |\,y- x\,|\leqq 2\sqrt{3}$$
Q.E.D. The equality condition occurs when $z= 1\,\therefore\,x+ y= 2\,\therefore\,x= 1- \sqrt{3},\,y= 1+ \sqrt{3}$.
Say it (Added)
The @user10354138's solution is so amazing, I try writing the inequality into the homonogeous form, then find $t\!=\!constant$ such that $3(\!y- z\!)^{\!2}\leqq 2t(\!x+ y+ z\!)^{\!2}+ 2(\!1- t\!)(\!x^{\!2}+ y^{\!2}+ z^{\!2}\!)$. That will lead to:
$${\rm discriminant}= 0\,\therefore\,t= -\,2,\,-\,\frac{1}{2},\,1$$
The coefficients of $y^{2}$ and $z^{2}$ both are negative there, I can't make the form like the solution as follow !
| Let $y-x=d$
$z=3-x-y=3-x-(x+d)=3-2x-d$
$9=x^2+(x+d)^2+(3-d-2x)^2$
Rearrange to form a quadratic equation in $x$
As $x$ is real, the discriminant must be $\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of integral with different roots I am supposed to calculate integral:$$\int \frac{4x+5\sqrt{x+2}}{\sqrt[3]{\left ( x+2 \right )^{2}}}dx$$ but I do not know, how to use the substitution.
Can anyone help me?
| Hint. One may write
\begin{align}
\frac{4x+5\sqrt{x+2}}{\sqrt[3]{\left ( x+2 \right )^{2}}}&=\frac{4x+8-8}{\sqrt[3]{\left ( x+2 \right )^{2}}}+\frac{5\sqrt{x+2}}{\sqrt[3]{\left ( x+2 \right )^{2}}}
\\\\&=(x+2)^{1/3}-\frac{8}{(x+2)^{2/3}}+\frac{5}{(x+2)^{1/6}}
\end{align} then just perform the change of variable $t=x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\cos (\theta)$ and $\sin (\theta)$, find $2\theta$ I am working on my scholarship exam. I worked through almost final step but got my answer wrong. Could you please have a look?
If $\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ with
$0\leq\theta<2\pi$, it follows that $2\theta = ..... \pi$
What I have got is below:
$\sin(2\theta)=2\sin\theta\cos\theta$
Then, $\sin(2\theta)=-\frac{1}{\sqrt{2}}$
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$ (quadrant 3 or 4)
$\theta=\frac{5\pi}{8}$ or $\frac{7\pi}{8}$
Since $\cos\theta$ is positive and $\sin\theta$ is negative, $\theta$ should be in quadrant 4 but my $\theta$'s are not. So I cannot use my $2\theta$ as a final answer.
However, the answer key provided is $\theta=\frac{15\pi}{4}$, Why do you think that is the case? How did they get to this answer? Please help.
| If $0\le \theta<2\pi$, then $0\le 2\theta<4\pi$. You should have $4$ possible values of $\theta$.
$\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ imply that $\sin(2\theta)=-\dfrac1{\sqrt2}$, but not the other way round.
Note that $\displaystyle \tan\theta=\frac{-\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}}{\sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}}=-\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=1-\sqrt{2}$.
So $\theta=n\pi-\dfrac{\pi}{8}$, ($n\in\mathbb{Z}$).
[Note: $\frac{\tan(-\frac\pi8)}{1-\tan^2(-\frac\pi8)}=\tan2(-\frac\pi8)=-1$ implies that $\tan(-\frac\pi8)=1-\sqrt2$.]
As $\theta\in[0,2\pi)$, $\cos\theta>0$ and $\sin\theta<0$, we have $\theta=2\pi-\dfrac{\pi}{8}$ and hence $2\theta=\dfrac{15\pi}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Systems of equations involving linear and quadratic terms Can we solve for $y$ in this system using algebra?
$$\left\{
\begin{aligned}
x^2 - yz &= 3 \\
y^2 - xz &= 4 \\
z^2 - xy &= 5
\end{aligned}
\right.$$
I’ve tried to evaluate it using elimination and it just gives another equation with unknowns.
First I've tried to multiply the first equation by $y$, second by $z$ and third by $x$. I get $x^2 - y^2z = 3y, y^2z - xz^2 = 4z,$ and $z^2x - x^2y=5x$. Simplifying I get $5x + 4z + 3y = 0$. I've tried it again by multiplying the 1st and 3rd equation by $z, x$ and $y$ respectively. I get $5y + 4x + 3z = 0$. I don't know where to get my third equation.
| \begin{equation}
x^2-yz=3\hspace{2cm}(1)\\
y^2-xz=4\hspace{2cm}(2)\\
z^2-xy=5\hspace{2cm}(3)
\end{equation}
$(2)-(1)\implies$
$$(y-x)\cdot(x+y+z)=1\hspace{2cm}(4)$$
$(3)-(2)\implies$
$$(z-y)\cdot(x+y+z)=1\hspace{2cm}(5)$$
$(5)-(4)\implies$
$$(x+y+z)\cdot(2y-x-z)=0\hspace{2cm}(6)$$
$\implies$
$$x=-(y+z)$$ or $$x=(2y-z)$$
Now suppose $(x+y+z)=0$,Then
$(1)\implies$ $$(y+z)^2-yz=3
\implies y^2+z^2+yz=3$$
$(2)+(3)\implies$
$$y^2+z^2+(y+z)(y+z)=9$$
$\implies$
$$2(y^2+z^2+yz)=9 \implies 2\cdot 3=9 \hspace{2cm}\Rightarrow\Leftarrow
$$
So $x=2y-z$
$(2)\implies$
\begin{align}
y^2-(2y-z)z=4 \implies (y-z)^2=4\implies y=z\pm 2\hspace{2cm}(7)\\
\end{align}
Also since $z=2y-x$
\begin{align}
y^2-(2y-x)x=4 \implies (x-y)^2=4 \implies x=y\pm 2\hspace{2cm}(8)\\
\end{align}
If $x=z$, then
$(1)\implies$ $$x^2-xy=3$$
$(3)\implies$ $$x^2-xy=5 \hspace{2cm} \Rightarrow\Leftarrow $$
Therefore $x\neq z$ and the possible combinations are $(x,x+2,x+4)$ and $(x,x-2,x-4)$[from $(7)\&(8)$]
Assuming $y=z+2$ and solving $(3)$, we get $$x=\frac{11}{6}, y=\frac{-1}{6}, z=\frac{-13}{6}$$
Assuming $y=z-2$ and solving $(3)$, we get $$x=\frac{-11}{6}, y=\frac{1}{6}, z=\frac{13}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Minimize value of the function $a^2+b^2+c^2+2\sqrt{3abc}$ Let $a,b,c$ be the positive real numbers such that $a+b+c=1$. Find Minimize of $$P=a^2+b^2+c^2+2\sqrt{3abc}$$
WA says that $P$ gets only a local minimum. But i think it must be maximum value of $P$.
Then by AM-GM: $$\text{L.H.S}= a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}$$
$$\le a^2+b^2+c^2+2(ab+bc+ca)$$
$$=(a+b+c)^2=1$$
| For $c\rightarrow0^+$ and $a=b=\frac{1}{2}$ we obtain a value $\frac{1}{2}.$
We'll prove that it's an infimum.
Indeed, we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$9u^2-6v^2+6\sqrt{uw^3}\geq\frac{9}{2}u^2$$ and we see that it's enough to prove out inequality for a minimal value of $w^3$,
which happens in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
In this case we need to prove that
$$a^2+b^2\geq\frac{1}{2}(a+b)^2$$, which is true by C-S:
$$a^2+b^2=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2;$$
2. Two variables are equal.
Since the inequality $a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2$ is homogeneous and symmetric, we can assume $b=c=1$ and we need to prove that
$$a^2+2+2\sqrt{3a(a+2)}\geq\frac{1}{2}(a+2)^2$$ or
$$4\sqrt{3(a+2)}\geq\sqrt a(4-a),$$ which is obvious for $a>4$, but for $0<a\leq4$ we need to prove that
$$48(a+2)^2\geq a(4-a)^2$$ or $$a^3-8a^2-32a-96\leq0$$ or
$$a^3-64-(8a^2+32a+32)\leq0,$$ which is obviously true.
Done!
The maximal value is indeed, $1$.
For which we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\leq(a+b+c)^2$$ or
$$ab+ac+bc\geq\sqrt{3abc(a+b+c)}$$ or after squaring of the both sides
$$\sum_{cyc}c^2(a-b)^2\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$x^2-7x+m=0$, $x_1^2+4 x_2^2=68$, $m$=? $x_1, x_2$ are solutions to the equation and are whole numbers
$x^2-7x+m$
I dont see how to represent $x_1^2+4 x_2^2=68$ only with $x_1 + x_2$ and $x_1 x_2$.I could try to use the fact that $x_1, x_2$ are whole numbers and try to guess $m$
$$x_{1,2} = \dfrac{7\pm\sqrt{49-4m}}{2}$$
But i think that is not the proper way to solve this
| 1)
If the two solutions to $ax^2 + bx + c = 0$ are $\frac {-b\pm \sqrt{b^2 - 4ac}}{2a}$ then the sum of these solutions is $\frac {-b+ \sqrt{b^2 - 4ac}}{2a}+\frac {-b- \sqrt{b^2 - 4ac}}{2a} =-\frac ba$.
And the product is $(\frac {-b+\sqrt{b^2 - 4ac}}{2a})(\frac {-b-\sqrt{b^2 - 4ac}}{2a}) = \frac {b^2 - (b^2 -4ac)}{4a^2} = \frac ca$.
So if $x_1, x_2$ are the solutions to $x^2 -7x + m = 0$ then $x_1 + x_2 = 7$ and $x_1x_2 = m$.
2)
Forget all that.
If $x_1$ and $x_2$ are whole numbers there are only finitely many pairs where $x_1^2 + 4x_2^2= 68$.
$4x_2^2 \le x_1^2 + 4x_2^2 = 68$ so $x_2^2 \le 17$ so $|x_2| \le 4$.
Just test all $|x_2|$ from $0$ to $4$.
Or if you feel clever reason that:
$4x_2^2$ and $68$ are even so $x_1^2$ is even so $x_1$ is even.
If we let $x_1=2c$ then $4c^2 + 4x_2^2 = 68$ or $c^2 + x_2^2 = 17$.
Trial and error give us $0 + x_2^2 = 17$ is impossible $1 + 4^2 = 17$ is possible. $4 + x_2^2 = 17$ is impossible. $9 + x_2^2 =17$ is impossible and $4 + 1^2 = 17$ are possible and nothing else is.
So $(c, x_2) = \{(\pm 1, \pm 4), (\pm 4, \pm 1)\}$ and
$(x_1,x_2) = (2c,x_2) = \{(\pm 2,\pm 4)(\pm 8, \pm 1)\}$ are the only possible answers.
.....
Now back to 1)
But if $x_1 + x_2 = 7$ the only possible options are $x_1 = 8$ and $x_2 =-1$.
So $x_1x_2 = -8$.
And so as $m= x_1x_2 = -8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find the remainder of the division of $a$ by $18$ knowing that $\gcd(a^{226} +4a +1, 54)=3$ Let's define $b:= a^{226} +4a +1$. We know that $b$ is odd because $54$ is even and the gcd is odd. But if $a$ were odd, $b$ would be even; so $a$ is also even.
We also know that $3\nmid a$ (since $3\nmid b$, if it did divide $a$, it would be a divisor of $1$, which is absurd.) Applying Fermat's theorem, we have
$$a^{226} + 4a + 1 = (a^2)^{113} + 4a + 1 \equiv 4a + 2 \equiv 0 \pmod 3$$
This means that $a \equiv 1 \pmod 3$. We infer the following congruences:
$$a \equiv 0 \pmod2 \\ a \equiv 1 \pmod 3$$
If I had a $\pmod 9$ instead of a $\pmod 3$ in the last congruence, I'd be able to aply the Chinese Remainder Theorem.
How can I bound the values of $r_{9}(a)$, given that $r_{3}(a)=1$?
| You already got $a\equiv 1\bmod 3$, which means $a\equiv 1,4\hbox{ or }7\bmod 9$. We would like to check which of them is the one that works.
Now, you can look $\mod 9$ by using Euler's theorem. Since $\gcd(a,9)=1$ you can apply Euler's theorem. We have $\varphi(9)=6$, so $$a^{226}+4a+1=(a^6)^{38}a^{-2}+4a+1\equiv \overline{a}^2+4a+1\mod 9$$
where $\overline{a}$ denotes the inverse of $a\bmod 9$.
Now let's check:
If $a\equiv 1\bmod 9$, then $\overline{a}^2+4a+1\equiv 6\mod 9$
If $a\equiv 4\bmod 9$, then $\overline{a}^2+4a+1\equiv 12\mod 9$
If $a\equiv 7\bmod 9$, then $\overline{a}^2+4a+1\equiv 0 \mod 9$
As you can see, if $a\equiv 7\bmod 9$, we have $a^{226}+4a+1$ is divisible by $9$ hence in this case $\gcd(54,a^{226}+4a+1)$ is divisible by $9$. This case is discarded.
We conclude that
$$a\equiv 0\bmod 2$$
$$a\equiv 1 \hbox{ or }4\bmod 9$$
By chinese remainder theorem we conclude
$$a\equiv 10 \hbox{ or }4\bmod 18$$
Edit: @Bill Dubuque showed an easier way to compute $a^{226}+4a+1\bmod 9$ in this case. Check that if $a\equiv 1,4,7\bmod 9$ then $a^3\equiv 1\bmod 9$. Hence, $a^{226}+4a+1\equiv a+4a+1=5a+1\bmod 9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove the Napier Logarithm rules Question:
The Napier logarithm can be defined as follows:
$$Nap.log~x = 10^7 \ln \frac{10^7}{x}$$
Prove the following rules:
a) $$Nap.log~xy = Nap.log~x + Nap.log~y - Nap.log~1$$
b) $$Nap.log~\frac{x}{y} = Nap.log~x - Nap.log~y + Nap.log~1$$
c) $$Nap.log~x^a = a \cdot Nap.log~x + (1-a) Nap.log~1$$
Attempted answer:
The basic approach is to use the definition of Napier logarithm to convert to natural logarithm, rearrange them and show that the results can be converted back to Napier logarithm to fulfill the rules to be proven.
a)
$$Nap.log~xy = 10^7 \ln \frac{10^7}{xy} = 10^7 \ln \frac{10^7}{x} + 10^7 \ln \frac{10^7}{y} - 10^7 \ln 10^7$$
$$ = Nap.log~x + Nap.log~y - Nap.log~1$$
b) For this law, I will just use the trick that x divided by y is the same as multiplying x with the reciprocal of y:
$$Nap.log~\frac{x}{y} = Nap.log~(x \cdot \frac{1}{y}) = Nap.log~x + Nap.log~\frac{1}{y} =$$
$$ Nap.log~x - Nap.log~y + Nap.log~1$$
c) This one is a bit tricky for me. I can tell it must involve using the standard logarithm for exponent and moving it down as a factor, but not entirely clear on how to apply it. Here is what I have so far:
$$Nap.log~a^x = 10^7 ln \frac{10^7}{x^a} = 10^7 \ln 10^7 - 10^7 \ln x^a = 10^7 \ln 10^7 - a \cdot 10^7 \ln x $$
...but this does not seem to easily work out. How do I wrap up this last part of the question?
| For c), adding $$10^7a\ln 10^7-10^7a\ln 10^7\ (=0)$$ works.
$$\begin{align}Nap.log~x^a&=10^7\ln\frac{10^7}{x^a}
\\\\&=10^7\ln 10^7-10^7a\ln x
\\\\&=10^7\ln 10^7-10^7a\ln x+10^7a\ln 10^7-10^7a\ln 10^7
\\\\&=(10^7a\ln 10^7-10^7a\ln x)+(10^7\ln 10^7-10^7a\ln 10^7)
\\\\&=10^7a\ln\frac{10^7}{x}+(1-a)10^7\ln 10^7
\\\\&=a \cdot Nap.log~x + (1-a) Nap.log~1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Looking for solutions of the following differential equation I obtained the following, apparently clean-looking differential equation, while solving a problem, but cannot find any efficient way to solve it analytically. The equation is of the form,
$$[A - B\cos (q\phi)]y'' + \frac{qB}{2}\sin(q\phi)y' + [C + D\cos(q\phi)]y = 0$$
Where, $q$ is an integer, and $0 \lt \phi \lt 2\pi$, and $y = y(\phi)$. The only boundary condition I know would be, $y(0) = y(2\pi)$.
My first idea for simplifying this was to consider the function to be of the form,$y = G\psi$, and substitute it, to remove the first-order derivative. Taking $F = [A-B\cos(q\phi)]$, this results in $G = \frac{ln F}{4}$ and the modified differential equation looks like,
$$[\frac{F\ddot{F} - \dot{F}^2}{4F} + \frac{\dot{F}^2}{8F} + (C + D\cos(q\phi))ln F]\psi + F\frac{lnF}{4}\ddot{\psi} =0$$
This isn't helpful. Any other ideas, how to tackle this equation? Hints or ideas are welcome. Also, if there is any link to existing literature on this type of equation, please share. I could'nt find any.
| Hint.
Making
$$
\sin t = \sum_{k=0}^N(-1)^k\frac{t^{2k+1}}{(2k+1)!}\\
\cos t = \sum_{k=0}^N(-1)^k\frac{t^{2k}}{(2k)!}\\
y = \sum_{k=0}^N a_k t^k
$$
and substituting into the DE
$$
(\alpha - \cos t)y'' + \frac{\sin t}{2} y' + (\beta + \gamma \cos t)y = 0
$$
we get a set of linear equations in the form
$$
M a = b
$$
like the following (for $N = 7$)
$$
\left(
\begin{array}{cccccccc}
\beta +\gamma & 0 & 2 (\alpha -1) & 0 & 0 & 0 & 0 & 0 \\
0 & \beta +\gamma -\frac{1}{2} & 0 & 6 (\alpha -1) & 0 & 0 & 0 & 0 \\
-\frac{\gamma }{2} & 0 & \beta +\gamma & 0 & 12 (\alpha -1) & 0 & 0 & 0 \\
0 & \frac{1}{12} (1-6 \gamma ) & 0 & \beta +\gamma +\frac{3}{2} & 0 & 20 (\alpha -1) & 0 & 0 \\
\frac{\gamma }{24} & 0 & \frac{1}{12} (1-6 \gamma ) & 0 & \beta +\gamma +4 & 0 & 30 (\alpha -1) & 0 \\
0 & \frac{1}{240} (10 \gamma -1) & 0 & -\frac{\gamma }{2} & 0 & \beta +\gamma +\frac{15}{2} & 0 & 42 (\alpha -1) \\
-\frac{\gamma }{720} & 0 & \frac{1}{360} (15 \gamma -2) & 0 & \frac{1}{6} (-3 \gamma -1) & 0 & \beta +\gamma +12 & 0 \\
0 & \frac{1-14 \gamma }{10080} & 0 & \frac{1}{240} (10 \gamma -1) & 0 & \frac{1}{12} (-6 \gamma -5) & 0 & \beta +\gamma +\frac{35}{2} \\
\end{array}
\right)\left(
\begin{array}{c}
a_0 \\
a_1 \\
a_2 \\
a_3 \\
a_4 \\
a_5 \\
a_6 \\
a_7 \\
\end{array}
\right) = 0
$$
This is a homogeneous linear system with the trivial solution $a = 0$ as expected. So to have a nontrivial solution we need $\det M = 0$ which means that the parameter $\beta$ can be considered as an eigenvalue and the eigenvalues can be obtained by solving
$$
\det M(\beta) = 0
$$
This approach makes it possible to obtain the system's transient response whereas the Fourier series approximation only gives us the response in permanent regime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| Hint: Just equate the coefficients of the two polynomials (by FTA they have to be equal),
and solve the resulting system.
Thus, $\begin {cases} A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2\end{cases}$.
To solve, you could row-reduce the following augmented matrix: $\left(\begin {array}{rrrrr|r}1&1&0&0&0&1\\0&-1&1&0&0&-1\\4&2&-1&1&0&2\\0&-2&2&-1&1&-1\\4&0&-2&0&-1&2\end{array}\right) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Can we conclude $0^\circMy attempt
Based on the sine rule and the graph of $\sin A = k a$ (where $k$ is a constant) in interval $(0,\pi)$,
increasing $a$ up to $1/k$ will either
*
*increase $A$ up to $90^\circ$.
*decrease $A$ up to $90^\circ$.
So I cannot conclude that increasing $a$ will increase $A$.
Now I use the cosine rule (it is promising because the cosine is decreasing in the given interval).
\begin{align}
A &= \cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)\\
B &= \cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\\
C &= \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)\\
\end{align}
It is hard to show that $0^\circ<A\leq B\leq C<180^\circ$ for any $\triangle ABC$ with $0<a\leq b\leq c$. Could you show it?
It means that I need to show that
$$
-1<\frac{a^2+b^2-c^2}{2ab}\leq \frac{a^2+c^2-b^2}{2ac} \leq \frac{b^2+c^2-a^2}{2bc}<1
$$
for $0<a\leq b\leq c$.
| I would just split the problem in two cases, depending if $c^2$ is greater or smaller than $a^2+b^2$. If $c^2\ge a^2+b^2$, then $C\ge 90^\circ$, so then $A$ and $B$ are less then $90$, therefore less then $C$. After that, apply the law of sines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.