Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$ sin^2 x = \frac{1}{2}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sin x= \lvert\frac{1}{\sqrt2}\rvert$$
$$ \sin x= \frac{\sqrt2}{2} ; \sin x= -\frac{\sqrt2}{2}$$
So if I look for all the values of $x$ that solve the above I should get not only $ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°) but also $ x= \frac{3\pi}{4}$ (135°) ; $ x= \frac{7 \pi}{4}$ (315°).
What am I doing wrong?
| The problem is a pretty generic one. By squaring both sides of the equation earlier on, you introduced extraneous solutions. It's sort of like how in solving the equation
$$x+\sqrt{x+9} = 2$$
you can get to
$$\sqrt{x+9} = 2-x$$
and then square both sides to get
$$x+9 = 4 - 4x + x^2 \implies x^2 - 5x - 5 = 0$$
You should get
$$x = \frac{5 \pm 3\sqrt 5}{2}$$
Not the cleanest expression, but you'll notice that, if you substitute the positive root into the original equation, it won't work, though the negative root does. This applies beyond the context of polynomials. For example, if we have
$$x = x$$
we can square and get
$$x^2 = x^2$$
Obvious, right? But a problem: if you wanted to work backward, how would you know whether either is positive or negative? After all, $x^2 = (-x)^2$ in the usual arithmetic. Same problem here:
$$(\sin(x))^2 = (-\sin(x))^2 = (\cos(x))^2 = (-\cos(x))^2$$
Extra solutions galore.
This doesn't make your method of solution invalid, necessarily, but you have to be conscious of extraneous solutions, and always substitute each solution back in to check its validity. In fact, substituting both $x=3\pi/4$ and $x = 7\pi/4$ into the original equation, you should notice that the two differ by $\sqrt 2$ and $-\sqrt 2$, respectively, so you don't have equality in those cases!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without L'Hospital's Rule. I need to solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without using L'Hospital's Rule. Using that rule I found the equation becomes $\lim_{x \to 0}(\frac{1}{\sqrt{1+2x}} - \frac{2}{\sqrt{1-4x}}) = \frac{-1}{\sqrt{1}}$.
However, I'm not sure how to solve this without L'Hospital's Rule. The only tool I know of is multiply both numerator and denominator by a conjugate, but multiplying by $\frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$ doesn't seem to get me the same answer.
$$\lim_{x \to 0}\frac{\sqrt{1+2x} - \sqrt{1-4x}}{x} \cdot \frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$= \lim_{x \to 0} \frac{1+2x-(1-4x)}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6x}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$=\frac{6}{\sqrt{1} + \sqrt{1}}$$
But $\frac{6}{\sqrt{1} + \sqrt{1}} \neq \frac{-1}{\sqrt{1}}$.
| There is a mistake, it must be $$\frac{1}{\sqrt{1+2x}}+\frac{2}{\sqrt{1-4x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the equation of a hyperbola given the asymptote, equation of axis and a point Given that a hyperbola has asymptote $y=0$, passes through the point $(1,1)$ and has axis $y=2x+2$, determine its equation.
The answer arrived at is $\displaystyle{4xy+3y^2+4y-11=0}$. However, I have had no success in reaching it.
I first tried to relate $a$ and $b$ using the point $(1,1)$ to get $$ \frac 1{a^2} - \frac{1}{b^2} = 1 = \frac{a^2b^2}{a^2b^2}$$
Then I changed the subject of the formula $$ x=\frac{a\left(4a+b\sqrt{b^2+4-4a^2}\right)}{b^2-4a^2},\:x=\frac{a\left(4a-b\sqrt{b^2+4-4a^2}\right)}{b^2-4a^2};\quad \:b^2-4a^2\ne \:0 $$
But I didn't find any helful use of that information.
Next, I located the centre $(-1,0)$, which is the intersection of the axis and the asymptote. I then related the vertices $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (-1,0)$$ but that gives two equations in four unknowns. I tried to substitute these equations in the canonical equation of a hyperbola but always ended up with a more complex equation that with more than one variable. At this point, I'm out of ideas. How can I approach this problem?
| As pointed out in the comment by Jan-MagnusØkland, solving the problem requires using the property that the axis bisects the angle between the two asymptotes of the hyperbola. We are given one asymptote $y=0$ thus we can find the second one by reflecting $y=0$ across the axis $y=2x+2$. Let's rewrite it in the standard form $2x-y+2=0$.
Let's pick the point $(0,0)$ to reflect. The distance between the axis and the point $(0,0)$ is $$d=\left| \frac{Ax_0 + By_0 +C}{\sqrt{A^2 + B^2}} \right| = \left| \frac{2x_0 -y_0 +2}{\sqrt{(2)^2 + (-1)^2}} \right| = \left| \frac{2\cdot0 -1\cdot0 +2}{\sqrt{5}} \right| = \left| \frac{2}{\sqrt{5}} \right|$$
Therefore, the distance from the symmetric point $(x_0, y_0)$ is also $\frac{2}{\sqrt{5}}$. This gives the equation:
$$ \left| \frac{2x_0 -y_0 +2}{\sqrt{5}} \right| = \frac{2}{\sqrt{5}} \implies \left| {2x_0 -y_0 +2} \right| = 2 \qquad \qquad \qquad (*)$$
Since the point symmetric to $(0,0)$ with respect to the axis $2x-y+2=0$ lies on the perpendicular to $2x-y+2=0$, we can find the gradient of the perpendicular $$ m_1m_2=-1 \implies m_2=\frac{-1}2$$
Thus, together with $(*)$ we have our second (or third) equation $$\frac{y_2-y_1}{x_2-x_1} = \frac{y_0 - 0}{x_0 - 0} = \frac{-1}2 \implies x_0 = -2y_0 \qquad \qquad \qquad (**)$$.
The equation in $(*)$ gives us two cases:
$$\left| {2x_0 -y_0 +2} \right| = 2 \implies \begin{cases}
2x_0 -y_0 +2 = 2,\\
2x_0 -y_0 +2 = -2
\end{cases}$$
Solving the fist case of $(*)$ simultaneously with $(**)$, we get $(x_0,y_0)$ = $(0,0)$ which is our initial point. So the second case gives us our symmetric point:
$$\begin{cases}
2x_0 -y_0 +2 = -2\\
x_0 = -2y_0
\end{cases} \implies (x_0,y_0) = \left(\frac{-8}{5}, \frac45 \right)$$
That gives us the reflection of the point $(0,0)$ across the line $2x-y+2=0$.Next, we know that the asymptote intersects the axis at the centre of the hyperbola which gives us $c=(-1,0)$. Using $c$ and the symmetric point we just calculated, we find that the equation of the second asymptote is $$y=\frac{-4}{3}x - \frac43 \equiv 3y+4x+4=0 $$
Using the property that the equation of a hyperbola can be given by its asymptotes $$(Ax+By+C)(A_1x+B_1y+C_1)=k$$
We have $$y(3y+4x+4)=k$$
Since the point $(1,1)$ lies on the hyperbola, we get that $k=11$ giving the final answer to be $$4xy+3y^2+4y-11=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find two independent orthogonal eigenvectors We are given that $C$ is an $n\times n$ complex matrix, and $C$ is Hermitian ($C = C^*$). Now we define $C = A + Bi$ where $A$ and $B$ are real matrices.
Let $M$ be defined by
\begin{bmatrix}
A & -B \\
B & A
\end{bmatrix}
Given that $\lambda$ is an eigenvalue of $C$ with eigenvector $z = x + iy$, we want to find two independent orthogonal eigenvectors of $M$ with eigenvalue $\lambda$.
I found that
\begin{bmatrix}
z \\
-iz
\end{bmatrix}
is one eigenvector. How do I find the other one?
| Given a Hermitian matrix
$C = C^\dagger \in M_{n \times n}(\Bbb C), \tag 1$
which has an eigenvalue $\lambda$ with associated eigenvector $z$:
$Cz = \lambda z, \tag 2$
it is well-known that
$C^\dagger = C \Longrightarrow \lambda \in \Bbb R; \tag 3$
since the entries of $C$ lie in $\Bbb C$, with real $\lambda$, we will generally have
$z \in \Bbb C^n; \tag 4$
that is,
$z = x + iy, \tag 5$
with
$x, y \in \Bbb R^n; \tag 6$
thus we may write
$C(x + iy) = \lambda(x + iy); \tag 7$
now (1) implies we also have
$C = A + iB, \; A, B \in M_{n \times n}(\Bbb R); \tag 8$
thus (7) becomes
$(A + iB)(x + iy) = \lambda x + i \lambda y, \tag 9$
or
$(Ax - By) + i(Ay + Bx) = \lambda x + i \lambda y; \tag{10}$
equating the real and imaginary parts of either side yields
$Ax - By = \lambda x, \tag{11}$
$Ay + Bx = \lambda y. \tag{12}$
We can now in fact find two real orthogonal eigenvectors for
$M = \begin{bmatrix} A & -B \\ B & A \end{bmatrix}, \tag{13}$
each corresponding to the eigenvalue $\lambda$; set
$w = \begin{pmatrix} x \\ y \end{pmatrix}; \tag{14}$
$Mw = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} Ax - By \\ Bx + Ay \end{pmatrix} = \begin{pmatrix} \lambda x \\ \lambda y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix} = \lambda w; \tag{15}$
now set
$v = \begin{pmatrix} -y \\ x \end{pmatrix}; \tag{16}$
$Mv = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} \begin{pmatrix} -y \\ x \end{pmatrix} = \begin{pmatrix} -Ay - Bx \\ -By + Ax \end{pmatrix} = \begin{pmatrix} -\lambda y \\ \lambda x \end{pmatrix} = \lambda \begin{pmatrix} -y \\ x \end{pmatrix} = \lambda v; \tag{17}$
furthermore,
$\langle v, w \rangle = \begin{pmatrix} -y \\ x \end{pmatrix}^T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y^T & x^T \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = -y^Tx + x^Ty; \tag{18}$
$y^T x, x^Ty \in \Bbb R \Longrightarrow y^Tx = (y^Tx)^T = x^Ty, \tag{19}$
thus,
$\langle v, w \rangle = 0. \tag{20}$
Note: It may be worth pointing out that the condition on $C$
$C^\dagger = (\bar C)^T = C; \tag{21}$
has implitions for $A$ and $B$ as in (8), for then
$\bar C = A - iB, \tag{22}$
$(\bar C)^T = A^T - iB^T; \tag{23}$
thus,
$A = A^T, \tag {24}$
$B = -B^T; \tag{25}$
we see that $A$, the real part of $C$, must be symmetric, whilst $B$, the imaginary part, must be skew-symmetric. Finally we observe that the matrix $M$ (13) is in fact symmetric as well:
$M^T = \begin{bmatrix} A & -B \\ B & A \end{bmatrix}^T = \begin{bmatrix} A^T & B^T \\ -B^T & A^T \end{bmatrix} = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} = M, \tag{26}$
by virtue of (24)-(25). End of Note.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1).
Since $x=y^2$, we have $\frac{dx}{dy} = 2y$.
$$\begin{align}
L &= \int_{0}^{1} \sqrt{1+(\frac{dx}{dy})^2}dy \\
&= \int_{0}^{1} \sqrt{1+(2y)^2}dy
\end{align}$$
Here is where I am getting stuck. I am supposed to make a trigonometric where $y = \frac{1}{2}\tan\theta$, but based off the trig substitution equation I have, $\sqrt{a^2+y^2} \rightarrow y = \tan\theta$ I am getting confused how I am supposed to get $y = \frac{1}{2}\tan\theta$. I keep thinking it should be $ y = \frac{1}{4}\tan\theta$.
Here is where I get my $y$ value from:
$$\sqrt{a^2 + y^2} = \sqrt{1+ 2^2 y^2} \\
= \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$
Here is a picture of the example in the textbook. I am not able to follow the math :(
| When you say:
"Here is where I get my $y$ value from: $$\sqrt{a^2 + y^2} = \sqrt{1+
> 2^2 y^2} \\
> = \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$"
From here you want not $y = \frac{1}{4} \tan\theta$ but $y^2 = \frac{1}{4} \tan^2\theta$. Taking square roots gives $y = \frac{1}{2} \tan\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 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 $y$?
| By false position:
Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification
$$3x'+2y'=0,\\5x'+4y'=2.$$
We easily eliminate $y'$ (using $4y'=-6x'$) and get
$$-x'=2.$$
Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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"answer_id": 8
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Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
| Using Newton's identities
$$
\begin{aligned}
e_{1}&=p_{1}\\
2e_{2}&=e_{1}p_{1}-p_{2}\\
3e_{3}&=e_{2}p_{1}-e_{1}p_{2}+p_{3}\\
4e_{4}&=e_{3}p_{1}-e_{2}p_{2}+e_{1}p_{3}-p_{4}\\
5e_{5}&=e_{4}p_{1}-e_{3}p_{2}+e_{2}p_{3}-e_{1}p_{4}+p_{5}\\
\end{aligned}
$$
with $p_1=1,p_2=2,p_3=3,e_4=0, e_5=0$, we get $p_5 = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Series of Beta Function Solve:
$$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx$$
I tried
$$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx=\sum_{k=0}^\infty(-1)^k\int_0^\frac{\pi}{2}(\sin{x})^\frac{1}{2}(\cos{x})^{k+\frac{1}{2}}=\sum_{k=0}^\infty(-1)^kB\left(\frac{3}{4},\frac{k}{2}+\frac{3}{4}\right)$$ but it did not work, did it?
| We are given:
$$ \int_0^{\pi/2} \frac{\sqrt{\sin x\cos x}\,dx}{1+\cos x} $$
Put $t=\tan\tfrac{x}{2}$; then
$$
\sin x = \frac{2t}{1+t^2}
\quad\quad
\cos x = \frac{1-t^2}{1+t^2}
\quad\quad
dx = \frac{2\,dt}{1+t^2}
$$
Then
$$ \sqrt{\sin x\cos x} = \frac{2t(1-t^2)}{1+t^2} $$
and
\begin{align}
1 + \cos x &= 1 + \frac{1-t^2}{1+t^2} \\
&= \frac{1+t^2+1-t^2}{1+t^2} = \frac{2}{1+t^2}
\end{align}
so that
$$ \int_0^{\pi/2}\frac{\sqrt{\sin x\cos x}\,dx}{1+\cos x} = \int_0^1\frac{\sqrt{2t(1-t^2)}\,dt}{1+t^2} $$
This is an elliptic integral. Here's my attempt at expressing it in terms of the Legendre canonical forms: put $t^2=v$; then $dt=2v\,dv$ and the integral becomes
\begin{align}
\int_0^1 \frac{\sqrt{2t(1-t^2)}\,dt}{1+t^2} &= \int_0^1 \frac{\sqrt{2v^2(1-v^4)}(2v\,dv)}{1+v^4} \\
&= 2\sqrt2 \int_0^1 \frac{v^2(1-v^4)}{1+v^4}\frac{dv}{\sqrt{1-v^4}} \\
&= 2\sqrt2 \int_0^1 \left(\frac{i}{1+iv^2}-\frac{i}{1-iv^2}-v^2\right)\frac{dv}{\sqrt{1-v^4}} \\
&= 2\sqrt2(i\Pi(-i,i)-i\Pi(i,i)-E(i)+K(i))
\end{align}
where $K,E,\Pi$ are the first, second and third complete elliptic integrals, in terms of the modulus $k$.
Wolfram Alpha gives an exact value
$$ 2\sqrt 2(K(i)-E(i)) = -\frac{4\pi^{3/2}}{\Gamma(\tfrac{1}{4})^2} $$
which, given Peter Foreman's comment, looks right. But W|A does not give an exact value for the term involving the $\Pi$'s, so I think I made a mistake there. I'll look into that later. I'll also try and see if I can prove these values for the integrals.
Update: In
$$ E(i) - K(i) = \int_0^1\frac{v^2\,dv}{\sqrt{1-v^4}} $$
put $r=v^4$; then $dv=v^{-3}\,dr/4$ and the integral becomes
\begin{align}
\int_0^1 \frac{v^2\,dv}{\sqrt{1-v^4}} &= \frac{1}{4}\int_0^1 (1-r)^{-1/2}r^{1/2}r^{-3/4}\,dr \\
&= \frac{1}{4}{\rm B}(\tfrac{1}{2},\tfrac{3}{4}) = \frac{1}{4}\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{3}{4})}{\Gamma(\tfrac{5}{4})} \\
&= \frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{3}{4})}{\Gamma(\tfrac{1}{4})}
\end{align}
By the reflection formula,
$$ \Gamma(\tfrac{1}{4})\Gamma(\tfrac{3}{4}) = \frac{\pi}{\sin\tfrac{\pi}{4}} = \pi\sqrt 2 $$
Using next $\Gamma(\tfrac{1}{2})=\sqrt\pi$ we put everything together yielding
$$ E(i) - K(i) = \frac{\pi^{3/2}\sqrt2}{\Gamma(\tfrac{1}{4})^2} $$
I'll look into the case of the $\Pi$'s next.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the median in a triangle with trigonometry In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50º$. Let $M$ be the midpoint of $BC$. Determinate $AM$.
My try
I applied law of cosines $3$ times, first to find $BC$, then I let $\angle BCA=\alpha$ and $\angle ABC=130-\alpha$, and applied law of cosines in triangle $ACM$ and $ABM$, but it didn't work.
Any hints?
|
\begin{align}
\triangle ADC:\quad
|AD|=4\cos50^\circ
,\quad
|CD|&=4\sin50^\circ
,\\
\triangle BCD:\quad
|BC|&=\sqrt{(4\sin 50^\circ)^2+(7-4\cos 50^\circ)^2}
\\
&=\sqrt{16\sin^2 50^\circ
+49-56\cos 50^\circ+16\cos^2 50^\circ}
\\
&=\sqrt{65-56\cos50^\circ}
,\\
\triangle BCD,\triangle BME:\quad
|ME|&=\tfrac12|CD|
=2\sin 50^\circ
,\\
|DE|&=\tfrac12|BD|
=\tfrac72-2\cos 50^\circ
,\\
|AE|&=|AD|+|DE|
=\tfrac72+2\cos 50^\circ
,\\
\triangle AME:\quad
|AM|&=\sqrt{|AE|^2+|ME|^2}
\\
&=\sqrt{(\tfrac72+2\cos 50^\circ)^2
+(2\sin 50^\circ)^2
}
\\
&=\sqrt{\tfrac{49}4+14\cos 50^\circ
+4\cos^2 50^\circ+4\sin^2 50^\circ}
\\
&=\tfrac12\sqrt{65+56\cos 50^\circ}
.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3189380",
"timestamp": "2023-03-29T00:00:00",
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Show with induction that $\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$ Show with induction that
$\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$
n = 1
$LHS = \frac{1}{2}$
$RHS = 6 - \frac{1+4+6}{2} = \frac{1}{2}$
n = p
$LHS_{p} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + ... \frac{p^{2}}{2^{p}}$
$RHS_{p} = 6 - \frac{P^2+4p+6}{2^{p}}$
n = p + 1
$LHS_{p+1} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + ... \frac{p^{2}}{2^{p}} + \frac{(p+1)^{2}}{2^{(p+1)}}$
$RHS_{p+1} = 6 - \frac{(p+1)^2+4(p+1)+6}{2^{(p+1)}}$
Show that, $RHS_{p+1} = RHS + \frac{(p+1)^{2}}{2^{(p+1)}}$
$6 - \frac{(p+1)^2-4(p+1)+6}{2^{(p+1)}} = 6 - \frac{P^2+4p+6}{2^{p}} + \frac{(p+1)^{2}}{2^{(p+1)}}$
| I will assume the correct formula is:
$$\sum_{k=1}^n \dfrac{k^2}{2^k} = 6-\dfrac{n^2+4n+6}{2^n}$$
Assume this is true for $n=p$ and try to prove for $n=p+1$:
$$\begin{align*}\sum_{k=1}^{p+1} \dfrac{k^2}{2^k} & = \sum_{k=1}^p \dfrac{k^2}{2^k} + \dfrac{(p+1)^2}{2^{p+1}} \\ & = 6-\dfrac{p^2+4p+6}{2^p}+\dfrac{(p+1)^2}{2^{p+1}} \qquad \text{ by the induction assumption}\\& = 6-\dfrac{p^2+4p+6}{2^p}\cdot \dfrac{2}{2} - \left(-\dfrac{p^2+2p+1}{2^{p+1}}\right) \\ & = 6 - \dfrac{2p^2+8p+12}{2^{p+1}} -\left(- \dfrac{p^2+2p+1}{2^{p+1}}\right) \\ & = 6 - \left( \dfrac{2p^2+8p+12}{2^{p+1}} - \dfrac{p^2+2p+1}{2^{p+1}}\right) \\ & = 6 - \dfrac{2p^2+8p+12-(p^2+2p+1)}{2^{p+1}} \\ & = 6 - \dfrac{p^2+6p+11}{2^{p+1}} \\ & = 6 - \dfrac{(p^2+2p+1)+(4p+4)+6}{2^{p+1}} \\ & = 6 - \dfrac{(p+1)^2+4(p+1)+6}{2^{p+1}}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Drawing without replacement: why is the order of draw irrelevant? I am trying to wrap my head around this problem:
Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.
What I remember from my college days that the probability is found by this formula:
$$P(A)=\frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}}=\frac{5}{21}$$
Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $\frac{6}{10}$ so the probability for the second draw becomes $\frac{5}{9}$ for red and $\frac{4}{9}$ for green. But if the first picked ball is green, the probability for the second draw becomes $\frac{6}{9}$ for red and $\frac{3}{9}$ for green. What am I missing?
| You can comprehend the calculation in a simpler way with smaller numbers.
Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
is the probability that he picks $2$ red and $2$ green if balls are drawn
without replacement.
Indeed we have to regard the order. There are $\frac{4!}{2!\cdot 2!}=6$ ways to draw 2 red and 2 green balls:
$$\color{green}g\color{green}g\color{red}r\color{red}r, \color{green}g\color{red}r\color{green}g\color{red}r, \color{green}g\color{red}r\color{red}r\color{green}g, \color{red}r\color{green}g\color{green}g\color{red}r, \color{red}r\color{green}g\color{red}r\color{green}g, \color{red}r\color{red}r\color{green}g\color{green}g$$
Each way has the same probability: $\frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2} \quad (ggrr)$
Multiplying with 6 (ways) we get $6\cdot \frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{3}5=0.6 $
Using binomial coefficients we get $\frac{\binom{3}{2}\cdot \binom{2}{2}}{\binom{5}{4}}=\frac{3\cdot 1}{5}=\frac35=0.6$
And we get the same result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find function $f(x)$ satisfying $\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$ I am looking for a non-trivial function $f(x)\in L_2(0,\infty)$ independent of the parameter $n$ (a natural number) satisfying the following integral equation: $$\displaystyle\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$$
or prove that $f(x)=0$ is the only solution.
The similar question is here but there are no parameters in the integral and the answer is based upon hit and trial method.
I want to know what would be the nice approach to tackle this problem.
EDIT: Such a function may exist, here is an example due to Stieltjes. A function $f(x) = \exp(-x^{1/4}) \sin x^{1/4}$ satisfies $\int_0^{\infty} f(x) x^n dx = 0$ for all integers $n \ge 0$.
We use the substitution $x=u^4$ to write
$I_n = \int_0^{\infty} f(x) x^n dx = 4 \int_0^{\infty} e^{-u} \sin(u) u^{4n+3} du$;
then integrate by parts four times (differentiating the power of $u$, and integrating the rest) to show that $I_n$ is proportional to $I_{n-1}$,
and finally check that $I_0=0$.(the Edit copied from here)
| The issue condition can be written in the form of
$$\int_0^\infty f\left(\dfrac zn\right)\dfrac{dz}{1+e^z} = 0.\tag1$$
Let the Maclaurin series of $f(x)$ exists, then
$$f(z) = \sum\limits_{k=0}^\infty a_kz^k.$$
Taking in account the integral representation of the Riemann zeta function in the form of
$$\int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} =
(1-2^{-k})k!\zeta(k+1)\tag3$$
(see also Wolfram Alpha),
one can get
$$I_n = \sum\limits_{k=0}^\infty \dfrac{a_k}{n^k} \int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} = \sum\limits_{k=0}^\infty \dfrac{1-2^{-k}}{n^k}k!\zeta(k+1) = 0,$$
$$I_n = \sum\limits_{k=0}^\infty \dfrac{b_k}{n^k} = 0\quad \forall (n\in \mathbb N),\tag4$$
where
$$b_k = (1-2^{-k})k!\zeta(k+1).\tag5$$
The system $(4)$ is linear relatively to the coefficients $b_k.$ On the other hand,
determinant of this system is
$$ \Delta =
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\
1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dfrac1{2^4} & \dots & \dfrac1{2^k} &\dots\\
1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dfrac1{3^4} & \dots & \dfrac1{3^k} &\dots\\
1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dfrac1{4^4} & \dots & \dfrac1{4^k} &\dots\\
&&&&\dots\\
1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dfrac1{n^4} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix}\\[6pt]
=
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\
0 & -\dfrac12 & -\dfrac1{2^2} & -\dfrac1{2^3} & -\dfrac1{2^4} & \dots & -\dfrac1{2^k} &\dots\\
1 & -\dfrac23 & -\dfrac2{3^2} & \dfrac2{3^3} & \dfrac2{3^4} & \dots & -\dfrac2{3^k} &\dots\\
1 & -\dfrac34 & \dfrac3{4^2} & -\dfrac3{4^3} & -\dfrac3{4^4} & \dots & -\dfrac3{4^k} &\dots\\
&&&&\dots\\
1 & -\dfrac{n-1}n & -\dfrac{n-1}{n^2} & -\dfrac{n-1}{n^3} & -\dfrac{n-1}{n^4} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix}\\[6pt]
\sim
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\
0 & 1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dots & \dfrac1{2^k} &\dots\\
0 & 1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dots & \dfrac1{3^k} &\dots\\
0 & 1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dots & \dfrac1{4^k} &\dots\\
&&&&\dots\\
0 & 1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix} \not = 0.$$
If the homogenius linear system has unzero determinant, then zero solution is single.
Looks that consideration of the determinant sequences allows to assume that $\textbf{the issue task has only the trivial solution}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 6,
"answer_id": 4
} |
Evaluating $\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} \,dx \,dy \,dz$
I am trying to solve this triple integral problem , but I am having some issues.
$$\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} dx dy dz$$
I tried with the 2 different approaches.
*
*I converted the variables to spherical coordinates, and I think it went well, until the moment to evaluate the integral using those limits $(-1 , -2)$. I couldn't pass that.
*Conversion to cylindrical coordinates, but no luck at all there.
So, could someone be so kind in giving me some light around this? Please. Not sure if I am follow a right approach.
| Following @kimchilover's suggestion, suppose $X,Y,Z$ are independent and identically distributed random variables having the uniform distribution on $(-2,-1)$.
So the pdf of $(X,Y,Z)$ is just $$f(x,y,z)=\mathbf1_{-2<x,y,z<-1}$$
Now,
$$E\left(\frac{X^2+Y^2+Z^2}{X^2+Y^2+Z^2}\right)=1$$
Or,
$$E\left(\frac{X^2}{X^2+Y^2+Z^2}\right)+E\left(\frac{Y^2}{X^2+Y^2+Z^2}\right)+E\left(\frac{Z^2}{X^2+Y^2+Z^2}\right)=1\tag{*}$$
By symmetry, $(X,Y,Z),(Y,Z,X)$ and $(Z,X,Y)$ have the same distribution, so that in turn $\frac{X^2}{X^2+Y^2+Z^2}, \frac{Y^2}{X^2+Y^2+Z^2}$ and $\frac{Z^2}{X^2+Y^2+Z^2}$ also have the same distribution.
Therefore $(*)$ reduces to
$$3E\left(\frac{X^2}{X^2+Y^2+Z^2}\right)=1$$
That is, $$\iiint_{-2<x,y,z<-1} \frac{x^2}{x^2+y^2+z^2}\,dx\,dy\,dz=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$f: \mathbb{R} \to \mathbb{R},\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space$find $f(3)$ in terms of $f(0)$.
$f: \mathbb{R} \to \mathbb{R},\space\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space\space\space$
Find $f(3)$ in terms of $f(0)$.
My approach:
$$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$$
$$\Rightarrow f(\frac{x}{2})-2f(\frac{x}{4})+f(\frac{x}{8})=\frac{x^2}{4}$$
$$\Rightarrow f(\frac{x}{4})-2f(\frac{x}{8})+f(\frac{x}{16})=\frac{x^2}{16}$$
$$\vdots$$
$$\Rightarrow f(\frac{x}{2^{n-1}})-2f(\frac{x}{2^n})+f(\frac{x}{2^{n+1}})=\frac{x^2}{4^{n-1}}$$
Summing up these when n $\rightarrow$ $\infty$
$$f(x)-f(\frac{x}{2})=x^2\cdot\frac{4}{3}$$
I am unable to proceed further.
| Making $y=2^x$ we have
$$
F(y)-2F(y-1)+F(y-2)=4^y
$$
which is a linear difference functional equation with solution
$$
F(y) = y \Phi_1(y)+\Phi_2(y)+\frac{4^{y+2}}{9}
$$
Here $\Phi_1(y),\Phi_2(y)$ are generic periodic functions with period $1$ Assuming $\Phi_1(y) = C_1,\Phi_2(y)= C_2$ we have
$$
f(x) = C_1\log_2 x + C_2 + \frac{16}{9}x^2
$$
then for the feasibility of $f(0)$ we have $C_1 = 0$ and
$$
f(0) = C_2
$$
and
$$
f(3) = f(0)+16
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a Probability for a Gamma Distributed Random Variable Below is a problem from the book "Probability and Statistics". It is one of the Schaum's books. I am getting the
wrong answer and I believe that I am doing the integration incorrectly.
Thanks
Bob
Problem:
A random variable $X$ is gamma distributed with $\alpha = 3$, $\beta = 2$. Find $P( X \leq 1 )$.
Answer:
The density function for a gamma distribution is:
$$f(x) = \begin{cases}
\frac{x^{\alpha-1} \, e^{-\frac{x}{\beta}}}{ \beta^\alpha \Gamma(\alpha)} & \text{for }x > 0 \\
0 & \text{for }x <= 0 \\
\end{cases} \\
$$
In this case, the density function is:
$$f(x) = \begin{cases}
\frac{x^{2}e^{-\frac{x}{2}}}{2^3 \Gamma(3)} & \text{for }x > 0 \\
0 & \text{for }x <= 0 \\
\end{cases} \\
$$
Now we can simplify the density function.
\begin{align*}
\Gamma(3) &= 2 \Gamma(2) \\
\Gamma(2) &= 1 \Gamma(1) = 1 \\
\Gamma(3) &= 2 \\
\end{align*}
Hence the density function is:
$$f(x) = \begin{cases}
\frac{x^{2}e^{-\frac{x}{2}}}{16} & \text{for }x > 0 \\
0 & \text{for }x <= 0 \\
\end{cases} \\
$$
\begin{align*}
P( X \leq 1 ) &= \int_0^1 \frac{x^{2}e^{-\frac{x}{2}}}{16} \, dx \\
\text{Let } u &= -\frac{x}{2} \\
x &= -2u \\
du &= -\frac{1}{2} \, dx \\
P( X \leq 1 ) &= \int_{0}^{ -\frac{1}{2} } -\frac{4u^2e^{u}}{16} \, du \\
P( X \leq 1 ) &= \int_{0}^{ -\frac{1}{2} } -\frac{u^2e^{u}}{4} \, du \\
\end{align*}
Now let's consider the integral:
$$ \int u^2e^u \, du$$
To evaluate this integral, we will use integration by parts twice.
\begin{align*}
\int u^2e^u \, du &= -u^2e^{-u} - \int -2ue^{-u} \, du \\
\int u^2e^u \, du &= -u^2e^{-u} + \int 2ue^{-u} \, du \\
\int u^2e^u \, du &= -u^2e^{-u} -2ue^{-u} - 2 \int -e^{-u} \, du \\
\int u^2e^u \, du &= -u^2e^{-u} -2ue^{-u} + 2 \int e^{-u} \, du \\
\int u^2e^u \, du &= -u^2e^{-u} - 2ue^{-u} - 2e^{-u} + C \\
P( X \leq 1 ) &= \left( -\frac{1}{4} \right) \left( -u^2e^{-u} - 2ue^{-u} - 2e^{-u} \Big|_{0}^{-\frac{1}{2}} \right) \\
\end{align*}
\begin{align*}
P( X \leq 1 ) &= \left( \frac{1}{4} \right) \left( u^2e^{-u} + 2ue^{-u} + 2e^{-u} \Big|_{0}^{-\frac{1}{2}} \right) \\
P( X \leq 1 ) &= \left( \frac{ u^2e^{-u} }{4} + \frac{ue^{-u}}{2} + \frac{e^{-u}}{2} \Big|_{0}^{-\frac{1}{2}} \right) \\
P( X \leq 1 ) &=
\frac{e^{\frac{1}{2}}}{16} - \frac{ \frac{1}{2} e^{\frac{1}{2}} }{2} + \frac{e^{\frac{1}{2}}}{2} - \frac{e^0}{2} \\
P( X \leq 1 ) &= \frac{e^{\frac{1}{2}}}{16} - \frac{ e^{\frac{1}{2}} }{4} + \frac{e^{\frac{1}{2}}}{2} - \frac{1}{2} \\
%
P( X \leq 1 ) &= \frac{e^{\frac{1}{2}}}{16} + \frac{e^{\frac{1}{2}}}{4} - \frac{1}{2} \\
P( X \leq 1 ) &= \frac{5 e^{ \frac{1}{2} } }{16} - \frac{1}{2} \\
\end{align*}
However, the book's answer is:
$$ 1 - \frac{13}{8\sqrt{e}} \doteq 0.0143877 $$
An online integral calculator found the book's answer.
| Until here everything is perfect:
$$P( X \le 1 ) = \int_0^1 \frac{x^{2}e^{-\frac{x}{2}}}{16} \, dx $$
Now when you substitute $-\frac{x}{2}=u\Rightarrow x=-2u\Rightarrow dx=-2du\,$ don't forget to add the coefficient of $du$, like this:
$$P( X \le 1 ) = \int_0^1 \frac{x^{2}e^{-\frac{x}{2}}}{16} \, dx=\int_0^{-1/2}\frac{(-2u)^2e^u}{16}(-\color{red}{2})du=\int_0^{-1/2}-\frac{u^2e^u}{2}du$$
The next issue would be when you integrate by parts. Note that the rule is:
$$(fg)'=f'g+fg'\Rightarrow fg=\int f'g +\int fg'\Rightarrow \boxed{\int f'g=fg-\int fg'}$$
So when you chose $f'=e^u$ then $f=e^u$ not $e^{-u}$. I am talking about the first row from here:
$$\int u^2e^u \, du = -u^2e^{-u} - \int -2ue^{-u} \, du$$
The correct way would be:
$$\int u^2 (e^u)'du=u^2 e^u -\int 2u e^udu$$
But I think the previous factor of $e^{-x/2}$ made you see an $e^{-u}$.
Anyway the rest is fine, I'll finish it for you so you can verify.
$$\int u^2 e^u du=u^2 e^u -2\int u e^{u} du$$
$$=u^2 e^u -2 \left(ue^u - \int e^u du\right)$$
$$=u^2e^u -2 ue^u +2 e^u+C$$
$$P(X\le 1)=-\frac12e^u\left(u^2 -2 u +2 \right)\bigg|_0^{-1/2}$$
$$=-\frac12e^{-1/2}\left(\frac14+1 +2\right)+\frac12(0-0+2)$$
$$=-\frac12 \frac{1}{\sqrt{e}} \cdot \frac{13}{4} +1=1-\frac{13}{8\sqrt{e}}$$
As a side note, I would recommend you to write
$$\int_0^{-1/2} \color{red}- u^2 e^u du=\int_{-1/2}^0 u^2 e^u du$$
But there is no mistake in letting it your way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derive Distance of skew lines with methods from analysis It is well known, that given two skew lines
\begin{align*}
g: x &= a + \lambda b \quad \\
h: y &= c + \mu d
\end{align*}
the distance is given by
$$D = \frac{|(a - c) \cdot b \times d|}{|b \times d|}$$
See e.g. here
I thought, it would be a nice exercise to derive the distance using an optimization
method, but I get stuck. Short version: Why is
\begin{equation*}
\begin{aligned}
\left\| \left(a - c\right) - \frac{|b|^2\cdot |d|^2}{|b \times d|^2} \left[\vphantom{\frac{b}{b}}\right.\right.
& \left.\left. \left(\left(a - c\right)\cdot \left(\frac{b}{|b|} - \frac{b \cdot d}{|b|\cdot|d|} \cdot \frac{d}{|d|}\right)\right) \cdot \frac{b}{|b|} \right.\right. + \\
& \left.\left. \left(\left(a - c\right)\cdot \left(\frac{d}{|d|} - \frac{b \cdot d}{|b|\cdot|d|} \cdot \frac{b}{|b|}\right)\right) \cdot \frac{d}{|d|}
\right] \right\| = \frac{|(a - c) \cdot b \times d|}{|b \times d|}
\end{aligned}
\end{equation*}
It seems clear to me, that if you subtract the orthogonal projections of $a - c$ in
direction of $b$ and $d$, the result must be proportional to $b \times d$.
On the other hand, there must be a clear line of arguments, why
$\frac{|b|^2\cdot |d|^2}{|b \times d|^2}$ exactly does the job.
Long version. For notation and reference see e.g. here:
The difference vector between two of those skew lines is
$$D = |y - x| = |c - a + \mu d - \lambda b|$$
and the local extrema $(\lambda, \mu)$ is found to be at
$$\frac{1}{\mbox{det } A}
\left(\begin{matrix}
d^2 & d\cdot b \\
d \cdot b & b^2
\end{matrix}\right) \cdot \left(\begin{matrix}
(c-a) \cdot b \\
-(c-a) \cdot d
\end{matrix}\right)$$
where $$\mbox{det A} = b^2 d^2 - (d \cdot b)^2 = |b \times d|^2$$
If you insert above solution into $D$, you get
\begin{equation*}
\begin{aligned}
D = |c - a + \mu d - \lambda b|
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
D = \left\| \left(a - c\right) - \left[\vphantom{\frac{b}{b}}\right.\right.
& \frac{|d|^2}{|b \times d|^2} \left.\left. \left(\left(a - c\right)\cdot \left(b - \frac{b \cdot d}{|b|\cdot|d|^2} \cdot d\right)\right) \cdot b \right.\right.+\\
& \frac{|b|^2}{|b \times d|^2} \left.\left. \left(\left(a - c\right)\cdot \left(d - \frac{b \cdot d}{|b|^2\cdot|d|} \cdot b\right)\right) \cdot d
\right] \right\|
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
D = \left\| \left(a - c\right) - \frac{|b|^2\cdot |d|^2}{|b \times d|^2} \left[\vphantom{\frac{b}{b}}\right.\right.
& \left.\left. \left(\left(a - c\right)\cdot \left(\frac{b}{|b|} - \frac{b \cdot d}{|b|\cdot|d|} \cdot \frac{d}{|d|}\right)\right) \cdot \frac{b}{b} \right.\right.+\\
& \left.\left. \left(\left(a - c\right)\cdot \left(\frac{d}{|d|} - \frac{b \cdot d}{|b|\cdot|d|} \cdot \frac{b}{|b|}\right)\right) \cdot \frac{d}{d}
\right] \right\|
\end{aligned}
\end{equation*}
Then I don't see how to proceed...
As pointed out by @ancientmathematician, we could set $u=a-c$, and assume $|b|=|d|=1$ without loss of generality, then
\begin{equation*}
\begin{aligned}
D = \left\| u - \frac{1}{|b \times d|^2} \left[\vphantom{\frac{b}{b}}\right.\right.
& \left.\left. \left(u\cdot \left(b - (b \cdot d) \cdot d\right)\right) \cdot b \right.\right.+\\
& \left.\left. \left(u\cdot \left(d - (b \cdot d) \cdot b\right)\right) \cdot d \vphantom{\frac{b}{b}} \right] \right\|
\end{aligned}
\end{equation*}
and $|b \times d|^{-2}= \sin(\alpha)^{-2}$. But the occurence of the angle between $b$ and $d$ at that place is quite puzzling to me. Do I miss something obvious?
Initially I thought
\begin{equation*}
\begin{aligned}
D = \left\| u - \left[\right.\right.
\left.\left. \left(u\cdot \left(b - (b \cdot d) \cdot d\right)\right) \cdot b \right.\right.+
\left.\left. \left(u\cdot \left(d - (b \cdot d) \cdot b\right)\right) \cdot d
\right] \right\|
\end{aligned}
\end{equation*}
which is wrong...
| Without loss of generality, let $\ c=0,\ $ but we do not let $\ |b|=|d|=1\ $
because we would lose the valuable homogeneous property.
We recall the standard scalar
Triple product identity:
$$ \Delta := (a \cdot (b \times d))^2 = \begin{vmatrix} a\cdot a & a \cdot b & a\cdot d \\
b \cdot a & b\cdot b & b\cdot d \\
d \cdot a & d\cdot b & d\cdot d
\end{vmatrix}. \tag{1} $$
$$ \textrm{Let }\quad b_v := \frac{b}{|b|} \!-\! \frac{b\cdot d}{|b|\,|d|} \frac{d}{|d|},
\quad d_v := \frac{d}{|d|} \!-\! \frac{b\cdot d}{|b|\,|d|} \frac{b}{|b|},
\quad \textrm{ and} \tag{2} $$
$$ a_v := a\ |b\times d|^2 - |b|^2|d|^2 \Big(
a\cdot b_v\frac{b}{|b|} + a\cdot d_v\frac{d}{|d|}\Big),
\quad D := a_v/|b\times d|^2. \tag{3} $$
A bit of simplification of equation $(2)$ gives us
$$ |b|\ |d|^2\ b_v = b\ |d|^2 - d\ (b\cdot d), \quad
|b|^2\ |d|\ d_v = d\ |b|^2 - b\ (b\cdot d). \tag{4} $$
Substituting equation $(3)$ in equation $(4)$ gives us
$$ a_v = (a_0)a + (b_0)b + (d_0)d \quad \textrm{ with } \tag{5} $$
$$ a_0 :=|b\!\times\!d|^2,\quad
b_0 := -a\!\cdot\!\big(b\ |d|^2 \!-\! d\ (b\!\cdot\! d)\big),\quad
d_0 := -a\!\cdot\!\big(d\ |b|^2 \!-\! b\ (b\!\cdot\! d)\big). \tag{6} $$
Taking the dot product of equation $(5)$ we get
$$ a_v\!\cdot\!a_v = (a\!\cdot\!a)a_0^2 \!+\! (b\!\cdot\!b)b_0^2
\!+\! (d\!\cdot\!d)d_0^2 \!+\! 2(a\!\cdot\!b)a_0b_0 \!+\!
2(a\!\cdot\!d)a_0d_0 \!+\! 2(b\!\cdot\!d)b_0d_0. \tag{7}$$
Using $\ |b\!\times\!d|^2 = |b|^2|d|^2\!-\!(b\!\cdot\!d)^2,\ $
and equations $(1)$ and $(6)$ and factoring equation $(7)$ we get
$$ |a_v|^2 = a_v\cdot a_v = |b\times d|^2 \Delta =
|b\times d|^2 |a\cdot (b\times d)|^2 \tag{8}$$
Solving for $\ |a_v|\ $ and using equation $(3)$ we get
$$ |a_v| = |b\times d|\ |a\cdot b\times d|,\quad
|D| = |a\cdot b\times d|/|b\times d| \tag{9} $$
which is what we wanted to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$, $\lim_{n\to \infty}a_n$ is Given $a_1,a_2,n\in \mathbb N$
$$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$$
Then $\lim_{n\to \infty}a_n$ is
(A) $2(a_2-a_1)+a_1e^{-1}$
(B) $2(a_1-a_2)e^{-1}+a_2$
(C) $2(a_1-a_2)e^{-1}+a_1$
(D) $2(a_2-a_1)e^{-1}+a_1$
My attempt,
$a_1,a_2\in \mathbb N$
$$a_3=(1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1}$$
$$a_4=(1-\frac{1}{4})a_{3}+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1}$$
$$a_5=(1-\frac{1}{5})a_{4}+\frac{1}{5}a_{3}$$ $$=(1-\frac{1}{5})((1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1})+\frac{1}{5}((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})$$
But I am not able to conclude from here, I couldn't able to generalise anything from here.
| Multiplying both sides of the equation by $n$ and rearranging, you get that: $$n(a_n-a_{n-1}) = -(a_{n-1}-a_{n-2})$$
Setting $b_n = a_n-a_{n-1}, n\geq 2$ and $b_1 =a_1$, this becomes: $$b_n = \frac{-1}{n}b_{n-1} = ... = (-1)^n\frac{2}{n!}b_2 = (-1)^n\frac{2}{n!}(a_2-a_1),\text{ for } n \geq2.$$
So, $a_n = a_n-a_{n-1} + a_{n-1} - a_{n-2}+...-a_1+a_1 = \sum \limits_{i=1}^n b_i = a_1 + 2(a_2-a_1)\sum \limits_{i=2}^n \frac{(-1)^n}{n!}.$
It remains to calculate $\lim\limits_{n\rightarrow \infty}a_n = a_1+2(a_2-a_1)\sum \limits_{i=2}^\infty \frac{(-1)^n}{n!}.$ But this infinite series is the Taylor expansion of $e^x$ evaluated at $x=-1$.
So, the final answer is: $$\lim\limits_{n\rightarrow \infty}a_n = a_1+2(a_2-a_1)e^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How do i show$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$ without using integration? Without using integration show that $$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.
| If we let $a_n=\frac{[(n+1)(n+2)...(2n)]}{n^n}=\frac{(2n)!}{n!n^n}$, then $\frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=(a_n)^{\frac{1}{n}}$. Then $\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{(2n)!}×\frac{n!}{(n+1)!}×\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)}×\frac{n^n}{(n+1)^{n+1}}=\frac{2(2n+1)}{(n+1)}×\frac{n^n}{(n+1)^{n+1}}=\frac{4n(1+\frac{1}{2n})}{n(1+\frac{1}{n})}×(1+\frac{1}{n})^{-n}\rightarrow4×e^{-1}$.
Edit:explaining the last step explicitly... note that as $n\rightarrow \infty, \frac{1}{n},\frac{1}{2n}\rightarrow 0$, so we can replace $\frac{1}{2n}$ with $\frac{1}{n}$. Now hence, $\frac{4n(1+\frac{1}{2n})}{n(1+\frac{1}{n})}×(1+\frac{1}{n})^{-n}=4×(1+\frac{1}{n})^{-n} \rightarrow 4× e^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to do it regular way and I looked for the maximum/maximas finding partial derivatives and setting it to zero and I got $$
\begin{cases}
0=-4x^3+-2x+3x^2y-y^3 \\ 0=-4y^3-2y-3y^2x-x^3\\
\end{cases}
$$
easy to see (0,0) is a solution however I do not know how to prove that it is the only one
| divide by $x^2y^2$if $x\neq0$ and $y\neq0$
$$\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}+\frac{x}{y}+\frac{y}{x}\geq 0$$
$$(\frac{x}{y}+0.5)^2+(\frac{y}{x}+0.5)^2+\frac{1}{y^2}+\frac{1}{x^2}-0.5\geq 0$$
now if $x\geq y$ then $(\frac{x}{y}+0.5)^2>1$
or if $y\geq x$ then $(\frac{y}{x}+0.5)^2>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve $2x^2-5x+2=$ $\frac{5-\sqrt{9+8x}}{4}$ Solve $2x^2-5x+2$= $\frac{5-\sqrt{9+8x}}{4}$
I simply do square both sides solve it and I get two value of x one is 2 and other is $\frac{3-√5}{2}$ but this approach it take more time so is there any approach for solving this equation.
| Taking Jerry Chang's observation (which amounts to the fact that the expression on the right-hand side is what you get when you plug the coefficients of the quadratic into the quadratic formula, just choosing the minus sign for the square root) and setting $y=2x^2-5x+2$ so that $x=2y^2-5y+2$ we can subtract one of these from the other to obtain:
$$y-x=2(x^2-y^2)-5(x-y)$$
Which yields $y=x$ ; or
$1=5-2(x+y)$ ie $x+y=2$
Then the problem splits as $2x^2-5x+2=x$ or $2x^2-5x+2=2-x$
The solutions of these equations have to be plugged back into the original for checking to see which belongs to which choice of sign of the square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}*\frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
| Here is an alternate approach which avoids working with fractions
\begin{eqnarray}
ax^2+bx+c&=&0\\\
&&\\\
\text{subtract }c&&\\\
ax^2+bx&=&-c\\\
&&\\\
\text{multiply by }4a&&\\\
4a^2x^2+4abx&=&-4ac\\\
&&\\\
\text{add }b^2&&\\\
4a^2x^2+4abx+b^2&=&b^2-4ac\\\
&&\\\
\text{factor the left side}&&\\\
(2ax+b)^2&=&b^2-4ac\\\
&&\\\
\text{take square root of both sides}&&\\\
2ax+b&=&\pm\sqrt{b^2-4ac}\\\
&&\\\
\text{solve for }x&&\\\
x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Eliminating $\theta$ from the system $x\sin\theta-y\cos\theta=-\sin4\theta$, $x\cos\theta+y\sin\theta=\frac52-\frac32\cos4\theta$
Eliminate $\theta$ from the system of equations.
$$\begin{align}
x\sin\theta-y\cos\theta&=\phantom{\frac52\frac32}-\sin4\theta \\
x\cos\theta+y\sin\theta&=\frac52-\frac32\cos4\theta
\end{align}$$
I am stuck at this question after squaring and adding.
| My believe that one of the coefficients of $x,y$ in Eliminating $\theta$ from trigonometric system will be negative has been corroborated by the current question.
Solving for $x,y$
$$\dfrac x{-\cos t(3\cos4t-5)-2\sin t\sin4t}=\dfrac y{2\cos t\sin4t-\sin t(3\cos4t-5)}=\dfrac12$$
$$4x=10\cos t-5\cos3t-\cos5t\iff x=5c-4c^5$$
$$4y=-\sin5t+5\sin3t+10\sin t\iff y=5s-4s^5$$ where $c=\cos t,s=\sin t$ using this and this
$$x^2+y^2=25-40(1-2c^2s^2)+16(1-3c^2s^2)$$
$$=1+32c^2s^2=1+8\sin^22t=1+4(1-\cos4t)=5-4\cos4t$$
$$\cos4t=\dfrac{5-x^2-y^2}4\ \ \ \ (1)$$
Again squaring & adding $$4(x^2+y^2)=4\sin^24t+(5-3\cos4t)^2=25+4(1-\cos^24t)+9\cos^24t-30\cos4t\ \ \ \ (2)$$
Replace the value of $\cos4t$ in $(2)$ from $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the generator $(g)$ that generates $(29, \sqrt{-5} ± 13)$
Find the generator $(g)$ that generates $(29, \sqrt{-5} + 13)$. The ring is $\mathbb{Z}\left[\sqrt{-5}\right]$.
The fact I used was that $\text{Norm}(g)$ must divide both 29 and $\text{Norm}(\sqrt{5} + 13)$, which means solving the Pell equation $x^2 + 5y^2 = 29$. The solutions to this Pell equation are $(\pm 3, \pm 2)$. I don't know what to do with the $\pm$ signs so let's just refer to $(3, 2)$ for simplicity.
Because ideals are subsets, proving ideal equality is proving two subset inclusions.
Show that $(29, \sqrt{-5} + 13) \subseteq (3 + 2\sqrt{-5})$ and that $(3 + 2\sqrt{-5}) \subseteq (29, \sqrt{-5} ± 13)$. (Because I do not know what to do with the signage, this may not be true.
How do I continue?
| To show that $(29, 13+\sqrt{-5}) \subseteq (g)$ we have to find integers $x,y,z,t$ such that $g(x + y\sqrt{-5})=29$ and $g(z + t\sqrt{-5})=13+\sqrt{-5}$.
If $g=3 + 2\sqrt{-5}$ then
$\cases{3z-10t=13\\
2z+3t=1}.$
That is $z=\tfrac{49}{29}$ and $t=-\tfrac {23}{29}$, which is impossible.
Let's try $g=3 - 2\sqrt{-5}$ (the other sign choices are reduced to considered by multiplication of $g$ by $-1$). Then
$\cases{3x+10y=29\\
-2x+3y=0}$
$\cases{3z+10t=13\\
-2z+3t=1}$
That is $x=3$, $y=2$, $z=1$, and $t=1$.
To show that $(3 - 2\sqrt{-5}) \subseteq (29, 13+\sqrt{-5})$ we have to find integers $x,y,z,t$ such that $$(x + y\sqrt{-5})29+(13+\sqrt{-5})(z + t\sqrt{-5})=3-2\sqrt{-5}.$$ That is
$\cases{29x+13z-5t=3\\
29y+z+13t=-2}$
$-13z=29x-5t-3=13(29y+13t+2).$
If $y=0$ then $29(x-1)=174t$ that is $x-1=6t$. Thus we can put $x=1$, $y=0$, $z=-2$, and $t=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $2x^2-3xy+y^2=0$ is transitive and anti-symmetric or symmetric and reflexive. Let $R$ be the binary relation defined on $\mathbb{R}$ by $xRy$ iff $2x^2-3xy+y^2=0$
For reflexive we get $2x^2=2x^2\implies-x=x$ which means reflexive on $xRx$
$2x^2-3xy+y^2=0$ tried going for $2y^2-yz+z^2=0$ then adding them together but now I'm stuck with a long useless equation any tips of proving this transitive as for anti symmetric i know $2y^2-3yx+x^2=0$ in case $x=y$ so it should be anti symmetric but I don't know how to say it.
| $\text{R}$ is a binary relation on $\mathbb{R}$ defined as $x\text{R}y\iff 2x^2-3xy+y^2=0$
Reflexive:
If $x=y$, then $2x^2-3xy+y^2=2x^2-3x^2+x^2= 0$
Hence, we have $x\text{R}x\ \ \forall \ x \in\mathbb{R}$ and the relation is Reflexive
Now using a technique called Completing the Square we have:
$$\begin{align}x\text{R}y &\iff 2x^2-3xy+y^2=0\\
&\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}=0\\
&\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}+\big(\frac{3y}{4}\big)^2-\big(\frac{3y}{4}\big)^2=0\\
&\iff\big(x-\frac{3y}{4}\big)^2-\big(\frac{y}{4}\big)^2=0\\
&\iff \big(x-\frac{y}{2}\big)\big(x-y\big)=0\\
&\iff x=\frac{y}{2} \ \ \text{Or} \ \ x=y \ \ \ \ \ \ \ \ \ \ \ \ -(1)
\end{align}$$
Antisymmetric:
We will proceed by contradiction. Let us assume $\text{R}$ is not antisymmetric. Then $\exists \ x,y\in\mathbb{R}$ such that $x\neq y$ and both $x\text{R}y$, $y\text{R}x$ are satisfied. Since $x\neq y$, by $(1)$ we have the following $$x\text{R}y\Rightarrow x=\frac{y}{2}$$ $$y\text{R}x\Rightarrow y=\frac{x}{2}$$
Thus,
$$x= y=0$$
Which leads to a contradiction. Hence, $\text{R}$ is antisymmetric.
Transitive:
The relation is not transitive because $3\text{R}6$ and $6\text{R}12$ but $3\text{R}12$ does not hold.
Symmetric:
The relation is not symmetric because $6\text{R}12$ but $12\text{R}6$ does not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Invertibility of a matrix in portfolio optimization Let $A$ be an $n\times n$ symmetric matrix with non-negative entries. Let $\mathbf{1}$ be the column vector of dimension $n$ with all entries being $1$.
Consider the $(n+1)\times (n+1)$ matrix
$$ B=
\begin{bmatrix}
A & \mathbf{1} \\
\mathbf{1}^T & 0
\end{bmatrix}
$$
Question: what is the condition for $A$ so that $B$ is invertible?
Remark: This matrix is related to portfolio optimization problems in finance. I note that when $A$ is a constant matrix, the determinant of $B$ is $0$ and thus $B$ is not invertible.
| If $\det(A) \neq 0$ so that $A^{-1}$ exists
and the scalar $\alpha = \mathbf{1}^T A^{-1} \mathbf{1} \neq 0$, then we have
$$B^{-1} = \begin{bmatrix}
A^{-1} - \alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1} & \alpha^{-1}A^{-1}\mathbf{1} \\
\alpha^{-1}\mathbf{1}^TA^{-1} & -\alpha^{-1}
\end{bmatrix}$$
Note that
$$\begin{align}BB^{-1}&= \begin{bmatrix}
AA^{-1} - A\alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1}+ \mathbf{1}\alpha^{-1}\mathbf{1}^TA^{-1} & A\alpha^{-1}A^{-1}\mathbf{1}-\alpha^{-1}\mathbf{1} \\
\mathbf{1}^TA^{-1} - \mathbf{1}^T\alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1} + 0\alpha^{-1}A^{-1}\mathbf{1} & \mathbf{1}^T\alpha^{-1}A^{-1}\mathbf{1}-0\alpha^{-1}
\end{bmatrix} \\ \\&= \begin{bmatrix}
I -\alpha^{-1}\mathbf{1}\mathbf{1}^TA^{-1}+ \alpha^{-1}\mathbf{1}\mathbf{1}^TA^{-1} & \alpha^{-1}\mathbf{1}-\alpha^{-1}\mathbf{1} \\
\mathbf{1}^TA^{-1} - \alpha^{-1}\alpha\mathbf{1}^TA^{-1} & \alpha^{-1}\alpha \end{bmatrix}\\ \\ &= \begin{bmatrix}
I & \mathbf{0} \\
\mathbf{0}^T & 1 \end{bmatrix}\end{align} $$
Addendum
In general, for a block matrix
$$B = \begin{bmatrix}
A & C \\
E & D
\end{bmatrix},$$
if $A^{-1}$ exists , then the Schur complement
of $A$ is $D- EA^{-1}C$ and
$$\det(B) = \det(A) \det(D- EA^{-1}C)$$
Thus, $\det(B) \neq 0$ and $B^{-1}$ exists if and only if $\det(D- EA^{-1}C) \neq 0$.
In this case, the Schur complement reduces to a scalar $-\mathbf{1}^TA^{-1} \mathbf{1}$, and the condition $\mathbf{1}^TA^{-1}\mathbf{1} \neq 0$ is necessary and sufficient for $B$ to be invertible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Comparing two variable expressions quickly Let us say $S_1=2xy^2+3xy$, and $S_2=3y^2+7x+7y+8$. Then, can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?
I think yes, but the usual quadratic function method is taking too much time, although I found that the discriminant of the final quadratic is $\ge0$ as I assume $x\ge5$. Actually, I am trying to prove $$4\left\lfloor\frac{xy}{2(y+1)}\right\rfloor\ge \left\lceil\frac{x}{y}\right\rceil\left(\frac{y-1}{2}\right)+\frac{x}{2}+y$$. Any other to prove this, or, is the inequality wrong? Any hints? Thanks beforehand.
|
can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?
No, we cannot.
If $y=1$, then $S_1-S_2=-2x-18\lt 0$ for $x\ge 3$.
If $y=2$, then $$S_1-S_2=7x-34\ \begin{cases}\lt 0&\text{for $x=4$}\\\\ \gt 0&\text{for $x\ge 5$}\end{cases}$$
If $y\ge 3$, then we have $2y^2+3y-7\gt 0$, so
$$\begin{align}S_1-S_2&=x (2 y^2 + 3 y - 7) - 3 y^2 - 7 y - 8
\\\\&\ge (y+2)(2y^2+3y-7)-3y^2-7y-8
\\\\&=2\{(y^3-11)+2y(y-2)\}\gt 0\end{align}$$
The inequality
$$4\left\lfloor\frac{xy}{2(y+1)}\right\rfloor\ge \left\lceil\frac{x}{y}\right\rceil\left(\frac{y-1}{2}\right)+\frac{x}{2}+y$$
does not hold when $(x,y)=(5,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I should show that
$$\frac{x}{\sin{x}} \le \frac{3x}{2} $$
but last inequality is not true...
| $$(f=)\ \frac{x}{\sin\ x} = \frac{x}{x-x^3/6+\cdots }
=\frac{1}{1-x^2/6+\cdots }=1+Cx^2+\cdots ,\ C>0$$ so that
$\frac{x}{\sin\ x}$ is a positive increasing convex function on
$[{\pi/6}, {\pi/2}]$
By drawing a graph of $f$, the integral $\int\ f(x)\ dx$ is smaller than an area of
trapezoid i.e. $$ \int\ f(x)\ dx \leq \frac{ f( \pi/6 ) + f(\pi/2)
}{2}\cdot [ -\pi/6 + \pi/2 ] = \frac{5\pi^2}{36} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that for all $n \in \mathbb{N}$, either 3 or 13 divides $3^n + 13n^2 + 38$ Let $a\in \{3,13\}.$ I'm having trouble with this proof. I know that
$$3^{n+1} + 13(n+1)^2 + 38 = (3^n + 13n^2 + 38) + (2\cdot 3^n + 26n + 13)$$
But I can't prove that $a \mid 2\cdot3^n + 26n + 13$. I know that 13 doesn't divide this because $13 \nmid 2\cdot3^n$. How can I prove that $3 \mid 26n + 13$?
| If $n\equiv1$ or $2\pmod3$ then $3^n+13n^2+38\equiv n^2+2\equiv0 \pmod 3$
because $3^n\equiv0$ and $13\equiv1$ and $38\equiv2 \pmod 3$, so $3|3^n+13n^2+38$.
If $n\equiv0\pmod3$ then $3^n+13n^2+38\equiv 1+12\equiv0\pmod { 13}$
because $3^3=27\equiv1 $ and $13\equiv0$ and $38\equiv12 \pmod {13}$, so $13|3^n+13n^2+38$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}{8 a^{3/2}} + \frac{x^3}{16 a^{5/2}} - \frac{5 x^4}{128 a^{7/2}} + \frac{7 x^5}{256 a^{9/2}} + O(x^6)$$
which (I think) equals:
$$\sqrt{a} \left( \frac{1}{2} \left(\frac{x}{a}\right) - \frac{1}{8} \left(\frac{x}{a}\right)^2 + \frac{1}{16} \left(\frac{x}{a}\right)^3 - \frac{5}{128} \left(\frac{x}{a}\right)^4 + \frac{7}{256} \left(\frac{x}{a}\right)^5 + O\left(\left(\frac{x}{a}\right)^6\right) \right)$$
How quickly do the coeffients decrease?
How many terms are needed to reach $53$ bits of accuracy (IEEE double precision) in the result given that $10^{-300} < \left|\frac{x}{a}\right| < 1$ is known?
Alternatively, what are the threshold values of $\left|\frac{x}{a}\right|$ where the number of terms changes?
What about rounding errors, assuming each value is stored in double precision?
| The Taylor series is
$$ \sqrt{a+x} - \sqrt{a} = \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k)!}{(k!)^2(2k-1)} 4^{-k} a^{1/2-k} x^k$$
If $|x/a| < 1$, the absolute values of the terms decrease, since if $c_k = (2k)!/((k!)^2 (2k-1) 4^k)$,
$$ \frac{c_{k+1}}{c_k} = \frac{2k-1}{2k+2} < 1$$
Thus if $a > x > 0$ the absolute value of the error is always less than that of the next term.
However, if $x/a$ is close to $1$ the convergence is rather slow:
$$c_k \sim \frac{1}{2 \sqrt{\pi} k^{3/2}}$$
so that won't be less than $2^{-53}$ unless $k > 1.862 \times 10^{10}$ approximately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime number and square problem How many pairs of natural numbers, not bigger than 100, are such that difference between that pair is a prime number, and their product is a square of a natural number.
My attempt: I tried writing relationship such as $x-y=p$ and $xy=n^2$ but I can't seem to find any pattern to enumerate it.
| If $x - y$ is prime, then there can be at most one number that divides both $x$ and $y$, and it has to be that prime. Suppose the prime $p$ divides both $x, y$. Thus we can write $x = np$, and $y = (n-1)p$. Then we would have that $n(n-1)p^2$ is a square, hence so is $n(n-1)$. But $n, n-1$ are coprime, so this would mean that both $n, n-1$ are squares. This is impossible, so $p$ does not divide $x$ or $y$. It follows that $x, y$ are coprime.
Because $xy$ is also a square, it follows that $x, y$ must be squares individually -- so write $x = a^2, y = b^2$. Now a square is a sum of odd numbers,
$$
a^2 = \sum_{k=1}^a 2k-1,
$$
so
$$
a^2 - b^2 = \sum_{k=b+1}^a 2k-1.
$$
But we run into a restriction: if we have a sequence of odd numbers of more than one term, say $5, 7, 9$ or $25, 27$, then they are a multiple of their average. The first sum equals $3 \times 7$ and the second one $2 \times 26$. Thus we must have that the sum consists of at most one term, and we have $a = b+1$, and the sum is prime if and only if $2a-1$ is prime.
In the end your numbers are the $x, y$ such that $x = \left(\frac{p+1}{2}\right)^2$ and $y = \left(\frac{p-1}{2}\right)^2$, where $p$ is an odd prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\int_{0}^{\pi}\frac{1}{1 + \frac{1}{\sec^2x}}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{\sec^2x + 1}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{2 + \tan^2x}dx
$$
Let $\tan x = u$, then $du = \sec^2x dx$
Then,
$$\int \frac{\sec^2x}{2+\tan^2x}dx = \int \frac{du}{2 + u^2} = \frac{1}{\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}+c = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\tan x}{\sqrt{2}}+c
$$
Therefore,
$$
2I = \frac{\pi}{\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\
\Rightarrow I = \frac{\pi}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\= \frac{π}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan \pi}{\sqrt{2}} - \tan^{-1}\frac{\tan 0}{\sqrt{2}}\right]\\=\frac{\pi}{2\sqrt{2}}[\tan^{-1}0 - \tan^{-1}0] \\= 0\\$$
But the answer given in the book is $\frac{\pi^2}{2\sqrt{2}}$
What am I doing wrong ?
| Hint For a slightly more theoretically demanding solution you can start out like this:
Let $$h(x) = f(x)\cdot g(x)\\ f(x) = x\\ g(x) = \frac{1}{1+\cos(x)^2}$$
Now we seek $$\int_0^{\pi}h(x)dx$$
But with Fourier analysis we know:
$$\hat h(0) = \int h(x)dx$$
And furthermore we know $${\hat {(f\cdot g)}} = \hat f * \hat g$$
Your integral can be viewed as DC component (frequency 0 component of Fourier transform) of product between on interval $x\in [0,\pi]$.
These functions $\hat f, \hat g$ should be raaather nice to describe in Fourier domain and we can leave the rest as an exercise for the curious student.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
Linear dependence of 3 vectors in $\mathbb{R}^4$ Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are linearly dependent on $\mathbb{R}^4$? I know that the usual method of solving this is to show that there exists scalars $x_1,x_2,x_3$ not all zero such that
\begin{align}
x_1\vec{v_1}+x_2\vec{v_2}+x_3\vec{v_3}=\vec{0}.
\end{align}
Doing this would naturally lead us to the augmented matrix
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
1& b & 1 &0\\
-2 & 2 & c &0\\
\end{pmatrix}
Doing some row reduction would lead us to the matrix
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
I'm not quite sure how to proceed after this. Do I take cases on when whether $b+1$ or $c+2$ are zero and nonzero?
| The matrix
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
$$
is not in row echelon form. Sum to the second column the first multiplied by $-4$, getting
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
0 & a+4 & -3 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
$$
Still not row echelon form, but “almost”. Anyway, computing the rank is easy.
If $b+1\ne0$, you can swap the second and third row, then continue the Gaussian elimination getting
$$
\begin{pmatrix}
1 & -1 & 1 & 0\\
0 & 1 & 0 & 0\\
0& 0 & 1 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
$$
which has rank $3$. Thus a necessary condition so the vectors are linearly dependent is $b=-1$. Now the matrix becomes
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
0 & a+4 & -3 &0\\
0& 0 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
$$
If $a+4\ne0$ and $c+2\ne0$, a row echelon form is (swapping the third and fourth rows)
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
0 & a+4 & -3 &0\\
0 & 0 & c+2 &0\\
0& 0 & 0 &0\\
\end{pmatrix}
$$
so the matrix has rank $3$. In order the rank is less than $3$ we need either $a=-4$ or $c=-2$.
You can so state that the vectors are linearly dependent if and only if $b=-1$ and either $a=-4$ or $c=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explained in the textbook. The long division I need to do is this one:
$$\frac{3x^3-x^2-13x-13}{x^2-x-6}$$
The book states that the result is:
$$3x+2+\frac{7x-1}{x^2-x-6}$$
But the mechanics of the long division are never explained. Can I please get some help with this one?
| Polynomial long division is exactly same as long division of decimals.
With respect to your posed problem here are steps.
Step1: We have to divide $3x^3-x^2-\dots$ with $x^2$, so we first multiply the divisor $(x^2-x-6)$ by $3x$ to give $3x^3$. THen we subtract and copy the remaining terms.
Step2: Again now we have $2x^2+5x-13$ and we have to divide by $x^2-x-6$. So we multiply our divisor by $2$.
Finally, we get $3x+2$ as dividend and $7x-1$ as remainder. Now recall if we had $\frac{17}{3}$ we would write it as $5+\frac{2}{3}$. Here $\bf{2}$ is remainder.
Similarly we write our result of long division as $$3x+2+\frac{7x-1}{x^2-x-6}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Evaluate the Limit Without L'Hopital Rule Let $\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$, find the value of $\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$ for $a\neq0$.
Using L'Hopital rule I found that the answer is $a^4-aL$. My question is how to solve this problem without using L'Hopital rule.
Here's my attempt using L'Hopital.
$$
\begin{aligned}
\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}&=L\\
\lim_{x\to a}\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2&=-L...(1)\\
\end{aligned}
$$
Let the numerator equal to zero.
$$
(x^2-x\left | x \right |-a^2)^2=a^4...(2)
$$
Ergo
$$\begin{aligned}
\lim_{x\to a}\frac{(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}&=\lim_{x\to a}(x^2-x\left | x \right |-a^2)^2+x\cdot\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2\\
&=a^4-aL
\end{aligned}
$$
| Case 1: $a>0$. Since $x \to a$, we can assume that $x>0.$ Then show that $\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=0.$ Hence $L=0.$
Case 2: $a<0$. Since $x \to a$, we can assume that $x<0.$ Then show that $\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=-4x^2(x+a).$ Hence $L=-8a^3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that hat if $a,b \ge 2$ then $ab \ge a+b$ Prove that if $a,b \ge 2$ then $a+b \le ab$
so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$
$(a-1)(b-1) \ge b-1$
$(a-1)(b-1) \ge 1$
$(a-1)(b-1) - 1 \ge 0$
$ab -a -b\ge 0$
$ab \ge a + b$
Thanks in advance, is this valid?
| Let $a\equiv x+ 2,\,b\equiv y+ 2$
$$\therefore\,ab- a- b= xy+ x+ y\geqq 0$$
$$\because\,x,\,y\geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculus of $\sum_{n=1}^{\infty}( \frac{1}{2^n})$, $n$ is even I want to prove that $\sum_{n=1}^{\infty}( \frac{1}{2^n}) = \frac{1}{3}$, where $n$ can only be an even integer.
That is, how can I use linear algebra to prove that $\frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \frac{1}{2^8} + \cdots = \frac{1}{3}$?
I heard some people saying that linear algebra help them to solve the aforementioned problem. Can anyone explain to me in detail how to get the solution $\frac{1}{3}$?
Any help is very appreciated.
| Use the fact that the set of all even (non-zero) integers is $2\mathbb N^*$. This gives the following sum to compute
$$\sum_{k=1}^{\infty} \dfrac{1}{2^{2k}}=\sum_{k=1}^{\infty}\dfrac{1}{4^k}=\dfrac{1}{4}\dfrac{1}{1-1/4}=\dfrac{1}{4}\dfrac{4}{3}=\dfrac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving a problem with generating functions This is a problem from a course of MIT.
Find the coefficients of the power series $y = 1 + 3 x + 15 x^2 + 184 x^3 + 495 x^4 + \cdots $ satisfying
$$ (27 x - 4)y^3 + 3y + 1 = 0 . $$
This is an interesting problem, but I am really cluelss (My naive approaches all failed). Can anyone give me a hint? Or an answer?
| The given coefficients of the power series
\begin{align*}
y(x)=1+3x+15x^2+\color{blue}{84}x^3+495x^4+\cdots
\end{align*}
indicate they are $\binom{3n}{n}$ resulting in
\begin{align*}
y(x)=\sum_{n=0}^\infty \binom{3n}{n}x^n \tag{1}
\end{align*}
We prove (1) by showing it fulfils the functional equation
\begin{align*}
(27 x-4)y^3+3y+1=0\tag{2}
\end{align*}
We do so by emplyoing the Lagrange inversion following the paper Lagrange Inversion: when and how by R. Sprugnoli (et al.). We also use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We note the coefficients $\binom{3n}{n}$ from (1) are the coefficients of
\begin{align*}
\binom{3n}{n}=[x^n](1+x)^{3n}\tag{3}
\end{align*}
Let us suppose that a formal power series $w=w(x)$ is implicitely defined by a relation $w=x\Phi(w)$, where $\Phi(x)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:
$$[x^n]w(x)^k=\frac{k}{n}[x^{n-k}]\Phi(x)^n$$
There are several variations of the LIF stated in the paper. We use in the following formula $G6$ (with $F(x)=1)$:
Let $w=x\Phi(w)$ as before, then the following is valid:
\begin{align*}
[x^n]\Phi(x)^n=\left[\left.\frac{1}{1-x\Phi'(w)}\right|w=x\Phi(w)\right]\tag{4}
\end{align*}
The notation $[\left.f(w)\right|w=g(x)]$ is a linearization of $\left.f(w)\right|_{w=g(x)}$ and denotes the substitution of $g(x)$ to each occurrence of $w$ in $f(w)$ (that is, $f(g(x))$). In particular, $w=x\Phi(w)$ is to be solved in $w=w(x)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.
We obtain from (4) with $\Phi(x)=(1+x)^3$ and using
\begin{align*}
x\Phi^{\prime}(w)=3x(1+w)^2=\frac{3x\Phi(w)}{1+w}=\frac{3w}{1+w}
\end{align*}
\begin{align*}
\binom{3n}{n}&=[x^n](1+x)^{3n}\\
&=[x^n]\left[\left.\frac{1}{1-x\Phi^{\prime}(w)}\right|w=x\Phi(w)\right]\\
&=[x^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=x\Phi(w)\right]\\
&=[x^n]\left[\left.\frac{1+w}{1-2w}\right|w=x\Phi(w)\right]\tag{5}\\
\end{align*}
It follows from (5)
\begin{align*}
y(x)=\sum_{n=0}^\infty\binom{3n}{n}x^n=\frac{1+w(x)}{1-2w(x)}
\qquad\text{resp.}\qquad w(x)=\frac{y(x)-1}{2y(x)+1}
\end{align*}
Since $w=x\Phi(w)=x(1+w)^3$, we finally obtain
\begin{align*}
\frac{y(x)-1}{2y(x)+1)}=x\left(1+\frac{y(x)-1}{2y(x)+1}\right)^3
\end{align*}
from which
\begin{align*}
\color{blue}{(27x-4)y(x)^3+3y(x)+1=0}
\end{align*}
follows, showing the relationship (2) is fulfilled by (1).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $ $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $.
Exam problem which stayed unproven for me. I tried induction/binomial properties together and separately.I also tried to find a power serie which could use these binomials. Could someone provide at least a hint?
| Induction always works for hypergeometric sum identities.
$r=0$: $$\sum_{k=0} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$$
$r > 0$: $$\begin{eqnarray}
\textrm{LHS} &=& \binom{m+n}{n} \binom{m+n+(r-1)+1}{m+n+1} + \binom{n+r}{r} \binom{m+n+r}{n+r} \\
&=& \binom{m+n}{n} \frac{(m+n+r)!}{(m+n+1)!(r-1)!} + \frac{(m+n+r)!}{n!r!m!} \\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!}{(m+n+1)!(r-1)!} + \frac{(m+n+r)!}{(m+n)!r!} \right] \\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!r + (m+n+r)!(m+n+1)}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!(r+m+n+1)}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r+1)!}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}
\end{eqnarray}$$
Alternatively, for a combinatorial proof, both sides count the number of ways of colouring $m+n+r$ objects in a line such that $m$ are red, $n$ are blue, the rest are green or yellow, and all the yellow ones are together at the end of the line.
On the LHS, $k$ is the number of greens and the term inside the sum is a simple multinomial.
On the RHS, we take $m+n+r+1$ slots and choose $m+n+1$ of them. Colour the slots to the right of the last chosen one yellow, and the unchosen slots to its left green. Then delete that slot. From the other $m+n$ chosen slots, choose $m$ to colour red.
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of $f(x,y) = {\sqrt{x+y-1 \over x-y+1}}$ How I can get the domain of the function
$$ f(x,y) = \sqrt{x+y-1 \over x-y+1}?$$
I know that is: $x+1 \neq y$ and $x^2 \ge (y-1)^2$
But I don't know how to get the second condition.
| We need $\dfrac{x+y-1}{x-y+1} \ge 0$.
1) $x+y-1 \ge 0$, and $x-y+1 >0$
$(x+(y-1)) \cdot (x-(y-1)) \ge 0$;
$x^2 \ge (y-1)^2$;
2) $x+y-1\le 0$, and $x -y +1 <0;$
Proceed likewise, note that multiplying
$(x+y-1) \le 0$ by $(x-y+1) <0$
changes the sign of the inequality.
$(x +(y-1))(x-(y-1)) \ge 0.$
Same result as in 1).
| {
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Prove or disprove: If $a+b \leq \frac{1}{2}$, then $\frac{(1-a)(1-b)}{ab} \geq 1$ for positive $a,b$
Let $a,b$ be two positive numbers. Prove or disprove the statement:
If $a+b \leq \frac{1}{2}$, then $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$.
True. Assume $a+b \leq \frac{1}{2}$. Then
$$\dfrac{1-a}{a} \dfrac{1-b}{b}=\dfrac{1-b-a+ab}{ab}=\dfrac{1}{ab}-\dfrac{a+b}{ab}+1=-\dfrac{a+b}{ab}+\dfrac{1}{ab}+1\geq \dfrac{-1}{ab}+\dfrac{1}{ab}+1=1. $$
Can you check my answer?
| You have succeeded in proving that $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$
if $a,b>0$ and $a+b \leq 1.$
Therefore, since $\frac12<1$, certainly $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$ if $a+b\leq\dfrac12.$
Here is another way to write the proof:
$\dfrac{1-a}a\dfrac{1-b}b\ge1\iff(1-a)(1-b)\ge ab\iff 1-a-b\ge0\iff1\ge a+b.$
| {
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find all values of y such that $(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$ One of the questions on my Algebra 1 homework was to find all values of $y$ such that $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$$
This was listed under "Lecture 13: Quadratic Equations"
I can not figure out how to approach this problem? Should I expand it? I also noticed that one of the terms used $x$.
| $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0 \\ (y-2)(y+3)(x^2-6y+9)-2(y-3)(y+3)=0 \\ (y-2)(y+3)(x^2-6y+9)=2(y-3)(y+3) \\ (y-2)(x^2-6y+9)=2(y-3) \\ x^2-6y+9=\frac{2(y-3)}{y-2} \\ x^2=\frac{2(y-3)}{y-2}+6y-9 \\ x^2= \frac{2(y-3)}{y-2}+\frac{6y(y-2)}{y-2}-\frac{9(y-2)}{y-2} \\x^2= \frac{2(y-3)+6y(y-2)-9(y-2)}{y-2} \\ x^2=\frac{2y-6+6y^2-12y-9y+18}{y-2}=\frac{6y^2-19y+12}{y-2} \\ x^2=\frac{6y^2-19y+12}{y-2}$$
That's the simplest form for the equation you've written above; However there isn't any unique root (solution) for the equation; as the equation has two variables and needs two equations to find unique $(x,y)$, so solving for $x$ and $y$ can't be; because there isn't enough given. The equation above is said to be a "Relation between $x$ and $y$" Unless the $x$ in the equation become $y$.
Graphed using: Geogebra.
| {
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Proof by induction when the first element is defined in an interval Given that, for a sequence $x_n$:
*
*$1<x_1<2$
*$x_{n+1} = 1 + x_n - \frac{1}{2}(x_n)^2$ for $n \ge 2$
Prove that, for $n\ge3$, $|x_n - \sqrt2| < \frac{1}{2^n}$
So, at first I tried to verify the base case by first calculating $x_2$.
Thus, I got $|x_2 - \frac{3}{2}| < \frac{1}{2}$
But surprisingly, for $x_3$:
$\frac{1}{2} < |x_3 - \frac{3}{2}| < 2$
I wonder where does that $\sqrt2$ comes from.
Later, while working on the induction step, I've found that
$x_{k+2} = 2 - \frac{1}{2}x_{k+1}$
It seems that I'm stuck on this problem but close to the actual proof.
Could you guys give me any elucidation?
| This problem can be solved by letting $a_n = x_n - \sqrt 2$. Rewriting the condition, we obtain $a_{n+1} + \sqrt 2 = 1 + a_n + \sqrt 2 - \frac{1}{2}(a_n + \sqrt 2)^2$, and expanding and rearranging gives $a_{n+1} = (1-\sqrt 2)a_n - \frac{1}{2} a_n^2 = (1- \sqrt 2 - \frac{a_n}{2})a_n$.
Now if $|a_n| < 3 - 2 \sqrt 2$, we have $1- \sqrt 2 - \frac{a_n}{2}> 1 - \sqrt 2 - \frac{3}{2} + \sqrt 2= \frac{-1}{2}$. Similarly, we have $1- \sqrt 2 - \frac{a_n}{2} < 1 - \sqrt 2 + \frac{3}{2} - \sqrt 2 = \frac{5}{2} - 2 \sqrt 2 < 0$.
So now, using $a_{n+1} = (1- \sqrt 2 - \frac{a_n}{2})a_n$, we get that $|a_{n+1}| < \frac{1}{2} |a_n|$. Note that as soon as $|a_n| < 3 - 2 \sqrt 2$ holds for some $n$, it holds for all later $n$ as the absolute value is at least halved each time. So now all that remains is to prove the base case $a_3$; this is a fairly straightforward computation.
| {
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For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equiv 3\pmod 4$
Let $n = k+1$ such that
$${(k+1)^2 + 2(k+1)}\equiv {k^2 + 2k + 1 + 2k + 2}\pmod 4$$
Then I substitute $k^2 + 2k$ with both $0$ or $3$ from the earlier assumption.
So we have
$$(0 + 2k + 3) \equiv (2k + 3) \pmod 4$$
Or,
$$(3 + 2k + 3) \equiv (2k + 6) \pmod 4$$
Where do I go from here?
| $$n^2+2n=(n+1)^2-1$$
Now $n+1\equiv0,\pm1,2\pmod4$
$\implies(n+1)^2\equiv0,1\pmod4$
$(n+1)^2-1\equiv?,?$
| {
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as perfect squares to get the least value? a Hint is much appreciated. Thanks!
| It is $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6$$ and now use that $$x+\frac{1}{x}\geq 2$$ for all $$x>0$$
You can also use that $$\frac{a^2 b+ab^2+b^2c+c^2b+a^2c+ac^2}{6}\geq \sqrt[6]{(abc)^6}=abc$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of the sequence $x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$
$\left\{x_n\right\}$ is a convergent sequence where $x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$.
What is $\lim\limits_{n\to\infty}x_n?$
Here are my two approaches:
Using Euler's constant($\gamma$):
$$\begin{align}\lim_{n\rightarrow\infty}x_n
&=\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n})\\
&=\lim_{n\rightarrow\infty}[(1+\frac{1}{2}+\dots+\frac{1}{2n})-(1+\frac{1}{2}+...+\frac{1}{n})]\\
&=\lim_{n\rightarrow\infty}[(\gamma_{2n}+\log 2n)-(\gamma_n+\log n)]\\
&=\lim_{n\rightarrow\infty}(\gamma_{2n}-\gamma_n+\log 2)\\
&=\log(2)
\end{align}$$
because $\lim_{n\rightarrow\infty}\gamma_{2n}=\gamma$ and $\lim_{n\rightarrow\infty}\gamma_n=\gamma$.
2nd method:
$$\begin{align}x_n&=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\\
&=\frac{1}{n}(\frac{n}{n+1}+\frac{n}{n+2}+...+\frac{n}{2n})\\
&=\frac{1}{n}\sum_{i=1}^{n}\frac{n}{n+i}
\end{align}$$
Now, as $n\rightarrow\infty$, each $\frac{n}{n+i}\rightarrow1$, hence $\sum_{i=1}^{n}\frac{n}{n+i}\rightarrow n$. Hence $\lim_{n\rightarrow\infty}x_n=1$.
Thus, from the two methods, I end up having two different answers. I suspect that the second method is wrong, but I cannot identify where the mistake is. And I also want to know a method to solve the sum without using Euler's constant. So if anyone can point out the mistake in the 2nd method, and modify it, then it will be of great help to me.
Thanks in advance.
| Observe that you can rewrite $x_n$ as,
$$x_n =\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{2n}=\left(1+\frac{1}{2}+\frac{1}{3}+\dots \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}\right)$$
$$=\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2n}\right)$$
$$=\left(1-\frac{1}{2}+\frac{1}{3}+\dots -\frac{1}{2n}\right)$$
$$\lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots -\frac{1}{2n}\right) = \ln 2$$
The sequence limit exists otherwise the Taylor expansion of $\ln 2$ wouldn't exist.
| {
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Nice olympiad inequality Let's go for an olympiad inequality :
let $a,b,c>0$ then we have :
$$\sum_{cyc}\frac{ab}{a+b}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{abc}{a+b+c}}$$
My proof :
$$\sum_{cyc}\frac{ab}{a+b}=\frac{(a^2 b^2 + 3 a^2 b c + a^2 c^2 + 3 a b^2 c + 3 a b c^2 + b^2 c^2)}{(a+b)(b+c)(c+a)}=\frac{(ab + ac + bc)^2+abc(a+b+c)}{(ab+bc+ca)(a+b+c)-abc}$$
So if we use the $uwv's$ method we get :
$$\frac{9v^4+w^3(3u)}{3v^2(3u)-w^3}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{w^3}{3u}}$$
But I can't conclue by myself so can you help me ?
| By your work we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=3v^4+uw^3+\frac{1}{2}(w^3-9uv^2)\sqrt{\frac{w^3}{u}}.$$
We see that $f$ increases, which says that it's enough to prove our inequality for a minimal value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$.
In this case our inequality is obvious;
*Two variable are equal.
Since our inequality is homogeneous and symmetric, it's enough to assume $b=c=1$,
which gives
$$\frac{5a+1}{a+1}\geq3\sqrt3\sqrt{\frac{a}{a+2}},$$ which is wrong for $a\rightarrow+\infty.$
| {
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Simplification of proof for $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$ Question: Consider function $f(x,y)=|x+y|$ for $x,y\in \mathbb{R}$. Show that $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$.
My proof:
i) if $|x|,|y|\leq 1$, then $|x+y|\leq 2\leq 2(1+\sqrt{x^2+y^2})$;
ii): if $|x|\leq 1$ and $|y|\geq 1$, then $|x+y|\leq 1+|y|\leq 1+\sqrt{x^2+y^2}$;
iii): if $|x|,|y|\geq 1$ such that $|x|\geq |y|$, then $|x+y|\leq2|x|\leq2(1+\sqrt{x^2})\leq 2(1+\sqrt{x^2+y^2})$.
Hence, we conclude, and $c=2$ works for the claim.
However, I feel like there should be a simpler proof which does not prove in a case-by-case way like I did.
| Notice that
$$|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}\quad\text{and}\quad|y|=\sqrt{y^2}\leq\sqrt{x^2+y^2}$$
and so
$$|x+y|\leq|x|+|y|\leq 2\sqrt{x^2+y^2}<2\left(1+\sqrt{x^2+y^2}\right)$$
| {
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Minimize a function in two variables with constraint I have to minimize this:
\begin{align*}
\min&\quad{ (x-3)^2+(y-1)^2} \\
s.t.& \quad 2x+y \leq 2 \\
&\quad x^2 + 2y = 3\\
&\quad x, y \geq 0
\end{align*}
Can I isolate $y$ in the second constraint and substitute it in the first?
| Let $$(x-3)^2+(y-1)^2=z~~~(1)$$ Let us put $x^2=(3-2y)$ in (1), we get $$y^2-4y-6x+13-z=0.~~~(2)$$ $z$ will attain optimum value if when the line $y=2-2x$ touches the curve (2)
which is a parabola. Let us but this line in (2)
$$4x^2-6x+9-z=0.~~~~(3)$$
Now demand $B^2=4AC$.This gives $z=27/4,$ the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge n(n-1)/2 $ , prove that $a_{1} + ... + a_{n} \ge n$ for $n \ge 2$ using AM-QM Let $a_{1}, a_{2}, ...$ be a sequence of positive real numbers. Let the following relation holds:
$$ a_{k+1} \ge \frac{k a_{k}}{a_{k}^{2} + (k-1)}, \:\: k \ge 1$$
Prove that $ S_{n} = a_{1} + a_{2} + ... + a_{n} \ge n $, for $n \ge 2$.
Solution:
I have posted this question before and solved it using 2 approaches, (Given $ a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$, prove $ S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2 $).
this time I would like to solve it using another approach, the hint is that $ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge n(n-1)/2 $ and then use AM-QM inequality. Here is my attempt:
$$ \sum_{1 \le i < j \le n} a_{i} a_{j} = \sum_{j=2} S_{j-1} a_{j} $$
by using a result in my previous post that $S_{m} \ge m/a_{m+1}$ we have
$$ \sum_{1 \le i < j \le n} a_{i} a_{j} = \sum_{j=2}^{n} S_{j-1} a_{j} \ge \sum_{j=2}^{n} (j-1) = n(n-1)/2 $$
the hint is proved, then:
$$ \sum_{j=2}^{n} S_{j-1} a_{j} \le S_{n-1} \sum_{j=2}^{n} a_{j} $$
then by AM-QM:
$$S_{n-1} \sum_{j=2}^{n} a_{j} \le S_{n-1} \sqrt{n \sum_{i=2}^{n} a_{i}^{2}} \le S_{n-1} \sqrt{n} \sqrt{(S_{n}^{2})} = \sqrt{n} S_{n-1} S_{n}$$
so $$ \sqrt{n} S_{n-1} S_{n} \ge n(n-1)/2$$
if we use induction, by assuming $S_{n-1} \ge n-1$ then we can have
$$ S_{n} \ge \sqrt{n}/2$$
This is as far as i have gone.
| Since
$$(a_1+a_2+\cdots +a_n)^2=(a_1^2+a_2^2+\cdots +a_n^2)+2\sum_{1 \le i < j \le n} a_{i} a_{j}$$
we can write
$$\sum_{1 \le i < j \le n} a_{i} a_{j}=\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}$$
So, the hint $$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge \frac{n(n-1)}{2} $$
is equivalent to
$$\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}\ge \frac{n(n-1)}{2},$$
i.e.
$$a_1^2+a_2^2+\cdots +a_n^2\le S_n^2-n(n-1)\tag1$$
By AM-QM inequality, we get
$$\frac{a_1^2+a_2^2+\cdots +a_n^2}{n}\ge \left(\frac{a_1+a_2+\cdots +a_n}{n}\right)^2,$$
i.e.
$$a_1^2+a_2^2+\cdots +a_n^2\ge \frac{S_n^2}{n}\tag2$$
It follows from $(1)(2)$ that
$$\frac{S_n^2}{n}\le S_n^2-n(n-1)$$
from which
$$S_n\ge n$$
follows for $n\ge 2$.
| {
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value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^2\alpha\bigg)$$
$$\tan^{-1}\bigg(\frac{\frac{2\csc\alpha}{1-\csc^2\alpha}+2\sin\alpha\sec^2\alpha}{1-\frac{2\csc\alpha}{1-\csc^2\alpha}2\sin\alpha\sec^2\alpha}\bigg)$$
How do i solve it Help me please
| Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan p=\begin{cases} \arctan\dfrac{2p}{1-p^2} &\mbox{if } p^2<1 \\
\pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } p>1\\-\pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } p<-1\end{cases} $$
So, if $2m\pi\alpha>0>(2m-1)\pi,\csc\alpha<0\implies\csc\alpha<-1$
Consequently, $$2\arctan(\csc\alpha)=-\pi+\arctan\dfrac{2\csc\alpha}{1-\csc^2\alpha}$$
$$=-\pi+\arctan\left(-\dfrac{2\csc\alpha}{\cot^2\alpha}\right)$$
$$=-\pi-\arctan\left(\dfrac{2\csc\alpha}{\cot^2\alpha}\right)$$
$$=-\pi-\arctan\left(\dfrac{2\sin\alpha}{\cos^2\alpha}\right)$$
Here $-1<\alpha<0,\implies\csc\alpha<0$
| {
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Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.
| We have that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}} \ (m \in \mathbb N)$$
$$\iff m - \sqrt{n + 2} = \sqrt{n + \sqrt{n + 2}} \iff (m - \sqrt{n + 2})^2 = n + \sqrt{n + 2}$$
$$\iff m^2 - (2m - \sqrt{n + 2})\sqrt{n + 2} = (\sqrt{n + 2} + 1)\sqrt{n + 2} - 2$$
$$\iff m^2 + 2 = (2m + 1)\sqrt{n + 2} \iff \sqrt{n + 2} = \frac{m^2 + 2}{2m + 1}$$
As an addition, $\dfrac{m^2 + 2}{2m + 1} \in \mathbb Q^+, \forall m \in \mathbb N \implies \sqrt{n + 2} \in \mathbb Q^+$
$\implies \sqrt{n + 2} \in \mathbb N \implies \dfrac{m^2 + 2}{2m + 1} \in \mathbb N \iff \dfrac{4(m^2 + 2) - (2m + 1)(2m - 1)}{2m + 1} \in \mathbb N$
$\iff \dfrac{9}{2m + 1} \in \mathbb N \iff 2m + 1\mid 9 \iff 2m + 1 \in \{1, 3, 9\} \iff m \in \{0, 1, 4\}$
We can set up a table for different value of $m$ and $\sqrt{n + 2}$.
$$\begin{matrix} m&& 0&& 1&& 4\\ \sqrt{n + 2} = \dfrac{m^2 + 2}{2m + 1}&& 2&& 1&& 2 \end{matrix}$$
$\iff n \in \{-1, 2\}$.
Plugging $n \in \{-1, 2\}$ in $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$, we have that $(m,n) = (1, -1)$ and $(m, n) = (4, 2)$ is the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\iint\limits_D(x^2-2xy+y^2)\,\mathrm{d}x\,\mathrm{d}y$ where $4\leqslant x^2-y^2 \leqslant 16,~-5\leqslant y-\frac{x}{2}\leqslant 1$
Evaluate
$$I=\iint\limits_D(x^2-2xy+y^2)\,\mathrm{d}x\,\mathrm{d}y$$
where $D$ is bounded by $$x^2-y^2=4,~x^2-y^2=16,~y=\frac{x}{2}-5,~y=\frac{x}{2}+1.$$
Attempt. The obvious transformation $u=x^2-y^2,~v=2y-x$ has Jacobian determinant
$$det\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{det\frac{\partial(u,v)}{\partial(x,y)}}=\frac{1}{4x-2y},$$
where $x=x(u,v)$ and $y=y(u,v)$. So:
$$I=\frac{1}{2}\iint\limits_{[4,16]\times[-10,2]}\frac{x^2-2xy+y^2}{|4x-2y|}\,\mathrm{d}u\,\mathrm{d}v,$$
but the function inside the integral seems compicated enough, in order to turn into a simple function of $u,\,v.$
Thanks for the help.
| The problem with $u = x^2 - y^2, v = 2y-x$ is that it is difficult to express the objective as a function of $u,v$
Which gets me thinking about
$u = x+y\\
v = x-y\\
du dv = 2 dx dy$
This works nice with the objective and one of the functions on the boundary, but not the other.
Our transformed objective is now $\frac 12 v^2$
and the boundaries are $uv = 4, uv = 16, u - 3v = -20, u-3v = 4$
Suppose we try:
$s = uv\\
t = u\\
v = \frac {s}{t} = v\\
u = t$
$du dv = \left|\begin{array}{} \frac{1}{t} &-\frac{s}{t^2}\\1 & 0\end{array}\right| = \frac {s}{t^2} ds dt$
After this transformation.
objective $\frac {s^3}{2}$
and the boundaries are $s = 4, s = 16, \frac{s}{t} - 3t = -20, \frac {s}{t}-3t = 4$
Can we express isolate $t$ in the following:
$ \frac {s}{t} - 3 t = -20\\
3t^2 - 20 t - s\\
t = \frac {10 \pm \sqrt {100 + 3s}}{3}$
and
$t = \frac {2 \pm \sqrt {4 + 3s}}{3}$
At this point it is worth noting that if we sketch the region as it was originally defined there are, in fact two regions.
$\int_4^{16}\int_{\frac{2 - \sqrt{4+3s}}{3}}^{\frac{10 - \sqrt{100+3s}}{3}} s^3 \ dt\ ds + \int_4^{16}\int_{\frac{2 + \sqrt{4+3s}}{3}}^{\frac{10 + \sqrt{100+3s}}{3}} s^3 \ dt\ ds$
Rather than the two step transformation we could have said
$u = (x-y)(x+y)\\
v = x+y$
at the start. But, I didn't know that is what we needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Prove $e^{x \cos(x)}=1+x+\frac{x^2}{2} - \frac{x^3}{3}-\frac{11x^4}{24}- \frac{x^5}{5} + \cdots$ How do we do this question using Maclaurin's Series. I tried expanding it by putting $x \cos x$ in place of $x$ in Maclaurin's expansion of $e^x$, and then using multinomial theorem to open the squares, cubes etc of $\cos x$. Is my approach correct?
| There's a sign error in your $x^3$ coefficient. With $\equiv$ denoting equality up to $x^5$ terms, $$\exp(x\cos x)\equiv\exp\left(x-\frac{x^3}{2}+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1-\frac{x^3}{2}\right)\left(1+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{29x^5}{120}\right)\left(1+\frac{x^5}{24}\right)\\\equiv 1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{x^5}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Integrating $\int^2_0 xe^{x^2}dx$ Well what I was thinking was to integrate the indefinite integral first.
$u=x^2$, $x=\sqrt u$
$du=2xdx = 2\sqrt {u} dx$
$dx= \frac{1}{2\sqrt{u}}du$
$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$
Now I can evaluate $\frac{1}{2}e^{x^2}\Big|_0^2= \frac{1}{2} e^{4} -\frac{1}{2} e^0 =\frac{1}{2}e^4-1$
so my answer should be $$\frac{1}{2}e^4-1$$
Is this correct? It's been a while since I've done stuff like this.
| Also, one might set
$g(x) = e^{x^2}; \tag 1$
then
$g'(x) = 2xe^{x^2}; \tag 2$
then
$\displaystyle \int_0^2 xe^{x^2} \; dx = \dfrac{1}{2} \int_0^2 g'(x) \; dx = \dfrac{1}{2}(g(2) - g(0))$
$= \dfrac{1}{2}(e^{2^2} - e^0) = \dfrac{1}{2} (e^4 - 1) = \dfrac{1}{2}(e^4 - 1) = \dfrac{1}{2}e^4 - \dfrac{1}{2}. \tag{3}$
If one wants to use indefinite integrals, we write
$\displaystyle \int xe^{x^2} \; dx = \dfrac{1}{2} \int g'(x) \; dx = \dfrac{1}{2}g(x) + C = \dfrac{1}{2}e^{x^2} + C, \tag 4$
and then proceed to take
$g(2) - g(0) = \dfrac{1}{2}e^4 - \dfrac{1}{2}; \tag 5$
the constant of integration $C$ of course has been cancelled out of this expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the sum of series?? What is the sum of first 50 terms of the series
$$(1\times3)+(3\times5)+(5\times7)+\ldots$$
I had tried to solve this question
It seems that it is mixed arithmetic series
In which first $A.p$ is $1,3,5\ldots$
Nd other one is $3,5,7\ldots$
But i don't know how to solve it together
Please give me a better solution or approach to solve this problem?
Tell me a best way to solve
Such problems
| You have better go through Falling Factorials
$$
\eqalign{
& \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)} = 4\sum\limits_{n = 0}^{49} {\left( {n + 1/2} \right)\left( {n + 3/2} \right)} = \cr
& = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } } \cr}
$$
because then the sum telescopes nicely.
In fact we have
$$
\eqalign{
& \left( {n + 3/2} \right)^{\,\underline {\,2\,} } = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right)^{\,\underline {\,3\,} } - \left( {n + 3/2} \right)^{\,\underline {\,3\,} } } \right) = \cr
& = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right)\left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)
- \left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)\left( {n - 2 + 3/2} \right)} \right) = \cr
& = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right) - \left( {n - 2 + 3/2} \right)} \right)\left( {\left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)} \right) = \cr
& = \left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right) \cr}
$$
so that the sum becomes
$$
\eqalign{
& \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)} = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } } = \cr
& = {4 \over 3}\sum\limits_{n = 0}^{49} {\left( {\left( {n + 1 + 3/2} \right)^{\,\underline {\,3\,} } - \left( {n + 3/2} \right)^{\,\underline {\,3\,} } } \right)} = \cr
& = {4 \over 3}\left( {\left( {50 + 3/2} \right)^{\,\underline {\,3\,} } - \left( {3/2} \right)^{\,\underline {\,3\,} } } \right) = \cr
& = {4 \over 3}\left( {\left( {50 + 3/2} \right)\left( {49 + 3/2} \right)\left( {48 + 3/2} \right) - \left( {3/2} \right)\left( {1/2} \right)\left( { - 1/2} \right)} \right) = \cr
& = {4 \over {3 \cdot 2^3 }}\left( {103 \cdot 101 \cdot 99 + 3} \right) = {{1029900} \over 6} = 171650 \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\\
8 + 16x^2 + x &= 4x - x^2 + 22\\
16x^2 + x &= 4x - x^2 + 14\\
16x^2 &= 3x - x^2 + 14\\
17x^2 &= 3x + 14
\end{align}
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
| You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $\frac{-14}{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Solution Of Diophantine Equations. Find all positive integers $x$ and $y$ such that $(x+y)(xy+1)=2^z$
Now as $2$ is a prime number we get $x+y=2^a$ and $xy+1=2^b$ for some natural numbers $a$ and $b$. Now as $xy+1>x+y$ for every $x, y>1$. So we get $2^b>2^a=> b>a=> 2^a|2^b$. So $x+y|xy+1$. Now let $p$ be a common prime factor of $x$ and $y$, then $p$ divides $x+y$, so $p$ divides $xy+1$ but as $p$ divides $x$ and $y$ so $p$ divides $xy$. So we get $p$ divides $1$, so there is no common factor of $x$ and $y$.
Now what i do from here?
| You have $xy=2^k-1$ for some integer $k> 0$
Note that with $x=1$ and $y=2^k-1$ you get $$(x+y)(xy+1)=2^{2k}$$
So we always have that solution.
We also have other solutions if we can factor $2^k-1$ into two integers whose sum is a power of $2$
For example with $k=4$ we get $2^{k}-1=3(5)$ where $3+5=2^3$ so $(x+y)(xy+1)=2^{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx$ Evaluate:
$I=\int\limits_0^1 \frac{\sqrt{1+x^2}}{1+x}dx$
My try:
Let $x=\tan y$ then $dx=(1+\tan^{2} y)dy$
As for the integration limits: if $x=0$ then $y=0$ and if $x=1$ then $y=\frac{π}{4}$
So:
$I=\int\limits_0^{\frac{π}{4}}\frac{1+\tan^{2} y}{(1+\tan y)\cos y}\,dy$
$I=\int\limits_0^{\frac{π}{4}}\frac{1}{\cos^{3} y+\cos^{2} y\sin y}\,dy$
But I don't know how to continue.
| \begin{align}
\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx
= &\int\limits_0^{1}\frac{x-1}{\sqrt{1+x^2}}+\frac2{\sqrt{1+x^2}(1+x)}\ {dx}\\
=&\ \left(\sqrt{1+x^2}-\sinh^{-1}x -\sqrt2\sinh^{-1}\frac{1-x}{1+x} \right)_0^1\\
=& \ \frac{\sinh^{-1}1+1}{\sqrt2+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate the integral from $0$ to $2\pi$ I get that it's $0$ because $\tan\left(2x\right)$ evaluates to $0$ at both $x=0$ and $x=2\pi$. Is there another way to solve this integral and get the correct answer ($2\pi\sqrt2$)?
| You've got many methods here to find the anti-derivative, but hey, why not one more?
Here
\begin{align}
I &= \int \frac{1}{\sin^4(x) + \cos^4(x)}\:dx = \int \frac{1}{\cos^4(x)\left(\tan^4(x) + 1\right)}\:dx \nonumber \\
&= \int \sec^2(x)\sec^2(x) \cdot\frac{1}{\tan^4(x) + 1}\:dx = \int \sec^2(x) \cdot\frac{\tan^2(x) + 1}{\tan^4(x) + 1}\:dx
\end{align}
Let $t = \tan(x)$:
\begin{align}
I &= \int \frac{t^2 + 1}{t^4 + 1}\:dt = \int \frac{t^2 + 1}{\left(t^2 + \sqrt{2}t + 1\right)\left(t^2 - \sqrt{2}t + 1\right)}\:dt \nonumber \\
&= \int \frac{1}{2}\left[\frac{1}{t^2 + \sqrt{2}t + 1} + \frac{1}{t^2 - \sqrt{2}t + 1} \right]\:dt \nonumber \\
&= \frac{1}{2}\left[ \int \frac{1}{t^2 + \sqrt{2}t + 1}\:dt + \int \frac{1}{t^2 - \sqrt{2}t + 1}\:dt \right] = \frac{1}{2}\left[ J_1 + J_2\right]
\end{align}
We now address $J_1$ and $J_2$ separately. Here we begin with $J_1$:
\begin{equation}
J_1 = \int \frac{1}{t^2 + \sqrt{2}t + 1}\:dt = \int \frac{1}{\left(t + \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}}\:dt = \frac{1}{\sqrt{\frac{1}{2}}}\arctan\left(\frac{t + \frac{1}{\sqrt{2}}}{\sqrt{\frac{1}{2}}}\right)= \sqrt{2}\arctan\left(\sqrt{2}t + 1\right)
\end{equation}
Note I've omitted the constant of integration here. Similarly we find that $J_2$ is
\begin{equation}
J_2 = \sqrt{2}\arctan\left(\sqrt{2}t - 1\right)
\end{equation}
Thus our integral $I$ becomes:
\begin{align}
I &= \frac{1}{2}\left[ J_1 + J_2\right] = \frac{1}{2}\bigg[\sqrt{2}\arctan\left(\sqrt{2}t + 1\right) + \sqrt{2}\arctan\left(\sqrt{2}t - 1\right)\bigg] + C = \frac{1}{\sqrt{2}}\bigg[ \arctan\left(\sqrt{2}t + 1\right) + \arctan\left(\sqrt{2}t - 1\right)\bigg] + C \nonumber \\
&=\frac{1}{\sqrt{2}}\bigg[ \arctan\left(\sqrt{2}\tan(x) + 1\right) + \arctan\left(\sqrt{2}\tan(x) - 1\right)\bigg] + C
\end{align}
Where $C$ is the constant of integration. Thus,
\begin{equation}
\int \frac{1}{\sin^4(x) + \cos^4(x)}\:dx =\frac{1}{\sqrt{2}}\bigg[ \arctan\left(\sqrt{2}\tan(x) + 1\right) + \arctan\left(\sqrt{2}\tan(x) - 1\right)\bigg] + C
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}
Now i think it gets more complicated , any help would be appreciated
| It can be seen that the radical of the form $\sqrt{x-\sqrt{x+..}}$ is undefined for $x<1$. Since $\sin{x}$ is on the interval $[-1,1]$. This means that it is only defined for $x=\arcsin(1)$. Or generalizing this, we get;
$$x=2n\pi+\frac{\pi}{2}, \tag{for n ∈ Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Differentiating from the first principle How do I convincingly differentiate $\sin 3x$ or any trig function that comes in the form $\tan nx$, $\sin nx$, $\cos nx$ from the first principle?
I have tried expanding some forms using some trig identities but the $n$ coefficient isn't just showing up
| If you allow the trigonometric functions to be defined by their McLauren series, then you can differentiate these term-by-term.
For example:
$$\cos nx = 1 - \frac{(nx)^2}{2!} + \frac{(nx)^4}{4!}-\frac{(nx)^6}{6!}+\cdots$$
$$
\frac{d\cos nx}{dx} = - \frac{2n(nx)}{2!} + \frac{4n(nx)^3}{4!}-\frac{6n(nx)^5}{6!}+\cdots$$
$$
\frac{d\cos nx}{dx} = - n(nx) + \frac{n(nx)^3}{3!}-\frac{n(nx)^5}{5!}
+\cdots= -n \sin nx$$
Also
$$ \tan nx = nx + \frac{1}{3} (nx)^3 + \frac{2}{15} (nx)^5 + \frac{17}{315}(nx)^7+\cdots$$
$$\frac{d \tan nx}{dx} = n + n(nx)^2 + \frac{2n}{3} (nx)^4 + \frac{17n}{45} (nx)^6 + \cdots = n\sec^2 nx$$
since the McLauren series for $\sec nx$ is
$$\sec nx = 1 + \frac{1}{2} (nx)^2 + \frac{5}{24} (nx)^4 + \frac{61}{720} (nx)^6 +\cdots$$
Using Euler's identity:
$$e^{inx} = \cos nx + i \sin nx$$
$$\frac{d e^{inx}}{dx}=n i e^{i n x} = n ( i \cos nx - \sin nx)$$
Equating the real and imaginary parts:
$$\frac{d \cos nx}{dx} = -n \sin nx, \quad \frac{d \sin nx}{dx} = n\cos nx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral in terms of $u$ and simplified it as much as possible.
$$\implies \int\frac{1}{5+3(\frac{1-u^2}{1+u^2})} \cdot \frac{2du}{1+u^2}= \int \frac{1}{5+\frac{3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2} = \int \frac{1}{\frac{5+5u^2+3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}$$
$$=\int \frac{1}{\frac{8+2u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1}{\frac{2(4+u^2)}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1+u^2}{2(4+u^2)} \cdot \frac{2du}{1+u^2}=\int \frac{1}{4+u^2} \cdot du$$
I know that $\int \frac{1}{1+x^2}= \text{arctan(x)}$ but in my case I have a $4$ and not a $1$. My guess is that I have to use $u$-subsitution again but I can't figure out what $v$ is going to be. I've seen a lot of videos from BlackPenRedPen about $u$-substitution and I've noticed that he's able to see what the end result should be and choose his $u$ based on that, however I have no such talent. Therefore I'm confused as to what my new $u$ which will actually be a $v$ since I've already used a $u$, will be.
| You should know by heart the formulæ
\begin{align}\int\frac{\mathrm dx}{a^2+x^2}&=\frac 1a\,\arctan\frac xa,&\int\frac{\mathrm dx}{a^2-x^2}&=\frac 1{2a}\,\ln\biggl|\frac{a+x}{a-x}\biggr|\\
&&&=\frac 1{a}\arg\tanh\frac xa\quad\text{ if } |x|<a.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_{0}^{1} dx \int_{- \sqrt{x- x^2 }}^{\sqrt{x- x^2}}\sqrt{1- x^2 - y^2} dy $ I am having problems with integration.
I have a function $f(x, y) = \sqrt{1- x^2 - y^2}$
And I need to integrate it on a circle $(x- \frac{1}{2} )^2 + y^2 = \frac{1}{4}$
I couldn't get far with usual coordinates:
$$
\int_{0}^{1} dx \int_{- \sqrt{x- x^2 }}^{\sqrt{x- x^2}}\sqrt{1- x^2 - y^2} dy
$$
So I tried with polar coordinates.
$$
x= r \cdot \cos\phi \\\
y= r \cdot \sin \phi \\\
\phi \in (-\frac{\pi}{2}, \frac{\pi}{2}) \\\
r \in (0, \cos\phi) \\\
\int_{-\pi / 2}^{\pi / 2} d\phi \int_{- \sqrt{x- x^2 }}^{\sqrt{x- x^2}} \sqrt{1- r^2} r dr
$$
I tried with a new variable $u = 1- r^2$,
$du= -2r dr$.
Then I got.
$$
\int_{-\frac{\pi}{ 2}}^{\frac{\pi}{2}} \frac{-1}{3} ((1-\cos{\phi}^2)^{3/2}-1) d\phi =
\\
\int_{-\pi / 2}^{\pi / 2} \frac{-1}{3} (\sin{\phi}^3 - 1) d\phi =
\\
\frac{-1}{3} \bigg( \int_{-\pi / 2}^{\pi / 2} \sin{\phi}^3 d\phi - \int_{-\pi / 2}^{\pi / 2} d\phi\bigg) = \\
0 + \frac{-1}{3} (\pi / 2 + \pi /2)
$$
But it's not right. $\sin$ function is odd and on a symmetric interval its integral should be zero, right?
| Near the end of your work, it should be $(1-\cos{\phi}^2)^{3/2}=|\sin{\phi}|$ (with the absolute value!) and therefore
$$\frac{-1}{3} \bigg( \int_{-\pi / 2}^{\pi / 2} |\sin{\phi}|^3 d\phi - \int_{-\pi / 2}^{\pi / 2} d\phi\bigg)=-\frac{2}{3} \int_{0}^{\pi / 2} \sin^3{\phi}\, d\phi +\frac{\pi}{3}=-\frac{4}{9}+\frac{\pi}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\end{array}
$$
Continue this pattern until the box is $10 \times 10$.
Then add all the numbers together.
So this was my attempt.
Set the square $1\times1$ as $a_1$ square $2 \times 2$ as $a_2$ and so on. $a_1$ to $a_2$'s d is $(1 \cdot 2)+2 a_2$ to $a_3$'s d is $(1 \cdot 2)+(2 \cdot 2)+3$ and so on. So then I can the individual ds from each square to the next square.
And because
$$
1+(1+2)+(1+2+3)\ldots+(1+2+3+\ldots+10)
= \sum_{k=1}^{1} 1 + \sum_{k=1}^{2} 1+ \ldots +\sum_{k=1}^{10} 1=\sum_{k=1}^{10} \frac{n(n+1)}{2}
$$
so
$$
(1\cdot2)+(1\cdot2)+(2\cdot2)\ldots+(1\cdot2)+(2\cdot2)\ldots+(10\cdot2)
= \sum_{k=1}^{1} 2k + \sum_{k=1}^{2} 2k+\ldots+\sum_{k=1}^{10} 2k
$$
which also means
$$
\sum_{k=1}^{10} 2 \cdot \frac{n(n+1)}{2}
=\sum_{k=1}^{10} {n^2+n}
$$
And these are my calculations
$$\frac{10 \cdot 11 \cdot 21}{6}+\frac{10 \cdot 11}{2}$$
$$385+55=440$$
$$1+2+3+4+\ldots+10=55$$
$$440+55=495$$
It all seems right to me but the answer says its 385 is there any thing I did wrong?
| Another way to think of it is by summing diagonals: A $n \times n$ square has the following diagonal sums
$$
1, \\ 1 + 2 , \\ \vdots \\ 1 + \ldots + n - 1, \\ 1 + \ldots + n, \\ 1 + \ldots + n - 1, \\ \vdots \\ 1 + 2, \\ 1.
$$
Therefore, the formula for the sum is
$$
2 \sum_{k = 1}^{n - 1}\sum_{j = 1}^{k} j + \sum_{j = 1}^{n} j
= 2 \sum_{k = 1}^{n - 1} \frac{k(k + 1)}{2} + \sum_{j = 1}^{n} j
= \frac{(n - 1) n (n + 1)}{3} + \frac{n(n + 1)}{2}
= \frac{n (n + 1) (2 n + 1)}{6}.
$$
We notice that this is also the formula for the sum of the first $n$ square numbers. This is because if you consider the sum of all entries of the and mirrored $L$-shape (not counting the one element that is in the row and column simultaneously twice) you get
$$
\sum_{j = 1}^{k} j + \sum_{j = 1}^{k - 1} j
= \frac{k(k + 1)}{2} + \frac{k(k - 1)}{2}
= \frac{k^2 + k + k^2 - k}{2}
= k^2.
$$
To explain the $L$-shape better: In the first square the first L is just the lower left number, one. The next $L$ consist of the adjacent numbers, $1 + 2 + 1 = 2^2$ and the next one captures the remaining ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Division of 11 people into 3 groups with at least 2 people in each
How many ways can 11 people be divided into three teams where each
team must have at least two members?
We are supposed to use multinomial coefficients and number of integer solutions. I have tried this
The number of ways to divide 5 people into three groups
and this
Arranging $10$ people in $2$ lanes. Each lane has to have at least $2$ people. and more specifically this Number of ways to divide n people into k groups with at least 2 people in each group, but the last one in particular I could not understand at all (since I do not know what the brackets { } mean).
I have tried several things so far, but I am not convinced any of them are correct. Here is my best attempt:
If each group has at least two people, I initially choose 6 from 11:
$$\binom{11}{6}=42$$
Then, these 6 people need to be put in the groups of 2 people:
$$\frac{6!}{2!2!2!}=90$$
But, since order does not matter, we must divide by 3!:
$$\frac{6!}{2!2!2!3!} = 15$$
So for the choice of the two people in each group we have 90*15 = 1350 possibilities.
Now we need to consider the 5 remaining people. Let $x_1,x_2,x_3$ be the number of people in each group. Then we have a total of
$$\binom{5+3-1}{3-1} = \binom{7}{2}=21$$
(non negative) integer solutions for $x_1+x_2+x_3=5$.
However, the possible cases are:
$$(0,1,4),(0,2,3),(0,0,5),(1,1,3),(1,2,2),$$
where the first two triplets appear a total of 3! each (order does not matter) and the last three apear $\frac{3!}{2!} =3$ times (due to permutation of terms with same amount of people).
Case $(0,1,4)$:
$\binom{5}{1}\binom{4}{4}=5,$ giving a total of $3!5 = 30$ possibilities.
Case $(0,2,3)$:
$\binom{5}{2}\binom{3}{3}=10,$ giving a total of $3!10 = 60$ possibilities.
Case $(0,0,5)$:
$\binom{5}{5}=5,$ giving a total of $3\times 5 = 15$ possibilities.
Case $(1,1,3)$:
$\binom{5}{1}\binom{4}{1}\binom{3}{3}=2,0$ giving a total of $3\times 20 = 60$ possibilities.
Case $(1,2,2)$:
$\binom{5}{1}\binom{4}{2}\binom{2}{2}=6,$ giving a total of $3\times 6 = 18$ possibilities.
Then we will have a total of
$1350(30+60+15+60+18) = 247050$
possibilities.
Can anyone help with the logical reasoning here to see if this is correct? In case it is wrong, where am I going wrong?
| It is ill-advised to treat the compulsory two people in each group separately. Rather, the partitions of $11$ into $3$ with each part at least $2$ should be looked at directly, which as you have worked out are
$$(7,2,2),(6,3,2),(5,4,2),(5,3,3),(4,4,3)$$
These lead to the following counts for each partition.
*
*$(7,2,2)\to\binom{11}7\binom42/2=990$ ways
*$(6,3,2)\to\binom{11}6\binom53=4620$
*$(5,4,2)\to\binom{11}5\binom64=6930$
*$(5,3,3)\to\binom{11}5\binom63/2=4620$
*$(4,4,3)\to\binom{11}3\binom84/2=5775$
The division by $2$ in three of the cases accounts for the indistinguishability of groups with the same people but different positions. Adding the counts up, there are $22935$ admissible partitions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solutions modulo square of a number. Let $p$ be a prime such that $$p^n\equiv 5 \mod{6}$$ for each odd positive integer $n$. Then how to find the values of $p^n$ modulo $36$, i.e. I need to find $$p^n\equiv ?\mod{36}.$$ Please help.
| If $p^n \equiv 5 \pmod 6$ for every odd $n$ then $p^1 \equiv 5 \pmod 6$.
(Which is okay; that means $p^{2k+1} \equiv (p^2)^k*p \equiv (25^k)*5 \equiv 1^k*5 \equiv 5 \pmod p$.
So $p = 5 + 6M$ for some $M$ and $p \equiv 5 + 6m\pmod {36}$ where $m \equiv M \pmod 6$
$p^2 \equiv (5+6k)^2 \equiv 25 + 60m + 36m^2 \equiv 1+ 24m \pmod {36}$.
So $p^n = p^{2k + 1} = p*(p^2)^k\equiv 5*(1+24m)^k\pmod {36}$
Now $(1 + 24m)^k = 1 + k*24m + {k\choose 2}*24^2m^2 + ..... $ be notice that all the $24^i; k\ge 2$ are $24^i = 3^i2^{3i}=36*3^{i-1}2^{3i - 2} \equiv 0 \pmod {36}$ so
$(1 + 24m)^k \equiv 1 + k*24m\pmod {36}$.
So $p^n \equiv 5(1 + k*24m) \equiv 5 + k*120*m \equiv 5 + 12km \pmod {36}$
Where $n = 2k + 1$ and $p \equiv 5 + 6m \pmod {36}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inverse Function Verification
Are functions $f(x)=\frac{7x+4}{x+6}$ and $g(x)=\frac{6x-4}{7-x}$ inverses of each other?
I'm experiencing a very strange issue with this problem. When I find the inverse of $g$ I get $f(x)$. However, when I do the same thing for $f$ I do not get $g(x)$.
\begin{align*}
f(x)&=\frac{7x+4}{x+6} \equiv y=\frac{7x+4}{x+6} \\ x&=\frac{7y+4}{y+6} \\ x(y+6)&=7y+4 \\ xy-7y&=-6x+4 \\ y(x-7)&=-6x+4\\ y&=\frac{-6x+4}{x-7} \\ \therefore f^{-1}(x)&=\frac{-6x+4}{x-7}
\end{align*}
It's very close, only the signs deviate. Yet when I try this method on $g(x)$ I get
\begin{align*}
x&=\frac{6y-4}{7-y} \\ 7x-xy&=6y-4 \\ -y&=(\frac{-7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}
\end{align*}
Which says the functions are inverses of each other. What am I getting wrong with finding the inverse of function $f$?
| It should be
\begin{align*}
-y&=-(\frac{7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}=\ f(x)
\end{align*}
And you already have $$f^{-1}(x)=\frac{-6x+4}{x-7}=\frac{6x-4}{7-x}=g(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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What is the problem here (all integers are irrational proof...I think so)? Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime, let us simplify the fraction so this it is (I don't know how to talk like mathematicians).
Which implies, $$a=b/c$$ (where $b$ and $c$ are coprime integers). Squaring on both sides,
\begin{align}
a^2&=b^2/c^2\\
a^2c^2&=b^2
\end{align}
So $a^2$ is a factor of $b^2$, and also of $b$, due to the uniqueness of the fundamental theorem of arithmetic.
So,
\begin{align}
b &=a^{2}d \tag{where $d$ is an integer}\\
b^2 &= a^{4}d^{2}
\end{align}
But $b^2=a^2c^2$ So,
\begin{align}
a^2c^2 &= a^4d^2\\
c^2 &= a^2d^2
\end{align}
So, $a^2$ is a factor of $c^2$ and $c$ due to the fundamental theorem of arithmetic. So $b$ and $c$ have $a^2$ as a common factor. But this contradicts the fact that $b$ and $c$ are coprime. This is because we have taken $a$ as a rational integer, so $a$ cannot be a rational integer.
What's wrong here (genuinely asking)?
| I think you are confusing that if $p$ is prime and $p$ divides $b^k$ then $p|b$. That is true if $p$ is prime.
Actually it's also true for a composite $a|b^k$ then $a|b$ if $a$ has no square factors. But if $a$ as any prime factors to a power greater than $1$ it need not be true.
And in fact its obviously not true as $a^2$ divides $a^2$ but $a^2$ doesn't divide $a$ (unless $a = 1$).
Read on....
It most certainly is not true if $a|b^k$ that $a|b$ It means that the prime factors of $a$ are prime factors of $b$. And it means that the powers of those prime factors of $a$ are at most equal to $k$ times the powers of the same prime factors of $b$ but because $k$ is larger than .....
Oh let me put it this way.
Suppose $a = \prod p_i^{m_i}$ be the prime factorization of $a$. Suppose $a|b^k$. Then that means that $p_i$ are prime factors of $b$ and that $b = d\prod p_i^{j_i}$. And it means that $b^k = d^k \prod p_i^{k*j_i}$.
And as $a|b^k$ that means each $m_i \le k*j_i$. But that does not mean $m_i \le j_i$ which would mean $a|b$.
You statement $a|b^k$ means $a|b$ if $a$ has square free and all the prime factor powers were $1$ but not other wise.
Simple example if $a = 12 = 2^2*3$ and $b= 90 = 2*3^2*5$. Now $a|b^2 = 8100 = 2^2*3^4*5^2$.
This means the prime factors of $a$ ($2,3$) are also prime factors of $b$. And it means that the powers of the prime factors of $a$ ($2\mapsto 2; 3\mapsto 1$) are less or equal to $2$ times the powers in $b$ ($2\mapsto 1$ and $2 \le 2*1$ and $3\mapsto 2$ and $1 \le 2*2$) but it doesnt mean the are less than or equal to the powers of $b$. (In $a; 2\mapsto 2$ but in $b; 2\mapsto 1$ and $2 \not \le 1$).
So $12 \not \mid 90$.
It's certainly can't be the case that $a|b \implies a^2| b^2 \implies a^2|b$! That would mean every time you have $a|b$ you can just keep squaring and reducing to get $a^{m}|b$ for any power of $m$.
That would mean if $3|6$ then $3^2|6$ and $3^4|6$ and $3^{2048}|6$ and so on.
Or in this case as $a = b$ (and $c=1$.... because $a$ is an integer) you would have $a|a$ so $a^2|a$? And $a^4|a$. That's .... simply not true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.
Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 + 3xy((\sqrt{2}+2)^{\frac{1}{3}} -1) - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?
| Using Dr. Sonnhard Graubner's answer, multiply everything by $27x^3$ and you face an awful cubic equation in $x^3$ for which $\Delta=-1062882$ which means that there is only one real root.
Use the hyperbolic methid for that case and you will end with the "beautiful"
$$x^3=6 \left(a+b \cosh \left(\frac{1}{3}\cosh (c)\right)\right)$$ where
$$a=1+\sqrt[3]{2+\sqrt{2}}$$
$$b=\sqrt{3 \left(3+2 \sqrt[3]{2+\sqrt{2}}+\left(2+\sqrt{2}\right)^{2/3}\right)}$$
$$c=\frac{\sqrt{3} \left(8+\sqrt{2}+6 \sqrt[3]{2+\sqrt{2}}+4
\left(2+\sqrt{2}\right)^{2/3}\right)}{2 \left(3+2
\sqrt[3]{2+\sqrt{2}}+\left(2+\sqrt{2}\right)^{2/3}\right)^{3/2}}$$ Evaluated, this gives $$x^3=44.9381694189876 \implies x= ??? \implies y= ???$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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If at least one of the selected items must be vegetarian, how many different dinners could Jane create?
Problem: Jane must select $3$ different items for each dinner she will serve. The items are to be chosen from $5$ different vegetarian and $4$ different meat selections. If at least one of the selected items must be vegetarian, how many different dinners could Jane create?
So my attempt to solve this problem is as follows. It seems that the solution is just a slightly varied application of the multiplication principle
So if we assume the first item is the one mandatory vegetarian one, then the first one has $5$ possible choices. The second item has $(5-1)+4 = 8$ choices (since at this point it doesn't matter if we choose vegetarian or meat) and the third has $((5-1)+4)-1 = 7$ choices. Applying the multiplication princple we see that there are
$$5\cdot 8\cdot 7 = 280$$
possible dinners.
However the correct answer is $80$ dinners. What have I done wrong? Also while any answer would be specific to the question above, more generally is there a counting principle that I have missed here or used incorrectly?
| You are counting meals with more than one vegetarian item multiple times. Also, the order in which the courses are selected does not matter.
Method 1: Direct count.
Since there are five different vegetarian options and four different meat options, the number of ways of selecting $k$ vegetarian options and $3 - k$ meat options is
$$\binom{5}{k}\binom{4}{3 - k}$$
Since at least one of the three courses must be vegetarian, the number of ways Jane could select at least one vegetarian course when selecting three courses is
\begin{align*}
\sum_{k = 1}^{3} \binom{5}{k}\binom{4}{3 - k} & = \binom{5}{1}\binom{4}{2} + \binom{5}{2}\binom{4}{1} + \binom{5}{3}\binom{4}{0}\\
& = 5 \cdot 6 + 10 \cdot 4 + 10 \cdot 1\\
& = 30 + 40 + 10\\
& = 80
\end{align*}
Method 2: Complementary counting.
The number of ways of selecting at least one vegetarian course can be found by subtracting the number of ways of selecting only meat courses from the total number of ways of selecting three courses. There are $\binom{5 + 4}{3} = \binom{9}{3}$ ways to select three of the nine courses. There are $\binom{4}{3}$ ways to select three of the four meat courses. Hence, the number of ways of selecting at least one vegetarian course is
$$\binom{9}{3} - \binom{4}{3} = 84 - 4 = 80$$
Your errors:
You are counting each meal with two vegetarian courses twice, once for each way you could designate one of the two courses as the vegetarian course. You are counting each meal with three vegetarian courses three times, once for each way you could designate one of the three courses as the vegetarian course. Observe that
$$\binom{5}{1}\binom{4}{2} + \color{red}{\binom{2}{1}}\binom{5}{2}\binom{4}{1} + \color{red}{\binom{3}{1}}\binom{5}{3}\binom{4}{0} = 140$$
Then why did you obtain $280$?
Suppose that the meal Jane selects consists of vegetarian chili, avocado and tomato salad, and green salad. If you designate vegetarian chili as the vegetarian course, you get the same meal whether you select avocado and tomato salad for the second course and green salad as the third course or whether you select green salad as the second course and avocado and tomato salad as the third course. Therefore, by taking the order in which Jane selects the additional courses into account, you have counted each pair of additional courses twice, which doubles your meal count. Notice that
$$\color{red}{2}\left[\binom{5}{1}\binom{4}{2} + \color{red}{\binom{2}{1}}\binom{5}{2}\binom{4}{1} + \color{red}{\binom{3}{1}}\binom{5}{3}\binom{4}{0}\right] = 280$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove that $\left\lceil \frac{4N}{3} \right\rceil \bmod 4 \ne 1$? Ok this is for Base $64$ padding.
The length of Base $64$ representation of a sequence of bytes is represented by
$\left\lceil \frac{4N}{3} \right\rceil$
The padding should add $M$ characters such that $$\frac{\left(\left\lceil \frac{4N}{3} \right\rceil + M \right)}{4}=0.$$
How can I prove that $M\ne 1$ for all values of $N > 0$?
I created the sequence in Excel and the sequence is $2,3,4,6,7,8,10$ which means the difference is $1-1-2$ that repeats itself. It seems that the number skipped is always $4N+1$ but how to come to this conclusion?
| For any $n$, we have either $n=3m$, $n=3m+1$, or $n=3m+2$. Then we just have three cases to consider:
\begin{align*}
\left\lceil \frac{4n}{3} \right\rceil
&=
\left\lceil 4m \right\rceil
= 4m \equiv 0 \pmod 4
\\
\left\lceil\frac{4n}{3} \right\rceil
&=
\left\lceil 4m+\frac{4}{3} \right\rceil
=4m+2 \equiv 2 \pmod 4
\\
\left\lceil \frac{4n}{3} \right\rceil
&=
\left\lceil 4m+\frac{8}{3} \right\rceil
=4m+3 \equiv 3 \pmod 4
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291545",
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"source": "stackexchange",
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Finding $\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$ For $a,b>-1$
$$\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$$
I am really confused in this one. I tried to calculate it but the answer comes out to be $1$ as I divided the numerator and denominator with $n^a$ and $n^b$ respectively. But the answer is wrong.
Please help.
| Another similar solution using generalized harmonic numbers
$$S_a=\sum_{i=1}^n i^a=H_n^{(-a)}$$ Using asymptotics
$$S_a=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$
$$n^{-a} S_a=\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
n}+O\left(\frac{1}{n^3}\right)+n^{-a}\zeta (-a)$$
$$n^{b-a}\frac{S_a}{S_b}=\frac{n^{-a}S_a}{n^{-b}S_b}\sim \frac{\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
n} }{\frac{n}{b+1}+\frac{1}{2}+\frac{b}{12
n} }=\frac{b+1}{a+1}+\frac{(b+1) (a-b)}{2 (a+1)
n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $x+y+z \ge xy+yz+zx$ Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+ca-abc=0$$
$$ab+bc+ca-abc=2(a+b+c)$$
Where $a, b,c \le 1$
is there a clue here?
| Suppose otherwise $x+y+z<xy+xz+yz$. Let $x=ka$, $y=kb$ and $z=kc$, such that $k>0$ and $a+b+c=ab+ac+bc$.
Thus, $$k(a+b+c)<k^2(ab+ac+bc),$$ which gives $$k>1$$ and
$$4=k(a+b+c)+k^3abc>a+b+c+abc,$$ which is a contradiction because we'll prove now that
$$a+b+c+abc\geq4.$$
Indeed, we need to prove that
$$\frac{(ab+ac+bc)^2}{a+b+c}+abc\geq\frac{4(ab+ac+bc)^3}{(a+b+c)^3}$$ or
$$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3+3a^4bc-a^3b^2c-a^3c^2b-a^2b^2c^2)\geq0,$$ which is true by Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding constant $a$ so that differential equation becomes correct Question:
Find the value of $a$ so that the function
$$y = \sqrt{x} \ln{x}$$
is a solution to the differential equation
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}}$$
Attempted solution:
My basic approach would be to take the derivative of y, put y and y' into the differential equation and solve for a.
The derivative of y is gotten with the product rule and then the chain rule for the square root:
$$y' = \sqrt{x} \cdot (\ln{x})' + (\sqrt{x})' \cdot \ln{x} = \frac{\sqrt{x}}{x} + \frac{\ln{x}}{2 \cdot \sqrt{x}}$$
Solving for a:
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}} \Rightarrow a = \frac{y' - \frac{1}{\sqrt{x}}}{y} = \frac{xy' - \sqrt{x}}{y}$$
Putting y and y' into the expression for a gives:
$$a = \frac{x(\frac{\sqrt{x}}{x} + \frac{\ln {x}}{2\sqrt{x}})}{\sqrt{x}\ln {x}} = \frac{1+\frac{\ln{x}}{2}}{\ln{x}} = \frac{1}{\ln{x}} + \frac{1}{2}$$
The expected answer is just $\frac{1}{2}$, so for some reason I have gotten an extra $\frac{1}{\ln{x}}$ somwhere.
| You forgot in the last line the following:
$$a = \frac{x(\frac{\sqrt{x}}{x} + \frac{\ln {x}}{2\sqrt{x}}) \color{blue}{-\sqrt{x}}}{\sqrt{x}\ln {x}} = \frac{1+\frac{\ln{x}}{2}\color{blue}{-1}}{\ln{x}} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:
$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$
Putting it equal to zero and solving for x:
$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$
...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$
So I must have gone wrong somewhere.
| Looks like you have a problem with the differentiation. You should have.
You dropped a factor of 3 in the right-hand term. It should be $(\frac 32 \sin 3x)(\sin 3x)$ in the first line.
You have brought it back by the time you get to.
$3\cos 3x + 1.5\cos^2 3x + \frac 12 \sin^2 3x = 0$
But then $1.5$ becomes $1$ in the next line.
$3\cos 3x + 1.5 = 0$
Solving for x:
$\cos3x = -\frac 12$
$3x = \pm\frac {2\pi}{3} + 2n\pi\\
x = \pm \frac {2\pi}{9} + \frac {2n\pi}{3}$
$x = \frac {2\pi}{9} + \frac {2n\pi}{3}$ are the maxima
and $x = -\frac {2\pi}{9} + \frac {2n\pi}{3}$ are the minima
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to find Sum of Sum of Partial Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $? How do I find the Sum of the Sum of Finitely Many Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $?
According to maple it is:
$\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} = \left( \left( m+1 \right) ^{2}-m-4 \right) \left( \Psi \left( m+2
\right) +\gamma \right) + \left( - \left( m+1 \right) ^{2}+2\,m+5
\right) \left( \Psi \left( m+3 \right) +\gamma \right) -3
$
| Let us use
$$H_k=- \sum_{m=1}^{k}(-1)^k\frac {{k \choose m}}{m},~~ {n \choose r}={n-1 \choose r-1}+{n-1 \choose r},~~ \sum_{k=m}^{n} {k \choose m}= {n+1 \choose m+1}.$$ Then
$$S_n=\sum_{k=1}^{n} H_k=- \sum_{k=1}^{n} \sum_{m=1}^{n} (-1)^m \frac{{n \choose m}}{m}= -\sum_{m=1}^{n} (-1)^{m} \frac{{n+1 \choose m+1}}{m}
$$ $$\Rightarrow S_n= -\sum_{m=1}^{n} (-1)^m \frac{{n \choose m}}{m}- \sum_{m=1}^{n}(-1)^m \frac{{n \choose m+1}}{m}=H_n- \sum_{m=1}^{n} (-1)^m \frac{n}{m(m+1)} {n-1 \choose m+1}$$ $$\Rightarrow S_n=H_n-n\sum_{m=0}^{n} (-1)^m \left( \frac{1}{m}-\frac{1}{m+1} \right) {n-1 \choose m}$$
$$\Rightarrow S_n= H_n-n\sum_{m=1}^{n-1} (-1)^m \frac{{n-1 \choose m}}{m}-\sum_{m=1}^{n}(-1)^m \frac{n}{m+1} {n-1 \choose m}=H_n+nH_{n-1}-\sum_{m=1}^{n} (-1)^m {n \choose m+1}$$
$$ \Rightarrow S_n= H_n+nH_{n-1}+\sum_{p=2}^{n} (-1)^p {n \choose p}=H_n+n(H_n-\frac{1}{n})+1-n=(n+1)H_n-n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof Verification: $4x^2+6xy+4y^2>0$ unless both $x$ and $y$ are equal to $0$ This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 16(b).
Using the fact that
$x^2 + 2xy + y^2 = (x + y)^2 > 0$,
show that $4x^2 + 6xy + 4y^2 >0$ unless $x$ and $y$ are both $0$.
This is my proof:
Firstly, as any non-zero number $a$ satisfies the inequality $a^2>0$ (I have proved that already), $(x+y)^2>0 \Rightarrow 3(x+y)^2$, $x^2>0$ and $y^2>0$
$\therefore 3(x+y)^2+x^2+y^2=4x^2+6xy+4y^2>0$
Does this make sense?
| More generally,
since
$\begin{array}\\
ax^2+2bxy+ay^2
&=a(x^2+y^2)+b(x^2+2xy+y^2)-b(x^2+y^2)\\
&=(a-b)(x^2+y^2)+b(x+y)^2\\
\end{array}
$
if
$a > b$ then
$ax^2+2bxy+ay^2 \ge 0$
with equality if and only if
$x = y = 0$.
Your case is
$a=4, b=3$.
You can handle the more general case of
$ax^2+2bxy+cy^2$
by looking at how
the quadratic formula
is derived.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find the minimum value of $(\sin x+ \csc x)^2 + (\sec x + \cos x)^2$ I made it till $5+\tan^2x + \cot^2x$, but I don’t know how to proceed further.
| I don't think your function is $5+\tan^2x+\cot^2x$, actually. With $c:=\cos x,\,s:=\sin x$ your sum is $$\left(s+\frac{1}{s}\right)^2+\left(c+\frac{1}{c}\right)^2=5+\frac{1}{s^2}+\frac{1}{c^2}=5+\frac{1}{c^2s^2}=5+4\csc^22x\ge 9,$$with equality iff $\sin2x=\pm 1$, as occurs e.g. if $x=\frac{\pi}{4}$ for which we get$$\left(s+\frac{1}{s}\right)^2+\left(c+\frac{1}{c}\right)^2=2\left(\frac{1}{\sqrt{2}}+\sqrt{2}\right)^2=2\left(\frac{3}{\sqrt{2}}\right)^2=9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$
$ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
$ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator?
is there also alternative proof using jensen inequality?
| Jensen's inequality is $\rm{avg~}f(x_i) \le f(\rm{avg~} x)$ (for concave up functions). So put in $f(z) = 1/z$:
$$\frac{1}{n} \sum_{i} \frac{1}{x_i} \le \frac{1}{\frac{1}{n} \sum_i x_i}$$
Note that this assumes $x_i > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| The terms in the development are of the form $x^{2a}(-x)^b2^c\propto x^3$, and the exponents must satisfy
$$\begin{cases}a+b+c=10,\\2a+b=3.\end{cases}$$
The solutions are $0,3,7$ and $1,1,8$.
Hence by the multinomial formula
$$\frac{10!}{0!3!7!}1^0(-1)^32^7+\frac{10!}{1!1!8!}1^1(-1)^12^8=-120\cdot128-90\cdot256=-38400.$$
By Barry's method, condensed,
$$(p^{10})'=10p^9p',\\(p^{10})''=90p^8p'^2+10p^9p'',\\(p^{10})'''=720p^7p'^3+270p^8p'p''$$
(there is no $p'''$), gives, evaluating at $0$,
$$\frac{720\cdot2^7\cdot(-1)^3+270\cdot2^8\cdot(-1)\cdot2}6=-38400.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.
| The given matrix is equal to $A+B$ where
$$
A=\begin{bmatrix}
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1
\end{bmatrix},\quad
B=\begin{bmatrix}
1&0&0&0&-1&0&0&0&\\
0&1&0&0&0&-1&0&0&\\
0&0&1&0&0&0&-1&0&\\
0&0&0&1&0&0&0&-1&\\
-1&0&0&0&1&0&0&0&\\
0&-1&0&0&0&1&0&0&\\
0&0&-1&0&0&0&1&0&\\
0&0&0&-1&0&0&0&1&
\end{bmatrix}.
$$
These matrices have rank $1$ and $4$ respectively, so their sum has rank at most $1+4$.
| {
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"url": "https://math.stackexchange.com/questions/3305273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$.
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=6$, $\gcd \neq 1$ but I couldn't.
Can someone help me with this problem?
| I think that I just solve it:
Let $d=gcd(n^2+3, (n+1)^2+3).$ By the property $\gcd(a,b)=\gcd(a,b-a)$, we obtain:
$$d=\gcd(n^2+3,2n+1).$$
Since $d \mid n^2+3,$ we have $n^2+3=ds$ for some integer $s$.
Similarly, $2n+1=dr$ for some integer $r$, and by this relation we obtain: $4n^2=d^2 r^2 -2dr +1 \iff 4ds=4(n^2+3)=d^2 r^2 -2dr +13$.
Hence, it must be clear that $13=dq$ for some integer $q$. Then $d \mid 13$.
So $d=13$ or $d=1$ since $13$ is prime.
It is correct?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$
Is there an elegant proof for this identity for all real $s \neq -1, -1/2$?
$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$
Where $H_x$ represents harmonic numbers.
I found it in a very roundabout way for $s=0,1,2,3,\dots$, (see this answer and this answer) and I don't actually have the proof for the other identity I used:
$$\sum_{n=0}^\infty \sum_{k=1}^{n+1}\frac{H_{k-\frac{1}{2}}}{k} x^n= \frac{2}{x(1-x)} \left( \operatorname{arctanh}^2 \sqrt{x}+\log 2 \log (1-x) \right)$$
So whichever one you can prove works fine for me, but I think the first series should be easier to prove.
Using Ali Shather's comment, we can simplify:
$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{2H_{2n}-H_n}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$
One of the interesting consequences:
$$\frac{1}{x} \arcsin^2 \sqrt{x}=\sum_{l=0}^\infty \frac{(4 x)^l}{(2l+1) (l+1) {2l \choose l}}$$
| Partial solution to the second double sum:
Lets work on the inner sum first
\begin{align}
S_n=\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}=\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}+\frac{H_{n+\frac12}}{n+1}
\end{align}
by substituting $\ H_{k-\frac12}=2H_{2k}-H_k-2\ln2$, to have
\begin{align}
\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}&=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\sum_{k=1}^{n}\frac{H_k}{k}-2\ln2\sum_{k=1}^{n}\frac{1}{k}\\
&=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n\\
\end{align}
Therefore
$$S_n=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n+\frac{H_{n+\frac12}}{n+1}\tag{1}$$
Which follows that
$$\sum_{n=0}^\infty\left(\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}\right)x^n=2\underbrace{\sum_{n=0}^\infty\sum_{k=1}^{n}\frac{H_{2k}}{k}x^n}_{\Large S_1}-\underbrace{\sum_{n=0}^\infty\left(H_n^2+H_n^{(2)}\right)x^n}_{\Large S_2}-2\ln2\underbrace{\sum_{n=0}^\infty H_nx^n}_{\Large S_3}+\underbrace{\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n}_{\Large S_4}$$.
Starting with $S_2$ and by using the following identity: ( proved by SuperAbound here)
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=0}^\infty \left(H_n^2-H_n^{(2)}\right)x^n\tag{2}$$
Add $\ \displaystyle2\sum_{n=0}^\infty H_n^{(2)}x^n=\frac{2\operatorname{Li}_2(x)}{1-x}$ to both sides of $(2)$, we get
$$\boxed{S_2=\frac{2\operatorname{Li}_2(x)+\ln^2(1-x)}{1-x}}$$
$$\boxed{S_3=-\frac{\ln(1-x)}{1-x}}$$
and the last sum
\begin{align}
S_4&=\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n\\
&=\sum_{n=1}^\infty\frac{H_{n-\frac12}}{n}x^{n-1}\\
&=\frac1x\sum_{n=1}^\infty \left(\frac{2H_{2n}-H_n-2\ln2}{n}\right)x^n\\
&=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x\sum_{n=1}^\infty \frac{H_{n}}{n}x^n-\frac{2\ln2}x\sum_{n=1}^\infty \frac{x^n}{n}\\
&=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x(\operatorname{Li}_2(x)+\frac12\ln^2(1-x))-\frac{2\ln2}x(-\ln(1-x))\\
\end{align}
The remaining sum can be simplified as follows:
\begin{align}
\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n&=2\sum_{n=1}^\infty \frac{H_{2n}}{2n}(\sqrt{x})^{2n}\\
&=\sum_{n=1}^\infty \frac{H_{n}}{n}(\sqrt{x})^n(1+(-1)^n)\\
&=\operatorname{Li}_2(\sqrt{x})+\frac12\ln^2(1-\sqrt{x})+\operatorname{Li}_2(-\sqrt{x})+\frac12\ln^2(1+\sqrt{x})\\
&=\frac12\operatorname{Li}_2(x)+\frac12\ln^2(1-\sqrt{x})+\frac12\ln^2(1+\sqrt{x})
\end{align}
Thus
$$\boxed{S_4=\frac1x\left(\ln^2(1-\sqrt{x})+\ln^2(1+\sqrt{x})-\frac12\ln^2(1-x)+2\ln2\ln(1-x)\right)}$$
I hope someone will take care of $S_1$ and I hope you find my attempt helpful.
| {
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"url": "https://math.stackexchange.com/questions/3309728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
| \begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{\large x\ \mapsto\frac1x}{=}-I\\
2I&=0\\
I&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote
$y = -2x - 3/2$.
How we get this asymptote?
| Note that if $ax+b$ is an asymptote for the function $f(x)$ in $-\infty$, then$$a=\lim_{x\to -\infty}{f(x)\over x}\\b=\lim_{x\to -\infty}{f(x)-ax}$$therefore$$a{=\lim_{x\to -\infty}{\sqrt{x^3\over x-3}-x\over x}\\=\lim_{x\to -\infty}{-\sqrt{x\over x-3}-1}\\=-2}$$and $$b{=\lim_{x\to -\infty}\sqrt{x^3\over x-3}-x+2x\\=\lim_{x\to -\infty}\sqrt{x^3\over x-3}+x\\=\lim_{x\to -\infty}{{x^3\over x-3}-x^2\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{{3x^2\over x-3}\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{3x^2\over -\sqrt{x^3(x-3)}-x(x-3)}\\=\lim_{x\to -\infty}{3\over -\sqrt{x-3\over x}-{x\over x-3}}\\=-{3\over 2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ holds.
My work.In fact, I want to solve another problem ( Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ ) This another problem has already been solved, but I want to solve it by the method that the author of the problem intended. The fact is that in the original problem there was another inequality ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). It seems to me that the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ is proved as the author of the problem wanted. I think that the author of the problem wanted us to prove inequality $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ on the basis of inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$. To this end, it suffices to prove the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ I checked this inequality by numerical methods on a computer. This inequality holds for all $n$. Interestingly, if $n \to \infty$ then this inequality (in the limit) is an equality.
Perhaps this will help to solve the problem: let $x_k=\frac{k+1}{k}$. Then $1<x_k \le 2$ and inequality takes the form $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$
| We have$$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}{=\sum \limits_{k=1}^n (2k-1)\left\{1+{1\over k}\right\}\\=\sum \limits_{k=1}^n 2k-1+\sum \limits_{k=1}^n {2k-1\over k}\\=n^2+2n-\sum_{k=1}^{n}{1\over k}\\=n^2+n-H_n}$$where $H_n$ is the famous Harmonic Number. Similarly$$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3-3H_{n+1}$$and$$n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)=n^2(n+H_n)(n+1-H_{n+1})$$which by substituting, leave us with the following inequality:
$$(n^2+n-H_n)(n^2-2n-3-3H_{n+1})\le n^2(n+H_n)(n+1-H_{n+1})$$
after expanding the terms we obtain$$LHS\le RHS$$where$$LHS=n^4-4n^2+3n^2H_{n+1}+6nH_{n+1}-3n^2-6n-n^2H_{n}+2nH_{n}-3H_{n}H_{n+1}+3H_{n}\\RHS=n^4+n^3-n^3H_{n+1}+n^3H_{n}+n^2H_{n}-n^2H_{n}H_{n+1}$$by rearranging the terms we can rewrite
$$-n^3-7n^2-6n+H_n(-n^3-2n^2+2n+3)+H_{n+1}(n^3+3n^2+6n)+H_nH_{n+1}(n^2-3)\le0$$
or equivalently
$$3H_n+H_{n+1}(n^2+8n)+H_nH_{n+1}(n^2-3)\le n^3+6n^2+5n+{3n\over n+1}$$
For $n=1$ and $n=2$, this inequality obviously holds and for $n\ge 3$ we divide the two sides of the inequality to $n^2-3$ therefore
$${3H_n\over n^2-3}+H_{n+1}{n^2+8n\over n^2-3}+H_nH_{n+1}\le n+6+{8\over n}+{21n+24\over n^2-3}$$
since for $n\ge 3$ we have ${3H_n\over n^2-3}<1$ we only need to show that
$$H_{n+1}{n^2+8n\over n^2-3}+H_nH_{n+1}\le n+5+{8\over n}+{21n+24\over n^2-3}$$
Also this inequality holds for $3\le n\le 8$ by simple hand calculations and for $n\ge 9$:$${n^2+8n\over n^2-3}<2$$and we only show that
$$2H_{n+1}+H_nH_{n+1}\le n+5+{8\over n}$$
which is obvious since the RHS and LHs are convex and concave functions respectively and the inequality holds for $n=9$. Therefore the proof is complete $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$.
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$
Let $P_{n} = a^{2^n} + b^{2^n} + c^{2^n}$, we have that $$4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$$
$$ = 4(abc)^{2^m} \cdot \left(a^{2^m} + b^{2^m} + c^{2^m}\right) + \left(a^{2^{m + 1}} + b^{2^{m + 1}} + c^{2^{m + 1}}\right)^2 - \left(a^{2^{m + 2}} + b^{2^{m + 2}} + c^{2^{m + 2}}\right)$$
$$ = 4 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right] + 2 \cdot \left[(ab)^{2^{m + 1}} + (bc)^{2^{m + 1}} + (ca)^{2^{m + 1}} \right]$$
$$ = 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$$
So now we need to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$
$\iff \dfrac{1}{2}(a^4 + b^4 + c^4) \mid 4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$. But I don't know about whether it is sufficient to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid P_{n}$ for $\forall n \ge 2$.
And why does $(a, b, c)$ have to be a Pythagorean triple?
| HINT.-Given that $a, b, c$ are Pythagorean triples, we can deduce that the problem is equivalent to demonstrating that for every pair of integers $a, b$ the quotient
$$\frac{(a^{2^m}+b^{2^m})(a^2+b^2)^{2^{m-1}}+a^{2^m}b^{2^m}}{(a^2+b^2)^2-a^2b^2}$$ is always an integer for $m\ge2$ (which can be verified numerically).
Trying to apply induction we have for the minimum value $m=2$ the verification (where the numerator is exactly the square of the denominator)
$$\frac{(a^{4}+b^{4})(a^2+b^2)^{2}+a^{4}b^{4}}{(a^2+b^2)^2-a^2b^2}=(a^2+b^2)^2-a^2b^2$$
I cannot for now (lack of time) complete the inductive demonstration so I invite someone who wishes to do so.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the surface area of a region which is generated by revolving a curve around a line The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is early on in the book
so I would expect / hope any integral would be easy to solve.
Problem:
Find the area of the surface generated by revolving the following curve about the line $y = -1$. The curve is
$y = \frac{x^3}{3} + \frac{1}{4x}$ for $1 \leq x \leq 3$.
Answer:
Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as
revolving it around $y = 0$. The format of the integral for surface area revolved around the y-axis is:
$$ S = \int_a^b 2\pi x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \,\, dx $$
Now we need to find the bounds on $y$.
\begin{align*}
y(1) &= \frac{1^3}{3} + \frac{1}{4} = \frac{1}{3} + \frac{1}{4} \\
y &= \frac{7}{12} \\
y(3) &= \frac{3^3}{3} + \frac{1}{4(3)} = 9 + \frac{1}{12} \\
y(3) &= \frac{109}{12} \\
y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\
\frac{dy}{dx} &= x^2 - \frac{x^{-2}}{4} \\
\frac{dy}{dx} &=\frac{4x^2 - x^{-2}}{4} \\
\frac{dx}{dy} &= \frac{4}{4x^2 - x^{-2}} \\
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \left( \frac{4}{4x^2 - x^{-2}} \right)^2} \,\, dx
\end{align*}
Now we need to integrate.
\begin{align*}
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \frac{16}{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{ \frac{\left(4x^2 - x^{-2}\right)^2 + 16 }{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\
\end{align*}
This does not seem right to me.
Based upon the comments from the group, I updated my solution.
Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as
revolving it around $y = 0$. Let $S$ be the surface area we are trying to find.
\begin{align*}
y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\
y' &= x^2 - \frac{x^{-2}}{4} \\
S &= \int_1^3 2 \pi \left(y+1 \right) \sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\
S &=
\int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right)
\sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\
S &=
\int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right)
\sqrt{ \frac{16x^4 + 8 + x^{-4}}{16} } \,\, dx \\
\end{align*}
This does not seem right to me. How do I complete this integration?
| The integral formula for $S$ In the post is questionable. It is good for surfaces revolving around $x$, not $y$,
The following expression should be used, instead, to integrate surfaces around $y$,
$$2\pi \int_1^3 (y+1)\sqrt{1 + \left( \frac{dy}{dx} \right)^2} \,\, dx $$
Otherwise, the integrand goes to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $3n^2+3n-1=0\pmod{5}$?
How do I solve $3n^2+3n-1=0\pmod{5}$?
I know the answer should be $1 \pmod 5 $ and $3 \pmod5$, but am unsure how to get there. Thanks!
| The quadratic formula still applies, as the integers modulo $5$ is a field (where division by $2$ is allowed), like the rational numbers and the real numbers (this is because $5$ is a prime number):
$$
n\equiv\frac{-3\pm\sqrt{3^2+4\cdot3}}{2\cdot3}\\
=\frac{-3\pm\sqrt{21}}{6}
$$
Now, $-3$ is the same as $2$, $21$ is the same as $1$ and $6$ is the same as $1$. The square root (or rather a square root) of $1$ is $1$. So we simplify and get
$$
n\equiv\frac{2\pm1}{1}
$$
which is easily calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove $(x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
if $x,y,z \ge 0$
I made this problem using Holders inequality, notice that
$$ (1+x)x^{3} + (1+y)y^{3} + (1+z)z^{3} \le \left[ (x^{3})^{2} + (y^{3})^{2} + (z^{3})^{2} \right]^{1/2} \left[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} \right]^{1/2} $$
then squaring both sides we get
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
My question is, will this inequality be obvious to be solved without Holders inequality? if so then what is the alternative solution?
| Yes, you can!
A full expending gives
$$\sum_{cyc}x^6\sum_{cyc}(x+1)^2-\left(\sum_{cyc}(x^4+x^3)\right)^2=$$
$$=\sum_{cyc}(x^6y^2+x^6z^2-2x^4y^4+2x^6y-2x^4y^3-2x^4z^3+2x^6-2x^3y^3)=$$
$$=\sum_{cyc}x^2y^2(x^2-y^2)^2+2\sum_{cyc}xy(x+y)(x^2+xy+y^2)(x-y)^2+\sum_{cyc}(x^3-y^3)^2\geq0.$$
Actually, the inequality that you named Holder's inequality, named Cauchy-Schwarz inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$ Problem:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
Using two different methods I am getting two different answers and have trouble finding why.
Method 1:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$
$$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$x+2=u$$
$$dx=du$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$
Method 2:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$
$$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$
$$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$
$$2x+4=u$$
$$dx=\frac{du}{2}$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$
| Both answers are the same, mind the constant:
$$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Do all unitary matrices belong to a one-parameter unitary group (for Stone's theorem). Background. Per Stone's theorem, a one-parameter unitary matrix group $U_t$ corresponds to a Hermitian matrix $H$:
$$U_t=e^{iHt}$$
Example. The group of unitary matrices
$$U_t=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&\cos t&i\sin t \\ 0&0&i\sin t&\cos t \\
\end{matrix}
\right)
$$
corresponds to the matrix
$$H=
\left(
\begin{matrix}
0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \\
\end{matrix}
\right)
$$
and then for $t=\pi/2$ we have
$$
U_{\pi/2}=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&i \\ 0&0&i&0 \\
\end{matrix}
\right)
$$
Question. For any given unitary matrix $U$, does it belong to a one-parameter group, such that Stone's theorem applies?
I've been trying to find $H$ for the matrix below, but haven't yet managed to:
$$
U_{?}=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \\
\end{matrix}
\right)
$$
Note that may not be relevant: it's unitary, though with $det~U_?=-1$.
| It is indeed true that every unitary matrix $U$ is an element of some one-parameter subgroup. Equivalently, the map $H \mapsto \exp(iH)$ is a surjective map from the Hermitian matrices to the unitary matrices. One can say that this is a consequence of the fact that the unitary matrices form a connected, compact Lie group; if you prefer, this can also be proven as a consequence of the construction suggested below.
The $U$ which you have given can be diagonalized as
$$
U = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0} = \\
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{1\\&1\\&&1\\&&&-1}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
$$
so, it belongs to the one-parameter subgroup
$$
U_t = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{1\\&1\\&&1\\&&&e^{it}}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
\\
= I + \frac{e^{it} - 1}{2}\pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&-1\\0&0&-1&1},
$$
which corresponds to the generator
$$
H = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{0\\&0\\&&0\\&&&1}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
\\ = \frac 12 \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&-1\\0&0&-1&1}.
$$
Your $U$ is attained at $t = \pi$.
As far as the determinant goes, you might find it interesting that
$$
\det U_t = \det \exp(iHt) = \exp (\operatorname{trace}(iHt)) = e^{it}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $0 < x < y$, then $x^n < y^n$ The problem asks to prove that if $0 < x < y$ then $x^{n} < y^{n}$, where $n$ is a positive integer, so I started by assuming that $0 < x < y$. I then wrote this chain of inequalities: $x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$. It is true that $x^{n} < x^{n-1}y^{1}$, since dividing both sides by $x^{n-1}y^{0}$ yields $x < y$. Similarly, dividing both sides of $x^{n-1}y^{1} < x^{n-2}y^{2}$ by $x^{n-2}y^{1}$ yields $x < y$. It seems to me that this pattern continues for all the inequalities, but I haven't been able to prove this. Does that make the proof invalid? Also, how can I prove that this pattern always holds true?
| Your proof is absolutely correct. But to make the proof clearer and standard I would recommend you to write these first.
$\because$ $0<x<y$.
$\implies \frac{y}{x}>1$ .....$(i)$
Multiplying both sides by $x^n$ we get,
$x^{n-1}y^1>x^n$
Again Multiplying both sides of $(i)$ by $x^{n-1}y$ we get,
$x^{n-2}y^2>x^{n-1}y$
Continuing this process of multiplying LHS of the obtained inequality to $(i)$, we finally obtain,
$y^n>x^1y^{n-1}$
Now you can write
$x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$.$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice? Question
Let's say I have $n$ fair 6-sided dice. What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice?
Example
$n=8$:
Suppose $8$ dice are rolled. If the first roll is $[1, 3, 6, 2, 3, 4, 3, 1]$, there are two 1's, one 2, three 3's, one 4, and one 6.
Rolling the same dice again, what is the probability that the next roll will also have two 1's, one 2, three 3's, one 4, and one 6?
The above example uses 8 dice, but I'm curious about the probability for any positive $n$.
Edit
My question is different than Two dice throw probability since I'm looking for the probability of $n$ dice having the same result in two consecutive throws
| You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for example, this answer on mathoverflow.
For example, if we have $n = 2$, we need to sum over all variants to partitioning $2$ into $6$ non-negative terms. There are $15$ of them with $2$ ones and $4$ zeroes: $1+1+0+0+0+0$, $1+0+1+0+0+0$, ..., $0+0+0+0+1+1$ and $6$ with $1$ two and $5$ zeroes: $2+0+0+0+0+0$, $0+2+0+0+0+0$, ...
First variant will give in our sum $15$ terms equal to $\left(\frac{2!}{1!1!0!0!0!0!} \cdot\frac{1}{36}\right)^2 = \frac{1}{18^2}$. Second will give $6$ terms equal to $\left(\frac{2!}{2!0!0!0!0!0!} \cdot \frac{1}{36^2}\right)^2 = \frac{1}{36^2}$. So the answer is $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.
| {
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"timestamp": "2023-03-29T00:00:00",
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a hint for this Taylor series$ \frac{\cos\left(2x\right)-1}{x^2}$ Compute the first three terms (nonzero) $\frac{\cos\left(2x\right)-1}{x^2}$
the first term is $\cos \left(2\right)-1$ but in the answer, the first term that I have to choose is... $-2$ or $2$ or $-1/2$
UPDATE: Use the substitution method
| Assuming that you want to find the Maclaurin series which is the Taylor series expansion about $0$, you can start from
$$\cos(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+ \dots$$
therefore
\begin{align}\cos(2x)=\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n}}{(2n)!}&=1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!}+ \dots\\&=1-2x^2+\frac{2x^4}{3}-\frac{4x^6}{45}+\dots\end{align}
and
\begin{align}\frac{\cos(x)}{x^2}=\sum_{n=0}^{\infty}(-1)^n\frac{2^{2n}(x)^{2n-2}}{(2n)!}&=\frac{1}{x^2}-\frac{(2x)^2}{2!(x^2)}+\frac{(2x)^4}{4!(x^2)}-\frac{(2x)^6}{6!(x^2)}+ \dots \\&=\frac{1}{x^2}-\frac{2^2}{2!}+\frac{2^4 x^2}{4!}-\frac{2^6 x^4}{6!}+ \dots
\\&=\frac{1}{x^2}-2+\frac{2x^2}{3}-\frac{4x^4}{45}+ \dots\end{align}
will allow you to analyze terms of the Taylor series for $\frac{cos(2x)-1}{x^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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General rule for integrals of irrational functions involving $r = \sqrt{a^2 + x^2}$ Browsing the List of integrals of irrational functions I found out that if
$$r=\sqrt{(a^2+x^2)}$$
then
$$\int{rdx}=\frac12(xr+a^2\ln{(x+r)})$$
$$\int{r^3dx}=\frac14xr^3+\frac38a^2xr+\frac38a^4\ln(x+r)$$
$$\int{r^5dx}=\frac16xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln(x+r)$$
Is there any general rule for defining $$\int{r^ndx}, \quad n=2k+1, \quad k\in\mathbb{N}$$
If so, then how to derive the formula? Thank you.
| If we denote the integral by
$$I_k=\int\left(a^2+x^2\right)^{(2k+1)/2}\mathrm{d}x$$
then we can apply integration by parts to get
$$\begin{align}
I_k
&=x\left(a^2+x^2\right)^{(2k+1)/2}-(2k+1)\int x^2\left(a^2+x^2\right)^{(2k-1)/2}\mathrm{d}x\\
&=x\left(a^2+x^2\right)^{(2k+1)/2}-(2k+1)\int \left(a^2+x^2\right)^{(2k+1)/2}-a^2\left(a^2+x^2\right)^{(2k-1)/2}\mathrm{d}x\\
&=x\left(a^2+x^2\right)^{(2k+1)/2}-(2k+1)I_k+a^2(2k+1)I_{k-1}\\
\end{align}$$
Then we have the reduction formula
$$I_k=\frac1{2(k+1)}\left(x\left(a^2+x^2\right)^{(2k+1)/2}+a^2(2k+1)I_{k-1}\right)$$
So we then only need the solution to $I_1$ in order to calculate any other $I_k$. In fact it's easier to calculate $I_{-1}$ which is
$$I_{-1}=\int\frac{\mathrm{d}x}{\sqrt{a^2+x^2}}=\ln{\left(x+\sqrt{a^2+x^2}\right)}+C=\text{arsinh}\left(\frac{x}a\right)+C$$
| {
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Find sum $\sum _{k=0}^nF_kF_{n-k}$
Find sum $\sum _{k=0}^nF_kF_{n-k}$
My try
Let $$ a_n = \sum _{k=0}^nF_kF_{n-k} // \cdot x^n
\\ a_n x^n = \sum _{k=0}^nF_k x^k F_{n-k} x^{n-k} // \sum_n \\
A(x) = (F(x))^2 + \cdots + (F(x))^2 = n\cdot (F(x))^2$$
I have some doubts:
*
*I am not sure that summing $F_k x^k$ by $n$ gives $F(x)$ (generating functions for $f_n$ sequence)
*If it is true, how can it be finished? There is weird situation. I know that $a_n$ is a element from $\sum _{k \ge 0} F_k \cdot \sum _{k \ge 0} F_k$ multiplied by $n$ but it is not obvious for me how it can be extracted from there.
| I guessed the formula, then prove it by induction.
Let $s_n = \sum_0^n F_k F_{n-k}$
From identity: $F_m F_n = F_m F_{n+1} + F_{m-1} F_n$
$s_n = (n-1) F_{n-1} - s_{n-2}$
$s_n = (n-1) F_{n-1} - (n-3) F_{n-3} + (n-5) F_{n-5} - \cdots$
Simplify the sum, so that $s_n = a F_{n} + b F_{n-1}$, and build a table:
$$\begin{matrix}
n & s_n & a & b \cr
2 & 1 & 1/5 & 4/5 \cr
3 & 2 & 2/5 & 6/5 \cr
4 & 5 & 3/5 & 8/5 \cr
5 & 10 & 4/5 & 10/5 \cr
6 & 20 & 5/5 & 12/5
\end{matrix}$$
Guessed formula for $s_n = \frac{(n-1)F_n + 2 n F_{n-1}}{5}$
The prove is in pairs, check n=1,2; assumed true for n=k; check n=k+2:
For n=1, $\frac{(1-1)(1) + (2)(1)(0)}{5} = 0$
For n=2, $\frac{(2-1)(1) + (2)(2)(1)}{5} = 1 = F_1^2$
For n=k+2, use recurrence formula to build $s_{k+2}$:
$$(k+1)F_{k+1} - \frac{(k-1)F_k + 2 k F_{k-1}}{5} = \frac{((k+2)-1)F_{k+2} + 2 (k+2) F_{(k+2)-1}}{5}$$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
The question is: How to obtain this integral by other methods not using the trigonometric substitution.
| Because we have the radical $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}$ (and because we know the anser), we may try the Euler substitution $$
t=\sqrt{\frac {1+x}{1-x}}\ .
$$
In
https://en.wikipedia.org/wiki/Euler_substitution
it is the third Euler substitution:
$$
\sqrt{(1+x)(1-x)}=t(1-x)\ .
$$
Then $t^2=(1+x)/(1-x)$, $t^2(1-x)=1+x$, $x(t^2+1)=t^2-1$,
$$
x =\frac{t^2-1}{t^2+1}=
1-\frac2{t^2+1}\ ,
$$
so we can express $x$ rationally in terms of $t$, so also $t(1-x)$, which is the radical.
Putting all together:
$$
\begin{aligned}
\int\frac 1{(1-x)^2\sqrt{1-x^2}}\; dx
&=
\int\frac 1{\frac4{(t^2+1)^2}\cdot \frac {2|t|}{t^2+1}}\;\frac{4t}{(t^2+1)^2}\; dt
\\
&\qquad\text{(and i will assume $t>0$...)}
\\
&=\int \frac 12(t^2+1)\; dt
\\
&=\frac 16t^3+\frac 12 t+\text{locally constant function .}
\\
&=\frac 16\cdot t\cdot(t^2+3)+\text{locally constant function .}
\end{aligned}
$$
This matches, since $t^2+3$ is $2(x-2)/(x-1)$.
| {
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Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$
I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence being subtracted is a partial sum of its Taylor series. I believe you use the alternating series test to explain, but all I get from it is that the series will converge on $[-1,1]$.
| you can actually get a stronger result such as $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 1/11 $$
because the error in a convergent alternating series is less than the absolute value of the first missing term which in this case is $1/11.$
The alternating series is simply the Taylor series of $\tan ^{-1} x $ evaluated at $x=1$ and it converges by the alternating series test.
| {
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Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7x^4(1-x^2)^{-9/2}$
From this I get $f'(0)=2 $, $f''(0)=0 $, $f^{(3)}(0)=2 $, $f^{(4)}(0)=0 $
But I don't know how to find a general term. Can this be solved in easier steps?
| As David has pointed out, $\cos^{-1}(1-2x^2)$ is not differentiable at $x=0$, so it does not have a Maclaurin series in a strict sense.
But now let $x\ge0$ and $\cos^{-1}(1-2x^2)=u$. We have
$$1-2x^2=\cos u$$
$$\sqrt{\frac{1-\cos u}2}=x=\sin\frac u2$$
$$\sin^{-1}x=\frac u2$$
$$u=2\sin^{-1}x=\cos^{-1}(1-2x^2)$$
Now the Maclaurin series of $2\sin^{-1}x$ can be derived easily:
$$2\sin^{-1}x=\sum_{n=0}^\infty\frac1{2^{2n-1}(2n+1)}\binom{2n}nx^{2n+1}$$
Therefore $\cos^{-1}(1-2x^2)$ may be written as a "quasi-Maclaurin series", using $|x|$ and not just $x$:
$$\cos^{-1}(1-2x^2)=\sum_{n=0}^\infty\frac1{2^{2n-1}(2n+1)}\binom{2n}n{|x|^{2n+1}}\qquad{|x|\le1}$$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.