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What's the derivative of $f(x)=(\frac{x^2 + 1}{x^2 + 3})^{\sin(2x)}$ I'm trying to calculate the derivative of $f(x)=(\frac{x^2 + 1}{x^2 + 3})^{\sin(2x)}$ using the difference quotient but somehow I don't succeed.
Help and elegant solution appreciated!
|
Hint
$$\log f(x) = \sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)$$
$$\begin{align} \implies f'(x) &= f(x)\frac{d}{dx}\left[\sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]\\&=\left(\frac{x^2 + 1}{x^2 + 3}\right)^{\sin(2x)}\frac{d}{dx}\left[\sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]\end{align}$$
I will now focus on the remaining derivative
$$\frac{d}{dx}\left[\sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]$$
$$=\left[2\cos(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)+\sin(2x)\frac{d}{dx}\log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]$$
If we now just focus on the logarithm; using the substitution $u=\frac{x^2 + 1}{x^2 + 3}$ we get
$$\frac{d}{dx}\log\left(\frac{x^2 + 1}{x^2 + 3}\right)$$
$$=\frac{1}{u}\frac{du}{dx}$$
$$=\frac{x^2 + 3}{x^2 + 1} \frac{4 x}{(x^2+3)^2}$$
$$=\frac{4 x}{(x^2 + 1)(x^2+3)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$? I have to compare the numbers $2^{39}$, $5^{19}$ and $52^7$. I don't know how to do that because their exponents don't have anything in common.
|
Hint: you can establish the right order between $5^{19}$ and $2^{39}$ as follows
\begin{align}
5^{19} &= 5^{20-1}\\
&=\frac{5^{20}}{5}\\
&=\frac{(5^2)^{10}}{5}\\
&=\frac{25^{10}}{5}\\
&>\frac{16^{10}}{2}\\
&=\frac{(2^4)^{10}}{2}\\
&=2^{39}
\end{align}
then note that
\begin{align}
52^7 &= 52^{10-3}\\
&=\frac{52^{10}}{52^3}\\
&=\frac{26^{10}\cdot 2^{10}}{26^{3}\cdot 2^{3}}\\
&<\frac{25^{10}\cdot 2^{10}}{26^{3}\cdot 2^{3}}\\
&<\frac{25^{10}\cdot 2^{10}}{26^{3}\cdot 5}\\
&=\frac{5^{20-1}\cdot 2^{10}}{26^{3}}\\
&=5^{19} \cdot \frac{2^{10}}{26^3}\\
&<5^{19}
\end{align}
and now you have to find if $52^7$ is somewhere in between $5^{19}$ and $2^{39}$ or under $2^{39}$.
|
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|
least value of expression What is the least value of $\csc^2(x)+36\sec^2(x)$ ?
So I differentiated and when simplifying we get $\tan^4(x)=1/36$, but that is giving me max value i think. Can someone just help me . Or is there any other efficient way. Thanks !
|
As you stated, yes taking derivative equal to zero you get $$\tan^4(x)=\frac{1}{36} \rightarrow tan^2(x)=\frac{1}{6} $$
More over $$ \frac{1}{\cos^2(x)}= 1+tan^2(x) = 1+ \frac{1}{6} =\frac{7}{6}$$ so the minimum is at such $x$. Substitute in the original formula :
$$\frac{1}{\sin^2 x} + \frac{36}{\cos^2 x} = \frac{1}{\cos^2(x)} \Big( \frac{cos^2(x)}{\sin^2 (x)} + 36\Big) =\frac{1}{\cos^2(x)} \Big( \frac{1}{\tan^2 (x)} + 36\Big) = \frac{7}{6} ( 6+36) =49 $$
By the way, you may see that also graphically:
|
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|
Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$ Let $a,b,c$ are $3$ edge of a triangle. Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$. My try: I suppose $c=\min\{a,b,c\}$ but I don't know what next.
|
I give a correct answer here (in the last one there was a wrong factor $4$ instead of the correct $16$).
Basically I use here the following two facts:
(a) Stewart’s theorem.-It is applied to triangles with a cevian. In the figure below we apply this theorem twice, with the cevian $d=OG$ in $\triangle DOB$ and with the median (which is a cevian of course) $BD$ in $\triangle ABC$.
(b) The fact that $A-M\ge G-M$, applied to
$$ \frac{\frac ab+\frac ba+\frac bc+\frac cb+\frac ac+\frac ca}{6}\ge \sqrt[6]{1\cdot 1\cdot 1}=1$$ hence $$\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}\ge 6\qquad (1)$$
Taking in account the formulas $$S=\sqrt{p(p-a)(p-b)(p-c)}=\frac {abc}{4R}$$
giving the area $S$ of a triangle $\triangle {ABC}$, the proposed inequality becomes
the equivalent $$\frac{a^2+b^2+c^2}{R^2}\le (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\qquad (2)$$
We’ll prove $(2)$.
We have $$h^2=R^2\frac{b^2}{4}\qquad (3)$$
$$\text{(Stewart)}\space R^2m+h^2(2m)=(m+2m)(d^2+2m\cdot m)$$ i.e.
$$R^2+2h^2=3d^2+6m^2\qquad (4)$$ $(3)$ in $(4)$ gives
$$3R^2-\frac{b^2}{2}=3d^2+6m^2\qquad (5)$$
$$\text{(Stewart again)}:\space\space c^2(\frac b2)+a^2(\frac b2)=b(3m)^2+\frac{b^2}{4}$$ i.e.
$$c^2+a^2=18m^2+\frac{b^2}{2}\qquad (6)$$
Eliminating $m$ in $(5)$ and $(6)$ one gets
$$9R^2-(a^2+b^2+c^2)=9d^2\ge 0\Rightarrow \frac{a^2+b^2+c^2}{R^2}\le 9$$
$$\color{green}{\text{is it}\space 9\le (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)?}$$
If YES, we have finish. In fact it is; because of $(1)$ one has
$$(a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)=\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}+3\ge 6+3$$
This ends the proof.
|
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|
Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where to start. It’s trivial to construct some solution of the form $\int u'v dx = uv - \int uv' dx$ to the integral on the left, but I can’t see how to get at this exact one.
EDIT:
I have tried to solve it by splitting it up,
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx
\end{equation}
but as far as I can tell this results in something rather different from where I am supposed to end up:
\begin{equation}
\int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx = \frac{1}{a}arctan \Big(\frac{x}{a}\Big) \cdot \frac{1}{(x^2 + a^2)^{n-1}} + \frac{2(n-1)}{a} \int{\frac{x^2 arctan \big(\frac{x}{a}\big)}{(x^2 + a^2)^{n}}}dx
\end{equation}
It is quite possible that I have made a very obvious mistake, so apologies in advance.
I have also tried this:
\begin{equation}
\int{(x^2 + a^2)^{-n}dx} = \int{\Big(1 \cdot (x^2 + a^2)^{-n}\Big) dx}
= x(x^2 + a^2)^{-n} + n\int{\frac{2x^2}{(x^2 + a^2)^{n+1}} dx}
\end{equation}
Again, it doesn’t seem to lead me nearer the specific solution I need.
|
Hint:
Use integration by parts with $\;\displaystyle \int\dfrac{\mathrm d\mkern1mu x}{(x^2+a^2)^{n-1}}$, setting
$$u=\frac1{(x^2+a^2)^{n-1}},\quad\mathrm d\mkern1mu v=\mathrm d\mkern1mu x.$$
|
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|
Say for what values of $a \in \mathbb {R} $ this matrix system has solutions
Let $a \in \mathbb{R}$ and
$$
A_a =
\begin{pmatrix}
1 & a & 1 \\
a & 2 & 3 \\
2 & 3 & 4
\end{pmatrix}.
$$
*
*Say for each values of $a \in \mathbb{R}$ the system:
$$
A_a
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix}
$$
accepts one solution only and for what values have no solutions.
*Resolve the system for $a = 1$.
I got this problem from this image.
I think I know how to resolve system of equations using inverse matrices $3 \times 3$, $2 \times 2$ and more by finding the determinant. Then, I can reflect the matrix about the diagonal and multiply it with the inverse to find the solution. However, here, I get a little lost, because it seems like I should work backwards. What should I do?
|
You can do the exercise in one swoop by using Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
a & 2 & 3 & 0\\
2 & 3 & 4 & 1
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
0 & 2-a^2 & 3-a & -a\\
0 & 3-2a & 2 & -1
\end{array}\right] && \begin{matrix}R_2\gets R_2-aR_1\\R_3\gets R_3-2R_1\end{matrix}
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & a & 1 & 1 \\
0 & 3-2a & 2 & -1 \\
0 & 2-a^2 & 3-a & -a
\end{array}\right] && R_2\leftrightarrow R_3
\\[6px]\color{red}{(3-2a\ne0)}\quad&\to
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
0 & 1 & 2/(3-2a) & 1/(2a-3)\\
0 & 2-a^2 & 3-a & -a
\end{array}\right] && R_2\gets (3-2a)^{-1}R_2
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & a & 1 & 1 \\
0 & 1 & 2/(3-2a) & 1/(2a-3) \\
0 & 0 & \frac{(4a-5)(a-1)}{3-2a} & \frac{(a-2)(a-1)}{3-2a}
\end{array}\right] && R_3\gets R_3-(2-a^2)R_2
\end{align}
If $a=1$, the last row is zero, which means the system has infinitely many solutions; the matrix $A_a$ has rank $2$.
If $a=5/4$, the matrix $A_a$ has rank $2$, and the system has no solution.
If $a\notin\{1,5/4,3/2\}$, the matrix $A_a$ has rank $3$ and the system has a single solution.
It remains to see the case when $a=3/2$.
\begin{align}
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
0 & 2-a^2 & 3-a & -a\\
0 & 3-2a & 2 & -1
\end{array}\right]
&=
\left[\begin{array}{ccc|c}
1 & 3/2 & 1 & 1\\
0 & -1/4 & 3/2 & -3/2\\
0 & 0 & 2 & -1
\end{array}\right]
\end{align}
It's clear that the matrix $A_a$ has rank $3$ and the system has a single solution.
The solution for $a=1$ is readily available, because the last matrix in the elimination process is
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\
0 & 1 & 2 & -1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
Do the transformation $R_1\gets R_1-R_2$ to get
$$
\left[\begin{array}{ccc|c}
1 & 0 & -1 & 2 \\
0 & 1 & 2 & -1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
so the system can be written
$$
\begin{cases}
x=2+z\\[4px]
y=-2-2z
\end{cases}
$$
so the solutions are of the form
$$
\begin{bmatrix}2\\2\\0\end{bmatrix}+
t\begin{bmatrix}1\\-2\\1\end{bmatrix}
$$
|
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|
Prove that $f(n) = 3n^5 + 5n^3 + 7n$ is divisible by 15 for every integer $n$
*
*So far I have only been able to complete the base case for which I got the following:
$$f(n) = 3n^5 + 5n^3 + 7n$$
$$f(n) = 3(1)^5 = 5(1)^3 + 7(1)$$
$$f(n) = 3 + 5 + 7$$
$$15/15 = 1$$
From here I got a bit confused with the inductive step in terms of number manipulation, a hint that my professor gave me was $f(-n) = -f(n)$. Any further help is appreciated.
|
Although you asked for induction, it's much easier to simply split this into cases. We'll first prove that it's divisible by $3$.\begin{align}
n\equiv 0\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 0+0+0\equiv 0\mod 3\\
n\equiv 1\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3+5+7\equiv 0\mod 3\\
n\equiv 2\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3(-1)^5+5(-1)^3+7(-1)\equiv 0\mod 3
\end{align}
And so $f(n)\equiv 0\mod 3$ for every $n$, so $3|f(n)$. I suppose you try $5|f(n)$ as an excercise for yourself, as I'm sure you're able to. Let me know if you encounter any problems.
I should probably also note Fermat's Little Theorem, which states that for any prime $p$, the equivalence $a^p\equiv a\mod p$ holds for all integer $a$. This means you can reduce your polynomial to (let's take $\mod 3$ again) to \begin{align}
f(n)&=3n^5+5n^3+7n\\
&\equiv 3\cdot n^3\cdot n^2+5n^3\cdot n^3\cdot n+7n\\
&\equiv 3\cdot n\cdot n^2+5n\cdot n\cdot n+7n\\
&\equiv 3n^3+5n^3+7n\\
&\equiv 3n+5n+7n\\
&\equiv 15n \mod 3
\end{align}
and from $f(n)\equiv 15n\mod 3$ we can easily see that $$f(n)\equiv 3\cdot 5n\equiv 0\mod 3$$
Hope this helped!
|
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|
I need some help with Geometry. Is this a correct answer to this problem? Good day,
I have a question regarding geometry. I don't know whether my answer is correct because the answer in my book uses a totally different method for solving this particular problem.
Here's the problem:
Given is a triangle, ABC, in which the middle of AB is also the middle of the circle drawn outside of the triangle. Finally, we draw a line from C to M where M signifies the point in the middle of AB. Now you have angle C sub 1 and angle C sub 2 ( together: angle C sub 12.)
We have created 2 equal-sided triangles, within the triangle ABC, within the circle, namely:
triangle ACM and triangle BCM.
Proof that angle C sub 12 is equal to 90 degrees.
My answer:
Given:
AM = MC = MB (2 equal-sided triangles)
To Proof:
$$\angle C_{12} = 90^{\circ}$$
Proof:
$$
\left.\begin{matrix}
& & & & & \\
& & & & & \\
& & & & & \\
\angle A + \angle M_{1} + \angle C_{1} = 180^{\circ}\\
\angle B + \angle M_{2} + \angle C_{2} = 180^{\circ}\\
\angle M_{2} + \angle 2B = 180^{\circ} \\
\angle M_{1} + \angle 2A = 180^{\circ} \\
\angle M_{1} + \angle M_{2} = 180^{\circ}\\
\angle A + \angle B + \angle C = 180^{\circ} (= Q)\\
\\
\end{matrix}\right\}general
\\
$$
Calculate for angle A and angle B:
$$ \angle M_{2} + \angle 2B = 180^{\circ}\\
\angle 2B = 180^{\circ} - \angle M_{2}\\
\angle B = 90^{\circ} - (\angle M_{2} / 2) $$
$$and$$
$$ \angle M_{1} + \angle 2A = 180^{\circ}\\
\angle 2A = 180^{\circ} - \angle M_{1}\\
\angle A = 90^{\circ} - (\angle M_{1} / 2) $$
Also:
$$
\angle M_{2} = 180^{\circ} - \angle M_{1}\\
\angle M_{1} = 180^{\circ} - \angle M_{2} (= U)\\
$$
Now we can put angle A and angle B in the sum of angles (indicated by Q in general section.):
$$(90^{\circ} - (\angle M_{1} / 2)) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}$$
Now we substitute $$ \angle M_{1}$$ inside $$((\angle M_{1} / 2)) $$ with U:
$$(90^{\circ} - (\frac{180^{\circ} - \angle M_{2}}{2}) + ((90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\
(90^{\circ} - (\frac{180^{\circ}}{2}-\frac{\angle M_{2}}{2})) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\
90^{\circ} - \frac{180^{\circ}}{2}+\frac{\angle M_{2}}{2} + 90^{\circ} -\frac{\angle M_{2}}{2} + \angle C = 180^{\circ}\\
\Rightarrow 90^{\circ} + \angle C = 180^{\circ}\\
Conclusion: \angle C_{12} = 90^{\circ}
$$
(not the prettiest proof but hopefully you get the idea.)
Here's the more concise ( and beautiful ) proof from the answers in my book:
$$\angle A + \angle B + \angle C = 180^{\circ} \Rightarrow \angle C_{1} + \angle C_{2} + \angle C_{12} = 180^{\circ} \Rightarrow \angle 2C_{12} = 180^{\circ} \Rightarrow \angle C_{12} = 90^{\circ} $$
If you need more information please ask and I'll provide the necessities.
Thanks!
Picture:
|
It is straightforward.
$\angle A+\angle B+\angle C=\pi$
But given that $CM=AM=BM$, hence creating two isocele triangles $AMC$ and $BMC$ you have
$\angle A+\angle B=\angle C$
Hence the same as in your book $2\angle C=\pi$...
|
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|
For which $a$ and $b$ the function $\frac{ax+b}{a^2x+b^2}$ is increasing?
For which $a$ and $b$ the function $$\frac{ax+b}{a^2x+b^2}$$ is increasing?
I know that function is increasing if $x_1 > x_2 \implies f(x_1)>f(x_2)$ but how can I find $a$ and $b$ for which this statement will be true?
|
Divide both polynomials:
$$\frac{ax+b}{a^2x+b^2}=\frac{a}{a^2}\left(1+\frac{b/a-b^2/a^2}{x+b^2/a^2}\right)=\frac{a}{a^2}+\frac{{a/a^2(b/a-b^2/a^2)}}{x+{b^2/a^2}}$$
This is function $1/x$ scaled and moved. The scale is
$$k=\frac{a}{a^2}\left(\frac{b}{a}-\frac{b^2}{a^2}\right)$$
You know that $c/x$ is decreasing when $c>0$ and increasing when $c<0$. In your case $k<0$:
$$\frac{a}{a^2}\left(\frac{b}{a}-\frac{b^2}{a^2}\right)<0$$
$$a(ab-b^2)<0$$
$$a^2b-b^2a<0$$
$$a^2b<b^2a$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Area stacked between common tangent and circles Is there any way to find area of shaded region?
The radii of circles are $4$ and $12$ units.
|
By similar triangles, $CD=8$.
By Pythagoras' Theorem, $CG=EF=8\sqrt{3}$
Area of trapezoid $ACEF=\frac{(4+12)\times 8\sqrt{3}}{2}=64\sqrt{3}$
$\angle BAF=\cos^{-1} \left( \frac{1}{2} \right)=60^{\circ}$ and
$\angle BCE=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore$ area of sector $BAF
=\pi (12)^{2} \times \frac{60^{\circ}}{360^{\circ}}
=24\pi$
$\therefore$ area of sector $BCE
=\pi (4)^{2} \times \frac{120^{\circ}}{360^{\circ}}
=\frac{16\pi}{3}$
The required area $=64\sqrt{3}-24\pi-\frac{16\pi}{3}
=64\sqrt{3}-\frac{88\pi}{3}$
|
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|
Prove that $|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$
Prove for any $a, b \in [-1, 1]$ that $$|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$$
I'm sure there is a solution using the Cauchy-Swartz inequality. Thus i tried to prove $$\Big(a\sqrt{(1-b)(1+b)}+b\sqrt{(1-a)(1+a)}-\sqrt{3(1-a)(1+a)(1-b)(1+b)}+\sqrt{3}ab\Big)^2 \le 4$$
Can we use C-S to prove that?
|
Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that
$$
\lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta|
=\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2,
$$
that is,
$$
\left|\frac12\sin(\alpha+\beta)-\frac{\sqrt3}2\cos(\alpha+\beta)\right|\le1.
$$
Of course, the last one is immediate.
|
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|
Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$ How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$
without using induction? Note that clearly $n\neq 0$
Thanks for any help!!
|
The reverse inequality is true for $n≥1$.
If $1≤k≤n$ then $\frac{1}{n+k}≥\frac{1}{n+n}$. Therfore, you get :
$$\frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n} ≥ \frac{1}{n+n}+\frac {1}{n+n}+\frac{1}{n+n}...+\frac{1}{n+n} = n \cdot \frac{1}{n+n} = \frac{1}{2}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\frac{n}{n+1} < \frac{n+1}{n+2}$ by induction? I have the inequality
$\frac{n}{n+1} < \frac{n+1}{n+2}$
I'm not sure how to go about proving it. I've started by testing with n = 1, which results in
$\frac{1}{2} < \frac{2}{3}$ which is true
I then assume true for n = k and have to prove that it is true for n = k + 1, but I don't know how to start manipulating
$\frac{k}{k+1} < \frac{k+1}{k+2}$
to become
$\frac{(k+1)}{(k+1)+1} < \frac{(k+1)+1}{(k+1)+2}$
How do I go about doing this?
|
if you need induction:
\begin{gather*}
n(n+2)<(n+1)^2 \\
\begin{aligned}
(n+1)(n+3)={}&n(n+2+1)+(n+3)=(n(n+2)+n+n+3)<(n+1)^2+2n+3={} \\
{}={}&(n^2+4n+4)=(n+2)^2.
\end{aligned}
\end{gather*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
The equation represents the line of intersection of two planes. Using augmented matrix
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
2 & 3 & 4 & 6
\end{bmatrix}$$
$R_2\rightarrow R_2-2R_1$
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 0
\end{bmatrix}$$
$$y+2z=0$$
$$x+y+z=3$$
Equation of line is
$$\frac{x-0}{1}=\frac{y-0}{-2}=\frac{z-0}{1}$$
Angle made by line and normal is $$\cos(90-\theta)=\frac{1}{\sqrt{6}}$$
where $\theta$ is the angle made by line and plane.
How should I proceed?
|
Note:
I get
$$
y + 2z = 0 \Rightarrow z = - y/2 \\
3 + z = (x + y + z) + z = x + y + 2z = x
$$
So
$$
\frac{x - 3}{1} = \frac{y-0}{-2} = \frac{z - 0}{1}
$$
or
$$
(x, y, z) = (3,0,0) + t (1, -2, 1)
$$
Hint: The projection onto $z=0$ should affect what coordinate?
Spoiler:
The $z$-coordinate is forced to $0$. So we have $(x,y,z) = (3,0,0) + t (1,-2,0)$. We have $y = - 2 t$ and $x - 3 = t$ so $y = -2(x-3) = -2x + 6$ is the equation of the projected line within the $x$-$y$-plane.
|
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|
integrate $\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx=\int \frac{3\left(\frac{16}{9}-x^2\right)^{\frac{3}{2}}}{x^6}dx$$
$x=\frac{4}{3}\sin\theta$
$dx=\frac{4}{3}\cos\theta d\theta$
$$\int \frac{3\left(\frac{16}{9}-\frac{16}{9}\sin^2\theta\right)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(1-\sin^2\theta)^{\frac{3}{2}}}{x^6}d\theta=\int \frac{4(\cos\theta)^3}{x^6}d\theta=4\int \left(\frac{\cos\theta}{\frac{16}{9}\sin\theta}\right)^3d\theta$$
How should I continue?
|
Another way using reduction formula:
$$\int(a^2-x^2)^nx^mdx$$
$$=(a^2-x^2)^n\int x^m\ dx-\int\left(\dfrac{d\{(a^2-x^2)^n\}}{dx}\int x^m\ dx\right)dx$$
$$\implies\int(a^2-x^2)^nx^mdx=(a^2-x^2)^n\cdot\dfrac{x^{m+1}}{m+1}+\dfrac{2n}{m+1}\int(a^2-x^2)^{n-1}x^{m+2}\ dx$$
Setting $n=\dfrac32, m=-6$
$$\int(a^2-x^2)^{3/2}x^{-6}dx=-(a^2-x^2)^{3/2}\cdot\dfrac{x^{-5}}5-\dfrac35\int(a^2-x^2)^{1/2}x^{-4}\ dx$$
$n=\dfrac12,m=-4\implies$
$$\int(a^2-x^2)^{1/2}x^{-4}\ dx=-(a^2-x^2)^{1/2}\cdot\dfrac{x^{-3}}3-\dfrac13\int\dfrac1{x^2\sqrt{a^2-x^2}}\ dx$$
Now,
$y=\arcsin\dfrac xa,x=a\sin y\implies\cot y=\dfrac{\sqrt{a^2-x^2}}x$
$$\int\dfrac1{x^2\sqrt{a^2-x^2}}\ dx=\int\dfrac{dy}{a^2\csc^2y}dy=c-\dfrac{\cot y}{a^2}$$
|
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|
Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible. If $Det(A)=k$,and $Det(A+kadj(A))=0$ Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible.
If $Det(A)=k$,and
$Det(A+kadj(A))=0$,
then find the value of $Det(A-kadj(A))$
My attempt:
$Det(A+kadj(A))=0$
$Det(A)Det(A+kadj(A))=0$
$Det(A^2+kAadj(A))=0$
$Det(A^2+k^2I)=0$
Using same steps,I get $Det(A-kadj(A))=\frac{1}{k}Det(A^2-k^2I)$
but can't understand what to do next
|
Since you're given that $A$ is a $2\times 2$ you could brute force it...
Calculating the determinant of
$$
\begin{align*}
A^2 + k^2I =
\begin{pmatrix}
a + k^2 & b \\
c & d + k^2
\end{pmatrix}
\end{align*}
$$
and setting it equal to zero gives that $a+d = -(1+k^2)$. Now solving for the determinant of
$$
\begin{align*}
A^2 - k^2I =
\begin{pmatrix}
a - k^2 & b \\
c & d - k^2
\end{pmatrix}
\end{align*}
$$
gives $\frac 1k \det(A^2 -k^2I) = 2k(1 + k^2)$.
|
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|
If $ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$, Prove that $x+y+z$ $=0$
Question If $$ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} $$ Prove that $x+y+z$ $=0$
I've attmempted this question by cross multiplying so that
$$ x(c-a)(a-b) = y(b-c)(a-b) = (b-c)(c-a)z $$
but that did not work
I also tried splitting the equation so
$ \frac{x}{b-c} = \frac{y}{c-a}$ and $\frac{y}{c-a} = \frac{z}{a-b}$
and tried forming systems of equation but that didn't lead me to the proof as well...
Could someone please help!
|
\begin{align}
x+y+z&=\frac{b-c}{a-b}z+\frac{c-a}{a-b}z+z\\
&=z\left(\frac{b-c+c-a+a-b}{a-b}\right)\\
&=z\cdot 0\\
&=0
\end{align}
|
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|
To evaluate integral using Beta function - Which substitution should i use? $$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$
I have to use some kind of substitution but i do not understand what i use and why ?
Thanks
|
Let's try the substitution
$$x=\frac{1-y}{1+cy}$$
so that $1-x=(1+c)\frac{y}{1+cy}$.
Then, when $x=0$, $y=1$ and when $x=1$, $y=0$. We also have $dx=-(1+c)\frac{1}{(1+cy)^2}\,dy$. Then, we can write
$$\begin{align}\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\int_0^1 \frac{\left(\frac{1-y}{1+cy}\right)^{m-1}\left((1+c)\frac{y}{1+cy}\right)^{n-1}}{\left(a+b\frac{1-y}{1+cy}\right)^{m+n}}\,(1+c)\frac{1}{(1+cy)^2}\,dy\\\\
&=\frac{(1+c)^n}{(a+b)^{n+m}}\int_0^1\frac{y^{n-1}(1-y)^{m-1}}{\left(1-\frac{b-ac}{a+b}y\right)^{m+n}}
\end{align}$$
Choosing $c=b/a$, we obtain
$$\begin{align}
\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\frac{1}{a^n(a+b)^m}\int_0^1 y^{n-1}(1-y)^{m-1}\,dy\\\\
&=\frac{1}{a^n(a+b)^m}B(m,n)
\end{align}$$
And we are done!
|
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|
Understanding basis algorithm result I've a matrix ${\bf A}$ defined as
A = \begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
2 & 1 & -3 & 1 & 1\\
\end{pmatrix}
And ${W_1}$ is the solution space associated to the homogenous system ${\bf A}$. I am asked to find a basis for ${W_1}$.
I re-wrote the matrix as linear combination of ${c_1}$, i.e.:
\begin{align}
x = c_1 + 2 c_2 \\
y = -2c_1+ c_2 \\
z = -3c_2 \\
v = 3c_1 + c_2 \\
w = 7c_1 + c_2 \\
\end{align}
I use these equations to build another matrix, I solve it and then i reach to:
\begin{align}
x - \frac{2}{3}y = c_1 \\
x - \frac{-1}{3}z = c_2 \\
2x + y \frac{-5}{3}z = 0 \\
x - y - v = 0 \\
3x - 2y - 2 = 0
\end{align}
So, the last three equations are the general ones for $W_1$ and from there I can describe the basis, is this right ?
|
A standard Gaussian elimination leads to a reduced row echelon form:
\begin{align}
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
2 & 1 & -3 & 1 & 1
\end{pmatrix}
&\to
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
0 & 5 & -3 & -5 & -13
\end{pmatrix}
&&R_2\gets R_2-2R_1
\\
&\to
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
0 & 1 & -3/5 & -1 & -13/5
\end{pmatrix}
&&R_2\gets \frac{1}{5}R_2
\\
&\to
\begin{pmatrix}
1 & 0 & -6/5 & 1 & 9/5\\
0 & 1 & -3/5 & -1 & -13/5
\end{pmatrix}
&&R_2\gets \frac{1}{5}R_2
\end{align}
This means the system is equivalent to
$$
\begin{cases}
x_1=\dfrac{6}{5}x_3-x_4-\dfrac{9}{5}x_5 \\[6px]
x_2=\dfrac{3}{5}x_3+x_4+\dfrac{13}{5}x_5
\end{cases}
$$
Now it's just a matter of choosing values for $x_3$, $x_4$ and $x_5$ to obtain linearly independent vectors; the choices, in order to also remove denominators, can be
$$
x_3=5,\quad x_4=0,\quad x_5=0\\
x_3=0,\quad x_4=1,\quad x_5=0\\
x_3=0,\quad x_4=0,\quad x_5=5
$$
leading to the three vectors
$$
\begin{pmatrix}6\\3\\5\\0\\0\end{pmatrix},\quad
\begin{pmatrix}-1\\1\\0\\1\\0\end{pmatrix},\quad
\begin{pmatrix}-9\\13\\0\\0\\5\end{pmatrix}.
$$
These three vectors form a basis for $W_1$.
|
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|
Prove the inequality regarding complex numbers If $\theta_i\in [0,\pi/6],i=1,2,3,4,5$.And $$\sin \theta_1\ z^4 + \sin\theta_2 \ z^3 + \sin\theta_3 \ z^2 + \sin\theta_4 \ z + \sin\theta_5=2$$ Prove that $|z|\gt \frac{3}{4}$.
|
$\theta_i\in [0,\pi/6]$ implies that $0 \le \sin \theta_i \le \frac 12$.
So it follows from your equation that
$$
2 \le \frac 12 \left( 1 + |z| + |z|^2 + |z|^3 + |z|^4 \right)
$$
or, with $r = |z|$,
$$
f(r) = 1+ r+r^2 +r^3 +r^4 - 4 \ge 0
$$
$f(r)$ is monotonically increasing in $r$ and $f(3/4) < 0$,
therefore a solution must satisfy $r > 3/4$.
A numerical computation with PARI/GP
? solve(r=0, 1, 1+r+r^2+r^3+r^4-4)
%1 = 0.88817966758531001825082741056489343552
shows that $r > 0.888$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\right) $$
Now no terms cancel. We get 0 in numerator and denominator too.
Ans: My teacher told me that the limit is $+\infty$, but didn't tell how.
|
Given limit does not tend to $\infty$. This is the graph of $f(x)=\frac{x^2+9}{\sqrt{x^2+16}-5}$ in WolframAlpha.
As you see,
$$
\lim_{x\to -3+0}f(x)=-\infty
$$
and
$$
\lim_{x\to -3-0}f(x)=\infty.
$$
|
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|
How many ways we can choose items from different boxes I searched through the internet but couldn't find my answer, which can either be a very simple or a hard one.
Assume there are $3$ boxes, which carry, respectively, $1$, $4$, $2$ items.
My question is how many ways we can select $3$ items from these boxes.
I am looking for a formula rather than a solution for these specific values.
If I choose (take away) $3$ items by trying one by one.
\begin{array}{c c c}
0 & 3 & 1\\
0 & 2 & 2\\
0 & 4 & 0\\
1 & 1 & 2\\
1 & 3 & 0\\
1 & 2 & 1
\end{array}
Items remain each time, so the answer seems to be $6$ different ways. But I am not sure.
|
Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
A more combinatorial argument: The number of ways to write $n$ as a sum of $k$ nonnegative integers is
$$
\binom{n+k-1}{k-1}
$$
and a discussion can be found here.
So, in this case, the number of ways that $a_1+a_2+a_3=3$ (without restrictions) is
$$
\binom{3+3-1}{3-1}=\binom{5}{2}=10.
$$
This, however, counts too many possible sums. Suppose that we take too many from box $1$, this means that we take at least $2$ from box $1$. In this case, we can write $a_1=2+b_1$ where $b_1$ is nonnegative. Then, the initial sum becomes $b_1+a_2+a_3=1$. Using the same formula, this results in
$$
\binom{1+3-1}{3-1}=\binom{3}{2}=3
$$
impossible ways. Continuing, there is no way to take too many objects from the second box, but it is possible to take too many objects from box $3$. In this case, one must take $3$ objects from box $3$, so we write $a_3=3+b_3$ where $b_3$ is nonnegative. This results in the equation $a_1+a_2+(3+b_3)=3$. There are
$$
\binom{0+3-1}{3-1}=1
$$
ways for this sum to occur. We should now use the inclusion/exclusion principle to see if we've over-counted. This could happen if we take more than $1$ item from box $1$ and more than $2$ items from box $3$. Then, we have $(2+b_1)+b_2+(3+b_3)=3$, but this has no solutions as a sum of nonnegative integers cannot be negative.
Therefore, out of the original $10$ possibilities, $3+1=4$ are impossible, leaving the $6$ that we've found.
A dynamic programming-type solution: Let $N(b,s)$ be the number of ways to use the first $b$ boxes to sum to $s$. Also, write $m_i$ for the number of objects in box $i$. In your case:
\begin{align*}
N(1,0)&=1\\
N(1,1)&=1\\
N(1,2)&=0\\
N(1,3)&=0.
\end{align*}
Then, the values in the second box can be computed as follows:
$$
N(b+1,s)=\sum_{i=0}^{\min\{s,m_{b+1}\}}N(b,s-i).
$$
Using this formula:
\begin{align*}
N(2,0)&=N(1,0)=1\\
N(2,1)&=N(1,0)+N(1,1)=2\\
N(2,2)&=N(1,0)+N(1,1)+N(1,2)=2\\
N(2,3)&=N(1,0)+N(1,1)+N(1,2)+N(1,3)=2.
\end{align*}
Continuing for the third column,
\begin{align*}
N(3,0)&=N(2,0)=1\\
N(3,1)&=N(2,0)+N(2,1)=3\\
N(3,2)&=N(2,0)+N(2,1)+N(2,2)=5\\
N(3,3)&=N(2,1)+N(2,2)+N(2,3)=6.
\end{align*}
We are interested in the value $N(3,3)=6$.
|
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|
Consecutive strings of heads problem So the question asks:
We toss a fair coin $n$ times and record the outcome as a sequence of H and T. We say that there is a run of heads if there is a consecutive string H...H which starts either at the first toss or after the coin lands tails and which ends either at the last toss or before the coin lands tails. For example, the sequence HTHHTHHHHTTHH has four runs of heads. Find the expected number of runs of heads.
So far I have:
Suppose the probability that all $K$ tosses have the same type of toss equal to the probability of all heads in $K$ tosses + probability of all tails in $K$ tosses.
Sets of subsequent $K$ tosses $=(n-k+1)$
Expected number of runs of heads equals the number of subsequent $K$ tosses times the probability that one set of $K$ tosses is a streak $= (n-k+1)\times 2\times (1/2)^k$
But I am really not sure about my solution, is this the right way to do this kind of probability problem?
|
With $z$ representing tails and $w$ representing heads and $u$
counting runs we get the generating function
$$\frac{1}{1-z}
\left(\sum_{q\ge 0}
\left(u \frac{w}{1-w} \frac{z}{1-z}\right)^q\right)
\left(1+u\frac{w}{1-w}\right).$$
Actually we are not interested in the distinction between heads and
tails once runs have been counted so we may use
$$\frac{1}{1-z}
\left(\sum_{q\ge 0}
\left(u \frac{z^2}{(1-z)^2}\right)^q\right)
\left(1+u\frac{z}{1-z}\right).$$
This is
$$\frac{1-z+uz}{(1-z)^2} \frac{1}{1-uz^2/(1-z)^2}
= (1-z+uz) \frac{1}{(1-z)^2 - uz^2}.$$
Note that putting $u=1$ we obtain
$$\frac{1}{1-2z}$$
which is good news because it shows we have counted all strings.
Now for the expected number of runs differentiate with respect to $u$
and set $u=1$ first getting
$$\frac{z}{(1-z)^2 - uz^2}
- (1-z+uz) \frac{1}{((1-z)^2-uz^2)^2} (-z^2).$$
and then
$$\frac{z}{1-2z}
+ \frac{z^2}{(1-2z)^2}
= \frac{z(1-2z)+z^2}{(1-2z)^2}
= \frac{z-z^2}{(1-2z)^2}.$$
We thus get for the expectation
$$\frac{1}{2^n} [z^{n-1}] \frac{1}{(1-2z)^2}
- \frac{1}{2^n} [z^{n-2}] \frac{1}{(1-2z)^2}
\\ = \frac{1}{2^n} n 2^{n-1}
- \frac{1}{2^n} (n-1) 2^{n-2}
= \frac{1}{2} n - \frac{1}{4} (n-1)
= \frac{1}{4} (n+1).$$
|
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|
Is there any integral for the Golden Ratio? I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:
$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$
$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$
$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$
Is there an interesting integral* (or some series) whose result is simply $\phi$?
* Interesting integral means that things like
$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$
are not a good answer to my question.
|
Here is a collection of the series with reciprocal binomial coefficients.
$$\sum_{n=0}^\infty (-1)^n \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{4}{5} \left(1-\frac{\sqrt{5}}{5} \ln \phi \right)$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=-\frac{2\sqrt{5}}{5} \ln \phi$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=-2 \ln^2 \phi$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{4\sqrt{5}}{5} \ln \phi$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{8\sqrt{5}}{5} \ln \phi-4 \ln^2 \phi$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{n-1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{3\sqrt{5}}{5} \ln \phi-\frac{1}{2}$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{(n-1)^2} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=1-\sqrt{5} \ln \phi+ \ln^2 \phi$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2(n^2-1)} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=4\ln^2 \phi-\frac{\sqrt{5}}{2} \ln \phi-\frac{3}{8}$$
A one with $\pi$:
$$\sum_{n=0}^\infty \left( \begin{matrix} 4n \\ 2n \end{matrix} \right)^{-1}=\frac{16}{15}+\frac{\sqrt{3}}{27} \pi-\frac{2\sqrt{5}}{25} \ln \phi $$
Source here
|
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"timestamp": "2023-03-29T00:00:00",
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|
sum of floor ("weighted" divisor summary function) is there a closed form for
$$
\sum_{i=1}^N\left\lfloor\frac{N}{i}\right\rfloor i
$$
Or is there any faster way to calculate this for any given N value?
|
This answer does not give closed form formula (for reasons mentioned in comments), but it gives a way to calculate the sum more efficiently than just going through all $i\leq N$.
The trick is that values of $\left\lfloor\frac{N}{i}\right\rfloor$ do not change for long ranges of $i$, for example:
$$
\left\lfloor\frac{10}{6}\right\rfloor = \left\lfloor\frac{10}{7}\right\rfloor = \dots = \left\lfloor\frac{10}{10}\right\rfloor = 1.
$$
This can be used to skip over these and sum them all in one iteration. Since you have
$$
\left\lfloor\frac{N}{i}\right\rfloor = k \Leftrightarrow \frac{N}{k+1} < i \leq \frac{N}{k}
$$
all you need to do is iterate from current $i$ to next different value which will be $\left\lfloor\frac{N}{k}\right\rfloor+1$.
For example one can define
\begin{align*}
k_n&=\left\lfloor\frac{N}{i_{n-1}}\right\rfloor\\
i_n&=\left\lfloor\frac{N}{k_n}\right\rfloor+1\\
s_n&=s_{n-1}+\sum_{i=\left\lfloor\frac{N}{k_n+1}\right\rfloor+1}^{\left\lfloor\frac{N}{k_n}\right\rfloor} k_n = s_{n-1}+\frac{1}{2}k_n\Bigl(\left\lfloor\frac{N}{k_n+1}\right\rfloor+\left\lfloor\frac{N}{k_n}\right\rfloor+1\Bigr)\Bigl(\left\lfloor\frac{N}{k_n}\right\rfloor-\left\lfloor\frac{N}{k_n+1}\right\rfloor\Bigr)
\end{align*}
with $k_0=0,s_0=0,i_0=1$. Then just iterating while $i_n \leq N$, resulting sum is then $s_{n+1}$.
Example for $N=10$:
\begin{align*}
k_0&=0&s_0&=0&i_0&=1\\
k_1&=10&s_1&=10&i_1&=2\\
k_2&=5&s_2&=20&i_2&=3\\
k_3&=3&s_3&=29&i_3&=4\\
k_4&=2&s_4&=47&i_4&=6\\
k_5&=1&s_5&=87&i_5&=11\\
\end{align*}
So result is $87$ and only $5$ iterations were needed. For $N=100$ it takes $19$ iterations, for $N=1000$ it takes $62$, etc...
|
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|
Integration of $\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx$ How can we integrate:
$$
\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx
$$
Using simple algebraic identities I deduced it to
$$
\int\frac{1-2\sin^2x\cdot\cos^2x}{(\sin x+\cos x)(1-\sin x\cdot\cos x)}dx
$$ but can't proceed further. Please provide some directions?
|
Noting that
$$
\sin ^4 x+\cos ^4 x= \left(\sin ^3 x+\cos ^3 x\right)(\cos x+\sin x)-\sin x \cos x\left(\sin ^2 x+\cos^2x \right),
$$
we have
$$
\begin{aligned}
I&=\int(\cos x+\sin x) d x-\int \frac{\sin x \cos x}{\sin ^3 x+\cos^3 x} d x \\
& =\sin ^2 x-\cos x-\int \frac{\sin x \cos x}{\sin ^3 x+\cos ^3 x} d x \\
& =\sin x-\cos x-\frac{1}{3} \underbrace{ \int \frac{\sin x+\cos x}{\sin ^2 x-\sin x \cos x+\cos ^2 x} d x}_{J} +\frac{1}{3} \underbrace{\int \frac{1}{\sin x+\cos x} d x}_K \\
&
\end{aligned}
$$
$$
\begin{aligned}
J & =\int \frac{2 d(\sin x-\cos x)}{1+(\sin x-\cos x)^2} \\
& =2 \tan ^{-1}(\sin x-\cos x)+C_1
\end{aligned}
$$
Using Weierstrass substitution $t=\tan \frac{x}{2} $, we get
$$
\begin{aligned}
K& =\int \frac{1}{\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} \cdot \frac{2 d t}{1+t^2} \\
& =\sqrt{2} \tanh ^{-1}\left(\frac{\tan \frac{x}{2}-1}{\sqrt{2}}\right)+C_2
\end{aligned}
$$
Now we can conclude that
$$\boxed{I=\sin x-\cos x-\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+\frac{\sqrt{2}}{3} \tanh ^{-1}\left(\frac{\tan \frac{x}{2}-1}{\sqrt{2}}\right)+C }$$
|
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|
Find all functions $f:\mathbb R\to \mathbb R$ such that $f(a^2+b^2)=f(a^2-b^2)+f(2ab)$ for every real $a$,$b$ I guessed $f(a)=a^2$ and $f(a)=0$, but have no idea how to get to the solutions in a good way.
Edit: I did what was suggested:
from $a=b=0$
$f(0)=0$
The function is even, because from $b=-a$
$f(2a^2)=f(-2a^2)$.
|
Define $g : \Bbb{R} \to \Bbb{R}$ by $g(x) = f(\sqrt{x})$ and $g(-x) = -g(x)$ for $x \geq 0$. ($g$ is well-defined since $f(0) = 0$.) We claim that
Claim. $g$ solves the Cauchy functional equation
$$ g(x+y) = g(x) + g(y), \quad x, y \in \Bbb{R} \tag{1}. $$
Proof. Let $x, y \in \Bbb{R}$.
*
*If $x, y \geq 0$. then we can pick $a\geq b\geq 0$ such that $a^2 - b^2 = \sqrt{x}$ and $2ab = \sqrt{y}$. (This becomes transparent if we write $(a, b)$ in polar coordinates.) Then we have
\begin{align*}
g(x+y)
&= f(\sqrt{x+y}) = f(a^2 + b^2) \\
&= f(a^2 - b^2) + f(2ab) = g(x) + g(y).
\end{align*}
*If $x, y \leq 0$, then we have $|x| = -x$ and $|y| = -y$ and thus
$$g(x+y) = -g(|x|+|y|) = -g(|x|) - g(|y|) = g(x) + g(y)$$
by the definition and $\text{(1)}$.
*If $x \leq 0$ and $0 \leq |x| \leq y$, then from $g(y) = g(y-|x|) + g(|x|)$, we have
$$ g(x) = -g(|x|) = -(g(y) - g(y-|x|)) = g(y+x) - g(y). $$
Rearrange this to get $g(x+y) = g(x) + g(y)$.
*If $x \leq 0$ and $0 \leq y \leq |x|$, then from $g(|x|) = g(|x|-y) + g(y)$ we have
\begin{align*}
g(x)
&= -g(|x|) = -(g(|x|-y) + g(y)) \\
&= -g(-x-y) - g(y) = g(x+y) - g(y).
\end{align*}
Rearrange this to get $g(x+y) = g(x) + g(y)$.
*Interchanging the role of $x$ and $y$, the identity $\text{(1)}$ also holds when $y \leq 0 \leq x$.
These cover all the possible sign combinations of $(x, y)$. Therefore $g$ solves $\text{(1)}$. ////
Conversely, for any $g$ solving the Cauchy functional equation, $f(x) = g(x^2)$ solves the problem. So we obtain a 1-1 correspondence between the solution of
$$ f(a^2 + b^2) = f(a^2 - b^2) + f(2ab), \quad a, b \in \Bbb{R} \tag{2} $$
and the solution of the Cauchy functional equation $\text{(1)}$. Now assuming the Axiom of Choice, the equation $\text{(1)}$ has solutions which is not of the form $g(x) = cx$, which means that $\text{(2)}$ also has solutions which is not of the form $f(a) = ca^2$.
|
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|
My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.
Is the book right or me ?
|
I think your working is perfect. If the book is correct then
$$1-\frac4{3x}+\frac2{x^2}=2\implies 3x^2-4x+6=6x^2\implies3x^2+4x-6=0$$
whereas according to what is written there it must be $\;3x^2+4x-6=4x^2\;$, so I'd say the book's answer is wrong...or you forgot a coefficient $\;3\;$ for the quadratic term.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that if $n \in \mathbb{Z}[\sqrt{2}]$ has an even norm, then $\sqrt{2} \mid n$ Aside from multiplying and dividing some specific numbers in this ring, e.g., $(1 + \sqrt{2})\sqrt{2}$ I have not really done anything productive on this question. I either go around in circles or jump to conclusions, such as that it follows from the fact that $\langle \sqrt{2} \rangle$ is a prime ideal with prime norm.
Though I do have this hunch that a very similar argument can be used to prove something similar in $\mathbb{Z}[\sqrt{-2}]$...
|
If the norm $N(a+b\sqrt 2)=a^2-2b^2$ is even, then $a^2$ is even, and so is $a$. Then $a=2a'$ for some integer $a'$. Therefore :
$$a+b\sqrt 2 = 2a'+b\sqrt 2=\sqrt 2 (b+a'\sqrt 2)$$ From this you conclude that $\sqrt 2 \mid a+b\sqrt 2$ in $\Bbb Z[\sqrt 2]$, provided that the norm of $a+b\sqrt 2$ is even.
More generally, if $d \in \Bbb Z$ is square-free, and if $d \mid N(a+b\sqrt d)$ (in $\Bbb Z$), then $\sqrt d \mid a+b\sqrt d$ in $\Bbb Z[\sqrt d]$.
Indeed, if $N(a+b\sqrt d)= a^2-db^2$ is a multiple of $d$, then $a^2$ is a multiple of $d$. As $d$ is square-free, you have that $d \mid a$ (this is not true in general, for instance $d=2 \cdot 4^2=32$ divides $a^2=64=8^2$, but $32$ doesn't divide $a=8$).
Then you can write $a=da'$ for some integer $a'$. Therefore $$a+b\sqrt d = \sqrt d (a'\sqrt d + b)$$ is a multiple of $\sqrt d$ as desired.
The converse also holds (if $d$ is square-free). Suppose that $x=a+b\sqrt d$ is a multiple of $\sqrt d$ (in $\Bbb Z[\sqrt d]$).
Then you can write $$a+b\sqrt d = \sqrt d(m+n\sqrt d)=nd+m\sqrt d$$
for some integers $m$ and $n$.
The norm of $x$ is
$$N(x)=N(nd+m\sqrt d)=n^2 d^2 - dm^2=d(dn^2-m^2).$$
In particular the norm is a multiple of $d$.
Here is an example. Let $d=6$ and $x=2+\sqrt 6$.
The norm of $x$ in $\Bbb Z[\sqrt 6]$ is $N(2+\sqrt 6)=4-6=-2$ which is not a multiple of $d=6$. In particular, $x$ is not a multiple of $\sqrt 6$ in $\Bbb Z[\sqrt 6]$.
It is possible to prove this without using norms.
Indeed, suppose that we could write
$$x=2+\sqrt 6 = \sqrt 6(m+n\sqrt 6) = 6n + m\sqrt 6 \in \Bbb Z[\sqrt 6]$$
for some integers $m$ and $n$.
Then recall that if $a+b\sqrt d = 0$ then $a=b=0$ (where $d$ is square-free). Suppose for a contradiction that $b≠0$. Then $\sqrt d =-a/b$ would be rational, which is impossible. Then $b=0$, and $a=0$.
In particular, $2+\sqrt 6 = 6n + m\sqrt 6$ is equivalent to $2-6n+\sqrt 6 (1-m)=0$. This yields $2-6n=0$ and $1-m=0$. But $3$ divides the integer $6n$ and $6n=2$... this is not possible.
Hence, we proved that $x=2+\sqrt 6$ is not a multiple of $\sqrt 6$ in $\Bbb Z [\sqrt 6]$.
(Notice that saying "$\sqrt d$ is irrational" is equivalent to saying $N(a+b\sqrt d)=0 \iff a=b=0$).
|
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|
Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but can not proceed further :(
|
This technique comes from A Course in Pure Mathematics by Margaret M. Gow:
$E=(a+b+c)^{3}-(b+c-a)^{3}-(c+a-b)^{3}-(a+b-c)^{3}$
$c=0 \implies E=(a+b)^{3}-(b-a)^{3}-(a-b)^{3}-(a+b)^{3}=0$
$\therefore c$ is a factor of $E$.
By symmetry, $a, b$ also are factors of $E$.
Since $E$ is cubic, let $E=kabc$ where $k$ is a constant.
Take $a=b=c=1$, $E=24=k$.
$\therefore E=24abc$
|
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|
Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.
$\bullet~$ I can show that $x^2_n \geqslant 2$ by induction:
*
*Basis step: $x_1=2, x_1^2 = 4 >2$.
*Inductive step:
Assume we have some $k$ for which $x^2_K \geqslant 2$, then
\begin{align*}
(x_{k+1})^2 =&~ \bigg(\dfrac{1} {2} \bigg(x_k + \dfrac{2} {x_k}\bigg)\bigg)^2\\
=&~ \dfrac{1} {4} x_k^2 + 1 + \dfrac{1} {x_k^2}\\
\geqslant&~ \dfrac {1} {4} x_k^2 + 1 \geqslant 2\\
&~\dfrac {1} {4} x_k^2 \geqslant 1 \quad \forall~ n \geqslant 2
\end{align*}
How do I use this to show that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$
and hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$?
Thanks!!
|
Note the fact that $x_{n+1} =\frac{1}{2}(x_n+\frac{2}{x_{n}}) \le \frac{1}{2}(x_n+x_{n})=x_{n}$ from $x_{n}^2 \ge 2$ (or $x_{n} \ge \frac{2}{x_{n}}$)
Thus, $x_{n}-x_{n+1}\le0$.
This implies that $x_{n}$ is a decreasing sequence that is greater than $\sqrt{2}$, implying that $x_{n}$ is convergent.
Let $\lim{x_n}=a$.
Note the fact that $a=\frac{1}{2}(a+\frac{2}{a})$. This implies that $a=\sqrt{2}$.
|
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|
Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$ For every $a,b,c>0$ such that $abc=1$ prove the inequality
$$\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$$
My work so far:
$abc=1 \Rightarrow \frac 13 (a+b+c)\ge 1$
Then $$c^2+a+b \le c^2+(a+b)\frac 13(a+b+c)$$
Or more
$$ (\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a})((c^2+a+b)+(a^2+b+c)+(b^2+c+a)) \ge $$
$$\ge(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2$$
|
As we know:
$$2ab \le a^{2}+b^{2} $$, and
$$3a^{\frac{1}{3}} = 3(a^{2}bc)^{\frac{1}{3}} \le a^{2} + b + c $$ (because of $abc =1$)
Now we have :
$$2(\frac{ab}{3c^{1/3}} + \frac{ac}{3b^{1/3}} +\frac{cb}{3a^{1/3}})$$
Now using Cauchy we could prove our inequality.
|
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|
Remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$. I am currently preparing for a certain quiz show when I encountered this question:
What is the remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$?
I know for a fact that $x^3-1=(x-1)(x^2+x+1)$.
And I got
$$x^{2016}+x^{2013}+\cdots+x^6+x^3=x^3(x^{2013}+x^{2010}+\cdots+x^3+1)$$
$$=[(x-1)(x^2+x+1)+1](x^{2013}+x^{2010}+\cdots+x^3+1)$$
$$=(x-1)(x^2+x+1)
(x^{2013}+x^{2010}+\cdots+x^3+1)+(x^{2013}+x^{2010}+\cdots+x^3+1)$$
I know that the remainder comes from the term not containing $x^2+x+1$ but I do not know to get it since it involved so many terms.
Any help please? Thanks in advanced!
|
\begin{align}
\sum_{k=1}^{672}x^{3k}&=\sum_{k=1}^{672}\left((x-1)(x^2+x+1)+1\right)^{k}\\
&=\sum_{k=1}^{672}\left(\sum_{j=0}^{k}{k \choose j}\left((x-1)(x^2+x+1)\right)^{j}\right)\\
&=\sum_{k=1}^{672}\left((x^2+x+1)P_k(x)+1\right)\\
&=(x^2+x+1)P(x)+\sum_{k=1}^{672}1
\end{align}
|
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|
Using the ratio test for the series $\sum \frac1{3^n-2^n}$, I can't compute the limit $$\sum_{n=1}^{\infty} \frac{1}{3^n-2^n}$$
I know this series is convergent and using the ratio test.
But I can't conclude the proving.
$$\begin{align}\lim_{n\to\infty} \left| \frac{\frac{1}{3^{n+1}-2^{n+1}}}{ \frac{1}{3^n-2^n}}\right|
&= \lim_{n\to\infty} \left| \frac{3^n-2^n}{ 3^{n+1} - 2^{n+1}} \right|\\
&= \lim_{n\to\infty} \left| \frac{1-(\frac{2}{3})^n}{ 3(1-2^{n+1}/3^{n+1})}\right| \end{align}$$
By using calculation, the limit is 0.
But I can't compute this limit by myself without using calculation. Any help?
|
You've almost got this one solved
$\lim_{n\to\infty} {(\frac{2}{3})}^n = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} .... = 0.\dot{6} \times 0.\dot{6} \times 0.\dot{6} ... = 0$
so your solution will come out to $\frac{1}{3}$
|
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|
How to find Ratio And proportion If $\frac{4x+3y}{4x-3y}=\frac{7}{5}$.Find the Value of the $\frac{2x^2-11y^2}{2x^2+11y^2}$.
Okay I just want hint how to solve this problem.
|
In your first equation $\frac{4x+3y}{4x-3y}=\frac{7}{5}$, try solving x in terms of y or vice versa.
Then you will find that $x=\frac{9}{2}y$ or $y=\frac{2}{9}x$,
so now substitute one of these to $\frac{2x^2-11y^2}{2x^2+11y^2}$
if you choose $x=\frac{9}{2}y$ you'll have,
$$\frac{2x^2-11y^2}{2x^2+11y^2}=\frac{\frac{81}{2}y^2-11y^2}{\frac{81}{2}y^2+11y^2}$$
cancelled all y^2, finally you have $\frac{81-22}{81+22}=\frac{59}{103}$
also if you choose $y=\frac{2}{9}x$, same process and you'll have $\frac{59}{103}$ as an answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1668727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \rightarrow \infty} (x_n-n)=\frac{1}{4}$
|
Using that $\frac{k}{n^2} \xrightarrow[n\to\infty]{} 0$ for all $1\leq k\leq n$, one can use the Taylor expansion of $\sqrt{1+x}$, $$\sqrt{1+x} = 1+\frac{x}{2} - \frac{x^2}{8} + o(x^2)$$ when $x\to 0$:
$$\begin{align}
x_n &= \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}
= \sum_{k=1}^n \left(1+\frac{k}{2n^2}-\frac{k^2}{8n^4}+o\left(\frac{k^2}{n^4}\right)\right)
\\
&= n+ \frac{1}{2n^2}\cdot\frac{n(n+1)}{2} - \frac{1}{8n^4}\cdot\frac{n(n+1)(2n+1)}{6} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{1}{4n} - \frac{1}{8n^4}\cdot\frac{2n^3}{6} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{6}{24n} - \frac{1}{24n} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{5}{24n} + o\left(\frac{1}{n}\right)
\end{align}$$
giving you $n(x_n - n - \frac{1}{4}) = \frac{5}{24} + o(1)$.
Note: Now, while the above is a valid and legit method, one should stop and at least consider why the manipulation and summation of the $o(\cdot)$'s is indeed acceptable. (Specifically: while it's implicit, what each $o(\cdot)$'s is taken with regard to? $n\to\infty$, or $\frac{k}{n}\to0$ each for a different $k$? And why is it still OK?)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1669217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluation of $\int_{0}^{\pi} \frac{\pi}{1-\cos(x)\sin(x)}dx$ How can I evaluate
$$\int_{0}^{\pi} \frac{\pi}{1-\cos(x)\sin(x)}dx$$
I tried it using identity $\sin(x)=\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and $\cos(x)=\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ but it is making question quite calculative. Is there any better approach to solve this?
|
This problem can also solve by complex method. Let $t=2x$, we have
$$
\int_0^\pi\frac{\pi}{1-\cos(x)\sin(x)}dx=\int_0^\pi\frac{\pi}{1-\frac{1}{2}\sin(2x)}dx\\
=\pi\int_0^{2\pi}\frac{dt}{2-\sin(t)}dt:=\pi I
$$
Now, let $z=e^{it}$, then
$$
\sin(t)=\frac{z-\frac{1}{z}}{2i}
$$
and
$$
dt=\frac{dz}{iz}
$$
Plug in, we have
$$
I=\int_{|z|=1}\frac{2dz}{4iz-z^2+1}=-\int_{|z|=1}\frac{2dz}{(z-(2+\sqrt{3})i)(z-(2-\sqrt{3})i)}
$$
Notice in the unit circle of complex plane, the only singular point (1st order) is $z=(2-\sqrt{3})i$, and
$$
\mathrm{Res}(\frac{2}{4iz-z^2+1},z=(2-\sqrt{3})i)=-\frac{i}{\sqrt{3}}
$$
Thus, we have
$$
I=2\pi i(-\frac{i}{\sqrt{3}})=\frac{2\pi}{\sqrt{3}}
$$
Finally, we have
$$
\int_0^\pi\frac{\pi}{1-\cos(x)\sin(x)}dx=\pi I=\frac{2\pi^2}{\sqrt{3}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1671370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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|
Convergence of a sum, which test should I use? I have to show if this sum
$$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}$$
converges or not.
I tried to do something like
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$$
but I got nothing, because the best I could make it was:
$$\frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+2}}\cdot\frac{1}{\sqrt{n+1}}\cdot\frac{\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}\cdot\frac{\sqrt{n}}{1}$$
And I really don't know how to simplify this expression.
|
Three methods, increasingly involved (but increasingly generalizable):
To use a very simple trick:
As for any $n\geq 1$ we have
$$
0 \leq \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
\leq \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
$$
we can immediately conclude by comparison, since the series $\sum_n \left( \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$ converges (it is a telescopic series):
$$
\sum_{n=1}^N \left( \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} \right)
= 1 - \frac{1}{\sqrt{N+1}} \xrightarrow[N\to\infty]{} 1.
$$
To use only simple manipulations:
$$\begin{align}
0 \leq \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
&=
\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
=
\left(\frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{n}\sqrt{n+1}}\right)\\
&=
\frac{1}{n}\left(\frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}}\right)\\
&\leq
\frac{1}{n}\left( \sqrt{1+\frac{1}{n}}-1 \right)\\
&=\frac{1}{n}\left( \frac{1+\frac{1}{n}-1}{\sqrt{1+\frac{1}{n}}+1} \right)
=\frac{1}{n^2}\left( \frac{1}{\sqrt{1+\frac{1}{n}}+1} \right)\\
&\leq \frac{1}{2n^2}
\end{align}$$
and conclude by comparison.
To use Taylor series: since $\sqrt{1+u} = 1+\frac{u}{2} + o(u)$ and $\frac{1}{1+u} = 1-u + o(u)$ when $u\to0$,
$$
\sqrt{n+1} = \sqrt{n}\sqrt{1+\frac{1}{n}} = \sqrt{n}\left(1+\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)
$$
so
$$\begin{align}
\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} &= \frac{1}{\sqrt{n}}\left( 1-\frac{1}{1+\frac{1}{2n}+o\left(\frac{1}{n}\right)} \right)
= \frac{1}{\sqrt{n}}\left( 1-\left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right) \right)\\
&= \frac{1}{\sqrt{n}}\left( \frac{1}{2n}+o\left(\frac{1}{n}\right) \right) \\
&= \frac{1}{2n^{3/2}}+o\left(\frac{1}{n^{3/2}}\right)
\end{align}$$
and finally
$$
\left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
= \frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)
$$
and you can again conclude by comparison.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1672004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
limit $ \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $ Calculate the limit $ \displaystyle \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $
I now the answer, it is $ \displaystyle e^\frac{\log^2z}{2} $, but I don't know how to prove it. It seems like this notable limit $\displaystyle \lim \limits_{x \to \infty} {\left(1 + \frac{c}{x}\right)^x} = e^c$ should be useful here. For example I tried this way: $$ (z^{1/\sqrt n} + z^{-1/\sqrt n}) = (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 + 2 $$
$$ \displaystyle \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} = \displaystyle \lim \limits_{n \to \infty} {\left(1 + \frac{(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2}{2}\right)^n} $$
where $ (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 $ seems close to $ \frac{\log^2 z}{n} $.
Also we can say that $$ \left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n = e^{n \log {\left(1 + \frac{\left(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)}\right)^2}{2}\right)}}$$ and $ \log {\left(1 + \frac{(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2}{2}\right)} $ can be expand in the Taylor series. But I can't finish this ways.
Thanks for the help!
|
If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{\log\left(1 + \dfrac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}} - 2}{2}\right)}{\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}}\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\
&=\frac{1}{2}\lim_{n \to \infty}n\left(\frac{z^{1/\sqrt{n}} - 1}{z^{1/2\sqrt{n}}}\right)^{2}\notag\\
&= \frac{1}{2}\lim_{n \to \infty}\{\sqrt{n}(z^{1/\sqrt{n}} - 1)\}^{2}\notag\\
&= \frac{(\log z)^{2}}{2}\notag
\end{align}
Hence $L = \exp\left\{\dfrac{(\log z)^{2}}{2}\right\}$.
|
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"url": "https://math.stackexchange.com/questions/1674455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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|
Radius of an e-circle in terms of triangle's sides and area. Following is the derivation given in my textbook:
I can't figure out that how do the areas of triangles $ABI_1$ and $ACI_1$ can be given by $\frac{1}{2} cr_1$ and $\frac{1}{2} br_1$ when $r_1$ is the radius of e-circle and $b,c$ are sides of triangle $ABC$ opposite to the vertices $B$ and $C$ respectively.
$$Area = base \times height$$
So areas of triangles $ABI_1$ and $ACI_1$ should be given by $\frac{1}{2}.(\overline{AI_1}).(\overline{BP_1})$ and $\frac{1}{2}.(\overline{AI_1}).(\overline{CP_1})$ respectively. And of course (from the figure) it is clear that $$\overline{AI_1} \neq b$$ $$\overline{AI_1} \neq c$$ $$\overline{CP_1} \neq r_1$$ $$\overline{BP_1} \neq r_1$$
|
Let the excircle at side AB touch at side AC extended at G, and let this excircle's radius be $ r_{c}$ and its center be $ {\displaystyle J_{c}}$.
Then $ {\displaystyle J_{c}G}$ is an altitude of $ {\displaystyle \triangle ACJ_{c}}$, so $ {\displaystyle \triangle ACJ_{c}}$ has area $ {\tfrac {1}{2}}br_{c}$. By a similar argument, ${\displaystyle \triangle BCJ_{c}}$ has area ${\tfrac {1}{2}}ar_{c}$ and $ {\displaystyle \triangle ABJ_{c}}$ has area $ {\tfrac {1}{2}}cr_{c}$. Thus the area $ \Delta$ of triangle ${\displaystyle \triangle ABC}$ is
$$ \Delta ={\frac {1}{2}}(a+b-c)r_{c}=(s-c)r_{c}.$$
So, by symmetry, denoting $r$ as the radius of the incircle,
$${\displaystyle \Delta =sr=(s-a)r_{a}=(s-b)r_{b}=(s-c)r_{c}}.$$
By the Law of Cosines, we have
$$\cos A={\frac {b^{2}+c^{2}-a^{2}}{2bc}}.$$
Combining this with the identity $\sin ^{2}A+\cos ^{2}A=1$, we have
$ \sin A={\frac {\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}}}{2bc}}$
But $\Delta ={\tfrac {1}{2}}bc\sin A$, and so
\begin{aligned}\Delta &={\frac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}}}\\&={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\&={\sqrt {s(s-a)(s-b)(s-c)}},\end{aligned}
which is Heron's formula.
Combining this with $sr=\Delta $, we have
$$ r^{2}={\frac {\Delta ^{2}}{s^{2}}}={\frac {(s-a)(s-b)(s-c)}{s}}.$$
Similarly, $(s-a)r_{a}=\Delta$ gives
$$r_{a}^{2}={\frac {s(s-b)(s-c)}{s-a}}$$
and
$$r_{a}={\sqrt {\frac {s(s-b)(s-c)}{s-a}}}.$$[24]
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving correctness of a recursive function of multiplication by induction. Define multiplication of two numbers y and z as:
$$m_c(y,z)=\begin{cases}0&z=0\\m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)&z\neq 0\end{cases}\tag{$\forall c\geq 2$}$$
Now I need to show that this is correct using induction.
It is obviously true for the case $z=0$. Otherwise:
$$\begin{align}\forall z\neq 0,\;m_c(y,z)=&m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)\\=&m_c\left(c^2y,\left\lfloor\frac 1c\left\lfloor \frac zc\right\rfloor\right\rfloor\right)+y(z\mod c)+cy\left(\left\lfloor\frac zc\right\rfloor\mod c\right)\\=&m_c\left(c^3y,\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor \frac zc\right\rfloor\right\rfloor\right\rfloor\right)+y(z\mod c)+cy\left(\left\lfloor\frac zc\right\rfloor\mod c\right)+\\&c^2y\left(\left\lfloor\frac 1c\left\lfloor\frac zc\right\rfloor\right\rfloor\mod c\right)\\=&\ldots\\=&m_c\left(c^ny,\underbrace{\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor\frac 1c\ldots\left\lfloor\frac zc\right\rfloor\ldots\right\rfloor\right\rfloor\right\rfloor}_{\text{n times c}}\right)\\&+\sum_{r=0}^{n-1} c^ry\left(\underbrace{\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor\frac 1c\ldots\left\lfloor\frac zc\right\rfloor\ldots\right\rfloor\right\rfloor\right\rfloor}_{\text{n-1 times c}}\mod c\right)\end{align}$$
Now for induction, where do I start? From n=1 , then k then k+1. Or something else? A hint is enough.
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Base Case: If $z=0$ then it is true.
Assumption: true for $z\le n,c\ge 2,y\ge 0$
Induction: $$\begin{align}m(y,n+1)&=m\left(cy,\left\lfloor\frac {n+1}c\right\rfloor\right)+y((n+1)\mod c)\\
&=cy.\left\lfloor\frac {n+1}c\right\rfloor+y((n+1)\mod c)\tag{$\left\lfloor\frac{n+1}c\right\rfloor<n+1\forall c\ge 2$}
\\&=y\left(c\left\lfloor\frac{n+1}c\right\rfloor+(n+1\mod c)\right)
\\&=y(n+1)\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1676893",
"timestamp": "2023-03-29T00:00:00",
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|
Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
|
Since $n>a+c \ge 2$ then $n \ge 3$. Moreover, $a<b$ and $c<d$, because $$\frac ab + \frac cd <1$$.
Consider the following cases.
a) Let $b \ge n$ and $d \ge n$, then $$\frac ab + \frac cd \le \frac an + \frac cn=\frac {a+c}{n} \le \frac {n-1}{n}=1-\frac 1n<1- \frac{1}{n^3}.$$
b) Let $b \le n$ and $d \le n$, then
$$\frac ab + \frac cd <1 \Rightarrow ad+bc<bd \Rightarrow ad+bc+1 \le bd$$
From here
$$\frac ab + \frac cd \le 1- \frac{1}{bd} \le 1- \frac {1}{n^2}<1- \frac{1}{n^3}$$
c) Let $b < n<d$. If $d \le n^2$. then $bd \le n^3$ and then
$$\frac ab + \frac cd \le 1- \frac{1}{bd} <1- \frac{1}{n^3}$$
If $d>n^2$, then
$$\frac cd \le \frac{n-2}{n^2}=\frac 1n- \frac {2}{n^2},$$
because $c<n-a \le n-1$, that $c \le n-2$. Suppose
$$\frac ab + \frac cd \ge 1 - \frac {1}{n^3}.$$
Then
$$1-\frac ab \le \frac cd + \frac {1}{n^3} \le \frac 1n - \frac {2}{n^2}+\frac{1}{n^3}< \frac 1n. $$
Hence, what $b>n(b-a)\ge n$ (here we take into account that $a<b$), which contradicts inequality $b<n<d$.
d) Let $d < n<b$. If $b \le n^2$, then $bd < n^3$, and then
$$\frac ab +\frac cd \le 1 - \frac {1}{bd} < 1- \frac {1}{n^3}$$
If $b>n^2$, then
$$\frac ab \le \frac {n-2}{n^2}=\frac 1n - \frac {2}{n^2},$$
because $a<n-c \le n-1$, that $a \le n-2$. Suppose
$$\frac ab + \frac cd \ge 1 - \frac {1}{n^3}.$$
Then
$$1 - \frac cd \le \frac ab + \frac {1}{n^3} \le \frac 1n - \frac {2}{n^2}+\frac{1}{n^3}< \frac 1n.$$
$c<d \Rightarrow d> n(d-c) \ge n$. Got a contradiction with inequality $d<n<b$.
So inequality $$\frac ab + \frac cd < 1 - \frac {1}{n^3}$$ is done in all cases that had.
prove.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I suppose that I can't do it in this way. Why?
EDIT:
This is my solution:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)
\Longleftrightarrow
\sin(2x) - \sin(x) = 0
\Longleftrightarrow
2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0
\Longleftrightarrow
\cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$
Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
|
There are two errors in your solution.
First, the transformation $$
\frac{\sin x}{\cos x} = 2\sin x \iff \sin x = 2\sin x \cos x
$$ is false. The correct inference is $$
\left( \frac{\sin x}{\cos x} = 2\sin x \text{ and } \cos x \neq 0 \right) \text{ or } \cos x = 0 \iff (\sin x = 2\sin x \cos x \text{ and } \cos x \neq 0) \text{ or } \cos x = 0
$$ otherwise, going to the left, you divide by zero, and going to the right, you (might) delete the content of your equation by multiplying both sides by zero. (The hazards are $0 = 0 \implies 0 x = 0 y \not\Rightarrow x = y$ and $x \neq y \not\Rightarrow 0x \neq 0y \implies 0 \neq 0$.) Starting with either of your equations, you infer the entire other side of the more complicated bi-implication. Since this can introduce spurious solutions, all solutions must be checked in the original equation at the end of the process.
This suggests a general rule: Do not cancel. Instead subtract and factor. This highlights your second error. From $$
\sin x = 2 \sin x \cos x \text{,}
$$ subtract and factor to get $$
(1-2\cos x)\sin x = 0 \text{.}
$$ The first factor gives your $\cos x = \frac{1}{2}$. The second factor gives $\sin x = 0$. (This is because the product of several things being zero means at least one of them is.)
By proper inference, you should have \begin{align*}
& & \tan x &= 2 \sin x \\
&\iff & \frac{\sin x}{\cos x} &= 2 \sin x \\
&\iff & \sin x &= 2 \sin x \cos x \text{ or } \cos x = 0 \\
&\iff & (1-2\cos x)\sin x &= 0 \text{ or } \cos x = 0 \\
&\iff & \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{ or } \cos x = 0 \text{.}
\end{align*} Checking for spurious solutions, all solutions of $\cos x = \frac{1}{2}$ and all solutions of $\sin x = 0$ are solutions of the given equation, but none of the solutions of $\cos x = 0$ are, so the final solution is $\cos x = \frac{1}{2} \text{ or } \sin x = 0$.
Somewhat shorter uses subtract and factor earlier, avoiding the complicated bi-implication altogether: \begin{align*}
& & \tan x &= 2 \sin x \\
&\iff & \frac{\sin x}{\cos x} &= 2 \sin x \\
&\iff & \left( \frac{1}{\cos x} - 2 \right) \sin x &= 0 \\
&\iff & \frac{1}{\cos x} - 2 &= 0 \text{ or } \sin x = 0 \\
&\iff & \frac{1}{\cos x} &= 2 \text{ or } \sin x = 0 \\
&\iff & \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{.}
\end{align*}
|
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|
Integral $I=\int \frac{dx}{(x^2+1)\sqrt{x^2-4}} $ Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such subtitution that will make this integral completely the same as this standard integral so i could use substitution i mentioned, so i tried two following substitutions
$x^2-4=t^2 \Rightarrow x^2=t^2+4 \Rightarrow x=\sqrt{t^2+4}$ then i had to determine $dx$
$2xdx=2tdt \Rightarrow dx=\frac{tdt}{\sqrt{t^2+4}}$
from here i got:
$\int\frac{dt}{(t^2+5)\sqrt{(t^2+4)}}$ but i have no idea what could i do with this, so i tried different substitution
$x^2+1=t^2$ and then, by implementing the same pattern i used with the previous substitution i got this integral
$\int\frac{dt}{t\sqrt{(t^2-1)(t^2-5)}}$ but again, i don't know what to do with this, so i could use some help.
|
The curve $t^2=x^2-4$ is a hyperbola, which can be parametrized by a single value. Rewrite this equation as $(x+t)(x-t)=4$, and set $y=x+t$. Then $4/y=x-t$, and we have
$$
x=\frac{y+\frac{4}{y}}{2},\;\;\;\;t=\frac{y-\frac{4}{y}}{2}.
$$
Compute $dx=(1/2-2 y^{-2})dy$, and the integral becomes
$$
\int \frac{dx}{(x^2+1)\sqrt{x^2-4}}=\int \frac{\frac{1}{2}-\frac{2}{y^2}}{\left(\left(\frac{y+\frac{4}{y}}{2}\right)^2+1\right)\left(\frac{y-\frac{4}{y}}{2}\right)}dy=\int \frac{4y\, dy}{y^4+12 y^2+16}.
$$
This can be computed using partial fractions.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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|
Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{2\pi}{14}\sin\frac{4\pi}{14}\sin\frac{6\pi}{14}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{\pi}{14}\sin\frac{2\pi}{14}\sin\frac{3\pi}{14}\sin\frac{4\pi}{14}\sin\frac{5\pi}{14}\sin\frac{6\pi}{14}$$
How can I compute $$\prod\limits_{r=1}^{n}\sin\frac{r\pi}{c}$$
|
Use $\sin\left(\dfrac\pi2-A\right)=\cos A$ and $\sin2B=2\sin B\cos B$
to find the product $$=\dfrac{\sin2t\sin4t\sin6t}8$$ where $7t=\pi$
Now use Method to find $\sin (2\pi/7)$, to find
$$\sin2t\sin4t\sin6t=\sqrt{\dfrac7{64}}$$
|
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|
Pascal's triangle induction proof I am trying to prove
$$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$ for each $k \in \{1,...,n\}$ by induction. My professor gave us a hint for the inductive step to use the following four equations:
\begin{align*}
\binom{n + 1}{k} & = \binom{n}{k} + \binom{n}{k - 1}\\
\binom{n + 1}{k - 1} & = \binom{n}{k - 1} + \binom{n}{k - 2}\\
\binom{n}{k} & = \binom{n}{k - 1}\frac{n - k + 1}{k}\\
\binom{n}{k - 1} & = \binom{n}{k - 2}\frac{n - k + 2}{k - 1}
\end{align*}
I keep getting stuck in the inductive step. I was hoping someone could help me.
|
Assume that $$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$
We need to prove:
$$\binom{n}{k+1} = \binom{n}{k}\frac{n-k}{k+1}$$
We will have
$$\begin{equation}\begin{aligned}
\binom{n}{k+1} &= \frac{n!}{(k+1)!(n-k-1)!} \\
&= \frac{n!}{k!\times(k+1)\times1\times2\times...\times{(n-k)\div(n-k)}} \\
&= \frac{n!}{k!(n-k)!}\times\frac{n-k}{k+1} \\
&=\binom{n}{k-1}\times\frac{n-k+1}{k}\times\frac{n-k}{k+1}\\
&=\frac{n!}{(n-k+1)!(k-1)!}\times\frac{n-k+1}{k}\times\frac{n-k}{k+1}\\
&=\frac{n!(n-k+1)}{(n-k)!\times(n-k+1)\times(k-1)!\times k}\times\frac{n-k}{k+1}\\
&=\frac{n!}{k!(n-k)!}\times\frac{n-k}{k+1}\\
&=\binom{n}{k}\frac{n-k}{k+1}
\end{aligned}\end{equation}$$
If $P(k)$ is true then $P(k+1)$ is also true, we can easily prove that $P(1)$ is true, so the induction is complete.
|
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|
Finding the inverse of a matrix given an equation So I've been given this equation:
$A\begin{bmatrix}
2&3&1&5\\
1&0&3&1\\
0&2&-3&2\\
0&2&3&1
\end{bmatrix} = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$
I'm supposed to find the inverse of A using this but I'm not really sure where to start. Any ideas?
|
If $AB=C$, then $A^{-1}=BC^{-1}$.
Now if we interpret the right-hand side matrix as a change of basis matrix, such that
$$u_1=e_4,\enspace u_2=e_1,\enspace u_3=e_3, \enspace u_4=e_2,$$
then, conversely
$$ e_1=u_2,\enspace e_2=u_4,\enspace e_3=u_3,\enspace e_4=u_1,$$
so that
$$\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}^{-1}=\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}$$
and finally
$$BC^{-1}=\begin{bmatrix}
0&2&3&1\\
2&3&1&5\\
0&2&-3&2\\
1&0&3&1
\end{bmatrix}.$$
|
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|
How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$? How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$
where
$$f(x,y)=x^4+y^4-x^2-2xy-y^2$$
Answer:
$$f_x=4x^3-2x-2y$$
$$f_y=4y^3-2y-2x$$
Critical points are $(0,0), (1,1), (-1,-1)$.
All of these critical points are in D.
$$f(0,0)=0$$
$$f(1,1)=-2$$
$$f(-1,-1)=-2$$
Now we can find the extreme values of $f$ on the boundary of the region $x^2+y^2=4$
$$x^2=4-y^2$$
Therefore we have,
$$g(y)=2y^4-8y^2-2y \sqrt{4-y^2}+16$$
$$g'(y)=8y^3-16y-2 \sqrt{4-y^2}+2y^2(4-y^2)^{-1/2}=0$$
$\color{red}{problem:}$
I could do only up to here. I don't know how to solve the above equation to get the values for $y$.
After finding extreme values of this, we can compare those values with the above and we can find the maximum and minimum values.
|
We have on the boundary $x^2+y^2=4$:
$$f=x^4+y^4-(x^2+y^2)-2xy=(x^2+y^2)^2-2x^2y^2-(x^2+y^2)-2xy=12-2((xy)^2+xy)$$
Note that $|xy| \leq \frac{x^2+y^2}{2}=2$, and the extremal values of $t \mapsto t^2+t$ for $|t| \leq 2$ are $6$ (for $t=2$) and $-\frac{1}{4}$ (for $t=-\frac{1}{2}$).
This shows
$$0 \leq f=12-2((xy)^2+xy) \leq \frac{25}{2}$$
and the values are admitted. $x=y=\pm \sqrt{2}$ yields $f=0$ and $f=\frac{25}{2}$ is admitted for the solutions of $xy=-\frac{1}{2}, x^2+y^2=4$: https://www.wolframalpha.com/input/?i=xy%3D-0.5,+x%5E2%2By%5E2%3D4 (Of course you can compute the solutions by hand, but that is not my job :) )
|
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|
integral for range in x axis Calculate the area in the shaded region:
${f(x) = x^3 -2x + 7}$
${{{\int_{-1}^2}} f(x) dx = x^4 - x^2 + 7x}$
$= {[(2^4) -(2^2) + 14] - [(-1)^4 - (-1)^2 + 7(-1)]}$
$= [16 - 4 + 14] - [- 7] = 19$
But the answer in the book is $21{3\over 4}$.
|
HINT:
$$\int ax^3 + bx^2 + cx^1 + d \, \mathrm{d}x$$
is
$$ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^3 + dx + C $$
You have miscalculated your first term. Everything else is fine.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with steps.
|
By using the chain rule, we have
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\left(\frac{{2x}}{{1+x^2}}\right)'\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ or
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\frac{2 \left(1-x^2\right)}{\left(1+x^2\right)^2}\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ giving
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\frac{2(1-x^2)}{1+6 x^2+x^4}.
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if a and b have the same sign then |a| + |b| = |a + b| We start with a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result :
$|a| + |b| = a+b$
$(|a| + |b|)^2 = (a+b)^2$
$\sqrt{(|a| + |b|)^2} = \sqrt{(a+b)^2}$
$a+b =|a+b|$
Thus : |a+b|= |a| + |b|
For the case where a, b < 0, we use the same technique but with minuses before the constants Opinion ?
|
If $a\geq 0$ and $b\geq 0$ then
$$|a|+|b|=a+b=|a+b|.$$
If $a\leq 0$ and $b\leq 0$ then
$$|a|+|b|=-a-b=-(a+b)=|a+b|.$$
|
{
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|
Equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ Find the equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and passes through the point $(1,1,1)$.
First I thought required line passes through point on intersection of given lines, but as I checked, lines are non-intersecting. So doesn't that mean that there will be infinite lines which satisfy the given conditions?
|
write the first equation in the form $$ [x,y,z]=[1,2,3]+t[2,3,4]$$ and the second one in the form $$[x,y,z]=[-2,3,-1]+s[1,2,4]$$
|
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|
Work out the area enclosed? I am doing a simple exercice and I think that either the book's solution is wrong or I misunderstood the problem.
Here is the problem,
平面上で次の曲線又は直線で囲まれる図形の面積を求めよ。
極座標系について、曲線 $r = a(1+2\cos \theta)$, $(0 ≦ \theta ≦ \frac{2\pi}{3})$, $(a > 0)$と直線 $\theta = 0$.
and its translation:
In the plane, please find the area enclosed by the curves or straight lines.
Consider the curve $r = a(1+2\cos \theta)$, $(0 \leq \theta \leq \frac{2\pi}{3})$, $(a > 0)$ and the straight line $\theta = 0$.
The book's answer is:
$$S = \frac{a^2}{2}(\pi + \frac{7}{8}\sqrt{3})$$
Here is what I have done:
The line $\theta = 0$ is just the same as the $x$-axis so I can safely ignore it in further calculations according to the graph of the function $r$ (that can be seen here).
So I just need to calculate:
$$ I = \frac{2S}{a^2} = \int_0^{\frac{2\pi}{3}}(1+2\cos\theta)^2\mathrm{d}\theta$$.
Which gives:
$$ S = \frac{a^2}{2}(2\pi + \frac{3}{2}\sqrt{3}) $$
Here is how I worked out the integral:
$$ \begin{align} I &= \int_0^{\frac{2\pi}{3}} 1 + 4cos\theta + 4\cos^2\theta\mathrm{d}\theta \\
I &= \frac{2\pi}{3} + 4\left[ \sin\theta \right]_0^{\frac{2\pi}{3}} + 2\left[ \theta + \frac{\sin2\theta}{2} \right]_0^{\frac{2\pi}{3}} \\
I &= \frac{2\pi}{3} + 4(\frac{\sqrt{3}}{2}) + 2(\frac{2\pi}{3}-\frac{\sqrt{3}}{4}) = 2\pi + \frac{3}{2}\sqrt{3}
\end{align}$$
|
This only shows what the book's answer means. It does not mean that the op's answer is wrong because the given curve is $\left[0,\dfrac{2 \pi}{3}\right]$ (blue one), the red one is $\left[\pi,\dfrac{4 \pi}{3}\right]$
|
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|
Multiple Nested Radicals $\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$
I have no idea how to unnest radicals, can anyone help?
|
Work from the inside out.
Let $\sqrt{3 - 2\sqrt{2}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides yields
$$3 - 2\sqrt{2} = a + b - 2\sqrt{ab}$$
Then
\begin{align*}
a + b & = 3 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{2} \tag{2}
\end{align*}
Dividing equation by $-2$ yields
$$\sqrt{ab} = \sqrt{2}$$
Squaring both sides of the equation yields
$$ab = 2$$
Hence,
$$b = \frac{2}{a}$$
Substituting for $b$ in equation 1 yields
\begin{align*}
a + \frac{2}{a} & = 3\\
a^2 + 2 & = 3a\\
a^2 - 3a + 2 & = 0\\
(a - 1)(a - 2) & = 0
\end{align*}
Hence, $a = 1$ or $a = 2$. If $a = 1$, then $b = 2$. However, $\sqrt{1} - \sqrt{2} = 1 - \sqrt{2} < 0$, but $\sqrt{3 - 2\sqrt{2}} > 0$. Thus, $a = 2$ and $b = 1$. Hence,
$$\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$$
Substituting $\sqrt{2} - 1$ for $\sqrt{3 - 2\sqrt{2}}$ yields
$$\sqrt{10 + 4\sqrt{3 - 2\sqrt{2}}} = \sqrt{10 + 4(\sqrt{2} - 1)} = \sqrt{6 + 4\sqrt{2}}$$
Let $\sqrt{10 + 4\sqrt{2}} = \sqrt{c} + \sqrt{d}$, where $c$ and $d$ are rational numbers. Continue.
|
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|
Bivariate normal distribution $X$ and $Y$ I need help figuring out the following.
Let $X$ and $Y$ have the bivariate normal distribution
$$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$
Show that $X$ and $Z=\frac{Y−ρX}{\sqrt{1−ρ2}}$ are independent standard normal random variables.
For $X$, is it asking to show that $f(x)$ is equal to the density function of the standard normal? If yes, how can I do that with $Z$? And how could I prove they are independent given the information provided? I would really appreciate any help here.
I also have to proof $P(X>0,Y>0) = \frac 14 + \frac{1}{2\pi}\arcsin \rho$. How can I do this using the answer to the above? I'm kind of new trying to understand this concept.
|
I assume your initial distribution is
$$
f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right)
$$
Apply the transformation
$Y = Z\sqrt{1-\rho^2} + \rho X$
Including the Jacobian determinant $\sqrt{1-\rho^2}$, the pdf for $X,Z$
$$
f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2(1-\rho^2) +2\rho zx\sqrt{1-\rho^2} +\rho^2x^2 - 2\rho (zx\sqrt{1-\rho^2} + \rho x^2)}{2(1-\rho^2)} \right)
$$
You can show that this is equivalent to the pdf for independent standard normals.
$$
f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right)
$$
|
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|
Moment generating function, determine distribution If $Y$ is a random variable such that $\mathrm{E}[Y^k] = \frac{1}{4} + 2^{k-1}$ for $k = 1,2,{...},$ I'm supposed to determine the distribution. I know the answer is $P(X=0) = \frac{1}{4}, P(X=1) = \frac{1}{4}, P(X=2) = \frac{1}{2},$ but I don't know how to reach that answer.
So $\mathrm{E}[Y] = \frac{5}{4}$ and $\mathrm{E}[Y^2] = \frac{9}{4},$ but that is about as far as I've made it.
Is it all about guessing and confirming by computing the moments or is there a method to this?
|
Knowing those moments, the moment generating function is then $$M_X(t) = 1 + (\tfrac{1}{4} + \tfrac{1}{2} 2^{1})t+(\tfrac{1}{4} +\tfrac{1}{2} 2^{2})\frac{t^2}{2!}+(\tfrac{1}{4} + \tfrac{1}{2}2^{3})\frac{t^3}{3!}+\cdots $$
and rewriting $1 = \tfrac{1}{4} +(\tfrac{1}{4} +\tfrac{1}{2} 2^{0})t^0$ we have $$M_X(t)= \tfrac{1}{4}+\tfrac{1}{4}e^t+\tfrac{1}{2}e^{2t}.$$
Since $M_X(t) = E\left[e^{tX}\right]$, this is clearly the moment generating function of a discrete random variable $X$ with $P(X=0)=\frac14$, $P(X=1)=\frac14$, $P(X=2)=\frac12$
|
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|
Good substitution for this integral What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$
|
$$I=\int \frac{4t}{1-t^4}dt$$
Let $t=\sqrt{\sin\theta}$. Then, $dt=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.
Then,
$$I=\int \frac{4\sqrt{\sin\theta}\cos\theta d\theta}{2\sqrt{\sin\theta}\cos^2\theta}=2\int\sec\theta d\theta=2\ln|\sec\theta+\tan\theta|+c=2\ln\left|\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right|+c=2\ln\left|\frac{t^2+1}{\sqrt{1-t^4}}\right|+c=\ln\left|\frac{(t^2+1)^2}{(1+t^2)(1-t^2)}\right|+c=\ln\left|\frac{t^2+1}{1-t^2}\right|+c$$
|
{
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|
find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$
I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
|
Starting from where you left off: $S = \displaystyle \sum_{k=2}^\infty \left(\left(\dfrac{1}{2(k-1)} - \dfrac{1}{2k}\right)+\left(\dfrac{1}{2k} - \dfrac{1}{2(k+1)}\right)\right)$. Can you find the $2$ telescoping sums and their values?
|
{
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|
Find $ \lim\limits_{\substack{x\to\infty\\y\to0\,\,}}\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}$ I have to solve the following limit:
\begin{align*}
\lim_{\substack{x\to\infty\\y\to0\,\,}}\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}
&=
\lim_{\substack{x\to\infty\\y\to0\,\,}}\left\{\left[\left(1+\frac{1}{x}\right)^x\;\right]^\frac{x^2}{x+y}\right\}^\frac{1}{x}\\
&=
\lim_{\substack{x\to\infty\\y\to0\,\,}} e\cdot\frac{x}{x+y}\\
&=e
\end{align*}
Paint.
So, this is the solution I got. Is everything correct?
.....And something else. I really don't have a clue about this one:
$$
\lim_{\substack{x\to\infty\\y\to\infty}}\frac{\exp\left\{-\dfrac{1}{x^2+y^2}\right\}}{x^4+y^4}.$$
Paint2.
Can you guys help me find a solution?
Thanks!
|
$$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right]\right]=\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\ln\left(\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right)\right]\right]\right]=$$
$$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+y}\right]\right]\right]=\lim_{x\to\infty}\left[\exp\left[\lim_{y\to0}\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+y}\right]\right]=$$
$$\lim_{x\to\infty}\exp\left[\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+0}\right]=\exp\left[\lim_{x\to\infty}\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x}\right]=$$
$$\exp\left[\lim_{x\to\infty}\frac{x\ln\left(1+\frac{1}{x}\right)}{1}\right]=\exp\left[\lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)\right]=$$
Applying l'Hôpital's rule, to get:
$$\exp\left[\lim_{x\to\infty}\frac{1}{1+\frac{1}{x}}\right]=\exp\left[\frac{1}{1+0}\right]=\exp\left[1\right]=e^1=e$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute all possible values of $\cot{\theta} - \frac{6}{z}$. Note that $ 0 < \theta < \frac{\pi}{2}$.
Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.
Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.
$\bf{My\; Try::}$ Let $z-3i=3e^{i\theta}\Rightarrow z=3i+3\cos \theta+3i\sin \theta = 3\cos \theta+3i(1+\sin \theta)$
So we get $$z=3\sin \left(\frac{\pi}{2}-\theta\right)+3i\left[1+\cos\left(\frac{\pi}{2}-\theta\right)\right]$$
So we get $$z=3i\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)e^{-i\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$$
Now how can i solve after that, Thanks
|
WLOG $z-3i=3(\cos2t+i\sin2t)$
$\iff z=3\{\cos2t+i(1+\sin2t)\}$
$=6\sin\left(\dfrac\pi4-t\right)\left[\cos\left(\dfrac\pi4-t\right)+i\sin\left(\dfrac\pi4-t\right)\right]$
$\dfrac6z=\cdots=\dfrac{\cos\left(\dfrac\pi4-t\right)-i\sin\left(\dfrac\pi4-t\right)}{\sin\left(\dfrac\pi4-t\right)}=\cot\left(\dfrac\pi4-t\right)-i$
$\tan\theta=\cdots=\tan\left(\dfrac\pi4-t\right)$
Can you take it from here?
|
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|
Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would be really helpful if someone could point out the method.
we are to use $$z - \frac1{z} = 2i\sin x$$
$$z^n + (\frac1{z})^n = 2\cos nx$$
|
The problem is simple but you may have made a calculation mistake. Let's start with the identity $$\sin^{2}x = \frac{1 - \cos 2x}{2}$$ and then cube it to get $$\sin^{6}x = \frac{1}{8}(1 - 3\cos 2x + 3\cos^{2}2x - \cos^{3}2x)$$ and then we need to note that $$\cos^{2}2x = \frac{1 + \cos 4x}{2},\,\cos 6x = 4\cos^{3}2x - 3\cos 2x$$ and using these we get $$\sin^{6}x = \frac{1}{8}\left(1 - 3\cos 2x + \frac{3 + 3\cos 4x}{2} - \frac{\cos 6x + 3\cos 2x}{4}\right)$$ and on simplifying we get $$\sin^{6}x = \frac{10 - 15\cos 2x + 6\cos 4x - \cos 6x}{32} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the transformation matrix given a set of points The map $A:ℝ^3→ℝ^3$ satisfies:
$A(2,−1,2)=(−2,−9,−30)$
$A(−1,2,2)=(−23,12,−9)$
$A(2,2,−1)=(13,6,−6)$
So if I understand it correctly there is a matrix $A$ such that
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
and
$ [A] \begin{bmatrix}-1\\2\\2\end{bmatrix} = \begin{bmatrix}-23\\12\\-9\end{bmatrix}$
and
$ [A] \begin{bmatrix}2\\2\\-1\end{bmatrix} = \begin{bmatrix}13\\6\\-6\end{bmatrix}$
I'm not sure where to start.
|
An answer for noobs.
You said you had this:
(1)
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
(2)
$ [A] \begin{bmatrix}-1\\2\\2\end{bmatrix} = \begin{bmatrix}-23\\12\\-9\end{bmatrix}$
(3)
$ [A] \begin{bmatrix}2\\2\\-1\end{bmatrix} = \begin{bmatrix}13\\6\\-6\end{bmatrix}$
$ A = \begin{bmatrix}A_{00} A_{01} A_{02}\\ A_{10} A_{11} A_{12} \\ A_{20} A_{21} A_{22} \end{bmatrix}$
So, the first equation is actually:
$ 2A_{00}-1A_{01}+2A_{02}+0A_{10}+0A_{11}+0A_{12}+0A_{20}+0A_{21}+0A_{22}=-2$
$ 0A_{00}+0A_{01}+0A_{02}+2A_{10}-1A_{11}+2A_{12}+0A_{20}+0A_{21}+0A_{22}=-9$
$ 0A_{00}+0A_{01}+0A_{02}+0A_{10}+0A_{11}+0A_{12}+2A_{20}-1A_{21}+2A_{22}=-30$
Doing this for all 3 equations, you get the matrix M in:
$ M \begin{bmatrix}A_{00} \\ A_{01} \\ A_{02}\\ A_{10} \\ A_{11} \\ A_{12} \\ A_{20} \\ A_{21} \\ A_{22} \end{bmatrix} = \begin{bmatrix}-2 \\ -9 \\ -30 \\ -23 \\ 12 \\ -9 \\ 13 \\ 6 \\ -6 \end{bmatrix}$
Then solve for vector A.
|
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|
Rusty at complex integrals, finding error in steps I am looking for some help spotting where I messed up some computations, it's been awhile... I'll show my work to my answer and then say what the correct answer is...
Integrate:
$$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta$$
Let $z=e^{i\theta}$ for $0\leq \theta \leq 2\pi$. Then $|z|=1$ and
$$\cos(3\theta)=\frac{z^3+z^{-3}}{2}=\frac{z^6+1}{2z^3}$$ and
$$\cos(\theta)=\frac{z+z^{-1}}{2}$$ implies
$$\frac{\cos(3\theta)}{5-4\cos(\theta)}=\frac{z^6+1}{2z^3}\cdot \frac{z}{(-2z^2+5z-2)}$$
$$=\frac{z^6+1}{2z^2(-2z^2+5z-2)}=-\frac{z^6+1}{4z^2(z^2-\frac{5}{2}z+1)}=-\frac{z^6+1}{4z^2(z-1/2)(z-2)}.$$
Thus by the change of variable,
$$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta=\oint_{|z|=1} -\frac{z^6+1}{4z^2(z-1/2)(z-2)} \frac{\mathrm{d}z}{iz}$$
$$=-\frac{1}{4i}\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z.$$
Now obviously the poles of $f(z)=\frac{z^6+1}{z^3(z-1/2)(z-2)}$ are $z_0=0$, of order 3, and the simple poles $z_1=1/2$ and $z_2=2$. Only $z_0,z_1\in int \,\gamma$ where $\gamma=\{z\in \mathbb{C} : |z|=1\}$ (and $int \cdot$ is the interior of a set) thus by the Cauchy-residue theorem, we have
$$\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z=2\pi i (Res(f,0)+Res(f,1/2)).$$
Now
$$Res(f,0)=\frac{1}{2!} \lim_{z\to 0}\frac{\mathrm{d}^2}{\mathrm{d}z^2}z^3f(z)=\frac{1}{2}\lim_{z\to 0}\frac{\mathrm{d}^2}{\mathrm{d}z^2} \frac{z^6+1}{(z-1/2)(z-2)}=\frac{21}{2}$$
by an elementary yet tedious calculus computation. I used wolfram alpha to be honest. Next,
$$Res(f,1/2)=\lim_{z\to 1/2} (z-1/2)f(z)=\lim_{z\to 1/2} \frac{z^6+1}{z^3(z-2)}=\frac{(1/2)^6+1}{(1/8)(-3/2)}=-\frac{65}{12}$$
Thus
$$\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z=2\pi i\frac{61}{12}.$$
But,
$$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta=-\frac{1}{4i}\oint_{|z|=1} \frac{z^6+1}{z^3(z-1/2)(z-2)} \mathrm{d}z$$
$$=-\frac{1}{4i}2\pi i\frac{61}{12}=-\frac{61}{24}\pi.$$
However wolfram says the integral (in its original form) is equal to $\pi/12$. Where did I mess up?
|
Community wiki answer so the question can be marked as answered:
As DavidP commented, the residue at $0$ is $\frac{21}4$. Apparently you forgot to multiply the Wolfram|Alpha result by the factor $\frac12$.
|
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|
Can every parabola be written in the form of a quadratic $y=ax^2+bx+c$ or $x=dy^2+ey+f$? I understand that the graph of any equation of the form $y=ax^2+bx+c$ is a parabola (please correct me if I am mistaken).
My question is about the converse: Can every parabola be written in the form of an equation $y=ax^2+bx+c$ or $x=dy^2+ey+f$?
|
A parabola is just the locus of points (a collection) that is equidistant from a point and a straight line. Let $(x,y)$ be the set of points, then for let's say you have a fixed point $(p,q)$ and a line $ax+by+c=0$,
The distance between $(x,y)$ and the straight line is given by:
$$\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$$
We want to set that to be equal to the distance between $(p,q)$ and $(x,y)$
$$ \frac{|ax+by+c|}{\sqrt{a^2+b^2}}=\sqrt{(x-p)^2+(y-q)^2}$$
$$\frac{a^2x^2+b^2y^2+2abxy+2acx+2bcy}{a^2+b^2}=x^2-2px+p^2+y^2-2qx+q^2$$
$$b^2x^2+a^2y^2-(2p(a^2+b^2)+2ac)x-(2q(a^2+b^2)+2bc)y-2abxy+(p^2+q^2)(a^2+b^2)=0$$
So the formula doesn't look pleasant, but we know from that, any parabola will be in the form of $ax^2+by^2+cx+dy+exy+f=0$
|
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|
Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so
$$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get
$$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$
So
$$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$
Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so
$$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$
Need a hint to proceed further.
|
Notice, when $\Re[n]>0$:
*
*$$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x=\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}$$
*$$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x=\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}$$
So:
$$\frac{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x}=\frac{\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}}{\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}}=\frac{2+n+n^2}{n+n^2}$$
Now, when $n=100$:
$$\frac{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100}\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100+1}\space\text{d}x}=\frac{2+100+100^2}{100+100^2}=\frac{10102}{10100}=\frac{5051}{5050}$$
|
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|
Calculate Tangent Points to Circle Problem
Given a circle with radius $r = 2$ and center $C = (4,2)$ and a point $P = (-1,2)$ outside the circle.
How do I calculate the coordinates of two tangent points to the circle, given that the tangents both have to go through $P$?
My (sad) try
All I can think of is finding the equation for the circle, which is
\begin{equation}
(x-4)^{2} + (y-2)^{2} = 4.
\end{equation}
I have no idea what to do next. (I don't even if finding the circle's equation is relevant.)
Update
After using Dhanush Krishna'a answer, I can (easily) find the two intersection points:
\begin{equation}
(x_{1,2}, y_{1,2}) = \frac{2}{5}(8, 5\pm\sqrt{21}).
\end{equation}
|
There is another way to find the that may strike you as funny. Use homogeneous coordinates for the points in the place so the exterior point is located at $P = (-1,2,1)$.
The homogeneous coordinates of a circle located at $(c_x,c_y)$ with radius $r$ are
$$ C = \begin{vmatrix} 1 & 0 & -c_x \\ 0 & 1 & -c_y \\ -c_x & -c_y & c_x^2+c_y^2-r^2 \end{vmatrix} $$
such that the expression $$\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{vmatrix} 1 & 0 & -c_x \\ 0 & 1 & -c_y \\ -c_x & -c_y & c_x^2+c_y^2-r^2 \end{vmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$ simplifies to the standard equation for a circle
$$ (x-c_x)^2 + (y-c_y)^2 - r^2 = 0 $$
Ok, back to the problem. If you take the product $L=C P$, then $L$ is a line with coordinates $L=[a,b,c]$ such as $a x+b y + c =0$ is the equation of the line. This line connects the two tangent points.
In your case
$$ C = \begin{vmatrix} 1 & 0 & -4 \\ 0 & 1 & -2 \\ -4 & -2 & 16 \end{vmatrix} $$
so
$$ L = \begin{vmatrix} 1 & 0 & -4 \\ 0 & 1 & -2 \\ -4 & -2 & 16 \end{vmatrix} \begin{pmatrix} -1 \\ 2 \\1 \end{pmatrix} = \begin{bmatrix} -5 \\ 0 \\ 16 \end{bmatrix} $$
or $$ \left. (x,y,1) \cdot [-5,0,16] = 0 \right\} \left. 16-5 x =0 \right\} x = \frac{16}{5} $$
Plug the line equation in the circle equation to get the other coordinate
$$ \left. \left( \frac{16}{5} \right)^2 - 8 \left( \frac{16}{5} \right) + y^2 -4 y + 16 =0 \right\} y =2 \pm \frac{2\sqrt{21}}{5} $$
The coordinates are thus
$$ \begin{pmatrix} \frac{16}{5} \\ 2 - \frac{2\sqrt{21}}{5} \end{pmatrix} \mbox{ or } \begin{pmatrix} \frac{16}{5} \\ 2 + \frac{2\sqrt{21}}{5} \end{pmatrix}$$
NOTE: The relationship between the point $P$ and the line $L$ is that of pole and polar
|
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|
Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$.
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?
|
$$I_2=\frac{-1}{2} \int_0^a \frac{d(a^2-x^2)}{\sqrt{a^2-x^2}}$$
let $a^2-x^2=t$. If $x=0,$ then $t=a^2$ and if $x=a,$ then $t=0$.
Hence $$I_2=\frac{-1}{2} \int_{a^2}^0 \frac{dt}{\sqrt{t}}$$
$$\Rightarrow I_2= \int_{0}^{a^2} \frac{dt}{2\sqrt{t}}$$
$$\Rightarrow I_2=a-0=a$$
|
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|
Using induction, show ${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$ is an even integer. I'm having serious difficulties with that task, so it should be nice, if there is someone that can help!
The task says:
Prove that the number
$${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$$ is an even integer for all $n\in\mathbb{N}$
|
If $a_n=pr^n+qs^n$, with $r\ne s$, the sequence $a_n$ satisfies the recurrence relation
$$
a_0=p+q,\qquad a_1=pr+qs,\qquad a_{n+2}-(r+s)a_{n+1}+rsa_n
$$
Indeed,
\begin{align}
a_{n+2}-(r+s)a_{n+1}+rsa_n
&=pr^{n+2}+qs^{n+2}-(r+s)(pr^{n+1}+qs^{n+1})+rs(pr^n+qs^n) \\
&=p(r^{n+2}-r^{n+2}-sr^{n+1}+sr^{n+1}) \\
&\qquad+q(s^{n+2}-rs^{n+1}-s^{n+2}+rs^{n+1})\\
&=0
\end{align}
In our case $p=q=1$, $r=(1+\sqrt{2})^2=3+2\sqrt{2}$, $s=(1-\sqrt{2})^2=3-2\sqrt{2}$; thus $r+s=6$ and $rs=1$.
Therefore the recurrence relation is
$$
a_0=2,\qquad a_1=6, \qquad a_{n+2}-6a_{n+1}+a_n=0
$$
Since $a_0$ and $a_1$ are even (base of the induction), if we assume $a_k$ even for every $k$ with $0\le k<n+2$, then
$$
a_{n+2}=6a_{n+1}-a_n
$$
is even too.
A similar strategy is considering $b_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n$. The recurrence relation is
$$
b_0=2,\qquad b_1=2,\qquad b_{n+2}-2b_{n+1}-b_n=0
$$
We want to see that $b_{2n}$ is even; the statement is true for $n=0$. Suppose $b_{2n}$ is even; then
$$
b_{2n+2}=2b_{n+1}+b_n
$$
is even as well.
|
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|
When will the sequence $k \mapsto A + Bk + k^2$ yield a perfect square? Consider the following sequence:
$$a(k) = A + Bk + k^2 ,$$
where $A$ and $B$ are both integers, and $A < B$ ($k$ is of course an integer variable, B is even).
Problem:
For which $k^*$ is $a(k^*) = r^2$ (in other words for which $k$ the given sequence will yield a perfect square)?
Thanks in advance for any advice.
|
By repeatedly replacing $k$ by either $k-1,$ or repeatedly by $k+1,$ we can demand that $0 \leq B < 2.$ This will alter $A,$ let us now call it $C.$ This is simply Gauss reduction, if you think in terms of the quadratic form $f(k,j)=k^2 + B k j + A j^2.$
Two sub-problems only:
$$k^2 + C = u^2$$
$$ k^2 + k + C = v^2 $$
$k^2 + C = u^2$ becomes $u^2 - k^2 = C.$ There is at least one solution as long as $C \neq 2 \pmod 4.$ Then write $C = mn,$ with $m \equiv n \pmod 2,$ and then $u-k=m, u+k = n$
$ k^2 + k + C = v^2 $ becomes $4 k^2 + 4k + 4C = 4v^2, $ $4 k^2 + 4k + 1 +(4C - 1 ) = 4v^2, $ $ (2k+1)^2 + (4C-1) = 4v^2. $ Or $ (4C-1) = 4v^2 - (2k+1)^2 $ and. Write all factorings, including negative, for $4C-1 = mn,$ so that $m,n$ are both odd, agree $\pmod 2,$ so we can then solve $2v-2k-1 = m,$ $2v+2k+1 = n.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Point on a plane closest to the origin Consider the plane with equation $x-y+z=1.$ Find the point on the plane closest to the origin.
I think I need to find a function for the distance between the plane and the origin and then find it's derivative and set it equal to zero. I'm not sure exactly how to go about doing this though. Any ideas?
|
The point on the plane $ax+by+cz=d$ nearest the origin is $$\biggl(\frac{da}{\sqrt{a^2+b^2+c^2}}, \frac{db}{\sqrt{a^2+b^2+c^2}}, \frac{dc}{\sqrt{a^2+b^2+c^2}}\biggr)$$
Here, the point on the plane $x-y+z=1$ nearest the origin is $$\Biggl(\frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times-1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}\Biggr) = \biggl(\frac{1}{\sqrt3}, \frac{-1}{\sqrt3}, \frac{1}{\sqrt3}\biggr)$$
This is an extension into three dimensions of the two dimensional nearest point to the origin on the line $ax+by=c$ is $\Bigl(\frac{ca}{\sqrt{a^2+b^2}}, \frac{cb}{\sqrt{a^2+b^2}}\Bigr)$, which is the point of intersection of $ax+by=c$ and $bx-ay=0$, perpendicular and passing through the origin.
|
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|
Critical points of $g(x, y) = x^3+3xy^2-12x-6y$
Find the critical points of $g(x, y) = x^3+3xy^2-12x-6y$.
$\nabla g(x, y) = (3x^2+3y^2-12, 6xy-6)$ so $\nabla g(x, y) = 0 \implies (3x^2+3y^2-12, 6xy-6) = (0,0)$
So $x^2+y^2 -4 = 0$ and $xy = 1$; combining and writing it in two different ways we get:
$\begin{aligned} &\begin{cases} x^2+2xy+y^2 = 6xy \implies (x+y)^2 = 6 \implies x+y = \pm \sqrt{2}\sqrt{3} \\ x^2-2xy+y^2 = 2xy \implies (x-y)^2 = 2 \implies x-y = \pm \sqrt{2} \end{cases} \\& \implies (x,y) = (\pm \dfrac{1}{\sqrt{2}}(1+\sqrt{3}), \pm \frac{1}{\sqrt{2}}(\sqrt{3}-1)) ~ \text{are the critical points}.\end{aligned}$
This disagrees with the answer that wolfram alpha is giving me. What did I do wrong?
|
Your method can't find two other critical points. You successfully found that $x+y=\pm \sqrt{6}$ and $x-y=\pm \sqrt{2}$, but plus and minus signs don't need to coincide. Then four possibilities occur:
*
*$x=\dfrac{\sqrt{6}+\sqrt{2}}{2}$
*$x=\dfrac{\sqrt{6}-\sqrt{2}}{2}$
*$x=\dfrac{-\sqrt{6}+\sqrt{2}}{2}$
*$x=\dfrac{-\sqrt{6}-\sqrt{2}}{2}$
Here 2 and 3 are missing in your attempt.
|
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|
If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3? It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
Regarding $n^4-1$: If $n>1$, we factor $n^4-1$ as $(n+1)(n-1)(n^2+1)$. We claim that in fact, every prime factor of $n^2+1$ is either 2 or is congruent to 1 mod 4. If $p$ is an odd prime that divides $n^2+1$, then $-1$ is a square mod $p$, but the odd primes for which $-1$ is a square mod $p$ are precisely the primes congruent to 1 mod 4. It remains just to show that $n^2+1$ cannot be a power of 2. If $n$ is even this is obvious, and if $n=2k+1$ is odd, then $n^2+1=(2k+1)^2+1=4k^2+4k+2$ is 2 more than a multiple of 4.
Regarding $n^5-1$, I don't have a proof, but based on experimenting with a few dozen numbers, I conjecture that in fact, every prime factor of $n^4+n^3+n^2+n+1$ is either 5 or is 1 more than a multiple of 5.
|
The case $n=1$ is uninteresting, so let $n\gt 1$. We show that $n^2+n+1$ has a prime factor congruent to $1$ modulo $3$, by showing that $4(n^2+n+1)$ has such a prime factor.
Note that $n^2+n+1$ is odd. We first show that $3$ is not the only odd prime that divides $n^2+n+1$. For suppose to the contrary that $(2n+1)^2+3=4\cdot 3^k$ where $k\gt 1$. Then $2n+1$ is divisible by $3$, so $(2n+1)^2+3\equiv 3\pmod{9}$, contradiction.
Now let $p\gt 3$ be a prime divisor of $(2n+1)^2+3$. Then $-3$ is a quadratic residue of $p$. A straightforward quadratic reciprocity argument shows that $p$ cannot be congruent to $2$ modulo $3$.
Remark: By considering $n$ of the form $q!$ for possibly large $q$, we can use the above result to show that there are infinitely many primes of the form $6k+1$.
|
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|
Differential forms should be invariant under coordinate transformations I am wondering why, if we transform the following differential form, it does not seem to be invariant under the coordinate transformation. The $1$-form on $\mathbb R^2$ is
$$
\omega = \sqrt{x^2 + y^2} dx + 0\cdot dy
$$
and as the coefficient function for $dx$ is radially symmetric, it should be easier to express this form in polar coordinates. Then with
$$
dx = \cos\theta dr - r\sin\theta d\theta \quad
dy = \sin\theta dr + r\sin\theta d\theta
$$
and $f(x(r,\theta), y(r,\theta)) = r$ for $f(x,y) = \sqrt{x^2+y^2}$ we have
$$
\omega = r(\cos\theta dr - r\sin\theta d\theta)
= r\cos\theta dr - r^2\sin\theta d\theta.
$$
So, now it might be reasonable to expect that the value of this differential form is not changed by the coordinate transformation. So consider the point $(1,1)$ in cartesian coordinates, and the tangent vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to that point, then evaluating $\omega$ in cartesian coordinates we have
$$
\omega(1,1)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
= \sqrt{2} \cdot dx\begin{pmatrix} 1 \\ 0 \end{pmatrix}
= \sqrt{2}
$$
as $dx\begin{pmatrix} x \\ y \end{pmatrix} = x$. Now evaluating the form in polar coordinates, the same point is $(\sqrt 2, \pi/4)$ expressed in polar coordinates, so evaluating $\omega$ in polar coordinates we get
\begin{align*}
\omega(\sqrt 2, \pi/4)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
& = \sqrt 2 \cos(\pi/4) dr\begin{pmatrix} 1 \\ 0 \end{pmatrix}
- 2\sin(\pi/4) d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
& = 1\cdot dr\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \sqrt 2 \cdot d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
& = 1 \cdot 0 - \sqrt 2 \cdot 0 \\
& = 1
\end{align*}
which is different (again using that $dr, d\theta$ maps on the first and second component). So maybe the tangent vector has to be transformed too, so
$$
\begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\theta \\ \sin\theta\end{pmatrix}
= \frac{1}{2}\begin{pmatrix}\sqrt 2 \\ \sqrt 2 \end{pmatrix}.
$$
So now doing the second computation again:
\begin{align*}
\omega(\sqrt 2, \pi/4)\left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right)
& = \sqrt 2 \cos(\pi/4) dr \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) - 2\sin(\pi/4) d\theta \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) \\
& = \frac{1}{2}\sqrt 2 - \sqrt 2 \frac{1}{2}\sqrt 2 \\
& = \frac{1}{2}\sqrt 2 - 1
\end{align*}
which is again different.
So whats wrong here? Have I done any transformations wrong? The form should yield the same value for the same point and tangent vector, regardless of the coordinate system in which it is written, or not?
|
The problem with your computation is that the polar coordinates of the basis tangent vector $\partial_x$ at $(1,1)$ are $(\sqrt 2 /2,-1/2)$:
$$v = \partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)} = \frac{\sqrt 2}{2}\partial_r\rvert_{(1,1)} - \frac{1}{2}\partial_\theta\rvert_{(1,1)}.$$
Therefore we find $\omega_{(1,1)}(v) = dr_{(1,1)}(v) - \sqrt 2 d\theta_{(1,1)}(v) = \frac{\sqrt2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$.
Viewing $\mathbb{R}^2 - 0$ as a manifold, you have a polar coordinate chart $\phi = (r,\theta)$ which maps some neighborhood of $(1,1)$ onto a neighborhood of $(\sqrt 2,\pi/4)$. The basis vectors $\partial_r\rvert_{(1,1)}$ and $\partial_\theta\rvert_{(1,1)}$ of the tangent space at $(1,1)$ in terms of this coordinate chart are defined as pushforwards of the standard basis vectors of the tangent space at $(\sqrt 2,\pi/4)$:
$$\partial_r\rvert_{(1,1)} = (\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}), \quad \partial_\theta\rvert_{(1,1)} = (\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}).$$
By definition of the pushforward, we have $$\phi_*(\partial_x\rvert_{(1,1)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_x\rvert_{(\sqrt 2, \pi/4)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_y\rvert_{(\sqrt 2, \pi/4)}$$
and so
$$\partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}) + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)}.$$
|
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|
Finding when $y=1+2\sin(x)$ and $y=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$ are equal How would you solve the following equation?
$$1 +2\sin(x) = 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \quad\text{for }−2\pi\leq x \leq 2\pi.$$
Steps I tried:
\begin{align}
1+2\sin(x) &= 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \\
2\sin(x) -2\sin(\tfrac{x}{2}) &= 2\cos(\tfrac{x}{2}) -1 \\
2[\sin(x) -\sin(\tfrac{x}{2})] &= 2\cos(\tfrac{x}{2}) -1 \\
\sin(x) -\sin(\tfrac{x}{2}) &= \cos(\tfrac{x}{2}) -\tfrac{1}{2} \\
\sin(x) +\tfrac{1}{2} &= \cos(\tfrac{x}{2}) +\sin(\tfrac{x}{2})
\end{align}
Now, squaring both sides,
$$\sin^2(x) +\tfrac{1}{4} = \cos^2(\tfrac{x}{2}) +\sin^2(\tfrac{x}{2}).$$
Using $\sin^2(x) +\cos^2(x) = 1$,
\begin{align}
\sin^2(x) &= 1 -\tfrac{1}{4} \\
\sin(x) &= \mp\sqrt\frac{3}{4}.
\end{align}
When I follow through, I get the solutions below the $x$-axis but not above. Can anyone please explain what's wrong with my solution?
|
Examining your attempt
Your first five lines do nothing but divide the original equation by $2$.
$$1+2\sin(x)=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$$
$2\sin(x)-2\sin(\frac{x}{2})=2\cos(\frac{x}{2})-1$
$2[\sin(x)-\sin(\frac{x}{2})]=2\cos(\frac{x}{2})-1$
$\sin(x)-\sin(\frac{x}{2})=\cos(\frac{x}{2})-\frac{1}{2}$
$$\sin(x)+\frac{1}{2}=\cos(\frac{x}{2})+\sin(\frac{x}{2})$$
You then make a classic mistake when squaring. This line is incorrect.
$$\sin^2(x)+\frac{1}{4}=\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})$$
When squaring both sides of an equation you should square the entire left hand side and the entire right hand side not just each individual bit. If you could square each bit then we could write $1+1=2\implies1^2+1^2=2^2$ which is clearly not true. Instead if you squared your step would look like:
$$\left(\sin(x)+\frac{1}{2}\right)^2=\left(\cos(\frac{x}{2})+\sin(\frac{x}{2})\right)^2$$
Solution
If we expand the brackets above we get:
$$\sin^2(x)+\sin(x)+\frac14=\cos^2(\frac{x}{2})+2\cos(\frac{x}{2})\sin(\frac{x}{2})+\sin^2(\frac{x}{2})$$
$$\sin^2(x)+\sin(x)+\frac14=1+\sin(x)$$
$$\sin^2(x)=\frac34$$
This is the same as what you had but it is a coincidence. Your working is still incorrect (or at least very very brief and skipping some important working out).
$$\sin(x)=\pm\frac{\sqrt{3}}{2}$$
$$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\text{ for }0\leq x\leq2\pi$$
We need to substitute these back in and check each answer as taking the square root could introduce extraneous answers of which we find one. Hence we get:
$$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\text{ for }0\leq x\leq2\pi$$
Obviously there exists other solutions which are these plus integer multiple of $2\pi$.
|
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|
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
|
$$\int x^3\sqrt{4-x^2}\space\text{d}x=$$
Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:
$$\frac{1}{2}\int u\sqrt{4-u}\space\text{d}u=$$
Substitute $s=4-u$ and $\text{d}s=-\space\text{d}u$:
$$\frac{1}{2}\int(s-4)\sqrt{s}\space\text{d}s=\frac{1}{2}\int\left[s^{\frac{3}{2}}-4\sqrt{s}\right]\space\text{d}s=$$
$$\frac{1}{2}\left[\int s^{\frac{3}{2}}\space\text{d}s-4\int\sqrt{s}\space\text{d}s\right]=\frac{1}{2}\left[\int s^{\frac{3}{2}}\space\text{d}s-4\int s^{\frac{1}{2}}\space\text{d}s\right]=$$
Use:
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$
$$\frac{1}{2}\left[\frac{2s^{\frac{5}{2}}}{5}-4\cdot\frac{2s^{\frac{3}{2}}}{3}\right]+\text{C}=\frac{1}{2}\left[\frac{2s^{\frac{5}{2}}}{5}-\frac{8s^{\frac{3}{2}}}{3}\right]+\text{C}=\frac{s^{\frac{5}{2}}}{5}-\frac{4s^{\frac{3}{2}}}{3}+\text{C}=$$
$$\frac{(4-u)^{\frac{5}{2}}}{5}-\frac{4(4-u)^{\frac{3}{2}}}{3}+\text{C}=\frac{(4-x^2)^{\frac{5}{2}}}{5}-\frac{4(4-x^2)^{\frac{3}{2}}}{3}+\text{C}$$
|
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|
What does the diagonalized matrix say about a Transformation? I have a matrix given:
$$A=\begin{pmatrix}
7 & -2 \\
-1 & 8
\end{pmatrix}
$$
I have found its characteristic polynomial: $\lambda^2 - 15\lambda +54 = 0$,
which gave me $\lambda = 6, 9$.
Now, I used that to find the Diagonal matrix.
$$\begin{pmatrix}
6 & 0\\
0 & 9
\end{pmatrix}$$
Now, to describe the transformation $ T: \mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $T(\vec{x}) = A\vec{x}$ geometrically I would think the following, but I am uncertain.
The diagonal matrix shows the transformation upon the elementary matrix. So, in this case we could say the transformation of a vector in $\mathbb{R}^2$ eg: $\begin{pmatrix}2\\1\end{pmatrix}$ would give us the following vector: $\begin{pmatrix}12\\9\end{pmatrix}$ is this correct?
My reasoning is because I think the x-value of a vector in $\mathbb{R}^2$ is multiplied by magnitude of 6 and the y-value by 9.
|
Did you do the computation $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}2 & 1 \end{pmatrix}$? That gives $\begin{pmatrix}7(2)-2(1) \\ -1(2)+ 8(1)\end{pmatrix}= \begin{pmatrix}12 \\ 6 \end{pmatrix}$ not $\begin{pmatrix}12 \\ 9 \end{pmatrix}$. You can multiply the matrix $\begin{pmatrix}6 & 0 \\ 0 & 9 \end{pmatrix}$ by a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ only when the given vector can be written as x times the unit vector in the direction of the eigenvector corresponding to eigenvalue 6 and y times then unit vector in the direction of the eigenvector corresponding to eigenvalue 9. To find those solve $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}6x \\ 6y\end{pmatrix}$ and $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}9x \\ 9y\end{pmatrix}$
|
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|
Simplification of surds $\frac{x}{\sqrt{x^2 - x^4}}$
$$\frac{x}{\sqrt{x^2 - x^4}}$$
I believe that I can factor out the $x^2$ in the square root to get
$$\frac{1}{\sqrt{1-x^2}} .$$
However, Wolfram Alpha doesn't do the simplification, hence my confusion.
|
$$\frac{x}{\sqrt{x^2 - x^4}}=
\Bigg\{\begin{array}{c}
\frac{1}{\sqrt{1-x^2}}, \ \ 0<x<1 \\
-\frac{1}{\sqrt{1-x^2}}, \ \ -1<x<0
\end{array}
$$
|
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|
Roots of quadratic equation are given by $b \pm \sqrt{b^2 - c}$ I was reading slides about the cancellation error in quadratic equations and it's written:
The roots of the quadratic equation:
$$x^2 - 2bx + c = 0$$
with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.
Fact that let me perplexed, since I always thought that the roots can be found using the following formula:
$$\frac{b \pm \sqrt{b^2 - 4ac}}{2a}$$
What the relation between one and the other?
|
Never mind, they can be found by that formula only for this quadratic equation $$x^2 - 2bx + c = 0$$
Indeed if we use the usual formula:
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
to the equation above we obtain:
$$\frac{-(-2b) \pm \sqrt{(-2b)^2 - 4*1*c}}{2*1}$$
$$\frac{2b \pm \sqrt{4b^2 - 4c}}{2}$$
$$\frac{2b \pm \sqrt{4(b^2 - c)}}{2}$$
$$\frac{2b \pm 2\sqrt{b^2 - c}}{2}$$
$$\frac{2(b \pm \sqrt{b^2 - c})}{2}$$
$$b \pm \sqrt{b^2 - c}$$
|
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|
Proving a congrent modular How do I show that $3^{1974}+5^{1974}\equiv 0 \pmod {13}$? I have tried feeding the values onto a calculator but they are so big to be computed. What is the best approach?
|
since Fermat's little theorem, implies $3^{12}\equiv5^{12}\equiv 1\pmod{13}$ we have:
$$
3^{1974}+5^{1974}=3^{12*164+6}+5^{12*164+6}=(3^{12})^{164}3^6+(5^{12})^{164}5^6\equiv 3^6+5^6 \pmod{13}
$$
Now, you can easily see that:
$$3^3=27\equiv 1\pmod{13}\Rightarrow 3^6=1\pmod{13}$$
and:
$$5^2=25\equiv -1\pmod{13}\Rightarrow5^6=-1\pmod{13}$$
thus, summing the above:
$$
3^{1974}+5^{1974}\equiv 3^6+5^6 \equiv 1-1 =0 \pmod{13}
$$
|
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|
Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f $ must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$
This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.
From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
|
Because $f$ is divisible by $x^3$, we must have $f(x)=r_3x^3+r_4x^4+r_5x^5$.
The second condition gives, comparing coefficients of powers of $x$, that
$$
f(x)=6x^5 - 15x^4 + 10x^3.
$$
Edit: Comparing coefficients gives me the linear equations $r_4 + 3r_5 - 3=0, r_3 - 3r_5 + 8$ and $r_5=6$.
|
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|
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$.
Thanks.
|
The continuous function
$$
f(a, b, c, d) = abc+abd+acd+bcd-abcd
$$
has a maximum on the compact set
$$
\{ (a, b, c, d) \in \Bbb R^4\mid a, b, c, d \ge 0, a+b+c+d = 1 \}
$$
which is obtained at some point $(A, B, C, D)$. If we can show
that $A=B=C=D$ then it follows that the maximum is
$$
M = f(\frac 14,\frac 14,\frac 14,\frac 14) = \frac{15}{256} \, .
$$
First observe that at most one of $(A, B, C, D)$ can be zero,
since otherwise $f(A, B, C, D) = 0$, and $0$ is not the maximum.
Now assume that $A=B=C=D$ does not hold. Without loss of generality,
assume that $A \ne B$. From
$$
f(a, b, c, d) = ab(c+d-cd) + (a+b)cd
$$
it follows that
$$
f(\frac{A+B}2, \frac{A+B}2,C, D) - f(A, B, C, D) =
\left( \frac{(A+B)^2}4 - AB \right) (C + D - CD) \\
= \frac{(A-B)^2}4 (1 - (1-C)(1-D))
$$
and that expression is $> 0$ because $A \ne B$, and $C, D$ are
not both zero. So we have
$$
f(\frac{A+B}2, \frac{A+B}2,C, D) > f(A, B, C, D) \, ,
$$
contradicting the assumption that $f$ obtains its maximum
at $(A, B, C, D)$.
|
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|
How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?
|
To find $$\int\frac{x^2dx}{(x^2+1)^2}$$ you may want consider the trigonometric subsitution $x=\tan t$ and $dx=\sec^2t dt$
This gives:
$$\int\frac{\tan^2t \sec^2t dt}{(\tan^2t+1)^2}=\int\frac{\tan^2t \sec^2t dt}{\sec^4t}=\int\frac{\tan^2t dt}{\sec^2t}=\int \frac{\sin^2t\cos^2t}{\cos^2t}dt=\int \sin^2tdt$$
The last integral is easily evaluated with the reduction formula for $\sin^2t= \dfrac {1 - \cos 2 t} 2$.
|
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|
Sum of number-of-divisors function equals $\sum_{j=1}^{n} \lfloor n/j \rfloor$. I am trying to prove the identity
$$t(1) + t(2) + \cdots + t(n) = \Big\lfloor \dfrac{n}{1} \Big\rfloor + \Big\lfloor \dfrac{n}{2} \Big\rfloor + \cdots + \Big\lfloor \dfrac{n}{n} \Big\rfloor,$$
where $t(j)$ is the number of divisors of $j$.
Attempt: The highest power of $k \in \{2, 3, \ldots, n\}$ that divides $n!$ is given by
$$H_{k} := \Big\lfloor \dfrac{n}{k}\Big\rfloor + \Big\lfloor \dfrac{n}{k^{2}}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{k^{N_{c}}}\Big\rfloor,$$
where $N_{c}$ is the maximal integer satisfying $k^{N_{c}} \leq n$. With a little thought, one can see that
\begin{equation*}
\begin{split}
\Big\lfloor\dfrac{n}{1}\Big\rfloor + \Big\lfloor\dfrac{n}{2}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{n}\Big\rfloor &= H_{2} + H_{3} + \cdots + H_{n} + \Big\lfloor\dfrac{n}{1}\Big\rfloor \\
&= H_{2} + H_{3} + \cdots + H_{n} + n.
\end{split}
\end{equation*}
Each $H_{k}$ is a counted in the sum $t(1) + t(2) + \cdots + t(n)$, since each $t(k^{j})$ is a summand for $j \in \{1, 2, \ldots, N_{k}\}$, and also each $t(m)$ is a summand, where $m$ is a multiple of $k$ less than or equal to $n$.
This is where I'm a little stuck: I think that the only other summand in $t(1) + t(2) + \cdots + t(n)$ would be $1 + 1 + \cdots + 1 = n$, since each $t(j)$ counts $1$ as a divisor. My problem is showing that these are the only extra terms in $t(1) + t(2) + \cdots + t(n)$, which would show equality.
Any hints would be appreciated, or even suggestions of going about it a different way.
|
It really helps to write $t(1)+t(2)+\dotsb+t(n)$ in a triangular array:
1 +
1 + 1 +
1 + 0 + 1 +
1 + 1 + 0 + 1 +
1 + 0 + 0 + 0 + 1 +
1 + 1 + 1 + 0 + 0 + 1 +
1 + 0 + 0 + 0 + 0 + 0 + 1 +
1 + 1 + 0 + 1 + 0 + 0 + 0 + 1 +
1 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 1 + ...
Row $k$ is $t(k)$ written in a way that indicates which numbers from 1 to $k$ divide $k$. What do you see vertically?
As a final hint, note that $\lfloor n/d \rfloor$ counts the number of multiples of $d$ not exceeding $n$.
Since you have resolved your issue, let me explain the above.
We have
$$\sum_{k=1}^n t(k) = \sum_{k=1}^n \sum_{\substack{1 \le d \le n \\ d|k}} 1$$
and we want to swap the order of summation. Well, as $k$ and $d$ range over $1$ to $n$, you add $1$ every time $d$ divides $k$. This is the same as adding $1$ every time $k$ is a multiple of $d$. For a fixed $d$, how many multiples of $d$ are there between $1$ and $n$? Precisely $\lfloor n/d \rfloor$ (they are $d$, $2d$, $3d$, ..., $md \le n < (m+1)d$).
So $$\sum_{k=1}^n \sum_{\substack{1 \le d \le n \\ d|k}} 1 = \sum_{d=1}^n \sum_{\substack{1 \le k \le n\\ k = md}} 1 = \sum_{d=1}^n \Big\lfloor \frac{n}{d} \Big\rfloor$$
and the proof is complete.
|
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|
Solve $\lfloor x \rfloor = ax+1$ for integral $a$ Solve $\lfloor x \rfloor = ax+1$, where $a$ is an integer.
I have found the values of $x$ for $a=0$, $a=1$ and $a=-1$. But I don't know how to continue. How can I find the solutions for integral $a\leq-2$ and $a\geq2$ ?
|
When $a=0$ there are infinite solutions , they are clearly $x\in [1,2)$. We now assume $a\neq 0$.
Since $\lfloor x \rfloor$ is an integer we conclude $ax+1$ is an integer, so $x=\frac{k}{a}$ for some $k$.
We must therefore find all the integer values for $k$ such that $\lfloor \frac{k}{a} \rfloor=k+1$.
It is clear that if $k$ is non-negative then $\lfloor \frac{k}{a} \rfloor < k+1$.
Now, if $k$ is negative we have $\lfloor \frac{k}{a} \rfloor = - \lceil \frac{-k}{a} \rceil$.
So we must find the non-negative integer values of $l$ so that $\lceil \frac{l}{a} \rceil= l-1$
$l=0$ never works
$l=1$ works if and only if $a$ is negative and $a\neq -1$.
$l=2$ works if and only if $a$ is positive and $a\neq1$.
$l=3$ works if and only if $a=2$.
$l=4$ never works.
Finally $l> 4$ does not work for any $a$, clearly not for $a=1$, and if $a\neq 1$ we have: $\lceil\frac{l}{a}\rceil\leq \frac{l}{a}+1\leq \frac{l}{2}+1< l-1$.
We conclude the solutions are as follows:
If $a<-1$ we have $x =\frac{-1}{a}$.
If $a=-1$ we have no solutions.
If $a=0$ we have $x\in(1,2]$.
If $a=1$ we have no solutions.
if $a=2$ we have $x\in \{\frac{-2}{a},\frac{-3}{a}\}$.
If $a>2$ we have $x = \frac{-2}{a}$.
|
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|
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.
|
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$?
It is the latter.
So, having $\tan\alpha=-6/8=-3/4$ gives $\alpha=\arctan(-3/4)\approx -36.9^\circ$.
Hence, for $n\in\mathbb Z$,
$$\begin{align}10\cos(x-(-36.9^\circ))=5&\Rightarrow \cos(x+36.9^\circ)=1/2\\&\Rightarrow x+36.9^\circ=\pm 60^\circ+360^\circ n\\&\Rightarrow x=-36.9^\circ\pm 60^\circ+360^\circ n\\&\Rightarrow x=23.1^\circ,\quad 263.1^\circ\end{align}$$
since $0^\circ\le x\le 360^\circ$.
|
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|
Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we get $x^4+a^2-2ax^2=a-x\Rightarrow x^4-2ax^2+x+a^2-a=0$
Now How can i solve it after that, Help me
Thanks
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It factors!
$x^4-2ax^2+x+(a^2-a)=(x^2+x-a)(x^2-x+(1-a))$
How do I get that? Well, there will be an extraneous root with the wrong signs for the inner square root, thus:
$x=\sqrt{a-\sqrt{a-x}}$
$-x=-\sqrt{a-\sqrt{a+(-x)}}$
The second extraneous equation is consistent with the simpler recursion
$y=-\sqrt{a+y}$
with $y=-x$. So upon squaring the last equation we find that $y^2-y-a=x^2+x-a=0$ meaning the quadratic expression must be a factor of the quartic.
But perforce it is the wrong factor. The intended roots must be in the complementary quadratic factor
$(x^2-x+(1-a))=0$.
|
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|
Block matrix of order $m$ with three block matrices How to find eigenvalues of following block matrices?
$M=\begin{bmatrix}
A & B & O & O & O & O & O & \cdots & O & O\\
B & A & B & O & O & O & O & \cdots & O & O\\
O & B & A & B & O & O & O & \cdots & O & O\\
O & O & B & A & B & O & O & \cdots & O & O\\
O & O & O & B & A & B & O & \cdots & O & O\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \cdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\
O & O & O & O & O & O & O & O & A & B\\
O & O & O & O & O & O & O & O & B & A\\
\end{bmatrix}_m$
Where,
$A=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 &\cdots & 0 & 1 \\
1 & 0 & 1 & 0 & 0 & \cdots & 0 & 0\\
0 & 1 & 0 & 1 & 0 & \cdots & 0 & 0\\
0 & 0 & 1 & 0 & 1 & \cdots & 0 & 0\\
0 & 0 & 0 & 1 & 0 & \ddots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \vdots\\
0 & 0 & 0 & 0 & 0 & \ddots & 0 & 1\\
1 & 0 & 0 & 0 & 0 & \cdots & 1 & 0\\
\end{bmatrix}_n$
$B=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 &\cdots & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
\end{bmatrix}_n$
$O=\begin{bmatrix}
0 & 0 & 0 & 0 & 0 &\cdots & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\
\end{bmatrix}_n$
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You can find a computation method for the eigenvalues and eigenvectors in the following recent document: "Eigendecomposition of Block Tridiagonal Matrices" by
A. Sandryhaila and J.M.F. Moura (arxiv.org/pdf/1306.0217)
|
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|
A seemingly-trivial divisibility conjecture While working on another problem, I stumbled on the following divisibility claim.
Conjecture: No integers $a,b,c,d$ satisfy all of the following conditions:
*
*$a^2+b^2-c^2-d^2 = 2(ad-bc)-1$;
*$\gcd(ac+bd,a^2+b^2-c^2-d^2) > 1$;
*$\gcd(ac+bd+1,ad-bc) > 1$;
*$(ac+bd)-(ad-bc) = \gcd(ac+bd,ad-bc)^2 > 1$.
There is also the following weaker version.
Conjecture (Weak Version): No integers $a,b,c,d$ satisfy all of the following conditions:
*
*$a^2+b^2-c^2-d^2 = 2(ad-bc)-1$;
*$\gcd(ac+bd,a^2+b^2-c^2-d^2) > \sqrt{ac+bd}$;
*$\gcd(ac+bd+1,ad-bc) > 1$;
*$(ac+bd)-(ad-bc) = \gcd(ac+bd,ad-bc)^2 > ad-bc$.
It seems like either (and especially the Weak Version) should be easy to prove… but I've had no luck. Note that all four restrictions seem necessary, as removing any one allows solutions to be found.
Anyone have any hints for me?
EDIT #1:
We have the identity
\begin{align*}
b^2(a^2+b^2-c^2-d^2) - (ac+bd)(ac-bd) = (a^2+b^2)(b^2-c^2),
\end{align*}
so $\gcd(a^2+b^2-c^2-d^2\!,\, ac+bd) \mid (a^2+b^2)(b^2-c^2).$ We also have
\begin{align*}
a(ac+bd)-b(ad-bc) = c(a^2+b^2)
\end{align*}
so $\gcd(ac+bd,\, ad-bc) \mid c(a^2+b^2)$. Finally, we have
\begin{align}
a^2\bigl((ac+bd+1)-(a^2+b^2-c^2-d^2+1)\bigr) - (ad-bc)(ad+bc+ab) = (a^2+b^2)(c^2+ac-a^2),
\end{align}
so $\gcd(ac+bd+1,\, ad-bc)\mid (a^2+b^2)(c^2+ac-a^2)$; alternatively [and more efficiently],
\begin{align}
a(ac+bd+1)-a(ad-bc) = c(a^2+b^2)+a,
\end{align}
so $\gcd(ac+bd+1,ad-bc) \mid \bigl(c(a^2+b^2)+a\bigr)$. But I can't seem to take it to the goal line.
EDIT #2:
If we write $g=ac+bd$ and $f=ad-bc$, replace Conditions 2-4, and “absorb” Condition 1, we have
*
*$\gcd(g,2f-1) > 1$;
*$\gcd(g+1,f) > 1$;
*$g-f = \gcd(g,f)^2 > 1$.
This system has many solutions! That is to say, not only can Condition 1 not be eliminated entirely, it cannot even be “absorbed” into the other conditions.
|
Not an answer, but just to share if it could be a way, on the track of your edit2.
Put $z = a + i\,b\quad w = c + i\,d$
then:
$$
a^{\,2} + b^{\,2} - c^{\,2} - d^{\,2} = \tilde z\,z - \tilde w\,w
$$
$$
\left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = \tilde z\,z - \tilde z\,w + \tilde w\,z - \tilde w\,w = \tilde z\,z - \tilde w\,w - 2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)
$$
$$
ad - bc = {\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)\quad ac + bd = {\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right)
$$
and
$$
\left\{ \matrix{
\tilde z\,z - \tilde w\,w - 2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) = \left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = - 1 \hfill \cr
\gcd \left( {{\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) - 1} \right) > 1 \hfill \cr
\gcd \left( {1 + {\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)} \right) > 1 \hfill \cr
{\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right) - {\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) = \gcd \left( {{\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)} \right)^{\,2} > 1 \hfill \cr} \right.
$$
|
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Problem about find the extreme of a function (Multipliers of Lagrange) Good morning. I have a problem with this:
Find the maximum and minimum distances from the origin to the curve* $$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$
I have done this:
Function to optimize:$f\left(x,y\right)=x^{2}+y^{2}$
Restriction: $g\left(x,y\right)=5x^{2}+6xy+5y^{2}=8
$
Applying Lagrange multipliers:
$\nabla f\left(x,y\right)=\lambda\nabla g\left(x,y\right)$
Then,
$\nabla f\left(x,y\right)=2x\hat{i}+2y\hat{j}$ and $\lambda\nabla g(x,y)=\lambda(2x+\frac{6}{5}y)\hat{i}+\lambda\left(2y+\frac{6}{5}x\right)\hat{j}
$
Making the ecuation system:
$\begin{cases}
2x=(2x+\frac{6}{5}y)\lambda\\
2y=(2y+\frac{6}{5}x)\lambda\\
x^{2}+\frac{6}{5}xy+y^{2}=8
\end{cases}$
But I have serious problem solving the system. Any suggestions?
|
It turns out that this system of Lagrange equations doesn't really warrant the use of Mathematica. We are faced with
$$ \ 2x \ = \ (2x \ + \ \frac{6}{5}y)\lambda \ \ , \ \ 2y \ = \ (2y \ + \ \frac{6}{5}x)\lambda \ \ . $$
Since bringing all the terms to one side in each equation won't help us to factor the expressions, we might instead solve each equation for $ \ \lambda \ $ to obtain
$$ \lambda \ \ = \ \ \frac{x}{x \ + \ \frac{3}{5} y} \ \ = \ \ \frac{y}{y \ + \ \frac{3}{5} x} \ \ . $$
Assuming for the moment that neither denominator is equal to zero (it will turn out that they are not), we can "cross-multiply" the ratios to produce
$$ xy \ + \ \frac{3}{5} x^2 \ \ = \ \ xy \ + \ \frac{3}{5} y^2 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ \ \Rightarrow \ \ y \ = \ \pm x \ \ . $$
We have two cases now for insertion into the curve equation:
$$ \mathbf{y = x :} \quad 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ \ \rightarrow \ \ 5x^2 \ + \ 6·x·x \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 16x^2 \ = \ 8 $$ $$ \Rightarrow \ \ x^2 \ = \ \frac{1}{2} \ \ ; $$
$$ \mathbf{y = -x :} \quad \quad \quad \rightarrow \ \ 5x^2 \ + \ 6·x·(-x) \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 4x^2 \ = \ 8 \ \ \Rightarrow \ \ x^2 \ = \ 2 \ \ . $$
If we only need the "distance-squared" from the origin for the extremal points, then the cases are $ \ \mathbf{y = x :} \ \ x^2 \ + \ y^2 \ = \ \frac{1}{2} \ + \ \frac{1}{2} \ = \ 1 \ \ $ and $ \ \ \mathbf{y = -x :} \ \ x^2 \ + \ y^2 \ = \ 2 \ + \ 2 \ = \ 4 \ , $ making the extremal distances from the origin $ \ 1 \ $ and $ \ 2 \ . $ [The points at these distances are $ \ \left( \pm \frac{1}{\sqrt{2}} \ , \ \pm \frac{1}{\sqrt{2}} \right) \ $ and $ \ \left( \pm \sqrt{2} \ , \ \mp \sqrt{2} \right) \ , $ respectively.
The constraint curve $ \ 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ $ has symmetry about the origin, so it is reasonably to expect that there should be a critical point in each quadrant (with extremal points of the same type in opposite quadrants). The curve is a "rotated ellipse" for which we have found the lengths of its semi-major and semi-minor axes and the locations of their endpoints.
|
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|
Logarithm question with base change If $\log_{12} 27 = a$ then find the value of $\log_6 16$.
|
$a=\log_{12}27$ is equivalent to $3^3=12^a=2^{2a}3^a$. So $2^{2a}=3^{3-a}$. Hence $2^{3+a}=2^{3-a}3^{3-a}=6^{3-a}$. So $16=2^4=6^b$ where $b=\frac{4(3-a)}{3+a}$. Hence $\log_616=b=\frac{4(3-a)}{3+a}$.
|
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"url": "https://math.stackexchange.com/questions/1761837",
"timestamp": "2023-03-29T00:00:00",
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Please prove the following: Given $ƒ(x) = e^x$, verify that $\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$. Given $ƒ(x) = e^x$, verify that
$$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$ and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.
|
Recall that all we need to show is that $\lim_{h\to 0} \frac{e^h - 1}{h} = 1$. The only things we need in this case are the definition of $e$ given by $e^h = \lim_{n\to \infty}\left ( 1 + \frac{h}{n} \right )^n$, and the binomial theorem. Observe that
\begin{eqnarray*}
\lim_{h\to 0} \frac{e^h -1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \left ( 1 + \frac{h}{n} \right )^n - \frac{1}{h} \\
& = & \lim_{h\to 0}\lim_{n\to \infty} \frac{1}{h} \sum_{k=0}^n \binom{n}{k}\left ( \frac{h}{n}\right )^k - \frac{1}{h}.
\end{eqnarray*}
If we factor out $\frac{1}{h}$ from both terms, we see that we can simply start the summation from the $k=1$ term, hence we have
\begin{eqnarray*}
\lim_{h\to 0} \frac{e^h-1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \sum_{k=1}^n \binom{n}{k} \left ( \frac{h}{n} \right )^k \\
& = & \lim_{h\to0} \lim_{n\to \infty}\frac{1}{h} \left (n \frac{h}{n} + \frac{n(n-1)}{2}\frac{h^2}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{h^3}{n^3} + \ldots + \frac{h^n}{n^n} \right ) \\
& = & \lim_{h\to 0} \lim_{n\to\infty} \left( 1 + \binom{n}{2}\frac{h}{n} + \binom{n}{3} \frac{h^2}{n^3} + \ldots + \frac{h^{n-1}}{n^n} \right ).
\end{eqnarray*}
No matter what the form of the series is in the limit we'll have that there are terms in various powers of $h$, which in the limit as $h\to 0$ will all vanish. The only term that remains is the value of $1$ at the beginning of the summation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the first order ordinary differential equation $y'(x)=2x \cos^2 y(x)$
Solve $$y'(x) =2x \cos^2 y(x) .$$
\begin{align}
\frac{dy}{dx} &= \ 2x \cos^2 y(x) \\
\frac{dy}{\cos^2 y(x)} &=2x \, dx \\
\tan y(x) &=x^2+k, \qquad\qquad k \in \mathbb{R} \\
y(x) &=\arctan (x^2+k)
\end{align}
Is is correct?
Thanks!
|
You can check if your proposed solution actually solves the equation. Let's calculate the derivative:
$$
y(x) = \arctan(x^2 + k)\\
y'(x) = \frac{2x}{(x^2+k)^2+1}
$$
This should equal $2x\cos^2(y(x))$ if $y(x)$ solved the original equation:
$$
2x\cos^2(y(x)) = 2x\cos^2(\arctan(x^2 + k)) = 2x\left(\frac{1}{\sqrt{(x^2+k)^2+1}}\right)^2 = \frac{2x}{(x^2+k)^2+1}
$$
Since those are the same quantity, you can say that what you derived solves the equation. Some comments and other answer underscore that that this may not be a unique solution since the inverse tangent of the tangent of an angle is not necessarily the original angle.
|
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|
Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ however I think this will make the proving even more harder. I also attempted applying the Cauchy-Schwarz inequality and Hölder's inequality but to no avail.
Could someone please show me how to prove it using the inequalities above or another method.
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Note that $$\sum_{cyc}\frac{a}{bc+2}=\sum_{cyc}\frac{a^2}{2a+abc}=\sum_{cyc}\frac{a^2}{2a+1} \ge \frac{(a+b+c)^2}{2a+2b+2c+3} $$
Now one can use that $$\sum_{cyc}a \ge 3\sqrt[3]{abc}=3$$This establishes that $$(a+b+c-1)^2 \ge 4 \Leftrightarrow (a+b+c)^2 \ge 2(a+b+c)+3$$Thus, our proof is done, with equality at $a=b=c$.
|
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|
Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displaystyle \int \sec^3\theta \sqrt{\tan^2\theta}\sec\theta\tan\theta d\theta}$
*${\displaystyle \int \sec^3\theta \tan\theta \cdot \sec\theta \tan\theta d\theta}$
*${\displaystyle \int \sec^4\theta \cdot \tan^2\theta d\theta}$
*${\displaystyle \int \sec^2\theta \sec^2\theta\tan^2\theta d\theta}$
*$u=\tan\theta$, $du= \sec^2\theta d\theta$
*${\displaystyle \int u^2(u^2 + 1)}$
*${\displaystyle \int u^4 + u^2}$
*${\displaystyle = \frac{1}{5}u^5 + \frac{1}{3}u^3 + C}$
here is where I'm even more shaky:
*${\displaystyle \frac{1}{5}\tan^5 + \frac{1}{3}\tan^3 + C}$
*since $\tan= \sin/\cos$, $\tan$ also $= \sec/\csc$, and if $\sec\theta=x$, per the substitution, couldn't $\tan$ also be written $x/(1/x)=x^2$, right?
*I replaced all $\tan$s in the final answer with $x^2$, leaving a final answer of ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$
|
How about this:
\begin{align}
u & = \sqrt{x^2-1} \\
u^2 & = x^2 - 1 \\
u^2+1 & = x^2 \\
2u\,du & = 2x\,dx \\
\end{align}
\begin{align}
\int x^3 \sqrt{x^2-1} \, dx & = \int x^2 \sqrt{x^2-1} \Big( x\,dx\Big) \\
& = \int (u^2+1) u \left( \frac 1 2 \, du \right) \\
& = \frac 1 2 \int (u^4 + u^2) \, du \\
& \phantom{{}={}} \text{etc.}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the $\lim_{n\to \infty} {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}$ Because $(1+\frac{k}{n})\leq (1+\frac{1}{n})^k$ $$ {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}\leq\sqrt[n]{(1+\frac{1}{n})^{n(n+1)/2}}=(1+\frac{1}{n})^{(n+1)/2}$$ and so $$\lim_{n\to \infty}{\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}<\sqrt{e}$$
But what is the exact limit?
Added: I Gave this to Wolfram and the answer was $\infty$ ! but why?
|
You can apply Cauchy's limit theorem
$$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$
with
$$a_n = (1+1/n)(1+2/n)…(1+n/n) = \frac{(2n)!}{n! n^n}.$$
Then
$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!} \\ = \frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}} \\ = 2\frac{(2 + 1/n)}{(1+1/n)}\frac{1}{(1+1/n)^n},$$
and
$$\lim_{n \to \infty}\sqrt[n]{a_n} = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = 2 \cdot 2 \cdot \frac{1}{e} = \frac{4}{e}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Induction Proof $2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{ 2}$ I am looking for an induction proof...
$$2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{2}$$
when $n \geq 1$.
I know there are $3$ steps to this.
1) Check
2) Do $n = k$
3) Do $n = k + 1$
Problem is, I can't seem to get past the first step, when I check:
$n = 1: (9(1) - 1) = 8$ but isn't it suppose to be come to $2$?
|
For $n=1$, Ok. It's easy to check
Now assume it works for $n$, the we have to show it works for $n+1$.
(note that $9(n+1)-1=9n+8$)
we have to show the sentence below is true:
$$2+5+8+...+9n-1+9n+2+9n+5+9n+8=\frac{3(n+1)(9(n+1)+1)}{2}=\frac{27n^2+27n+30}{2}$$
Knowing it works for $n$, then we have:
$$2+5+8+...+9n-1+9n+2+9n+5+9n+8$$
$$=\frac{3n(9n+1)}{2}+9n+2+9n+5+9n+8$$
$$=\frac{27n^2+3n}{2}+27n+15$$
$$=\frac{27n^2+57n+30}{2}$$ $$=\frac{3(n+1)(9(n+1)+1)}{2}.$$
|
{
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"url": "https://math.stackexchange.com/questions/1773501",
"timestamp": "2023-03-29T00:00:00",
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|
Find the value of integral $ \int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } $ Integral to get the result?
$$ \int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } $$
with Solutions ,tnx.
|
With partial fraction decomposition you have :
$$f(z)=\frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } = 2\frac{1}{z-\frac{1}{3}}+\frac{1}{2}\frac{1}{z-\frac{1}{2}}+\frac{1}{2}\frac{1}{(z-\frac{1}{2})^2}$$
We will try to use Residue theorem now :
$Res(f,\frac{1}{3})=Res(2\frac{1}{z-\frac{1}{3}})=2$
you should also find $Res(f,\frac{1}{2})=\frac{1}{2}$ (you have to justify a bit this calculation).
So finally : $$\int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } dz=2\pi i(2+\frac{1}{2})=\pi i5$$
|
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|
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.
|
I don't think the best bound can be solved for analytically.
using Lagrange multipliers , I tried to maximize $x^2y+y^2z+z^2x$ $\quad$ subject to the constraint $x+y^2+z^3=1$
maximize $g=x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$
$\frac {\partial g}{\partial x}=0:2xy+z^2+\lambda=0$
$\frac {\partial g}{\partial y}=0:x^2+2yz+2\lambda y=0$
$\frac {\partial g}{\partial z}=0:y^2+2zx+3\lambda z^2=0$
$\frac {\partial g}{\partial \lambda}=0:x+y^2+z^3-1=0$
$x\frac {\partial g}{\partial x}=0:2x^2y+z^2x+\lambda x=0$
$y\frac {\partial g}{\partial y}=0:x^2y+2y^2z+2\lambda y^2=0$
$z\frac {\partial g}{\partial z}=0:y^2z+2z^2x+3\lambda z^3=0$
$3(x^2y+y^2z+z^2x)=-\lambda (x+2y^2+3z^3)=(2xy+z^2)(x+2y^2+3z^3)$
If the global maximum occur on $ x, y, z \ge 0 $ then we can solve $(2xy+z^2)(x+2y^2+3z^3) \ge 0 $ for $x, y, z \ge 0 $
If no local maximum occur at $ x, y,z \ge 0 $ then we have to search for farthest points from the point of global minimum.
|
{
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"url": "https://math.stackexchange.com/questions/1775498",
"timestamp": "2023-03-29T00:00:00",
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|
If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{1-x}+1\right]$$
Now $$f(a) = \frac{a^2+1}{a^2+2} = \frac{a^2+2-1}{a^2+2} = 1-\frac{1}{a^2+2}$$
So $$f(x) = 1-\frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2}$$
Now put $1-x=t\;,$ Then we get $$f(t) =1- \frac{1}{\left(t-\frac{4}{t}+1\right)^2+2}$$
Now How can I maximize $\displaystyle \frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2, }\;,$ Help Required, Thanks
|
Write
$$
\frac{a^2+1}{a^2+2}=1-\frac{1}{a^2+2}
$$
Minimizing this is the same as maximizing $1/(a^2+2)$ which, in turn, is the same as minimizing $a^2+2$ or, as well, minimizing $a^2$.
Since
$$
a=-\frac{x^2-3x-2}{x-1}
$$
the minimum value for $a^2$ is obtained when $x^2-3x-2=0$.
|
{
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"url": "https://math.stackexchange.com/questions/1776723",
"timestamp": "2023-03-29T00:00:00",
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|
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
|
Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{b}{c}))^2+\frac{c}{13b}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{c}{a}))^2\geq \frac{1+\frac{a}{b}+\frac{c}{b}}{18}$$
Now we put $\sqrt{\frac{13}{5}}\frac{a}{b}=x$, $\sqrt{\frac{13}{5}}\frac{b}{c}=y$, $\sqrt{\frac{13}{5}}\frac{c}{a}=z$, your inequality is equivalent to:
$$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}$$$$\geq \dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
with the condition $xyz=(\sqrt{\frac{13}{5}})^3$.
We study the following function:
$$f(x)=\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}-\dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=\sqrt{\frac{3\sqrt{77}}{17}-\frac{26}{17}}=\alpha$. So with the condition $xyz=(\sqrt{\frac{13}{5}})^3$ becomes $yz=\frac{(\sqrt{\frac{13}{5}})^3}{\alpha}=\beta$. So we have this inequality just with $y$:
$$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(\alpha)^3}{(\alpha^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(\frac{\beta}{y})^2}{((\frac{\beta}{y})^2+1)}\geq\dfrac{1+\sqrt{\dfrac{5}{13}}\alpha+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
|
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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.